# 2.61 Internal Combustion Engines Design Project Solution. Table 1 below summarizes the main parameters of the base engine. Table 1 Base Engine Summary

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1 .6 Internal Combustion Engines Design roject Solution Here is a possible solution for the design problem.. Base Engine Table below summarizes the main parameters of the base engine Table Base Engine Summary Displacement (m3 Cylinders 6 Bore (m 0.36 Stroke (m 0.36 Compression Ratio 8 Connecting Rod Length (m max torque (ka 895 max power (ka 835 Maximum torque (N-m 783 Maximum power (kw 73 Maximum engine speed at maximum power(rm 6 There are two possible methods to size the engine, and they should be consistent with each other: Method : ηmη f, iηv ( N VdQHV ρ a, oφ( F / A stoich ( Method : Assume a bmep based on practical limits and fuel-air cycle charts, and solve for the power output: ( bmep( Vd ( N ( 000

2 Using the first method, we must determine η m,, η f, i, ηv, ρ a o, φ, N Calculate Engine Speed (N Sp max N find L (3 L Assume B ~ L, so V d ( ( ( 3 πb L 6πL 6 cylinders ( V 4(0.0 3 d L 0. 36m 6π 6π so: 4 N 0m/s ( revs / sec, or 60 RM Determine φ, and η f,i Chose r c 8 (maybe a bit high, and φ0.7 (smoke limit, maximum possible fuel we can get in per mass of air. Using Fuel-air cycle results (Fig. 5-9, Heywood p. 8, then η f,i Applying a correction factor of around 80%, actual η f,i 46%. The correction factor can be between 80% and 85%; For this case, I chose 80% so that Method, and, as explained above, are consistent with each other. Determine IME For phi0.7, and r c 0.8, we get imep i ( ( so imep, (5 Note that i is not atmospheric pressure. At WOT, there is a pressure loss in the intake system, due to frictional losses that scale with speed. i will be less than atmospheric. Likewise, the exhaust pressure (e is not atmospheric; a higher than atmospheric pressure is needed to pump the gases through the exhaust system. Once the gases leave the exhaust system and reach ambient conditions, they will expand to atmospheric pressure. Additionally, depending on the opening timing of the exhaust valves, the gases might exit at a higher pressure than what is required to overcome the pumping loss in the

3 exhaust system. To get an idea, of the value of i, look at Figure 3-3 in the text (Heywood. 75. For a piston speed of 0 m/s, pmep (e - i 0.4x(Sp 0.4(0 40 (6 Now allocate this pumping loss between e and i. At high speeds around 8% of the loss is on the intake side, and the remaining 8% on the exhaust side. This will be consistent with volumetric efficiency as explained below. So: i 0 ka 0.8(40 ka 93.8 ka e0 ka + 0.8(40 ka 33.8 ka We can now calculate an imep: ( 93.8 ka( ka imep Determine Mechanical Efficiency η m imep tfmep tfmep η m ; (7 imep imep where tfmpe total friction mep fmep (rubbing friction and auxiliary mep + pmep From figure 3-7 (Heywood p 7, fmep for a fired engine at 60 rpm 40 ka. So tfmpe ( ηm 8.7% 985 ka 80ka; and Determine Volumetric Efficiency and ρ a, o Using figure 6-8 (Heywood p. 7, assume a volumetric efficiency of 90% for a piston speed of 0 m/s. Note that this volumetric efficiency measures the efficiency of the entire intake system. Also note that we have chosen the right pressure loss allocation for the intake system (as calculated in the imep section, consistent with volumetric efficiency. The air density ρ a, o, is just calculated from ideal gas law, at ambient conditions. The value is.7 kg/m 3

4 Fuel-to-Air Ratio & Heating Value From table D.4 in the text (Heywood p. 95 we get the stoichiometric Fuel-to-Air ratio of gasoline as , and its heating value of 43. MJ/kg ower calculation With the estimates for each value, we can now calculate the power (0.87(0.46(0.90(37.7m / sec(0.0m 73kW 3 ( 43.e3kJ / kg(.7kg / m (0.7( We also use method to check for consistency. Rearranging equation.9b (Heywood p50, we get: ( bmep( Vd ( N ( 805 ka( dm3( 37.7rev / sec 67kW the methods are close For low loads, follow the same procedure, with lower pumping loss, due to lower speed (see figure 3-3, Heywood, and lower rubbing and auxiliary friction (see figure 3-7 Heywood; additionally, the allocation of pressure losses is different, and must be consistent with volumetric efficiency.. Boost, Turbo-machinery and Intercooler Boost pressure: To find the boost pressure required, we use equation, and replace the volumetric efficiency for the entire inlet system with the volumetric efficiency for the valves only ( η v ~ 94%. We also replace the ambient air density with the air density right before the valves, ρ. This density can be determined from the ideal gas law, knowing the pressure a, i (which is approximately cylinder pressure divided by volumetric efficiency, and the temperature (about the same as the cylinder temperature. Thus, we can vary the cylinder pressure until we get the required power level, as defined by equation. ηmη η ( N V Q ρ φ( F / A f, i v d HV a, i stoich Note that as we vary the cylinder pressure, and consequently the density, the mechanical efficiency (as defined by equation 7 above will also change because the pumping loss will change.

5 mep exhaust - intake Thus the solution to this problem is iterative, and can easily be done with a spreadsheet. After varying the cylinder pressure, determining the corresponding air density at the valves through the ideal gas law, and calculating the mechanical efficiency, we get the following target power: (0.98(0.46(0.944(37.7m / sec(0.0m 360kW 3 ( 43.e3kJ / kg(.065kg / m (0.7( For this case the pressure that gives a density of.065 is 76 ka, as dictated by the ideal gas law: cylinder kJ / kmolek (.065kg / m ( 34 (0.944 ρ a, i( beforevalves RT( beforevalves * ηv _ valves K 8.97kg / kmole which gives cylinder76 ka. The pressure that must come out of the compressor is approximately: η 76ka cylinder comp 86 v ka Thus the desired boost is 85 ka. That is we have to compress 85 ka above atmospheric. Note that to relate pressure before the valves, and after the valves, as a first approximation I have used the volumetric efficiency. Turbo-machinery Knowing the desired boost, the turbo-machinery can now be sized to generate the required pressure. This is done using the insentropic relationships for the compressor and turbine. First we must size the compressor by finding the work required to compress the gas to the desired pressure. Second, we must size the turbine to produce the work that drives the compressor. To determine the amount of work that is required to compress the gas we do an energy balance assuming an adiabatic compressor: W c mc & ( T T (8 p a where, T a Actual compressor exit temperature T Compressor inlet temperature (300K

6 We calculate Ta using the compressor efficiency, and isentropic relationships: T T T s a + T ηc and: γ γ (9 T s T (0 T,,, γ, and η c are all known, so Ta, and consequently the compressor work can be calculated. Knowing the compressor work, we now size the turbine using the following equations: Wc Wt ( η m where ηm is the mechanical efficiency for the turbine and compressor system. 95% is reasonable estimate for this number Wt T 5 a T 4 ( Cp where: T5aActual turbine exhaust temperature T4 Turbine inlet temperature (engine exhaust, given at 900K To find the required turbine pressure ratio: 4 5 γ γ T5s r (3 T4 where: T T T 5a 4 5s + T4 ηt (4 Thus, enough equations for enough unknowns. Values for the temperatures, pressures, and compressor work, are show in table. Intercooler Adding an intercooler to lower the intake temperature, will increase the density of the gas, and consequently decrease the required boost, as reflected in table. For a given pressure rise we get a higher change in density (due to lower gas temperatures going into the engine. To size the intercooler you can select a coolant, and based on adequate

7 estimates for inlet and outlet coolant temperatures, you can determine the required mass flow-rate that is needed to achieve a certain temperature change in the air. You must use the definition for heat exchanger to determine the allowed change in air temperature: For the coolant I used water (Cp4. kj/kg K, and assumed that it goes in at 300 K, and I want it to leave at 380 K. To find the exit temperature of the air, and to determine the required flow rate of water, I use the definition for heat exchanger effectiveness in conjunction with an energy balance: For effectiveness we have: ε mcp & coolant, or air mcp & min ( T ( T in in T T out out _ max coolant or air and for the energy balance we have: ( m& Cp T coolant ( mcp & T air For this case, I chose the air and water to have about the same capacitance (mcp. Using the effectiveness equation I can solve for Tout air. Note that the capacitances will cancel out in the equation, and the maximum change in temperature occurs when Tout air Twater in, thus: ( Tin Tout airr ε Tair _ out Tair _ in ε ( T ( T T in _ air water _ in air _ in T water _ in Assuming, water temperature increases from 300 to 357, then Tair in is 34 K. Values for the heat exchanger temperatures and flow-rates are also shown in table. Figure Schematic of Turbocharged Engine with Intercooler 3 Heat Exchanger Engine compressor turbine 5

8 Table Turbocharged Engine with Intercooler: Operating arameters No intercooling With intercooling State Temperature (K ressure (ka Temperature (K ressure (ka N/A N/A N/A N/A Work Compressor 40 kw Work Compressor 9 kw Intercooler m_dot water m_dot air kg/sec kg/sec 3. Brake Efficiency At half maximum speed, and full load at that speed, I kept the same boost, but lowered the fmep, per figure 3-7 in the text. The BSFC came out to be 88 g/kw-hr. This number is actually quite good for industry standards. Other people perhaps got lower (around 75, however, as I previously explained, I was more conservative in my efficiency estimate from fuel air cycle tables, to be consistent with different ways of calculating power. My calculation is shown below: bsfc m& f ( g / hr ower( kw kg / sec(3600sec/ hr(000g / kg 93 88g / kw hr Note that if we directly use break engine efficiency, we should get the same answer: bsfc 88g / kw hr η ( Q η ( Q 0.96(0.460(43. f, b HV mη f, i HV Ways to decrease bsfc include raising compression ratio, and reducing frictional losses. 4. Emissions NOx

9 Figure Schematic of Turbocharged Engine with Intercooler 7 4 Venturi/ Mixer Engine Intercooler 3 Intercooler 5 compressor turbine 6 The schematic shows how EGR will be driven from the engine. There are a few ways of doing this; one way is to use a Venturi system, as shown above. Another way is to optimize the system so that the pressures at the air and EGR intersection are about the same. It is necessary for these pressures to be equal, otherwise there will be backflow in the direction of lower pressure. Overall, however, the addition of EGR will impact the fuel economy of the engine. This is the price that we must pay to have lower emissions. The first step of this problem is to define the amount of EGR that is needed to meet EA emissions levels. The emissions requirements along with their safety levels are shown in table 3 below. Engine Mode 5% max torque, 600 rpm Maximum power NOx Standard g/bhp-hr NOx safety target g/bkwhr Table 3 NOx Safegy target g/bkwhr EGR Timing (CA from TDC Hit in Fuel economy (percentage points % % % 0.5 %

10 Using the figures provided, there are various possibilities for selecting EGR, depending on the hit on fuel economy. Figure 3 below is an example that shows that there is a range of timings and EGR levels that will give the proper amount of NOx Figure 3 Acceptable operating area for low load Must operate below dashed line Once EGR has been calculated, the loss in engine efficiency can be assessed, as well as the required boost. Again this is an iterative process. There are many variables affecting engine power, and they are all related as well, thus at least a spreadsheet must be setup. For example, boost affects engine power, but it also affects mechanical efficiency, which in turn affects engine power, thus all these variables must be connected when solving the system. One important implication of adding EGR, is that the pressure in the cylinder chamber must increase if we are to maintain constant mass of fresh fuel and air; this is what we should desire if we are to maintain the same power output from the engine as the case without EGR. The total pressure is equal to the sum of the partial pressures of air and the EGR:

11 + T air EGR Assuming the molecular weights of both Air and EGR are about the same, then the mole fraction is approximately equal to the mass fraction of each mixture, and T can be expressed as: T air EGR Additionally, since there is a pressure loss of around 6 to 0 ka associated with the venturi, a higher boost is still needed. To reach the target power output, a total boost of 94 ka is required, for total pressure of 95 ka. This is a high boost, higher than industry standard for this size engines. erhaps a more practical boost is 50 ka ( T 5 ka, or less. However this limits the maximum power to 30 kw. If you recognized the practical limitations, this is a perfectly acceptable answer. Table 4 With intercooling State Temperature (K ressure (ka Table 5 Intercooler Mass flowrate (kg/sec Tin (K Tout (K EGR Compressor

12 5. Emissions articulates (a The particulate emissions corresponding to the chosen EGR level, can be obtained from the data provided (M levels vs injection timing for various EGR fractions. At 4% EGR, the M coming out of the engine is approximately 0. g/bkw-hr (b To size the trap, use the space velocity that will minimize the volume of the trap, in this case 8,000 hr -. This is evident from the relationship for space velocity: V& V Space _ velocity where V & is the volume flow-rate of the gases going through the trap, and V is the volume of the trap. For a smaller trap volume we get a higher space velocity. Solving for the volume of the trap we get V V& Space _ velocity Using the ideal gas law to solve for V & m& air + m& fuel + m& egr RTexhaust V&.0m ( 3 exhaust Solving for Volume / sec V 3.0m /sec 0.4m 8,000/ 3600sec 3 4 L Values used are shown in table 6 below Table 6 Mdot_air&fuel (kg/sec Mdot_egr (kg/sec Texhaust (K exhaust (ka Space Velocity (/sec Volume flowrate (m3/sec Trap Volume (L As shown in figure 4 of SAE , The maximum pressure loss through the trap, at a space velocity of 8,000/hr, is 6 ka. This is a very small percentage of the total exhaust pressure (~.5%, and the effect on turbo-machinery is small.

13 Emissions requirements Using the same safety factor as in part 6, the table below shows the new emissions that must be met: Table 7 NOx Standard g/bhp-hr NOx safety target g/bhp-hr NOx Safegy target g/bkw-hr Current Engine out g/bkw-hr Required efficiency (catalytic M Standard g/bhp-hr M safety target g/bhp-hr M Safegy target g/bkw-hr Current Engine out g/bkw-hr Required Efficiency (trap Where catalytic converter efficiency is defined as: η cat m& m& pollu tan t, out pollu tan t, in As shown in table 7, a catalytic converter with 9% efficiency will be needed to meet 007 NOx emissions requirements. The required particulate trap efficiency is fairly high (97% but the trap presented in the Ford paper seems to have efficiencies of around 99%, so it should work fine for 007 emissions requirements.

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