Heat Engine Cycles. Chapter 2

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1 Chapter 2 Heat Engine Cycles 2.1) An engine cylinder contains kg of fuel with a heat of combustion, q c, of 45,000 kj/kg. The volume V 1 at top dead center is m 3, and the volume V 2 at bottom dead center is m 3. The air-fuel ratio is 16:1, and the mixture temperature T 1 at the start of compression is 300 K. Modeling the compression and combustion as an ideal gas (γ = 1.4,c v = 0.71 kj/kg K) Otto cycle, (a) what is the maximum temperature T 3 and pressure P 3, (b) What is the pressure P 1 at the start of compression? 3 Pressure (P/P 1 ) 2 Volume (V/V 2 ) 4 1 Figure 2.1: Problem 2-1 P-V plot a) The heat input is given by: Q in = m f q c = mq in = ( ) (45,000) Q in = 3.15kJ The compression ratio is given by: r = V 1 = V r = 10 1

2 2 CHAPTER 2. HEAT ENGINE CYCLES Process 1-2: isentropic compression T 2 = T 1 r γ 1 = (300)(10) T 2 = 753K Process 2-3: constant volume heat addition T 3 = Q in mc v +T 2 The mass of gas in the cylinder: m = m a +m f = m f (1+AF) = ( ) (1+16) m = kg The maximum temperature: T 3 = Q in T 2 = ( mc ) v (0.71) +753 T 3 = 4481K The maximum pressure: P 3 = mrt ( ) (0.287)(4481) = V P 3 = 10,203kPa b) Working backwards P 2 = P 3T 2 = (10,203)(753) T P 2 = 1714kPa P 1 = P 2 r γ = P 1 = 68.2kPa (Slightly throttled from P atm = 101 kpa)

3 3 2.2) The Lenoir air cycle is composed of three processes: 1-2 constant volume heat addition, 2-3 isentropic expansion, and 3-1 constant pressure heat rejection. This cycle is named after Jean Lenoir ( ), a Belgian engineer who developed an internal combustion engine in It is a cycle in which combustion occurs without compression of the mixture. (a) Draw the Lenoir cycle on p V and T s diagrams, (b) Assuming the working fluid is an ideal gas with constant properties, derive an expression for the thermal efficiency of the Lenoir air cycle, and (c) Compare the Lenoir cycle thermal efficiency to the Otto cycle efficiency for standard atmospheric pressure and temperature, Q in = 1000 J, r = 8, m = 1.0 g, and γ = 1.4. The p V and T s diagrams are shown below: 2 Pressure 1 3 Volume Figure 2.2: Problem 2-2 p V plot 2 Temperature 3 1 Entropy Figure 2.3: Problem 2-2 T s plot

4 4 CHAPTER 2. HEAT ENGINE CYCLES Process 1-2: Constant volume heat addition T 2 = Q 1 2 mc v +T 1 = 1000 (1)(0.72) +298 = 1687K Process 2-3: Isentropic expansion T 3 = T 2 (1/r) γ 1 = 1687(1/8) 0.4 = 734K The Lenoir cycle thermal efficiency is η = W out Q in = 1 Q out Q in = 1 mc p(t 3 T 1 ) mc v (T 2 T 1 ) = 1 γ (T 3 T 1 ) (T 2 T 1 ) = ( ) ( ) = 0.56 The Otto cycle efficiency is η = 1 r 1 γ = 0.565, slightly greater than the Lenoir cycle.

5 5 2.3) (a) Show for an Otto cycle that T 3 /T 2 = T 4 /T 1. (b) Derive the Otto cycle efficiency equation, Equation (2.10). 3 Pressure (P/P 1 ) 2 Volume (V/V 2 ) 4 1 Figure 2.4: Problem 2-3 P-V plot a) From the cycle analysis of Section 2.2; for an isentropic compression: T 2 = r γ 1 T 1 ( T 4 1 = T 3 r ) ( T2 T 1 T 3 T 2 = T 4 T 1 ) γ 1 ( T4 T 3 ) = 1 b) The thermal efficiency is η = 1 Q in Q out Q out = mc(t 4 T 3 ) Q in = mc(t 3 T 2 ) Substituting we obtain ( ) η = 1 Q in = 1 T 4 T T T4 1 1 T 1 1 = 1 ( ) Q out T 3 T 2 T T3 2 T 2 1 η = 1 T 1 T 2

6 6 CHAPTER 2. HEAT ENGINE CYCLES 2.4) Derive the Otto and Diesel cycle imep equation, Equation (2.11). The imep is defined as: imep = W V d W = η Q in r = V c +V d V c = V d V c r 1 ( ) r V 1 = V c +V d = V d r 1 ( )( )( ) imep 1 η Qin r = P 1 P 1 V 1 r 1

7 7 2.5) (a) Show that for a Diesel cycle (T 3 /T 2 ) γ = T 4 /T 1. (b) Derive Equation (2.15) and Equation (2.16). a) From the cycle analysis of Section 2.3, and with reference to Figure 2.3, T 2 T 1 = r γ 1 β = V 3 = T 3 T 3 = βt 2 = βr γ 1 T 1 V 2 T 2 ( ) γ 1 ( ) γ 1 T 4 V3 β = = T 4 = β γ T 1 T 3 r V 4 ( T3 T 2 ) γ = T 4 T 1 = β γ b) The thermal efficiency is: η = 1 Q out Q in = 1 c v(t 4 T 1 ) c p (T 3 T 2 ) = 1 (β γ T 1 T 1 ) γ(βr γ 1 T 1 r γ 1 T 1 ) = 1 β = T 3 T 2 Application of the first law from 2 3: T 3 = T 2 + Qin mc p m = P 1V 1 RT 1 R = γ 1 c p γ Substituting back in we obtain: ( )( )( ) γ 1 Qin 1 β = 1+ γ P 1 V 1 r γ 1 ( )( 1 β γ ) 1 r γ 1 γ(β 1)

8 8 CHAPTER 2. HEAT ENGINE CYCLES 2.6) What does the compression ratio of a Diesel cycle need to be to have the same thermal efficiency of an Otto cycle engine that has a compression ratio, r = 9? Assume the specific heat ratio γ = 1.3, and Q in /P 1 V 1 = 30. The Otto cycle (o) thermal efficiency is η o = 1 r 1 γ o = = The Diesel cycle (d) thermal efficiency is η d = 1 r 1 γ d where β is defined as β = 1+ γ 1 γ = β γ 1 γ(β 1) Q in r 1 γ d P 1 V r 0.3 d = r 0.3 d setting η d = η o, and solving for r d, we find r d = 20.7

9 9 2.7) A Diesel cycle has a compression ratio of 20, and the heat input q in to the working fluid is 1600 kj/kg. The Diesel cycle is unthrottled, so at the start of compression P 1 = 101 kpa and T 1 = 298K. Assuming the working fluid is air with constant properties, what is the maximum pressure and temperature in the cycle, the cycle efficiency, and imep? Assume γ = 1.3 and c p = 1.0 kj/kg K. 1-2: Isentropic compression T 2 T 1 = r γ 1 = = 2.46 T 2 = 732K P max = P 2 = P 1 r γ = (101) = 4962kPa 2-3: Constant P heat addition T max = T 3 = T 2 +q in /c p = (732)+1600/1.0= 2332K so β = T3 T 2 = 2332/732= 3.19 and the thermal efficiency η is: η d = 1 r 1 γ βγ 1 γ(β 1) The imep is: = (3.19 1) = imep = w net v 1 v 2 The specific net work is w net = ηq in = (0.497)(1600) = 795kJ/kg v 1 = RT 1 /P 1 = (0.287)(298)/(101)= 0.84m 3 /kg v 2 = RT 2 /P 2 = (0.287)(732)/(4962)= m 3 /kg so the imep = 795/( ) = 997 kpa.

10 10 CHAPTER 2. HEAT ENGINE CYCLES 2.8) Show that for the Otto cycle as r 1, imep/p 1 Q in (γ 1)/P 1 V 1 (use L Hopital s rule). The dimensionless imep is imep = Q ( )( in r 1 1 ) P 1 P 1 V 1 r 1 r γ 1 = Q ( )( in r r γ 1 ) 1 P 1 V 1 r 1 r γ 1 = Q ( in r γ ) r P 1 V 1 r γ r γ 1 As r 1, this blows up, so using L hopital s rule a(r) lim r ) ( r γ r lim r 1 r γ r γ 1 Substituting back in b(r) = lim r = da(r) dr db(r) dr γ 1 γ (γ 1) = γ 1 imep P 1 = Q in P 1 V 1 (γ 1) as the limit r 1

11 11 2.9) A engine is modeled with a Limited Pressure cycle. The maximum pressure is to be 8000 kpa. The compression ratio is 17:1, the inlet conditions are 101 kpa and 320 K, the non-dimensional heat input Q in /P 1 V 1 = 30, and γ=1.3. Find the thermal efficiency and the values of α and β. The limited pressure parameters are α and β α = 1 ( )( ) P r γ = P α = 1.99 β = 1+ γ 1 αγ β = η = 1 1 r γ 1 η = 0.53 [ Qin P 1 V 1 [ ] 1 α 1 rγ 1 γ 1 αβ γ ] 1 α 1+αγ(β 1) = [ ( ) 1 30 (1.99)(1.3) (1.99 1) ] (1.33 1) = 1 1 [ (1.99)(2.104) (1.99 1)+(1.3)(1.99)( ) ]

12 12 CHAPTER 2. HEAT ENGINE CYCLES 2.10) (a) Derive the equation for the Miller cycle efficiency, Equation (2.23). (b) Derive the equation for the Miller cycle imep, Equation (2.24). We need to get the η and imep as a function of λ and γ. Using a state by state cycle analysis 1-2 T 2 = T 1 r c γ Q in = mc v (T 3 T 2 ) 3-4 T 4 = T 3 r e 1 γ 4-1 Q out = mc v (T 4 T 5 )+mc p (T 5 T 1 ) a) Miller cycle efficiency derivation: λ = r e r c = V 5 V 3 V 1 = V 5 = P 5V 5 = RT 5 = T 5 T 5 = λt 1 V V 1 P 1 V 1 RT 1 T 1 2 η = 1 Q out = 1 mc v(t 4 T 5 )+mc p (T 5 T 1 ) = 1 (T 4 T 5 )+γ(t 5 T 1 ) Q in Q Q in in mc v T 3 = T 2 + Q in mc v = T 1 r c γ 1 + Q in mc v T 4 = T 1 λ 1 γ + Q in mc v r e 1 γ Solving for efficiency: [ ] η = 1 r 1 γ λ 1 γ λ+γ(λ 1) e Q in = 1 (λr c ) 1 γ c [ vt 1 ] η = 1 (λr c ) 1 γ λ 1 γ λ(1 γ) γ Q in P 1V 1 (γ 1) b) Miller cycle imep derivation: [ ] λ 1 γ λ(1 γ) γ Q in c vt 1 imep = W V d W = η Q in V d = V 5 V 2 = r e V1 V ( ) 1 λrc 1 = V 1 r c r c r c ( )( ) imep Qin rc = (η) P 1 P 1 V 1 λr c 1

13 ) For Otto and Miller cycles that have equal compressionratios, r c = 10, what are the respective thermal efficiencies and non-dimensional imeps? Assume that the parameter, λ is equal to 1.5 for the Miller cycle, the specific heat ratio γ = 1.3, and Q in /P 1 V 1 = 30. a) The Otto thermal efficiency is: η otto = 1 r c 1 γ = 1 (10) 0.3 η otto = 0.50 The Miller thermal efficiency is: ( λ η miller = 1 (λr c ) 1 γ 1 γ )( ) λ(1 γ) γ P1 V 1 γ 1 Q in ( 1.5 = 1 (15) ) 1.5( 0.3) 1.3 (0.3)(30) η miller = 0.55 The Miller cycle has a 10% greater efficiency b) The Otto imep is: ( ) imep P 1 ( ) imep P 1 otto otto The Miller imep is: ( ) imep P 1 ( ) imep P 1 miller miller ( )( ) Qin rc = η otto = (0.50)(30) P 1 V 1 r c 1 = 16.6 ( ) 10 9 ( )( ) ( ) Qin rc 10 = η miller = (0.55)(30) P 1 V 1 λr c 1 14 = 11.8 The Miller cycle has a 29% lower imep.

14 14 CHAPTER 2. HEAT ENGINE CYCLES 2.12) Develop a complete expansion cycle model in which the expansion stroke continues until the pressure is atmospheric. Derive an expression for the efficiency in terms of γ, α = V 4 /V 3, and β = V 1 /V 4. a) With reference to Figure 2.7, the thermal efficiency is defined as η = 1 Q out Q in = 1 c p(t 4 T 1 ) c v (T 3 T 2 ) Since heat addition is at constant volume and heat rejection is at constant pressure [ ] T4 (1 T1 η = 1 γ T 4 ) T 3 (1 T2 T 3 ) For the isentropic expansion from 1 to 2 and from 3 to 4 T 2 T 1 = T 4 T 3 = Simplifying ( V1 V 2 ( V3 V 4 T 2 T 1 T4 T 3 = β γ 1 T 2 T 3 = β γ 1 ) γ 1 ( ) γ 1 V1 = V4 = (βα) γ 1 V 4 V 2 ) γ 1 ( ) γ 1 1 = α ( T1 T 4 Since P 1 =P 4, from the ideal gas law ) T 1 V 1 = T 4 V 4 T 1 T 4 = V 1 V 4 = β So the thermal efficiency can be expressed as η = 1 ( γ α γ 1 ) (1 β) (1 β γ )

15 ) If a four-cylinder, four-stroke engine with a 0.1 m bore and an 0.08 m stroke operating at 2000 rpm has the same heat/mass loss parameters as Example 2.3, how much indicated power (kw) would it produce? What if it were a two-stroke engine? Theenginespeedω =N *2π/60=2000rev/min*2π/60=209.4rad/s. UsingtheFiniteHeatMassLoss.m program as shown below, imep/p 1 =11.55, so imep= e+5 kpa. The displacement volume V d = n c (π/4)b 2 s = 4(π/4)(0.1) 2 (0.08) = 2.51e 3 m 3. The indicated power Ẇi is Ẇ i = imepv d N/2 = (11.55e+5)(2.51e 3) = 48,320W = 48kW If the engine is two stroke, the indicated power would double to 96 kw. The program input is: function [ ] = FiniteHeatMassLoss( ) % Gas cycle heat release code with and w/o heat transfer % data structure for engine parameters clear(); thetas = -20; % start of heat release (deg) thetad = 40; % duration of heat release (deg) r =10; % compression ratio gamma = 1.4; % gas const Q = 20.; % dimensionless total heat release h = 0.2; % dimensionless ht coefficient tw = 1.2; % dimensionless cylinder wall temp beta = 1.5; % dimensionless volume a = 5; % weibe parameter a n = 3; % weibe exponent n omega =209.4; % engine speed rad/s c = 0.8; % mass loss coeff The program output is: Weibe Heat Release with Heat and Mass Loss Theta_start = Theta_dur = P_max/P1 = = 11.0 P_tdc/P_max = 0.72 Net Work/P1V1 = Heat Loss/P1V1 = 4.45 Mass Loss/m = Efficiency = Eff./Eff. Otto = Imep/P1 = 11.55

16 16 CHAPTER 2. HEAT ENGINE CYCLES 2.14) Using the program BurnFraction.m,and assuming that a = 5, the beginning of heat addition is -10, and the duration ofheat addition is 40, (a) Plot the Weibe heat releasefraction curvefor the following form factor values: n = 2, 3, and 4. (b) At what crankangle is 0.10, 0.50, and 0.90of the heat released? a.) The problem data is entered into the program BurnFraction.m and the resulting cumulative and rate of burn fraction plots are shown below. Cumulative Burn Fraction n=2 n=3 n= Crank Angle (deg) Figure 2.5: Problem 2-15 Burn fraction plot n=2 n=3 n=4 Burn Rate (1/deg) Crank Angle (deg) Figure 2.6: Problem 2-15 Rate of burn fraction plot b.) As the form factor increases, the crank angles for the 0.1, 0.5, and 0.9 burn fractions all increase. For n=2, the 0.1 fraction is at -3.9 degrees, the 0.5 fraction at 4.9 degrees, and the 0.9 fraction at 17.5 degrees. For n=3, the 0.1 fraction is at 1.3 degrees, the 0.5 fraction at 11.0 degrees, and the 0.9 fraction at 21.1 degrees. For n=4, the 0.1 fraction is at 3.4 degrees, the 0.5 fraction at 14.6 degrees,

17 17 and the 0.9 fraction at 23.1 degrees. The program, modified for Problem 2.15 is: function [ ]=BurnFraction2_15( ) % this program computes and plots the cumulative burn fraction % and the instantanous burnrate clear(); a = 5; % Weibe efficiency factor thetas = -10; % start of combustion thetad = 40; % duration of combustion theta=linspace(thetas,thetas+thetad,100); %crankangle theta vector dum=(theta-thetas)/thetad; % theta diference vector i1=0; n = 2; % Weibe form factor temp=-a*dum.^n; xb1=1.-exp(temp); %burn fraction dxb1=n*a*(1-xb1).*dum.^(n-1); %element by element vector multiplication n = 3; % Weibe form factor temp=-a*dum.^n; xb2=1.-exp(temp); %burn fraction dxb2=n*a*(1-xb2).*dum.^(n-1); n = 4; % Weibe form factor temp=-a*dum.^n; xb3=1.-exp(temp); %burn fraction dxb3=n*a*(1-xb3).*dum.^(n-1); %get heat release angles for i=1:100 test=xb3(i); if test > 0.1 i1 = i; break end end %get heat release angles for i=1:100 test=xb3(i); if test > 0.5 i2 = i; break end end %get heat release angles for i=1:100 test=xb3(i); if test > 0.9 i3 = i; break end end fprintf ( crank angle for 0.1 burn fraction is %6.1f degrees \n, theta(i1)); fprintf ( crank angle for 0.5 burn fraction is %6.1f degrees \n, theta(i2)); fprintf ( crank angle for 0.9 burn fraction is %6.1f degrees \n, theta(i3)); %plot results plot(theta,xb1, -,theta,xb2, --,theta,xb3, :, linewidth,2); set(gca, fontsize, 18, linewidth,2); %grid xlabel( Crank Angle (deg), fontsize, 18); ylabel( Cumulative Burn Fraction, fontsize, 18); legend( n=2, n=3, n=4 );

18 18 CHAPTER 2. HEAT ENGINE CYCLES figure(); plot(theta,dxb1, -,theta,dxb2, --,theta,dxb3, :, linewidth,2); set(gca, fontsize, 18, linewidth,2); %grid xlabel( Crank Angle (deg), fontsize, 18); ylabel( Burn Rate (1/deg), fontsize, 18); legend( n=2, n=3, n=4 ); end

19 ) Using the program FiniteHeatRelease.m, determine the effect of heat release duration on the net work, power, mean effective pressure, and thermal efficiency for a four stroke engine with heat release durations of 40, 30, 20, 10, and 5 degrees. Assume that the total heat addition Q in = 2500 J, the start of heat release θ s remains constant at -10 atdc, a = 5, n = 3, and γ = 1.4. The engine bore and stroke are m, the compression ratio is 9:1, and engine speed is 3000 rpm. The table below showsthe effect ofheat releaseduration. Foraheat releasemodel, the work, imep, and efficiency will increase as the energy release duration decreases, and the cycle approaches a constant volume Otto cycle. The Figure shows the greater peak pressure with the shorter 20 degree duration of Engine 1 relative to the longer 40 degree duration of Engine 2. However, if the heat release duration occurs only during the compression stroke, as for the 10 and 5 degree duration points, this will not be the case. θ d W i/p 1V 1 imep/p 1 η Table for Problem Engine 1 Engine Pressure (bar) Theta (deg) Figure 2.7: Problem 2-16 Pressure profile plot The input to and output of the program FiniteHeatRelease.m is shown below. function[ ]=FiniteHeatRelease( ) % Gas cycle heat release code for two engines % engine parameters clear(); thetas(1,1)= -10; % Engine1 start of heat release (deg) thetas(2,1)= -10; % Engine2 start of heat release (deg) thetad(1,1) = 20; % Engine1 duration of heat release (deg)

20 20 CHAPTER 2. HEAT ENGINE CYCLES thetad(2,1) = 40; % Engine2 duration of heat release (deg) r=9; %compression ratio gamma= 1.4; %gas const q= 33.0; % dimensionless total heat release Qin/P1V1 a= 5; %weibe parameter a n= 3; %weibe exponent n The program output is Engine 1 Engine 2 Theta_start Theta_dur P_max/P_ Theta_max Net Work/P1V Efficiency Eff. Ratio Imep/P

21 ) If a four cylinder unthrottled Otto cycle engine is to generate 100 kw at an engine speed of 2500 rpm, what should its bore and stroke be? Assume a square block engine with equal bore and stroke and a compression ratio of 10:1. Assume Q in /P 1 V 1 = 20, the start of heat release θ s remains constant at -15 atdc, the combustion duration is 40, a = 5, n = 3, and γ = 1.4. Use the single cylinder program FiniteHeatRelease.m, and solve for a power output of 25 kw. The program input and output is shown below, resulting in W i /P 1 V 1 = Therefore, V 1 = W i P = = m 3 Since V 1 = V d r/(r 1), b = r 1 4 V 1 r π = π = 0.105m function [ ]=FiniteHeatRelease( ) % Gas cycle heat release code for two engines % engine parameters clear(); thetas(1,1)= -15; % Engine1 start of heat release (deg) thetas(2,1)= -15; % Engine2 start of heat release (deg) thetad(1,1) = 40; % Engine1 duration of heat release (deg) thetad(2,1) = 40; % Engine2 duration of heat release (deg) r=10; %compression ratio gamma= 1.4; %gas const q= 20; % dimensionless total heat release Qin/P1V1 a= 5; %weibe parameter a n= 3; %weibe exponent n Engine 1 Engine 2 Theta_start Theta_dur P_max/P_ Theta_max Net Work/P1V Efficiency Eff. Ratio Imep/P

22 22 CHAPTER 2. HEAT ENGINE CYCLES 2.17) Develop a four-stroke Diesel cycle model (along the lines used in Example 2.4).Compute the net cycle efficiency and the net imep for an engine with r = 22,γ = 1.3,T i = 300K, P i = 101kPa,P i /P e = 0.98,M = 29, and q in = 2090 kj/kg gas. The program FourStrokeOtto.m was modified into FourStrokeDiesel.m for this problem. The constant volume heat addition was changed to a constant pressure heat addition, and the efficiency η redefined accordingly, as shown below. The output is given below. With a compression ratio r of 22, the residual fraction f = 0.015, and the thermal efficiency η = Four Stroke Diesel Cycle State Pressure (kpa): Temperature (K): Ideal Thermal Eff.= Net Thermal Eff.= Exhaust Temp. (K)= Volumetric Eff.= 1.00 Residual Fraction Net Imep (bar)= % Four stroke Diesel cycle model % input parameters Ti = 300; % inlet temperature, K Pi = 101; % inlet pressure, kpa Pe = 103; % exhaust pressure, kpa r = 22; % compression ratio qin = 2090; % heat input, kj/kg (mixture) gamma = 1.3; % ideal gas specific heat ratio R = 0.287; % gas constant kj/kg K cp= 1.24; %const vol specific heat, kj/kg K f=0.05;% guess value of residual fraction f Te = 1000; % guess value of exhaust temp, K tol=0.0001; % tolerance for convergence err = 2*tol; %error initialization gam=(gamma -1)/gamma; while (err > tol) %while loop for cycle calc %intake stroke T1=(1-f)*Ti + f*(1 - (1- Pi/Pe)*gam)*Te; P1=Pi; %isentropic compression P2=P1*r^gamma; T2=T1*r^(gamma-1); %const p heat addition T3=T2 + qin*(1-f)/cp; P3=P2; %isentropic expansion beta = T3/T2; P4=P3*(beta/r)^gamma; T4=T3*(beta/r)^(gamma-1); %isentropic blowdown P5=Pe; T5=T4*(P4/Pe)^(-gam); %const p exhaust stroke Te=T5; fnew=(1/r)*(pe/p4)^(1/gamma); %new residual fraction err=abs(fnew-f)/fnew; f=fnew;

23 23 end %cycle parameters eta= 1-1/r^(gamma-1)*(beta^gamma -1)/(gamma*(beta-1)); imep = P1*(qin*(1-f)/(R*T1))/(1-(1/r))*eta; pmep=pe-pi; etanet= eta*(1-pmep/imep); imepnet= (imep-pmep)/100.; voleff=1-(pe/pi -1)/(gamma*(r-1)); %output calcs fprintf( \nfour Stroke Diesel Cycle \n ) fprintf( State \n ); fprintf( Pressure (kpa): %6.1f %6.1f %6.1f %6.1f \n,p1,p2,p3,p4); fprintf( Temperature (K): %6.1f %6.1f %6.1f %6.1f \n,t1,t2,t3,t4); fprintf( Ideal Thermal Eff.= %6.3f Net Thermal Eff.= %6.3f \n,eta, etanet); fprintf( Exhaust Temp. (K)= %6.1f Volumetric Eff.= %6.2f \n,te, voleff); fprintf( Residual Fraction %6.3f Net Imep (bar)= %6.2f \n,f, imepnet); Four Stroke Diesel Cycle State Pressure (kpa): Temperature (K): Ideal Thermal Eff.= Net Thermal Eff.= Exhaust Temp. (K)= Volumetric Eff.= 1.00 Residual Fraction Net Imep (bar)= 12.60

24 24 CHAPTER 2. HEAT ENGINE CYCLES 2.18) Using the program FourStrokeOtto.m, plot the effect of inlet throttling from 100 kpa to 25 kpa on the peak pressure, P 3, and the volumetric efficiency η v. Assume the following conditions: T i = 300K,r = 9,γ = 1.3, and q in = 2400 kj/kg gas. The inlet pressure was varied as shown in the program input below, and the peak pressure and volumetric efficiency plotted below. There is a linear relationship between peak and inlet pressure. The volumetric efficiency rises in a non-linear fashion from 0.62 to Peak Pressure (kpa) Inlet Pressure (kpa) Figure 2.8: Problem 2-19 Peak pressure plot Volumetric Efficiency Inlet Pressure (kpa) Figure 2.9: Problem 2-19 Volumetric Efficiency plot % Four stroke Otto cycle model % input parameters Ti = 300; % inlet temperature, K Pi = 28; % inlet pressure, kpa Pe = 100; % exhaust pressure, kpa

25 r = 9; % compression ratio qin = 2400; % heat input, kj/kg (mixture) gamma = 1.3; % ideal gas specific heat ratio R = 0.287; % gas constant kj/kg K cv= R/(gamma-1); %const vol specific heat, kj/kg K f=0.05;% guess value of residual fraction f Te = 1000; % guess value of exhaust temp, K tol=0.0001; % tolerance for convergence err = 2*tol; %error initialization gam=(gamma -1)/gamma; 25

26 26 CHAPTER 2. HEAT ENGINE CYCLES 2.19) In Example 2.3, T e is the exhaust temperature during the constant pressure exhaust stroke. It is not the same as the average temperature of the gases exhausted. Explain. Gases at temperatures greater than T e are also exhausted during the blowdown phase. The average temperature of the gases leaving the engine depend on the definition of average. A time average, a mass flow rate average, and enthalpy average are all different, since the mass and enthalpy flow varies with crank angle.

Solutions Manual Internal Combustion Engines: Applied Thermosciences

Solutions Manual Internal Combustion Engines: Applied Thermosciences Professor Allan T. Kirkpatrick Mechanical Engineering Department Colorado State University Fort Collins, CO January 23, 2017 2 Chapter