Chapter 15.7, Problem 83P

Reconsider Prob. 15–82. The combustion products are expanded in an isentropic turbine to 140 kPa. Calculate the work produced by this turbine in kJ/kg fuel.

To determine

The work produced by an isentropic turbine.

Answer to Problem 83P

The work produced by an isentropic turbine is $\text{6365}\text{kJ}/\text{kg}\text{fuel}$.

Explanation of Solution

Express the balanced equation.

$\left[\begin{array}{l}0.7215\text{C}+0.2535{\text{H}}_{\text{2}}+0.01618{\text{O}}_{\text{2}}+0.00711{\text{N}}_{\text{2}}+0.00177\text{S}\\ +1.2509\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.7215{\text{CO}}_{\text{2}}+0.2535{\text{H}}_{\text{2}}\text{O}+0.00177{\text{SO}}_2\\ +4.710{\text{N}}_{\text{2}}+0.4170{\text{O}}_{\text{2}}\end{array}\right]$ (I)

Here, carbon is $\text{C}$, hydrogen is ${\text{H}}_{\text{2}}$, oxygen is ${\text{O}}_{\text{2}}$, nitrogen is ${\text{N}}_{\text{2}}$, sulfur is $\text{S}$, carbon dioxide is ${\text{CO}}_{\text{2}}$, water is ${\text{H}}_{\text{2}}\text{O}$ and sulfur oxide is ${\text{SO}}_{\text{2}}$.

Express the change in entropy of the mixture during the mixture when there is no entropy change during an isentropic process.

${\displaystyle \sum y}\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)=R_a\ln \frac{P_2}{P_1}\left(N_{{\text{CO}}_{\text{2}}}+N_{{\text{H}}_{\text{2}}\text{O}}+N_{{\text{SO}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}\right)$ (II)

Here, final entropy is ${\overline{s}}_2^{\text{o}}$, initial entropy is ${\overline{s}}_1^{\text{o}}$, universal gas constant of air is $R_a$, final and initial pressure is $P_2\text{and}{P}_1$, number of moles of carbon dioxide, water, sulfur dioxide, nitrogen and oxygen is $N_{{\text{CO}}_{\text{2}}},N_{{\text{H}}_{\text{2}}\text{O}},N_{{\text{SO}}_{\text{2}}},N_{{\text{N}}_{\text{2}}},\text{and}{N}_{{\text{O}}_{\text{2}}}$ respectively.

Express the change in entropy of the mixture at temperature of $\text{1000}\text{K}$.

${\displaystyle \sum y}\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)=\left[\begin{array}{l}{N}_{{\text{CO}}_{\text{2}}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{CO}}_{\text{2}}}+N_{{\text{H}}_{\text{2}}\text{O}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{H}}_{\text{2}}\text{O}}\\ +N_{{\text{N}}_{\text{2}}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{N}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{O}}_{\text{2}}}\end{array}\right]$ (III)

Express the change in entropy of the mixture at temperature of $\text{1200}\text{K}$.

${\displaystyle \sum y}\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)=\left[\begin{array}{l}{N}_{{\text{CO}}_{\text{2}}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{CO}}_{\text{2}}}+N_{{\text{H}}_{\text{2}}\text{O}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{H}}_{\text{2}}\text{O}}\\ +N_{{\text{N}}_{\text{2}}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{N}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)}_{{\text{O}}_{\text{2}}}\end{array}\right]$ (IV)

Express the work produced during the isentropic expansion of combustion gases.

$\begin{array}{l}{\overline{w}}_{\text{out}}={\displaystyle \sum y}\left({\overline{h}}_1-{\overline{h}}_2\right)\\ =\left[\begin{array}{c}{N}_{{\text{CO}}_{\text{2}}}{\left({\overline{h}}_1-{\overline{h}}_2\right)}_{{\text{CO}}_{\text{2}}}+N_{{\text{H}}_{\text{2}}\text{O}}{\left({\overline{h}}_1-{\overline{h}}_2\right)}_{{\text{H}}_{\text{2}}\text{O}}\\ +N_{{\text{N}}_{\text{2}}}{\left({\overline{h}}_1-{\overline{h}}_2\right)}_{{\text{N}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{\left({\overline{h}}_1-{\overline{h}}_2\right)}_{{\text{O}}_{\text{2}}}\end{array}\right]\end{array}$ (V)

Here, initial and final enthalpy is ${\overline{h}}_1\text{and}{\overline{h}}_2$ respectively.

Express molar mass of the product gases.

$M_m=\frac{N_{{\text{CO}}_{\text{2}}}{M}_{{\text{CO}}_{\text{2}}}+N_{{\text{H}}_{\text{2}}\text{O}}{M}_{{\text{H}}_{\text{2}}\text{O}}+N_{{\text{SO}}_{\text{2}}}{M}_{{\text{SO}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}{M}_{{\text{N}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{M}_{{\text{O}}_{\text{2}}}}{N_{{\text{CO}}_{\text{2}}}+N_{{\text{H}}_{\text{2}}\text{O}}+N_{{\text{SO}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}}$ (VI)

Here, molar mass of carbon dioxide, water, sulfur dioxide, nitrogen and oxygen is $M_{{\text{CO}}_{\text{2}}},M_{{\text{H}}_{\text{2}}\text{O}},M_{{\text{SO}}_{\text{2}}},M_{{\text{N}}_{\text{2}}},\text{and}{M}_{{\text{O}}_{\text{2}}}$ respectively.

Express the work produced per kg of the fuel.

$w_{\text{out}}=\frac{{\overline{w}}_{\text{out}}}{M_m}$ (VII)

Conclusion:

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

$\begin{array}{l}{N}_{{\text{CO}}_{\text{2}}}=0.7215\text{kmol}\\ N_{{\text{H}}_{\text{2}}\text{O}}=0.2535\text{kmol}\\ N_{{\text{SO}}_{\text{2}}}=0.0018\text{kmol}\end{array}$

$\begin{array}{l}{N}_{{\text{N}}_{\text{2}}}=4.71\text{kmol}\\ N_{{\text{O}}_{\text{2}}}=0.4170\text{kmol}\end{array}$

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

$\begin{array}{l}{M}_{{\text{CO}}_{\text{2}}}=44\text{kg}/\text{kmol}\\ M_{{\text{H}}_{\text{2}}\text{O}}=18\text{kg}/\text{kmol}\\ M_{{\text{SO}}_{\text{2}}}=64\text{kg}/\text{kmol}\end{array}$

$\begin{array}{l}{M}_{{\text{N}}_{\text{2}}}=28\text{kg}/\text{kmol}\\ M_{{\text{O}}_{\text{2}}}=32\text{kg}/\text{kmol}\end{array}$

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

$R_a=8.314\text{kJ}/\text{kmol}\cdot \text{K}$

Substitute $8.314\text{kJ}/\text{kmol}\cdot \text{K}$ for $R_a$, $\text{140}\text{kPa}\text{and}\text{1380}\text{kPa}$ for $P_2\text{and}{P}_1$ respectively, $0.7215\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$, $0.2535\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $0.0018\text{kmol}$ for $N_{{\text{SO}}_{\text{2}}}$, $4.71\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $0.4170\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ in Equation (II).

$\begin{array}{c}{\displaystyle \sum y}\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)=\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left[\ln \frac{\text{140}\text{kPa}}{\text{1380}\text{kPa}}\right]\left[\begin{array}{l}0.7215\text{kmol}+0.2535\text{kmol}\\ +0.0018\text{kmol}+4.71\text{kmol}\\ +0.4170\text{kmol}\end{array}\right]\\ =\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(-2.288\right)\left(6.1038\text{kmol}\right)\\ =-116.1\text{kJ}/\text{K}\end{array}$

Refer Table A-18, A-19, A-20, A-21, A-22, and A-23, and write the initial and final entropy at temperature of $\text{1000}\text{K}$ as in Table (1).

Components	${\overline{s}}_1^{\text{o}}$ $\left(\text{kJ}/\text{kmol}\cdot \text{K}\right)$	${\overline{s}}_2^{\text{o}}$ $\left(\text{kJ}/\text{kmol}\cdot \text{K}\right)$
${\text{CO}}_{\text{2}}$	307.992	269.215
${\text{H}}_{\text{2}}\text{O}$	263.542	232.597
${\text{N}}_{\text{2}}$	251.424	228.057
${\text{O}}_{\text{2}}$	267.891	243.471

Substitute $0.7215\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$, $0.2535\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $4.71\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.4170\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and values from Table (I) in Equation (III) to get,

$\begin{array}{l}{\displaystyle \sum y}\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)=\left[\begin{array}{c}\left(0.7215\right)\left(269.215-307.992\right)+\left(0.2535\right)\left(232.597-263.542\right)\\ +\left(4.71\right)\left(228.057-251.242\right)+\left(0.417\right)\left(243.471-267.891\right)\end{array}\right]\\ =-155.2\text{kJ}/\text{K}\end{array}$

Refer Table A-18, A-19, A-20, A-21, A-22, and A-23, and write the initial and final entropy at temperature of $\text{1200}\text{K}$ as in Table (2).

Components	${\overline{s}}_1^{\text{o}}$ $\left(\text{kJ}/\text{K}\right)$	${\overline{s}}_2^{\text{o}}$ $\left(\text{kJ}/\text{K}\right)$
${\text{CO}}_{\text{2}}$	307.992	297.307
${\text{H}}_{\text{2}}\text{O}$	263.542	240.333
${\text{N}}_{\text{2}}$	251.242	234.115
${\text{O}}_{\text{2}}$	267.891	249.906

Substitute $0.7215\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$, $0.2535\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $4.71\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.4170\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and values from Table (II) in Equation (IV) to get,

$\begin{array}{l}{\displaystyle \sum y}\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)=\left[\begin{array}{c}\left(0.7215\right)\left(297.307-307.992\right)+\left(0.2535\right)\left(240.333-263.542\right)\\ +\left(4.71\right)\left(234.115-251.242\right)+\left(0.417\right)\left(249.906-267.891\right)\end{array}\right]\\ =-101.8\text{kJ}/\text{K}\end{array}$

Perform the interpolation method to obtain the final temperature.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (VIII)

Here, the variables denote by x and y is entropy change and final temperature respectively.

Show the final temperature corresponding to entropy change as in Table (3).

Entropy change $\begin{array}{l}{\displaystyle \sum y}\left({\overline{s}}_2^{\text{o}}-{\overline{s}}_1^{\text{o}}\right)\\ \left(\text{kJ}/\text{K}\right)\end{array}$	Final temperature $T_2\left(\text{K}\right)$
$-155.2$ $\left(x_1\right)$	1000 $\left(y_1\right)$
$-116.1$ $\left(x_2\right)$	$\left(y_2=?\right)$
$-101.8$ $\left(x_3\right)$	1200 $\left(y_3\right)$

Substitute $-155.2\text{kJ}/\text{K}\text{,}-116.1\text{kJ}/\text{K}\text{and}-101.8\text{kJ}/\text{K}$ for $x_1,x_2\text{and}{x}_3$ respectively, $\text{1000}\text{K}$ for $y_1$ and $\text{1200}\text{K}$ for $y_3$ in Equation (VIII).

$\begin{array}{c}{y}_2=\frac{\left[\left\{-116.1-\left(-155.2\text{kJ}/\text{K}\right)\right\}\text{kJ}/\text{K}\right]\left[\left(1200-\text{1000}\right)\text{K}\right]}{\left[\left\{-101.8-\left(-155.2\text{kJ}/\text{K}\right)\right\}\text{kJ}/\text{K}\right]}+\text{1000}\text{K}\\ =1146\text{K}\\ =T_2\end{array}$

Thus, the final temperature is,

$T_2=1146\text{K}$

Refer Table A-18, A-19, A-20, A-21, A-22, and A-23, and write the initial and final enthalpy during the isentropic expansion of combustion gases from $\text{1956}\text{K}\text{to}\text{1146}\text{K}$ as in Table (4).

Components	${\overline{h}}_1$ $\left(\text{kJ}/\text{kmol}\right)$	${\overline{h}}_2$ $\left(\text{kJ}/\text{kmol}\right)$
${\text{CO}}_{\text{2}}$	98,160	50,189
${\text{H}}_{\text{2}}\text{O}$	80,352	42,037
${\text{N}}_{\text{2}}$	63,236	34,961
${\text{O}}_{\text{2}}$	66,223	36,527

Substitute $0.7215\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$, $0.2535\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $4.71\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.4170\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and values from Table (III) in Equation (V) to get,

$\begin{array}{l}{\overline{w}}_{\text{out}}=\left[\begin{array}{c}\left(0.7215\right)\left(98,160-50,819\right)+\left(0.2535\right)\left(80,352-42,037\right)\\ +\left(4.71\right)\left(63,236-34,961\right)+\left(0.417\right)\left(66,223-36,527\right)\end{array}\right]\\ =189,428\text{kJ}/\text{kmol}\text{fuel}\end{array}$

Substitute $0.7215\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$, $0.2535\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $4.71\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.4170\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.0018\text{kmol}$ for $N_{{\text{SO}}_{\text{2}}}$, $44\text{kg}/\text{kmol}$ for $M_{{\text{CO}}_{\text{2}}}$, $18\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$, $64\text{kg}/\text{kmol}$ for $M_{{\text{SO}}_{\text{2}}}$, $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (VI).

$\begin{array}{c}{M}_m=\frac{\left[\left(0.7215\times 44\right)+\left(0.2535\times 18\right)+\left(0.0018\times 64\right)+\left(4.71\times 28\right)+\left(0.4170\times 32\right)\right]\text{kg}}{\left(0.7215+0.2535+0.0018+4.71+0.417\right)\text{kmol}}\\ =29.76\text{kg}/\text{kmol}\end{array}$

Substitute $189,428\text{kJ}/\text{kmol}\text{fuel}$ for ${\overline{w}}_{\text{out}}$ and $29.76\text{kg}/\text{kmol}$ for $M_m$ in Equation (VII).

$\begin{array}{c}{w}_{\text{out}}=\frac{189,428\text{kJ}/\text{kmol}\text{fuel}}{29.76\text{kg}/\text{kmol}}\\ =\text{6365}\text{kJ}/\text{kg}\text{fuel}\end{array}$

Hence, the work produced by an isentropic turbine is $\text{6365}\text{kJ}/\text{kg}\text{fuel}$.