CHAPTER I GAS POWER CYCLES

Transcription

1 CHAPTER I GAS POWER CYCLES 1.1 AIR STANDARD CYCLES Air standard cycles are used for comparison of thermal efficiencies of I.C engines. Engines working with air standard cycles are known as air standard engines. Air is used, as the working fluid and the efficiency calculated to these engines are known as air standard efficiencies ASSUMPTIONS The assumptions made are, (i) (ii) (iii) (iv) Air is the working fluid, assumed to be a perfect gas. Effect of calorific values of fuels is neglected by using hot and cold body contacts with the engine cylinder head for addition and rejection of heat respectively. Frictionless. No heat is either gained or lost during the cycle except during the contact of hot body and cold body with the cylinder head IMPORTANT TERMINOLOGIES USED IN AIR STANDARD CYCLES (i) Bore It is diameter of the cylinder measured internally and is denoted by D. (ii) Stroke Stroke is defined as the displacement of the piston from its Top Dead Centre to the Bottom Dead Centre and is denoted by L (iii) Clearance volume It is the minimum volume of a gas inside the cylinder when the piston is at Top Dead Centre (T.D.C) and is denoted by V (iv) Swept or stroke volume It is the volume of a cylinder when the piston is moving from T.D.C to B.D.C and is denoted by V V π 4 D L 1

2 (v) Cylinder volume It is the volume of a gas inside the cylinder when the piston is at Bottom Dead Centre (B.D.C) and is denoted by V. VV V (vi) Compression ratio It is defined as the ratio of cylinder volume to the clearance volume and is denoted by γ. r V V V V V (vii) Air standard efficiency It is defined as the ratio of work done by the engine to the heat supplied. It is also known as ideal efficiency of the cycle. It is denoted by η work done heat supplied heat rejected η heat supplied heat supplied (viii) Mean effective pressure It is defined as the average pressure required for developing same power as that of the same engine operating under different pressures under same operating conditions. It can be calculated from P-V diagram. It is also defined as the ratio of work output to the stroke volume. 1.2 CARNOT CYCLE P work output stroke volume This cycle was first invented by a French engineer sadi Carnot in 1824 for perfect gas undergoing cyclic and reversible processes. It has two reversible iso-thermal frictionless processes and two reversible frictionless adiabatic processes. Figure 1.1 P-v and T-s diagram of Carnot Cycle 2

3 1-2 Isothermal Expansion 2-3 Isentropic Expansion 3-4 Isothermal compression 4-1 Isentropic compression. Isothermal expansion Process 1-2 The gas is expanded from state 1 to state 2 isothermally. Actually in anexapnsion process gas temperature decreases, here to maintain the temperature isothermally heat is supplied continuously. Heat supplied Q P V ln V mrt V ln V V Isentropic Expansion Process 2-3 The gas is expanded further at isentropic process from state 2 to state 3. During the process gas temperature decrease from T to T Isothermal Compression Process 3-4 The gas is compressed from state 3 to state 4 isothermally. Actually in a compression process gas temperature increases, here to maintain the temperature isothermally heat is removed continuously. Heat rejected Q P V ln V mrt V ln V V Isentropic Compression process 4 1 The gas is compressed further at isentropic process from state 4 to state 1. During the process gas temperature increases from T to T Expression for efficeincy η work done heat supplied heat rejected heat supplied heat supplied Heat supplied Q P V ln V V mrt ln V V Heat rejected Q P V ln V mrt V ln V V V V r V V Work done mrt T ln r 3

4 η work done heat supplied mrt T ln r mrt ln r η 1 T T T T T 1.3 OTTO CYCLE / CONSTANT VOLUME CYCLE Otto cycle was invented by German scientist Nichloas Otto in In this cycle he proposed constant volume heat addition and rejection instead of isothermal heat addition and rejection used in Carnot cycle. All petrol, Spark ignition (S.I.) and gas engines are working under this cycle. Figure 1.2 P-v and T-s diagram of Otto Cycle 1-2 Adiabatic / Isentropic compression 2-3 Heat addition at constant volume 3-4 Adiabatic / Isentropic Expansion 4-1 Heat rejection at constant volume Isentropic compression process 1-2 Air is compressed isentropically from state 1 to state 2. During this process pressure, temperature of air increases and volume decreases. Constant volume heat addition process 2-3 Heat is added at constant volume from state 2 to state 3. During this process temperature, pressure increases and volume remains constant. Heat supplied Q mc T T 4

5 Isentropic expansion process 3-4 Air is expanded isentropically from state 3 to state 4. During the process pressure, temperature decreases and volume increases. Constant volume heat rejection process 4 1 Heat is rejected from the air at constant volume from state 4 to state 1. During the process pressure, temperature decreases and volume remains constant. Heat rejected Q mc T T In this cycle combustion takes place fully at constant volume process, so this cycle is also called as constant volume cycle. Expression for air standard efficiency η work done heat supplied heat rejected heat supplied heat supplied Heat supplied Q mc T T 1 Heat supplied Q mc T T (2)... Work done = Heat Supplied Heat Rejected = mc T T mc T T η work done heat supplied mc T T mc T T mc T T η 1 T T T T 3 For isentropic process 1-2, T V T V For process 3-4, T T V 4 V T V T V 5

6 T T V 5 V From P-V diagram, V V,V V So (5) changes to T T V 6 V We know that, Compression ratior V V Substitute (4) & (6) in (3) we get, η 1 T T T V V T V V 1 T T V V T T η 1 1 r 7 From the above equation, air standard efficiency of Otto cycle increases with increase in compression ratio and vice-versa. Compression ratio is maintained 7 10 for better performance in engines operating under this cycle. If it increases more than 10 knocking will take place due to this life of the cylinder is reduced. Expression for mean effective pressure Mean effective pressure is the average pressure in Newton s per unit area which acts on the piston throughout the cycle. It is given by the breadth of rectangle whose length is equal to the swept volume. work done Mean effective pressure swept volume Work done = force x distance moved = pressure x area x length = P x A x L = pressure x volume Swept volume = V V V from P-V diagram. 6

7 Work done = Area of the curve Then, P V P V γ1 P V P V γ1 mean effective pressure P V P V P V P V V V mr T T mrt T γ 1V V mr T T T 1 T T T 1 γ 1V V mr T V 1T V V 1 V γ 1V V 1 V Mean effective pressure Compression ratior V V,V V,V V mr T V 1T V V 1 V γ 1V V 1 V mrt r 1 T r 1 γ 1V r 1 r 1P V P V γ 1V r 1 Where, r P P pressure ratio mrr 1T T γ 1V r 1 r 1V P P γ 1V r 1 rr 1P P 1 P P rr 1r 1 γ 1r 1 γ 1r 1 Mean effective pressurep P rr 1r 1 γ 1r DIESEL CYCLE / CONSTANT PRESSURE CYCLE Diesel cycle was invented by Rudolph Diesel in All Diesel, Compression Ignition (C.I.) engines are working under this cycle. 7

8 Figure 1.3 P-v and T-s diagram of Diesel Cycle 1-2 Adiabatic / Isentropic compression 2-3 Constant pressure heat addition 3-4 Adiabatic / Isentropic expansion 4-1 Constant volume heat rejection Isentropic Compression Process 1-2 Air is compressed isentropically from state 1 to state 2. During this process pressure, temperature of air increases and volume decreases. Heat addition Process 2-3 Heat is added at constant pressure from state 2 to state 3. During this process temperature, volume increases and pressure remains constant. Heat supplied Q mc T T Expansion Process 3-4 Air is expanded isentropically from state 3 to state 4. During the process pressure, temperature decreases and volume increases. Heat rejection process 4 1 Heat is rejected from the air at constant volume from state 4 to state 1. During the process pressure, temperature decreases and volume remains constant. Heat rejected Q mc T T In this cycle combustion takes place fully at constant pressure process, so this cycle is also called as constant pressure cycle. 8

9 Expression for air standard efficiency η work done heat supplied heat rejected heat supplied heat supplied Heat supplied Q mc T T 1 Heat supplied Q mc T T (2)... Work done = Heat Supplied Heat Rejected = mc T T mc T T η work done heat supplied mc T T mc T T mc T T η 1 C T T C T T 1 T T γt T 3 For isentropic process 1-2, T V r T V T T V T V r 4 For constant pressure process 2-3, V T V T cut off ratio r V T V T Cutoff ratio (r c ): It is the ratio of the final volume to the initial volume during constant pressure combustion. Therefore T T r T r r 5 For process 3-4, 9

10 T V T V Where, T T V V T r 6 expansion ratio r V V Substitute (5) in (6) we get, T T r r r T r r T T r r T r 7 η 1 T r T γ T r r T r 1 T r 1 γt r r r η 1 1 γr r 1 r 1 Expression for Mean Effective Pressure Mean effective pressure work done heat supplied heat rejected swept volume swept volume mc T T mc T T V V mc γt T T Substituting (5) & (7) in above equation we get, 1 T T T V 1 V V 1 P mc γt r 1 T r 1 V 10

11 mc r γt r 1 T r 1 RT r1 P P C T r γ T T r 1 r 1 T C C r 1 P C T r γ T r T 1 r 1 C T γ 1r 1 Substitute (4) in above equation we get, P r γr r 1 r 1 γ 1r1 1.5 DUAL CYCLE / LIMITED PRESSURE CYCLE Figure 1.4 P-v and T-s diagram of Dual Cycle 1-2 Isentropic Compression 2-3 Constant Volume Heat addition 3-4 Constant Pressure Heating 4-5 Isentropic Expansion 5-1 Constant Volume Cooling Isentropic Compression Process 1-2 Air is compressed isentropically from state 1 to state 2. During this process pressure, temperature of air increases and volume decreases. Heat addition Process

12 Heat is added at constant volume from state 2 to state 3. During this process temperature, pressure increases and volume remains constant. Heat supplied Q mc T T Heat addition Process 3-4 Heat is added at constant pressure from state 3 to state 4. During this process temperature, volume increases and pressure remains constant. Heat supplied Q mc T T Expansion Process 4-5 Air is expanded isentropically from state 4 to state 5. During the process pressure, temperature decreases and volume increases. Heat rejection process 5-1 Heat is rejected from the air at constant volume from state 5 to state 1. During the process pressure, temperature decreases and volume remains constant. Heat rejected Q mc T T In this cycle conbustion takes place partially at constant volume and partially at constant pressure, so this cycle is called as Dual cycle. Expression for air standard efficiency work done heat supplied heat rejected η heat supplied heat supplied Heat supplied Q mc T T mc T T 1 Heat rejected Q mc T T 2 work done heat supplied heat rejected η mc T T mc T T mc T T mc T T mc T T 1 1 mc T T mc T T mc T T T T T T γt T 3 12

13 For isentropic compression (process 1-2), T V r T V T T V T V r For constant volume heat addition (process 2-3) Where, r is pressure ratio For constant pressure process 3-4, T T P P r T T r T r r For isentropic expansion process 4-5, Substitute (4), (5), (6) & (7) in (iii) we get, η 1 V V T T cut off ratio r V T V T T T r T r r r T V 1 T V r T T 1 T T r r T r r T r r r r r T r r T T r r T r γt r r r T r r 1 r r 1 r r 1 γr r 1 Pressure ratio (r p ): It is the ratio of the final pressure to the initial pressure during constant volume combustion. Expression for mean effective pressure 13

14 Mean effective pressure work done heat supplied heat rejected swept volume swept volume P mc T T mc T T mc T T V V P mc T r r T r mc T r r r T r r mc T r r T V 1 V V P mc T r r T r mc T r r r T r r mc T r r T V P mc T r r T r mc T r r r T r r mc T r r T RT P rc T r r T r C T r r r T r r C T r r T T C C r 1 Taking C,T commonly outside P P rc T r r r γr r r r r r r 1 C T γ 1r 1 P rr r 1 γr r 1r r 1 γ 1r1 1.6 BRAYTON CYCLE The Brayton cycle was first proposed by George Brayton in the reciprocating oil-burning engine around This cycle finds application in Gas turbine engines, jet engines and in airplanes. It is also known as Joule s cycle. The first Ericsson cycle is similar to the Brayton cycle. Working Principle Figure 2.5 shows the schematic arrangement of closed Braytoncycle.Low-pressure air is drawn into a compressor (state1) from a heat exchanger where it is compressed to a higher pressure (state 2). The compressed high pressure air is mixed with a fuel in a combustion chamber and burnt.after the combustion is over, hot gases enter into the turbine (state 3) and 14

15 expand to state 4. After expansion of hot gases, remaining gases are drawn into the heat exchanger where it is cooled to the atmospheric conditions by circulating the cool water outside the heat exchanger tube. Figure 2.6 shows P-v and T-s diagram of the closed loop cycle. Figure 1.5 Closed Brayton Cycle Figure 1.6 P-v and T-s diagram of closed loop cycle The cycle consists of the following processes, isentropic compression constant pressure heat addition isentropic expansion constant pressure heat rejection 15

16 Isentropic Compression Process 1-2 Air is compressed isentropically from state 1 to state 2 in a compressor. During this process pressure, temperature of air increases and volume decreases. Heat addition Process 2-3 Heat is added at constant pressure from state 2 to state 3. During this process temperature, volume increases and pressure remains constant. Heat supplied Q mc T T Expansion Process 3-4 Air is expanded isentropically from state 3 to state 4 in a turbine. During the process pressure, temperature decreases and volume increases. Heat rejection process 4 1 Heat is rejected from the air at constant volume from state 4 to state 1. During the process pressure, temperature decreases and volume remains constant. Heat rejected Q mc T T Expression for air standard efficiency η work done heat supplied heat rejected heat supplied heat supplied Heat supplied Q mc T T 1 Heat supplied Q mc T T (2)... Work done = Heat Supplied Heat Rejected = mc T T mc T T η work done heat supplied mc T T mc T T mc T T η 1 T T T T 3 For isentropic process 1-2, T T P P 16

17 For process 3-4, From P-V diagram, So (5) changes to We know that, from P-V diagram T T P T r P 4 T T P P T T P T r P 5 T T P P P P,P P 6 Compression ratio r P P Substitute (4) & (6) in (3) we get, η 1 T T T P T P P P 1 η 1 1 T T P T P T r 7 From the above equation, air standard efficiency of Brayton cycle increases with increase in compression ratio and vice-versa.fig. 1.7 shows the variation of efficiency with pressure ratio. 17

18 Fig. 1.7 Variation of Efficiency Vs Pressure ratio 1.7 COMPARISON OF OTTO, DIESEL AND DUAL CYCLES The important variable factors which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat addition, heat rejection and the network. In order to compare the performance of the Otto, Diesel and Dual combustion cycles, some of the variable factors must be fixed. In this section, a comparison of these three cycles is made for the same compression ratio, same heat addition, constant maximum pressure and temperature, same heat rejection and net work output. This analysis will show which cycle is more efficient for a given set of operating conditions. Case 1: Same Compression Ratio and Heat Addition The Otto cycle , the Diesel cycle 1-2-3'-4'-1 and the Dual cycle are shown in p-v and T-s diagram in Fig (a) and (b) respectively for the same compression ratio and heat input. 18

19 Fig. 1.8 From the T-s diagram, it can be seen that Area = Area 5-2-3'-6 = Area 5-2-2"- 3"-6" as this area represents the heat input which is the same for all cycles. All the cycles start from the same initial state point 1 and the air is compressed from state 1 to 2 as the compression ratio is same. It is seen from the T-s diagram for the same heat input, the heat rejection in Otto cycle (area ) is minimum and heat rejection in Diesel cycle (5-1-4'-6') is maximum. Consequently, Otto cycle has the highest work output and efficiency. Diesel cycle has the least efficiency and Dual cycle having the efficiency between the two. One more observation can be made i.e., Otto cycle allows the working medium to expand more whereas Diesel cycle is least in this respect. The reason is heat is added before expansion in the case of Otto cycle and the last portion of heat supplied to the fluid has a relatively short expansion in case of the Diesel cycle. Case 2: Same Compression Ratio and Heat Rejection Fig. 1.9 Efficiency of Otto cycle is given by [Figs (a) and (b)], η O 1 Q Q Where, Qs is the heat supplied in the Otto cycle and is equal to the area under the curve 2-3 on the T-s diagram [Fig (b)]. The efficiency of the Diesel cycle is given by, η D 1 Q Q 19

20 Where Q s is heat supplied in the Diesel cycle and is equal to the area under the curve 2-3' on the T-s diagram [Fig (b)]. From the T-s diagram in Fig.4.7.2, it is clear that Qs > Q s i.e., heat supplied in the Otto cycle is more than that of the Diesel cycle. Hence, it is evident that, the efficiency of the Otto cycle is greater than the efficiency of the Diesel cycle for a given compression ratio and heat rejection. Case 3: Same Peak Pressure, Peak Temperature and Heat Rejection: Figures (a) and (b) show the Otto cycle and Diesel cycle 1-2'-3-4 on p-v and T- s coordinates, where the peak pressure and temperature and the amount of heat rejected are the same. The efficiency of the Otto cycle, η O 1 Q Q Fig Where, Qs in the area under the curve 2-3 in Fig (b). The efficiency of the Diesel cycle, 1-2-3'-3-4 is, η D 1 Q Q 20

21 It is evident from Fig that Qs > Q s. Therefore, the Diesel cycle efficiency is greater than the Otto cycle efficiency when both engines are built to withstand the same thermal and mechanical stresses. Case 4: Same Maximum Pressure and Heat Input: Fig For same maximum pressure and heat input, the Otto cycle ( ) and Diesel cycle ( ) are shown on p-v and T-s diagrams in Fig (a) and (b) respectively. It is evident from the figure that the heat rejection for Otto cycle (area on T-s diagram) is more than the heat rejected in Diesel cycle ( ). Hence Diesel cycle is more efficient than Otto cycle for the condition of same maximum pressure and heat input. One can make a note that with these conditions, the Diesel cycle has higher compression ratio than that of Otto cycle. One should also note that the cycle which is having higher efficiency allows maximum expansion. The Dual cycle efficiency will be between these two. Case 5: Same Maximum Pressure and Work Output The efficiency, η can be written as η work done heat supplied work done work done heat rejected Refer to T-s diagram in Fig (b). For same work output the area (work output of Otto cycle) and area (work output of Diesel cycle) are same. To achieve this, the 21

22 entropy at 3 should be greater than entropy at 3.It is clear that the heat rejection for Otto cycle is more than that of diesel cycle. Hence, for these conditions, the Diesel cycle is more efficient than the Otto cycle. The efficiency of Dual cycle lies between the two cycles. 1.8 THEORETICAL AND ACTUAL CYCLES FOUR STROKE PETROL ENGINE Fig Fig (a) and (b) shows the actual p-v diagram and theoretical p-v diagram of four stroke Petrol engine. The line 5-1 represents the suction stroke in which the charge enters into the cylinder. The suction of mixture is possible only if the pressure inside the cylinder is below atmospheric pressure. That s the reason line 5-1 lies below the atmospheric pressure line.the burnt gases can be pushed out only if the pressure of the exhaust gas isabove atmospheric pressure. This is represented by the line 1-5. The area in between the process 5-1 and 1-5 in the indicator diagram gives pumping loss of the engine. The compression stroke is shown by the line 1-2 which shows that the inlet valve closes (IVC) a little beyond 1. At the end of this stroke, there is an increase in pressure inside the engine cylinder. Before the end of compression stroke, the charge is ignited (IGN) with the help of spark plug. Thus, the pressure and temperature of the cylinder increase. But the volume remains constant as shown by line 2-3. Expansion is shown by the line 3-4. The exhaust valve opens (EVO) little before 4. The burnt gases are exhausted to atmosphere. 22

23 It has been found practically, that the actual pressure rise in such an engine is only half of the theoretical value. The corners are rounded off because both inlet and exhaust valves do not open and close suddenly. 1.9 THEORETICAL AND ACTUAL CYCLES FOUR STROKE DIESEL ENGINE Fig (a) and (b) shows the actual p-v diagram and theoretical p-v diagram of four stroke Diesel engine. Fig The line 5-1 represents the suction stroke in which the air enters into the cylinder. The suction of mixture is possible only if the pressure inside the cylinder is below atmospheric pressure. That s the reason line 5-1 lies below the atmospheric pressure line. The burnt gases can be pushed out only if the pressure of the exhaust gas isabove atmospheric pressure. This is represented by the line 1-5. The air is compressed adiabatically in the cylinder during 1-2 process which takes place after inlet valve closed. Before the end of compression stroke, fuel is injected through the fuel injector (FVO). The fuel is ignited due to the temperature of highly compressed air inside the cylinder. The combustion takes place at constant pressure as shown in line 2-3. Actually, combustion at constant pressure is not possible as the fuel will not burn completely as it is introduced into the cylinder. Then the charge is expanded adiabatically is shown by the line 3-4. The exhaust valve opens (EVO) little before 4. The burnt gases are exhausted to atmosphere. Theoretically, the compression and expansion are followed adiabatically. But in actual cycle it is not so. Because of heat and pressure losses are involved. 23

24 1.10 THEORETICAL AND ACTUAL CYCLES FOR TWO STROKE PETROL ENGINE Fig (a) and (b) shows the actual p-v diagram and theoretical p-v diagram of two stroke petrol engine. Fig The suction stroke is carried out from transfer ports open (TPO) and transfer port close (TPC). During half of the suction stroke, exhaust port is also opened. Now, the volume of air fuel mixture is entered into the cylinder. This happens as the piston moves from TDC to BDC. During second half of the suction stroke (i.e. BDC to TPC), the air and fuel mixture is compressed and burnt gases pushed out. A little beyond TPC, the exhaust port closes (EPC) at 1. These processes are represented in theoretical cycle as and Now, the air fuel mixture is compressed isentropically in the cylinder. This is shown by line 1-2. A little before the end of compression, the charge is ignited (IGN) with the help of spark plug as shown in the above diagram. Combustion of air fuel mixture increases the pressure and temperature of the products of combustion. During this process, volume remains constant. This represented by the line 2-3. The expansion process is shown by line 3-4. The end of the expansion stroke, exhaust port opens (EPO) at 4 and burnt gases are pushed out to the atmosphere. During this, the pressure falls suddenly to the atmospheric pressure. As the piston is moving towards BDC, the volume of burnt gases increases from 4 to 5. At 5, transfer port opens (TPO) and the suction starts. 24

25 1.11 THEORETICAL AND ACTUAL CYCLES FOR TWO STROKE DIESEL ENGINE Fig (a) and (b) shows the actual p-v diagram and theoretical p-v diagram of two stroke Diesel engine. Fig The suction stroke is carried out from transfer ports open (TPO) and transfer port close (TPC). During half of the suction stroke, exhaust port is also opened. Now, the volume of air is entered into the cylinder. This happens as the piston moves from TDC to BDC. During second half of the suction stroke (i.e. BDC to TPC), the air is compressed and burnt gases pushed out. A little beyond TPC, the exhaust port closes (EPC) at 1. These processes are represented in theoretical cycle as and Now, the air is compressed isentropically in the cylinder. This is shown by line 1-2. A little before the end of compression, the fuel is admitted into the cylinder by means of fuel injector (INJ).Combustion of fuel increases the pressure and temperature of the products of combustion. During this process, pressure remains constant. This represented by the line 2-3. The expansion process is shown by line 3-4. The end of the expansion stroke, exhaust port opens (EPO) at 4 and burnt gases are pushed out to the atmosphere. During this, the pressure falls suddenly to the atmospheric pressure. As the piston is moving towards BDC, the volume of burnt gases increases from 4 to 5. At 5, transfer port opens (TPO) and the suction starts. 25

26 SOLVED PROBLEMS 1. In a Brayton cycle, air enters at 100 kpa and 25 C. The pressure leaving the compressor is 3 bar and the temperature of turbine inlet is 650 C. Determine per Kg of air i) Cycle efficiency, ii) Heat added, iii) Work available, iv) Heat rejected in the cooler at the shaft, v) Temperature of air leaving the turbine. Given data: T 1 = 25 C = = 298 K P 1 = 100 kpa = 1 bar P 2 =P 3 = 3 bar T 3 = 650 C = = 923 K For air assume C J,γ1.4 K Solution: η1 1 η r r P P % Heat added C T T T T P P T K Heat added = 1.005( ) = kj/kg Work available = Turbine work compressor work Turbine work = C T T T T P P T T P P K 3. Turbine work = C T T kj/kg Compressor work = C T T kj/kg

27 Work available = = kj/kg Heat rejected Q C T T kj/kg Temperature of air leaving the turbine K 2. An air standard dual cycle has a compression ratio of 16, and compression begins at 1 bar, 50 C. The maximum pressure is 70 bar. The heat transferred to air at constant pressure is equal to that at constant volume. Estimate, i) The pressures and temperatures at all cardinal points of the cycle, ii) The cycle efficiency, iii) mean effective pressure of the cycle. Given data: r = 16 = V V T 1 = 50 C = = 323 K P 1 = 100 kpa = 1 bar P 3 =P 4 = 70 bar Heat transfer at constant volume = heat transfer at constant pressure For air assume C J,C K J,γ1.4 K Solution: P V P V P P V bar V T T V T V r K V RT P m V m 16 P T P T T T P K P V V m 27

28 Heat supplied Q mc T T mc T T 2C T T Given that heat transfer at constant volume is equal that of constant pressure Heat supplied Q 2C T T kj kg Heat supplied Q C T T C T T T T K V V T T T V V m T T T V V T K V V P P V V bar Heat rejected Q C T T kj/kg Efficiency η 1 Q % Q work done Mean effective pressure P swept volume heat suppliedq heat rejectedq swept volume V V kpa bar The compression ratio in an air-standard Otto cycle is 8. At the beginning of compression process the pressure is 1 bar and the temperature is 300K. The heat transfer to the air per 28

29 cycle is 1900 kj/kg of air calculate. i) Pressure and temperature at the end of each process of the cycle, ii) Thermal efficiency, iii) Mean effective pressure. Given data Compression ratior 8 T 1 = 300 K P 1 = 1 bar Heat supplied Q 1900 kj/kg For air assume C J,γ 1.4 K Solution: Isentropic compression process 1-2 T T V T V r K P P V P V r bar V RT P m V m 8 Heat supplied Q 1900kJ/kg C T T T T K P T P T T P P bar T T T V V K P P V V bar η 1 1 r % 8. 29

30 Work done heat supplied heat rejected Mean effective pressure P kj/kg work done swept volume kpa bar 4. An air standard diesel cycle has a compression ratio of 18, and the heat transferred to the working fluid per cycle is 1800 kj/kg. At the beginning of the compression stroke, the pressure is 1bar and the temperature is 300k. Calculate i) Pressure and temperature at each point in the cycle, ii) Thermal efficiency, iii) Mean effective pressure. Given data: Compression ratior 18 T 1 = 300 K P 1 = 1 bar Heat supplied Q 1800 kj/kg For air assume C J K,C J K,γ1.4 Solution: Isentropic compression process 1-2 T T V T V r K P P V P V r bar V RT P m V m 18 Heat supplied Q 1900kJ/kg C T T T T K V V T T 30

31 T V V m T T T V V K P P V V bar Heat rejected Q C T T kj/kg Work done heat supplied heat rejected η kj/kg work done % heat supplied 1800 Mean effective pressure P work done swept volume kpa bar 5. In an air standard diesel cycle, the pressure and volume at the beginning of compression are 100 kpa and 0.03m 3 respectively. Pressure after isentropic compression is 4.2 MPa, and after isentropic expansion is 200 kpa. Determine i) Compression ratio, ii) Cut-off ratio, iii) expansion ratio and iv) Cycle efficiency. Given data: Compression ratior 18 V 1 = 0.03m 3 P 1 = 100 kpa = 1 bar P 2 = 4.2 MPa = 42 bar P 4 = 200 kpa = 2 bar Solution Isentropic compression process (1-2) P P V V compression ratior V P. 42 V P 1 Isentropic compression process (3-4)

32 P P V V expansion ratio r V V V P. 42 V P expansion ratio r V V 8.8 cut off ratio r V V P P compression ratior V V V V V V r r V V r r r η 1 1 γ r r 1 r % 6. Consider a stationary power plant operating on an ideal Brayton cycle. The pressure ratio of the cycle is 8, And the gas temperature at the compressor inlet and the turbine inlet are 27 C and 1027 C respectively. Determine the following i) Gas temperature at the compressor and turbine exit, ii) Back work ratio, iii) Thermal efficiency. Given data: r P 8 P T 1 =27 C = 300 K T 3 =1027 C = 1300 K Solution Isentropic expansion process (1 2) igas temperature at compressor exit T T P T r P K Isentropic expansion process (3 4) 32

33 T T P P Gas temperature at turbine exit T T P P K Compressor work W mc T T kj/kg Turbine work W T mc T T kj/kg iiback work ratio W W T iiiη or 44.79% 8. r 7. In an Otto Cycle, air at 17 C and 1 bar is compressed adiabatically until the pressure is 15bar. Heat is added at constant volume until the pressure rises to 40 bars. The swept volume is m 3. Calculate the air standard efficiency, the compression ratio and the mean effective pressure. Given data: Vs = V 1 V 2 = 0.711m 3 P 1 = 1 bar, T 1 =17 C = 290 K P 2 = 15 bar P 3 = 40 bar Solution Isentropic compression process (1-2) T T P P T T P P 1 Constant volume process (2-3) K V RT P m P P T T 33

34 T T P K P 15 Swept volume V V V m V V m T T V V K Heat supplied Q C T T kj/kg Heat rejected Q C T T kj/kg a efficiency η Q Q Q b compression ratio r V V or 53.62% c Mean effective pressure P Q Q V kpa or 5.67 bar 8. In an air standard dual cycle, the compression ratio is 12 and the maximum pressure in the cycle is 70 bar. The lowest pressure and temperature of the cycle are 1bar and 300K. Heat is added during constant pressure process upto 3% of the stroke. Taking diameter 25cm and stroke 30 cm, determine a) The pressure and temperature at the end of compression, b) The thermal efficiency and c) The mean effective pressure. Given data: Compression ratior 12 P 1 = 1 bar, T 1 = 300 K P 3 =P 4 =70 bar, diameter d 0.25 m, stroke l 0.30 m For air assume C J,C K J,γ1.4 K Solution: Isentropic compression process 1-2 T T V T V r K 34

35 P P V P V r bar V RT P m V m 12 P T P T T P T K P x V V 100 V V 3 V V m V V T T T T V K V T T V V T K V V P P V V bar Heat supplied Q C T T C T T kj/kg Heat rejected Q C T T kj/kg Work done heat supplied heat rejected

36 η kj/kg work done % heat supplied Mean effective pressure P work done swept volume kpa bar OBJECTIVE QUESTIONS 1) Otto cycle consists of processes (a) adiabatic and constant volume (b) adiabatic and constant pressure (c) isothermal and constant pressure (d) isothermal and constant volume 2) Carnot engine is irreversible due to (a) friction between moving parts (b)losses from working fluid (c) high speed (d) both (a) and (b) of the above 3) Air standard efficiency of an Otto cycle is equal to (a) 1-(1/r γ+1 ) (b) 1-(1/r γ-1 ) (c) 1+(1/r γ+1 ) (d)1+(1/r γ-1 ) r-is compression and expansion ratio γ-ratio of two specific heats 4) Otto cycle is a theoretical cycle, on which (a) only petrol engine (b) only diesel engine (c) only gas engine (d)petrol and gas engine 5) Compression ratio for petrol engine is (a) 3 to 6 (b) 5 to 8 (c) 15 to 20 (d) 20 to 30 6) Diesel engine consists of (a) two adiabatic and two constant volume process (b) two adiabatic and two constant pressure process (c) two adiabatic and one constant pressure and one constant volume process (d) two isothermal and one constant pressure and one constant volume process 7) The efficiency of a diesel cycle increases with (a) increase in cut-off (b) decrease in cut-off (c) constant cut-off (d) none of the above 8) For the same compression ratio, the efficiency of diesel cycle as compared to Otto 36

37 cycle is (a) less (b) more (c) equal (d) none 9) Efficiency of diesel cycle approach Otto cycle efficiency with (a) increase in cut-off (b) decrease in cut-off c) zero cut-off (d) constant cut-off 10) Compression ratio for diesel engine is (a) 3 to 6 (b) 5 to 8 (c) 15 to 20 (d) 20 to 30 11) In Carnot cycle, heat is rejected at constant (a) Volume (b) pressure (c) Temperature (d) Entropy 12) Area of P-V Diagram for a Carnot cycle represents (a)heat supplied (b) Heat rejected (c) workdone (d) Temperature drop 13) In Carnot cycle, the algebraic sum of the entropy changes for the cycle is (a) positive (b)negative (c)zero (d) none of the above 14) In Carnot cycle, the process carried at extremely slow speed is (a) Isothermal compression (b) adiabatic compression (c) adiabatic expansion (d) above all 15) The dual combustion cycle consists of two adiabatic processes ) For the same compression ratio, the efficiency of dual combustion cycle as compared to Diesel cycle is (a) more (b) less (c) equal (d) none 17) For the same compression ratio, the efficiency of dual combustion cycle as compared to Otto cycle is (a) more (b) less (c) equal (d) none 18) The cycle used for gas turbines is (a)rankine cycle (b)carnot cycle (c)otto cycle (d) Brayton cycle 19) The ideal efficiency of simple gas turbine cycle depends upon 37

38 (a) cut-off ratio (b) pressure ratio (c) both (a) and (b) (d) none of the above 20) If the air after practical expansion on the turbine is reheated, the efficiency of the gas turbine cycle (a)decreases (b)increases (c)remains constant (d) first increases then decreases 21) The thermal efficiency of a gas turbine with regenerator is maximum when pressure ratio is (a)less than 1.3 (b) more than 1.0 (c) equal to 1.0 (d) zero 22) By having multistage compressor with intercoolers the efficiency of a gas turbine cycle (a)decreases (b)increases (c)remains constant (d) first increases then decreases 23) When the pressure ratio is equal to 1.0 the thermal efficiency of a gas turbine with regenerator is (a)less than Carnot cycle (b) more than Carnot cycle (c) equal to Carnot cycle (d) none of the above 24) If the exhaust from the gas turbine is utilized in heating the compressed air, the efficiency of the ga turbine cycle (a)decreases (b)increases (c)remains constant (d) first increases then decreases TWO MARK QUESTIONS 1. What is air standard cycle? 2. State the assumptions made in the air standard cycles? 3. Draw the P-V and T-S diagram for Otto cycle? 4. Draw the P-V and T-S diagram for Diesel cycle? 5. Draw the P-V and T-S diagram for Dual cycle? 6. Draw the P-V and T-S diagram for Brayton cycle? 7. Draw the layout of a closed cycle Gas turbine power plant? 8. What is mean effective pressure? 9. Determine the mean effective pressure of an engine of stroke volume mm 3 producing 600 KJ of work. 38

39 10. Define Compression ratio. 11. What is Cut-off Ratio? 12. Why actual work done is always less than the theoretical work done of an I.C Engine? 13. A Brayton cycle works between the pressure limits 60 bar and 2 bar, what is the thermal efficiency? 14. The compression ratio of an Otto cycle is 8. Determine its air standard efficiency? 15. For the same compression ratio and heat rejection which cycle will be efficient? 16. For the same compression ratio and heat input which cycle will be efficient? 17. For the same maximum pressure and temperature which cycle will be efficient? 18. Define work ratio of gas turbine. DESCRIPTIVE QUESTIONS 1. Derive an expression for the air standard efficiency of Otto cycle. 2. Derive an expression for the mean effective pressure of Diesel cycle. 3. Derive an expression for the air standard efficiency of a Dual cycle and from it deduce the expression for the air standard efficiencies of Otto cycle and Diesel cycles. 4. Prove that the efficiency of Carnot engine working between the temperature limits T 1 T1 T 2 and T 2 is equal to. T1 5. Prove that the air standard efficiency of Diesel cycle is 1 rc r rrc 1 6. Derive the expression for the thermal efficiency of Brayton Cycle. 7. Draw theoretical and actual P-V diagram for a 4 Stroke S.I engine and explain the reason for the difference. 8. Using T-S diagram, prove that for the same compression ratio and same heat input the otto < Dual < Diesel 9. Prove that the mean effective pressure of Otto cycle is given by m 1 p Pr 1 r 1 r 1 P 1r 1 39

40 EXERCISE PROBLEMS 1. In an Otto cycle air at 15 C and 1.02 bar is compressed until the pressure is 12.5 bar. Heat is added at constant volume until the pressure raises to 35 bar absolute. Calculate the compression ratio, the air standard efficiency and the mean effective pressure. Take γ = An engine equipped with a cylinder having a bore of 150mm and a stroke of 450mm operated on an Otto cycle. If clearance volume is 2000 cm 3. Compute the air standard efficiency. 3. In an air standard Otto Cycle the compression ratio is 7 and compression begins at 35 C, 0.1MPa. The maximum temperature of the cycle is 1100 C. Find the temperature and pressure at all corner points of the P-V diagram, the heat supplied per kg of air, the work done per kg of air, the cycle efficiency and mean effective pressure? 4. An Engine working on Otto cycle has an air standard cycle efficiency of 56% and rejects 444kJ/kg of heat. The pressure and temperature at the beginning of compression is 0.1MPa and 60 C respectively. Find the temperature and pressure at all points, compression ratio, the work done per kg of air and mean effective pressure. 5. An engine working on Otto cycle has a volume of 0.5 m 3, 1 bar pressure and 27 C temperature at the beginning of compression. The pressure at the end of compression is 10bar 220 KJ of heat is added at constant volume. Calculate the volume, pressure and temperature at salient points in the cycle. Also find the percentage clearance, the work done per cycle, efficiency, and mean effective pressure. Also calculate the power developed in kw if the working cycles are 400 per minute. 6. An air standard Diesel cycle has a compression ratio of 16 and a maximum cycle temperature of 1600 K the compression begins at 1.0 bar and 15 C. Calculate the values of pressure, volume, temperature the end of each process, heat supplied, heat rejected, network and the thermal efficiency. 7. In a Diesel engine, the compression ratio 13: 1 and the fuel is cut-off at 8% of the stroke. Find the air standard efficiency of the engine. Take = In an air standard Diesel cycle the compression ratio is 15. Compression begins at 0.1MPa, 40 C. The heat added is MJ/kg. Calculate the air standard efficiency and mean effective pressure of the cycle. 40

41 9. In an ideal Diesel cycle, the temperature at the beginning and end of compression is 57 C and 603 C respectively. The temperature at the beginning and end of expansion is 1950 C and 870 C respectively. Determine the ideal efficiency of the cycle. =1.4. If the compression ratio is 14 and the pressure at the beginning of the compression is 1 bar, calculate the maximum pressure in the cycle. 10. In a Diesel cycle the pressure and temperature of air at the beginning of isentropic compression is 1barabsolute and 15 C respectively.the compression ratio is The Expansion ratio is 7. Calculate the cut off ratio, air standard efficiency and the mean effective pressure of the cycle. Take C p = kj/kg K, C v = kj/kg K. 11. In a Dual combustion cycle air at the beginning of compression is at 20 C and 1 bar absolute. The Heat supplied is 1250 kj/kg of which half of heat supplied is at constant volume and half at constant pressure. If the compression ratio is 15. Calculate the air standard efficiency. Take C p = 0.24kJ/kg K, C v = kj/kg K. 12. In a Dual combustion cycle the pressure and temperature at the beginning of the cycle is1.0bar and 15 C. The specific volume at the end of isentropic compression is m 3 / kg of the working substance.the highest temperature reached in the cycle is 1440 K. The heat supplied at constant volume is 230kJ/kg. Calculate the compression ratio, cut off ratio and thermal efficiency of the cycle. 13. In compression ignition engine, working on a dual combustion cycle, pressure and temperature at the start of compression are 1 bar and 300K respectively. At the end of compression, pressure reaches a value of 25bar. 420kJof heat is supplied per kg of air during constant volume heating and pressure becomes 2.8 bar at the end of isentropic expansion. Estimate the ideal thermal efficiency. Take C p =1.005kJ/kgK and C V =0.712kJ/kgK 14. An engine working on a dual combustion cycle has a pressure of 1bar and 50 o C before compression. The air is then compressed isentropically to 1/5 th of its original volume. The maximum pressure is twice the pressure at the end of isentropic compression. If the cut-off ratio is 2, determine the temperature at the end of each process and the ideal efficiency of the cycle. = An ideal dual cycle engine works with a stroke volume of 10lit of air with a compression ratio of 16. The pressure and temperature of air before isentropic compression is 1 bar and 300K respectively. If the heat is added at a constant pressure of 70bar and for 5% of stroke, Determine, 41

42 a. a) pressure ratio b)cut off ratio c) mass of air contained b. d) heat added e) heat rejected f) work done c. g) thermal Efficiency and h) mean effective Pressure. d. Take C p =1 kj/kgk and C V =0.714kJ/kgK 16. An air standard limited pressure cycle has a compression ratio is 15 and Compression begins at 0.1MPa, 40 C. The maximum pressure is limited to 6MPa and heat added is MJ/kg. Calculate per kg of air heat supplied at constant volume, heat supplied at constant pressure, the work done, the cycle efficiency, mean effective pressure, temperature at the end of the constant volume heating and the cut off ratio. 17. In a gas turbine plant, the intake temperature and pressure are 18 C and 1 bar respectively. The air is then compressed to a pressure of 4.1 bar by a compressor, whose isentropic efficiency is 80%. The temperature of the gas whose properties may be assumed to resemble with those of air, is raised to 645 C in the combustion chamber where there is a pressure drop of 0.1bar. Expansion to atmospheric pressure then occurs. If the thermal efficiency of the plant is to be 19%. What must be the isentropic efficiency of the turbine? Neglect mass of fuel. = Determine the efficiency of a gas turbine plant fitted with a heat exchanger of 75% effectiveness. The pressure ratio is 4:1 and the compression is carried out in two stages of equal pressure ratio with inter cooling back to initial temperature of 290K. The maximum temperature is 925K. The turbine isentropic efficiency is 88% and each compressor isentropic efficiency is 85%. For air =1.4. andc p =1.005kJ/kg. K 19. A gas turbine plant has temperature limits 1080 o C and 10 o Ccompression in compressor and expansion in the turbine are isentropic. Determine e. the pressure ratio which will give the maximum network output f. the maximum net specific work output g. the thermal efficiency at maximum work output h. the work ratio at maximum work output i. Take =1.41. C p =1.007 kj/kgk. 20. Air enters the compressor of an open cycle gas turbine at 100kN/m 2 and 27 o C. The pressure of air after compression is 400kN/m 2. The isentropic efficiencies of compressor and turbine are 78% and 84% respectively. The air fuel ratio is 75:1. The rate of flow of air is 2.5kg/s. Determine the power developed and thermal efficiency 42

43 of the cycle. Take =1.41. C p =1.005 kj/kgk both for air and gases. The calorific value of fuel used is kj/kg. 43

44 CHAPTER II INTERNAL COMBUSTION ENGINES 2.1 INTRODUCTION Heat Engines are otherwise called Thermal Engines. It is a machine which converts heat energy into useful mechanical work. Heat engines develop more than 80% of energy generated in the world. They are broadly classified into two types: 1. Internal Combustion Engines 2. External Combustion Engines INTERNAL COMBUSTION ENGINES In the Internal Combustion Engine, the chemical energy of the fuel is released as a heat by the way of combustion inside the engine cylinder where power is produced. The heat produced is nothing but the products of combustion. By expansion of this hot medium inside the cylinder, heat energy is converted into useful work. The name Internal Combustion Engine is a misnomer since the fuel is burnt internally EXTERNAL COMBUSTION ENGINES They are steam engines and steam turbines. In these, heat energy is produced during the combustion of fuel in a boiler furnace. This energy is used to produce the steam under the pressure in boiler. The steam expands in turbine and thereby does work. The name External combustion engine is a misnomer since the fuel is burnt externally. 2.2 CLASSIFICATION CLASSIFICATION BASED ON IGNITION (i) Spark Ignition Engines (S.I engines) In this type of engine, combustible mixture is sucked into the engine cylinder. This mixture is compressed. The compression ratio is about 5:1 to 7:1. At the 44

45 end of compression, the mixture exists in the cylinder as high pressure and temperature. The Electric spark ignites this mixture. The burning of mixture produces greater pressure and temperature. The product of combustion expands and produced power. Then the products are expelled out. (ii) Combustion Ignition Engines (C.I Engines) In this type, air alone is sucked into the engine cylinder the air is compressed. The compression ratio is about 14:1 to 17:1. The heat of compression in the air is much greater due to high compression. At the end of compression the fuel is injected in the form of fine spray into the engine cylinder. The compression heat ignites the fuel and causes in to burn. Combustion of fuel produces high pressure and temperature. The product of combustion expands and thereby produces power. The combustion products are then exhausted. (iii) Precombustion chamber Engines The mixture is ignited by a spark in a special small anti-chamber, while the takes in the main chamber CLASSIFICATION BASED ON NUMBER OF STROKES: (i) Four stroke Engine In this engine, four strokes of the piston is required to complete a working cycle. In this engine, two revolution of the crankshaft is used to complete the cycle of operation. (ii) Two Stroke Engine In this engine, two strokes of the piston is required to complete a working cycle. In this engine, one revolution of the crankshaft is used to complete the cycle of operation CLASSIFICATION BASED ON CYCLE OF OPERATION (i) Otto cycle (ii) Diesel cycle (iii) Dual cycle CLASSIFICATION BASED ON THE TYPE OF FUEL USED: 45

46 (i) Engines using Light Liquid Fuels ---- Petrol Engines (ii) Engines using Heavy Liquid Fuels ---- Diesel Engines (iii) Engines using Gaseous Fuels ---- Gas Engines (iv) Mixed Fuel Engines (v) Multi Fuel Engines CLASSIFICATION BASED ON THE MODE OF CONVERSION OF HEAT ENERGY INTO MECHANICAL WORK: (i) Reciprocating Engines: In this, heat energy gets converted into mechanical energy in the inside of engine cylinder while the piston reciprocates. (ii) Rotary Engines: In this heat energy is converted in to mechanical energy in the inside of engine cylinder while the specially shaped casing and a rotor rotates inside thecasing. (iii) Gas Turbine: The heat energy in the hot gases is converted into mechanical work onthe rotating blades of the gas turbine. (iv) Combination Engine: In this the heat energy is converted into mechanical work partly in thecylinder of reciprocating engine and partly in the blades of turbine CLASSIFICATION BASED ON THE METHOD OF MIXTURE FORMATION: (i) External Mixture Formation Engines: 46

47 Used in spark ignition engines, and gas engines in which the fuel is injected intothe intake pipe or intake manifold and mixes with air externally to the cylinder, i.e., in the carburetor. (ii) Internal Mixture Formation Engines: Used in diesel engines, with injection of fuel into the cylinders and in gas engines in which the gas is fed into the cylinder at the beginning of compression CLASSIFICATION BASED ON THE METHOD OF COOLING (i) Air cooled engines (ii) Water cooled engines CLASSIFICATION BASED ON THE METHOD OF GOVERNING (i) Quantity governing (ii) Quality governing (iii) Hit and miss governing 2.2.9CLASSIFICATION BASED ON THE ARRANGEMENT OF CYLINDERS (i) Inline Engine All cylinders are arranged in a line and the power is taken from a singlecrankshaft. This arrangement is used in automobiles. (ii) V-Type It is a combination of two inline engines set at an angle of V varies from 30 to 75. (iii) Opposed piston engines The piston reciprocates in a common cylinder having common combustionchamber at the centre. (iv) Radial Engine 47

48 All the cylinders are set along the radius of a circle. The connecting rods point towards the centre of the circle. The connecting rods of all pistons work on a single crank pin which rotates around the centre of the circle. This occupies little floor space and simplifies the balancing problems. This is popular in aircrafts. (v) Rotary Engine The engine consists of three sided converse type of piston rotating in a cylinder. This engine is known as Wankel engine. It is of high speed, lighter weight andworks on spark ignition system CLASSIFICATION BASED ON THE METHOD OF CONTROL OF CHARGE UNDER VARIABLE LOAD (i) Quality control engines In which the composition of the mixture which undergoes combustion is changedby admitting more or less quantity of fuel in accordance with the variation ofload. Air quantity remains almost constant. All diesel engines are quality controlengines. (ii) Quantity control engines In which the composition of the mixture which remains almost constant when theload varies and the quantity of the mixture admitted is changed. All Petrol engines are quantity control engines. (iii) Combination control engines In which both quality and quantity of the mixture are varied depending on the load CLASSIFICATION BASED ON THE PURPOSE (OR) APPLICATION (i) Stationary Engines (ii) Mobile Engines (iii) Aero Engines (iv) Marine Engines 48

49 2.3 I.C. ENGINES COMPONENTS AND FUNCTIONS For effective functioning of the internal combustion engine every components of the engine has to work properly. The following components of the engine are, Cylinder It is a cylindrical space (or) container in which piston reciprocates. The working substance contained within the cylinder is subjected to different thermodynamics processes. The cylinder is supported in cylinder block.material: Grey cast iron, Aluminium Piston It is a reciprocating cylinder component which is fitted in to the cylinder. The power generated by the working substance during the expansion stroke is transmitted into the piston; hence it forms the first link in transmitting the gas force to crankshaft. Material: C.I, Aluminium alloy, Cast steel Components of I.C Engine 49

50 Fig. 2.1 I.C. Engine Piston Ring These piston rings are fitted into the slots around the piston, provide a tight seal between piston and cylinder wall, thus preventing leakage of combustion gases Combustion chamber It is the space enclosed in the upper part of the cylinder, below the cylinder head and above the top of the piston surface during the combustion process. The combustion of the fuel takes place with in this space. Connecting Rod The connecting rod inter connects the piston and the crank-shaft and transmits the gas forces from the piston to the crankshaft. It has two ends called small end and big end. The small end of the connecting rod is connected with piston by using a pin called gudgeon pin. The big end of the connecting rod is connected with crank pin by using a pin called crank pin.material: Plain carbon steel, Aluminum alloys Crankshaft It converts the reciprocating motion of the piston in to useful rotary motion of the output shaft. The crankshaft is enclosed within crankcase. The crankshaft is attached with big end of the connecting rod.material: Alloys steel. Spark plug It is usually mounted on the cylinder head. It is a component which initiates the combustion process in spark ignition engines. Fuel injector This component is present in the case of combustion ignition (CI) engines. This component atomizes the fuel into fine droplets, thus injecting it at correct timing, in correct proportion during the working cycle. Inlet Manifold 50

51 It is a piping system which connects the intake system to the inlet opening. Air, as in the case of CI engine (or) air fuel mixture, as in the case of SI engine, will follow through the inlet manifold. Inlet Valve It is mounted on the cylinder head. It is used to regulate the charge (either air or air fuel mixture) coming into the cylinder.material: Nickel chrome. Exhaust Manifold It is a piping system which connects the exhaust system with the exhaust (or) outlet opening. Products of combustion from the cylinder will escape into the atmosphere through this system. Exhaust Valve It is also mounted on the cylinder head. It is used to control and regulate the discharge of combustion products from the cylinder into the atmosphere. In general the exhaust valve is subjected to higher temperature and corrosive atmosphere than the intake or inlet valve.material: Nickel chrome, Stainless steel etc. Cam Shaft The cam shaft is driven by crank shaft through timing gears having gear ratio of 2. The cam shaft is used to control the opening and closing of inlet and exhaust valves.material: Alloys steel Cam These are integral parts of the cam shaft. They are designed in such a way to open the valve at the correct timing and keep them in the same position for necessary duration and to close it. Flywheel It is mounted on the crank shaft and its function is to maintain the speed of the engine as a constant. It is done by storing excess energy during the power stroke and is utilized during remaining strokes of operation. 2.4 WORKINGOF TWO STROKE PETROL ENGINE The two stroke cycle engine requires two strokes of the piston or one revolution of the crankshaft to complete the cycle. In two stroke engines, ports are used instead of valves. The 51

52 exhaust gases are sent out from the engine cylinder by the fresh charge of the fuel entering the cylinder. In this engine the suction and exhaust strokes are eliminated. In case of petrol engine, the mixture of air and petrol is ignited by means of an electric spark produced at the spark plug. The two strokes of the engine are: (a) first stroke (b) second stroke. a) First stroke Assume that the piston is at its BDC position. During this stroke, the piston moves upwards from BDC to TDC. It closes the transfer port and the exhaust port. The charged air-petrol mixture which is already there in the cylinder is compressed. Due to upward movement of the piston, a partial vacuum is created in the crankcase and a fresh charge is drawn into the crankcase through the uncovered inlet port. At the end of the stroke, the piston reaches the TDC position. 52

53 b) Second stroke The compressed charge is ignited in the combustion chamber by means of electric spark produced by the spark plug, slightly before the completion of the compression stroke. Due to the combustion of the air-petrol mixture, the piston is acted on by a large force and is pushed in the downward direction producing the useful power. During this stroke, the inlet port is covered by the piston and the fresh charge is compressed in the crankcase. Further downward movement of the piston uncovers the exhaust port and then the transfer port. The expanded gases start escaping through the exhaust port and the same time fresh charge which is already compressed in the crankcase is forced into the cylinder through the transfer poet. The charge strikes the deflector on the piston crown, rises to the top of the cylinder and pushes out of the exhaust gases. The piston is now at the BDC position. The cylinder is completely with the fresh charge, although it is somewhat diluted with exhaust gases. The cycle of events is then repeated. 2.5 WORKING OF TWO STORKE DIESEL ENGINE As the piston moves down on the power stroke, it first uncovers the exhaust port, and the cylinder pressure drops to atmospheric pressure as the products of combustion come out from the cylinder. Further downward movement of the piston uncovers the transfer port and slightly compressed air enters the engine cylinder from the crankcase. Due to deflector on the top of the piston, the air will move upto the top of the cylinder and sent put the remaining exhaust gases through the exhaust port. During this stroke, the piston moves upwards from BDC to TDC, first the transfer port and then the exhaust port closes. As soon as the exhaust port closes the compression of the air starts. As the piston moves up, the pressure in the crankcase decreases so that the fresh air drawn into the crankcase through the open inlet port as shown in figure below. 53

54 Just before the end of the compression stroke the fuel is forced under pressure in the form of fine spray into the cylinder through the fuel inject to into this hot air. At this movement, temperature of the compressed air is high enough to ignite the fuel. It suddenly increases the pressure and temperature of the products of combustion. The rate of fuel injection is such as to maintain the gas pressure constant during the combustion period. Due to increased pressure the piston is pushed down with a great force. Then the hot products of combustion expand. During expansion some of the heat energy produced is transformed into mechanical work. When the piston is near the bottom of the stroke it uncovers the exhaust port which permits the gases to flow out of the cylinder. This completes the cycle and the engine cylinder is ready to suck the sir ones again. 2.6 WORKINGOF FOUR STROKE PETROL ENGINE (a) Suction stroke During this stroke, piston is moving from TDC to BDC due to this vacuum is created, pressure inside the cylinder decrease to lower pressure than atmospheric pressure. Because of this inlet valve opens automatically and air-fuel mixture is admitted inside the cylinder. This happens until the piston reaches BDC. Exhaust valve is in closed condition throughout the stroke. (b) Compression stroke During this stroke, both valves remain closed. As the piston moves upwards, the airfuel mixture inside the engine gets compressed. The pressure and temperature of the 54

55 charge increases, continuously of about 8 bar. Just before the end of this stroke, the spark plug initiates the spark which ignites the mixture. The pressure and temperature inside the cylinder is further increased. This completes one revolution of the crank. (c) Power stroke During this stroke, both the valves remain closed. Due to the increase in pressure, a great force exerts by the gases pushes the piston downwards. This expansion of gases continues till the piston reaches the BDC. At this position exhaust valve is opened. (d) Exhaust stroke As soon as the exhaust valve opened, the pressure falls suddenly to atmospheric. This makes the piston to move from BDC to TDC by pushing out the combustion products through the exhaust valve. This completes the second revolution of the crankshaft and constitutes one cycle of operation and the cycle is repeated. 2.7 FOUR STROKE DIESEL ENGINE Diesel engines are known as compression ignition engines as the combustion takes place due to high pressure and temperature generated during the compression stroke. In this engines fuel injector is used instead of spark plug. 55

56 2.7.1 WORKING OF FOUR STROKE DIESEL ENGINE (a) Suction stroke During this stroke, piston is moving from TDC to BDC due to this vacuum is created, pressure inside the cylinder decrease to lower pressure than atmospheric pressure. Because of this inlet valve opens automatically and air alone is admitted inside the cylinder. This happens until the piston reaches BDC. Exhaust valve is in closed condition throughout the stroke. (b) Compression stroke During this stroke, both valves remain closed. As the piston moves upwards, the air inside the engine gets compressed. Since the compression ratio is high in this engine of about 20:1 the air is compressed to a high pressure and temperature of about 60 bar and 1000 respectively. Just before the end of this stroke, fuel is injected as a fine spray into the cylinder and the combustion starts instantaneously. The combustion is continued as long as the fuel is injected. This completes one revolution of the crank. 56

57 (c) Power stroke During this stroke, both the valves remain closed. Due to the increase in pressure of gases, a great force exerts by the gases pushes the piston downwards. This expansion of gases continues till the piston reaches the BDC. At this position exhaust valve is opened. (d) Exhaust stroke As soon as the exhaust valve opened, the pressure falls suddenly to atmospheric. This makes the piston to move from BDC to TDC by pushing out the combustion products through the exhaust valve. This completes the second revolution of the crankshaft and constitutes one cycle of operation and the cyle is repeated. 2.8 COMPARISON BETWEEN FOUR STROKE AND TWO STROKE CYLINDER ENGINES Four Stroke Cylinder Engine 1. For every two revolution of the crank shaft, there is one power stroke. 2. Because of the above, turning moment is not so uniform and hence heavier flywheel is needed. 3. For the same power more space is required. 4. Because of one power stroke in two revolutions, lesser cooling and lubrication requires. Lower rate of wear and tear. 5. Valves are required inlet and exhaust valves. 6. Because of heavy weight, complicated valve mechanism and water cooled, making it complicated design and difficult to maintain. 7. The air-fuel mixture is completely utilized Two Stroke Cylinder Engine 1. For every one revolution of the crank shaft, there is one power stroke. 2. Because of the above, turning moment is more uniform and hence a lighter flywheel is used. 3. For the same power less space is required. 4. Because of one power stroke for every revolution, greater cooling and lubrication requirements. Higher rate of wear and tear. 5. Ports are made in the cylinder walls inlet, exhaust, and transfer port. 6. Simple in design, light weight and air cooled and easy to maintain. 7. As inlet and exhaust port open 57

58 thus efficiency is higher. 8. Volumetric efficiency is high due to more time for induction. 9. Lower fuel consumption per horse power. 10. Used in heavy vehicles, e.g. Buses, lorries, trucks etc. 11. The engine cost is more. 12. The exhaust is less noisy. simultaneously, sometimes fresh charges may escape with exhaust gases. The exhaust gases are not always completely removed. This cause lower efficiency. 8. Volumetric efficiency is low due to lesser time for induction. 9. The fuel consumption per horse power is more because of fuel dilution by the exhaust gas. 10. Used in light vehicles, e.g. Motor cycle, scooter, etc. 11. The engine cost is less. 12. The exhaust is noisy due to short time available for exhaust. 2.9 COMPARISON OF S.I. AND C.I. ENGINES: S.I. Engines 1. The fuel used is gasoline (Petrol). 2. Air + Fuel mixture is taken during suction. 3. For mixing air and fuel a separate device called carburettor is required. 4. Since homogeneous mixture is produced in carburettor, no need of injector. 5. Pressure at the end of compression is about 10 bar. 6. A spark plug is used to ignite the air fuel mixture. 7. Self-ignition temperature of fuel is not attained. In other words, the fuel is not selfignited. 8. S.I. Engines works on otto cycle (i.e) C.I. Engines 1. Fuel used is Diesel. 2. Only air taken during suction. 3. No need of carburetor. 4. For atomizing and spraying the fuel inside the cylinder, fuel injector is necessary. 5. Pressure at the end of compression is about 35 bar. 6. Spark plug is not necessary. 7. The fuel gets ignited due to the high temperature of compressed air. 8. C.I. Engines works on diesel cycle (i.e) 58

59 combustion takes place at constant volume. 9. Compression ratio is around 6 to Cold starting of engine is easy. 11. These are very lighter. 12. Cost is comparatively low. 13. Running cost is high. 14. Less maintenance. 15. thernal is about 25%. 16. Overheating trouble is more. 17. Spark plug needs frequent maintenance. 18. These are high speed engines. 19. Noiseless operation due to less compression ratio. 20. Engine weight / kw is less. 21. Vibration is less. 22. Generally employed for light duty vehicles e.g. two wheeler, otto etc. combustion takes place at constant pressure. 9. Compression ratio is around 15 to Cold starting of engine is diffucult. 11. Heavier engine. 12. Cost is high. 13. Running cost is not high. 14. High maintenance is needed. 15. thernal is about 35 to 45%. 16. Overheating trouble is less. 17. Fuel injector needs less maintenance. 18. These are low speed engines. 19. Very noisy operation due to high compression ratio. 20. Engine weight / kw is more. 21. More vibration is there. 22. Generally employed for heavy duty vehicles e.g. truks, buses, etc VALVE TIMING DIAGRAM: The diagram representing the opening and closing of inlet and exhaust valves during one cycle of operation in a four stroke IC engine is known as valve timing diagram. The processes of the cycle are represented with respect to the movement of crank angles as shown in figure. Inlet valve opening (IVO) is advanced before TDC and the inlet valve closing (IVC) is delayed after BDC by few degrees to maximize the suction process. More the suction, higher the power developed by the engine. Fuel injection beginning and spark ignition are advanced to complete the combustion just after TDC. Exhaust valve opening (EVO) is advanced before BDC at the cost of power stroke and its closing (EVC) is delayed after TDC by few degrees to increase exhaust stroke to ensure maximum removal of exhaust gas. 59

60 VALVE TIMING DIAGRAM FOR FOUR STROKE PETROL ENGINE: VALVE TIMING DIAGRAM FOR FOUR STROKE DIESEL ENGINE: 60

61 2.11PORT TIMING DIAGRAM The representation of opening and closing of ports in one cycle of operation with respect to the rotation of the crankshaft is known as port timing diagram. One cycle of operation in a two stroke enginee is completed in one revolution of crankshaft or two strokes of piston. This is possible due to overlapping of chargingg in exhaust process. During this period, a portion of the fresh charge whichh helps to push the exhaust gas also escapes through the exhaust port. The fresh charge that goes out without combustion reduces the thermal efficiency of the two stroke engine significantly to four stroke engine in which overlapping is minimum. The duration of power stroke represented in the port timing diagram is less than that of four stroke engine. In case of petrol engine, spark ignition (SI) takes place few degrees before TDC such that maximumm pressure due to compression is created when the piston is at TDC. In the case of diesel engine, fuel injection begins few degrees before TDC such thatt the self-ignition of diesel takes place when piston reaches TDC and there after the flame produced ignites the injected diesel spontaneously to maintainn constant pressure till the fuel injection ends. The volume expansion inside the crankcase below the bottom side of the piston sucks the fresh charge into the crankcase PORT TIMING DIAGRAM FOR TWO STROKE PETROL ENGINE: 61

62 Port Timing Diagram for Two Stroke Diesel Engine: FUEL SUPPLY SYSTEM The functions of the fuel supply system are (i) To store the fuel (ii) To supply the fuel to the enginee to the required quantity and in proper conditionn (iii) To indicate the driver the fuel level in the fuel tank FUEL SUPPLY SYSTEM OF A PETROL ENGINE: The fuel feed system of a petrol engine consists of the following components: (i) Fuel Tank (ii) Fuel Pump (iii) Fuel Filter (iv) Carburettor (v) Intake manifold and (vi) Fuel Gauge Carburetor The functionn of the carburetor is to supply the proper fuel-air ratio to the engine cylinder during suction created by the downward movement of the piston. As the piston moves downward a pressure difference is created between the atmosphere and the cylinder 62

63 which leads to the suction of air in the cylinder. This sucked air will also carry with it some droplets of fuel discharged from a tube. The tube has an orifice called carburetor jet which is open to the path of sucked air. The rate at which fuel is dischargedd into the air will depend upon the pressure difference created. To ensure the atomization of fuel the suction effect must be strong and the fuel outlet should be small. Working of Simple Carburetor: To increase the suction effect the passage of air is made narrow. It is made in the form of venturi. The opening of the fuel jet is placedat the venturi, where the suction is greatest because the velocity of air will be maximum at that point. The fig. shows a simple carburetor consists of float chamber, nozzle, a venturi, a choke valve and a throttle valve. The narrow passage is called venturi. The opening of the fuel is normally placed a little below the venturi section. The atomized fuel and air is mixed at this place and then supplied to the intake manifold of the cylinder. The fuel is supplied to the fuel jet from the float chamber and the supply of the fuel to the float chamber is regulated by the float pivot and supply valve. As the fuel level in the chamber decreases the float pivot will open the supply of the fuel from fuel tank. As the air velocity of air passes through the venturi section will be maximumm correspondingly the pressure will be minimum. Due to the pressure differencee between the float chamber and the throat of the venturi, fuel is discharged from the jet to the air. To prevent the overflow of fuel from the jet, the level of fuel in the chamber is kept at a level slightly below the tip. The quantity of the fuel supplied is governed by the opening of the butterfly valve situated after the venturi tube. As the opening of the valve is small, a less quantity of fuel-air mixture is supplied to the cylinder which results in reduced power output. If the opening of the valve is more than an increased quantity of fuel is supplied to the cylinder which results in greaterr output 63

64 Types of Carburetors: 1. Solex Carburetor 2. Carter carburetor 3. S.U. Carburetor FUEL SUPPLY SYSTEM OF DIESEL ENGINE: The main difference between the fuel supply system of a diesel engine and that of a petrol engine is, the system in diesel engine consists of a fuel injector instead of a carburetor and the remaining elements are the same. So, the components of fuel supply system of a diesel engine includes (i) Fuel Tank, (ii) Fuel Filter, (iii) Injection Pump (or) Fuel Pump, (iv) Injector, (v) Pipings, (vi) Fuel Gauge. The fuel from tank directs to the main filter through a fuel pump. After filtered, the fuel proceeds to the inlet side of fuel injection pump. From the fuel injection pump the fuel under pressure flows, in the feed pipes to the fuel injector. From the fuel injector, the fuel gets injected into the cylinder in correct proportion FUEL INJECTION SYSTEM There are two methods of fuel injection in compression ignition engines: (a) air injection system (b) airless or solid or mechanical injection Air injection system This system was first developed by Dr. Rudolph Diesel. In this method, air is first compressed to a very high pressure. A blast of this air is then injected carrying the fuel along with it into the cylinder. The high pressure air requires a multi-stage compressor. The compressor consumes about 10 % of the power developed by the engine, thus decreasing the net output of the engine. This method of fuel injection is costly and complicated. Therefore it is now obsolete. 64

65 Solid or Airless or Mechanical injection In this system the fuel is supplied at a very high pressuree (150 bar) from the fuel pump to the fuel injector and from there it is injected to the combustion chamber. It burns due to heat of compressed air. This method requires a fuel pump. This method is used in all types of diesel engines. Air injection may be classified as (a) Individual Pump System, (b) Common Rail System, and (c) Distributor System (a) Individual Pump System In this system each cylinder has its own individual high pressure pump and metering unit as shown in figure. It is quite compact method and involves higher cost. The design of this type of pump should be very accurate because the volume of fuel injected per cycle at full load is 1/20000 of the enginee displacement and during idling it is 1/ of the enginee displacement. The time allowed for injecting such a small quantity of fuel is about 1/450 sec at 1500 rpm. The pressure required is about 10 MPa to 30 MPa. (b) Common Rail System In this system, the fuel is pumped b a multi cylinder pump into a common rail, in which the pressure is controlled by a relief valve. A metered quantity of fuel is supplied too each cylinder from the common rail. The advantages of common rail system are: (i) (ii) (iii) (iv) Only one pump is required for multi cyclinder engines. It fulfills the necessary of either (i) the constant speed with variable load or (ii) the constant load with variable speed. This system has a very simple arrangement and a low maintenance cost. The variaiton of the pump supply pressur will affect all the cyclinders unifromly. 65

66 The disadvantages of common rail system are: (i) Leaks may develop in the injection valve. (ii) More accurate workmanship and design are required. (c) Distributor System In this system, there is a single highh pressure pump an in common rail system. This pump is used for metering and compressing the fuel and after that the fuel is delivered to the common rotating distributor. The distributorr supplies the fuel to each cylinder. The function of the distributor is to select the cylinder to receive the fuel according the cam coming in contact with the distributor Essential Requirements of Fuel Injection System: (i) The fuel should be injected in a fined automized condition. (ii) The fuel should be properly distributed in the combustion chamber. (iii) The fuel injection timing should occur at correct moment. (iv) Quantity of fuel injected should meet the load conditionn of the engine. (v) The beginning and end of the injection should takes place sharply IGNITION SYSTEM There are two ignition systems usually employed in Petrol engines: (a) battery or coil ignition system (b) magneto ignition system. In both there ignitionn system supplies a very 66

67 high voltage up to 20,000 volts for igniting the compressed air fuel mixture by producing spark in the plug. The following are the requirements of an ignition system: (a) The voltage from the source must be stepped up to a very high value to produce spark. (b) The intensity of spark should lie within specified limit. (c) The high voltage should be supplied to each spark plug at the correct moment. (d) There should be no failure of spark Coil (or) Battery ignition system This system is used in cars and other vehicles using petrol engines. Figure 2.shows the circuit diagram of a battery or coil ignition system. The main components of this system are: (i) battery of 6 to 12 volts, (ii ignition switch, (iii) induction coil, (iv) circuit of contact breaker, (v) condenser, (vi) distributor and (vii) spark plugs. There are two circuits in this system one is the primary circuit and the other is the secondary circuit. The primary circuit consists of a battery, ignition switch, ammeter, primary winding in the induction coil, contact breaker and a condenser. The secondary circuit consists of induction coil which has large number of fine wire turns, distributor, rotor and spark plugs. The primary winding and secondary winding are wounded on a laminated soft iron core and are insulated each other. One end of secondary winding is earthed and the other end is connected to the distributor cap. The contact breaker is driven by a cam which rotates at half the engine speed. The condenser prevents the sparking at the contact breaker points. 67

68 Working of the battery or coil ignition system When the ignition switch is switched on and the contact breaker point touches a current flow from the battery through the switch to the primary winding of the coil to the contact breaker points and the circuit is completed through the ground. The current flows through the primary winding of the coil produce a magnetic field in the coil. When the primary circuit is opened by the contact breaker points, the magnetic field collapses. Electromotive force is induced in the secondary winding of the coil. A condenser is connected across the contact breaker in the primary circuit which helps to collapse the field very quickly and produces a very high voltage in the secondary coil as there are more turns of fine wire than in the primary coil. The voltage is increased up to 20,000 volts. One end of the secondary coil is connected to the ground and the other end is connected to the external terminal of the distributor. The distributor connects the secondary coil to the different spark plugs. The distributor directs this high voltage to the proper spark plug where it jumps the air gap of the spark plug electrodes and the charge in that cylinder is ignited. In a single cylinder engine the distributor is not required as in the case of motor cycle engine, scooter engine and a single cam is sufficient for giving that spark Magneto ignition system Figure shows the circuit diagram of a magneto ignition system. The system consists of a magneto instead of battery, which produces and supplies current in the primary winding. The magneto consists of a fixed armature having primary and secondary windings and a rotating assembly which is driven by the engine. It consists of a contact breaker, condenser, distributor rotor, distributor and spark plugs. As the magnet turns, a magnetic field is produced from a positive maximum to negative maximum and back again. As this value falls from a positive maximum value, an alternating current is induced in the primary winding. This current flow in the primary circuit till that contact points are closed. When the contact points are open, a very high voltage is induced in the secondary winding. This high voltage is then directed to the proper spark plug by the distributor. This ignition system is generally used in small spark ignition engines. 68

69 2.15 COOLING SYSTEM The peak temperature that occurs during combustion in internal combustion engines varies from 1500C to 2000C. This large amount of heat produced due to fuel combustion is absorbed by the piston, cylinder head and cylinder walls. The internal combustion engine at best can transform only 30% of the heat generated by burning the fuel in to useful work. About 30% has to be removed by the cooling system and the reminder by the exhaust and lubrication systems. Whatever may be the amount of heat carried away by the coolant, it must be noted that it is a dead loss, because not only no useful work can be obtained from it, but a part of engine power is also used to remove this heat. Therefore it goes without saying that heat loss must be kept minimum by the designer NECESSITY OF ENGINE COOLING i. The high temperature reduces the strength of the materials used for piston and piston rings. ii. The large temperature differences between the engines parts may cause unequal expansion, resulting in cracking of the parts and thereby the engine failure. iii. At high temperature, the lubricating oil may be heated up to such an extent heat decomposition of lubricating oil occurs and viscosity changes may render it unfit for effective lubrication. 69

70 iv. At high temperatures, the lubricating may even evaporate and burn, injuring position and cylinder surfaces. Piston seizure due to overheating, resulting from the failure of the lubrication is quite common. v. The overheating causes excessive thermal stresses in the engine parts, which may load to their distortion. vi. The overheating may cause burning of valves and valve seats. vii. In petrol engines, the pre-ignition of the charge is possible, if the ignition parts initially are at high temperature. viii. The overheating reduces the efficiency of the engine METHODS OF COOLING All the heat rejected from the engine ultimately goes to air. Nevertheless, two basic systems are used to cool the engine. These are: (i) Direct or air cooling (ii) Indirect or Water cooling or Liquid cooling Air Cooling In this method, the heat after being conducted through the cylinder walls is dissipated directly to the air. For this purpose fins and flanges are provided on the outer surfaces of the cylinder and cylinder head. An air current is flowing continuously over the heated surface of the engine from where heat is to be removed. The amount of heat dissipated depends upon the following factors: a. Surface area of metal in contact with air b. Rate of air flow c. Temperature difference between the heated surface and the air. d. Conductivity of the metal. 70

71 Water cooling In this method of cooling, the water is circulated through water jackets around each of the combustion chambers, cylinders, valve seats and valve stems. The circulating air while passing takes heat of the combustion. When it passes through the radiator it is cooled by air drawn through the radiator by a fan and by air flow developed by the forward motion of the vehicle. After passing through the radiator, the water is again circulated. Systems of water cooling There are two systems of water cooling: (a) Thermosiphon system or natural circulation systemin this system of cooling, the circulation of water is obtained due to the difference of densities of hot and cold regions of the cooling water. There is no pump to circulate the water. The hot water from the engine jacket rises up in the hose pipe as it is lighter and goes to the radiator from the top. Then it is cooled there and goes down to the bottom of the radiator. From there it goes again in engine jacket. This system is quite simple and cheap, but cooling is rather slow. To maintain continuity of the water flow, the water must be maintained up to a certain minimum head. If the water level falls down, the circulation will discontinue and the cooling system will fail. (b) Forced circulation system In this system of water cooling the circulation of water is obtained by a pump which is driven by a V-belt from a pulley on the engine crankshaft. This system is more effective. The circulation of water becomes faster as the engine speed increases. It is not necessary to maintain the water up to a particular level COMPARISON BETWEEN AIR COOLING AND WATER COOLINGSYSTEMS Air Cooling System Water Cooling System 1. It is a direct cooling system. 1. It is an indirect cooling system. 71

72 2. The design of this system is simple and less costly. 3. It does not depend on any coolant. 4. There is no danger of leakage of the coolant. 5. The installation is easier as it does not require radiator and water jacket. Hence size is small, causing reduction in weight. 6. It works smoothly and continuously. An air cooled engine can take up some degree of damage. A broken fin does not affect much. 7. Maintenance is easier. 8. used for small capacity engines. 9. Uniform cooling of cylinder, cylinder head and valve may not be possible. 2. The design of this system is complicated and more costlier. 3. It is dependent on supply of water. 4. There is danger of leakage of the coolant. 5. The installation is comparatively difficult; size of the engine is big with an increase in weight by about 20%. 6. If this system fails, it may cause serious damage to the engine within a short time. 7. It requires more maintenance. 8. Used for medium and large capacity engines. 9. Uniform cooling is possible with water cooling LUBRICATION SYSTEM Lubrication is the admittance of oil between two surfaces having relative motion. As a matter of fact, almost all the parts of the I.C. engines have relative motion and rub each other. Due to this, friction increases and the power is lost in the engine the parts are subjected to wear and tear and reduces the life of the engine. The lubrication is required to reduce the wear of parts earlier and to carry out the part of the heat generated inside the engine. Lubrication also reduces the power required to overcome the friction of the moving parts by introducing a thin film of lubrication between them Functions of Lubrication system (i) To reduce the friction and wear between the parts having relative motion (ii) To cool the surface by carrying away the heat generated due to friction (iii)to seal the clearance between two mating parts. (iv) To clean the surfaces by carrying away the carbon particles caused by wear. (v) To absorb shock between bearings and other parts as well as to reduce the nosie simultaneously 72

73 (vi) To decrease the power required to overcome the friction (vii) To control the corrosion and rusting of parts. (viii) To provide balance between high and low temperature oil thickening Properties of lubricant oil Viscosity Viscosity describes the flow behaviour of a fluid. The viscosity of lubricating oils diminishes as temperature rises and consequently is measured at a given temperature (e.g. 40 C). The viscosity of a lubricant determines the thickness of the layer of oil between metallic surfaces in reciprocal movement. The most widely used unit of measurement of viscosity is the centistoke (cst). Viscosity index The viscosity index is a characteristic used to indicate variations in the viscosity of lubricating oils with changes in temperature. The higher the level of the viscosity index, the lower the variation in viscosity at temperature changes. Consequently, if two lubricants with the same viscosity are considered at a temperature of 40 C, the one with the higher viscosity index will guarantee: better engine start up at low temperatures (lower internal friction) a higher stability of the lubricating film at high temperatures Pour Point The pour point refers to the minimum temperature at which a lubricant continues to flow. Below the pour point, the oil tends to thicken and to cease to flow freely. Flash point and Fire point The flash point is the minimum temperature at which an oil-vapour-air-mixture becomes inflammable. It is determined by progressively heating the oil-vapour-air-mixture in a standard laboratory receptacle until the mixture ignites. Fire point of oil is the lowest temperature at which the given oil gives sufficient vapour to form a continuous flame, when a flame is passed across the surface. Cloud point 73

74 It is the lowest temperature of the fluid at which it starts solidifying or the oil changes from the liquid to plastic or solid stage which makes the oil to appear to be cloudly Types of lubrication systems The various lubrication systems is used for internal combustion engines may be classified as (i) Wet sump lubrication system (ii) Dry sump lubrication system (iii) Petroil (mist) lubrication system (i) Wet sump lubrication system This system employs a large capacity sump at the bottom of the crankcase and oil is drawn and delivered to the different engine parts with a help of low pressure oil pump. There are three major categories of wet sump lubrication system. They are (a) Splash system (b) semipressure system (c) full pressure system. a) Splash system In this system, the connecting rod is always provided with a drilled hole through its body or a hollow pipe portion to transfer the oil from bigger end to the smaller end. The caps of the big end bearing of connecting rods are fitted with scoops and point towards the direction of the rotation of the crankshaft. When the piston is at BDC the scoops is just dipped in to oil troughs and thus transferring the oil through holes to the small end bearings from the bigger end bearings of the connecting rod. The level of the oil in trough is maintained constant by the oil sump. Due to the centrifugal action of the revolving crank, the oil is splashed in all directions and a thin layer of oil is coated on the surface of the cylinder and all other parts which need lubrication. The dripping from the cylinders is also connected in the sump. Figure shows the splash lubrication system for multi-cylinder engine. It may be noted that for a single cylinder engine, there is no oil sump as the crankcase acts as a sump and hence the oil filter and the oil pump are absent. 74

75 b) Semi pressure system This method is a combination of splash and pressure system. It is similar to splash system except that the oil trough is eliminated and the oil is directly supplied to the parts which need lubrication. Oil is drawn by a low pressure pump through oil filter from the oil sump and is supplied to the main bearings and camshaft bearings by means of a gear pump at a pressure of 1 bar. Oil is also delivered to pipes under pressure which directs a stream of oil to the big end of the connecting rod by means of a spray through nozzle and thus crankpin bearings are also lubricated. c) Full pressure system In this system oil is drawn by a gear pump through filter under a pressure of 2 to 5 bar. The oil is then led to the main gallery which is then supplied to the various parts such as bearings, piston and cylinder walls as shown in figure through diagonal holes. Some of the oil from the main lead is supplied at high pressure to the crankshaft main bearings from where the oil flows to the connecting rod big end bearings through the diagonal holes drilled along the section. It is then flows to the piston and cylinder walls. Camshaft is also lubricated along the path of the oil from the main lead to the cylinder. 75

76 (ii) Dry sump lubrication system In dry sump lubrication system, the sump does not contain any oil and all the oil required for lubrication remains in the circulation only. An auxiliary storage tank is required to supply the oil to the main bearings with the help of the pump. Storage tank is located outside the engine cylinder and it gets oil from the sump through a filter by means of sump pump. The pressure of the oil is used in this system lies between 3 to 8 bar. If the filter is clogged, the pressure relief valve opens allowing the oil to flow through it and reaches the supply tank. A separate oil cooler is used to cool the oil to remove the heat from the oil. (iii) Mist Lubrication system Mist lubrication system is also known as petroil lubrication system and hence most widely in the two stroke engines fitted to the light weight motor cycle and scooters. The lubrication is accomplished by mixing the petrol thoroughly with the lubricating oil (upto 6%) in the fuel tank. Now, when the mixture is flow through a carburetor, the fuel vaporizes and the oil is in the form of mist, lubricates the main and connecting rod 76

77 bearings while the remaining oil is flows through cylinder and lubricates the piston, piston rings and cylinder. Advantages of mist lubrication: 1. System is simple and low cost due to the absence of oil pump, strainer etc. 2. Formation of deposits and corrosion of bearings are eliminated. Disadvantages: 1. Burning of lubrication oil leads to heavy exhaust emissions and formation of heavy deposit on piston crown,ring grooves and exhaust port which affected engine efficiency. 2. The system requires proper mixing of lubrication oil with fuel for efficient lubrication. 3. Most of the engines are over oiled because of petrol vaporization FUELS The performance of an I.C engine is satisfactory only when the properties of fuel used is good and hence the fundamental knowledge of types of fuels and their characteristics is essential Requirements of fuel A good fuel should fulfill the following requirements. 1. High energy density. 2. Easy to handle and store. 3. High thermal stability. 4. Low toxility. 5. Easy availability. 6. Free from hazard and chemical reaction. 7. Low deposit forming tendency. 8. Good combustion qualities. 9. Products of combustion should be non-corrosive to the engine parts. 10. Air pollution effect should be minimum. 77

78 Types of fuels I.C engines are mostly operated on liquid or gaseous fuels. So, these two types are described briefly below Gaseous fuel Gaseous fuels are most widely used in S.I engines. The various gaseous fuels used in IC engines are natural gas, By-product gas, Bio gas plants. Natural gas is available with oil wells and it does not have any color or odour. Manufactured gases such as coal gas in manufactured by heating soft coal in closed vessel and water gas is formed by using steam. By-product gases are produced during the manufacture of other substances. The biogas produced from the cow dug which ia available in large quantities in India. It can be easily manufactured with any chemical reaction and therefore it is easy to manufacture and use locally. Advantages ofgaseous fuel 1. It can be easily carried through pipes. 2. It can be easily compressed and stored. 3. Less starting troubles and freezing problems are eliminated. 4. Engine can be operated on lean mixture. 5. Less deposits. Disadvantages: 1. Storage volume per unit energy is very comparatively large. 2. High cost and size and weight of the engine is comparatively large. 3. Purification cost is high. 4. Capital and running costs are also high. Liquid Fuels All over the world, 99% of the I.C engines employs liquid fuels such as petrol and diesel. All liquid fuels used in I.C engine are derived from crude petroleum which is naturally available in large quantities. The two most commonly used liquid fuels are petrol and diesel. Petrol gets vaporized easily in the combustion chamber during combustion. Diesel oil is injected directly into the combustion chamber which also vaporizes when it makes contact with air. Petroleum is a compound of paraffin, Naphthalene and aramatics. Sometimes olefins are added to this.they form gummy deposits and their percentages are kept as low as possible in the fuel used. 78

79 Advantages 1. Higher calorific value 2. Elimination of wear and tear of grate 3. Easy starting and stopping 4. Easy combustion control 5. Easy handling and supply 6. Less space and cleanliness of the surrounding 2.19 PERFORMANCE OF INTERNAL COMBUSTION ENGINE Indicated mean effective pressure: Area of Indicator DiagramScaleof the diagram Mean effective Pressure (mep) = Length of theindicator diagram = AS L Where, A is the area of the indicator diagram in mm 2 S is the scale of the indicator diagram in bar/mm L is the length of the indicator diagram in mm. Hence the mep will obtained in terms of bar Indicated Power: It is the power available inside the engine cylinder I.P. = Mean Effective Pressure X Stroke Volume LAn Stroke volume 60 Pm. LAn.. IP , /sec m Where L = Stroke length in m A = Area of the piston in m 2 79

80 n = Number of working strokes per min. = N for two stroke engines = N/2 for two stroke engines N = Speed in rpm P m = Mean Effective Pressure in N/m 2 and I.P= Indicated power in Watts. For multi cylinder engines, if K is the number of cylinders, the Indicated power is given by, KPm LAn IP Efficiencies of Internal Combustion Engine Brake Thermal Efficiency BP bte Heat in fuel BP3600 m CV f Where, m f = mass flow rate of fuel in kg/hr CV = Calorific Value of the Fuel in kj/kg BP = Brake Power developed in kw Indicated Thermal Efficiency: ite IP Heat Supplied IP 3600 m CV f 80

81 Mechanical Efficiency: m BP IP Relative Efficiency: Relative efficiency = Indicated Thermal Efficiency ite Air Standard Efficiency ase Volumetric Efficiency: Actual volume of air taken V vol = swept volume V a s V a = Actual Volume of air taken 2 = C d 2gh 4 d a a m 3 s V s = Swept Volume = K DLn m 3 s Where, D = Diameter of the cylinder in m L = Stroke length in m K = Number of the cylinder and n = Number of working cycles per min. 81

82 2.20 TESTING OF I.C. ENGINES Measurement of Indicated Power (I.P) The power developed inside the engine cylinder is known as indicated power and is designated as I.P. The indicated power is measured by indicator card with the help of an instrument called indicator. An indicator card is the graphical representation of the pressure-volume variation during the one working cycle. The indicator diagram has a positive loop and a negative loop. The area between the compression and expansion lines is called positive loop and the area between suction and exhaust lines is called negative loop. The positive loop represents the work gross work during the cycle and negative loop represents the pump loss due to admission of fresh charge and removal of exhaust gases. Net work done per cycle = area of positive loop area of negative loop. The areas of positive loop and negative loop are measured with the help of planimeter. Let, A area of positive loop in mm A area of negative loop in mm h = mean height of indicator diagram L = length of the indicator diagram in mm n = no. of explosions per minute S = spring scale in N/m / Then, A A A A N/m Generally area of the negative loop is negligible as compared with positive loop. Indicated power per cylinder Measurement of Brake Power (B.P) The power available at the engine crankshaft is known as brake power. The brake power is less than the indicated power because of different losses such as pump losses, mechanical losses in bearings, power required to drive the fuel pump, water pump and governor. The brake power is measured by coupling the brake dynamometer to the engine shaft. 82

83 Rarm length in m Nrpm of engine Ttorque in Nm Brake power Let, W load measured on the dynamometer in N WR2πN kw Measurement of I.P. of Multi-cylinder Engine The difference between the indicated power and the brake power is the friction power and it depends on engine speed. Friction power also depends on the indicated power as the increased pressures on piston increases friction Morse test Morse test is conducted to find indicated power of multi-cylinder engines. When a cylinder is disconnected by short circuiting spark plug or fuel supply, the brake power developed by it is reduced. The reduction in brake power corresponds to the indicated power of that cylinder if the friction power remains constant. For a four cylinder engine, when all cylinders are working,,,,,,,,, When cylinder no. 1 is disconnected, Indicated power,, Friction power,,, Brake power,, Subtracting, Indicated power Brake power,,, Brake power,, As the friction power is same in both cases even when cylinder 1 is not producing power. Similarly Indicated power Brake power,,, Brake power,, Indicated power Brake power,,, Brake power,, Indicated power Brake power,,, Brake power,, Indicated power of the engine IP IP IP IP In spark ignition engines, the morse test will be conducted by disconnecting the spark plug lead of each engine successively. The high voltages are dangerous and there will also be unburnt mixture entering the exhaust system. But in Multi-point fuel injection engines it is easy 83

84 to electrically isolate an injector. In diesel engines, the injectors can be disconnected, either mechanically 9by disconnecting the fuel supply) or electrically as appropriate Willans Line method The friction power of a single cylinder engine can be found out by plotting experimental values of fuel consumption and brake power as shown in fig.this line representing the linear variation of TFC with brake power is known as Willans line. It is assumed in this method that at constant speed the friction power is constant. Willans line can be extrapolated to zero fuel flow rate to determine the friction power. At zero power output, the fuel consumption is to overcome only the friction to run the engine at rated rpm. The equivalent power is obtained by extending the line on the left hand side to meet the power axis. Same scale is to be maintained on both sides of zero on the power axis Measurement of Air Consumption Orifice chamber method is used in laboratory for measuring the consumption of air. The arrangement of system is shown in fig. It consists of an air tight chamber in which sharp edged orifice is fitted. The orifice is situated away from the suction connection to the engine. A rubber diaphragm is provided to further reduce the pressure pulsations. There is a pressure depression due to the suction of the engine which causes the flow through orifice for obtaining steady flow; the volume of the chamber should be sufficiently large as compared with swept volume of the cylinder. Generally 500 to 600 times the swept volume. The pressure difference which causes the flow through the orifice is 84

85 measured with the help of manometer fitted the air box. The pressure difference is kept to 10 cm of water to make the compressibility effect negligible. Let, A area of orifice in m h head of water in cm d diameter of orifice in m ρ density of air in kg/m C Coefficient of discharge of orifice Head interms of air in m is given by H. ρ h 100 ρ H h ρ 10 h 100 ρ ρ The velocity of air passing through the orifice is given by V 2gH m/s The volume of air passing through the orifice is given by VAV C AC 2g10 h 14 C ρ A h m ρ s Mass of air passing through the orifice is given by, m m V ρ 14 C A ρ h ρ 14 C A ρ h The density of atmospheric air is given by, P V m R T ρ m V P R T Where, P is the atmospheric pressure in N m T atmospheric temperature in K The volumetric efficiency of the engine is given by, actual volume of air taken displacement volume 14 C A 60 η D LN No. of cylinders When the volumetric analysis of the exhaust gases is known, then the mass of air supplied per kg of fuel is given by, 85

86 m kgof fuel N C 33C C Where, N= Percentage of nitrogen by volume in exhaust gases. C 1 = Percentage of carbon dioxide by volume in exhaust gases. C 2 = Percentage of carbon monoxide by volume in exhaust gases. C = Percentage of carbon in fuel by weight Measurement of fuel consumption An arrangement for measuring the fuel consumption rate is shown in fig. A small glass tube is fitted to the main fuel tank. When the fuel consumption rate is to be measured the valve closed and the fuel is consumed from the burette as shown in figure. For a known value of fuel consumption, the time is measures and the fuel consumption rate is calculated as under, Fuel consumption kg hr V Sp. gravity of fuel 1000 t Measurement of heat carried away by cooling water The heat carried away by cooling water is generally measured by measuring the water flowrate through the cooling jacket and the rise in temperatures of the water during the flowthrough the engine. The inlet and out let temperatures of the water are measured by the thermometers insertingin the pockets provided at inlet to and outlet from the engine. The quantity of waterflowing is measured by collecting the water in a bucket for a specified period or directlywith the help of flow meter in case of large engine. The heat carried away by cooling wateris given by Where, m mass of water per minute C specific heat of water generally taken as kj/kg T outlet temperature of water T inlet temperature of wate Measurement of heat carried away by exhaust gases The mass of air supplied per kg of fuel used can be calculated by using the equation. 86

87 N C m 33C C The heat carried away by exhaust gas per kg of fuel is given by 1 Where, m 1 mass of exhaust gas formed per kg of fuel supplied C specific heat of exhaust gases T temperature of exhaust gases coming out from the engine T ambient temperature 2.21 HEAT BALANCE SHEET A heat balance sheet is an account of heat supplied and heat utilized in different ways in a system. The performance of the engine is obtained from the heat balance sheet. A heat balance account includes the following items. Heat supplied by the fuel to the engine m L.C.V Where, m = Total fuel consumption in kg/hr LCV = Lower Calorific value of the fuel in kj/kg (a) Heat equivalent of brake power = brake power x 60 kj/min (b) Heat lost to jacket cooling water = (c) Heat lost to exhaust gases = (d) The rest of the heat is lost by convection and radiation. This cannot be measured and so this is known as unaccounted loss. This is calculated by the difference of heat supplied and the sum of (a) + (b) + (c). Heat supplied per minute Heat supplied by the combustion of fuel kj % Heat expenditure per minute kj % 100 (a) heat equivalent of brake power (b) heat lost to jacket of cooling water (c) heat lost to exhaust gases

88 (d) unaccounted heat loss(heat supplied (a)+(b)+(c)) Total Total SOLVED PROBLEMS 1. The following observations were taken during trial on a single cylinder oil engine. During of test=1 hour, Total number of revolutions = 12000, load on the brake=2 KN, spring reading = 0.5 KN, mean effective pressure = 6 bar, fuel consumption = 7.6 kg, temperature of exhaust gases = 300 C, C.V of fuel =45000KJ/kg, total cooling water circulated =550 kg, inlet temperature of cooling water = 15 C, outlet temperature of cooling water =60 C, brake drum diameter = 1.83 m, Bore = 30cm, atmospheric temperature =20 C, air consumption =360 kg, C p for exhaust gases =1 KJ/kgK. Determine i) brake power, ii) mechanical efficiency, iii) indicated thermal efficiency of B.H.P basis and iv) draw up the heat balance sheet. Solution a) Brake power: b) Mechanical efficiency: c) Brake thermal efficiency: %

89 d) Heat balance sheet: % 1) Quantity of heat supplied. 2) Heat in B.P per minute ) Heat carried away by cooling water ) Heat carried away by exhaust gases, ) Unaccountable heat loss, Heat balance sheet Heat supplied % Heat distributed % Heat supplied by the fuel ) Heat in B.P 2) Heat carried by cooling water 3) Heat in dry exhaust gases 4) Heat in steam Heat in Heat out

90 2. A four cylinder, four stroke cycle stroke cycle petrol engine 82mm bore, 130 mm stroke develops KW brake power while running at 1500 rpm and using a 20% rich mixture. If the volume of mercury is 70% of the swept volume, the theoretical air fuel ratio is 14.8, the heating value of petrol used is 44000kJ/kg and the mechanical efficiency of the engine is 90%, find the indicated thermal efficiency. Take R =0.287 kj/kg K. Solution l For 20% rich mixture, S. F. C based on indicated power I. P % 90

91 3. A six cylinder gasoline engine operates on four-stroke cycle. The bore and stroke of each cylinder is 100 mm and 120 mm respectively. The clearance volume per cylinder is 80 cc. AT a speed of 4000rpm, the fuel consumption is 20 kg/hr and the torque developed is 200N-m. Calculate a) Brake power, b)brake mean effective pressure, c) brake thermal efficiency if C.V = 43MJ/kg, d) relative efficiency on brake power basis assuming the engine works on constant volume cycle. Solution a) Brake power: b) Brake mean effective pressure: c) Break thermal efficiency S. F. C based on brake power B. P d) Relative efficiency: %

92 4 l % % A test on a single cylinder four stroke oil engines having bore and stroke of 180mm and 360 mm gave the following results. Speed=290 rpm, brake torque =392 N-m, IMEP=7.2 bar,oil consumption =3.5kg/hr, coolant flow =270 kg/hr, cooling water temperature rise =35 C,A:F ratio =25, exhaust gas temperature =415 C, room temperature =21 C, C.V of fuel = KJ/kg and take C p for exhaust gases as KJ/KgK. Calculate a) indicated thermal efficiency, b) heat balance sheet in KJ/min basis. Solution Indicated thermal efficiency Heat balanced sheet: % ) Heat supplied by the fuel

93 2) Heat in B.P ) Heat carried away by cooling water, ) Heat carried away be exhaust gases, ) Unaccountable heat loss, Heat supplied Heat supplied by the fuel HEAT BALANCE SHEET % Heat distributed % ) Heat in B.P ) Heat carried by exhaust gases ) heat carried by

94 cooling water 4) Heat in steam Heat in Heat out An eight cylinder four stroke engine of 9 cm bore and 8 cm stroke has a compression ratio of 7 is tested at 4500 rpm on a dynamometer which has a 54 cm arm. During a ten minute test, the dynamometer load reading was 45 MJ/kg. Air at 27 C and 1 bar was supplied to the carburetor at the rate of 6 kg/min. Find for petrol engine i) brake mean effective pressure, ii) specific fuel consumption and specific air consumption, iii) Relative efficiency, iv) Volumetric efficiency. Given Data: D = 9 cm = 0.09 m L = 8 cm = 0.08 m r k = 7 N = 4500 rpm X = 54 cm = 0.54 m t = 10 min W = 42 kg = 42 x 9.81 = N C.V = 44 MJ/kg = KJ / kg T = 27 0 C = 300 K P = 1 bar = 100 Kpa a) Break mean effective pressure

95 b) SFC and SAC c) Relative efficiency % 7. d) Volumetric efficiency % % l % OBJECTIVE QUESTIONS 1. In a four stroke engine, the working cycle is completed in (a) One revolution of the crank shaft (b) two revolutions of the crank shaft 95

96 (c) three revolutions of the crank shaft (d) four revolutions of the crank shaft 2. A two stroke cycle engine gives the number of power strokes as compared to the four stroke cycle engine, at the same engine speed. (a) half (b) same (c) double (d) four times 3. The thermal efficiency of a two stroke cycle engine is a four stroke cycle engine. (a) equal to (b) less than (c) greater than 4. The theoretically correct mixture of air and petrol is (a) 10:1 (b) 15:1 (c) 20:1 (d) 25:1 5. The thermodynamic cycle on which the petrol engine works, is (a) Otto cycle (b) Joule cycle (c) Rankine cycle (d) Stirling cycle 6. A diesel engine has (a) one valve (b) two valves (c) three valves (d) four valves 7. If petrol is used in a diesel engine, then (a) low power will be produced (b) efficiency will be low (c) higher knocking will occur (d) black smoke will be produced 8. A petrol engine has compression ratio from (a) 6 to 10 (b) 10 to 15 (c) 15 to 25 (d) 25 to The function of a distributor in a coil ignition system of I.C. Engines is (a) to distribute the spark (b) to distribute the power (c) to distribute the current (d) to time the spark 10. Super charging the power developed by the engine. (a) has no effect on (b) increases (c) decreases 10. Supercharging the power developed by the engine. (a) has no effect on (b) increases (c) decrease 11. A carburetor is used to supply (a) Petrol, air and lubricating oil (b) air and diesel (c) Petrol and lubricating oil (d) petrol and air 12. A spark plug gap is kept from (a) 0.3 to 0.7mm (b) 0.2 to 0.8mm (c) 0.4 to 0.9mm (d) 0.6 to 1.0mm 13. The knocking tendency in spark ignition engines may be decreased by (a) Controlling the air fuel mixture (b) controlling the ignition timing (c) reducing the compression ratio (d) all of these 14. The violent sound pulsations within the cylinder of an I.C. Engine are due to 96

97 (a) detonation (b) turbulence (c) pre-ignition (d) none of these 15. Which of the following does not relate to a spark ignition engine? (a) Ignition coil (b) Spark plug (c) Distributor (d) Fuel Injector 16. In an internal combustion engine the fuel is (a) burnt outside the cylinder (b) inside the cylinder (c) not burnt anywhere (d) none of the above 17. The power produced inside the cylinder of an internal combustion engine is known as (a) brake power (b) indicated power (c) frictional power (d) none of the above 18. The power available at the shaft of an internal combustion engine is known as brake horse power and is equal to (a) total power produced frictional power (b) net I.P. - frictional power (c) net I.P. + frictional power (d) net I.P. / frictional power 19. In a two stroke engine, the working cycle is completed in the number of revolutions of the crank shaft equal to (a) one (b) two (c) three (d) four 20. The indicator of an internal combustion engine is used to determine (a) Temperature (b) B.P (c) Speed (d) I.P. and mean effective pressure 21. The pressure at the end of compression in petrol engine as compared to that of diesel engine (a) higher (b) lower (c) same (d) none of the above 22. Morse test is performed on internal efficiency engines to determine mechanical efficiency of (a) Single cylinder spark ignition engine (b) Single cylinder compression ignition engine (c) Multi cylinder engines (d) none of the above 23. For a four cylinder inline internal combustion engine, the most popular firing order is (a) (b) (c) (d) The increase in cut off ratio, of a diesel cycle with fixed compression ratio, would (a) decreasedm.e.p. (b) increasem.e.p. (c) keep same m.e.p. (d) none of the above 25. Fuel injector is used for a (a) Spark ignition engine (b) Compression ignition engine (c) Steam engine (d) Gas engine 97

98 TWO MARKS QUESTIONS 1. What is the difference between an I.C engine and E.C engine? 2. What is a four stroke engine? 3. What is as two stroke engine? 4. What is the material with which the following components are made up of? (i) Cylinder block (ii) Piston (iii) Connecting rod (iv) Crank Shaft 5. Draw the valve timing diagram of a four stroke petrol engine? 6. Draw the port timing diagram of a two stroke diesel engine? 7. What is scavenging? 8. What is carburetion? 9. What is fuel injection pressure of an air less injector? 10. Name two types of ignition systems used in petrol engine? 11. What is the function of a flywheel? 12. What is the necessasity for cooling an I.C engine? 13. What is lubrication? 14. Why pressure lubrication is prefered for muti cylinder engine? 15. What is Petro-oil lubrication? 16. What is the difference between Knocking and Detonation? 17. Draw the Actual P-V diagrams of four stroke Petrol and Diesel Engines 18. Define Brake power and Brake thermal efficiency. 19. Define Indicated Power and Indicated Thermal Efficiency. 20. Define Mechanical Efficiency and Volumetric Efficiency. DESCRIPTIVE QUESTIONS 1. What are the difference between S.I and C.I engines? 2. What are the merits and demerits of a two stroke engine over four stroke engine? 3. How are I.C engines classified? 4. Explain the working of four stroke petrol engine with neat sketches. 5. Describe the purpose of the different parts of a four stroke C.I engine with neat sketches 6. Explain the working of two stroke Diesel engine with neat sketches. 7. Describe the working of single Jet carburetor with neat sketch. What are its limitations? 98

99 8. What are the fundamental requirements of a fuel injection system used in Diesel engines? 9. Draw the critical diagram and describe the working of a battery ignition system used for a four cylinder petrol engine. 10. With neat sketches describe the working of water cooling system used for multi cylinder engines 11. Explain with neat sketch the working of a pressure lubrication system. 12. Explain the phenomena of knocking in S.I engine.what are the different factors which influence knocking? Describe the methods used to suppress it. EXERCISE PROBLEMS 1) A petrol engine uses per brake power hour 0.36kg of fuel of calorific value kj/kg. The mechanical efficiency is 78%, compression ratio is 5.6. Calculate 1) Brake thermal efficiency 2) Indicated thermal efficiency 3) Ideal air standard efficiency. Take γ = ) The output of an IC engine is measured by a rope brake dynamometer. The diameter of the brake pulley is 75 mm and rope diameter is 52mm. The dead load on the tight side of the rope is 410N and the spring balance reading is 50N. The engine consumes 4kg/h of fuel at rated speed of 1000rpm. The calorific value of fuel is 44150kJ/kg. Calculate brake specific fuel consumption and the brake thermal efficiency. 3) A compression ignition engine rated condition develops 7.5kW brake power. The mechanical losses are 1.5Kw. If the indicated thermal efficiency is 42%; air fuel ratio 22 and calorific value of fuel 43250kj/kg. Determine 1) Fuel consumption is kg/hr 2) air intake in kg/hr and 3) brake thermal efficiency. 4) A four cylinder, two stroke cycle petrol engine develops 30kw brake power at 2500rpm. The MEP on each piston is 8bar and the mechanical efficiency is 80%. Calculate the diameter and stroke of each cylinder if the stroke to bore ratio is 1.5. Also calculate the brake specific fuel consumption of the engine, if brake thermal efficiency is 28%. The calorific value of the fuel is 44150kJ/kg. 5) An engine is used on a job requiring 110kw B.P., the mechanical efficiency of the engine is 80% and the engine uses 50kg fuel/hr under the conditions of operation. A design improvement is made which reduces the engine friction by 5kw. Assuming the indicated thermal efficiency remains the same, how many kg of fuel per hour will be saved? 99

100 6) A four stroke, six cylinder gas engine with a stroke volume of 1.75litres develops 25kW at 480rpm. The mean effective pressure is 6 bar. Find average number of times each cylinder misfired in one minute. 7) An engine uses 6.5kg of oil per hour of calorific value 3000kj/kg. if the B.P of the engine is 22Kw and mechanical efficiency 85%, calculate, 1) indicated thermal efficiency 2) Brake thermal efficiency and 3) specific fuel consumption in kg/b.p/hr. 8) A four cylinder two stroke cycle petrol engine develops 23.5kw brake power at 2500rpm. The mean effective pressure on each piston is 8.5bar and the mechanical efficiency is 85%. Calculate the diameter and stroke of each cylinder, assuming the length of stroke equal to 1.5 times the diameter of cylinder. 9) Four stroke petrol engine has a piston displacement of 221cm 3. The compression ratio is 6.4. The fuel consumption is 0.13kg/min. The calorific value of fuel is 45000kJ/kg. The brake power developed while running at 2500rev/min is 50.25kW. Determine the brake mean effective pressure and the relative efficiency based on brake thermal efficiency. 10) A petrol engine is to deliver 6.5kW when running at 2000re/min with a mechanical efficiency of 80%. The stroke-bore ratio is 2.5. Expansion and compression are assumed to be adiabatic. The clearance volume is one-fifth the swept volume of the piston. The maximum pressure reached is three times the pressure at the end of the compression. The inlet pressure is 1.0bar. Calculate the diameter and stroke of the piston. 11) A four cylinder four stroke petrol engine working on the Otto cycle consumes 7kg of petrol per hour. The compression ratio of the engine is 5. The thermal efficiency of the engine is 62% of the air standard efficiency. Calculate the thermal and air standard efficiencies. The calorific value of the fuel is 45000kJ/kg. Determine the power developed per cylinder. 12) How do you measure the Brake power of an engine? A four stroke engine is loaded by means of a rope brake mounted on a brake drum of mean diameter 1.37m. The dead weight is 27.5kg and the spring balance reading is 1.82kg. The speed is 250rev/min., the bore and stroke are 150mm and 230mm respectively. The mean effective pressure based on indicated power is 7.0bar. Determine (i) mechanical efficiency (ii) thermal efficiency on the I.P basis, if the engine uses 1.22 kg of fuel per hour of calorific value 42000kJ/kg. 13) A Morse Test on a four cylinder engine resulted in the following data: Brake Power with all cylinders working = 21.7 kw Brake Power with No.1 cylinder cut-off = 15.5 kw Brake Power with No.2 cylinder cut-off = 15.6 kw 100

101 Brake Power with No.3 cylinder cut-off = 15.7 kw Brake Power with No.4 cylinder cut-off = 15.5 kw Calculate the mechanical efficiency and indicated thermal efficiency, if the engine uses 0.07 kg of fuel per minute. CV of fuel is kj/kg. 14) A stroke petrol engine 70mm bore and 90mm stroke, is tested at full throttle at constant speed condition. The fuel supply is found to be 0.065kg/min. and the plugs of the four cylinder are successfully short circuited without change of speed, the brake torque being correspondingly adjusted. The power measurements are, i. With all cylinders working = kw ii. With No.1 cylinder cut-off = kw iii. With No.2 cylinder cut-off = kw iv. With No.3 cylinder cut-off = kw v. With No.4 cylinder cut-off = kw Determine the mechanical efficiency and brake thermal efficiency if the calorific value of fuel is kj/kg. 15) The following observation were made during a trial of a constant speed compression ignition engine operating on the four stroke cycle: Brake wheel diameter = 600mm; Band thickness = 5mm; speed = 45rpm; Load on band = 206N; Spring balance reading = 3.0N; Area of indicator diagram = 14.15cm 2 ; Length of indicator diagram = 6.25cm; spring value = 108.0N/cm 2 ; Bore = 100mm; Stroke = 150mm; specific fuel consumption = 0.295kg/kW/hr; Heating value of fuel = kj/kg. Determine: (i) Mechanical efficiency and (ii) Brake Thermal efficiency. 16) A four cylinder four stroke diesel engine is to develop 30Kw at 1000re/min. The stroke is 1.4 times the bore and the indicated mean effective pressure is 6.0 bar. What should be the stroke and diameter of the engine? 17) In a test on a two stroke oil engine, the following results were obtained: Speed = 350 rev/min; Net brake load = 600N; MEP =2.66bar; Fuel consumption = 3.2kg/hr; cooling water used = 495kg/hr; temperatures of jacket water at inlet and outlet = 30 o C and 50 o C; Exhaust gases per kg of fuel = 32 kg; Temperature of exhaust gases = 432 o C; specific heat of exhaust gases = kj/kgk; Inlet air temperature = 32 o C. Draw up a heat balance for the engine if its cylinder diameter = 205mm and stroke = 275mm; Brake drum diameter = 1.0m; calorific value of fuel = kj/kg. 101

102 18) The following particulars refer to the full load test of a single cylinder, petrol engine working on the four cycle: Speed = 2500 rpm; Brake power = 118kW; Cylinder bore = 110mm; Cylinder stroke = 120mm; Lower calorific value of fuel = kj/kg; Petrol consumption = 40kg/hr; Jacket water rate = 2800kg/hr; Jacket water inlet temperature = 20 o C; Jacket water outlet temperature = 65.5 o C; Fuel-air ratio = 1:16; Room Temperature = 29 o C; Exhaust Temperature = 399 o C; Hydrogen in fuel = 15% by mass; Sp.heat of dry exhaust gases = kj/kgk; Sp.heat of water vapour = kj/kgk. Draw the heat balance sheet. Calculate the brake thermal efficiency and volumetric efficiency of the engine. 19) Calculate the brake specific fuel consumption, indicated thermal efficiency and obtain a heat balance sheet on minute basis from the following test data obtained in a four stroke two cylinder diesel engine: Duration of test = 1 hour; Brake power = 15kw; Total indicated power = 17.8kw; Fuel consumption = 4.24 litres of specific gravity 0.875; Lower calorific value of fuel = kj/kg, Jacket water circulated = 215kg; Inlet and outlet cooling water temperature = 32 o C and 80 o C. The heat in exhaust gases is measured by an exhaust gas calorimeter as 808kJ/minute. 20) Calculate the bore and stroke of a four stroke single cylinder oil engine to the following particulars. Brake power 18kw at 250rpm when running on oil having composition by mass C-85%, H-15%, and a lower calorific value of 42000kJ/kg. The oil is burnt with 25% excess air. The volumetric efficiency reckoned on atmospheric conditions of bar and 10 o C is 0.8. The mechanical efficiency is 0.9 and indicated thermal efficiency is Bore-Stroke ratio is 1:1.2. a. Take R = 0.287kJ/kgK. 21) A Six cylinder, four stroke S.I Engine, having a piston displacement of 700 x 10-6 m 3 per cylinder developed 78kw at 3200rpm and consumed 27 kg of petrol per hour. The CV of petrol is 44MJ/kg. Estimate 1) the volumetric efficiency of the engine if the air fuel ratio is 12 and the intake air is at 0.9bar and 32 o C; 2) the brake thermal efficiency and 3)the brake torque. For air R = 0.287KJ/kgK. 22) A four cylinder four stroke petrol engine produces 56kW indicated power when running at 4400rpm with a volumetric efficiency of 81.5%. the air-fuel ratio is 16:1 and the thermal efficiency is 35%.The fuel used has a calorific value of 44100kJ/kg. If the bore to stroke ratio is 1:1.04, calculate the cylinder dimensions. Assume the charge to have the density of air equal to 1.293kg/m 3 at N.T.P. 102

103 23) A diesel engine has a relative efficiency of 0.55 on the brake. If the compression ratio is 13.8 and the expansion ratio is 7.4 and the calorific value of fuel is kj/kg. Find the consumption of fuel oil in kg/kw/hr. 24) A twin cylinder oil engine coupled to an electric D.C generator, when tested forfuel consumption recorded the following data: Line current = 20Amps at 220volts; time for consumption of 10c.c of fuel = 17.8 seconds; RPM of the engine = 1000; Efficiency of generator = 79%; Specific gravity of fuel = 0.835; calorific value of the fuel oil = kj/kg. Calculate the specific fuel consumption in kg/kw/hr and brake thermal efficiency. 103

104 CHAPTER III STEAM NOZZLES 3.1 INTRODUCTION A steam nozzle is a passage of varying cross section, which converts heat energy of steam into kinetic energy. During the first part of the nozzle, the steam increases its velocity. But in its later part, the steam gains more in the volume than in the velocity. Since the mass of steam, passing through any section of the nozzle remains constant, the variation of the steam pressure in the nozzle depends upon the velocity, specific volume and dryness fraction of the steam. A well designed nozzle converts the heat Energy of steam into kinetic Energy. 3.2 TYPES OF STEAM NOZZLES Nozzles are classified into three types based on the shape. (1) Convergent nozzle. When the cross sectional area of a nozzle decreases continuously from entrance to exit, is called convergent nozzle. It is shown in Fig This nozzle is used where back pressure is equal or more than the critical pressure. (2) Divergent nozzle. When the cross sectional area of a nozzle increases continuously from entrance to exit, is called divergent nozzle. It is shown in Fig This nozzle is used where back pressure is less than the critical pressure. (3) Convergent divergent nozzle When the cross sectional area of a nozzle first decreases gradually from entrance to throat, and then increases from throat to exit, is called convergent-divergent nozzle. It is shown in Fig Throat is the area where cross sectional area is minimum, after throat area increases. This nozzle is used where back pressure is less than the critical pressure. 104

105 Fig. 3.1 Fig. 3.2 Fig FLOW OF STEAM THROUGH NOZZLE When steam is following through the nozzle, the following assumptions are considered i. No heat is supplied to or rejected from steam ii. No external work done during the flow iii. Frictionless adiabatic expansion of steam Velocity of steam Consider a flow of steam through nozzle as shown in fig At inlet pressure P 1, enthalpy h 1 and velocity V 1 is to be increased to a high velocity using a nozzle. Applying steady flow 105

106 energy equation between the entrance and exit where the pressure is P 2, enthalpy h 2 and velocity V 2, we get Mass of steam flowing is 1 kg/s, h V 2 Z h V 2 Z Z 1 = Z 2 because nozzle is kept horizontal h V 2 h V 2 V 2h h V V 2h h V V 1 is negligible when compared to V 2 h h V 2 V 2h h V 2h h Where h 1 and h 2 in kj/kg and V 1 and V 2 are in m/s 3.4 MASS OF STEAM DISCHARGED THROUGH NOZZLE We have already discussed that the flow of steam, through the nozzle is isentropic, which is approximately represented by the general law: Pv n = Constant We know that gain in kinetic energy V = n n 1 Heat drop = pv pv.(neglecting initial velocity of steam) Since gain in kinetic energy is equal to heat drop, therefore

107 2 V2 n 2 n 1 p v p v n p v pv 1 n pv 1 1 n we know that p1v1 p2v2 n 1 1 n n v p p v1 p2 p1 v1 Substituting the value of v 2 v 1 n p 2 2 v1 p1 V 2 V2 n p 2 p 2 pv n1 p1 p1 n p2 = pv n1 p1 n1 n 1 n n1 n n p pv n 1 p1 Now the volume of the steam flowing per second = Cross sectional area of nozzle velocity of steam = AV 2 and volume of 1kg of steam i.e., specific Volume of steam at pressure P 2 = V 2 m 3 /kg Mass of steam discharged through nozzle per second, volume of steam flowing per sec ond m = volumeof 1kg of steam at pressure p 2 107

108 AV2 A n p 2 2 pv v2 v2 n1 p1 n1 n 1 n1 n n A p 1 n p 2 2 pv v1 p2 n1 p 1 A v p n p 1 n1 p n n 2 2n p2 pv p p p A 1 p1 n1 v 1 p1 2 n1 n n 2 2n n1 n n 2n p 1 p 2 p 2 A n 1 v 1 p 1 p CRITICAL PRESSURE RATIO A nozzle is normally designed for maximum discharged by designing a certain throat pressure which produces this condition. Let P 1 = Initial pressure of steam in N/m 2 P 2 = Pressure of steam at throat in N/m 2 V 1 = Volume of 1kg of steam at pressure P 1 in m 3 V 2 = Volume of 1kg of steam at pressure P 2 in m 3, and A = cross sectional area of nozzle at throat, in m 2. 2 n1 n n 2n p 1 p 2 p 2 m A n 1 v 1 p 1 p 1 There is only one value of the ratio P 2 /P 1, which produces maximum discharge from the nozzle. This ratio P 2 /P 1 is obtained by differentiating the right hand side of the equation. We 108

109 see from this equation that except P 2 /P 1, all other values are constant. P 2 /P 1 is differentiated and equated to zero for maximum discharge. 2 n1 n n d p 2 p 2 p p 2 1 p1 d p 1 2 n1 1 1 n n 2 n p2 2 p 1 n p n p n n1 1 n n p 2 n p n p n p 2n 1 n n p p n1 n p p n 2 2n 1 n n p 2 p1 1n n 1 n 1 2 p 2 p1 n 1 2 n 1n p2 n1 n1 p n VALUES FOR MAXIMUM DISCHARGE THROUGH A NOZZLE: n n1 n 1n The relationship for maximum discharge through a nozzle is, 2 mmax A 2n p n n 1 v 1 n 1 The values of maximum discharge for the following three conditions: (i) When the steam is initially dry saturated: We know that for dry saturated steam, n=

110 Therefore maximum discharge is, p 1 mmax 0.637A v 1 (ii) When the steam is initially superheated: For superheated steam, n=1.3. Therefore maximum discharge is, p 1 mmax 0.666A v 1 (iii) For gases: For gases, n=1.4. Therefore maximum discharge is, 1 mmax 0.685A v VALUES FOR CRITICAL PRESSURE RATIO p p Critical pressure ratio n1 For the following conditions Critical pressure ratios are, (i) When the steam is initially saturated We know that for dry saturated steam, n=1.135 p2 p =0.577 ; P 2 = P 1 1 (ii) When the steam is initially superheated: We know that for superheated steam, n=1.3 p2 p =0.546 ; P 2 = P 1 1 (iii) When the steam is initially wet It has been experimentally found for wet steam, p2 p =0.582 ; P 2 = P 1 1 n n1 (iv) For gases: We know that for gases, n= p

111 p2 p =0.528 ; P 2 = P PHYSICAL SIGNIFICANCE OF CRITICAL PRESSURE RATIO Now consider two vessels A and B connected by a convergent nozzle as shown in Fig. 3.4 (a). Let the vessel A contains steam at a high and steady pressure (P 1 ), and the vessel B contains steam at another pressure (P 2 ) which may varied at will. Let the pressure P 2 in the vessel B be made equal to the pressure (P 1 ) in the vessel A. There will be no flow of steam through the nozzle. Now if the pressure (P 2 ) in the vessel B gradually reduced, the discharge through the nozzle will increase gradually as shown in Fig 3.4 (b). As the pressure (P 2 ) in the vessel B approaches the critical value, the rate of discharge will also approach its maximum value. If the pressure (P 2 ) in the vessel B is further reduced, it will not increase the rate of discharge. The ratio of exit pressure to the inlet pressure is called critical pressure ratio. We know that the velocity of steam at any section in the nozzle V n1 n n p pv n 1 p1 and the critical pressure ratio for maximum discharge, 111

112 n 1 2 n p2 n1 n p 2 2 or p1 n1 p1 n1 n 2 n n12 V2 2 pv pv 1 1 n1 n1 n1 n1 n 2n 1 = 2 pv 1 1 n 1 n 1 1 p Volume() v Density 1 we also know that for isentropic expansion, p v p v n n p p1 p 1 p p1 p 1 2 n n n p p n 1 1n n n p p p p p 2 p2 p1 2 p1 n 1n 1n p2 2 n 2 n 1 p n n p2 n1 = n Substituting the value of P 1 / in the V 2 equation, V 2 = 2n p2 n1 np2 n p2 2 n 1 2 This is the value of velocity of sound in the medium at pressure P 2 and it is known as sonic velocity. 112

113 3.6SUPERSATURATED FLOW OR METASTABLE FLOWTHROUGH NOZZLE The expansion of steam in an ideal nozzle is isentropic, which is accompanied by condensation process. If the steam is initially superheated, the condensation should start after it has become dry saturation. This is possible when the steam has proceeded through some distance in the nozzle and in a short interval of time. But from practical point of view, the steam has a great velocity. Equilibrium between the liquid and vapour phase is delayed and the steam continues to expand in a dry state. The steam in such a set of conditions, is said to be supersaturated or in metastable state. It is also called super cooled steam, as its temperature at any pressure is less than the saturation temperature corresponding to the pressure. The flow of supersaturated steam, through the nozzle is called supersaturated flow or metastable flow. Experiment of supersaturated flow of steam has shown that there is a limit to which the supersaturated flow is possible. This limit is represented by Wilson line on T-s diagram in 113

114 Fig. 3.6(a). The Wilsonline closely follows the 0.97 dryness fraction line. Beyond this Wilson line, there is no super saturation. Fig. 3.6(b) shows the isentropic expansion of steam in a nozzle. The point A represents the position of initial dry saturated steam at pressure P 1. The line AB represents the isentropic expansion of steam in the supersaturated region. The metastable state (point C) is obtained by drawing a vertical line through A to meet the Wilson line. At C, the steam condenses suddenly. The line CD represents the condensation of steam at constant enthalpy. The point D is obtained by drawing a horizontal line through C to meet the throat pressure (P 2 ) of the nozzle. The line DF represents the isentropic expansion of the steam in the divergent portion. 3.7 EFFECTS OF SUPERSATURATION The following effects in the nozzle, in which super saturation occurs, are, (i) Since the condensation does not take place during super saturation expansion, so the temperature at which the super saturation occurs will be less than the saturation temperature corresponding to the pressure. Therefore, the density of supersaturated steam will be more than for the equilibrium conditions, which gives the increase in mass of steam discharged. (ii) The super saturation increases the entropy and specific volume of the steam. (iii) The super saturation reduces the heat drop below that for thermal equilibrium. Hence the exit velocity of the steam is reduced. (iv) The super saturation increases dryness fraction of the steam. 3.8 FRICTION IN A NOZZLE AND NOZZLE EFFICIENCY When the steam flows through a nozzle, some loss in its enthalpy or takes place due to friction between the nozzle surface and the flowing steam. This can be understood with the help of h-s diagram and Mollier chart. 114

115 Fig. 3.7 (i) First of all, locate the point A for the initial condition of the steam. It is a point where the saturation line meets the initial pressure (P 1 ) line. (ii) Now draw a vertical line through A to meet the final pressure (P 2 ) line. This is done as the flow through the nozzle is isentropic, which is expressed by a vertical line AB. The heat drop (h 1 -h 2 ) isentropic heat drop. (iii) Due to friction in the nozzle the actual heat drop in the steam will be less than (h 1 -h 2 ). This heat drop is shown as AC instead of AB. (iv) As the expansion of the steam ends at the pressure P 2, therefore final condition of the steam is obtained by drawing a horizontal line through C to meet the final pressure (P 2 ) line at B. (v) Now the actual expansion of the steam in the nozzle is expressed by the curve AB (adiabatic expansion) Instead of AB (isentropic expansion). Actual heat drop (h 1 -h 3 ) is known as useful heat drop. Now the coefficient of nozzle or nozzle efficiency is defined as the ratio of useful heat drop to the isentropic heat drop. Useful heat drop AC h h K Isentropic heat drop AB h h SOLVED PROBLEMS 1. Dry saturated steam at a pressure of 11 bar expanded in a nozzle to 2 bar. If the flow is isentropic. Find (i) Velocity at throat, (ii) Exit velocity, (iii) Ratio of cross-sectional area of exit to throat (iv) Also find the area of throat and exit if the mass flow rate of steam is 15 kg/sec. Given data: p 11 bar 115

116 p 2 bar m 15 kg/sec Solution Critical pressure ratio, p 2 p n1. p Assume n for dry saturated steam, p 6.35 bar From Mollier chart, At p 11 bar, h 2785 kj kg At p 6.35 bar, h 2680 kj kg,v 0.35 m /kg At p 2 bar, h 2490 kj kg,v 0.8 m /kg i Velocity at throat C 2000 h h m/sec ii Exit Velocity C 2000 h h m/sec iiithroat area A, m A C v 15 A A m iv Exit area A, m A C v 15 A A m v Ratio of cross sectional area from exit to throat A A Steam enters through a convergent-divergent nozzle at 21 bar and 270 C. The discharge pressure is 0.07 bar the expansion is uniform throughout the operation. At convergent 116

117 portion there is no frictional loss. At divergent portion 10% friction is loss. Calculate the throat area and exit area of the nozzle if the mass flow rate of steam is 14kg/sec. Given data: p 21 bar,t 270 p 0.07 bar m 14 kg/sec Solution: From Steam tables, At 21 bar, saturated temperature is which is less than the inlet temperature, so the condition of steam is in superheated condition at inlet of Convergent-Divergent nozzle. Critical pressure ratio, p 2 p n1. p Assume n 1.3 for superheated steam, p bar From Mollier chart, At p 21 bar, T 270 h 2940 kj kg At p bar, h 2810 kj kg,v 0.18m /kg At p 0.07 bar, h 2040 kj kg,v 20 m /kg Velocity at throat C 2000 h h m/sec Exit Velocity C 2000η N h h ithroat area A, m/sec A m A C v 14 A m or cm ii Exit area A, m A C v 117

118 14 A A m or cm 3. Dry saturated steam at a pressure of 8 bar enters a Convergent-divergent nozzle and leave at 1.5 bar. If the flow is isentropic and the corresponding expansion index Find the ratio of Cross-sectional area at exit and throat for maximum discharge. Use steam tables only. Given data: p 8 bar p 1.5 bar n Solution: Critical pressure ratio,. p From steam tables, For isentropic flow, s s s p 2 p n1 p 4.62 bar At p 8 bar, h h kj kg &s s 6.66 kj kg s s At p 4.62 bar, s s x s,v v x v x x h h x h h kj/kg v v x v v m /kg At p 1.5 bar, x 5.79 x h h x h h kj/kg 118

119 v v x v v m /kg Velocity at throat C 2000 h h m/sec Exit Velocity C 2000 h h m/sec Mass flow rate m A C v A C v Ratio of cross sectional area from exit to throat for maximum discharge A A C v C v A A Steam at 10.5 bar and 0.95 dryness is expanded through a Convergent divergent nozzle. The steam pressure at nozzle. The steam pressure at nozzle is 0.85 bars. Find,i) the velocity of steam at throat for maximum discharge, ii) the area at the exit, iii) the steam discharge if the throat area is 1.2cm 2. Assume the flow is isentropic and there are no friction losses. Take n= Given data: p 10.5 bar, x 0.95 p 0.85 bar A 1.2 cm m n Solution Critical pressure ratio, p 2 p n1. p From Mollier chart, p 6.06 bar At p 10.5 bar and x 0.95, h 2680 kj kg 119

120 At p 6.06 bar, h 2580 kj kg,v 0.30 m /kg At p 0.85 bar, h 2275 kj kg,v 1.6 m /kg i Velocity at throat C 2000 h h m/sec ii Exit Velocity C 2000 h h m/sec iii Steam discharge m A C v m m kg/sec iv Exit area A, m A C v 0.18 A A m or 3.2 cm Dry saturated steam at 2.8 bar is expanded through a Convergent nozzle to 1.7 bar. The exit area is 3cm 2. Calculate the exit velocity and the mass flow rate, assuming i) isentropic expansion and ii)super saturated flow. Given data: p 2.8 bar p 1.7 bar A 3 cm 310 m Solution For isentropic expansion From Mollier Chart, At p 2.8 bar, h 2725 kj kg At p 1.7 bar, h 2640 kj kg,v 1.0 m /kg i Exit Velocity C 2000 h h m/sec ii Steam discharge m A C v 120

121 For supersaturated flow Assume n =1.3 m m kg/sec i Exit Velocity C 2000 n n1 p v 1 p p From Steam tables, v v m /kg iimass flow rate m p V p V V p / V p C A C V C 413 m/sec V / m /kg m m kg/sec 6. The inlet Condition of steam to a Convergent-divergent nozzle is 2.2MN/m 2 and 260 C. The exit pressure is 0.4 MN/m 2. Assuming frictionless flow upto the throat and a nozzle efficiency of 85 percent, determine i) flow rate for a throat area of 32.2 cm 2, ii) Exit area. Given data p 2.2 MN m 22 bar, T K p 0.4 MN 4 bar m A 32.2 cm m η N 85% Solution Critical pressure ratio, 121

122 p 2 p n1. p From Mollier chart, At p 22 bar, h 2920 kj kg p 12 bar At p 12 bar, h 2780 kj kg,v 0.16 m /kg At p 4 bar, h 2580 kj kg,v 0.46 m /kg Velocity at throat C 2000 h h m/sec Exit Velocity C 2000η N h h m/sec iii Flow rate m A C v m m kg/sec ii Exit area A, m A C v A A m or cm Steam at 3 bar with 10 C superheat is passed through a Convergent nozzle. The velocity of steam entering the nozzle is 91.5 m/sec. The back pressure is 1.5 bar. Assuming nozzle efficiency of 90%, determine the area of the nozzle at exit. Discharge through the nozzle is limited to 45 kg/sec. Take Cps(Superheated steam)=2.2 kj/kg C. Given data: p 3 bar, Degree of super heat 10 p 1.5 bar, m 0.45 kg/sec C m/sec, η N 90% 122

123 C ps (Superheated steam)=2.2 kj/kg C Solution: From steam tables, s s c ln T T at 3 bar, s kj kgk,t K T K, h kj kg s ln kj kgk s s s x s x x h h x h h kj/kg v v x v v m /kg h h c T T kj/kg C C 2000η N h h Mass flow rate m m/sec A C v 0.45 A A m or cm 8. A Convergent-divergent adiabatic steam nozzle is supplied with steam at 10 bar and 250 C. The discharge pressure is 1.2 bar. Assuming that the nozzle efficiency is 100% and initial velocity of seam is 50m/sec, find the discharge velocity. Given Data: p 10 bar, T 250 p 1.2 bar 123

124 C 50 m/sec Solution: From Mollier chart, At p 10 bar, h 2940 kj kg At p 1.2 bar, h 2530 kj kg Discharge Velocity C C 2000h h m/sec 9. Dry saturated steam at 10 baris expanded in a nozzle to 0.4 bar. The throat area is 7cm 2 and the inlet velocity is negligible. Determine the mass flow and exit area. Assume isentropic flow and take the index n=1.135 for dry saturated steam. Given Data: p 10 bar p 0.4 bar A 7 cm 710 m n = Solution: Critical pressure ratio, p 2 p n1. p From Mollier chart, At p 10 bar, h 2780 kj kg p 5.77 bar At p 5.77 bar, h 2675 kj kg,v 0.33 m /kg At p 0.4 bar, h 2250 kj kg,v 3.8 m /kg Velocity at throat C 2000 h h m/sec 124

125 Exit Velocity C 2000 h h m/sec i Steam discharge or mass flow rate m iv Exit area A, A m A C v m m A C v kg/sec A m or cm In a steam nozzle, steam expands from 16 bar to 5 bar with initial temperature of 300 C and mass flow rate of 1kg/sec. Determine the throat and exit area considering i)expansion to be frictionless and ii) friction loss of 10% throughout the nozzle. Given Data: p 16 bar, T K p 5 bar m 1 kg/sec Solution: At 16 bar, saturated temperature is which is less than the inlet temperature, so the condition of steam is in superheated condition at inlet of nozzle. n = 1.3 Critical pressure ratio, p 2 p n1 From Mollier chart, At p 16 bar, h 3035 kj kg. p p 8.73 bar At p 8.73 bar, h 2905 kj kg,v 0.26 m /kg 125

126 At p 5 bar, h 2790 kj kg,v 0.38 m /kg Velocity at throat C 2000 h h m/sec Exit Velocity C 2000 h h m/sec Steam discharge or mass flow rate m A C v Exit area A, m 1 A A m or 5.1 cm A C v A A m or 5.43 cm Velocity at throat C 2000η N h h m/sec Exit Velocity C 2000η N h h m/sec Steam discharge or mass flow rate m A C v Exit area A, m 1 A A m or cm A C v 1 A A m or 5.72 cm 11. A Convergent-divergent nozzle is to be designed in which steam initially at 14 bar and 80 C of superheat is to be expanded down to a back pressure of 1.05 bar. Determine the necessary throat and exit diameters of the nozzle for a steam discharge of 500kg/hr, assuming that the expansion is in thermal equilibrium throughout and friction reheat 126

127 amounting to 12% of the total isentropic enthalpy drop to be effective in the divergent part of the nozzle. Given Data: p 14 bar, T K p 1.05 bar m kg/sec,η N 88% Solution: n = 1.3 Critical pressure ratio, p 2 p n1 From Mollier chart, At p 14 bar, h 2980 kj kg. p p 7.64 bar At p 7.64 bar, h 2850 kj kg,v 0.3 m /kg At p 1.05 bar, h 2490 kj kg,v 1.6 m /kg Velocity at throat C 2000 h h m/sec Exit Velocity C 2000η N h h m/sec Steam discharge or mass flow rate m A C v A A m π 4 d d m or cm Exit area A, m A C v 127

128 A A m π d d m or cm OBJECTIVE QUESTIONS 1. The steam leaves the nozzle at a (a) High pressure and low velocity (c) High pressure and high velocity (b) low pressure and high velocity (d) low pressure and low velocity 2. The effect of friction in a nozzle dryness fraction of steam (a) Increase (b) decrease (c) neither increase nor decrease (d) Initially increase and then increase 3. The velocity of steam leaving the nozzle is given by (a) V=44.72 h d (b) V=44.72 kh d (c) V=44.72 h d k (d)v=44.72 k h d 4. The critical pressure ratio is a) P 2 /P 1 (b) (P 2 /P 1 ) (c) P 1 /P 2 (d) (P 1 /P 2 ) 5. Critical pressure ratio for dry saturated steam is (a) (b) (c) (d) The ratio of saturation pressure to the super saturation pressure is (a) Degree of under cooling (c) degree of super heat (b) Degree of super saturation (d) None of these 7. The difference of the super Saturated temperature and saturation at that pressure is known as a) Degree of under cooling (b) Degree of super saturation (c) degree of super heat (d) None of these 8. In a nozzle the effect of super saturation is to a) decrease the dryness fraction of the steam (b) decrease the specific volume of the steam (c) decrease the enthalpy drop (d) decrease the enthalpy 9. The cross section of a convergent divergent nozzle a) Goes on increasing (b) Goes on decreasing 128

129 (c) remains uniform (d) Initially decreases and then increases 10. The critical pressure ratio is given by p p 2 a) 2/ 1 n/ n n 1 2 (b) 2/ 1 n/ n n 1 1 p p 2 (c) 2/ 1 n/ n n 1 2 (d) 2/ 1 n/ n n In a nozzle isentropic heat is always the actual heat drop a) Less than (b) equal to (c) greater than (d) greater than or Less than 12. In a convergent divergent nozzle maximum mass flow rate occurs at a) The exit (b) the throat (c) the entrance (d) in between throat and exit 13. Nozzle efficiency is given by a) K=h a -h I h (b) K= I (c) K= h a * h a I a h I 14. The device used to increase the velocity of the steam is called (a) Diffuser (b) Turbine (c) Throttle Valve (d) Nozzle 15.Nozzle is used to increase of the steam (a) The pressure (b) Mass flow rate (c) the velocity (d) the density SHORT QUESTIONS 1. What is metastable flow? 2. What is a Nozzle? 3. Explain the term critical pressure ratio as applied to steam nozzles? 4. Define nozzle efficiency. 5. What are all the different types of Steam Nozzles? 6. Define Throat. 7. What are the effects of supersaturation in a steam nozzle? 8. Define the following (i) Sonic Flow (ii) Subsonic Flow (iii) Supersonic Flow 9. What is the steady flow energy equation as applied to steam nozzles? 10. Define the following (i) Degree of supersaturation (ii) Degree of under cooling p p 1 p p 1 DESCRIPTIVE QUESTIONS 1. Derive an expression for the velocity and mass flow through nozzles. 129

130 2. Prove that the critical pressure ratio of a steam nozzle is P2 P 1 2 n1 n n1 EXERCISE PROBLEMS 1. Steam is supplied to a nozzle at 3.5 bar and 0.96 dry. The steam enters the nozzle at 240m/sec. The pressure drops to 0.8 bar. Determine the velocity and dryness fraction of the steam when it leaves the nozzle. 2. Calculate the throat area of nozzle supplied with steam at 10bar and 200 o C. The rate of flow of steam is 1.2 kg/sec. Neglect friction and assumes the velocity at inlet to be small. 3. Dry air at a temperature of 27 o C and a pressure of 20 bar enters a nozzle and leaves at a pressure of 4 bar. Find the mass of air discharged if the area of nozzle is 200 m Steam enters of group of convergent- divergent nozzle at a pressure of 22 bar and with a temperature of 240 o C. The exit pressure is 4 bar and 9% of the total heat drop is lost in friction. The mass flow rate is 10kg/sec and the flow up to the throat may be assumed friction less. Calculate (i) the throat and exit velocities (ii) the throat and exit areas 5. Steam at 42 bar and 260 o C enters a nozzle and leaves at 28 bar. Neglecting initial kinetic energy and considering super-saturation, determine the discharge area for a flow of kg/hr and a nozzle velocity coefficient of 96%. 6. Compare the mass of discharge from a convergent-divergent nozzle expanding from 8 bar and 210 o C to 2 bar, when (i) the expansion takes place under thermal equilibrium (ii) the steam is in super-saturated condition during a part of its expansion Take area of nozzle as 2400mm 2 7. Steamexpands in a nozzle under the following conditions: Inlet pressure = 15 bar; Inlet temperature = 250 o C; Final pressure = 4 bar; Mass flow = 1 kg/sec. Calculate the required throat and exit areas, using Mollier diagram, when, (i) the expansion is frictionless (ii) the friction loss at any pressure amounts to 10percent of the total heat drop down to that pressure. 130

131 8. A gas expands in a convergent-divergent nozzle from 5 bar to 1.5 bar, the initial temperature being 700 o C and the nozzle efficiency is 90%. All the losses takes place after the throat. For 1 kg/sec mass flow rate of the gas, find throat and exit areas. 9. The throat diameter of a nozzle is 5 mm. If dry and saturated steam at 10 bar is supplied to nozzle, calculate mass flow rate. The exhaust pressure is 1.5 bar. Assume frictionless adiabatic flow and n = Five kg of steam per sec expands from 16 bar, 250 o C to a pressure of 3.5 barisentropically. If the expansion is superheated, determine the diameter of nozzle at exit, degree of under cooling, degree of superheating and % increase in mass flow rate due to metastable expansion to the thermal expansion. 131

132 CHAPTER IV 4.1 INTRODUCTION STEAM TURBINES A steam turbine is a prime mover in which rotary motion is obtained by the gradual change of momentum of the steam. In a steam turbine, the force exerted on the blades is due the velocity of steam. In general, a steam turbine, essentially, consists of the following two parts: (i) The nozzle in which the heat energy of high-pressure steam is converted into kinetic energy, so that the steam issues from the nozzle with a very high velocity. (ii) The blades, which change the direction of, steam issuing from the nozzle, so that a force acts on the blades due to change of momentum and propel them. Thus, the basic principle of operation of a steam turbine is the generation of high velocity steam jet by the expansion of high-pressure steam and then conversion of kinetic energy, so obtained into Mechanical work on rotor blades ADVANTAGES OF STEAM TURBINES OVER RECIPROCATING STEAM ENGINES Following are the important advantages of steam turbines over reciprocating steam engines: (i) A steam turbine may develop higher speeds and a greater steam range is possible. (ii) The efficiency of a steam turbine is higher. (iii) The steam consumption is less. (iv) Since all the moving parts are enclosed in a casing, the steam turbine is comparatively safe. 132

133 (v) A steam turbine requires less space and lighter foundations, as there are little vibrations. (vi) There is less frictional loss due to fewer sliding parts. (vii) The applied torque is more uniform to the driven shaft. (viii) A steam turbine requires less attention during running, Moreover, the repair costs are generally less. 4.2 CLASSIFICATION OF STEAM TURBINES: The steam turbines may be classified into the following types: (i) According to the mode of steam action Impulse turbine Reaction turbine Impulse-Reaction turbine (ii) According to the direction of steam flow Axial flow turbine Radial flow turbine (iii) According to exhaust condition of steam Condensing turbine Non condensing turbine (iv) According to pressure of steam High pressure turbine Medium pressure turbine Low pressure turbine (v) According to the number of stages Single stage turbine Multi stage turbine 4.3 IMPULSE TURBINE Impulse turbine consists of a set nozzle and set of moving blades are arranged as shown in figure. Expansion of steam takes in nozzle where steam is expanded to its final pressure because of this drop in pressure increases the velocity of steam. After the expansion, 133

134 high velocity steam from nozzle hits the set of moving blades mounted on the wheel. The velocity of the steam decreases and pressure remains constant. De-Laval turbine is the example for a simple impulse turbine. Fig. 4.1 Impulse turbine 4.4VELOCITY TRIANGLES FOR MOVING BLADE OFAN IMPULSE TURBINE We have already discussed that in an impulse turbine, the steam jet after leaving the nozzle impinges on one end of the blade. The jet then glide over the inside surface of the blade and finally leaves from other edge, as shown in the figure. It may be noted that the jet enters and leaves the blade tangentially for shockless entry and exit. Consider the steam jet entering a curved blade after leaving the nozzles. Now let us draw the velocity triangles at inlet and outlet tips of the moving blade, as shown in figure. Let, U Linear velocity of the moving blade m/s V Absolute velocity of steam entering the moving blade V Relative velocity of the jet to the moving blade. It is the vectorial difference between U and V. V Velocity of flow at entrance of the moving blade. It is the vertical component of V. V Velocity of whirl at entrance of the moving blade. It is the horizontal component of V. β Angle which the relative velocity of jet of the moving blade V makes with the direction of motion of the blade. Angle with the direction of motion of the blade at which the jet enters the blade. V,V,V,V,β, Corresponding values at exit of the moving blade 134

135 It may be seen from the above, that the original notations (V,V,V,V,β, stand for the inlet triangle. The notations with suffix 2V,V,V,V,β, stand for the outlet triangle. It may be noted that as the steam jet enters and leaves the blades without any shock (or in other words tangentially), therefore shape of the blades will be such that V and V will be along the tangents to the blades at inlet and outlet respectively. The angle β is called the blade at inlet and angle β is the blade at exit of the moving blade. Fig. 4.2 Velocity triangle of Impulse turbine Nozzle delivers the steam jet with a high velocity (V ) at an angle with the direction of motion of the blade. The jet impinges on a series of turbine blades mounted on the runner disc. The axial component ofv, which does no work on the blade, is known as velocity of flow (V ). It causes the steam to flow through the turbine and also an axial thrust on the rotor. The tangential component of V is known as velocity of whirl at inlet (V ). According to combined velocity diagram given below, the linear velocity or mean velocity of 135

136 the blade (U) is represented by AB in magnitude and direction. The length AC represents the relative velocity (V ) of the steam jet with respect to the blade. The jet now glides over and leaves the blade with relative velocityv, which is represented by AD. The absolute velocity of jet (V ) as it leaves the blade, is represented by BD inclined at an angle β with the direction of blade motion. The tangential component of V (represented by BF) is known as velocity of whirl at exit (V ). The axial component of V (represented by FD) is known as the velocity of flow at exit (V ) COMBINED VELOCITY TRIANGLE FOR MOVING BLADES (i) Draw a horizontal line, and cut off AB equal to velocity of blade (U) to some suitable scale. (ii) Now at B, draw a line BC equal to (V ) at an angle (i.e. velocity of steam jet at inlet of blade) to the scale. (iii) Join AC, which represents the relative velocity at inlet (V ). Now at A, draw a line AC at an angle with AB. (iv) Now with A as center and radius equal to AC, draw an arc meeting the line through A at D, such that AC = AD or V V (v) Join BD, which represents velocity of jet at exit (V 2 ) to the scale. (vi) From C and D, draw perpendiculars meeting the line AB produced at E and F respectively. (vii) Now BE and EC represent the velocity of whirl and velocity of flow at inlet (V w1 andv f1 ) to the scale. Similarly, BF and FD represents the velocity of whirl and velocity of flow at outlet (V w2 and V f2 ) to the scale Performance of an impulse turbine With the help of the combined velocity triangle, following performance parameter are easily calculated. Force Let m kg be the mass of steam passing through the moving blades in t seconds. Force in the tangential direction, F rate of change of momentum in tangential direction mass per ssecond change of velocity 136

137 F mv V,N Axial thrust or force in the axial direction, F rate of change of momentum in axial direction mass per ssecond change of velocity F mv V,N Work done Work done by steam on blades, W tangential force distance travelled per unit time in tangential direction Power developed Power developed by the turbine, WmV V U, Nm s P mv V U kw 1000 this developed power is known as rim power, different from the shaft power. Blade efficiency It is defined as the ratio of work done on the blade per second to the energy supplied to the blade per second. It is denoted by η. η mv V U mv 2V V U V Energy lost Due to the friction between blades some amount of energy loss is taking place in the form of heat generation. Energy lost due to blade friction mv V kw 1000 Stage efficiency It is the ratio of work done on the blade per kg of steam flowing through the stage. Blade Velocity coefficient η mv V U mh h V V U h where, h heat drop in the nozzle ring It is defined as the ratio of relative velocity of steam at outlet to the relative velocity of steam at inlet. 137

138 Blade velocity coefficient K V V It is mainly due to surface friction. The flow of steam over the surface blade is resisted by the surface friction. In general, there is a loss of 10 15% in relative velocity of steam while passing through the blades. Maximum diagram efficiency and maximum work From the combined velocity diagram the whirl velocity at inlet V V cosα V cosβ U1 Whirl velocity at outlet V V cosα V cosβ U2 Adding 1&2,we get, V V V cosβ U V cosβ U V cosβ V cosβ V cosβ 1 V cosβ V cosβ V cosβ 1KC K V V,C cosβ cosβ V V V cosα U1 KC work done per kg of steam per second V cosα U1 KCU Diagram efficinecy η mv V U mv V cosα U1 KCU V η 2ρcosα ρ 1 K.C. 3 where ρ U V blade speed ratio From equation (3), it may be noted that the diagram efficiency depends upon, (i) Nozzle angle α (ii) Blade speed ratio ρ (iii)blade angles at inletβ and at outlet β and (iv) Blade velocity y coefficient K The values of α, K and C are constant for a particular turbine. So the diagram efficiency depends mainly upon the value of blade speed ratioρ. 138

139 The optimum value of blade speed ratio can be obtained by differentiating (3) w.r.t ρ and equating it to zero. dη dρ The above equation (4) gives the condition for maximum diagram efficiency. Substituting (4) in (3) we get, If the blades are symmetrical and there is no friction, i.e., 1 1 4, 6 Maximum work done per kg of steam, From the above equation, we came to know that for the maximum efficiency or maximum work developed per kg of steam, the blade velocity would be approximately half of the absolute velocity of steam coming out of the nozzle. 4.5 REACTION TURBINES In a reaction turbine, the steam pressure gradually decreases from moving blades to fixed blades during expansion. Reaction turbine consists of large number of stages; each stage consists of a set of fixed and moving blades. The fixed blade directs the steam to enter the moving blades, at as a nozzle by increasing the velocity of the steam and also reduces the 139

140 pressure of steam. The pressure drop and heat drop takes place both in moving and fixed blades. The expansion of steam in moving blades creates a reaction on moving blades. So the turbine is called as reaction turbine. The moving blades are fixed in a rotor whereas fixed blades are attached with casing. In this turbine pressure drop is small; number of stages is more than that of a simple impulse turbine for the same capacity. A Parson sreaction turbine is the simplest type of reaction steam turbine, and is commonly used. A Parson s turbine is also known as 50% reaction turbine. 4.6 VELOCITY DIAGRAM OF A SIMPLE REACTION TURBINE Figure shows the velocity diagram for flow of steam on a moving blade of a simple reaction turbine with usual notations. In case of impulse turbine, the relative velocity remains constant or slightly reduced due to friction. But in reaction turbine the relative velocity of steam is increased due to continuous expansion of steam while flowing through the blades. Fig. 4.4 (a) Figure shows the combined velocity diagram for flow of stea on a moving blade of a simple reaction turbine with usual notations. 140

141 Fig. 4.4 (b) 4.6.1Performance of simple reaction turbine Degree of reaction It is defined as the ratio of heat drop over the moving blade of a stage to the total heat drop in that stage. heat drop in moving blades Degree of reaction R total drop in that stage Heat drop in moving blade is increase in relative velocity of steam flowing through the blade. increase in relative velocity V V 2 Total heat drop is nothing but the total work done by the seam in the stage Total work done =V V U V V Degree of reaction R V V U V V 2V V U 1 From the combined velocity diagram V V cosecβ V V cosecβ V V V cotβ V cotβ The flow velocity is almost constant while flowing through the blades i. e. V V V Substituting V,V,V V and V V V in (1), we get, 141

142 Degree of reaction R V cosec β cosec β 2 cotβ cotβ U V 1 cot β 1 cot β 2 cotβ cotβ U R V 2U cotβ cotβ 2 In Parson s reaction turbine, the degree of reaction is 50% which results α β and α β i. e., the moving and fixed blades are having same profile. For simple impulse turbine, 0 For pure reaction turbine, 1 Condition for maximum efficiency The following assumptions are made while deriving condition for maximum efficiency: (i) Degree of reaction is 50 % (ii) The fixed and moving blades are having symmetrical shape (iii)there is no velocity drop between the adjacent fixed and moving blades. The kinetic energy supplied to the fixed blade per kg of steam V The kinetic energy supplied to the moving blade per kg of steam V V Total energy supplied V 2 V V 2 Total energy supplied V V From the velocity diagram, V V U 2UV cosα Total energy supplied V V 2 V V Total energy supplied V V U 2UV cosα 2 The work done per kg of steam (W) =V V U From the velocity diagram, V V cosα and V V cosβ U 2 work done per kg of steam W V cosα V cosβ UU α β,v V and α β 2 2 work done per kg of steam W 2V cosα UU 142

143 We know that, Diagram efficiencyη 2V cosα UU V V U UV η 22V cosα UU V U 2UV cosα η workdone total energy suplied 2UV 2cosα U V V 1 U U V V 2ρ2cosα ρ 1 ρ 2ρcosα 3 where ρ U V blade speed ratio Diagram efficiency is maximum if the denominator value of equation (3) is minimum or d1 ρ 2ρcosα 0 dρ Sustotuting 4in 3,we get, Height of blades of reaction turbines 2ρ 2cosα 0 ρcosα 4 η 2cosα 2cosα cosα 1 cos α 2cos α η 2cos α 1cos α The steam enters the moving blades over the whole circumference. As a result of this, the area through which the steam flows is always full of steam. Now consider a reaction turbine whose end view of the blade ring is shown in figure. Let, D diameter of the drum h Height of blades 143

144 D mean diameter of the blade ring V V V velocity of flow at exit For negligible bade thickness, the are of steam flow A π D hπdhh volume of steam flowing through the turbine, V A V πdhh V We know that volume of 1 kg of steam at the given pressure is V g (from steam tables). Therefore mass of steam flowing, m V D V Where, x dryness fraction of steam &v specific volume of steam 4.7 COMPARISON BETWEEN IMPULSE AND REACTION TURBINE S. No. Impulse Turbine Reaction turbine 1 The steam flows through the nozzles and The steam first through guide mechanism impinges on the moving blades. and then through the moving blades. 2 The steam impinges on the buckets with The steam guides over the vanes with kinetic energy. pressure and kinetic energy. 3 The steam may or may not be admitted The steam must be admitted over the over the whole circumference. whole circumference. The steam pressure remains constant The steam pressure is reduced during its 4 during its flow through the moving blades. flow through the moving blades. The relative velocity of steam while The relative velocity of steam while 5 gliding over the blades remains constant gliding over the blades increases (assuming no friction). (assuming no friction). 6 The blades are symmetrical. The blades are not symmetrical. 7 The number of stages required are less The number of stages required are more for the same power developed. for the same power developed. 144

145 4.8 COMPOUNDING OF IMPULSE STEAM TURBINES (METHODS OF REDUCING ROTOR SPEEDS) In the recent years, high pressure (100 to 140 bar) and high temperature steam is used in the power plants to increase their thermal efficiency. If the entire pressure drop (from boiler pressure to condenser pressure (say from 125 bar to 1 bar) is carried out in one stage only, then the velocity of steam entering into the turbine will be extremely high. It will make the turbine rotor to run at a very high speed (even up to rpm). Form practical point of view, such a high speed of the turbine rotor is bound to have a number of disadvantages. In order to reduce the rotor speed, various methods are employed. All of these methods consist of a multiple system of rotors, in series, keyed to a common shaft and the steam pressure or the jet velocity is absorbed in stages as it flows over the rotor blades. This process is known as compounding. The following three methods are commonly employed for reducing the rotor speed: 1. Velocity compounding, 2. Pressure compounding and 3.pressure - velocity compounding Pressure Compounding (RATEAU) The expansion of steam from boiler pressure to exhaust pressure is carried out in a number of steps or stages. Each stage has a set of nozzles and a row of moving blades. The rows of moving blades are separated from each other by partitions or diaphragms, into which the nozzles are set. As only a portion of the velocity available is developed in each set of nozzles, the blade velocity is kept down to a reasonable amount. This type of compounding is known as the pressure compounding and the nozzle and blade arrangement for a pressure compounded impulse turbine is sketched. In this arrangement, the pressure of the steam drops in each set of nozzles as indicated by the pressure graph. The steam velocity is increased by each pressure drop and then decreases again in each row of moving blades, as the velocity graph shows. 145

146 Fig. 4.6 Pressure Compounding This amounts to a number of simple impulse turbines in series, the nozzle of one stage receiving steam discharged by the preceding ring of moving blades. The distribution of the total enthalpy drop to a number of sets of nozzle will be about the same and the velocity of steam at the outlet of each set of nozzles will be lower than the maximum velocity obtainable by single expansion.the work output of a pressure compounded turbine is the sum of that produced by theseparate impulse stages. However, the velocity of steam at entrance to nozzles particularly after the first stage must be taken into account VELOCITY COMPOUNDING (CURTIS) This design consists of one set of nozzles in which the steam is expanded from initial to exhaust pressure. The velocity of the steam resulting from this expansion is absorbed in two or more rows of moving blades.rows of fixed or guide blades, attached to the casing, are set between rows of moving blades and receive and redirect the steam to the next row of moving blades. As the velocity is absorbed in more than one row of moving blades, the blade speed is less than if the velocity was all absorbed in one row of blades. This type of compounding is known as the velocity compounding and the blade and nozzle arrangement for a velocity compounded impulse turbine is shown in fig

147 The pressure drops from inlet pressure to exhaust pressure in the single set of nozzles as the pressure graph shows. This large single pressure drop produces high steam velocity, which is absorbed in the two rows of moving blades. Note that there is no pressure or velocity drop in the fixed guide blades PRESSURE-VELOCITY COMPOUNDING This is a combination of the first two methods of compounding, namely pressure compounding and velocity compounding. The steam is expanded in two or more sets of nozzles in series, each set having velocity compounded blades to receive the steam issuing from the nozzles.the arrangement showed in Figure features two sets of nozzles. The steam pressure drops in each set of nozzles and the resulting velocity increase in each case is absorbed by in two rows of moving blades having a row of stationary blades in between them. The methods of reducing rotor speeds, namely, pressure compounding, velocity compounding, and pressure-velocity compounding have all applied to impulse turbines. In the case of the reaction turbine, it is not necessary to make special blade arrangements to reduce rotor speed. This is because the pressure drops across each row of moving blades as well as across each row of fixed blades and consequently the pressure drops in every stage and in small amounts all through the machine. 147

148 Fig. 4.8 Pressure - Velocity Compounding This requires, however, a large number of alternate rows of fixed and moving blades resulting in a long machine.therefore, in order to reduce the number of blade rows necessary, reaction turbines frequently have a velocity compounded impulse stage at the inlet end of the machine. 4.9 GOVERNING OF STEAM TURBINE The process of providing any arrangement, which will keep the speed constant (according to the changing load conditions), is known as governing of steam engines. Though there are many methods of governing steam turbines, yet the throttle governing is important from the subject point of view THROTTLE GOVERNING OF STEAM TURBINES: The throttle governing of a steam turbine is a method of controlling the turbine output by varying the quantity of steam entering into the turbine. This method is also known as servometer method, whose operation is given below: The centrifugal governor is driven form the main shaft of turbine by belt or gear arrangement. The control valve controls the direction of flow of the oil (which is pumped by gear pump) either in the pipe AA or BB. The servometer or relay valve has a piston whose motion (towards left or right depends upon the pressure of the oil flowing through the pipes AA or BB) is connected to a spear or needle which moves inside the nozzle, as shown in the figure. 148

149 Fig. 4.8 Throttle Governing We know that when the turbine is running at its normal speed, the positions of piston in the servo meter, control valve, fly balls of centrifugal governor will be in their normal positions as shown in the figure. The oil is pumped by the gear pump into the control valve, will come back into the oil sump as the mouths of both pipes AA or BB are closed by the tow wings of control valve. Now let us consider an instant, when load on the turbine increases. As a result of load increase, the turbine speed will be decreased. This decrease in the speed of the turbine wheel will also decrease the speed of the centrifugal governor. As a result of this, the fly ball will come down (due to decrease in centrifugal force) thus decreasing their amplitude. As the sleeve is connected to the central vertical bar of the centrifugal governor, therefore coming down of the fly balls will also bring down the sleeve. This down ward moment of the sleeve will raise the control valve rod, as the sleeve is connected to the control valve rod through a lever pivoted on the fulcrum. Now, a slight upward movement of the control valve rod will open the mouth of the pipe AA (still keeping the mouth of the pipe BB closed) ). Now the oil under pressure will rush from the control valve to the right side of the piston in the servo meter through the pipe AA. This oil, under pressure, will move the piston as well as spear towards left, which will open more area of the nozzle. This increase in the areaa of flow will increase the rate of steam of steam flow into the turbine. As a result of increasee in the steam flow, there will be an increase in increase in the turbine output as well as its speed. When speed of the turbine wheel will come up to its normal range, fly balls will move up. Now the 149

150 sleeve as well as control valve rod will occupy its normal position and the turbine will run at its normal speed. It may be noted that when load on the turbine decreases, it will increase the speed. As a result of this, the fly balls will go up (due to increasee in centrifugal force) and the sleeve will also go up. This will push the control valve rod downwards. This downward movement of the control valve rod will open the mouth of the pipe BB (still keeping the mouth of the pipe AA closed). Now the oil under pressure, will rush form the control valve to the left side of the piston in the servo meter through the pipe BB. This oil under pressure will move the piston and spear towards right, which will decrease the area of the nozzle. This decrease in the area of the flow will decrease the rate of steam flow into the turbine. As a result of decrease in the steam flow, theree will be a decrease in the turbine output as well as its speed. When the speed of the turbine is reduced to its normal range, the fly balls willl come down. Now the sleeve as well as control valve rod will occupy its normal positions, and the turbine will run its normal speed NOZZLE GOVERNING A diagrammatic arrangement of nozzle control governing is shown in figure. In this nozzles are grouped in 3 to 5 or more groups and each group of nozzle is supplied steam by controlled valves. The arc of admission is limited to 180 degree or less. The nozzle control governing is restricted to the first stage of the turbine, the nozzle area in other stages remaining constant. It is suitable for simple impulse turbine and for larger units which have an impulse stage followed b an impulse turbine. Fig. 4.8 Nozzle Governing 150

151 4.9.3 BY-PASS GOVERNING The high pressure impulse turbine, generally have a number of stages of small mean diameter of wheel. These turbines are generally designed for maximum efficiency at an economicc load whichh is about 80 % of the maximum continuous rating. Due to small heat drop in the first stage nozzle governing cannot be efficiently used. Secondly it is desirable to have fulll admission into high pressure stage at the rated economic load to eliminate the partial admission losses. In such case bypasss governing used. In this arrangement for high loads a bypass line is provided for the steam from the first stage nozzle box into late stage wheree work output increases. The bypass steam is automatically regulated by the lift of the valve. The bypass valve is under the control of the speed of the governorr for all loads within its range. In later stags though there is increase in work input, the efficiency is low due to throttling effect. Fig. 4.9 By-pass Governing OBJECTIVE QUESTIONS 1. The action of steam in a steam turbine is (a) static (b) dynamic (c) static and dynamic (d) neither static or dynamic 2. In an impulse turbine (a) the steam is expanded in nozzles only and there is a pressure drop and heat drop 151

152 (b) the steam is expanded both in fixed and moving blades continuously (c) the steam is expanded in moving blades only (d) the pressure and temperature of steam remains constant 3. In impulse turbines, when friction is neglected, the relative velocity of steam at outlet tip of the blade is the relative velocity of steam at inlet tip of the blade. (a) equal to (b) less than (c) greater than 4. The blade friction in the impulse turbine reduces the velocity of steam by while it passes over the blades. (a) 10 to 15% (b) 15 to 20% (c) 20 to 30% (d) 30 to 40% 5. In a reaction turbine (a) the steam is allowed to expand in the nozzle, where it gives a high velocity before it enters the moving blades (b) the expansion of steam takes place partly in the fixed blades and partly in moving blades (c) the steam is expanded from a high pressure to a condenser pressure in one or more nozzles (d) the pressure and temperature of steam remains constant 6. The Parson s reaction turbine has (a) only moving blades (b) only fixed blades (c) identical fixed and moving blades (d) fixed and moving blades of different shape 7. The degree of reaction is defined as the ratio of (a) heat drop in the fixed blades to the heat drop in the moving blades (b) heat drop in the moving blades to the heat drop in the fixed blades (c) heat drop in the moving blades to the total heat drop in the fixed and moving blades (d) total heat drop in the fixed and moving blades to the heat drop in the moving blades 8. In a reaction turbine, when the degree of reaction is zero, then there is (a) no heat drop in the moving blades (b) no heat drop in the fixed blades (c) maximum heat drop in the moving blades (d) maximum heat drop in the fixed blades 9. The maximum efficiency of a De-Laval turbine is (a) sin (b) cos (c) tan 10. The purpose of governing in steam turbine is to (d) 2 cot 152

153 (a) reduce the effective heat drop quality (c) completely balance against end thrust (b) reheat the steam and improve its (d) maintain the speed of the turbine TWO MARKS QUESTIONS 1. How turbines are classified? 2. What are the essential components of a steam turbine? 3. What is an Impulse turbine? 4. What is a Reaction turbine? 5. What are the differences between impulse and reaction turbines? 6. Draw the velocity triangles of an Impulse Turbine. 7. What is the effect of friction on Impulse Turbine? 8. Define Blade Velocity Coefficient. 9. What is meant by axial discharge in an Impulse turbine? 10. Define Degree of reaction of reaction turbine. 11. What do you understand by the term Height of Blades as applied to a Reaction Turbine? 12. Define the terms : (i) Diagram efficiency (ii) Stage Efficiency DESCRIPTIVE QUESTIONS 1. With the help of velocity diagram derive an expression for the work done by an impulse turbine. 2. State the advantages and disadvantages of velocity compounding 3. What are the methods of governing steam turbine? Describe any one method of governing steam turbines. EXERCISE PROBLEMS 1. The following data relates to a single stage impulse turbine. Steam velocity = 600 m/s. Blade speed = 250 m/s. Nozzle angle = 20. Blade outlet angle = 25. Neglecting effect of friction, calculate the absolute velocity of steam leaving the blade and the power developed by the turbine for the steam flow rate of 20 kg/s. Also calculate the axial thrust on the bearings. 2. The mean diameter of the blades of a single flow impulse turbine is 2 m. The speed is 3000 rpm. The nozzle angle is 18 and the blade to steam speed ratio at the inlet is The ratio of the relative velocity at the outlet to inlet of the blade is The blade outlet 153

154 angle is 3 less than the inlet angle. The steam flow rate is 7 kg/s. Draw the velocity diagram of the blade and calculates the power developed by the turbine and the axial thrust. 3. In a stage of impulse turbine provided with a single row wheel, the mean diameter of the blade ring is 800 mm and the speed of rotation is 3000 rpm. The steam issues from the nozzles with a velocity of 300 m/s and the nozzle angle is 20 C. The inlet and outlet blade angles are equal and the blade friction factor is What is the power developed in the blade when the axial thrust on the blade is 140 N. 4. Steam with absolute velocity of 400 m/s is supplied through a nozzle to a single stage impulse turbine. The nozzle angle is 25. The mean diameter of blade rotor is 1 m and it has a speed of 2000 rpm. Find suitable blade angles for zero axial thrust. If blade velocity coefficient is 0.9 and the steam flow rate is 10 kg/s. Calculate the power developed. 5. The steam supply to an impulse turbine with a single row of moving blades is 5 kg/s. The turbine developed 250 kw. The blade velocity is 100 m/s. The steam flows from a nozzle with a velocity of 300 m/s and velocity coefficient of blade is 0.9. Fine the nozzle angle, blade angle at the entry and exit, if the steam flows axially after passing over the blades. 6. A single row impulse turbine receives 3 kg/sec. of steam with a velocity of 425m/sec. The ratio of blade speed to jet speed is 0.4 stage output is 170Kw Determine the blade efficiency and blade coefficient if the internal losses due to friction is 15kw. The nozzle angle is 16 and the blade exit angle is A steam turbine running at 3600rpm takes 4.5 kg of steam per second. The nozzle angle is 16 and the mean diameter of the blade ring 1.2 meters. The blade outlet angle is 18 0 and the isentropic heat drop in the nozzle is 165kj/kg of steam. The shaft power is 485Kw. Assuming nozzle efficiency as 92% and blade velocity coefficient 0.85; find 1) Blade efficiency 2) Stage efficiency 3) Power lost in friction. 8. Steam is supplied to a simple impulse turbine at 10 bar, C. The pressure in the wheel casing is 1.2 bar and the nozzle efficiency is 91%. Determine the exit area required for a steam flow rate of 5000kg/hr. The nozzles are inclined to the plane of rotation at The blades are equiangular and have a velocity coefficient of 0.8. Assuming that the ratio of mean blade speed to steam at nozzle exit to be Calculate 1) the blade angles 2) the power developed 3) the stage efficiency. 9. A reaction turbine running at 360rpm consumes 5kg/sec of steam. The outlet angles of both moving and fixed blades are 20. The power developed is 4.8kw. Determine the drum diameter and blade height at a section where the pressure is 2 bar and dryness 154

155 fraction is Assume 10% of steam leaks from the stage. Take blade angle at inlet as A Parson s reaction turbine consumes 30tonnes of steam per hour and it runs at 400rpm. The steam at a certain stages is at 1.6 bar with dryness fraction of 0.9 and the stage develops 10kw. The axial velocity of flow is constant and is equal to 0.75 times the blade velocity. Find mean diameter and volume of steam flowing per second. Take blade angles at inlet and outlet as 35 and 20 respectively. 11. A reaction turbine runs at 300rpm and its steam consumption is 257kg/min. The pressure of steam at a certain point is 1.9 bar with 0.93 dry. The power developed by the pair is 3.5kw. The exit blade angle is 20 for both moving and fixed and the flow velocity is 0.72 times the blade velocity. If the leakage is 8%, determine drum diameter and blade height. 12. A reaction turbine with mean blade diameter of 1m runs at a speed of 50 rps. The blades are designed with exit angles of 50 and inlet angles of 30. If the turbine is supplied with steam at the rate of 20 kg/sec and stage efficiency of 85% determine power output, enthalpy drop in the stage. 155

156 CHAPTER V AIR COMPRESSORS 5.1 INTRODUCTION An air compressor is a machine to compress the air to raise its pressure. The air compressor sucks air from the atmosphere, compresses it and then delivers the same under a high pressure to a storage vessel; it may be conveyed by the pipe line to the place where the supply of air compressed air is required. Since the compression of air requires some work to be done on it, therefore the compressor must be driven by some prime mover. The compressed air is used for many purposes such as for operating pneumatic drills, riveters, road drills, paint spraying, in starting and supercharging of internal combustion engines, in gas turbine plants, Jet engines and air motors, etc. It is also utilized in the operation of lifts, rams, pumps and variety of devices. In industry, compressed air is used for producing blast of air in blast furnaces and Bessemer converters. 5.2 CLASSIFICATION OF AIR COMPRESSORS The air compressor may be classified into the following considerations: (a) According to the number of stages (i) Single stage: delivery pressure below 10 bar (ii) Multi stage : delivery pressure above 10 bar (b) According to the number of cylinders (i) Single cylinder (ii) Multi cylinder (c) According to working method (i) Reciprocating compressors (ii) Rotary compressors (d) According to the action of air (i) Single acting (ii) Double acting (e) According to the capacity (i) Low capacity: volume of air delivered 0.15 m 3 /sec or less (ii) Medium capacity:volume of air delivered 0.15 m 3 /sec to 5 m 3 /sec (iii) High capacity: volume of air delivered above 5 m 3 /sec 156

157 TERMINOLOGIES Inlet pressure It is the absolute pressure of air at the inlet of a compressor. Discharge Pressure It is the absolute pressure of air at the outlet of a compressor. Pressure Ratio It is the ratio of discharge pressure to the inlet pressure.since the discharge pressure is always more than the inlet pressure; therefore the value of compression ratio is more than unity. Compressor Capacity It is the volume of air delivered by the compressor and it expressed in m 3 /minor m 3 /sec. Free air delivery It is the actual volume delivered by the compressor when reduced to the normal temperature and pressure condition. The capacity of compressor is generally given in the terms of free air delivery. Swept Volume: It is the volume of air sucked by the compressor during its suction stroke. Mathematically, the swept volume or displacement of a single acting air compressor is given by, 4 2 Vs D L D = Diameter of cylinder bore, and L = Length of piston stroke. Mean effective pressure Air pressure on the compressor piston keeps on changing with the movement of the piston in the cylinder. The mean effective pressure of the compressor is equal to the ratio of working done per cycle to the stroke volume. 157

158 5.3 RECIPROCATING COMPRESSOR Construction of single acting reciprocating compressor The construction of a compressor is similar to that of the internal combustion engine except for the suction and delivery valves. Crank shaft connected to the machine, through connecting rod, produces reciprocating motion of the piston in the cylinder. Flywheel in the compressor maintains uniform turning moment throughout the operation. Water or air is circulated through the jackets of the cylinder to cool it. The valves are designed in such way that to operate automatically without any complicated mechanism. The various types of valves used in compressors are reed, plate and thimble and poppet valves Working of single acting, single stage reciprocating compressor When the piston moves downwards from T.D.C to B.D.C., the pressure inside the cylinder eventually falls below the atmospheric pressure. Due to this pressure difference inlet valve opens automatically and the atmospheric air is sucked into the cylinder. The suction of air takes place until the piston reaches the B.D.C. At this time the delivery valve is in closed position. When the piston starts to move in upward direction, a slight increase in pressure inside the cylinder leads to closure of inlet valve. Since both inlet and outlet valves are closed, the continuous upward movement of piston increases the pressure until it reaches the delivery pressure. At this stage, the delivery valve opened and the compressed air is delivered through this valve may be stored in the storage tank. 158

159 5.4 WORK DONEOF SINGLE STAGE COMPRESSOR (WITHOUT CLEARANCE) A Single stage reciprocating air compressor, in its simplest form, consists of a cylinder,piston, inlet and discharge valves. Fig Suction: Inlet valve is opened at pressure P Polytropic Compression Pv n = C 2 3 Delivery at pressure P 2 Outlet valve is opened Work done per cycle or work of compression or Indicated work is is obtained by calculating the area under the P V diagram W Area under 1 2 area under 2 3 area under 4 1 P V P V P n1 V P V P V P V 1 n1 1 n n1 P V P V W n n1 P V P V P V 1 we know that, for polytropic process P V P V W V P P V P P work done per cycle W n n1 P V P P 1 P P n n1 P V P 1 P 159

160 if m kg os air is compressed per cycle, then pv mrt work done per cycle W n n1 mrt T T 1 n n1 mrt P 1 P 5.5 WORK DONE OF SINGLE STAGE COMPRESSOR (WITH CLEARANCE) In order to avoid the damage to the piston when it moves upwards and not to strike the cylinder head a minimum amount of space is provided to accommodate valves, a clearance is necessary in reciprocating compressors. The minimum clearance is desirable and the ratio of clearance volume to the displacement volume is known as clearance ratio. It is denoted by C. Fig Suction 1 2 Polytropic Compression 2 3 Delivery 3 4 Polytropic Expansion of Clearance Volume Work done / cycle = Area = Area Area

161 161 = n n P n PV n P n n P n PV n P But P 3 = P 2, P 4 = P 1 Work done / cycle = n n P n PV PV n P = n n P n P V V n P Work done / cycle = n n a P n PV n P (1) Where 1 4 a V V V = Free air delivered Work done / cycle = n n a P n mrt n P Work done / kg of air = n n P n RT n P (2) Equation (4) is same as Equation (2), (i.e) the clearance will not affect the work of compression per kg of air. Clearance VOLUMETRIC EFFICIENCY: The actual volume of air passing through a reciprocating compressor with clearance is less than the stroke volume during the suction stroke. The ratio of actual volume of air to the swept or stroke volume is called volumetric efficiency of the compressor.

162 162 Fig. 5.4 It is the ratio of the actual volume of free air delivered to the displacement volume. (i.e) a vol s V V = 1 4 s V V V = 4 s c s V V V V = 4 s c s V V V V = 4 1 c s V V V = c c s V V V V = c s V V V V = n c s P V P V

163 vol = PV V 1 PV n n c P 2 P1 V s 1 n 1 V V n P P P3 P2; P4 P1 1 V n c P 2 = 1 1 V s P ISOTHERMAL EFFICIENCY: PV = C P 2 P PV n =C PV=C P V 2 Fig. 5.5 V 1 V Fig. 5.5 Isothermal Work done / cycle = Area = Area under Area under Area under 41 = V PV PV log e PV V2 163

164 1 1 PV, PV 1 1 Isothermal process V 1 = PV 1 1log e 1 V2 V P 1 2 = mrt1loge mrt 1 1loge V2 P1 P 2 (i.e) Work done / cycle = mrt1 log e P1 P 2 Work done / kg = RT1 log e P1 Iso Iso IsothermalWorkdone ActualWorkdone n P log 2 e P1 n1 n n P 2 1 P MULTI STAGE COMPRESSORS n RT log P 2 1 e P1 n1 n n P RT P1 1 A single stage air compressor suffers the following drawbacks: (i) The size of the cylinder will be too large. (ii) Due to compression, there is a rise in temperature of the air.it is difficult to reject heat from the air in the small time availableduring compression. (iii)the temperature of air, at the end of compression, is too high. Itmay heat up the cylinder head or burn the lubricating oil. In order to overcome the above mentioned difficulties, two or more cylinders are provided in series with intercooling arrangement between them. Such an arrangement is known as multistage compression. 164

165 5.8.1 Advantages of Multi Stage Compression Following are the main advantages of multistage compression over single stage compression: i. The work done per Kg of air is reduced in multistage compression withinter-cooling as compared to single stage compression for the same delivery pressure. ii. It improves the volumetric efficiency for the given pressure ratio. iii. The size of the cylinders may be adjusted to suit the volume and pressure ofthe air. iv. It reduces the leakage loss considerably. v. It gives more uniform torque, and hence a smaller size flywheel is required. vi. It provides effective lubrication because of lower temperature range. vii. It reduces the cost of compressor WORKING PRINCIPLE OF TWO STAGE COMPRESSOR WITH INTERCOOLER Air is sucked into the Low Pressure (L.P) cylinder from atmosphere, where it is compressed to an intermediate pressure and high temperature. After the compression process, air flows into an intercooler. In the inter-cooler, air is cooled to its atmospheric temperature or initial temperature. Then air is flowing to the High Pressure (H.P.) cylinder, air is compressed to the final required pressure. The compressed air from the H.P. cylinder is also passed through an after cooler to reduce the temperature of the compressed air before going to the storage receiver. Fig

166 The following assumptions are to be considered for a two stage compressor with intercooler. a. The effect of clearance is neglected b. Suction and delivery of air takes place at constant pressure c. No pressures drop in the intercooler d. Both cylinders follow. Case (i): when the inter-cooling is imperfect Figure shows the P-V diagram of a two stage air compressor with imperfect inter-cooling. 13 Fig. 5.7 If the inter-cooling is imperfect, the point 2 will not lie on the iso-thermal line Let, P suction pressure, N/m V Volume of L.P.cylinder,m P intermediate pressure, N m V Volume of H. P. cylinder, m P delivery pressure, N m n index of compression for both cylnders Work done in L.P. cylinder = area under curve Work done in H.P. cylinder = area under curve

167 Total work done W Workdone in L. P. cylinder Workdone in H. P. cylinder n n1 p v p 1 p n n1 p v p 1 p n n1 p v p n 1 p n1 p v p 11 p Indicated power of the compressor WN watt 60 If the compression is adiabatic should be substituted in place of in (1) Case 2: when the inter-cooling is perfect Figure shows the P-V diagram of a two stage air compressor with perfect inter-cooling. Fig. 5.8 If the inter-cooling is perfect, the point 2 will lie on the isothermal line, then. substituting this condition in equation (1), we get Total work done W n n1 p v p p 22 p p Indicated power of the compressor WN watts Condition for minimum work required in two stage compressor with intercooling 167

168 When the suction pressure and delivery pressure are fixed then the least value of inter cooler pressure can be obtained by differentiating equation (2) w. r.t. and equating to zero. W n n1 p v p p 2 p p n n1 k W kp v p p 2 p p For minimum work, dw 0 dp k p v is constant dw k p dp p k p p 0 k p p k p p p p p p p p p p p p 3 p p p 4 for minimum work or maximum efficiency p p p Substituting the value of p p in 2,we get, Minimum work required per cycle is given by W W n n1 p v p p 2 p p n n1 p v 2 p 2 p 2n n1 p v p 2n 1or p n1 mrt p 15 p If N number of stages in an air compressor then, 168

169 Minimum work required per cycle, W Nn n1 p v p N N Nn 1or p n1 mrt p N p N RATIO OF CYLINDER DIAMETERS Let, D diameter of L.P.cylinder D diameter of H.P.cylinder L length of stroke of cylinder V volume of L. P. cylinder V Volume of H. P. cylinder For perfect intercooling,p V P V P V D L P V D L D D D P / 7 D P For two stages compressor with perfect intercooling P P / 8 P P Sub 8in 7,We get, D P / D P 5.9 ROTARY AIR COMPRESSOR: In a rotary air compressor the air is entrapped between two sets of engaging surface and the pressure of the air is increased by squeezing action of back flow of air. Though there are many types of rotary air compressors, yet the following are important types: (i) Roots blower compressor (ii) Vane blower compressor (iii) Centrifugal blower compressor (iv) Axial flow compressor The first two compressors are popularly known as Positive displacement compressor, whereas the last two as Non- Positive displacement compressor. 169

170 5.9.1 ROOTS BLOWER COMPRESSOR Rotary Lobe type Air Compressor has two mating lobe-type rotors mounted in a case. The lobes are gear driven at close clearance, but without metal-to-metal contact. The suction to the unit is located where the cavity made by the lobes is largest. As the lobes rotate, the cavity size is reduced, causing compression of the vapor within. The compression continues until the discharge port is reached, at which point the vapor exits the compressor at a higher pressure Fig VANE BLOWER COMPRESSOR Rotary slide vane type Air Compressor has longitudinal vanes, sliding radially in a slotted rotor mounted eccentrically in a cylinder. The centrifugal force carries the sliding vanes against the cylindrical case with the vanes forming a number of individual longitudinal cells in the eccentric annulus between the case and rotor. The suction port is located where the longitudinal cells are largest. The size of each cell is reduced by the eccentricity of the rotor as the vanes approach the discharge port, thus compressing the air. Fig

171 5.9.3 CENTRIFUGAL COMPRESSOR Introduction Air compressors of various designs are used widely throughout DOE facilities in numerous applications. Compressed air has numerous uses throughout a facility including the operation of equipment and portable tools. Three types of designs include reciprocating, rotary, and centrifugal air compressors. Centrifugal Compressors The centrifugal compressor, originally built to handle only large volumes of low pressure gas and air (maximum of 40psig), has been developed to enable it to move large volumes of gas with discharge pressures up to 3,500 psig. Centrifugal compressors are now most frequently used for medium volume and medium pressure air delivery. One advantage of a centrifugal compressor is the smooth discharge of the compressed air. The air particles enter the eye of the impeller, designated D in Figure As the impeller rotates, air is thrown against the casing of the compressor. The air becomes compressed as more and more air is thrown out to the casing by the impeller blades. The air is pushed along the path designated A, B, and C in Figure The pressure of the air is increased as it is pushed along this path. Note in Figure 5.11 that the impeller blades curve forward, which is opposite to the backward curve used in typical centrifugal liquid pumps. Centrifugal compressors can use a variety of blade orientation including both forward and backward curves as well as other designs. There may be several stages to a centrifugal air compressorand the result would be the same; a higher pressure would be produced. The air compressor is used to create compressed or high pressure air for a variety of uses. Some of its uses are pneumatic control devices, pneumatic sensors, pneumatic valve operators, pneumatic motors, and starting air for diesel engines. Fig

172 5.9.4 AXIAL FLOW COMPRESSOR In axial flow compressor the air flows parallel to the axis. It consists of a number of rotating blades fixed to a rotating drum. Each stage consists of one row of moving blades and one row of fixed bades. As the air flows feom one set of stator and rotor to another, it gets compressed. The cuccessive compression of air in all the sets of stator and rotor increasses its pressure and the air is delivereed at a high pressure from the putlet. The blades are made of aerofoil section to reduce the losses caused by turbulence and boundary sepeartion. The number of stages may vary from 14 to 16. The pressure ratio per stage is from 1.2 to 1.3. Fig COMPARISION OF RECIPROCATING AND ROTARY AIR COMPRESSORS Following are the main points of comparison of reciprocating and rotary air compressor: S.No. Reciprocating Air Compressor Rotary Air Compressors 1. The maximum delivery pressure may be as high as 1000 bar. The maximum delivery pressure is10 bar only. 2. The maximum free air discharge is about 300 m 3 /min. The maximum free air discharge is as high as 3000 m3/min. 3. They are suitable for low discharge of air at very high pressure. They are suitable for large discharge of air at low pressure. 172

173 4. The speed of air compressor is low. The speed of air compressor is high. 5. The air supply is intermittent. The air supply is continuous. 6. The size of air compressor is large for the given discharge. The size of air compressor is small for the given discharge. 7. The balancing is a major problem. There is no balancing problem. 8. The lubricating system is complicated. The lubricating system is simple The air delivered is less clean, as it comes in contact with the lubricating oil. Isothermal Efficiency is used for all sorts of calculation. The air delivered is more clean, as it does not comes in contact with the lubricating oil. Isentropic Efficiency is used for all sorts of calculation COMPARISION OF CENTRIFUGAL AND AXIAL FLOW AIR COMPRESSORS Following are the main points of comparison of the centrifugal and axial flow air compressors: S. No Centrifugal Compressors The flow of air is perpendicular to the axis of compressor It has low manufacturing and running cost. It requires low starting torque. It is not suitable for multi- staging Axial Flow Compressors The flow of air is parallel to the axis of compressor It has high manufacturing and running cost. It requires high starting torque. It is suitable for multi- staging 173

174 5. It requires large frontal area for a given rate of flow. It requires less frontal area for a given rate of flow. It makes the compressor suitable for air crafts. SOLVED PROBLEMS 1. A single stage, single acting reciprocating compressor delivers 15m 3 of free air per minute, from 1 bar to 8 bar. The speed of the compressor is 300 rpm. If the clearance is th of the swept volume, determine the diameter and stroke of the compressor. Take = 1.5 and n=1.3. Given Data: 15 / 1 8, 1, ,1.3 Solution %, A small single acting compressor has a bore and stroke of 10 cm driven at 400 rpm. The clearance volume is 80 cm 3. The index of compression and expansion is 1.2. The suction pressure is 0.95 bar and the delivery is 8 bar. Calculate i) volume of freed air at 1.03bar and 20 C per minute, if the temp at the start of compression is 30 C, ii) mean effective pressure of the indicator diagram assuming constant suction and deliver y pressure. 174

175 Given Data: , Solution: , V %, i

176 3. A two stage single acting air compressor having capacity 4.5m 3 /min measured under free air conditions of bar and 15 C. the pressures during suction stroke is 0.98 bar. The temperature of air at the start of compressor in each stage is 27 C. The delivery pressure in 15 bar. The clearance volume in LP cylinder is 15% of the stroke. The index of compression and expansion is 1.3 and the speed is 140 rpm. The intercooler pressure is such that the work is shared equally between 2 cylinder V 1 =105 m 3 /min, V 2 =5 m 3 /min. Determine i) the indicated power, ii) Diameter and stroke of LP cylinder if bore is equal to stroke. Given Data: 4.5, , , , 1.3, 140, 0.15, 105 Solution:, / For isentropic compression process (1-2 ) C, % 176

177 , A two stage compressor delivers 2 cm 3 free air per minute. The temperature and Pressure of air at the suction are 27 C and 1 bar. The pressure at the delivery is 50 bar. The clearance is 5% of the stroke in LP cylinder as well as in HP cylinder. Assume perfect inter-cooling between the two stages, find i) The minimum power required to run the compressor at 200 rpm, ii) The diameters and strokes assuming the strokes of the cylinder are equal to the diameter of L.P cylinder. Law of compression and re-expansion in both the cylinders is PV 1.35 =constant. Also. Assume that the ambient air condition is same as suction condition. Given Data: , , 1.35, 200, 0.05, Solution:

178 %, n.. &.. & , / / A single acting two stage compressor with complete inter cooling delivers 5kg/min of air at a pressure of 15 bar. The intake state of air is 1bar and 15 C. The clearance volumes of L.P and H.P cylinders are 5% and 6% of the respective cylinder swept volumes. The speed of the compressor is 420 rpm. Assuming the compression and expansion processes are polytropic with n=1.3. Calculate i) The power required, ii) Isothermal efficiency, iii) Swept and clearance volumes of the L.P and H.P cylinders. Given Data: 5 / / 1 100, K , 1.3, ,

179 Solution: % % l % ,,

180 ,, A single stage double acting compressor has a free air delivery (F.A.D) of 14 m 3 /min measured at 1.013bar and 15 C. The pressure and temperature in the cylinder during induction are 0.95 bar and 32 C respectively. The delivery pressure is 7 bar and index of compression and expansion n=1.3. The clearance volume is 5% of the swept volume. Calculate the indicated power required and the volumetric efficiency. Given data: 14, , , , 1.3, 0.05 Solution:

181 % 7. Consider a single-acting 2- stage reciprocating air compressor running at 300 rpm. Air is compressed at a rate of 4.5 kg/min from bar and 288K through a pressure ratio of 9 to 1. Both the stages have the same pressure ratio and the index of compression and expansion in both stages is 1.3. Assume a complete inter cooling, find the indicated power and the cylinder swept volumes required. Assume that the clearance volumes of both stages are 5% of their respective swept volumes. Given Data: / / , , Solution: W %..,,

182 ..,, OBJECTIVEQUESTIONS 1. The volume of air delivered by the compressor is called (a) free air delivery (b) compressor capacity (c) swept volume (d) none of these 2. The volume of air sucked by the compressor during its suction stroke is called (a) free air delivery (b) compressor capacity (c) swept volume (d) none of these 3. The ratio of work done per cycle to the stroke volume of the compressor is known as (a) compressor capacity (b) compression ratio (c) compressor efficiency (d) mean effective pressure 4. inter cooling in a multi stage reciprocating compressor helps the process of compression come nearer to (a) isothermal (b) adiabatic (c) polytropic (d) constant volume 5. For a reciprocating compressor (a) isothermal> adiabatic (b) isothermal < adiabatic (c) compressor = isothermal (d) compressor = adiabatic 6. The clearance volume in a reciprocating compressor has no effect on (a) power of compressor (b) volumetric efficiency of compressor (c) discharge of compressor (d) intake of compressor 7. The volumetric efficiency of a reciprocating compressor normally lies between (a) 40 to 50% (b) 55 to 65% (c) 70 to 90% (d) 95 to 100% 8. Which process does not take place in reciprocating compressor? (a) constant temperature (b) constant volume (c) adiabatic (d) polytropic 9. For equal work in two cylinders of a compressor (a) pressure ratio of two cylinders must be same 182

183 (b) index of compression in two cylinders must be same (c) perfect inter cooling between the cylinders must be obtained (d) all the three conditions must be fulfilled 10. In conjunction with a gas turbine the compressor used is (a) reciprocating type (b) centrifugal type (c) axial flow type (d) screw type 11. The ideal process in a rotary compressor is (a) isothermal (b) polytropic (c) adiabatic (d) none of these 12. The rotary compressors are used for delivering (a) small quantities of air at high pressures (b) large quantities of air at high pressures (c) small quantities of air at low pressures (d) large quantities of air at low pressures 13. The maximum delivery pressure in a rotary compressor is (a) 10 bar (b) 20 bar (c) 30 bar (d) 40 bar 14. If the flow of air through the compressor is perpendicular to its axis, then it is a (a) reciprocating compressor (b) centrifugal compressor (c) axial flow compressor (d) turbo compressor 15. A compressor mostly used for super charging of I.C. engines is (a) radial flow compressor (b) axial flow compressor (c) roots blower (d) reciprocating compressor 16. Air is a mixture of gases (a) True (b) False 17. The positive displacement compressor is (a) roots blower compressor (b) vane blower compressor (c) centrifugal compressor (d) both (a) and (b) 18. In a reciprocating air compressor work done is minimum during compression (a) Isothermal (b) Isentropic (c) Polytropic (d) None of these 19. The pressure of air at the beginning of the compression is called as (a) Compressor pressure (b) back pressure (c) initial pressure (d) delivery pressure 20. Compression means (a) decreasing the volume at constant pressure (b) increasing the pressure at constant volume (c) increasing the pressure by decreasing the volume 183

184 (d) decreasing the volume and increasing the temperature 21. Swept volume is given by D L (b) 4 (a) 2 V S V S D 2 4 L VS 22. Volumetric efficiency is given by (a) D L (c) V D S (d) 4L (b) 1 1 (c) Clearance ratio does not have any effect on the (a) work done / cycle (c) work done / minute 24. The clearance ratio is given by (d) 1 (b) work done / kg of air (d) none of the above 1 (a) (b) (c) (d) Where V S = Stroke volume, V a = Actual volume, V C = Clearance volume 25. In a reciprocating air compressor one cycle is completed in (a) Each and every revolution of the crank (b) every two revolutions of the crank (c) Four strokes of the piston (d) one stroke of the piston TWO MARKS QUESTIONS 1. Mention any four uses of compressed air. 2. Define Compression. 3. What is the difference between a Reciprocating and a RotaryAir compressor? 4. What is meant by Free air delivered? 5. How is air compressors classified? 6. Define Compressor Capacity. 7. Define the following terms : 8. (i) Inlet pressure (ii) Discharge Pressure 9. Define the following terms : 10. (i) Pressure Ratio (ii) Swept volume. 11. What is the necessity for compressing air in multi stage air compressors? 12. What is Inter cooling? 184

185 13. What is the condition for minimum work required for two stage air compressors with perfect inter cooling? 14. Define Effectiveness of the Inter cooler. 15. Define the following, 16. (i) Clearance ratio, (ii) Volumetric efficiency, (iii) Isothermal Efficiency, (iv) Mechanical Efficiency. 17. Air is compressed from 1 bar to 7 bar absolute in a two stage air compressor. What pressure in the inter cooler will give the best efficiency? 18. Write the equation for the work required for a multi stage compressor with N stages? 19. What is the effect of clearance volume on the work done / Kg of air for a reciprocating air compressor 20. What is positive displacement compressor? 21. What is centrifugal compressor? 22. What is axial flow compressor? 23. What is the difference between the single acting and double acting reciprocating air compressor? DESCRIPTIVE QUESTIONS 1. Describe the working of a single stage reciprocating air compressor with neat sketch. 2. What are the advantages of multi stage compression? 3. Describe an expression for the work done of a single stage reciprocating air compressor with clearance. 4. Prove that the condition for minimum work required for a two stage reciprocating air compressor with perfect inter cooling i.e., P 2 = P1P 3 5. What are the classification of air compressor? 6. What are the difference between reciprocating and rotary air compressors? 7. Explain the working of a centrifugal compressor with neat sketch. EXERCISE PROBLEMS 1. Find the diameter and stroke of a single acting, single stage air compressor. The pressure at the inlet and discharge are 1.1atm and 7.7atm respectively. The speed of the compressor is 240rpm and the speed of the piston is limited to 150m/min. 185

186 The IP of the compressor is 10kW. Neglect the effect of clearance. Assume n=1.2 for compression and expansion. Also find the Mean Effective Pressure of the compressor cycle. 2. A two stage compressor delivers 10m 3 of free air per minute at 5ata to the receiver. The temperature and pressure of the ambient air are 15 o C and 1 atm. The temperature and pressure of the air entering the HP cylinder are 25 o C and 2atm. Assuming the suction and ambient conditions as same find the IP of the compressor. Assume the compression and expansion are adiabatic. What pressure in the intercooler would give the heat efficiency? Also determine the BP required to run the compressor if mechanical efficiency is 90%. 3. A single stage air compressor with 300mm bore and 400mm stroke is required to compress air from 1bar to 5bar. Find the power required to drive the compressor running at 200rpm, when the compression of air is (a) Isothermal (b) adiabatic with γ = A two stage air compressor takes in 22.5kg of air per minute at 15 o C and 1 bar and delivers it at 16.5bar. At the intermediate pressure it is cooled to initial temperature. Compression follows the law PV 1.2 =constant. Neglecting clearance, determine the intermediate pressure that gives least work. Also find heat rejected in the intercooler per minute and minimum power required to run the compressor. 5. A Compressor draws in 720 m 3 / hr of air at 1 bar and 20C and discharges at 5 bar and 90C. Heat rejected by air to the cooling medium is 30 kj/kg. Determine the shaft work done on air in kj/kg and mass of air handled. Neglect change in kinetic and potential energy. Assume R = and C V = 0.71 kj/kgk. 6. A two stage single acting air compressor having capacity 4.5 m 3 /min measured under free air conditions of and 15C. The pressure during the suction stroke is 0.98 bar. The temperature of air at the start of compression in each stage is 27C. The delivery pressure is 15 bar. The clearance volume in L.P. cylinder is 5 % of the stroke. The index of compression and expansion is 1.3 and aped is 140 rpm. The intercooler pressure is such that the work is shared equally between the two cylinders. Determine, (i) The indicated power and (ii) The diameter and stroke of the L.P. cylinder if bore is equal to thestroke. 186

187 7. A small single acting compressor has a bore and stroke both of 10 cm and is driven at 400 rpm. The clearance volume is 80 cm 3 and the index of compression and expansion is 1.2. The suction pressure is 0.95 bar and the delivery is 8 bar. Calculate (i) The volume of free air at 1.03 bar and 20C dealt with per minute, if the temperature at the start of compression is 30C and (ii) The mean effective pressure assuming constant suction and deliverypressure. A single stage, single acting reciprocating air compressor has a bore of 200 mm and a stroke of 300 mm. It receives air at 100 kpa and 15C and delivers it at 90 KPa. If index of polytropic compression is 1.3 and clearance volume is 4% of swept volume, determine (a) mean effective pressure and (b) power required to drive the compressor if it runs at 500 rpm. 8. A two stage double acting compressor compresses air to a delivery pressure of 20bar from atmospheric condition of 1 bar, 20 C. The condition of air at the inlet of compression stroke is 0.98bar, 32C. Take perfect inter cooling of air between stages. The size of low pressure cylinder is 400 mm bore with 500mm stroke. The clearance ratio of both cylinders is Speed = 240 rpm index of compression = Find power required to run the compressor and heat rejected in the intercooler. 9. A single stage double acting compressor runs at 300 rpm compress air upto a delivery pressure of 7bar. The condition of air at atmosphere is bar, 27C. The same at the end of suction stroke is 0.98 bar, 40C. Take C = 0.04 and n = 1.3, L/D = 1.3. Find the volumetric efficiency, indicated power, isothermal efficiency and cylinder dimensions. 187

188 CHAPTER VI REFRIGERATION 6.1 INTRODUCTION Refrigeration stands for the production of a cool confinement with respect to surroundings. It may be defined as The artificial withdrawal of heat, producing in a substance or within a space a temperature lower than which would exist under the natural influence of surrounding. Maintaining perishables at their required temperature is done by refrigeration. Not only perishables but today many human work spaces, in offices and factory buildings are airconditioned and refrigeration unit is the heart of the system. 6.2 REVERSED CARNOT CYCLE The most efficient refrigeration cycle is reversed Carnot cycle as shown in figure 6.1. The refrigerant absorbs heat at a constant temperature during the process 4 1 and is compressed isentropically during the process 1 2. Heat rejection from the refrigerant to the cooling medium takes place isothermally during the process 2 3 and then finally the refrigerant expanded back to its original position through an isentropic expansion process 3 4. In refrigeration cycle, output is to extract the maximum amount of heat in the evaporator for a net expenditure of work in compressor. The coefficient of performance (COP) of the reversed Carnot cycle is given as Heat absorbed COP = Work input Since the compression and expansion are isentropic and there is no heat transfer from the system to the surroundings, the work input is the difference between heat rejection and heat absorption. Heat absorbed COP = Heat rejected - Heat absorbed From the definition of entropy, dq T. ds 188

189 Since, Where and are the absolute temperature at which heat absorption and heat rejection takes place. Though reversed Carnot cycle gives the maximum COP for heat transfer between two temperatures, it is very difficult to achieve this in actual refrigeration systems. A combination of isentropic processes and isothermal processes is not possible to achieve in steady flow conditions under which the actual refrigeration systems work. The upper cycle temperature must be above the temperature of the atmosphere so that the required rate of heat rejection can be accomplished. Similarly the lower cycle temperature must be below the temperature of the cold chamber to enable the evaporator function and these necessary temperature differences place a limit on the coefficient of performance attainable Unit of refrigeration Refrigeration is the process of heat removal, to produce and maintain a low temperature that of the surroundings in a closed space. Unit of refrigeration is known as tonne of refrigeration (TR). One tonne of refrigertiaon is equal to 3.5 kw. One tonne of refrigeration is equivalent to the rate of heat removal required to produce 1000 kg of ice from water at 0 in one day. 1 tonne of refrigeration kcal min kw BELL COLEMAN CYCLE (REVERSED BRAYTON CYCLE) In the Bell-Coleman cycle, heat absorption and rejection follow constant pressure processes as shown in figure. The other two processes are isentropic. Sequence of process 4-1 constant pressure heat absorption 1-2 isentropic compression 2-3 constant pressure heat rejection 3-4 isentropic expansion 189

190 Air refrigeration system is used mostly in aircrafts and work on Bell-Coleman cycle. Low temperature air absorbs heat from the cold chamber during the process 4-1. This air is further compressed isentropically during the process 1-2. Compressed air rejects the heat absorbed and heat added during compression at constant pressure during the process 2-3. Isentropic expansion 3-4 brings back to the initial condition. The COP of the Bell-Coleman cycle is, COP heat absorbed orrefrigerating effect work input Compression and expansion are isentropic, 190

191 Heat rejected mc T T Heat absorbed mc T T Work input heat rejected heat absorbed mc T T mc T T COP mc T T mc T T mc T T T T T T T T 1 T T T T 1 1 T T T T T T 1 T T T T and T T T T T COP T T Non-condensing (air) working fluid is used in air-standard refrigeration cycle. This cycle is the reversed Brayton cycle and is used for the liquefaction of air, helium and also for aircraft cooling systems. 6.4 VAPOUR COMPRESSION REFRIGERATION SYSTEM (VCR) In a simple vapour compression system fundamental process are completed in one cycle. These are: Compression 2. Condensation 3. Expansion 4. Vapourisation The flow diagram of such a cycle is shown in fig. 191

192 The vapour at low pressure and temperature enters the compressor which it is compressed isentropically and subsequently its temperature and pressure increase considerably. The vapour after leaving the compressor enters the condenser where it is condensed into high pressure liquid and it is passes through the expansion valve, here it is throttled down to a lower pressure and has a low temperature. After finding its way through expansion valve it finally passes onto an evaporator where it extracts heat from the surroundings or circulating fluid being refrigerated and vaporizes to low pressure vapour FUNCTIONS OF PARTS OF SIMPLE VAPOUR COMPRESSION SYSTEM Here follows the brief description of various parts of a Simple Vapor Compression System shown in fig (i) Compressor The function of a compressor is to remove the vapour from the evaporator and to raise its temperature and pressure to a point such that it (vapour) can be condensed with available condensing media. (ii) Discharge Line (or Hot Gas Line) A hot gas line or discharge line delivers the high pressure, high temperature vapourform the discharge of the compressor to the condenser. (iii) Condenser The function of a condenser is to provide a heat transfer surface through which heat passes from the hot refrigerant vapour to the condensing medium. (iv) Receiver Tank A receiver tank is used to provide storage for a condensed liquid so that a constant supply of liquid is available to the evaporator as required. (v) Liquid Line A liquid line carries the liquid refrigerant from the receiver tank to the refrigerantflow control. 192

193 (vi) Expansion Valve (refrigerant flow control) Its function is to meter the proper amount of the refrigerant to the evaporator and to reduce the pressure of liquid entering the evaporatorr so that liquid will vapourize in the evaporator at the desired low temperature and take out sufficient amount of heat. (vii) Evaporator An evaporator provides a heat transfer surface through which heat can pass from the refrigerated space into the vapourising refrigerant. (viii) Suction Line The suction line conveys low pressure vapour form the evaporator to the suction inlet of the compressor THERMODYNAMIC ANALYSIS OF VAPOUR COMPRESSION CYCLE (i) When the evaporator is dry and saturated at the end of compression: The T-s diagram of the Vapour Compression refrigeration system is given below. At point 2 the vapour which is at low temperature (T 2 ) and low pressure enters the compressor cylinder and is compressed adiabatically to 3 when its temperaturee increases to the temperature T 1. It is then condensed in the condenser (line 3-4) where it gives up its latent heat to the condensing medium. It then undergoes throttling expansion while passing through the expansion valve and it again reduces to T 2, it is represented by the line 4-1. From the T-s diagram it may noted that due to this expansion the liquid partially evaporates, as its dryness fraction is represented by the ratio b 1 / b 2. At 1 it enters the evaporator where it is further evaporated at constant pressuree and constant temperature to the point 2 and the cycle is completed. Work done by the compressor = W = area b-2 193

194 Heat absorbed = areaa 2-1-g-f-2 C.O.P = (Heat extracted of refrigerating effect / work done) = [ (area 2-1-g-f-2) / (areaa b-2) ] C.O.P = (h 2 -h 1 ) / (h3-h 3 2 ) = (h 2 -h 4 4) / (h 3 -h 2 ) (h 1 = h 4 since during the throttling expansionn 4-1 the total heat content remains unchanged). (ii) When the vapour is superheated after compression: If the compression of the vapour is continued after it has become dry, the vapour will be superheated, its effect on T-s diagram is shown in figure. The vapour enters the compressor at condition 2 and it is compressed to 3 where it is superheated to temperature T sup. Then it enters the condenser. Here firstly superheated vapour cools the temperature T (represented by the line ) and then it condenses at constant temperature along the line 3 4; the remaining of the cycle; however is the same as before. T Now work done = area b-2 and heat extracted / absorbed = area 2-1-g-f-2 Heat extracted area '2-1-g-f-2' h C. OP. 1 work done ' b-2' h 2 3 h 1 h 2 In this case h 3 = h C P (T sup T sat ) and h 1 3 = enthalpy of dry and saturated vapour at the point 3 1. (iii) When the vapour is wet after compression: 194

195 Heat extracted C. OP. work done Work done by the compressor = area b-2 Heat extracted = area 2-1-g-f-2area '2-1-g-f-2' h '2-3-4-b-2' h 2 3 h 1 h 2 Note: If the vaopur is not superheated after compression, the operation is called WET COMPRESSION and if the vapour is superheated at the end of compression, it is known as DRY COMPRESSION, in actual practice is always preferred as it gives higher volumetric efficiency and mechanical efficiency and there are less changes of compressor damage EFFECT OF SUPERHEA ATING As may be seen from the figure the effect of superheating is to increase the refrigerating effect but this increase in refrigerating effect is at the cost of increase in amount of work spent to attain the upper pressure limit. Since the increase in work is more as compared to increase in refrigeratingg effect, therefore overall effect the superheating is to give a low value of C..O.P. 195

196 EFFECT OF SUB-COOLING OF LIQUID Sub-cooling is the process of cooling the liquid refrigerant below the condensing temperature for a given pressure. In the figure the process of sub-cooling is shown by As is evident from the figure the effect of sub-cooling is to increase the refrigerating effect. The sub-cooling results in increasee of C.O.P. provided that no further energy has to spent to obtain the extra cold coolant required. The sub-cooling or under cooling may be done by any of the following methods: (i) (ii) Inserting a special coil between the condenser and expansion valve. Circulating greater quantity of cooling water through the condenser. 196

197 6.5 PROPERTIES OF REFRIGERANT A substance which absorbs heat through expansion or vaporization is termed as a refrigerant. An ideal refrigerant should possess chemical, physical and thermodynamic properties which permit its efficient application in the refrigerating system. An ideal refrigerant should have the following properties: (i) Low boiling point (ii) High critical temperature (iii) High latent heat of vaporization (iv) Low specific heat of liquid (v) Low specific volume of vapour (vi) Non corrosive to metal (vii) Non-Flammable and Non- Explosive (viii) Non- toxic (ix) Easy to liquefy at moderate pressure and temperature (x) Easy of locating leaks by odour or suitable indicator (xi) Low Cost (xii) Mixes well with oil VAPOUR ABSORPTION REFRIGERATION SYSTEM (VAR) 197

198 In vapour absorption system, the compressor is replaced by an absorber, a pump and a generator. In the ammonia absorption system, low pressure ammonia leaving the evaporator is absorbed in water in an absorber. Heat is to be removed during the process by circulating the cold water. Thestrong ammonia solution is then pumped to the generator. The pump increases, pressure of strong ammonia solution to that desired in the condenser. Ammonia is liberated from the solution as heat is supplied to the generator. Ammonia vapour then goes to the condenser, where it is condensed. The high pressure liquid ammonia then passes through the expansion valve or throttle valve. The high pressure liquid is converted into a very wet vapour at low pressure and temperature during this process. The Low pressure liquid ammonia goes to the evaporator, where it extracts the latent heat of evaporation from the substance to be cooled. The ammonia vapor coming out of evaporator is fairly dry and enters the absorber where it mixes with the cold water. This completes the cycle. The hot weak ammonia solution left at the bottom of the generator is first throttled to low pressure by passing it through pressure reducing valve and then passed into the absorber. In actual practice, vapour absorption system is fitted with a heat exchanger, an analyzer and a rectifier to improve the efficiency of the system. The coefficient of performance of the absorption system is defined as, Refrigerating effect COP heat addition at generator pump work COP of the absorption system is always (=0.6) lower than that of the vapour compression system (=3). In actual practice, the vapour absorption system is fitted with heat exchanger, analyser and a rectifier as shown in Fig. 6.9 to improve the efficiency of the plant. 198

199 Heat exchanger The capacity of water at high temperature for absorbing ammonia vapour is low. So the hot weak solution coming from the generator to the absorber must be cooled. The heat removed from the weak solution may be used to raise the temperature of the strong solution coming from the absorber and going to the generator. The heat transfer is accomplished by replacing a counter flow heat exchanger between the pump and the generator. It increases the economy of the plant. Analyser The ammonia vapour contains water vapour while leaving the generator. The water vapour is to be removed before the ammonia vapour enters the condenser otherwise it will freeze at the throttle valve. The water vapour partly removed by passing the ammonia vapour through an analyser containing a series of trays. Rectifier The rectifier removes the remaining water vapour from the ammonia vapour coming out from the analyser by providing water cooling. The condensed liquid is returned to the upper part of the analyser by a drip return pipe. Rectifier is fitted before the condenser. Water can also be used as refrigerant in a vapour absorption refrigerator at the temperature levels which prevail in air-conditioning plant and heat pumps. Common absorbent in the case 199

200 is lithium bromide. It is a good absorber because it does not evaporate at the generator temperature. Water lithium bromide system is simpler and its COP is higher than ammoniawater absorption system Ammonia vapour absorption refrigerator has a COP of less than unity and is rarely used unless waste heat is available from existing source COMPARISON BETWEEN VAPOUR ABSORPTION AND VAPOUR COMPRESSIONREFRIGERATION SYSTEMS Energy:The vapour absorption system operates on thermal energy a comparatively much cheaper form of energy as compared to the mechanical or electrical energy required for the operation of the vapour compression system. Moving parts: The only moving part in the vapour absorption system is the pump for the delivery of the strong mixture. Since the size of the pump is quite small compared to the compressor, its cost is very low. However, the two major components are involved such as the absorber and generator which are not very costly. In addition the compressor causes vibration for which the strong foundation has to be provided. This is not so for the pump. Quality of refrigerant: The vapour temperature leaving the generator is slightly superheated and is controlled by the heat exchangers provided after the generator. But, for the vapour compression the degree of superheat is governed by the suction state of the vapour. Evaporator pressure: The capacity of the absorption system is controlled by adjusting the steam or generator temperature even if the evaporator pressure falls. On the other hand, the capacity of the compression system decreases rapidly with the reduced evaporator pressure. COP: The COP the absorption system is usually much lower in magnitude than the compression system. But this low value of the former is not of much importance since it uses the waste energy. Thus the selection of the above system should be based on the overall cost. 200

201 Effect of load variation on COP: The absorption system can be operated at the same COP as the designed value by the appropriate control of the generator temperature even at part load or reduced evaporator conditions. Capacity: The capacity of the absorption system may run into hundreds of tons in asingle unit since there is no bar like the inertial force of the reciprocating masses of the compression system. Hence for the large capacity compression systems multi-cylinder compressors are used with bore and stroke around 100 to 150 mm and operating sped around 900 to 1200 rpm. Thus the compression system becomes much more costly than the absorption system for a large tonnage. Maintenance: Since the only moving parts involved are pumping equipments, itsmaintenance is comparatively much less than the compression system. The additional maintenance of absorber, generator, etc., is much less due to static components. Wastage of refrigerants: There is less wastage of refrigerant in case of absorption system compared to the compression system due to leakage through the shaft seal. Space: For all units the absorption system is rather bulky compared to the compression system. The absorption system is much more compact and less bulky than the compression system for large capacity. SOLVED PROBLEMS 1. A dense air machine operates on a reversed Joule cycle and is required for a capacity of 12 TOR. The cooler pressure is 4.2 bar and the refrigerator pressures is 1.4 bar. The air is cooled in the cooler to a temperature of 47 and the temperature of air at the inlet of compressor is 17. Determine for the ideal cycle, a) COP, b) mass of air circulated per min, c) theoretical piston displacement of compressor, d) theoretical displacement of expander, e) net power per ton of refrigeration. Given Tonnage of refrigerationtor 12 TR kw T K P P 1.4 bar T K 201

202 P P 4.2 bar Solution Assume both compression and expansion is isentropic Compression process Expansion process COP Refrigerating Effect N m C m Refrigerating Effect C Theoretical displacement of compressor Theoretical displacement of expander kg sec m 140 sec m 140 sec Network done heat supplied heat rejected C C Network done kj kg kw kW ton 2. A refrigerator works between -7 and 27. The vapour is dry at the end of the adiabatic compression. There is no under cooling, the evaporation is done by throttle valve. Determine i) COP, ii) power of compressor to remove kj/hr. Given T K 202

203 T K Quantity of heat removed= kj/hr Solution The following details can be taken from the refrigeration tables Properties of refrigerant Temperature Sensible Heat Latent Heat( Entropy of Liquid Entropy of ( kj/kg kj/kg ( kj/kg K Vapour ( kj/kg K h h xh Vapour is dry saturated so x 1 h h xh kj kg s s xs Vapour is dry saturated so x 1 kj s s kg K s s s xs x x h h xh kj/kg h h kj/kg COP h h h h Quantity of heat removed kj kw hr 3600 Refrigerating Effect N m h h m 3.37 kg sec Power of the compressor m h h kw 203

204 APTITUTDE QUESTIONS 1. In a vapour compression refrigeration system, the refrigerant is in liquid state or very wet vapour between (a) compressor and condenser (b) condenser and expansion valve (c) expansion valve and evaporator (d) evaporator and condenser 2. Which component of the refrigeration system controls the flow of refrigerant? (a) expansion valve (b) condenser (c) compressor (d) evaporator 3. The highest temperature during the cycle, in a vapour compression refrigeration system occurs after (a) compression (b) condensation (c) expansion (d) evaporation 4. In a vapour compression refrigeration system, the lowest temperature during the cycle occurs after (a) compression (b) condensation (c) expansion (d) evaporation 5. In a domestic vapour compression refrigerator, the refrigerant commonly used is (a) CO 2 (b) Ammonia (c) Freon-12 (d) all of these 6. In a vapour compression refrigeration system, sub cooling the liquid refrigerant is to COP (a) increase (b) decrease 7. The leakage of refrigerant from a system is detected by (a) halide torch test (b) soap and water test (c) sulphur candle test (d) any of the above 8. An ordinary passenger aircraft requires a cooling system of capacity (a) 2 TR (b) 4 TR (c) 8 TR 9. The simple air cooling system is good for flight speeds. (a) low (b) high 10. A boot strap air cooling system has (a) one heat exchanger (b) two heat exchangers (c) three heat exchangers (d) four heat exchangers 11. The air cooling system mostly used in transport type aircraft is (a) simple air cooling system (b) simple evaporative air cooling system (c) boot strap air cooling system (d) all of the above DESCRIPTIVE QUESTIONS State the function of each of the components of a vapour compression refrigerating system with the help of a neat sketch. 204

205 1. Mention the advantages of vapour compression refrigeration system over air refrigeration system. 2. Describe with the help of a diagram the working of a Vapour absorption refrigeration system 3. What are the merits of vapour absorption refrigeration system over vapour compression refrigeration system? 4. State the important properties of good refrigerant. What are the normal refrigerants used? EXERCISE PROBLEMS 1. A Bell-Coleman cycle works between 1 bar and 5 bar. The adiabatic efficiency of compression is 85% and the expansion is 90%. Find the COP of the system and the ton of refrigeration when the air flow rate is 1 kg/sec. the ambient temperature is 27 and the refrigerator temperature is A Bell-Coleman cycle works between 1 bar and 6 bar. Compression follows PV 1.25 =constant and expansion follows PV 1.3 = constant. Find COP and capacity of unit in tons of refrigeration if the air flow is 0.5 kg/sec. assume compression and expansion begin at 7and 37 respectively. Neglect clearance. 3. In an aircraft cooling system, the air at 0.1 MPa and 4 is compressed to 0.3 MPa with an isentropic efficiency of compressor of 72% after being cooled to 55 at constant pressure in heat exchanger in a turbine to 0.1 MPa with an isentropic efficiency of 78%. The low temperature air absorbs a cooling load of 3 tons at a constant pressure before entering the compressor which is driven by the turbine. Determine a) COP of the refrigerator, ii) the driving power required, iii) mass flow rate. 4. A 5 ton refrigeration plant uses R12 as refrigerant. It enters the compressor at 5 as saturated vapour. Condensation takes place at 32 and there is no under-cooling of refrigerant liquid. Assuming isentropic compression, find i) COP of the plant, ii) mass flow of refrigerant, iii) power required to run the compressor in kw. Take Cp(saturated vapour) = kj/ kg K. 5. A vapour compression refrigeration system using R12 has a condensing temperature of 50 and evaporating temperature of 0. The refrigeration capacity is 7 tons. The 205

206 liquid leaving the condenser is saturated liquid and compression is isentropic. The vapour leaving the evaporator is dry saturated. Assume that the enthalpy at the end of compression is 210 kj/kg. determine, i) refrigeration flow rate, ii) power required to run the compressor, iii) heat rejected in the plant, iv) COP of the system. 6. A refrigeration system has working temperature of -30ºC and 40ºC. What is the maximum C.O.P. possible? If the actual C.O.P. is 75% of the maximum. Calculate the actual refrigerating effect produced per kw hour. 7. A Bell - Coleman refrigerator works between 4 bar and 1 bar pressure limits. After compression, the cooling water reduces the air temperature to 17ºC. What is the lowest temperature produced by the ideal machine? Compare the coefficient of performance of this machine with that of the ideal carnot cycle machine working between the same pressure limits, the temperature at the beginning of compression being -13ºC. 8. A dense air refrigerating system operating between pressures of 17.5 bar and 3.5 bar is to produce 10 tonne of refrigeration. Air leaves the refrigerating coils at -7ºC and it leaves the air cooler at 15.5ºC. Neglecting the losses and clearance, calculate the network done per minute and the co efficient of performance. For air, C P = kj/kg K and = An air refrigeration system is used for food storage provides 25 TR. The temperature of air entering the compressor is 7ºC and the temperature at exit of cooler is 27ºC. Find (a) C.O.P. of the cycle, and (b) power per tonne of refrigeration required by the compressor. The quantity of air calculated in the system is 3000 kg/hr. The compression and expansion both follows the law pv 1.3 = constant and take = 1.4 and C P = 1 for air. 10. The following data refers to a simple aircraft refrigeration system: Ram air temperature and pressure: 30ºC and 1 atm. Cabin air temperature and pressure: 27ºC and 1 atm. Pressure at the exit of main compressor: 4.5 bar. Cooling load: 21 kw 206

207 Determine (a) tonnage, (b) mass of the bled air from the main compressor for refrigeration, (c) Heat rejection, (d) power, (e) COP and (f) power supplied to the blower. 11. A refrigerator works between -7 C and 27 C. The vapour is dry at the end of compression. There is no under cooling and evaporation is by throttle valve. Determine: (i) The coefficient of performance. (ii) Power of the compressor to remove 180 kj/min The properties of the refrigerant are as under: Enthalpy (kj/kg) Entropy (kj/kgk) Temp. ( C) Liquid (h f ) Latent (h fg ) Liquid (S f ) Vapour (S g ) A F 12 vapour compression refrigerator system has a condensing temperature of 50 C and evaporating temperature of 0 C. The refrigeration capacity is 7 tonnes. The liquid leaving the condenser is saturated liquid and compression is isentropic. Determine: a. The refrigerant flow rate. b. The power required to run the compressor c. The heat rejected in the plant d. COP of the system Temp ( C) Pressure (bar) h f (kj/kg) h g (kj/kg) S f (kj/kgk) S g (kj/kgk) Take the enthalpy at the end of isentropic compression = 210 kj/kg. 13. An ammonia refrigerator produces 20 tonnes of ice per day from and at 0 C. The condensation and evaporation takes at 20 C and -20 C respectively. The temperature 207

208 of vapour at the end of isentropic compression is 50 C and there is no under cooling of the liquid. The actual COP is 70% of the theoretical COP. Determine: (i) The rate of NH 3 circulation. (ii) The size of single acting compressor when running at 240 rpm assuming L = D and volumetric efficiency of 80%. Take h fg (fusion of ice) = 335 kj/kg Use the properties of NH 3 as listed below. Enthalpy (kj/kg) Entropy (kj/kgk) Temp. ( C) h f h fg S f S g Take (V sup ) Specific volume of dry vapour at -20 C = m 3 /kg. Specific heat of superheated vapour = 2.8 kj/kgk. 14. A commercial refrigerator operates with R-12 between bar and bar. The vapour is dry and saturated at the compressor inlet. Assuming isentropic compression, determine the theoretical COP of the plant. The isentropic discharge temperature is 68ºC. If the actual COP is 80% of theoretical, calculate the power required to run the compressor to obtain a refrigerating capacity of 1TR. If the liquid refrigerant from the condenser is sub cooled through 10ºC. Calculate the power/tr. Assume specific heat of liquid is 1.05 kj/kg. 15. A refrigeration system of 10.5 tonnes capacity at an evaporator temperature of -12ºC and condenser temperature of 27ºC is needed in a food storage locker. The refrigerant ammonia is sub cooled by 6ºC before entering the expansion valve. The vapour is 0.95 dry as it leaves the evaporator coil. The compression is of adiabatic type. Using ph chart find: (i) Condition of vapour at outlet of the compressor. (ii) Condition of vapour at entrance to evaporator. (iii) C.O.P. and (iv) Power required in kw. Neglect valve throttling and clearance effect. 16. An R-134a refrigeration system operates between the pressure limits of 1.6 bar and 10 bar respectively. The heat transfer from the condenser is found to be 72 kj/min. The refrigerantvapour leaves the evaporator in the saturated state. The condensate leaves 208

209 the condenser in just saturated state. The refrigerant flow rate through the system is found to be 0.4 kg/min.obtain (a) COP, (b) capacity of the plant and (c) the energy input to the compressor. Pressure (bar) h f (kj/kg) h fg (kj/kg) h g (kj/kg) S f (kj/kgk) S g (kj/kgk)

210 7.1 PSYCHROMETRY CHAPTER VII AIR-CONDITIONING It deals with the state of atmosphere with respect to moisture content. On the other hand psychometrics deals with the thermal properties of air and the control and measurement s of the moisture content in air in addition to study the effects of atmospheric moisture on commodities and the human comforts IMPORTANT DEFINITIONS Dry bulb temperature (DBT) The temperature of an air or gas is measured by using an ordinary error free thermometer is called Dry Bulb Temperature. It is denoted by. Wet bulb temperature (WBT) The temperature of an air or gas is measured by using an ordinary error free thermometer whose temperature sensing bulb is covered by a wet cloth is called Wet Bulb Temperature. The air stream flows across the bulb with a velocity of 4.5 m/s. It is denoted by. Dew Point Temperature (DPT) It is defined as a temperature at which moisture present in the humid air starts condensing. This temperature is saturation temperature of water vapour at its partial pressure. It is denoted by. Humidity Ratio or Specific Humidity It is defined as the ratio of mass of water vapour to the mass of dry air in certain volume of mixture. Relative humidity

211 It is the ratio of the mass of water vapour in air in a given volume at given temperature to the mass of water vapour contained in the same volume at same temperature when the air is saturated. It is denoted by φ. Saturated air A mixture of dry air and water vapour in which the partial pressure of the vapourisequal to saturation pressure of water at same temperature of the mixture, is called saturated air. Degree of saturation It is defined as a ratio of the mass of the water vapour at a given temperature associated with the unit mass of dry air to the mass of the water vapour associated withthe unit weight of saturated air at same temperature. It is denoted by. 7.2 PSYCHROMETRIC CHARTS The psychrometric charts are plotted with dry bulb and specific humidity coordinates. The lines of constant wet bulb temperature, relative humidity, specific volume, etc are plotted. In some charts other details are also available to help boost the various calculations. The accurate empherical relations are employed to evaluate various quantities to a greater accuracy. 211

212 Unless the chart is made on the large sheets, it is not possible to show the constant enthalpy and wet bulb temperature lines separately. Hence, on those charts which represent both the constant wet bulb temperature and enthalpy by a single line give the slight error. Therefore the enthalpy deviation lines are provided. Psychrometric charts are also available which can be used to solve problems involving atmospheric pressures other than standard value. For this saturation lines for other atmospheric pressures are plotted on the standard chart. The following figure shows the lines of constant T wb for different pressures. The chart also gives the relation among T db, w, relative humidity, dew point temperature, enthalpy and atmospheric pressure T dp 7.3 Psychrometric Processes In the domestic and industrial air conditioning applications some psychrometric processes have to be performed on the air to change the psychrometric properties of air so as to obtain certain values of temperature and humidity of air within the enclosed space. Some of the common psychrometric processes carried out on air are: sensible heating and cooling of air, humidification and dehumidification of air, mixing of various streams of air, or there may be 5 combinations of the various processes. Illustrating and analyzing the psychrometric properties and psychrometric processes by using the psychrometric chart is very easy, convenient and time saving Sensible Cooling Cooling of the air is one of the most common psychrometric processes in the air conditioning systems. The basic function of the air-conditioners is to cool the air absorbed from the room or the atmosphere, which is at higher temperatures. The sensible cooling of air is the process in which only the sensible heat of the air is removed so as to reduce its temperature, and there is no change in the moisture content (kg/kg of dry air) of the air. During sensible cooling process the dry bulb (DBT) temperature and wet bulb (WBT) temperature of the air reduces, while the latent heat of the air and the dew point (DPT) temperature of the air remains constant. There is overall reduction in the enthalpy of the air. 212

213 In the ordinary window or the split air conditioner the cooling of air is carried out by passing it over the evaporator coil, also called as the cooling coil. The room air or the atmospheric air passes over this coil carrying the refrigerant at extremely low temperatures, and gets cooled and passes to the space which is to be maintained at the comfort conditions. In general the sensible cooling process is carried out by passing the air over the coil. In the unitary air conditioners these coils are cooled by the refrigerant passing through them and are called also called evaporator coils. In central air conditioners these coils are cooled by the chilled water, which is chilled by its passage through the evaporator of the large air conditioning system. In certain cases the coil is also cooled by the some gas passing inside it. The sensible cooling process is represented by a straight horizontal line on the psychrometric chart. The line starts from the initial DBT of the air and ends at the final DBT of the air extending towards the left side from high temperature to the low temperature (see the figure below). The sensible cooling line is also the constant DPT line since the moisture content of the air remains constant. The initial and final points on the psychrometric chart give all the properties of the air. 213

214 7.3.2 Sensible Heating Sensible heating process is opposite to sensible cooling process. In sensible heating process the temperature of air is increased without changing its moisture content. During this process the sensible heat, DB and WB temperature of the air increases while latent of air, and the DP point temperature of the air remains constant. Sensible heating of the air is important when the air conditioner is used as the heat pump to heat the air. In the heat pump the air is heated by passing it over the condenser coil or the heating coil that carry the high temperature refrigerant. In some cases the heating of air is also done to suit different industrial and comfort air-conditioning applications where large air conditioning systems are used. In general the sensible heating process is carried out by passing the air over the heating coil. This coil may be heated by passing the refrigerant, the hot water, the steam or by electric resistance heating coil. The hot water and steam are used for the industrial applications. Like the sensible cooling, the sensible heating process is also represented by a straight horizontal line on the psychrometric chart. The line starts from the initial DBT of air and ends at the final temperature extending towards the right (see the figure). The sensible heating line is also the constant DPT line. 214

215 7.3.4 Humidification Process The process in which the moisture or water vapor or humidity is added to the air without changing its dry bulb (DBT) temperature is called as humidification process. This process is represented by a straight vertical line on the psychrometric chart starting from the initial value of relative humidity, extending upwards and ending at the final value of the relative humidity. In actual practice the pure humidification process is not possible, since the humidification is always accompanied by cooling or heating of the air. Humidification process along with cooling or heating is used in number of air conditioning applications. (i) Cooling and Humidification Process Cooling and humidification process is one of the most commonly used air conditioning application for the cooling purposes. In this process the moisture is added to the air by passing it over the stream or spray of water which is at temperature lower than the dry bulb temperature of the air. When the ordinary air passes over the stream of water, the particles of water present within the stream tend to get evaporated by giving up the heat to the stream. The evaporated water is absorbed by the air so its moisture content, thus the humidity increases. At the same time, since the temperature of the absorbed moisture is less than the DBT of the air, there is reduction in the overall temperature of the air. Since the heat is released in the stream or spray of water, its temperature increases. One of the most popular applications of cooling and humidification is the evaporative cooler, also called as the desert cooler. The evaporative cooler is the sort of big box inside which is a small water tank, small water pump and the fan. The water from the tank is circulated by the pump and is also sprayed inside the box. The fan blows strong currents of air over the water sprays, thus cooling the air and humidifying it simultaneously. The evaporative cooler is highly effective cooling devise having very low initial and running cost compared to the unitary air conditioners. For cooling purposes, the cooling and humidification process can be used only in dry and hot climates like desert areas, countries like India, China, Africa etc. This cooling process cannot be used in hot and high humidity climates. The cooling and humidification process is also used in various industries like textile, where certain level of temperature and moisture content has to be maintained. In such cases large 215

216 quantity of water is sprayed, and large blowers are used to blow the air over the spray of water. During the cooling and humidification process the DBT of the air reduces, its WBT and the dew point temperature (DPT) increases, while its moisture content and thus the relative humidity also increases. Also, the sensible heat of the air reduces, while the latent heat of the air increases resulting in the overall increase in the enthalpy of the air. Cooling and humidification process is represented by an angular line on the psychrometric chart starting from the given value of the dry bulb temperature and the relative humidity and extending upwards toward left. (ii) Heating and Humidification Process In heating and humidification psychrometric process of the air, the dry bulb temperature as well as the humidity of the air increases. The heating and humidification process is carried out by passing the air over spray of water, which is maintained at temperature higher than the dry bulb temperature of air or by mixing air and the steam. When the ordinary air is passed over the spray of water maintained at temperature higher than the dry bulb temperature of the air, the moisture particles from the spray tend to get evaporated and get absorbed in the air due to which the moisture content of the air increase. 216

217 At the same time, since the temperature of the moisture is greater than the dry bulb temperature of the air, there is overall increase in its temperature. During heating and humidification process the dry bulb, wet bulb, and dew point temperature of the air increases along with its relative humidity. The heating and humidification process is represented on the psychrometric chart by an angular line that starts from the given value of the dry bulb temperature and extends upwards towards right Dehumidification The process in which the moisture or water vapor or the humidity is removed from the air keeping its dry bulb temperature (DBT) constant is called as the dehumidification process. This process is represented by a straight vertical line on the psychrometric chart starting from the initial value of relative humidity, extending downwards and ending at the final value of the relative humidity. Like the pure humidification process, in actual practice the pure dehumidification process is not possible, since the dehumidification is always accompanied by cooling or heating of the air. Dehumidification process along with cooling or heating is used in number of air conditioning applications. Let us see how these processes are obtained and how they are represented on the psychrometric chart. 217

218 This article describes psychrometric processes like dehumidification, cooling and dehumidification, and heating and dehumidification. The article describes how these processes are achieved and how they are represented on the psychrometric chart. (i) Cooling and Dehumidification Process The process in which the air is cooled sensibly and at the same time the moisture is removed from it is called as cooling and dehumidification process. Cooling and dehumidification process is obtained when the air at the given dry bulb and dew point (DPT) temperature is cooled below the dew point temperature. Let us understand the cooling and dehumidification process in more details. When the air comes in contact with the cooling coil that is maintained at the temperature below its dew point temperature, its DBT starts reducing. The process of cooling continues and at some point it reaches the value of dew point temperature of the air. At this point the water vapor within the air starts getting converted into the dew particles due to which the dew is formed on the surface of the cooling and the moisture content of the air reduces thereby reducing its humidity level. Thus when the air is cooled below its dew point temperature, there is cooling as well as dehumidification of air. The cooling and dehumidification process is most widely used air conditioning application. It is used in all types of window, split, packaged and central air conditioning systems for producing the comfort conditions inside the space to be cooled. In the window and split air conditioners the evaporator coil or cooling coil is maintained at temperature lower than the dew point temperature of the room air or the atmospheric air by the cool refrigerant passing through it. When the room air passes over this coil its Dry Bulb Temperature reduces and at the same time moisture is also removed since the air is cooled below its Dew Point Temperature. The dew formed on the cooling coil is removed out by small tubing. In the central air conditioning systems the cooling coil is cooled by the refrigerant or the chilled water. When the room air passes over this coil, it gets cooled and dehumidified. In the general the cooling and dehumidification process is obtained by passing the air over coil through which the cool refrigerant, chilled water or cooled gas is passed. During the cooling and dehumidification process the dry bulb, wet bulb and the dew point temperature of air reduces. Similarly, the sensible heat and the latent heat of the air also reduce leading to overall reduction in the enthalpy of the air. The cooling and 218

219 dehumidification process is represented by a straight angular line on the psychrometric chart. The line starts from the given value of the DBT and extends downwards towards left. (ii) Heating and Dehumidification Process The process in which the air is heated and at the same time moisture is removed from it is called as heating and dehumidification process. This process is obtained by passing the air over certain chemicals like alumina and molecular sieves. These elements have inherent properties due to which they keep on releasing the heat and also have the tendency to absorb the moisture. These are called as the hygroscopic chemicals. In actual practice the hygroscopic elements are enclosed in the large vessel and the high pressure air is passed inside the vessel through one opening. When the air comes in contact with the chemicals the moisture from the air is absorbed and since the chemicals emit heat, the DBT of the air increases. The hot and dehumidified air comes out from the vessel through other opening in the vessel. The inlet and outlet openings of the vessel are controlled by the valve. The heating and humidification process is commonly used for reducing the dew point temperature of air. There are number of automatic valves in the chemical plants that are operated by the compressed air at high pressure. If the dew point temperature of this air is high, there are chances of formation of dew inside the valves which can lead to their corrosion and also faulty their operation. Thus it is very important that the air passing to such automatic valves have very low dew point temperature. The heating and dehumidification process by using hygroscopic materials is used often in the air drying units. 219

220 During the heating and dehumidification process dry bulb temperature of the air increases while its dew point and wet bulb temperature reduces. On the psychrometric chart, this process is represented by a straight angular line starting from the given DBT conditions and extending downwards towards right to the final DBT conditions Adiabatic Saturation The thermodynamic wet bulb temperature or adiabatic saturation temperature is the temperature on which the air might be brought to saturation state, adiabatically, from the evaporation of water in the flowing air. The equipment utilized for the adiabatic saturation of air, in its easiest form, contain an insulated chamber having sufficient quantity of water. There is also an arrangement for additional water (known as make-up water) to flow into the chamber from its top, as illustrated in figure 220

221 Adiabatic Saturation of Air Consider the unsaturated air enters the chamber at section 1. Since the air flow through the chamber on a long sheet of water, the water evaporates that is carried with the passing stream of air, and the specific humidity of the air enhanced. The form water is added to the chamber at this temperature to do the water level constant. The air and water both are cooled up as the evaporation occur. This procedure continues till the energy transferred through the air to the water is equivalent to the energy needed to vaporise the water. While steady conditions are attained, the air passing on section two is saturated along water vapour. The temperature of the saturated air at section 2 is called as thermodynamic wet bulb temperature or adiabatic saturation temperature MIXING OF AIR STREAMS It is another important process where more than one stream of air are mixed such as the mixing of fresh and return air in the air conditioning, cold storage, etc. The following figure indicates the two streams air being mixed together. The states points are represented on the psychrometric chart. Using fundamental equations, it is written as: 221