Chapter 16.6, Problem 37P

(a)

To determine

The equilibrium composition of product gases.

Answer to Problem 37P

Thus, the equilibrium composition of mixture of ${\text{CO}}_{\text{2}}$, $\text{CO}$ and ${\text{O}}_{\text{2}}$ at 16 psia pressure and 3600 R is $\underset{\_}{0.9966{\text{CO}}_{\text{2}}+0.0034\text{CO}+0.1587{\text{O}}_{\text{2}}}$.

Explanation of Solution

Write the expression for the volume of oxygen used per lbmol of carbon monoxide $\left(v_{\text{CO}}\right)$.

$v_{\text{CO}}=\frac{RT}{P}$ (I)

Here, gas constant is R, temperature is T, and pressure is P.

Calculate the mass flow rate of carbon monoxide $\left({\dot{m}}_{\text{CO}}\right)$.

${\dot{m}}_{\text{CO}}=\frac{{\dot{v}}_{\text{CO}}}{v_{\text{CO}}}$ (II)

Here, volume flow rate of carbon monoxide is ${\dot{v}}_{\text{CO}}$.

Calculate the molar air fuel ratio $\left(\text{AF}\right)$.

$\text{AF}=\frac{N_{{\text{O}}_{\text{2}}}}{N_{\text{CO}}}$

$\text{AF}=\frac{{\dot{m}}_{{\text{O}}_{\text{2}}}/M_{{\text{O}}_{\text{2}}}}{{\dot{m}}_{\text{CO}}/M_{\text{CO}}}$ (III)

Here, number of moles of oxygen is $N_{{\text{O}}_{\text{2}}}$, number of moles of fuel is $N_{\text{fuel}}$, mass flow rate of oxygen is ${\dot{m}}_{{\text{O}}_{\text{2}}}$, molecular weight of oxygen is $M_{{\text{O}}_{\text{2}}}$, mass flow rate of fuel is ${\dot{m}}_{\text{CO}}$, and molecular weight of fuel is $M_{\text{CO}}$.

Express the stoichiometric reaction for the dissociation process.

${\text{CO}}_{\text{2}}\rightleftharpoons \text{CO}+\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}$ (IV)

From the stoichiometric reaction, infer that the stoichiometric coefficient for carbon monoxide $\left(v_{\text{CO}}\right)$ is 1, for oxygen $\left(v_{{\text{O}}_{\text{2}}}\right)$ is 0.5, and for carbon dioxide $\left(v_{{\text{CO}}_{\text{2}}}\right)$ is 1.

Express the actual reaction for the dissociation process.

$\text{CO}+0.657{\text{O}}_{\text{2}}\to x{\text{CO}}_{\text{2}}+\left(1-x\right)\text{CO}+\left(0.657-0.5x\right){\text{O}}_{\text{2}}$ (V)

From the actual reaction, infer that the equilibrium composition contains x amount of carbon dioxide $\left(N_{{\text{CO}}_2}\right)$, $\left(1-x\right)$ amount of carbon monoxide $\left(N_{\text{CO}}\right)$, and $\left(0.657-0.5x\right)$ amount of nitrogen $\left(N_{{\text{O}}_{\text{2}}}\right)$.

Express the formula for total number of moles $\left(N_{total}\right)$.

$N_{total}=N_{{\text{CO}}_{\text{2}}}+N_{\text{CO}}+N_{{\text{O}}_{\text{2}}}$ (VI)

Here, number of moles of carbon dioxide is $N_{{\text{CO}}_{\text{2}}}$, number of moles of carbon monoxide is $N_{\text{CO}}$, and number of moles of oxygen is $N_{{\text{O}}_{\text{2}}}$.

Write the expression for the equilibrium constant $\left(K_p\right)$ for the dissociation process.

$K_p=\frac{N_{\text{CO}}^{v_{\text{CO}}}{N}_{{\text{O}}_{\text{2}}}^{v_{{\text{O}}_{\text{2}}}}}{N_{{\text{CO}}_{\text{2}}}^{v_{{\text{CO}}_{\text{2}}}}}{\left(\frac{P}{N_{total}}\right)}^{\left(v_{\text{CO}}+v_{{\text{O}}_{\text{2}}}-v_{{\text{CO}}_{\text{2}}}\right)}$ (VII)

Conclusion:

Substitute $0.3831\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}$ for R, 560 R for T, and 16 psia for P in Equation (I).

$\begin{array}{c}{v}_{\text{CO}}=\frac{\left(0.3831\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\right)\left(560\text{R}\right)}{16\text{psia}}\\ =13.41{\text{ft}}^3/\text{lbm}\end{array}$

Substitute $12.5{\text{ft}}^{\text{3}}/\min$ for ${\dot{v}}_{\text{CO}}$, and $13.41{\text{ft}}^3/\text{lbm}$ for $v_{\text{CO}}$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{\text{CO}}=\frac{12.5{\text{ft}}^{\text{3}}/\min }{13.41{\text{ft}}^3/\text{lbm}}\\ =0.932\text{lbm}/\min \end{array}$

Substitute $0.7\text{lbm}/\min$ for ${\dot{m}}_{{\text{O}}_{\text{2}}}$, $0.932\text{lbm}/\min$ for ${\dot{m}}_{\text{CO}}$, $32\text{lbm}/\text{lbmol}$ for $M_{{\text{O}}_{\text{2}}}$, and $28\text{lbm}/\text{lbmol}$ for $M_{\text{CO}}$ in Equation (III).

$\begin{array}{c}\text{AF}=\frac{\left(0.7\text{lbm}/\min \right)/\left(32\text{lbm}/\text{lbmol}\right)}{\left(0.932\text{lbm}/\min \right)/\left(28\text{lbm}/\text{lbmol}\right)}\\ =0.657{\text{lbmol O}}_{\text{2}}/\text{lbmol fuel}\end{array}$

Substitute xfor $N_{{\text{CO}}_{\text{2}}}$, $\left(1-x\right)$ for $N_{\text{CO}}$, and $\left(0.657-0.5x\right)$ for $N_{{\text{O}}_{\text{2}}}$ in Equation (VI).

$\begin{array}{c}{N}_{total}=x+\left(1-x\right)+\left(0.657-0.5x\right)\\ =1.657-0.5x\end{array}$

Convert the temperature unit from Rankine to Kelvin.

$\begin{array}{c}T=3600\text{R}\\ =3600\text{R}\left(\frac{1\text{ K}}{\text{1}\text{.8}\text{R}}\right)\\ =2000\text{K}\end{array}$

Refer table A-28, “natural logarithm of equilibrium constants”, select the value of $\ln k_p$ for carbon dioxide as $-6.635$ at 2000 K. Hence, the value of equilibrium constant $\left(K_p\right)$ for carbon dioxide dissociation process is $1.313\times {10}^{-3}$.

Substitute $1.313\times {10}^{-3}$ for $k_p$, x for $N_{{\text{CO}}_{\text{2}}}$, $\left(1-x\right)$ for $N_{\text{CO}}$, $\left(0.657-0.5x\right)$ for $N_{{\text{O}}_{\text{2}}}$, 0.5 for $v_{{\text{O}}_{\text{2}}}$, 1 for both $v_{{\text{CO}}_{\text{2}}}$ and $v_{\text{CO}}$, 16 psia for P, and $1.657-0.5x$ for $N_{total}$ in Equation (VII).

$\begin{array}{l}1.313\times {10}^{-3}=\frac{\left(1-x\right){\left(0.657-0.5x\right)}^{0.5}}{x}{\left(\frac{16\text{psia}\frac{1\text{atm}}{14.7\text{psia}}}{1.657-0.5x}\right)}^{\left(1+0.5-1\right)}\\ \left(1.313\times {10}^{-3}\right)x=\left(1-x\right){\left(0.657-0.5x\right)}^{0.5}{\left(\frac{1.088}{1.657-0.5x}\right)}^{0.5}\\ 1.7125\times {10}^{-6}x=\left(1+x^2-2x\right)\left(0.657-0.5x\right)\left(\frac{1.088}{1.657-0.5x}\right)\end{array}$

Solve the equation and find the value of x as 0.9966.

Substitute 0.9966 for x in Equation (V).

$\begin{array}{l}\text{CO}+0.657{\text{O}}_{\text{2}}\to x{\text{CO}}_{\text{2}}+\left(1-x\right)\text{CO}+\left(0.657-0.5x\right){\text{O}}_{\text{2}}\\ \text{CO}+0.657{\text{O}}_{\text{2}}\to 0.9966{\text{CO}}_{\text{2}}+\left(1-0.9966\right)\text{CO}+\left(0.657-0.5\left(0.9966\right)\right){\text{O}}_{\text{2}}\\ \text{CO}+0.657{\text{O}}_{\text{2}}\to 0.9966{\text{CO}}_{\text{2}}+0.0034\text{CO}+0.1587{\text{O}}_{\text{2}}\end{array}$

Thus, the equilibrium composition of mixture of ${\text{CO}}_{\text{2}}$, $\text{CO}$ and ${\text{O}}_{\text{2}}$ at 16 psia pressure and 3600 R is $\underset{\_}{0.9966{\text{CO}}_{\text{2}}+0.0034\text{CO}+0.1587{\text{O}}_{\text{2}}}$.

(b)

To determine

The rate of heat transfer from the combustion chamber

Answer to Problem 37P

The rate of heat transfer from the combustion chamber is $\underset{\_}{2,602\text{Btu}/\text{min}}$.

Explanation of Solution

Write the expression for the energy balance equation for the combustion process.

$\begin{array}{c}-Q_{out}={{\displaystyle \sum N_P\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}}_R\\ =\left\{\begin{array}{c}{N}_{{\text{CO}}_{\text{2}}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{{\text{CO}}_{\text{2}}}+N_{\text{CO}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{\text{CO}}\\ +N_{{\text{O}}_{\text{2}}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{{\text{O}}_{\text{2}}}-N_{\text{CO}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{\text{CO}}\end{array}\right\}\end{array}$ (VII)

Here, heat released during combustion is $Q_{out}$, number of moles of product is $N_P$, number of moles of reactants is $N_R$, number of moles of ${\text{CO}}_2$ is $N_{{\text{N}}_2}$, number of moles of ${\text{CO}}_2$ is $N_{{\text{CO}}_2}$, number of moles of $\text{CO}$ is $N_{\text{CO}}$, enthalpy at reference state is ${\overline{h}}_f^{\circ }$, sensible enthalpy at specified state is $\overline{h}$, and sensible enthalpy at reference state is ${\overline{h}}^{\circ }$.

Write the expression for the mass flow rate of CO $\left(\dot{N}\right)$.

${\dot{N}}_{\text{CO}}=\frac{{\dot{m}}_{\text{CO}}}{M_{\text{CO}}}$ (VIII)

Write the expression for the rate of heat transfer $\left({\dot{Q}}_{out}\right)$.

${\dot{Q}}_{out}={\dot{N}}_{\text{CO}}{Q}_{out}$ (IX)

Conclusion:

Refer Table A-26, “Enthalpy of formation, Gibbs function of formation, and absolute entropy at

778F, 1 atm”, select the enthalpy of $\text{CO}$$\left({\left({\overline{h}}_f^{\circ }\right)}_{\text{CO}}\right)$ as $-47,540\text{Btu}/\text{lbmol}$.

Refer Table A-21, “Ideal-gas properties of carbon monoxide”, obtain the following properties of $\text{CO}$ gas from the table ‘Ideal gas properties of carbon monoxide’ in the text book.

Enthalpy of $\text{CO}$ gas at 3580 K, ${\left(\overline{h}\right)}_{\text{CO,}\text{3580}}=27,954\text{Btu}/\text{lbmol}$.

Enthalpy of $\text{CO}$ gas at 3620 K, ${\left(\overline{h}\right)}_{\text{CO, 3620}}=28,300\text{Btu}/\text{lbmol}$.

Enthalpy of $\text{CO}$ gas at 298 K, ${\left({\overline{h}}^{\circ }\right)}_{\text{CO}}=3,725.1\text{Btu}/\text{lbmol}$.

Use interpolation to get the Enthalpy of water vapor at 3600 K $\left({\left(\overline{h}\right)}_{\text{CO, 3600}}\right)$.

$\frac{{\left(\overline{h}\right)}_{\text{CO, 3600}}-{\left(\overline{h}\right)}_{\text{CO,}\text{3580}}}{{\left(\overline{h}\right)}_{\text{CO, 3620}}-{\left(\overline{h}\right)}_{\text{CO,}\text{3580}}}=\frac{3600\text{K}-3580\text{K}}{3620\text{K}-3580\text{K}}$

Here, Enthalpy of $\text{CO}$ gas at 3580 K is ${\left(\overline{h}\right)}_{\text{CO,}\text{3580}}$, and Enthalpy of $\text{CO}$ gas at 3620 K is ${\left(\overline{h}\right)}_{\text{CO,}\text{3620}}$.

Substitute $27,954\text{Btu}/\text{lbmol}$ for ${\left(\overline{h}\right)}_{\text{CO,}\text{3580}}$, and $28,300\text{Btu}/\text{lbmol}$ for ${\left(\overline{h}\right)}_{\text{CO,}\text{3620}}$.

$\begin{array}{l}\frac{{\left(\overline{h}\right)}_{\text{CO, 3600}}-27,954\text{Btu}/\text{lbmol}}{28,300\text{Btu}/\text{lbmol}-27,954\text{Btu}/\text{lbmol}}=\frac{3600\text{K}-3580\text{K}}{3620\text{K}-3580\text{K}}\\ {\left(\overline{h}\right)}_{\text{CO, 3600}}=28,127\text{Btu}/\text{lbmol}\end{array}$

Refer Table A-26, “Enthalpy of formation, Gibbs function of formation, and absolute entropy at

778F, 1 atm”, select the enthalpy of ${\text{CO}}_2$$\left({\left({\overline{h}}_f^{\circ }\right)}_{{\text{CO}}_2}\right)$ as $-163,300\text{Btu}/\text{lbmol}$.

Refer Table A-20, ‘Ideal gas properties of carbon dioxide’ find out the following enthalpies at different temperature.

Enthalpy of ${\text{CO}}_{\text{2}}$ gas at 3580 K, ${\left(\overline{h}\right)}_{{\text{CO}}_{\text{2}}}=43,121\text{Btu}/\text{lbmol}$.

Enthalpy of ${\text{CO}}_{\text{2}}$ gas at 3620 K, ${\left(\overline{h}\right)}_{{\text{CO}}_{\text{2}}}=43,701\text{Btu}/\text{lbmol}$.

Enthalpy of ${\text{CO}}_{\text{2}}$ gas at 298 K, ${\left({\overline{h}}^{\circ }\right)}_{{\text{CO}}_{\text{2}}}=4,027.5\text{Btu}/\text{lbmol}$.

Similarly, use interpolation and obtain the enthalpy of ${\text{CO}}_{\text{2}}$ gas at 3600 K as $43,411\text{Btu}/\text{lbmol}$.

Refer Table A-26, “Enthalpy of formation, Gibbs function of formation, and absolute entropy at

778F, 1 atm”, select the enthalpy of ${\text{O}}_2$$\left({\left({\overline{h}}_f^{\circ }\right)}_{{\text{O}}_2}\right)$ as $0$.

Refer Table A-19, ‘Ideal gas properties of oxygen’, choose the enthalpy at the following temperatures.

Enthalpy of ${\text{O}}_2$ gas at 3580 K, ${\left(\overline{h}\right)}_{{\text{O}}_2}=28,994\text{kJ}/\text{kmol}$.

Enthalpy of ${\text{O}}_2$ gas at 3620 K, ${\left(\overline{h}\right)}_{{\text{O}}_2}=29,354\text{kJ}/\text{kmol}$.

Enthalpy of ${\text{O}}_2$ gas at 298 K, ${\left({\overline{h}}^{\circ }\right)}_{{\text{O}}_2}=3,725.1\text{Btu}/\text{lbmol}$.

Similarly, use interpolation and obtain the enthalpy of ${\text{O}}_{\text{2}}$ gas at 3600 K as $29,174\text{Btu}/\text{lbmol}$.

Substitute $0$ for ${\left(\overline{h}\right)}_{{\text{O}}_2}$, $29,174\text{Btu}/\text{lbmol}$ for ${\left(\overline{h}\right)}_{{\text{O}}_2}$, $3,725.1\text{Btu}/\text{lbmol}$ for ${\left({\overline{h}}^{\circ }\right)}_{{\text{O}}_2}$, $-163,300\text{Btu}/\text{lbmol}$ for ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{CO}}_2}$, $43,411\text{Btu}/\text{lbmol}$ for ${\left(\overline{h}\right)}_{{\text{CO}}_2}$, $4,027.5\text{Btu}/\text{lbmol}$ for ${\left({\overline{h}}^{\circ }\right)}_{{\text{CO}}_2}$, $-47,540\text{Btu}/\text{lbmol}$ for ${\left({\overline{h}}_f^{\circ }\right)}_{\text{CO}}$, $28,127\text{Btu}/\text{lbmol}$ for ${\left(\overline{h}\right)}_{\text{CO}}$, $3,725.1\text{Btu}/\text{lbmol}$ for ${\left({\overline{h}}^{\circ }\right)}_{\text{CO}}$ 0.9966 mol for $N_{{\text{CO}}_{\text{2}}}$, 0.0034 mol for $N_{\text{CO}}$, and 0.1587 mol for $N_{{\text{O}}_2}$ in Equation (VII).

$\begin{array}{c}-Q_{out}=\left\{\begin{array}{l}\left(0.9966\text{mol}\right)\left(\begin{array}{l}-163,300\text{Btu}/\text{lbmol}\\ +43,411\text{Btu}/\text{lbmol}\\ -4,027.5\text{Btu}/\text{lbmol}\end{array}\right)\\ +\left(0.0034\text{mol}\right)\left(\begin{array}{l}-47,540\text{Btu}/\text{lbmol}\\ +28,127\text{Btu}/\text{lbmol}\\ -3,725.1\text{Btu}/\text{lbmol}\end{array}\right)\\ +\left(0.1587\text{mol}\right)\left(\begin{array}{l}0+29,174\text{Btu}/\text{lbmol}\\ -3,725.1\text{Btu}/\text{lbmol}\end{array}\right)\\ -1\left(\begin{array}{l}-47,540\text{Btu}/\text{lbmol}\\ +3,889.5\text{Btu}/\text{lbmol}\\ -3,725.1\text{Btu}/\text{lbmol}\end{array}\right)\end{array}\right\}\\ =-78,139\text{Btu}/\text{lbmol}\text{of}\text{CO}\end{array}$

$Q_{out}=78,139\text{Btu}/\text{lbmol}\text{of}\text{CO}$

Substitute $0.932\text{lbm}/\min$ for ${\dot{m}}_{\text{CO}}$, and $28\text{lbm}/\text{lbmol}$ for $M_{\text{CO}}$ in Equation (VIII).

$\begin{array}{c}{\dot{N}}_{\text{CO}}=\frac{0.932\text{lbm}/\min }{28\text{lbm}/\text{lbmol}}\\ =0.0333\text{lbmol}/\text{min}\end{array}$

Substitute $0.0333\text{lbmol}/\text{min}$ for ${\dot{N}}_{\text{CO}}$, and $78,139\text{Btu}/\text{lbmol}\text{of}\text{CO}$ for $Q_{out}$ in Equation (IX).

$\begin{array}{c}{\dot{Q}}_{out}=\left(0.0333\text{lbmol}/\text{min}\right)\left(78,139\text{Btu}/\text{lbmol}\text{of}\text{CO}\right)\\ =2,602\text{Btu}/\text{min}\end{array}$

Thus, the rate of heat transfer is $\underset{\_}{2,602\text{Btu}/\text{min}}$.