Engineering_Mechanics_Colin_R_Ferguson_A_Solved_Ch3

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<p>Chapter 3</p><p>Fuel, Air, and Combustion</p><p>Thermodynamics</p><p>3.1) What is the molecular weight, enthalpy (kJ/kg), and entropy (kJ/kg K) of a gas mixture at P = 1000</p><p>kPa and T = 500 K, if the mixture contains the following species and mole fractions?</p><p>a) A table of the given and computed parameters is:</p><p>i yi Mi ho</p><p>f hi − ho</p><p>f soi yi</p><p>(</p><p>soi − Ru ln(yi)</p><p>)</p><p>[kg/kmol] [MJ/kmol] [MJ/kmol] [kg/kmol−K] [kJ/kmol−K]</p><p>CO2 1 0.10 44.01 -393.52 33.40 269.30 28.84</p><p>H2O 2 0.15 18.01 -241.83 26.00 232.74 37.28</p><p>N2 3 0.70 28.01 0 21.46 228.17 161.79</p><p>CO 4 0.05 28.01 -110.53 21.69 234.54 12.97</p><p>The mixture molecular mass is:</p><p>M =</p><p>∑</p><p>yiMi = (0.10)(44.01) + (0.15)(18.01) + (0.70)(28.01) + (0.05)(28.01)</p><p>M = 28.11 kg/kmol</p><p>The specific mixture enthalpy is:</p><p>hf =</p><p>∑</p><p>yihi = (0.10)(−393.52+ 33.40) + (0.15)(−241.83+ 26.00)</p><p>+ (0.70)(0 + 21.46) + (0.05)(−110.53+ 21.69)</p><p>hf = −57.8MJ/kmol = −5.78× 104 kJ/kmol</p><p>The mixture enthalpy is:</p><p>h =</p><p>h</p><p>m</p><p>=</p><p>−5.78× 104</p><p>28.11</p><p>h = −2056 kJ/kg</p><p>The specific mixture entropy is:</p><p>s = −Ru ln</p><p>(</p><p>P</p><p>P0</p><p>)</p><p>+ yi (s</p><p>o</p><p>i −Ru ln(yi))</p><p>s = (−8.314) ln</p><p>(</p><p>1000</p><p>100</p><p>)</p><p>+ 28.84 + 37.28 + 161.79 + 12.97</p><p>s = 221.74 kJ/kmol−K</p><p>1</p><p>2 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS</p><p>The mixture entropy is:</p><p>s =</p><p>s</p><p>m</p><p>=</p><p>103.23</p><p>28.11</p><p>s = 7.888 kJ/kg−K</p><p>3</p><p>3.2) What is the enthalpy h (kJ/kg) and entropy s (kJ/kg-K) of a mixture of 30% H2 and 70% CO2 by</p><p>volume at a temperature of 3000 K ?</p><p>a) A table of the given and computed parameters is:</p><p>i yi Mi ho</p><p>f hi − ho</p><p>f soi yi</p><p>(</p><p>soi − Ru ln(yi)</p><p>)</p><p>[kg/kmol] [MJ/kmol] [MJ/kmol] [kg/kmol−K] [kJ/kmol−K]</p><p>CO2 1 0.70 44.01 -393.52 152.85 334.17 235.99</p><p>H2 2 0.30 2.016 0 88.72 202.90 63.87</p><p>The mixture molecular mass is:</p><p>M =</p><p>∑</p><p>yiMi = (0.70)(44.01) + (0.30)(2.016)</p><p>M = 31.41 kg/kmol</p><p>The specific mixture enthalpy is:</p><p>hf =</p><p>∑</p><p>yihi = (0.70)(−393.52+ 152.85) + (0.30)(0 + 88.73)</p><p>= −14.8MJ/kmol = −1.418× 105 kJ/kmol</p><p>The mixture enthalpy is:</p><p>h =</p><p>h</p><p>m</p><p>=</p><p>−1.418× 105</p><p>31.41</p><p>h = −4516 kJ/kg</p><p>The specific mixture entropy is:</p><p>s = −Ru ln</p><p>(</p><p>P</p><p>P0</p><p>)</p><p>+ yi (s</p><p>o</p><p>i −Ru ln(yi))</p><p>s = (−8.314) ln</p><p>(</p><p>2000</p><p>100</p><p>)</p><p>+ 235.99 + 63.87</p><p>s = 296.86 kJ/kmol−K</p><p>The mixture entropy is:</p><p>s =</p><p>s</p><p>m</p><p>=</p><p>296.86</p><p>31.41</p><p>s = 9.451 kJ/kg−K</p><p>4 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS</p><p>3.3) Using the Gordon and McBride equations, Equations (3.38) and (3.39), calculate the enthalpy h̄ and</p><p>entropy s̄o of CO2 and compare with the gas table values used in Example 3.1.</p><p>A Matlab program for calculating the enthalpy h̄ and entropy s̄o of CO2 is</p><p>t=1000; % temp in K</p><p>R=8.31451 % univ. gas const.</p><p>a1=2.4007797;</p><p>a2=8.73509757e-3;</p><p>a3=-6.60707878e-6;</p><p>a4=2.0021861e-9;</p><p>a5=6.3274039e-16;</p><p>a6=-4.8377527e4;</p><p>a7=9.6951457;</p><p>nondimh=a1+a2/2*t+a3/3*t^2+a4/4*t^3+a5/5*t^4+a6/t</p><p>h=nondimh*t*R</p><p>nondims=a1*log(t)+a2*t+a3/2*t^2+a4/3*t^3+a5/4*t^4+a7</p><p>s=nondims*R</p><p>The calculated h̄ = −3.6011 × 105 kJ/kmol and the entropy s̄o = 269.21 kJ/kmol-K. The gas table</p><p>values are h̄ = −3.6012× 105 kJ/kmol and the entropy s̄o = 269.30 kJ/kmol-K. The enthalpy values</p><p>agree to 4 figures, and the entropy values agree to 3 figures.</p><p>5</p><p>3.4) Using the program Fuel.m, at what temperature is the specific heat cp of methane CH4 = 3.0 kJ/kg-K ?</p><p>The program Fuel.m was modified to print out the cp values for methane with fuel id=1 and R =</p><p>0.5183 kJ/kg-K. The variable cpp is the dimensional specific heat, calculated using the formula:</p><p>cp = R× (a1 + a2T + a3T</p><p>2 + a4T</p><p>3 + a5T</p><p>4)</p><p>By trial and error, the temperature is found to be 521.1 K.</p><p>The modified program is listed below:</p><p>%function [ alpha, beta, gamma, delta, h, s, cp, mw, Fs, q ] = fuel3_4( id, T )</p><p>% [ alpha, beta, gamma, delta, h, s, cp, mw, Fs, q ] = fuel( id, T )</p><p>%</p><p>% Parameters</p><p>% id - 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</p><p>% T - Temperature (K) at which to eval 300</p><p>efficiency ?</p><p>a) The mass of fuel entering each cylinder per cycle for a four stroke engine</p><p>mf = ṁf</p><p>(</p><p>2</p><p>N</p><p>)(</p><p>1</p><p>nc</p><p>)</p><p>= 2.5</p><p>(</p><p>2 · 60</p><p>2000</p><p>)(</p><p>1</p><p>4</p><p>)</p><p>mf = 3.75× 10−2 g</p><p>b) Since the engine is port injected</p><p>ev =</p><p>ma +mf</p><p>ρivd</p><p>=</p><p>mf (AF + 1)</p><p>ρivd</p><p>=</p><p>mf (</p><p>AFs</p><p>φ + 1)</p><p>ρivd</p><p>From Table 3.5, the stoichiometric air-fuel ratio for octane is AFs = 15.03. Assume R = 0.287</p><p>ρi =</p><p>P</p><p>RT</p><p>=</p><p>101</p><p>(0.287)(298)</p><p>= 1.18 kg/m3 = 1180 g/m3</p><p>Solving for the volumetric efficiency:</p><p>ηv =</p><p>(3.75× 10−2)</p><p>(</p><p>15.03</p><p>0.9 + 1</p><p>)</p><p>(1180)</p><p>(</p><p>2.8 × 10−3</p><p>4</p><p>)</p><p>ηv = 0.80</p><p>10 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS</p><p>3.8) An engine cylinder has a 90 mm bore and a 85 mm stroke, and contains air and residual gases at 350</p><p>K and 1 bar. If the engine is to operate on diesel fuel and run lean with an overall equivalence ratio</p><p>of φ = 0.7, what is the mass of diesel fuel that needs to be injected during the compression stroke ?</p><p>(Assume R of the air-residual gas mixture is 0.29 kJ/kg K).</p><p>a)</p><p>m = ma +mr</p><p>We know that</p><p>f =</p><p>mr</p><p>m</p><p>Substituting we have</p><p>ma = (1− f)m</p><p>Since</p><p>φ =</p><p>FA</p><p>FAs</p><p>Substituting</p><p>mf = φ(FAs)(ma) = φ(FAs)(1− f)m</p><p>Using the ideal gas law</p><p>mf =</p><p>Pvd</p><p>RT</p><p>=</p><p>P</p><p>(</p><p>π</p><p>4</p><p>)</p><p>b2s</p><p>RT</p><p>=</p><p>100</p><p>(</p><p>π</p><p>4</p><p>)</p><p>(0.09)2(0.085)</p><p>(0.29)(350)(1000)</p><p>mf = 0.53 g</p><p>From Table 3.5, AFs=14.30 or FAs=0.0699</p><p>The mass of injected diesel fuel is:</p><p>mf = (0.7)(0.0699)(1− 0.015)(0.53)</p><p>mf = 2.5× 10−2 g</p><p>11</p><p>3.9) Using the low temperature (T</p><p>ratio input</p><p>T = 2000; % enter temperature (K) input</p><p>P = 750; % enter pressure (kPa) input</p><p>fuel_id =3;</p><p>% fuel_id - 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</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>.....</p><p>phi = 1.1; % enter equivalence ratio input</p><p>T = 2000; % enter temperature (K) input</p><p>P = 750; % enter pressure (kPa) input</p><p>fuel_id =3;</p><p>% fuel_id - 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</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>.....</p><p>Equilibrium Combustion Solver</p><p>Pressure (kPa) = 750.0</p><p>Temperature (K) = 2000.0</p><p>Fuel Air Equivalence ratio = 1.1</p><p>Mole Fractions</p><p>CO2 = 1.1647e-01</p><p>H2O = 1.2059e-01</p><p>N2 = 7.2421e-01</p><p>O2 = 3.1695e-06</p><p>CO = 3.1380e-02</p><p>H2 = 7.1772e-03</p><p>H = 5.0455e-05</p><p>O = 4.3468e-07</p><p>OH = 8.4277e-05</p><p>NO = 3.0302e-05</p><p>Mixture Properties</p><p>h(kJ/kg) = -569.9</p><p>u(kJ/kg) = -1153.7</p><p>s (kJ/Kg K) = 8.7980</p><p>v (m3/kg) = 0.7784</p><p>cp (kJ/Kg K) = 1.448</p><p>Molecular Mass = 28.48</p><p>dvdt = 3.8981e-04</p><p>dvdp = -1.0379e-03</p><p>b.) For an equivalence ratio of 1.25, the program output is</p><p>Equilibrium Combustion Solver</p><p>17</p><p>Pressure (kPa) = 750.0</p><p>Temperature (K) = 2000.0</p><p>Fuel Air Equivalence ratio = 1.2</p><p>Mole Fractions</p><p>CO2 = 9.0113e-02</p><p>H2O = 1.1899e-01</p><p>N2 = 6.9799e-01</p><p>O2 = 3.6224e-07</p><p>CO = 7.1816e-02</p><p>H2 = 2.0948e-02</p><p>H = 8.6198e-05</p><p>O = 1.4695e-07</p><p>OH = 4.8675e-05</p><p>NO = 1.0057e-05</p><p>Mixture Properties</p><p>h(kJ/kg) = -393.7</p><p>u(kJ/kg) = -993.6</p><p>s (kJ/Kg K) = 8.9958</p><p>v (m3/kg) = 0.7999</p><p>cp (kJ/Kg K) = 1.463</p><p>Molecular Mass = 27.72</p><p>dvdt = 4.0040e-04</p><p>dvdp = -1.0665e-03</p><p>c.) As the equivalence ratio increases, and the mixture becomes more rich, there is less oxygen in</p><p>the reaction. With the exception of CO, the oxygen containing species, CO2, H2O, O2, O, OH, NO</p><p>decrease in product concentration. The hydrogen atoms and molecules increase in concentration, to</p><p>balance the H2O decrease. The enthalpy, entropy, and cp increase since the extent of reaction is less.</p><p>18 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS</p><p>3.15) Using the program ecp.m, plot the product equilibrium mole fractions as a function of equivalence</p><p>ratio (0.5 3.125, there will be solid carbon in the products, since the carbon atoms are in</p><p>excess of those used to form CO.</p><p>21</p><p>3.18) At what temperature is the concentration of H2 a minimum for the combustion of gasoline and air at</p><p>φ = 1.2 and 4500 kPa? What is that minimum value of H2 ?</p><p>As shown below, the RunEcp program was modified to compute the mole fractions over a range of</p><p>temperature, plot the H2 mole fraction, and search and print out the minimum H2. The minimum</p><p>mole fraction is found to be yH2</p><p>= 0.0171 at a temperature of 2667 K. This minimum occurs in rich</p><p>mixtures only, and is due to the temperature dependent competing reactions for the formation of H2O</p><p>and H2.</p><p>1500 2000 2500 3000</p><p>0.016</p><p>0.018</p><p>0.02</p><p>0.022</p><p>0.024</p><p>0.026</p><p>0.028</p><p>Temperature (K)</p><p>H</p><p>2 m</p><p>ol</p><p>e</p><p>fr</p><p>ac</p><p>tio</p><p>n</p><p>Figure 3.2: Problem 3-18 H2 - Temperature plot</p><p>The modified RunEcp program is:</p><p>clear;</p><p>phi = 1.2; % enter equivalence ratio input</p><p>%T = 2800; % enter temperature (K) input</p><p>P = 4500; % 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>imax=100;</p><p>h2=zeros(1,imax);</p><p>temp=linspace(1500, 3000, imax);</p><p>for i=1:imax</p><p>T=temp(i);</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>h2(i)=Y(6);</p><p>end</p><p>[h2min,id]=min(h2);%find minimum</p><p>tempmin=temp(id);</p><p>fprintf(’ H2min = \t %1.4e at T(K) = %6.1f \n’, h2min,tempmin);</p><p>plot(temp, h2,’-’,’linewidth’,2);%plot concentration</p><p>set(gca,’fontsize’,18,’linewidth’,1.5);</p><p>xlabel(’Temperature (K)’,’fontsize’, 18);</p><p>ylabel(’H_2 mole fraction’,’fontsize’, 18);</p><p>22 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS</p><p>3.19) At what equivalence ratio is the concentration of OH a maximum for the combustion of diesel and air</p><p>at T = 2500 K and 4500 kPa? What is that maximum value of OH ?</p><p>As shown below, the RunEcp program was modified to compute the mole fractions over a range of</p><p>equivalence ratios, plot the OH mole fraction, and search and print out the maximum OH. The maxi-</p><p>mum mole fraction is found to be OH = 0.00363 at an equivalence ratio of 0.65. This maximum occurs</p><p>in lean mixtures only, and is due to the phi-dependent competing reactions for the formation of CO2</p><p>and H2O.</p><p>0 0.5 1 1.5</p><p>0.5</p><p>1</p><p>1.5</p><p>2</p><p>2.5</p><p>3</p><p>3.5</p><p>4</p><p>x 10</p><p>−3</p><p>Equivalence ratio</p><p>O</p><p>H</p><p>m</p><p>ol</p><p>e</p><p>fr</p><p>ac</p><p>tio</p><p>n</p><p>Figure 3.3: Problem 3-19 OH - Equivalence ratio plot</p><p>The modified RunEcp program is:</p><p>clear;</p><p>T = 2500; % enter temperature (K) input</p><p>P = 4500; % enter pressure (kPa) input</p><p>fuel_id =3;</p><p>% fuel_id - 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</p><p>imax=100;</p><p>oh=zeros(1,imax);</p><p>test=linspace(.1, 1.3, imax);</p><p>for i=1:imax</p><p>phi=test(i);</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>oh(i)=Y(9);</p><p>end</p><p>[ohmax,id]=max(oh);%find maximum</p><p>phimax=test(id);</p><p>fprintf(’ H2min = \t %1.4e at phi = %6.2f \n’, ohmax,phimax);</p><p>plot(test, oh,’-’,’linewidth’,2);%plot concentration</p><p>set(gca,’fontsize’,18,’linewidth’,1.5);</p><p>xlabel(’Equivalence ratio’,’fontsize’, 18);</p><p>ylabel(’OH mole fraction’,’fontsize’, 18);</p><p>23</p><p>3.20) At what temperature does the mole fraction of NO reach 0.010 for the equilibrium products resulting</p><p>from the combustion of gasoline and air at φ = 1.0 and 5000 kPa?</p><p>As shown below, the RunEcp program was modified to compute the NO mole fractions over a range</p><p>of temperature, plot the NO mole fraction, and search and print out the temperature where NO =</p><p>0.010. The temperature is found to be 2934 K. As shown in the plot, the NO mole fraction increases</p><p>monotonically with temperature.</p><p>2000 2200 2400 2600 2800 3000</p><p>0</p><p>0.002</p><p>0.004</p><p>0.006</p><p>0.008</p><p>0.01</p><p>0.012</p><p>Temperature (K)</p><p>N</p><p>O</p><p>m</p><p>ol</p><p>e</p><p>fr</p><p>ac</p><p>tio</p><p>n</p><p>Figure 3.4: Problem 3-20 NO - Temperature plot</p><p>The modified RunEcp program is:</p><p>phi = 1.0; % enter equivalence ratio input</p><p>P = 5000; % enter pressure (kPa) input</p><p>fuel_id =2; % fuel_id - 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</p><p>imax=100;</p><p>no=zeros(1,imax);</p><p>temp=linspace(2000, 3000, imax);</p><p>for i=1:imax</p><p>T=temp(i);</p><p>[ierr, Y, h, u, s, v, R, Cp, MW, dvdT, dvdP] = ecp( T, P, phi, fuel_id ); % call ecp function</p><p>no(i)=Y(10);</p><p>end</p><p>for i=1:imax %find the 0.01 value</p><p>nox=no(i);</p><p>tt=temp(i);</p><p>if nox > 0.01 break</p><p>end</p><p>end</p><p>fprintf(’ NO = \t %1.4e at T(K) = %6.1f \n’, nox,tt);</p><p>plot(temp, no,’-’,’linewidth’,2);%plot concentration</p><p>set(gca,’fontsize’,18,’linewidth’,1.5);</p><p>xlabel(’Temperature (K)’,’fontsize’, 18);</p><p>ylabel(’NO mole fraction’,’fontsize’, 18);</p><p>24 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS</p><p>3.21) At what temperature does the mole fraction of CO reach 0.080 for the equilibrium products resulting</p><p>from the combustion of methane and air at φ = 1.1 and 3000 kPa?</p><p>The equivalence ratio of the problem was increased to φ = 1.40 to increase the CO mole fraction and</p><p>keep T ¡ 3000 K, which is the maximum temperature for the NASA property curve fit. As shown below,</p><p>the RunEcp program was modified to compute the CO mole fractions over a range of temperature, plot</p><p>the CO mole fraction, and search and print out the temperature where CO = 0.080. The temperature</p><p>is found to be 2667 K. As shown in the plot, the CO mole fraction increases monotonically with</p><p>temperature.</p><p>1500 2000 2500 3000</p><p>0.06</p><p>0.065</p><p>0.07</p><p>0.075</p><p>0.08</p><p>0.085</p><p>Temperature (K)</p><p>C</p><p>O</p><p>m</p><p>ol</p><p>e</p><p>fr</p><p>ac</p><p>tio</p><p>n</p><p>Figure 3.5: Problem 3-21 CO - Temperature plot</p><p>The modified RunEcp program is:</p><p>phi = 1.4; % enter equivalence ratio input</p><p>P = 3000; % enter pressure (kPa) input</p><p>fuel_id =1; % fuel_id - 1=Methane, 2=Gasoline, 3=Diesel, 4=Methanol, 5=Nitromethane</p><p>imax=100;</p><p>co=zeros(1,imax);</p><p>temp=linspace(1500, 3000, imax);</p><p>for i=1:imax</p><p>T=temp(i);</p><p>[ierr, Y, h, u, s, v, R, Cp, MW, dvdT, dvdP] = ecp( T, P, phi, fuel_id ); % call ecp function</p><p>co(i)=Y(5);</p><p>end</p><p>for i=1:imax %find the 0.08 value</p><p>cox=co(i);</p><p>tt=temp(i);</p><p>if cox > 0.08 break</p><p>end</p><p>end</p><p>fprintf(’ CO = \t %1.4e at T(K) = %6.1f \n’, cox,tt);</p><p>plot(temp, co,’-’,’linewidth’,2);%plot concentration</p><p>set(gca,’fontsize’,18,’linewidth’,1.5);</p><p>xlabel(’Temperature (K)’,’fontsize’, 18);</p><p>ylabel(’CO mole fraction’,’fontsize’, 18);</p><p>25</p><p>3.22) What is the equilibrium and the frozen specific heat cp of the combustion products of gasoline at a</p><p>pressure of 2000 kPa and temperature of 2000 K burned at a.) an equivalence ratio of 1.1, and b.) an</p><p>equivalence ratio of 0.9 ?</p><p>The mixture specific heat is</p><p>cp =</p><p>1</p><p>M</p><p>[</p><p>n</p><p>∑</p><p>i=1</p><p>yic̄pi +</p><p>n</p><p>∑</p><p>i=1</p><p>h̄i</p><p>∂yi</p><p>∂T</p><p>−</p><p>h̄</p><p>M</p><p>n</p><p>∑</p><p>i=1</p><p>Mi</p><p>∂yi</p><p>∂T</p><p>]</p><p>The frozen specific heat cp,f is</p><p>cp,f =</p><p>1</p><p>M</p><p>n</p><p>∑</p><p>i=1</p><p>yic̄pi</p><p>As shown below, the RunEcp program was used to compute the mixture specific heat cp . The section</p><p>of the eco.m program that calculates the mixture properties was modified to include a calculation of</p><p>the frozen specific heat cp,f</p><p>At φ = 0.9, the mixture specific heat ( in units of kJ/kg-K) is cp = 1.4904, and cp,f = 1.4286, about</p><p>4% less. At φ = 1.1, the mixture specific heat ( in units of kJ/kg-K) is cp = 1.4732, and cp,f = 1.4585,</p><p>about 1% less. As one varies φ, one finds that the frozen specific heat increases with φ, however, the</p><p>mixture specific heat variation is more complex.</p><p>The RunEcp program is:</p><p>clear;</p><p>phi = 0.9; % enter equivalence ratio input</p><p>T = 2000; % enter temperature (K) input</p><p>P = 2000; % 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>% call ecp function</p><p>[ierr, Y, h, u, s, v, R, Cp, MW, dvdT, dvdP] = ecp( T, P, phi, fuel_id );</p><p>.....</p><p>The section of the ecp.m program that was modified is shown below. Note that the program line for</p><p>cpfroz has no semi-colon, so that the value of that variable is printed directly to the screen, without</p><p>having to modify the i/o portion of the program.</p><p>% compute cp,h,s</p><p>% initialize h, etc to zero</p><p>MW = 0;</p><p>Cp = 0;</p><p>Cpfroz=0;</p><p>h = 0;</p><p>s = 0;</p><p>dMWdT = 0;</p><p>dMWdP = 0;</p><p>for i=1:10, %sum over product species</p><p>cpo = AAC(i,1) + AAC(i,2)*T + AAC(i,3)*T^2 + AAC(i,4)*T^3 + AAC(i,5)*T^4;</p><p>ho = AAC(i,1) + AAC(i,2)/2*T + AAC(i,3)/3*T^2 + AAC(i,4)/4*T^3 +AAC(i,5)/5*T^4 + AAC(i,6)/T;</p><p>so = AAC(i,1)*log(T) + AAC(i,2)*T + AAC(i,3)/2*T^2 + AAC(i,4)/3*T^3 +AAC(i,5)/4*T^4 +AAC(i,7);</p><p>h = h + ho*Y(i); % h is dimensionless h/RT here</p><p>26 CHAPTER 3. FUEL, AIR, AND COMBUSTION THERMODYNAMICS</p><p>MW = MW + Mi(i)*Y(i);</p><p>dMWdT = dMWdT + Mi(i)*dydt(i);</p><p>dMWdP = dMWdP + Mi(i)*dydp(i);</p><p>Cp = Cp+Y(i)*cpo + ho*T*dydt(i); % Cp is dimensionless Cp/R here</p><p>Cpfroz = Cpfroz +Y(i)*cpo; % frozen Cp/R</p><p>if (Y(i)> 1.0e-37)</p><p>s = s + Y(i)*(so - log(Y(i)));</p><p>end</p><p>end</p><p>%</p><p>R = 8.31434/MW; %kJ/kgK</p><p>v = R*T/P;</p><p>Cp = R*(Cp - h*T*dMWdT/MW); %kJ/Kg K</p><p>Cpfroz = R* Cpfroz % kJ/KgK</p><p>h = h*R*T;</p><p>s = R*(-log(PATM) + s);</p><p>u=h-R*T;</p><p>dvdT = v/T*(1 - T*dMWdT/MW);</p><p>dvdP = v/P*(-1 + P*dMWdP/MW);</p>