Thermal Engines (Motores Térmicos)

Thermal Engines (Motores Térmicos) Tutorials Time schedule Hour/Day Monday Tuesday Wednesday Thursday Friday 11:00 12:00 MT MT MT 1

Tutorial 1 Engine Parts and Components Engine performance maps The reciprocating Internal Combustion Engine (ICE) https://www.youtube.com/watch?feature=player_embedded&v=o9tfifwlmz8 2

Engine parts and components Piston cylindrical shaped mass that reciprocates back and forth in the cylinder, transmitting the pressure forces in the combustion chamber to the rotating crankshaft Connecting rod Cross Head Type Engine Used in large engine The lower end of the piston rod is connected to a cross head, which slides up and down in guides Crankshaft (crank) it is a rotating shaft through which engine work output is supplied to external systems. Fixed with the cylinder block, and receives motion forces from the piston Cross section of piston showing cooling and lubrication, GM Engine parts and components Cylinder circular cylinders in the engine block inside which the piston reciprocate, and beneath which crank is fixed, thus confining the motion linkage. Camshaft rotating shaft used to push open valves at proper time in the engine cycle. The cam profile is made to give such desired movement to the valve. 3

Engine parts and components https://www.youtube.com/watch?v=xvdzj2edmhm Engine parts and components 4

Tutorial 2 Performance characteristics Engine performance maps Engine parameters Summary Piston travel distance x = f(θ) x θ = l + a a aosθ + R 2 sin 2 θ Piston swept volume V = f(θ) V θ = V d r v 1 + V d 2 R + 1 cosθ + R2 sin 2 θ Piston instantaneous speed U p = f(θ) U p θ = ω a sinθ 1 + cosθ R 2 sin 2 θ Piston acceleration a p = f(θ) a p θ = a ω 2 cosθ + a cos 2θ l Surface area A t = f(θ) A t = 2 πb 2 4 π B s 2 R + 1 cosθ + R2 sin 2 θ 1 2 5

Operating characteristics Mean Effective Pressure average (mean) pressure which, if imposed on the pistons uniformly from the top to the bottom of each power stroke, would produce the measured power output W= p mep V d But the power produced by the engine P is equal to the work done per operating cycle times the number of operating cycles per second. If N is the number of revolutions per second, and n c is the number of revolutions per cycle: P = W N / n c W = P n c / N p mep = W /V d p mep = Pn c V d N If T is the torque produced by the engine and w = 2pN is the angular velocity: P = 2πNT W = work per cycle (J) P = power output (W) P mep =mean effective pressure (Pascal) V d = displacement volume (m 3 ) N c = number of revolutions per cycle (for a 4-stroke engine n c = 2 ) N = number of revolutions per second (s -1 ) T = Torque (Nm) p mep = 2π Tn c V d Operating characteristics Mean effective pressure (MEP) is defined by the location measurement and method of calculation: Brake mean effective pressure (BMEP) -Mean effective pressure calculated from brake power Indicated mean effective pressure (IMEP) -Mean effective pressure calculated from in cylinder pressure, average in cylinder pressure over engine cycle, 720. Friction mean effective pressure (FMEP) -Theoretical mean effective pressure required to overcome engine friction, can be thought of as mean effective pressure lost due to friction. BMEP = IMEP FMEP P = p mepv d N n c Brake Power (BP) Indicated Power (IP) Friction Power (FP) -BP = IP FP P brake = p mep brake V d N n c P indicated = p mep indicated V d N n c P friction = p mep friction V d N n c η m = Brake Indicated 6

Operating Characteristics Specific Fuel Consumption Fuel consumption rate sfc = Power output sfc = m f P sfc mg/j = m f g/s P kw Thermal efficiency sfc g/kwh = m f g/h P kw Power output η th = Thermal energy released in combustion Excess Air P η th = m f Q LHV η Bth = P brake m f Q LHV η ind = P indicated m f Q LHV 1 η th = sfc Q LHV 1 η Bth = sfc brake Q LHV 1 η ind = sfc ind Q LHV λ = Air Fuel Air Fuel st for SI engines burning gasoline: 12 < A/F < 18 for CI engines burning diesel fuel: 18 < A/F < 70 Mechanical efficiency, h m η m = P brake P indicated Indicated Power Engine Losses = Indicated - Brake Friction losses Mechanical losses due to friction between all sliding surfaces, e g., con rod bearings; crankshaft bearing; camshaft bearings etc. Parasitic losses Loads required to operate engine auxiliaries, e g., air conditioner; oil pump; water pump; alternator; supercharger, etc Influencing parameters - Stroke to bore ratio - Engine size - Piston rings - Compression ratio - Engine speed - Load η m decreases with increasing N Typical values for modern automobile engines at full open throttle range from η m =90% at low speeds (1800 2400 rpm) to η m =75% at maximum speed. η m decreases as the engine is throttled 7

Indicated Power Power that develops inside the cylinder Theoretical power of a reciprocating engine if it is completely friction-less in converting the expanding gas energy (piston pressure displacement) in the cylinders. It is calculated from the pressure-volume diagram developed in the cylinders (called the indicator diagram), measured by a device called an engine indicator. P brake = η mec P ind P brake = 2πNT brake Brake Power (power available at the crankshaft ) =Indicated power Mechanical Losses Power measurements Engine Power before the loss in power caused by the gearbox and drive train. It is calculated from simultaneous measurements of the rotational speed and the Brake Torque measured with a dynamometer T brake = RF Illustration of an old device to measure torque (Prony brake) Electrical dynamometer setup showing engine, torque measurement arrangement and tachometer from: Marine Engineering Study Materials, www.marineengineeringonline.com The indicated power can be computed from the indicator diagram by measuring the area of the diagram (e g., with a planimeter) On modern engines the indicated diagram can be continuously taken by employing two transducers, one pressure transducer in the combustion space and other transducer on the shaft. Through the computer we can thus get on line indicated diagram and power of all cylinders. The output of the engine can also be measured at the driving wheels. These measurements give an indication of the engine power in real driving conditions, after losses in the drive train and gearbox. Note: In Europe the DIN standard tested the engine fitted with all ancillaries and exhaust system as used in the car. The American SAE system tests without alternator, water pump, and other auxiliary components such as power steering pump, muffled exhaust system, etc. so the figures are higher than the European figures for the same engine. Problem 1 Consider a four-cylinder automotive spark-ignition engine with a maximum brake torque of 150 N.m in the mid-speed range (~3000 rpm) at which the bmep is 925 kpa. Estimate a) the engine displacement; the cylinder bore; the engine speed at maximum piston speed; the maximum brake power Other data: R=Bore/Stroke=1; (MEP) max speed = 800 kpa; S pmax (maximum piston speed) = 15m/s Engine displacement: P brake = p mep brake V d N Substituting P = 2πNT in n c 2πNT = p mep brake V d N n c V d = 2πT n c 2 π 2 150 N m = = 2 10 3 m 3 = 2dm 3 p mep 925 1000 Pa Bore V d = 4 π B2 L 4 since B = L as for a small engine: B = V d π 1/3 = 0.086 m = 86 mm At the maximum piston speed of 15 m/s the engine speed is: Spmax = 2LN max N max = 87 rps = 5220 RPM Maximum brake power P brake = p mep brake V d N n c = 800 103 Pa 2 10 3 m 3 87 rps 2 =7000 W = 70 kw 8

A 4 cylinder, 4 stroke engine gave the following results on a test bed: Stroke L = 100 mm Bore B = 100 mm Shaft speed N = 2500 rpm Torque arm R = 0.4 m Net Brake Load F = 200N Fuel consumption m f = 2 g/s Calorific value Q LHV = 42 MJ/kg Area of indicator diagram A d = 300 mm 2 Pressure scale S p = 80 kpa/mm Base length of diagram Y = 60 mm Calculate: 1. Brake Power 2. Fuel Power 3. Indicated Mean Effective Pressure 4. Indicated Power 5. h THbrake 6. h THindicated 7. h mechanical 1. 2. 3. 4. 5. P brake = 2πN T brake P brake = 2π 2500 200 0.4 = 20.94 kw 60 P fuel = m f Q LHV P fuel = 0.002 kg/s 42000 kj/kg = 84 kw MEP ind = Area of indicator diagram Presure Scale Base length of diagram MEP ind = 300 80 = 400 kpa 60 P ind = MEP ind L A p N 1 N n cylinders c P ind = 400 0.1 π 0.12 2500 1 4 =26.18 kw 4 60 2 η thb = P brake 100% = 20.94 100% = 24.9% P fuel 84 Problem 2 6. 7. η thi = P ind 100% = 26.18 100% = 31.1% P fuel 84 η m = P brake 100% = 20.94 100% = 80% P ind 26.1 Problem 3 A 6 cylinder, 4 stroke spark engine gave the following results on a test: Stroke L = 80 mm Bore B = 90 mm Shaft speed N = 5000 rpm Fuel consumption m f = 0.3 dm 3 /min Fuel density m f = 0.3 kg/m 3 Calorific value Q LHV = 44 MJ/kg Net brake load F brake = 180 N Torque arm R = 0.5 m Area of indicator diagram A d = 720 mm 2 Pressure scale S p = 40 kpa/mm Base length of diagram Y = 60 mm Calculate: 1. Brake Power 2. Indicated Mean Effective Pressure 3. Indicated Power 4. h mechanical 5. h THbrake 1. 2. 3. 4. Brake Power P brake = 2πNT brake P brake = 2π 5000 90 = 47 123.9 W = 47.12 kw 60 T brake = F brake Torn Arm180 N 0.5 m = 90N m Mean effective pressure Net Indicated Area MEP Indicated = pressure scale Base Length of indicator diagram MEP Indicated = Indicated Power 720 mm2 40 kpa/mm = 480 kpa 60 mm P Indicated = MEP indicated V displaced N cilinders 1 n c P Indicated = MEP indicated πb2 4 L N cilinders 1 n c P Indicated = 480kPa π0.092 m 2 0.8m 6 1 4 2 P Indicated = 61.1 kw Mechanical efficiency η mechanical = P brake 100% = 47.12 100% = 77.2% P indicated 61.1 5. Brake thermal efficiency η Bth = P brake 47.12 47.12 100% = 100% = P Fuel ρ fuel Vfuel Q LHV 750 0.3 10 3 44 10 3 100%28.6% 9

Problem 4 A 4 cylinder, two stroke spark engine gave the following results on a test: Bore B = 100 mm Stroke L = 100 mm Shaft speed N = 2000 rpm Fuel consumption m f = 5 g/s Calorific value Q LHV = 46 MJ/kg Net brake load F brake = 500 N Torque arm R = 0.5 m Area of indicator diagram A d = 1500 mm 2 Pressure scale S p = 25 kpa Base length of diagram Y = 66 mm Calculate: 1. h THindicated (26.3 %) 2. H mechanical (87 %) 3. h THbrake (22.8 %) Problem 5 A 3 liter, six cylinders SI engine operates on a four stroke cycle and run at 3600 rpm. The compression ratio is 9.5 the length of connecting rod is 16.6cm, and the bore equal the stroke. Combustion ends at 20 o after TDC calculate: (1) Cylinder bore and stroke, (2) average piston speed, (3) clearance volume of one cylinder, (4) the distance piston has travelled from TDC at the end of combustion, (5) volume of the combustion chamber at the end of combustion. 10

Problem 6 The engine in Problem 5 is connected to a dynamometer which gives a brake output torque of 205 Nm at 3600 rpm. At this speed air enters the cylinder at 85 kpa and 60 o C, and the mechanical efficiency of the engine is 85%. Calculate: (1) Brake power, (2) indicated power, (3) bmep, (4) imep, (5 )fmep, (6) friction power, (7) engine specific volume. Problem 7 The engine in Problem 6 is running with A/F ratio =15,a fuel of heating value; 44000kJ/kg and a combustion efficiency of 97%. The atmosphere is at 1.013 kpa and 15 o C. Calculate: (1) the fuel flow rate (2) h BT, (3) h IT,(4) h V and (5) brake specific consumption. 11

Problem 8 A six-cylinder 4-stroke cycle petrol engine is to be designed to develop 300 kw of (b.p) at 2500 rpm. The bore / stroke ratio is to be 1:1.25. Assuming h m =83% and an indicated mean effective pressure of 9.5 bar, determine the required bore and stroke. If the compression ratio of the engine is to be 6.5 to 1, determine consumption of petrol in kg/h and in kg/bp.hr. Take the ratio of the indicated thermal efficiency of the engine to that of the constant volume air standard cycle as 0.55 and the calorific value of the petrol as 44770kJ/kg. Performance characteristics The performance characteristics of the engine are measured at OEMs for the complete range of operating conditions (engine speed) in special facilities, an engine test stand. the engine test stand The stand houses several sensors, data acquisition features and actuators to control the engine state. The most complete engine tests may include measurements of: crankshaft torque and angular velocity intake air and fuel consumption rates air-fuel ratio for the intake mixture Chemical composition of the exhaust gases (e g., carbon monoxide, different configurations of hydrocarbons and nitrogen oxides, sulfur dioxide) including particulate matter temperatures and gas pressures at several locations on the engine body (eg., engine oil, spark plug, exhaust gas, intake manifold pressure) atmospheric conditions (temperature, pressure and humidity) Engine Test Bed Rental AVL, Austria TÜV SÜD Accredited Laboratory 12

Power (kw) Torque (N-m) Power (kw) 24/10/2015 mep (kpa) sfc (g/kwh) Variable - Speed tests: Full-load tests allow determine maximum power and minimum s.f.c at each different speed Part-load tests allow determine variation in the s.f.c. Performance characteristics engine tests Full load test with SI engine: The throttle is fully opened and the lowest desired speed is maintained by brake load adjustment. The spark is adjusted to give maximum power at this speed. The test is started by the watch governing the fuel consumption, the test ended at the time the fuel- consumption test has been completed. During this interval of time, the average speed, brake load, temperatures, fuel weight etc., are recorded, then load is adjusted for the next run at different speed. Part load test: To run a part load test at variable speed, say 50% load, power reading of half the maximum power at each speed are obtained by varying the throttle and brake setting. 1100 200 imep 1000 Constant -Speed tests: Constant-Speed test is run with variable throttle from no load to full load in suitable steps of load to give smooth curves. Starting at zero load, the throttle is opened to give the desired speed. Then a load is put on the engine and the throttle is opened wider to maintain the same constant speed as before, and the second run is ready to start. The last run of the test is made at wide-open throttle. In a CI-engine test the last run would show smoke in the exhaust gas. For CI engines, the brake torque and mep vary only modestly with engine speed except at high speeds since the intake system of the compression-ignition engine can have larger flow areas than the intake system of the spark-ignition engines. The decrease in torque and bmep with increasing speed is mainly due to the increase in friction mean effective pressure with speed. The part load torque and the brake mean effective pressure characteristics have a shape similar to the full-load characteristics. Brake thermal efficiencies of 25 to 35% are usual with S.I. engines and may reach 50% in diesel engines. 180 160 140 120 100 80 60 bmep bsfc isfc 1000 1500 2000 2500 2750 Speed (rev/min) Adapted from Fundamentals of Internal Combustion Engines, H. N. Gupta, PHI Learning Pvt. Ltd., 2012 900 800 700 240 200 160 Performance characteristics Engine Maps The objective in designing an automobile engine is to flatten the torque-versusspeed curve so to have high torque at both high and low speed. CI engines generally have higher torque than SI engines. Large engines often have very high torque values with MBT (Maximum Break Torque) at relatively low speed. 200 175 150 125 V d = 3.2 liter V d = 2.5 liter 100 Compression Ratio Fuel Air Fuel Air st Manifold Absolute Pressure Maximum brake-mep Part-load efficiency (bar) (bar) (%) 300 200 SI engine 9.5 10.0 1.0 ~ 0.5 12 20-25 100 CI engine ~ 1.0 18 ~ 30 3000 4000 5000 6000 7000 Engine speed (rpm) from Fundamentals of Internal Combustion Engines, H. N. Gupta, PHI Learning Pvt. Ltd., 2012 13

Specific Fuel Consumption (s.f.c.) 24/10/2015 Performance characteristics Consumption loop test Test carried out at constant speed, constant throttle opening, and constant ignition setting. The specific fuel consumption is plotted to a base of "bmep". The A/F ratio is a minimum at A (i.e. richest mixture). As the A/F ratio is increased the "bmep" increases until a maximum is reached at B (usually for an A/F ratio between 10/1 and 13/1). Further increase in A/F ratio produce a decrease in "bmep" with increasing economy until the position of maximum economy is reached at D. Beyond D, for increasing A/F ratios, both "bmep" and consumption values are adversely affected. 18 E Near the point A the engine could be running unsteadily and there may be combustion of the mixture in the exhaust system. At E, with the weakest mixture, running will be unsteady and the combustion may be slow. Point C is the point of chemically correct A/F ratio. A D 10 B C 16 12 14.5 b.m.e.p. For multi-cylinder engines the consumption loops are less distinct, but are generally similar to that for the single cylinder engine. This is also true for tests made at part throttle opening. Adapted from Fundamentals of Internal Combustion Engines, H. N. Gupta, PHI Learning Pvt. Ltd., 2012 Performance characteristics Engine Maps SI Engine the bsfc increases upwards from point A where bsfc is maximum CI Engine In SI engines it is because of mixture enrichement from the action of the economizer and because of the poorer distribution at full throttle. In CI engines it is because of the increased fuel waste (smoke) associated with high fuel/air ratios at high loads. Moving to a lower bmep from point A, the bsfc increases due to reduced mechanical efficiency Moving to the left from point A to a lower piston speeds, the bsfc increases in SI engines because of the increased heat loss per cycle, poor distribution at low manifold velocities and lowered efficiency due to automatically retarded spark used for detonation control at low engine speeds. At very low speeds (not shown in the plot) the CI enmgines may also have increased bsfc because the injectioin equipment cannot be set to give completely satisfactory characetristics over the entire speed range. from Fundamentals of Internal Combustion Engines, H. N. Gupta, PHI Learning Pvt. Ltd., 2012 14

Tutorial 3 Engine Cycles Engine performance maps Problems P1 Estimate the temperature of the gas inside the cylinder at the ending of intake. Consider a normally aspirated engine, and if you assume it is a spark ignition engine consider the throttle is fully open. Assume values that seem realistic to the geometrical characteristics of the engine, temperature and gas pressure at the end of force exhaust, temperature and pressure at the end of intake. Remember there are variations of gas properties with temperature, but make the simplifications it deems appropriate. P2 For a given spark ignition engine of 9.5:1 compression ratio, calculate the temperature at the end of the compression (disregarding the effect of rising pressure that occurs due to the development of combustion). Consider that the temperature at the end of intake is T1 K, that the compression begins immediately at BDC (i.e., consider that the intake valve closes at BDC), that the properties of the mixture are approximately those of the air, and that the compression is adiabatic. Assume what you believe to be a realistic value for T1, and discuss the approximations listed above and those that you may make in solving this problem. P3 For the conditions of the previous problem, calculate the pressure at the end of the compression on the assumption that the pressure at the end of intake is p1 MPa. Assume a realistic value for p1, given that it is a spark ignition engine, normally aspirated, and the throttle is fully open. Discuss how you have solved this problem, and how you would have done it if you had not solved the problem 2). P4 For the conditions of problems 2 and 3, and considering a swept cylinder capacity of 400 cm3, calculate the work needed for the compression of the gas. P5 The same as problem 2) but for a Diesel engine with 22.0:1 compression ratio, and with the temperature T1 K at the end of intake. Assume again a value for T1 and compare it with the one assumed in problem 2). Discuss the approximations made and compare them and their suitability with those of problem 2) P6 The same as problem 3) but for the Diesel, and with the same pressure at the end of intake. P7 The same as problem 4) but for the Diesel (with the same cylinder capacity). 15

Engine boosting Superchargers and turbochargers Compressors mounted in the intake system Used to raise the pressure of the incoming air More air and fuel entering each cylinder during each cycle super-charging turbo-charging Supercharging Superchargers Mechanically driven directly off the engine - power is taken directly from the crankshaft driven by an accessory belt, which wraps around a pulley that is connected to a drive gear. Power needed to drive the supercharger is evaluated as W c = m a h out h in Where W sc = Power needed to drive the supercharger m a = air mass rate to the enguine h out = Enthalpy of the air at the outlet of the compressor h in = Enthalpy of the air at the inlet of the compressor Taking the air as a perfect gas: W c = m a C p T out T in Where T out = Temperature of the air at the outlet of the compressor T in = Temperature of the air at the inlet of the compressor power to drive the compressor is a parasitic load on the engine output, which is one of the disadvantages of superchargers 16

Isentropic efficiency: η is = W isc W real η is = m ac p T out,s T in m a C p T out T in If (T in and P in ) are known, and design output pressure is set: T out = T in p out p in γ 1 γ Supercharging Assuming constant C p : η is T out,s T in T out T in T out,s = T in p out,s p in γ 1 γ W c = m a C p T out T in W c = m ac p T in η is p out,s p in γ 1 γ 1 power needed to drive the supercharger η m = W ac W isc W ac = m ac p T in η is η m Mechanical efficiency of the compressor p out,s p in γ 1 γ 1 Power actually delivered by the engine Supercharging Additional requirements After-cooler cools the air temperature back to normal, thus reducing air density and improving volumetric efficiency. Generally C.I engines do not need after-coolers as there will be no concern about higher cycle temperature - engine water cooling (air-to-liquid heat exchanger) - air cooling (air-to-air heat exchanger) effectiveness of the after cooler: ε f = T c,in T c,out T c,in T coolant Multistage compression supercharger With more compressors to improve air/fuel delivery. 17

Turbocharging Turbocharging Principle of the Turbocharged Engine The turbocharger is bolted to the exhaust manifold of the engine and uses the exhaust flow to spin a turbine, which in turn spins a compressor. The advantage that the engine shaft output is not used to drive the compressor, and only waste energy in the exhaust is used. Main components: - A turbine (usually a radial inflow turbine) - A compressor (usually a centrifugal compressor) - A waste gate valve to divert the exhaust gases away from the turbine wheel. - An intercooler to cool the intake flow upstream the cylinder. 18

Turbocharging Thermodynamics η C = h 2s h 1 h 2 h 1 η T = h 3 h 4 h 3 h 4s η m = m 12C p12 T 2 T 1 m 34 C p34 T 3 T 4 Effect of overall turbocharger efficiency on the pressure ratio between engine inlet and exhaust manifold pressures, for a 2:1 compressor pressure ratio (p 1 /p 2 ) with different engine exhaust temperatures from R.Stone, Introduction to Internal Combustion Engines, 3rd edition, SAE Inc.,1999 Turbocharging Boost threshold engine speed (rpm) equivalent to the required exhaust gas flow for the turbo to produce boost, below this level the turbo simply will not produce boost and very little benefits will be achieved. Determined by: - the engine displacement - engine rpm - throttle opening and - size of the turbo Manual Pneumatic Electric (high-speed electrical motor to speed the turbocharger before exhaust gases are available) Hydraulic (drive system and over-speed clutch arrangement to accelerate the turbocharger) Turbo lag time delay between opening the engine s throttle valve and when the turbo accelerates and delivers positive pressure (boost) to the engine when engine speed is above the boost threshold. Determined by: - Inertia - friction and - compressor load The directly driven compressor in a supercharger does not suffer from this problem. Turbo lag is the most important parameter of a turbocharger when rapid changes in engine performance are required 19

Pressure ratio flow parameter adiabatic efficiency 24/10/2015 Turbine map Pressure ratio Example of a turbine map Compressor map Corrected mass flow Measured points Adiabatic efficiency Corrected shaft speed [rpm x 10-3 ] Example of a compressor map 20

Pressure 24/10/2015 Engine boosting summary Disadvantages and cautions Boost pressure is limited to keep the entire engine system, including the turbo, inside its thermal and mechanical design operating range. Over-boosting an engine frequently can cause: - pre-ignition - Overheating - Over-stressing the internal hardware of the engine To avoid engine knocking (pre-ignition or detonation) and the consequent damages to the engine, the intake manifold pressure must not get too high. Opening the waste-gate allows the energy for the turbine to bypass it and pass directly to the exhaust pipe. The turbocharger is forced to slow as the mass flow rate of exhaust gases to the turbine decreases. https://www.youtube.com/watch?v=vghlgphrbxa Slowing the turbine/compressor rotor produces less compressor pressure. Turbocharged Otto Cycle W Cycle = W 1 2 3 4 + W 5 6 7 1 For a turbocharged engine: it has to be A A-4-5-A > A 5-6-7-1-5 Thermodynamic cycle of the engine: 7 1 admission of air at the supercharging pressure (p 1 >p atm ) 1 2 isentropic compression 2 3 heat addition at constant volume 3 4 isentropic expansion 4-1-5 heat rejection at constant pressure (blowdown) 5 6 driving out exhaust at constant atmospheric pressure Volume Thermodynamic cycle of the supercharger: 6-5-0 admission of air at atmospheric pressure 0 1 isentropic compression to pressure p 1 1 7 delivery of supercharged air at constant pressure p 1 p 1 = supercharging pressure p o = exhaust pressure Area 0-5-6-7-1-0: supercharger work (mechanically driven) to supply air at a constant pressure p 1 Area 1-2-3-4-1: output of the engine Area 1-5-6-7-1: gain in work during the gas exchange process due to supercharging (Part of the work is recovered) Area 1-5-0-1: cannot be recovered and represents a loss of work 21

Efficiency, h 24/10/2015 Turbocharged Otto Cycle Defining b as the pressure ratio of the compressor: b is limited by the occurrence of knock : Where: β = p 1 p o = p 7 p 6 β = p 1 p o p 2,limit p o r v γ p 2,limit = p 1 r v,limit γ β γ 1 η = 1 1 + B γ 1 r v,limit γ 1 γ 1 B βγ 1 β 1 βγ + 1 + r v,limit B = Q LHV 1 with R T o 1 + Air Fuel Throttled turbocharged b = 0.22 Maximum efficiency Compression ratio, b Problem 8 Discuss the use of a turbocharger in the 6 cylinder SI engine of the previous lecture 22

Problem 8 (cont) efficiency of the turbine, h T = 65% efficiency of the compressor, h C = 70% N 3600 3600 3600 3600 3600 A/F 15 15 15 15 15 (r v) limit - 12 12 13 14 15 h Otto - 0.59 0.59 0.59 0.59 0.59 [h Otto] limit - 0.63 0.63 0.64 0.65 0.66 p 2,limit kpa 2755.96 2755.96 3082.76 3419.78 3766.58 T 1 K 333.00 333.00 333.00 333.00 333.00 (p 1) orig kpa 85.00 85.00 85.00 85.00 85.00 (Dp) admi kpa 16.30 p 1 kpa 85.00 117.89 131.86 146.28 161.11 p 2 kpa 1987.16 2755.96 3082.76 3419.78 3766.58 m ar/cylinder kg 4.972E-04 6.895E-04 7.713E-04 8.556E-04 9.424E-04 m fuel/cylinder kg 3.314E-05 4.597E-05 5.142E-05 5.704E-05 6.282E-05 m fuel kg/s 5.97E-03 8.27E-03 9.26E-03 1.03E-02 1.13E-02 m ar kg/s 8.95E-02 1.24E-01 1.39E-01 1.54E-01 1.70E-01 T 2 K 839.15 819.47 819.47 819.47 819.47 T 3 K 3505.82 3486.14 3486.14 3486.14 3486.14 p 3 kpa 8301.95 11724.21 13114.46 14548.19 16023.51 p 4 kpa 355.11 501.50 560.97 622.29 685.40 T 4 K 1424.62 1416.63 1416.63 1416.63 1416.63 T A K 995.53 896.98 868.71 843.34 820.38 W 4-A kw 19.10 32.08 37.84 43.92 50.31 p 7 kpa 249.88 X WG 0.13 0.16 0.18 0.19 (p 7=p 1) b - 2.47 1.16 1.30 1.44 1.59 Pressure at the sart of combustion may cause knock Waste gate valve opens β γ 1 η = 1 1 + B γ 1 r v,limit γ 1 γ B 1 B βγ Problem 8 (cont) 1 β 1 βγ + 1 + r v,limit 2.87E+06 6.17 4.26E+05 0.23 5.48E+06 h 0.52 23

N 4000 4000 4000 4000 4500 4500 4500 4500 3600 4000 4500 5000 A/F 15 15 15 15 15 15 15 15 15 15 15 15 (r v ) limit - 12 13 14 15 12 13 14 15 12 12 12 12 h Otto - 0.63 0.64 0.65 0.66 0.63 0.64 0.65 0.66 0.63 0.63 0.63 0.63 [h Otto ] limit - 0.59 0.59 0.59 0.59 0.59 0.59 0.59 0.59 0.59 0.59 0.59 0.59 p 2,limit kpa 2755.96 3082.76 3419.78 3766.58 2755.96 3082.76 3419.78 3766.58 2755.96 2755.96 2755.96 2755.96 T 1 K 333.00 333.00 333.00 333.00 333.00 333.00 333.00 333.00 333.00 333.00 333.00 333.00 (p 1 ) orig kpa 85.00 85.00 85.00 85.00 85.00 85.00 85.00 85.00 85.00 85.00 85.00 85.00 (Dp) admi kpa p 1 kpa 117.89 131.86 146.28 161.11 117.89 131.86 146.28 161.11 117.89 117.89 117.89 117.89 p 2 kpa 2755.96 3082.76 3419.78 3766.58 2755.96 3082.76 3419.78 3766.58 2755.96 2755.96 2755.96 2755.96 m ar /cylinder kg 6.90E-04 7.71E-04 8.56E-04 9.42E-04 6.90E-04 7.71E-04 8.56E-04 9.42E-04 6.90E-04 6.90E-04 6.90E-04 6.90E-04 m fuel /cylinde kg 4.60E-05 5.14E-05 5.70E-05 6.28E-05 4.60E-05 5.14E-05 5.70E-05 6.28E-05 4.60E-05 4.60E-05 4.60E-05 4.60E-05 m fuel kg/s 9.19E-03 1.03E-02 1.14E-02 1.26E-02 1.03E-02 1.16E-02 1.28E-02 1.41E-02 8.27E-03 9.19E-03 1.03E-02 1.15E-02 m ar kg/s 1.38E-01 1.54E-01 1.71E-01 1.88E-01 1.55E-01 1.74E-01 1.93E-01 2.12E-01 1.24E-01 1.38E-01 1.55E-01 1.72E-01 T 2 K 819.47 819.47 819.47 819.47 819.47 819.47 819.47 819.47 819.47 819.47 819.47 819.47 T 3 K 3486.14 3486.14 3486.14 3486.14 3486.14 3486.14 3486.14 3486.14 3486.14 3486.14 3486.14 3486.14 p 3 kpa 11724.21 13114.46 14548.19 16023.51 11724.21 13114.46 14548.19 16023.51 11724.21 11724.21 11724.21 11724.21 p 4 kpa 501.50 560.97 622.29 685.40 501.50 560.97 622.29 685.40 501.50 501.50 501.50 501.50 T 4 K 1416.63 1416.63 1416.63 1416.63 1416.63 1416.63 1416.63 1416.63 1416.63 1416.63 1416.63 1416.63 T A K 896.98 868.71 843.34 820.38 896.98 868.71 843.34 820.38 896.98 896.98 896.98 896.98 W 4-A kw 35.65 42.05 48.80 55.90 40.11 47.30 54.90 62.89 32.08 35.65 40.11 44.56 X WG kpa 0.13 0.16 0.18 0.19 0.13 0.16 0.18 0.19 0.13 0.13 0.13 0.13 b - 1.16 1.30 1.44 1.59 1.16 1.30 1.44 1.59 1.16 1.16 1.16 1.16 Problem 8 (cont) γ ex W t = m ex η t γ ex 1 R ex T ex 1 p t2 p t1 γ ex 1 γ ex γ c W c = m c η c γ c 1 R c T c p c2 p c1 γ c 1 γ c 1 p t1 = 1 C τ p c2 p c1 p t2 γ ex γ c 1 γ γ ex 1 τ = η t η c T ex T o C = λ A F γ c 1 + λ A γ c 1 γ ex 1 γ ex F From: Control problems in a turbocharged spark-ignition engine, W. Mitianiec and L. Rodak, J. of KONES Powertrain and Transport, Vol. 18, No. 3, 2011 24