Chapter 15.7, Problem 112RP

A steam boiler heats liquid water at 200°C to superheated steam at 4 MPa and 400°C. Methane fuel (CH₄) is burned at atmospheric pressure with 50 percent excess air. The fuel and air enter the boiler at 25°C and the products of combustion leave at 227°C. Calculate (a) the amount of steam generated per unit of fuel mass burned, (b) the change in the exergy of the combustion streams, in kJ/kg fuel, (c) the change in the exergy of the steam stream, in kJ/kg steam, and (d) the lost work potential, in kJ/kg fuel. Take T0 = 25°C.

To determine

The amount of steam generated per unit of fuel mass burned.

Answer to Problem 112RP

The amount of steam generated per unit of fuel mass burned is $\underset{\_}{18.72\text{kg}\text{steam}/\text{kg}\text{fuel}}$.

Explanation of Solution

Write the energy balance equation using steady-flow equation.

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

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 (I)

$\begin{array}{l}\left(0\right)-Q_{\text{out}}=\Delta U\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}\right)}}_R\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+{\overline{h}}_{500\text{K}}-{\overline{h}}_{298\text{K}}^{°}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}\right)}}_R\end{array}$ (II)

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 $N_R$.

Calculate the molar mass of the ${\text{CH}}_{\text{4}}$.

$M_{{\text{CH}}_{\text{4}}}=\left[\left({\text{N}}_{\text{C}}\right)\left(M_{\text{C}}\right)+\left({\text{N}}_{\text{H}}\right)\left(M_{\text{H}}\right)\right]$ (III)

Here, the number of carbon atoms is ${\text{N}}_{\text{C}}$, the molar mass of the carbon is $M_{\text{C}}$, the number of hydrogen atoms is ${\text{N}}_{\text{H}}$, the molar mass of the hydrogen is $M_{\text{H}}$.

Determine the amount of steam generated per unit mass of fuel burned from an energy balance.

$\frac{m_s}{m_f}=\frac{Q_{\text{out}}}{\Delta h_s}$ (IV)

Here, the mass of the steam is $m_s$, the mass of the fuel burned is $m_f$, and the change in the enthalpy of the steam.

Conclusion:

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

For air temperature enter in the machine,

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

For air temperature exit from the machine,

$\begin{array}{c}{T}_{\text{leave}}=227\text{°C}\\ =\left(227+273\right)\text{K}\\ =\text{500}\text{K}\end{array}$

Write the combustion equation of 1 kmol for ${\text{CH}}_{\text{4}}$.

$\left\{{\text{CH}}_{\text{4}}+3\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}{\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}+{\text{O}}_2\\ +11.28{\text{N}}_{\text{2}}\end{array}\right\}$ (V)

Here, liquid methane is ${\text{CH}}_{\text{4}}$, stoichiometric coefficient of air is $a_{\text{th}}$, oxygen is ${\text{O}}_{\text{2}}$, nitrogen is ${\text{N}}_{\text{2}}$, carbon dioxide is ${\text{CO}}_{\text{2}}$ and water is ${\text{H}}_{\text{2}}\text{O}$.

Refer Appendix Table A-18, A-19, A-20, and A-23, obtain the enthalpy of formation, at 298 K , and 500 K for ${\text{CH}}_4$, ${\text{O}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$, and ${\text{CO}}_{\text{2}}$ is given in a table (I) as:

Substance	$\begin{array}{l}{\overline{h}}_f^{°}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{298\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{500\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$
${\text{CH}}_{\text{4}}$	-74,850	---	---
${\text{O}}_{\text{2}}$	0	8682	14,770
${\text{N}}_{\text{2}}$	0	8669	14,581
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	-241820	9904	16,828
${\text{CO}}_{\text{2}}$	-393,520	9364	17,678

Refer Equation (V), and write the number of moles of reactants.

$\begin{array}{l}{N}_{R,{\text{CH}}_4}=1\text{kmol}\\ N_{R,{\text{O}}_{\text{2}}}=3\text{kmol}\\ N_{R,{\text{N}}_{\text{2}}}=11.28\text{kmol}\end{array}$

Here, number of moles of reactant methane, oxygen and nitrogen is $N_{R,{\text{CH}}_4},N_{R,{\text{O}}_{\text{2}}}\text{and}{N}_{R,{\text{N}}_{\text{2}}}$ respectively.

Refer Equation (V), and write the number of moles of products.

$\begin{array}{l}{N}_{P,{\text{CO}}_{\text{2}}}=1\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=2\text{kmol}\\ N_{P,{\text{O}}_2}=1\text{kmol}\\ N_{P,{\text{N}}_2}=11.28\text{kmol}\end{array}$

Here, number of moles of product carbon dioxide, water, oxygen and nitrogen is $N_{P,{\text{CO}}_{\text{2}}},N_{P,{\text{H}}_{\text{2}}\text{O}},N_{P,{\text{O}}_2}\text{and}{N}_{N,{\text{N}}_2}$ respectively.

Substitute the value of substance in Equation (II).

$\begin{array}{c}-Q_{\text{out}}=\left[\begin{array}{l}\left(1\right)\left(-393,520\text{kJ}/\text{kmol}+17,678\text{kJ}/\text{kmol}-9364\text{kJ}/\text{kmol}\right)\\ +\left(2\right)\left(-241,820\text{kJ}/\text{kmol}+16,828\text{kJ}/\text{kmol}-9904\text{kJ}/\text{kmol}\right)\\ +\left(1\right)\left(0+14,770\text{kJ}/\text{kmol}-8682\text{kJ}/\text{kmol}\right)\\ +\left(11.28\right)\left(0+14,581\text{kJ}/\text{kmol}-8669\text{kJ}/\text{kmol}\right)\\ -\left(1\right)\left(-74,850\text{kJ}/\text{kmol}\right)\end{array}\right]\\ =-707,373\text{kJ}/\text{kmol}\text{of}\text{fuel}\end{array}$

Therefore the heat transfer for ${\text{CH}}_{\text{4}}$ is $707,373\text{kJ}/\text{kmol}\text{of}\text{fuel}$.

Substitute 1 for ${\text{N}}_{\text{C}}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, 4 for ${\text{N}}_{\text{H}}$, $1\text{kg}/\text{kmol}$ for $M_{\text{H}}$ in Equation (III).

$\begin{array}{c}{M}_{{\text{CH}}_{\text{4}}}=\left[\left(12\right)+\left(1\right)\left(4\right)\right]\text{kg}/\text{kmol}\\ =\left[\left(12\right)+\left(4\right)\right]\text{kg}/\text{kmol}\\ =16\text{kg}/\text{kmol}\end{array}$

Calculate the heat loss per unit mass of the fuel.

$\begin{array}{c}{Q}_{\text{out}}=\frac{707,373\text{kJ}/\text{kmol}\text{of fuel}}{16\text{kg}/\text{kmol}\text{of fuel}}\\ =44,211\text{kJ}/\text{kg}\end{array}$

From the table A-4, “Saturated water-Temperature” obtain the value of the saturated enthalpy and entropy of liquid at the $200°\text{C}$ temperature as $852.26\text{kJ}/\text{kg}$ and $2.3305\text{}\text{kJ}/\text{kg}\cdot \text{K}$.

From the table A-6, “Superheated water” obtain the value of the enthalpy and entropy at the $400°\text{C}$ temperature and 4 MPa pressure as $3214.5\text{kJ}/\text{kg}$ and $6.7714\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $44,211\text{kJ}/\text{kg}\text{fuel}$ for $Q_{\text{out}}$, $3214.5\text{kJ}/\text{kg}$ for $h_2$, and $852.26\text{kJ}/\text{kg}$ for $h_1$ in Equation (IV).

$\begin{array}{c}\frac{m_s}{m_f}=\frac{44,211\text{kJ}/\text{kg}\text{fuel}}{\left(3214.5-852.26\right)\text{kJ}/\text{kg}\text{steam}}\\ =\frac{44,211\text{kJ}/\text{kg}\text{fuel}}{2362.24\text{kJ}/\text{kg}\text{steam}}\\ =18.715\text{kg}\text{steam}/\text{kg}\text{fuel}\\ \cong 18.72\text{kg}\text{steam}/\text{kg}\text{fuel}\end{array}$

Thus, the amount of steam generated per unit of fuel mass burned is $\underset{\_}{18.72\text{kg}\text{steam}/\text{kg}\text{fuel}}$.

To determine

The change in the exergy of the combustion steams, in $\text{kJ}/\text{kg}\text{fuel}$.

Answer to Problem 112RP

The change in the exergy of the combustion steams, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{-49,490\text{kJ}/\text{kg}\text{fuel}}$.

Explanation of Solution

Write the expression for entropy generation during this process.

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

Write the combustion equation of Equation (VI)

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

Here, the entropy of the product is ${\overline{s}}_P$, the entropy of the reactant is ${\overline{s}}_R$, the heat transfer for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $Q_{\text{out}}$, and the surrounding temperature is $T_{\text{surr}}$.

Determine the entropy at the partial pressure of the components.

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

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 exergy change of the combustion steam is equal to the exergy destruction.

$\begin{array}{c}\Delta X_{\text{gases}}=-X_{\text{des}}\\ =-T_0{S}_{\text{gen}}\end{array}$ (IX)

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

Conclusion:

Refer Equation (VIII) for reactant and product to calculation the entropy in tabular form as:

For reactant entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{CH}}_{\text{4}}$	1	---	186.16	---	186.16
${\text{O}}_2$	3	0.21	205.04	-12.98	654.06
${\text{N}}_{\text{2}}$	11.28	0.79	191.61	-1.960	2183.47
$S_R=3023.69\text{kJ}/\text{K}$

For product entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{CH}}_{\text{4}}$	1	0.0654	234.814	-22.67	257.48
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	2	0.1309	206.413	-16.91	446.65
${\text{O}}_2$	1	0.0654	220.589	-22.67	243.26
${\text{N}}_{\text{2}}$	11.28	0.7382	206.630	-2.524	2359.26
$S_P=3306.65\text{kJ}/\text{K}$

Substitute $3306.65\text{kJ}/\text{K}$ for $S_P$, $3023.69\text{kJ}/\text{K}$ for $S_R$, $298\text{K}$ for $T_{\text{surr}}$, and $707,373\text{kJ}/\text{kmol}$ for $Q_{\text{out}}$ in Equation (VI).

$\begin{array}{c}{S}_{\text{gen}}=\left(3306.65-3023.69\right)\text{kJ}/\text{K}+\frac{707,373\text{kJ}/\text{kmol}\cdot \text{K}}{298\text{K}}\\ =\left(282.96\right)\text{kJ}/\text{K}+\left(2373.735\right)\text{kJ}/\text{kmol}\\ =2656.695\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Substitute $2657\text{kJ}/\text{kg}$ for $S_{\text{gen}}$ and $298\text{K}$ for $T_{\text{0}}$ in Equation (IX).

$\begin{array}{c}\Delta X_{\text{gases}}=-\left(298\text{K}\right)\times \left(2657\text{kJ}/\text{kg}\right)\\ =-791,786\text{kJ}/\text{kmol}\text{fuel}\end{array}$

Calculate the exergy destruction per unit mass of the basis.

$\begin{array}{c}{Q}_{\text{out}}=\frac{-791,786\text{kJ}/\text{kmol}\text{of fuel}}{16\text{kg}/\text{kmol}\text{of fuel}}\\ =-49,490\text{kJ}/\text{kg}\end{array}$

Thus, the change in the exergy of the combustion steams, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{-49,490\text{kJ}/\text{kg}\text{fuel}}$.

To determine

The exergy change of the steam, in $\text{kJ}/\text{kg}\text{steam}$.

Answer to Problem 112RP

The exergy change of the steam, in $\text{kJ}/\text{kg}\text{steam}$ is $\underset{\_}{1039\text{kJ}/\text{kg}\text{steam}}$.

Explanation of Solution

Determine the exergy change of the steam stream.

$\begin{array}{c}\Delta X_{\text{steam}}=\Delta h-T_0\Delta s\\ =\left(h_2-h_1\right)-T_0\left(s_2-s_1\right)\end{array}$ (X)

Here, the final enthalpy is $h_2$, the initial enthalpy is $h_1$, the final entropy is $s_2$, and the initial entropy is $s_1$.

Conclusion:

Substitute $3214.5\text{kJ}/\text{kg}$ for $h_2$, $852.26\text{kJ}/\text{kg}$ for $h_1$, $6.7714\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, and $2.3305\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (X).

$\begin{array}{c}\Delta X_{\text{steam}}=\left(3214.5-852.26\right)\text{kJ}/\text{kg}-\left(298\text{K}\right)\left(6.7714-2.3305\right)\text{kJ}/\text{kg}\cdot \text{K}\\ =2362.24\text{kJ}/\text{kg}-\left(298\text{K}\right)\times 4.4409\text{kJ}/\text{kg}\cdot \text{K}\\ =1038.85\text{kJ}/\text{kg}\text{steam}\\ \cong 1039\text{kJ}/\text{kg}\text{steam}\end{array}$

Thus, the exergy change of the steam, in $\text{kJ}/\text{kg}\text{steam}$ is $\underset{\_}{1039\text{kJ}/\text{kg}\text{steam}}$.

To determine

The lost work potential, in $\text{kJ}/\text{kg}\text{fuel}$.

Answer to Problem 112RP

The lost work potential, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{30,040\text{kJ}/\text{kg}\text{fuel}}$.

Explanation of Solution

Determine the lost work potential is the negative of the net exergy change both streams.

$X_{\text{dest}}=-\left[\frac{m_s}{m_f}\Delta X_{\text{steam}}+\Delta X_{\text{gases}}\right]$ (XI)

Conclusion:

Substitute $18.72\text{kg}\text{steam}/\text{kg}\text{fuel}$ for $m_s/m_f$, $1039\text{kJ}/\text{kg}\text{steam}$ for $\Delta X_{\text{steam}}$, and $-49,490\text{kJ}/\text{kg}\text{steam}$ for $\Delta X_{\text{gases}}$ in Equation (XI).

$\begin{array}{c}{X}_{\text{dest}}=-\left[\left(18.72\text{kg}\text{steam}/\text{kg}\text{fuel}\right)\times \left(1039\text{kJ}/\text{kg}\text{steam}\right)+\left(-49,490\text{kJ}/\text{kg}\text{fuel}\right)\right]\\ =-\left[\left(19450.08\text{kJ}/\text{kg}\text{fuel}\right)+\left(-49,490\text{kJ}/\text{kg}\text{fuel}\right)\right]\\ =30,039.9\text{kJ}/\text{kg}\text{fuel}\\ \cong 30,040\text{kJ}/\text{kg}\text{fuel}\end{array}$

Thus, the lost work potential, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{30,040\text{kJ}/\text{kg}\text{fuel}}$.