Engineering_Mechanics_Colin_R_Ferguson_A_Solved_Ch4

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<p>Chapter 4</p><p>Fuel-Air Combustion Processes</p><p>4.1) Compute the higher, lower, and equilibrium heats of combustion for methanol CH</p><p>3</p><p>OH (l). The equi-</p><p>librium computation determines the quality of the water in the products at standard atmospheric</p><p>temperature and pressure.</p><p>As shown in the Appendix, the equilibrium quality χeq of methanol, using the chemical formula of the</p><p>fuel represented by CaHbOcNd, is</p><p>χeq =</p><p>nH2O,g</p><p>nH2O</p><p>=</p><p>(</p><p>9.52a+ 1.88b− 3.76c+ d</p><p>b</p><p>)(</p><p>Psat</p><p>P</p><p>/(1−</p><p>Psat</p><p>P</p><p>)</p><p>)</p><p>=</p><p>(</p><p>9.52(1) + 1.88(4)− 3.76 + 0</p><p>4</p><p>)(</p><p>3.17</p><p>101</p><p>/(1−</p><p>3.17</p><p>101</p><p>)</p><p>)</p><p>= 0.107</p><p>(4.1)</p><p>The stoichiometric combustion equation per kmol of methanol is</p><p>1 CH</p><p>3</p><p>OH+ 1.5O</p><p>2</p><p>−−→ 1CO</p><p>2</p><p>+ 2H</p><p>2</p><p>O</p><p>The energy equation, per kmol of fuel, is</p><p>q̄c =</p><p>∑</p><p>i</p><p>nr,ih̄r,i −</p><p>∑</p><p>i</p><p>np,ih̄p,i</p><p>=1× h̄fuel − nCO2</p><p>h̄CO2</p><p>− nH2O[h̄H2O − (1− χ)h̄fg,H2O]</p><p>= + 1 (−201.17)− 1 (−393.52)− 2 [(−241.83− (1 − χ)(43.99))]</p><p>= + 676.01 + (88.0)(1− χ) (MJ/kmol)</p><p>The molecular mass of methanol CH</p><p>3</p><p>OH is M = 32.04. Therefore, the heats of combustion are</p><p>qlhc =</p><p>q̄c(χ = 1)</p><p>M</p><p>= 676.01/32.04 = 21.09MJ/kg</p><p>qeq =</p><p>q̄c(χ = 0.208)</p><p>M</p><p>= 754.59/32.04 = 23.55MJ/kg</p><p>qhhc =</p><p>q̄c(χ = 0)</p><p>M</p><p>= 763.99/32.04 = 23.84MJ/kg</p><p>1</p><p>2 CHAPTER 4. FUEL-AIR COMBUSTION PROCESSES</p><p>4.2) The heat of combustion could have been defined without requiring complete conversion to carbon</p><p>dioxide and water. What would the lower heat of combustion be for the case φ = 1.4, fuel = C</p><p>8</p><p>H</p><p>18</p><p>(l)</p><p>octane, To = 298 K, Po = 1 atm? Assume the water quality χ = 1 and that the equilibrium constant</p><p>K = 9.95× 10−6.</p><p>Using Example 3.3 as a template, the combustion equation is</p><p>C</p><p>8</p><p>H</p><p>18</p><p>+</p><p>as</p><p>φ</p><p>(O</p><p>2</p><p>+ 3.76N</p><p>2</p><p>) −−→</p><p>n1CO2</p><p>+ n2 H2</p><p>O+ n3 N2</p><p>+ n5CO+ n6 H2</p><p>The calculation of the product mole fractions proceeds as follows:</p><p>a = 8, b = 18, c = d = 0</p><p>as = a+ b/4− c/2 = 12.5</p><p>d1 = 2 as(1− 1/φ) = 7.143</p><p>a1 = 1−K ≃ 1</p><p>b1 = b/2 + aK − d1(1−K) = 1.857</p><p>c1 = −ad1K = 5.69× 10−4</p><p>n1 = a− n5 = 8</p><p>n2 = b/2− d+ n5 = 1.857</p><p>n3 = d/2 + 3.76 as/φ = 33.571</p><p>n4 = 0</p><p>n5 = −K1/a1 ≃ 0</p><p>n6 = d1 − n5 = 7.143</p><p>N =</p><p>∑</p><p>ni = 56.167</p><p>(4.2)</p><p>The energy equation, per kmol of fuel, is</p><p>q̄c =</p><p>∑</p><p>i</p><p>nr,ih̄r,i −</p><p>∑</p><p>i</p><p>np,ih̄p,i</p><p>=1× h̄fuel − nCO2</p><p>h̄CO2</p><p>− nH2O[h̄H2O − (1− χ)h̄fg,H2O]</p><p>= + 1 (−249.96)− 8 (−393.52)− 1.857 [(−241.83− 0]</p><p>=3347.3 (MJ/kmol)</p><p>The molecular mass of octane C</p><p>8</p><p>H</p><p>18</p><p>is M = 114.22. The lower heat of combustion per kg of fuel is</p><p>qlhc =</p><p>q̄c(χ = 1)</p><p>M</p><p>= 3347.3/114.22 = 29.30MJ/kg</p><p>This is 34 % less than the stoichiometric lower heat of combustion. Since the combustion is rich, H2</p><p>is one of the product gases. However, since its enthalpy of formation is zero, it is not included in the</p><p>energy balance calculation.</p><p>3</p><p>4.3) With reference to Figure 4-2, explain why the heats of combustion at φ = 0.2 and φ = 1.2 are less then</p><p>those at φ = 1.0.</p><p>a.) For lean mixtures, all of the fuel is converted to H2O and CO2. Since there are extra moles of air in</p><p>the product gases, the partial pressure of the water will remain less than the saturated water pressure.</p><p>As φ increases beyond 0.2 at some point water will condense, increasing the heat of combustion.</p><p>b.) For rich mixtures, the combustion products contain both CO and H2, both of which can be reacted</p><p>to H2O and CO2, respectively. Thus the heat of combustion decreases as φ increases beyond φ = 1.0,</p><p>and increasing amounts of CO and H2 are produced.</p><p>4 CHAPTER 4. FUEL-AIR COMBUSTION PROCESSES</p><p>4.4) What is the residual mass fraction required to reduce the adiabatic flame temperature of gasoline,</p><p>diesel, methane, methanol, and nitromethane below 2000 K? Assume φ = 1.0 at 101 kPa and 298 K.</p><p>The program AdiabaticFlameTemp.m was modified with the addition of a loop, test for 2000 K, and</p><p>a plot over the residual fraction f . The results are:</p><p>Fuel f</p><p>Gasoline 0.182</p><p>Diesel 0.215</p><p>Methane 0.161</p><p>Methanol 0.163</p><p>Nitromethane 0.416</p><p>The residual fraction is between 0.16 and 0.20 for the first hydrocarbon fuels, and about 0.4 for the</p><p>nitromethane, due to its oxygen content.</p><p>The plot of adiabatic flame temperature versus residual fraction for gasoline is shown below.</p><p>0 0.1 0.2 0.3 0.4</p><p>1600</p><p>1700</p><p>1800</p><p>1900</p><p>2000</p><p>2100</p><p>2200</p><p>2300</p><p>Residual Fraction f</p><p>A</p><p>di</p><p>ab</p><p>at</p><p>ic</p><p>F</p><p>la</p><p>m</p><p>e</p><p>T</p><p>em</p><p>pe</p><p>ra</p><p>tu</p><p>re</p><p>(</p><p>K</p><p>)</p><p>Figure 4.1: Problem 4-4 Gasoline adiabatic flame temperature plot</p><p>The modified program AdiabaticFlameTemp.m is:</p><p>% Computes const pressure adiabatic flame temperature T2</p><p>% from first law: dh = q</p><p>% Inputs:</p><p>T1 = 298;% initial temperature (K)</p><p>P1 = 101; % initial pressure (kPa)</p><p>PHI = 1.0; % equivalence ratio</p><p>%f = .1; % residual fraction</p><p>ifuel=2; % 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</p><p>jmax=100;</p><p>ff= linspace(0, 0.4, jmax); % residual fraction vector</p><p>taf=zeros(1,jmax);</p><p>for j=1:jmax %loop over f vector</p><p>f=ff(j);</p><p>% use FARG for initial calc of H1</p><p>[~, H1, ~, ~, ~, ~, CP, ~, ~, ~] = farg( T1, P1, PHI, f, ifuel );</p><p>5</p><p>%fprintf(’ Initial Enthalpy H1 = %7.1f Initial CP = %7.3f \n’,H1,CP);</p><p>P2 = P1;</p><p>T2 = 2000; %initial guess of flame temp</p><p>MAXITS = 50;</p><p>TOL = 0.00001;</p><p>for i=1:MAXITS,</p><p>[~,~, H2, ~, ~, ~, ~, CP, ~, ~, ~] = ecp( T2, P2, PHI, ifuel );</p><p>%fprintf(’ Iterated Enthalpy H2 = %7.1f Iterated CP = %7.3f \n’,H2,CP);</p><p>DELT2 = (H1-H2)/CP;</p><p>T2 = T2 + DELT2;</p><p>%fprintf(’ Iterated Adiabatic Flame Temp (K) = %8.1f \n’,T2);</p><p>if ( abs(DELT2)/T2</p><p>= s1 = 8.919, we</p><p>find the final temperature T2=1235 K, and u2 = −1947.5. Therefore the work done is w1−2 = u1−u2</p><p>= -1075.5 - (-1947.5) = 872.0 kJ/kg.</p><p>Equilibrium Combustion Solver for State 1</p><p>Pressure (kPa) = 1000.0</p><p>Temperature (K) = 2000.0</p><p>Fuel Air Equivalence ratio = 0.9</p><p>Mixture Properties</p><p>h(kJ/kg) = -475.9</p><p>u(kJ/kg) = -1075.5</p><p>s (kJ/Kg K) = 8.919</p><p>v (m3/kg) = 0.600</p><p>cp (kJ/Kg K) = 1.545</p><p>Equilibrium Combustion Solver for State 2</p><p>Pressure (kPa) = 100.0</p><p>Temperature (K) = 1235.0</p><p>Fuel Air Equivalence ratio = 0.9</p><p>Mixture Properties</p><p>h(kJ/kg) = -1577.4</p><p>u(kJ/kg) = -1947.5</p><p>s (kJ/Kg K) = 8.919</p><p>v (m3/kg) = 3.701</p><p>cp (kJ/Kg K) = 1.357</p><p>9</p><p>4.7) Equilibrium combustion products of gasoline are expanded isentropically by a volume ratio of 10:1.</p><p>For φ = 1.1 and an initial state of T1 = 3000 K, P1 = 5000 kPa, find the final state (T2, P2) and the</p><p>work done. b.) Repeat the calculation for φ = 0.9. What is the effect of equivalence ratio ?</p><p>a.) Using the program RunEcp.m for State 1, the initial entropy s1 = 9.1281, and the initial volume</p><p>v1 = 0.1818. Revising RunEcp.m to iterate over temperature and pressure to find the state 2 where</p><p>s2 = s1 = 9.128, and v2 = 1.818, we find the final temperature T2= 1868 K, the final pressure P2 =</p><p>306 kPa, and u2 = −1344.8. Therefore the work done is w1−2 = u1−u2 = 343.2 - (-1344.8) = 1688.0</p><p>kJ/kg.</p><p>Equilibrium Combustion Solver for State 1</p><p>Pressure (kPa) = 5000.0</p><p>Temperature (K) = 3000.0</p><p>Fuel Air Equivalence ratio = 1.1</p><p>Mixture Properties</p><p>h(kJ/kg) = 1252.1</p><p>u(kJ/kg) = 343.2</p><p>s (kJ/Kg K) = 9.1281</p><p>v (m3/kg) = 0.1818</p><p>Revised RunEcp.m with P and T search loops</p><p>clear;</p><p>phi = 1.1; % enter equivalence ratio input</p><p>%T = 3000; % enter temperature (K) input</p><p>%P = 5000; % enter pressure (kPa) input</p><p>fuel_id =2;</p><p>% fuel_id - 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</p><p>smin = 9.1275;</p><p>smax = 9.1285;</p><p>vmin = 1.817;</p><p>vmax = 1.819;</p><p>for P=200:2000</p><p>for T=1000:2000</p><p>% call ecp function</p><p>[ierr, Y, h, u, s, v, R, Cp, MW, dvdT, dvdP] = ecp( T, P, phi, fuel_id );</p><p>if s>smin && svmin && vsmin && svmin && v</p><p>Enthalpy(kJ/kg)= -339.2 407.4 1014.2 -1431.1</p><p>Int.Energy(kJ/kg)= -444.5 115.8 115.8 -1892.6</p><p>Volume (m^3/kg) = 1.040 0.052 0.052 1.040</p><p>Cp (kJ/kg K) = 1.092 1.326 2.344 1.440</p><p>Work (kJ/kg)= 1448.1</p><p>Imep (kPa)= 1466.2</p><p>The file OttoFuel.m is modified with the addition of plotting commands to produce the two plots</p><p>below. The thermal efficiency increases as the compression ratio is increased. The ratio of the fuel-air</p><p>Otto cycle efficiency to the gas Otto cycle rises slightly from about 0.87 to 0.88 as the compression</p><p>ratio is increased.</p><p>5 10 15 20</p><p>0.35</p><p>0.4</p><p>0.45</p><p>0.5</p><p>0.55</p><p>0.6</p><p>0.65</p><p>compression ratio</p><p>th</p><p>er</p><p>m</p><p>al</p><p>e</p><p>ffi</p><p>ci</p><p>en</p><p>cy</p><p>Fuel−Air Otto</p><p>Gas Otto</p><p>Figure 4.3: Problem 4-17 plots</p><p>17</p><p>5 10 15 20</p><p>0.8</p><p>0.85</p><p>0.9</p><p>compression ratio</p><p>et</p><p>a</p><p>ra</p><p>tio</p><p>Figure 4.4: Problem 4-17 plots</p><p>18 CHAPTER 4. FUEL-AIR COMBUSTION PROCESSES</p><p>4.14) What compression ratio is required to have an imep of 1500 kPa with a methane fuel-air Otto cycle,</p><p>assuming T1 = 325 K, φ = 0.95, f = 0.05, and P1 = 101.3 kPa ?</p><p>Increasing the compression ratio r in the file OttoFuel.m to 18 will result in an imep = 1502 kPa.</p><p>Note that at this compression ratio the engine will be highly susceptible to knock.</p><p>Ottofuel input conditions: phi= 0.95 fuel= 1</p><p>State 1 2 3 4</p><p>Pressure (kPa)= 101.3 4808.9 15758.6 463.8</p><p>Temperature (K)= 325.0 881.6 2860.3 1487.8</p><p>Enthalpy(kJ/kg)= -347.8 308.2 910.4 -1362.7</p><p>Int.Energy(kJ/kg)= -445.4 43.5 43.0 -1809.5</p><p>Volume (m^3/kg) = 0.963 0.055 0.055 0.963</p><p>Cp (kJ/kg K) = 1.081 1.286 2.209 1.418</p><p>Work (kJ/kg)= 1364.1</p><p>Efficiency= 0.521</p><p>Imep (kPa)= 1501.7</p><p>19</p><p>4.15) What value of the equivalence ratio will maximize the imep for a gasoline fuel-air Otto cycle with a</p><p>compression ratio of 10? Assume T1 = 350 K, f = 0.05, and P1 = 101.3 kPa.</p><p>Revising Ottofuel.m by varying φ with a for-loop and adding a test comparing current and previous</p><p>imep values as shown below, the maximum value of imep is found to be 1425 kPa at φ= 1.04, slightly</p><p>rich.</p><p>clear;</p><p>T1 = 350; %Kelvin</p><p>P1 = 101.3; %kPa</p><p>%phi = 0.95; %equivalence ratio</p><p>f= .05; %residual fraction,</p><p>rc=10; %compression ratio</p><p>imepold=200;</p><p>for phi = 0.9:0.0025:1.2 %phi for-loop</p><p>.....</p><p>.....</p><p>imep=w/(v1-v2); %imep</p><p>test=imep-imepold; % value still increasing with positive test</p><p>imepold=imep; %reset imepold to latest value</p><p>if test</p><p>engine condi-</p><p>tions are the same as in Example 4.8. Using the program Homogeneous.m, a.) compute the pressure</p><p>and burned and unburned zone temperatures, imep, and thermal efficiency. b.) While keeping other</p><p>parameters constant, compute and plot the effect of varying the spark advance (vary θs from 0 to</p><p>-40◦ atdc), the equivalence ratio ( vary φ from 0.7 to 1.3), and the compression ratio (vary r from 5</p><p>to 15). Compare and discuss the computational results with the experimental results given in Section</p><p>4.7.</p><p>a.) For the CFR engine base case conditions, the thermal efficiency η = 0.379, and the imep = 1258</p><p>kPa. The pressure and temperature profiles are given below:</p><p>−100 −50 0 50 100</p><p>0</p><p>1000</p><p>2000</p><p>3000</p><p>4000</p><p>5000</p><p>6000</p><p>7000</p><p>8000</p><p>θ</p><p>P</p><p>re</p><p>ss</p><p>ur</p><p>e</p><p>(k</p><p>P</p><p>a)</p><p>Figure 4.9: Problem 4-17 Pressure profile</p><p>−100 −50 0 50 100</p><p>0</p><p>500</p><p>1000</p><p>1500</p><p>2000</p><p>2500</p><p>3000</p><p>θ</p><p>T</p><p>em</p><p>pe</p><p>ra</p><p>tu</p><p>re</p><p>(</p><p>K</p><p>)</p><p>Unburned</p><p>Burned</p><p>Figure 4.10: Problem 4-19 Temperature profile</p><p>b.) The Tables below list the effect of compression ratio, equivalence ratio, and spark angle. Analysis</p><p>26 CHAPTER 4. FUEL-AIR COMBUSTION PROCESSES</p><p>of the results indicate 1.) that the thermal efficiency increases with increasing compression ratio, and</p><p>is maximum for lean combustion; 2.) the imep also increases with increasing compression ratio, and</p><p>is maximized slightly rich of stoichiometric; and 3.) for a given combustion duration and other engine</p><p>parameters, there is an optimum spark advance that will optimize efficiency and imep.</p><p>r η imep</p><p>(kPa)</p><p>5 0.280 1072</p><p>6 0.314 1123</p><p>8 0.352 1200</p><p>10 0.380 1258</p><p>12 0.401 1304</p><p>14 0.417 1341</p><p>15 0.424 1357</p><p>Effect of compression ratio r on thermal efficiency and imep</p><p>φ η imep</p><p>(kPa)</p><p>0.7 0.405 945</p><p>0.8 0.399 1061</p><p>0.9 0.392 1172</p><p>1.0 0.380 1258</p><p>1.1 0.349 1269</p><p>1.2 0.316 1249</p><p>1.3 0.287 1227</p><p>Effect of equivalence ratio φ on thermal efficiency and imep</p><p>θs η imep</p><p>(kPa)</p><p>-40 0.372 1232</p><p>-35 0.380 1258</p><p>-30 0.383 1270</p><p>-25 0.382 1251</p><p>-15 0.368 1221</p><p>-10 0.356 1179</p><p>0 0.332 1069</p><p>Effect of spark angle θs on thermal efficiency and imep</p><p>27</p><p>4.20) Derive Equations (4.70),(4.74), (4.83) for the limited pressure fuel air cycle.</p><p>a.) We need to compute the thermodynamic properties at the end of the compression stroke from 1-2</p><p>of the limited pressure cycle. The gas mass is a mixture of a) residual gas composed of combustion</p><p>products and b) fresh air. We need to account for both the unburned air in the residual gases, and in</p><p>the fresh air in the intake mixture. The fuel-air ratio that produces the same amount of combustion</p><p>products as an equivalent fuel-air combustion process is</p><p>FA12 =</p><p>mf,r</p><p>ma +ma,r</p><p>where ma,r is the mass of air and mf,r is the mass of fuel used to create the residuals. The actual</p><p>fuel-air ratio FA expressed in terms of the residuals is</p><p>FA =</p><p>mf,r</p><p>ma,r</p><p>= φFAs</p><p>The mass of the residuals is the sum of the fuel and air mass that reacted to form the residuals,</p><p>mr = mf,r +ma,r</p><p>So</p><p>mf,r =</p><p>mrφFAs</p><p>φFAs + 1</p><p>ma,r =</p><p>mr</p><p>φFAs + 1</p><p>The residual fraction f at the end of the intake stroke is</p><p>f =</p><p>mr</p><p>ma +mr</p><p>and</p><p>1− f =</p><p>ma</p><p>ma +mr</p><p>Upon substitution,</p><p>FA12 =</p><p>mrφFAs</p><p>[ma(φ FAs + 1) +mr</p><p>=</p><p>f φ FAs</p><p>1 + (1− f)φ FAs</p><p>φ12 =</p><p>FA12</p><p>FAs</p><p>=</p><p>f φ</p><p>1 + (1− f)φFAs</p><p>Equation (4.70)</p><p>b.) The mass of fuel injected is mf , so the amount of mass in the cylinder at the end of the fuel</p><p>injection process, state 3, is</p><p>m3 = ma +mr +mf</p><p>The fuel-air ratio FA for the combustion process is</p><p>FA =</p><p>mf</p><p>ma</p><p>= φ FAs</p><p>Upon substitution,</p><p>mf</p><p>m3</p><p>=</p><p>φFAsma</p><p>ma +mr + φ FAsma</p><p>=</p><p>φFAs (1− f)</p><p>1 + (1− f)φ FAs</p><p>(Equation 4.74)</p><p>c.) Equation (4.83) follows by substitution of Equation (4.74) into the expression</p><p>η =</p><p>W</p><p>Q</p><p>=</p><p>wcv m3</p><p>mf qc</p>