Chapter 15.7, Problem 89P

A steady-flow combustion chamber is supplied with CO gas at 37°C and 110 kPa at a rate of 0.4 m3/min and air at 25°C and 110 kPa at a rate of 1.5 kg/min. Heat is transferred to a medium at 800 K, and the combustion products leave the combustion chamber at 900 K. Assuming the combustion is complete and T0 = 25°C, determine (a) the rate of heat transfer from the combustion chamber and (b) the rate of exergy destruction.

To determine

The rate of heat transfer from the combustion chamber.

Answer to Problem 89P

The rate of heat transfer from the combustion chamber is $\underset{\_}{3567\text{kJ}/\text{min}}$.

Explanation of Solution

Determine the volume of the CO in the combustion chamber.

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

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

Determine the mass flow rate of the CO in the combustion chamber.

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

Here, the volume flow rate of the CO in the combustion chamber is ${\dot{\nu }}_{\text{CO}}$.

Determine the molar air-fuel ratio.

$\begin{array}{c}\overline{\text{AF}}=\frac{N_{\text{air}}}{N_{\text{fuel}}}\\ =\frac{{\dot{m}}_{\text{air}}/M_{\text{air}}}{{\dot{m}}_{\text{fuel}}/M_{\text{fuel}}}\end{array}$ (III)

Here, the mass flow rate of the air is ${\dot{m}}_{\text{air}}$, the molar mass of the air is $M_{\text{air}}$, the mass flow rate of the fuel is ${\dot{m}}_{\text{fuel}}$, the molar mass of the fuel is $M_{\text{fuel}}$.

Write the energy balance equation using steady-flow equation.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$ (IV)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Substitute $0$ for $E_{\text{in}}$, $-Q_{\text{out}}$ for $E_{\text{out}}$, and $\Delta U$ for $\Delta E_{\text{system}}$ in Equation (IV)

$\begin{array}{l}\left(0\right)-Q_{\text{out}}=\Delta U\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+\overline{h}-\overline{h^{°}}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+\overline{h}-\overline{h^{°}}\right)}}_R\\ -Q_{\text{out}}={\displaystyle \sum N_P{\overline{h}}_{f,P}^{°}}-{\displaystyle \sum N_R{\overline{h}}_{f,R}^{°}}\end{array}$ (V)

Here, the enthalpy of formation for product is ${\overline{h}}_{f,P}^{°}$, the enthalpy of formation for reactant is ${\overline{h}}_{f,R}^{°}$, the mole number of the product is $N_P$, and the mole number of the reactant is $N_R$.

Determine the heat transfer per kg of CO.

$q_{\text{out}}=\frac{Q_{out}}{M_{\text{CO}}}$ (VI)

Here, the molar mass of the CO is $M_{\text{CO}}$.

Determine the rate of heat transfer.

${\dot{Q}}_{\text{out}}=\dot{m}{q}_{\text{out}}$ (VII)

Conclusion:

Perform unit conversion of temperature at state 1 from degree Celsius to Kelvin.

For air temperature enter in the combustion chamber,

$\begin{array}{c}{T}_{\text{air}}=25\text{°C}\\ =\left(25+273\right)\text{K}\\ =\text{298}\text{K}\end{array}$

For CO temperature enter in the combustion chamber,

$\begin{array}{c}{T}_{\text{CO}}=37\text{°C}\\ =\left(37+273\right)\text{K}\\ =\text{310}\text{K}\end{array}$

From the Table A-1 “Molar mass, gas-constant, and critical-point properties”, obtain the value of molar mass for air and carbon monoxide and universal gas constant of CO as:

$\begin{array}{l}{M}_{\text{air}}=29\text{kg}/\text{kmol}\\ M_{\text{CO}}=28\text{kg}/\text{kmol}\\ R_{\text{CO}}=0.2968\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.2968\text{kJ}/\text{kg}\cdot \text{K}$ for $R_{\text{CO}}$, $37°\text{C}$ for $T$, and $110\text{kPa}$ for $P$ in Equation (I).

$\begin{array}{c}{\nu }_{\text{CO}}=\frac{\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left(37°\text{C}\right)}{\left(110\text{kPa}\right)}\\ =\frac{\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\times \left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)\left(37°\text{C}+\text{273}\right)}{\left(110\text{kPa}\right)}\\ =\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(310\text{K}\right)}{\left(110\text{kPa}\right)}\\ =0.836{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $0.4{\text{m}}^{\text{3}}/\text{min}$ for ${\dot{\nu }}_{\text{CO}}$ and $0.836{\text{m}}^{\text{3}}/\text{kg}$ for ${\nu }_{\text{CO}}$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{\text{CO}}=\frac{0.4{\text{m}}^{\text{3}}/\text{min}}{0.836{\text{m}}^{\text{3}}/\text{kg}}\\ =0.478\text{kg}/\text{min}\end{array}$

Substitute $1.5\text{kg}/\text{min}$ for ${\dot{m}}_{\text{air}}$, $29\text{kg}/\text{kmol}$ for $M_{\text{air}}$, $0.478\text{kg}/\text{min}$ for ${\dot{m}}_{\text{fuel}}$ and $28\text{kg}/\text{kmol}$ for $M_{\text{fuel}}$ in Equation (III).

$\begin{array}{c}\overline{\text{AF}}=\frac{\left(1.5\text{kg}/\text{min}\right)/\left(29\text{kg}/\text{kmol}\right)}{\left(0.478\text{kg}/\text{min}\right)/\left(28\text{kg}/\text{kmol}\right)}\\ =3.03\text{kmol}\text{air}/\text{kmol}\text{fuel}\end{array}$

Here, the number of mole of oxygen used per mole of carbon monoxide is $3.03/4.76=0.637$.

Write the combustion equation of 1 kmol for $\text{CO}$.

$\left\{\text{CO}+0.637a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}{\text{CO}}_{\text{2}}+0.137{\text{O}}_{\text{2}}+\\ 2.40{\text{N}}_{\text{2}}\end{array}\right\}$ (VIII)

Here, stoichiometric coefficient of air is $a_{\text{th}}$, oxygen is ${\text{O}}_{\text{2}}$, nitrogen is ${\text{N}}_{\text{2}}$, and carbon monoxide is $\text{CO}$.

From the Table-21, 19, 18, and 20, obtain the enthalpy of formation, at 298 K, 310 K, and 900 K for ${\text{O}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, $\text{CO}$, and ${\text{CO}}_{\text{2}}$ is given in a tabular form as:

Substance	$\begin{array}{l}{\overline{h}}_f^{°}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{298\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{310\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{900\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$
$\text{CO}$	-110,530	8669	9014	---
${\text{O}}_{\text{2}}$	0	8682	---	27,928
${\text{N}}_{\text{2}}$	0	8669	---	26,890
${\text{CO}}_{\text{2}}$	-393,520	9364	---	37,405

Substitute the value of substance in Equation (V).

$\begin{array}{c}-Q_{\text{out}}=\left[\begin{array}{l}\left(1\right)\left(-393,520\text{kJ}/\text{kmol}+37,405\text{kJ}/\text{kmol}-9364\text{kJ}/\text{kmol}\right)\\ +\left(0.137\right)\left(0+27,928\text{kJ}/\text{kmol}-8682\text{kJ}/\text{kmol}\right)\\ +\left(2.4\right)\left(0+26,890\text{kJ}/\text{kmol}-8669\text{kJ}/\text{kmol}\right)\\ -\left(1\right)\left(-110,530\text{kJ}/\text{kmol}+9014\text{kJ}/\text{kmol}-8669\text{kJ}/\text{kmol}\right)\\ \left(0\right)-\left(0\right)\end{array}\right]\\ =-208,929\text{kJ}/\text{kmol}\end{array}$

Therefore the heat transfer for $\text{CO}$ is $208,929\text{kJ}/\text{kmol}$ for each kmol.

Substitute $208,929\text{kJ}/\text{kmol}$ for $Q_{\text{out}}$ and $28\text{kg}/\text{kmol}$ for $M_{\text{CO}}$ in Equation (VI).

$\begin{array}{c}{q}_{\text{out}}=\frac{208,929\text{kJ}/\text{kmol}}{28\text{kg}/\text{kmol}}\\ =7462\text{kJ}/\text{kg}\end{array}$

Substitute $7462\text{kJ}/\text{kg}$ for $q_{\text{out}}$ and $0.478\text{kg}/\text{min}$ for $\dot{m}$ in Equation (VII).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(0.478\text{kg}/\text{min}\right)\times \left(7462\text{kJ}/\text{kg}\right)\\ =3567\text{kJ}/\text{min}\end{array}$

Thus, the rate of heat transfer from the combustion chamber is $\underset{\_}{3567\text{kJ}/\text{min}}$.

To determine

The exergy destruction from the combustion chamber.

Answer to Problem 89P

The exergy destruction from the combustion chamber is $\underset{\_}{1610\text{kJ}/\text{min}}$.

Explanation of Solution

Write the expression for the relation of reversible work using the exergy balance on the combustion chamber.

$W_{\text{rev}}=\left[\begin{array}{l}{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}-T_0\overline{s}\right)}}_R-{{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}-T_0\overline{s}\right)}}_P\\ -Q_{\text{out}}\left(1-T_0/T_R\right)\end{array}\right]$

In the ideal gas table the value of entropy for 1 atm is equal to 101.325 kPa of pressure. Each reactant of the entropy and the product is to be calculated at the partial pressure of the components which is equal to:

$P_i=y_i{P}_{\text{total}}$

Here, the mole fraction of component $i$ is $y_i$, and

$\begin{array}{c}{P}_m=\frac{110}{101.325}\\ =1.0856\text{atm}\end{array}$.

Write the expression for entropy generation during this process.

$S_{\text{gen}}=S_P-S_R+\frac{Q_{\text{out}}}{T_{\text{surr}}}$ (VIII)

Write the combustion equation of Equation (VI)

$S_{\text{gen}}=S_P-S_R+\frac{Q_{\text{out}}}{T_{\text{surr}}}\to S_{\text{gen}}={\displaystyle \sum N_P{\overline{s}}_P}-{\displaystyle \sum N_R{\overline{s}}_R}+\frac{Q_{\text{out}}}{T_{\text{surr}}}$ (IX)

Here, the entropy of the product is ${\overline{s}}_P$, the entropy of the reactant is ${\overline{s}}_R$, the heat transfer for $\text{CO}$ is $Q_{\text{out}}$, and the surrounding temperature is $T_{\text{surr}}$.

Determine the entropy at the partial pressure of the components.

$\begin{array}{c}{S}_i=N_i{\overline{s}}_i\left(T,P_i\right)\\ =N_i{\overline{s}}_i^{°}\left(T,P_0\right)-R_u\ln \left(y_i{P}_m\right)\end{array}$ (X).

Here, the partial pressure is $P_i$, the mole fraction of the component is $y_i$, the total pressure of the mixture is $P_m$, and the universal gas constant is $R_u$.

Write the expression for rate of exergy destruction during this process.

$\begin{array}{c}{\dot{X}}_{\text{dest}}=T_0{\dot{S}}_{\text{gen}}\\ =T_0\dot{m}\left(S_{\text{gen}}/M\right)\end{array}$ (XI)

Here, the thermodynamic temperature of the surrounding is $T_0$

Conclusion:

The entropy calculation can be presented in tabular form as:

For reactant entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
$\text{CO}$	1	1.00	198.678	0.68	198.00
${\text{O}}_2$	0.637	0.21	205.04	-12.29	138.44
${\text{N}}_{\text{2}}$	2.400	0.79	191.61	-1.28	462.94
$S_R=799.38\text{kJ}/\text{kmol}\cdot \text{K}$

For product entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{CO}}_2$	1	0.2827	263.559	-9.821	273.38
${\text{O}}_{\text{2}}$	0.137	0.0387	239.823	-26.353	36.47
${\text{N}}_{\text{2}}$	2.400	0.6785	224.467	-2.543	544.82
$S_P=854.67\text{kJ}/\text{kmol}\cdot \text{K}$

Substitute $854.67\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_P$ and $799.38\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_R$ in Equation (VIII).

$\begin{array}{c}{S}_{\text{gen}}=\left(854.67-799.38\right)\text{kJ}/\text{kmol}\cdot \text{K}+\frac{208,929\text{kJ}}{800\text{K}}\\ =55.29\text{kJ}/\text{kmol}\cdot \text{K}+\frac{208,929\text{kJ}}{800\text{K}}\\ =316.5\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Substitute $25°\text{C}$ for $T_0$, $0.478\text{kg}/\text{min}$ for $\dot{m}$, $28\text{kg}/\text{kmol}$ for $M$, and $316.5\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_{\text{gen}}$ in Equation (XI).

$\begin{array}{c}{\dot{X}}_{\text{dest}}=\left(25°\text{C}\right)\left(0.478\text{kg}/\min \right)\left(316.5\text{kJ}/\text{kmol}\cdot \text{K}/28\text{kg}/\text{kmol}\right)\\ =\left(25°\text{C+273}\right)\left(0.478\text{kg}/\min \right)\left(316.5\text{kJ}/\text{kmol}\cdot \text{K}/28\text{kg}/\text{kmol}\right)\\ =1610\text{kJ}/\text{min}\end{array}$

Thus, the exergy destruction from the combustion chamber is $\underset{\_}{1610\text{kJ}/\text{min}}$.