Internal Combustion Engines Solution_v1

2.61 Internal Combustion Engines Design Project Solution Here is a possible solution for the design problem. 1. Base Engine Table 1 below summarizes the main parameters of the base engine Table 1 Base Engine Summary 11

Displacement (m3)

6

Cylinders Bore (m)

0.1326

Stroke (m)

0.1326 18

Compression Ratio Connecting Rod Length (m)

0.3315

Bmep @ max torque (kPa)

895

Bmep @ max power (kPa)

835

Maximum torque (N-m)

783

Maximum power (kW)

173

Maximum engine speed at maximum power(RPM)

2261

There are two possible methods to size the engine, and they should be consistent with each other: Method 1: P=

η mη f ,iη v (N )Vd QHV ρ a ,oφ (F / A)stoich 2

(1)

Method 2: Assume a bmep based on practical limits and fuel-air cycle charts, and solve for the power output: P=

(bmep )(Vd )(N ) 2000

(2)

Using the first method, we must determine η m , η f ,i , η v , ρ a ,o , φ , N

Calculate Engine Speed (N) N=

S p max find L 2L

(3)

Assume B ~ L, so Vd = (6 cylinders ) 1

(πB )L = (6πL ) 2

4

3

4

(4)

1

 4V  3  4(0.011)  3 L= d  =  = 0.1326m  6π   6π 

so: N=

10m/s = 37.71revs / sec, or 2260 RPM 2(0.1326)

Determine φ, and ηf,i Chose rc=18 (maybe a bit high), and φ=0.7 (smoke limit, maximum possible fuel we can get in per mass of air). Using Fuel-air cycle results (Fig. 5-9, Heywood p. 182), then ηf,i= 0.575. Applying a correction factor of around 80%, actual ηf,i= 46%. The correction factor can be between 80% and 85%; For this case, I chose 80% so that Method 1, and 2, as explained above, are consistent with each other. Determine IMEP For phi=0.7, and rc=0.8, we get imep = 10.5 so imep = (P1 )(10.5) , Pi

(5)

Note that Pi is not atmospheric pressure. At WOT, there is a pressure loss in the intake system, due to frictional losses that scale with speed. Pi will be less than atmospheric. Likewise, the exhaust pressure (Pe) 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

exhaust system. To get an idea, of the value of Pi, look at Figure 13-13 in the text (Heywood P. 725). For a piston speed of 10 m/s, pmep = (Pe - Pi) = 0.4 x ( S p) 2 = 0.4(10) 2 = 40

(6)

Now allocate this pumping loss between Pe and Pi. At high speeds around 18% of the loss is on the intake side, and the remaining 82% on the exhaust side. This will be consistent with volumetric efficiency as explained below. So: Pi= 101 kPa – 0.18(40 kPa) = 93.8 kPa Pe=101 kPa + 0.82(40 kPa) = 133.8 kPa We can now calculate an imep:

imep = (93.8 kPa )(10.5) = 984.9kPa Determine Mechanical Efficiency ηm  imep − tfmep  tfmep  = 1 − ; imep imep  

η m = 

(7)

where tfmpe = total friction mep = fmep (rubbing friction and auxiliary mep) + pmep From figure 13-7 (Heywood p 722), fmep for a fired engine at 2260 rpm ≈ 140 kPa. So

tfmpe = (140 + 40)kPa = 180kPa; and 180 ηm = 1 − = 81.7% 985 Determine Volumetric Efficiency and ρ a ,o Using figure 6-8 (Heywood p. 217), assume a volumetric efficiency of 90% for a piston speed of 10 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 1.17 kg/m3

Fuel-to-Air Ratio & Heating Value From table D.4 in the text (Heywood p. 915) we get the stoichiometric Fuel-to-Air ratio of gasoline as 0.0697, and its heating value of 43.2 MJ/kg Power calculation With the estimates for each value, we can now calculate the power

P=

(

)

(0.817)(0.46)(0.90)(37.7 m / sec)(0.011m )(43.2e3kJ / kg ) 1.17 kg / m 3 (0.7)(0.0697) 2

P = 173kW We also use method 2 to check for consistency. Rearranging equation 2.19b (Heywood p50), we get: P=

(bmep )(Vd )(N ) = (805 kPa )(11 dm3)(37.7rev / sec ) = 167 kW 2000

2000

the methods are close For low loads, follow the same procedure, with lower pumping loss, due to lower speed (see figure 13-13, Heywood), and lower rubbing and auxiliary friction (see figure 13-7 Heywood); additionally, the allocation of pressure losses is different, and must be consistent with volumetric efficiency. 2. Boost, Turbo-machinery and Intercooler Boost pressure: To find the boost pressure required, we use equation 1, 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 a ,i . This density can be determined from the ideal gas law, knowing the pressure (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 1. P=

ηmη f ,iηv (N )Vd QHV ρ a ,iφ (F / A)stoich 2

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.

Pmep= Pexhaust - Pintake 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: P=

(

)

(0.918)(0.46)(0.944)(37.7 m / sec)(0.011m )(43.2e3kJ / kg ) 2.065kg / m 3 (0.7)(0.0697) 2

P = 360kW

For this case the pressure that gives a density of 2.065 is 176 kPa, as dictated by the ideal gas law:  8.314kJ / kmoleK  (314 K )(0.944) Pcylinder = ρ a ,i ( beforevalves ) RT( beforevalves ) *η v _ valves = 2.065kg / m 3   28.97kg / kmole 

(

)

which gives Pcylinder=176 kPa. The pressure that must come out of the compressor is approximately: Pcomp =

Pcylinder

ηv

=

176kPa = 186kPa 0.944

Thus the desired boost is 85 kPa. That is we have to compress 85 kPa 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: Wc = m& C p (T2 a − T1 )

where, T2a= Actual compressor exit temperature T1= Compressor inlet temperature (300K)

(8)

We calculate T2a using the compressor efficiency, and isentropic relationships:

T2 a =

T2 s − T1

ηc

+ T1

(9)

and:  1−γ    γ 

 P   T2 s = T1  1   P2 

(10)

T1, P1, P2, γ, and ηc are all known, so T2a, and consequently the compressor work can be calculated. Knowing the compressor work, we now size the turbine using the following equations: W Wt = c (11)

ηm where η m is the mechanical efficiency for the turbine and compressor system. 95% is

reasonable estimate for this number W (12) T5 a = T4 − t Cp where: T5a=Actual turbine exhaust temperature T4 = Turbine inlet temperature (engine exhaust, given at 900K) To find the required turbine pressure ratio: T  P4 = Pr =  5 s  P5  T4 

 γ   1−γ

  

(13)

where:

T5 s =

T5 a − T4

ηt

+ T4

(14)

Thus, enough equations for enough unknowns. Values for the temperatures, pressures, and compressor work, are show in table 2. 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 2. 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

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 (Cp=4.2 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=

m& Cp coolant ,or air (Tin − Tout ) coolant or air m& Cp min (Tin − Tout _ max )

and for the energy balance we have:

(m& Cp∆T )coolant

= −(m& Cp∆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:

ε=

(Tin − Tout ) airr ⇒ Tair _ out = Tair _ in − ε (Tair _ in − Twater _ in ) (Tin _ air − Twater _ in )

Assuming, water temperature increases from 300 to 357, then Tair in is 314 K. Values for the heat exchanger temperatures and flow-rates are also shown in table 2. Figure 1 Schematic of Turbocharged Engine with Intercooler

2

3 Heat Exchanger

7

compressor

1

Engine

6

4

turbine

5

Table 2 Turbocharged Engine with Intercooler: Operating Parameters No intercooling

With intercooling

State

Temperature (K)

Pressure (kPa)

Temperature (K)

Pressure (kPa)

1

300

101

300

101

2

401

233

372

186

3

401

233

314

186

4

900

223

900

188

5

801

127

833

127

6 7

N/A N/A

N/A N/A

101 101

300 380

Work Compressor

40 kW

Work Compressor

29 kW

Intercooler m_dot water m_dot air

0..097 kg/sec 0.404 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 13-7 in the text. The BSFC came out to be 188 g/kw-hr. This number is actually quite good for industry standards. Other people perhaps got lower (around 175), 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 ) Power (kW )

=

0.010078kg / sec(3600 sec/ hr )(1000 g / kg ) = 188 g / kW − hr 193

Note that if we directly use break engine efficiency, we should get the same answer:

bsfc =

1 1 1 = = = 188 g / kW − hr η f ,b (QHV ) η mη f ,i (QHV ) 0.96(0.460)(43.2)

Ways to decrease bsfc include raising compression ratio, and reducing frictional losses. 4. Emissions – NOx

Figure 2 Schematic of Turbocharged Engine with Intercooler 7

2

Inter-

3

4 Venturi/ Mixer

cooler

InterEngine

cooler

5

compressor

turbine

1

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 EPA emissions levels. The emissions requirements along with their safety levels are shown in table 3 below. Table 3 NOx safety NOx Safegy NOx Standard Engine Mode target g/bkW- target g/bkWg/bhp-hr hr hr

EGR

Timing (CA from TDC)

Hit in Fuel economy (percentage points)

25% max torque, 1600 rpm

2.50

1.75

2.35

24%

1

2%

Maximum power

2.50

1.75

2.35

24%

0.5

2%

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:

PT = Pair + PEGR 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 PT can be expressed as: PT =

Pair 1 − EGR

Additionally, since there is a pressure loss of around 16 to 20 kPa associated with the venturi, a higher boost is still needed. To reach the target power output, a total boost of 194 kPa is required, for total pressure of 295 kPa. This is a high boost, higher than industry standard for this size engines. Perhaps a more practical boost is 150 kPa (PT =251 kPa), or less. However this limits the maximum power to 310 kW. If you recognized the practical limitations, this is a perfectly acceptable answer. Table 4 With intercooling State

Temperature (K)

Pressure (kPa)

1

300

101

2

434

295

3

327

295

4

349

279

5

990

241

6 7

864 438

121 241

Table 5 Intercooler

Mass flowrate (kg/sec)

Tin (K)

Tout (K)

EGR

0.171

300

380

Compressor

0.131

300

380

5. Emissions Particulates (a) The particulate emissions corresponding to the chosen EGR level, can be obtained from the data provided (PM levels vs injection timing for various EGR fractions). At 24% EGR, the PM coming out of the engine is approximately 0.2 g/bkW-hr (b) To size the trap, use the space velocity that will minimize the volume of the trap, in this case 28,000 hr-1. This is evident from the relationship for space velocity:

V& = Space _ velocity V 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& = = 1.10m 3 / sec Pexhaust Solving for Volume V=

1.10m 3 / sec = 0.141m 3 = 141 L 28,000 / 3600 sec

Values used are shown in table 6 below Table 6 Mdot_air&fuel (kg/sec)

Mdot_egr (kg/sec)

0.43

0.10

Texhaust (K) Pexhaust (kPa) 864.74

121.00

Space Velocity (1/sec)

Volume flowrate (m3/sec)

Trap Volume (L)

7.78

1.10

140.87

As shown in figure 4 of SAE 2003-01-0047, The maximum pressure loss through the trap, at a space velocity of 28,000/hr, is 6 kPa. This is a very small percentage of the total exhaust pressure (~2.5%), and the effect on turbo-machinery is small.

6. 2007 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 NOx safety Standard target g/bhp-hr g/bhp-hr 0.2000

0.1400

NOx Current Required Safegy Engine out efficiency target g/bkW-hr (catalytic) g/bkW-hr 0.1879

2.3490

0.9200

PM Standard g/bhp-hr 0.0100

PM safety PM Safegy Current Required target target Engine out Efficiency g/bhp-hr g/bkW-hr g/bkW-hr (trap) 0.0050

0.0067

0.2000

Where catalytic converter efficiency is defined as:

ηcat = 1 −

m& pollu tan t , out m& pollu tan t ,in

As shown in table 7, a catalytic converter with 92% efficiency will be needed to meet 2007 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 2007 emissions requirements.

0.966