Chapter 15.7, Problem 109RP

Repeat Prob. 15–112 using a coal from Utah that has an ultimate analysis (by mass) of 61.40 percent C, 5.79 percent H₂, 25.31 percent O₂, 1.09 percent N₂, 1.41 percent S, and 5.00 percent ash (noncombustibles).

To determine

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

Answer to Problem 109RP

The amount of steam generated per unit of fuel mass burned is $\underset{\_}{10.87\text{kg}\text{steam}/\text{kg}\text{fuel}}$.

Explanation of Solution

Express the number of moles of carbon.

$N_{\text{C}}=\frac{m_{\text{C}}}{M_{\text{C}}}$ (I)

Here, molar mass of carbon is $M_{\text{C}}$ and the mass of the carbon is $m_{\text{C}}$.

Express the number of moles of hydrogen.

$N_{{\text{H}}_{\text{2}}}=\frac{m_{{\text{H}}_{\text{2}}}}{M_{{\text{H}}_{\text{2}}}}$ (II)

Here, molar mass of hydrogen is $M_{{\text{H}}_{\text{2}}}$ and the mass of the hydrogen is $m_{{\text{H}}_2}$.

Express the number of moles of oxygen.

$N_{{\text{O}}_{\text{2}}}=\frac{m_{{\text{O}}_{\text{2}}}}{M_{{\text{O}}_{\text{2}}}}$ (III)

Here, molar mass of oxygen is $M_{{\text{O}}_{\text{2}}}$ and the mass of the oxygen is $m_{{\text{O}}_2}$.

Express the number of moles of nitrogen.

$N_{{\text{N}}_{\text{2}}}=\frac{m_{{\text{N}}_{\text{2}}}}{M_{{\text{N}}_{\text{2}}}}$ (IV)

Here, molar mass of nitrogen is $M_{{\text{N}}_{\text{2}}}$ and the mass of the nitrogen is $m_{{\text{N}}_2}$.

Express the number of moles of sulphur.

$N_{\text{S}}=\frac{m_{\text{S}}}{M_{\text{S}}}$ (V)

Here, molar mass of sulphur is $M_{\text{S}}$ and the mass of the sulphur is $m_{\text{S}}$.

Express the total number of moles.

$N_m=N_{\text{C}}+N_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}+N_{\text{S}}$ (VI)

Express the mole fraction of carbon.

$y_{\text{C}}=\frac{N_{\text{C}}}{N_m}$ (VII)

Express the mole fraction of hydrogen.

$y_{{\text{H}}_{\text{2}}}=\frac{N_{{\text{H}}_{\text{2}}}}{N_m}$ (VIII)

Express the mole fraction of oxygen.

$y_{{\text{O}}_{\text{2}}}=\frac{N_{{\text{O}}_{\text{2}}}}{N_m}$ (IX)

Express the mole fraction of nitrogen.

$y_{{\text{N}}_{\text{2}}}=\frac{N_{{\text{N}}_{\text{2}}}}{N_m}$ (X)

Express the mole fraction of sulphur.

$y_{\text{S}}=\frac{N_{\text{S}}}{N_m}$ (XI)

Write the energy balance equation using steady-flow equation.

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

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 (XII)

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

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 $N_R$.

Write the formula for the amount of steam generated per unit mass of fuel burned.

$\begin{array}{c}\frac{m_s}{m_f}=\frac{Q_{\text{out}}}{\Delta h_s}\\ =\frac{Q_{\text{out}}}{h_f-h_g}\end{array}$ (XIV)

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:

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

$\begin{array}{l}{M}_{\text{C}}=12\text{kg}/\text{kmol}\\ M_{{\text{H}}_2}=2\text{kg}/\text{kmol}\\ M_{{\text{O}}_{\text{2}}}=32\text{kg}/\text{kmol}\\ M_{\text{S}}=32\text{kg}/\text{kmol}\end{array}$

$\begin{array}{l}{M}_{\text{air}}=29\text{kg}/\text{kmol}\\ M_{{\text{N}}_{\text{2}}}=28\text{kg}/\text{kmol}\end{array}$

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

Substitute $\text{61}\text{.40}\text{kg}$ for $m_{\text{C}}$ and $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$ in Equation (I).

$\begin{array}{c}{N}_{\text{C}}=\frac{\text{61}\text{.40}\text{kg}}{12\text{kg}/\text{kmol}}\\ =5.117\text{kmol}\end{array}$

Substitute $\text{5}\text{.79}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$ in Equation (II).

$\begin{array}{c}{N}_{{\text{H}}_{\text{2}}}=\frac{\text{5}\text{.79}\text{kg}}{2\text{kg}/\text{kmol}}\\ =2.895\text{kmol}\end{array}$

Substitute $\text{25}\text{.31}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (III).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{\text{25}\text{.31}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.7909\text{kmol}\end{array}$

Substitute $\text{1}\text{.09}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (IV).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{\text{1}\text{.09}\text{kg}}{28\text{kg}/\text{kmol}}\\ =0.03893\text{kmol}\end{array}$

Substitute $\text{1}\text{.41}\text{kg}$ for $m_{\text{S}}$ and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (V).

$\begin{array}{c}{N}_{\text{S}}=\frac{\text{1}\text{.41}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.04406\text{kmol}\end{array}$

Substitute $5.117\text{kmol}$ for $N_{\text{C}}$, $2.895\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.7909\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.03893\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $0.04406\text{kmol}$ for $N_{\text{S}}$ in Equation (VI).

$\begin{array}{c}{N}_m=5.117\text{kmol}+2.895\text{kmol}+0.7909\text{kmol}+0.03893\text{kmol}+0.04406\text{kmol}\\ =8.886\text{kmol}\end{array}$

Substitute $5.117\text{kmol}$ for $N_{\text{C}}$ and $8.886\text{kmol}$ for $N_m$ in Equation (VII).

$\begin{array}{c}{y}_{\text{C}}=\frac{5.117\text{kmol}}{8.886\text{kmol}}\\ =0.5758\end{array}$

Substitute $2.895\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$ and $8.886\text{kmol}$ for $N_m$ in Equation (VIII).

$\begin{array}{c}{y}_{{\text{H}}_{\text{2}}}=\frac{2.895\text{kmol}}{8.886\text{kmol}}\\ =0.3258\end{array}$

Substitute $0.7909\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and $8.886\text{kmol}$ for $N_m$ in Equation (IX).

$\begin{array}{c}{y}_{{\text{O}}_{\text{2}}}=\frac{0.7909\text{kmol}}{8.886\text{kmol}}\\ =0.0890\end{array}$

Substitute $0.03893\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $8.886\text{kmol}$ for $N_m$ in Equation (X).

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{0.03893\text{kmol}}{8.886\text{kmol}}\\ =0.00438\end{array}$

Substitute $0.04406\text{kmol}$ for $N_{\text{S}}$ and $8.886\text{kmol}$ for $N_m$ in Equation (XI).

$\begin{array}{c}{y}_{\text{S}}=\frac{0.04406\text{kmol}}{8.886\text{kmol}}\\ =0.00496\end{array}$

Express the combustion equation.

$\left[\begin{array}{l}0.5758\text{C}+0.3258{\text{H}}_{\text{2}}+0.0890{\text{O}}_{\text{2}}+0.00438{\text{N}}_{\text{2}}+0.00496\text{S}\\ +1.5a_{\text{th}}\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.5758{\text{CO}}_{\text{2}}+0.3258{\text{H}}_{\text{2}}\text{O}+\\ 0.00496{\text{SO}}_2+0.5a_{\text{th}}{\text{O}}_2+0.5a_{\text{th}}\times 3.76{\text{N}}_{\text{2}}\end{array}\right]$ (XV)

Here, carbon dioxide, water, sulfur dioxide, nitrogen and oxygen is ${\text{CO}}_{\text{2}}{\text{,H}}_{\text{2}}{\text{O,SO}}_{\text{2}}{\text{,N}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}$ respectively, and stoichiometric coefficient of air is $a_{\text{th}}$.

Perform the species balancing according to the oxygen balance:

Oxygen balance:

$\begin{array}{l}0.0890+1.5a_{\text{th}}=0.5758+0.5\left(0.3258\right)+0.00496+0.5a_{\text{th}}\\ a_{\text{th}}=0.6547\end{array}$

Substitute $0.6547$ for $a_{\text{th}}$ in Equation (XV).

$\begin{array}{l}\left[\begin{array}{l}0.5758\text{C}+0.3258{\text{H}}_{\text{2}}+0.0890{\text{O}}_{\text{2}}+0.00438{\text{N}}_{\text{2}}+0.00496\text{S}\\ +1.5\left(0.6547\right)\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.5758{\text{CO}}_{\text{2}}+0.3258{\text{H}}_{\text{2}}\text{O}+\\ 0.00496{\text{SO}}_2+0.5\left(0.6547\right){\text{O}}_2+0.5\left(0.6547\right)\times 3.76{\text{N}}_{\text{2}}\end{array}\right]\\ \left[\begin{array}{l}0.5758\text{C}+0.3258{\text{H}}_{\text{2}}+0.0890{\text{O}}_{\text{2}}+0.00438{\text{N}}_{\text{2}}+0.00496\text{S}\\ +0.9821\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.5758{\text{CO}}_{\text{2}}+0.3258{\text{H}}_{\text{2}}\text{O}+\\ 0.00496{\text{SO}}_2+0.3274{\text{O}}_2+3.693{\text{N}}_{\text{2}}\end{array}\right]\end{array}$ (XVI)

Calculate the apparent molecular weight of the cool.

$\begin{array}{c}{M}_m=\frac{m_m}{N_m}\\ =\frac{\left(0.5758\times 12+0.3258\times 2+0.0890\times 32+0.00438\times 28+0.00496\times 32\right)\text{kg}}{\left(0.5758+0.3258+0.0890+0.00438+0.00496\right)\text{kmol}}\\ =\frac{10.69\text{kg}}{1.0\text{kmol}}\\ =10.69\text{kg}/\text{kmol}\text{coal}\end{array}$

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

Substance	$\begin{array}{l}{\overline{h}}_f^{°}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{298\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{500\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$
${\text{O}}_{\text{2}}$	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

Substitute the value of substance in Equation (XIII).

$\begin{array}{c}-Q_{\text{out}}=\left[\begin{array}{l}\left(0.5758\right)\left(-393,520\text{kJ}/\text{kmol}+17,678\text{kJ}/\text{kmol}-9364\text{kJ}/\text{kmol}\right)\\ +\left(0.3258\right)\left(-241,820\text{kJ}/\text{kmol}+16,828\text{kJ}/\text{kmol}-9904\text{kJ}/\text{kmol}\right)\\ +\left(0.3274\right)\left(0+14,770\text{kJ}/\text{kmol}-8682\text{kJ}/\text{kmol}\right)\\ +\left(3.693\right)\left(0+14,581\text{kJ}/\text{kmol}-8669\text{kJ}/\text{kmol}\right)\\ -\left(0\right)\end{array}\right]\\ =-274,505\text{kJ}/\text{kmol}\text{of}\text{fuel}\end{array}$

Calculate the heat loss per unit mass of the fuel.

$\begin{array}{c}{Q}_{\text{out}}=\frac{274,505\text{kJ}/\text{kmol}\text{of fuel}}{10.69\text{kg}/\text{kmol}\text{of fuel}}\\ =25,679\text{kJ}/\text{kg}\text{fuel}\end{array}$

Substitute $25,679\text{kJ}/\text{kg}\text{fuel}$ for $Q_{\text{out}}$, $3214.5\text{kJ}/\text{kg}$ for $h_f$, and $852.26\text{kJ}/\text{kg}$ for $h_g$ in Equation (XIV).

$\begin{array}{c}\frac{m_s}{m_f}=\frac{25,679\text{kJ}/\text{kg}\text{fuel}}{3214.5\text{kJ}/\text{kg}-852.26\text{kJ}/\text{kg}}\\ =\frac{25,679\text{kJ}/\text{kg}\text{fuel}}{2362.24\text{kJ}/\text{kg}\text{steam}}\\ =10.87\text{kg}\text{steam}/\text{kg}\text{fuel}\end{array}$

Thus, the amount of steam generated per unit of fuel mass burned is $\underset{\_}{10.87\text{kg}\text{steam}/\text{kg}\text{fuel}}$.

To determine

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

Answer to Problem 109RP

The change in the exergy of the combustion steams, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{-27,630\text{kJ}/\text{kg}\text{fuel}}$.

Explanation of Solution

Write the expression for entropy generation during this process.

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

Write the combustion equation of Equation (VI)

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

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.

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

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.

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

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

Conclusion:

Refer Equation (XIX) 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)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
$\text{C}$	0.5758	0.5758	5.74	-4.589	5.95
${\text{H}}_2$	0.3258	0.3258	130.68	-9.324	45.61
${\text{O}}_{\text{2}}$	0.0890	0.0890	205.04	-20.11	20.04
${\text{N}}_2$	0.00438	0.00438	191.61	-45.15	1.04
${\text{O}}_2$	0.9821	0.21	205.04	-12.98	214.12
${\text{N}}_2$	3.693	0.79	191.61	-1.960	714.85
$S_R=1001.61\text{kJ}/\text{K}$

For product entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{CO}}_{\text{2}}$	0.5758	0.1170	234.814	-17.84	145.48
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	0.3258	0.0662	206.413	-22.57	74.60
${\text{O}}_2$	0.3274	0.0665	220.589	-22.54	79.60
${\text{N}}_{\text{2}}$	3.693	0.7503	206.630	-2.388	771.90
$S_P=1071.58\text{kJ}/\text{K}$

Substitute $1071.58\text{kJ}/\text{K}$ for $S_P$, $1001.61\text{kJ}/\text{K}$ for $S_R$, $298\text{K}$ for $T_{\text{surr}}$, and $274,505\text{kJ}/\text{kmol}$ for $Q_{\text{out}}$ in Equation (XVII).

$\begin{array}{c}{S}_{\text{gen}}=\left(1071.58-1001.61\right)\text{kJ}/\text{K}+\frac{274,505\text{kJ}/\text{kmol}\cdot \text{K}}{298\text{K}}\\ =\left(69.97\right)\text{kJ}/\text{K}+\left(921.1577\right)\text{kJ}/\text{kmol}\\ =991.1\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Substitute $991.1\text{kJ}/\text{kg}$ for $S_{\text{gen}}$ and $298\text{K}$ for $T_{\text{0}}$ in Equation (XX).

$\begin{array}{c}\Delta X_{\text{gases}}=-\left(298\text{K}\right)\times \left(991.1\text{kJ}/\text{kg}\right)\\ =-295,347.8\text{kJ}/\text{kmol}\text{fuel}\\ \cong -295,348\text{kJ}/\text{kmol}\text{fuel}\end{array}$

Calculate the exergy destruction per unit mass of the basis.

$\begin{array}{c}{Q}_{\text{out}}=\frac{-295,348\text{kJ}/\text{kmol}\text{of fuel}}{10.69\text{kg}/\text{kmol}\text{of fuel}}\\ =-27,630\text{kJ}/\text{kg}\end{array}$

Thus, the change in the exergy of the combustion steams, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{-27,630\text{kJ}/\text{kg}\text{fuel}}$.

To determine

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

Answer to Problem 109RP

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.

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

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 $h_1$, $298\text{K}$ for $T_0$, $6.7714\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$, and $2.3305\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (XXI).

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

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 109RP

The lost work potential, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{16,340\text{kJ}/\text{kg}\text{fuel}}$.

Explanation of Solution

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

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

Conclusion:

Substitute $10.87\text{kg}\text{steam}/\text{kg}\text{fuel}$ for $m_s/m_f$, $1039\text{kJ}/\text{kg}\text{steam}$ for $\Delta X_{\text{steam}}$, and $-27,630\text{kJ}/\text{kg}\text{steam}$ for $\Delta X_{\text{gases}}$ in Equation (XXII).

$\begin{array}{c}{X}_{\text{dest}}=-\left[\left(10.87\text{kg}\text{steam}/\text{kg}\text{fuel}\right)\times \left(1039\text{kJ}/\text{kg}\text{steam}\right)+\left(-27,630\text{kJ}/\text{kg}\text{fuel}\right)\right]\\ =-\left[\left(11293.93\text{kJ}/\text{kg}\text{fuel}\right)+\left(-27,630\text{kJ}/\text{kg}\text{fuel}\right)\right]\\ =16336.1\text{kJ}/\text{kg}\text{fuel}\\ \cong 16340\text{kJ}/\text{kg}\text{fuel}\end{array}$

Thus, the lost work potential, in $\text{kJ}/\text{kg}\text{fuel}$ is $\underset{\_}{16,340\text{kJ}/\text{kg}\text{fuel}}$.