Chapter 10.9, Problem 80P

a)

To determine

The ratio of mass flow rate of air to mass flow rate of steam in the combined power cycle

Answer to Problem 80P

The ratio of mass flow rate of air to mass flow rate of steam in the combined power cycle is $\underset{\_}{6.426\text{kg air}/\text{kg steam}}$.

Explanation of Solution

Draw the T-s diagram for the combined gas-steam power cycle as shown in Figure (1).

Write the expression for the relation relative pressure and ideal pressure.

$P_{r9}=\frac{P_9}{P_8}{P}_{r8}$ (I).

Here, the relative pressure at state 9 is $P_{r9}$, and relative pressure at state 8 is $P_{r8}$.

Write the expression for the efficiency of compressor.

${\eta }_C=\frac{h_{9s}-h_8}{h_9-h_8}$ (II).

Refer to properties table of air, and interpret the value of $h_{9s}$ for a pressure of 19.40 kPa as $\text{635}\text{.5}\text{kJ}/\text{kg}$ .

Write the expression for the relation relative pressure and ideal pressure.

$P_{r11}=\frac{P_{11}}{P_{10}}{P}_{r10}$ (III).

Here, the relative pressure at state 11 is $P_{r11}$, and relative pressure at state 10 is $P_{r10}$.

Write the expression for the efficiency of turbine.

${\eta }_T=\frac{h_{10}-h_{11}}{h_{10}-h_{11s}}$ (IV).

Write the expression for the specific work input of pump I to the system/

$w_{\text{pI,in}}=v_1\left(P_2-P_1\right)$ (V).

Write the expression for the enthalpy of steam at state 2.

$h_2=h_1+w_{\text{pI,in}}$ (VI).

Write the expression for the specific work input of pump II to the system/

$w_{\text{pII,in}}=v_3\left(P_4-P_3\right)$ (VII).

Write the expression for the enthalpy of steam at state 4.

$h_4=h_3+w_{\text{pII,in}}$ (VIII).

Write the expression for the quality of steam at state 6s.

$x_{6s}=\frac{s_{6s}-s_f}{s_{fg}}$ (IX).

Here, specific entropy of wet steam at 0.6 MPa is $s_f$, and specific entropy of vaporization at 0.6 MPa is $s_{fg}$.

Write the expression for the specific enthalpy of steam at state 6s.

$h_{6s}=h_f+x_{6s}{h}_{fg}$ (X).

Here, specific enthalpy of wet steam at $0.6\text{MPa}$ is $h_f$, and specific enthalpy of vaporization at $0.6\text{MPa}$ is $h_{fg}$.

Write the expression for the efficiency of turbine.

${\eta }_T=\frac{h_5-h_6}{h_5-h_{6s}}$ (XI).

Write the expression for the quality of steam at state 7s.

$x_{7s}=\frac{s_7-s_f}{s_{fg}}$ (XII).

Here, specific entropy of wet steam at 20 kPa is $s_f$, and specific entropy of vaporization at 20 kPa is $s_{fg}$.

Write the expression for the specific enthalpy of steam at state 7s.

$h_{7s}=h_f+x_{7s}{h}_{fg}$ (XIII).

Here, specific enthalpy of wet steam at $20\text{kPa}$ is $h_f$, and specific enthalpy of vaporization at $20\text{kPa}$ is $h_{fg}$.

Write the expression for the efficiency of turbine.

${\eta }_T=\frac{h_5-h_7}{h_5-h_{7s}}$ (XIV).

Write the expression for the energy balance equation for the heat exchanger.

${\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}$ (XV).

Rewrite Equation (1) and rearrange the terms with mass and enthalpy terms.

${\dot{m}}_s{h}_5+{\dot{m}}_{\text{air}}{h}_{12}={\dot{m}}_s{h}_4+{\dot{m}}_{\text{air}}{h}_{11}$

${\dot{m}}_s\left(h_5-h_4\right)={\dot{m}}_{\text{air}}\left(h_{11}-h_{12}\right)$

$\frac{{\dot{m}}_{\text{air}}}{{\dot{m}}_s}=\frac{\left(h_5-h_4\right)}{\left(h_{11}-h_{12}\right)}$ (XVI).

Here, mass flow rate of steam is ${\dot{m}}_s$, and mass flow rate of air is ${\dot{m}}_{\text{air}}$.

Conclusion:

From the Table A-17, “Ideal-gas properties of air ”, select the relative pressure $P_{r8}$ and the enthalpy $h_8$ for air gas as 1.386 kPa, and $\text{300}\text{.19}\text{kJ}/\text{kg}$ respectively.

Substitute 1.386 kPa for $P_{r8}$, and 14 for $\frac{P_9}{P_8}$ in equation (I).

$\begin{array}{c}{P}_{r9}=14\times 1.386\\ =19.40\text{kPa}\end{array}$

From the Table A-17, “Ideal-gas properties of air ”, interpret the value of the the enthalpy $h_8$ for 19.40 kPa pressure as  $\text{635}\text{.5}\text{kJ}/\text{kg}$.

Substitute 0.82 for ${\eta }_C$, $\text{635}\text{.5}\text{kJ}/\text{kg}$ for $h_{9s}$, and $\text{300}\text{.19}\text{kJ}/\text{kg}$ for $h_8$ in equation (II)

$\begin{array}{l}0.82=\frac{635.5-300.19}{h_9-300.19}\\ h_9-300.19=408.91\\ h_9=709.1\text{kJ}/\text{kg}\end{array}$

From the Table A-17, “Ideal-gas properties of air ”, select the relative pressure $P_{r10}$ and the enthalpy $h_{10}$ for air gas as 450.5 kPa, and $\text{1,515}\text{.42}\text{kJ}/\text{kg}$ respectively.

Substitute 450.5 kPa for $P_{r10}$, and $\frac{1}{14}$ for $\frac{P_9}{P_8}$ in equation (III).

$\begin{array}{c}{P}_{r9}=\frac{1}{14}\times 450.5\\ =32.18\text{kPa}\end{array}$

From the Table A-17 “Ideal-gas properties of air , interpret the value of $h_{11s}$ for a pressure of 32.18 kPa as $\text{735}\text{.8}\text{kJ}/\text{kg}$.

Substitute 0.86 for ${\eta }_T$, $\text{1,515}\text{.42}\text{kJ}/\text{kg}$ for $h_{10}$, and $\text{735}\text{.8}\text{kJ}/\text{kg}$ for $h_{11s}$ in equation (IV).

$\begin{array}{l}0.86=\frac{1,515.42-h_{11}}{1,515.42-735.8}\\ 1,515.42-h_{11}=670.47\\ h_{11}=844.94\text{kJ}/\text{kg}\end{array}$

From the Table A-5, “saturated water-Pressure table”, select the enthalpy $h_1$ and the specific volume $v_1$ for the pressure of $20\text{kPa}$ as $\text{251}\text{.42}\text{kJ}/\text{kg}$ and $0.001017{\text{m}}^3/\text{kg}$ respectively.

Substitute $0.001017{\text{m}}^3/\text{kg}$ for $v_1$, $20\text{kPa}$ for $P_1$, and $600\text{kPa}$ for $P_2$ in equation (V).

$\begin{array}{c}{w}_{\text{pI,in}}=0.001017{\text{m}}^3/\text{kg}\left(600-20\right)\text{kPa}\left(\frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\right)\\ =0.59\text{kJ}/\text{kg}\end{array}$

Substitute $\text{251}\text{.42}\text{kJ}/\text{kg}$ for $h_1$, and $0.59\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in equation (VI)

$\begin{array}{c}{h}_2=251.42\text{kJ}/\text{kg}+0.59\text{kJ}/\text{kg}\\ =252.01\text{kJ}/\text{kg}\end{array}$

From the Table A-5, “saturated water-Pressure table”, select the enthalpy $h_1$ and the specific volume $v_1$ for the pressure of $600\text{kPa}$ as $\text{670}\text{.38}\text{kJ}/\text{kg}$ and $0.001101{\text{m}}^3/\text{kg}$ respectively.

Substitute $0.001101{\text{m}}^3/\text{kg}$ for $v_3$, $600\text{kPa}$ for $P_3$, and $8,000\text{kPa}$ for $P_4$ in equation (VII)

$\begin{array}{c}{w}_{\text{pI,in}}=0.001101{\text{m}}^3/\text{kg}\left(8,000-600\right)\text{kPa}\left(\frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\right)\\ =8.15\text{kJ}/\text{kg}\end{array}$

Substitute $\text{670}\text{.38}\text{kJ}/\text{kg}$ for $h_3$, and $8.15\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in equation (VIII).

$\begin{array}{c}{h}_4=670.38\text{kJ}/\text{kg}+8.15\text{kJ}/\text{kg}\\ =678.53\text{kJ}/\text{kg}\end{array}$

From the Table A-5, “Superheated water”, select the enthalpy $h_5$ and the entropy $s_5$ for the pressure and temperature of $8\text{MPa}$ and $400°\text{C}$ as $\text{3,139}\text{.4}\text{kJ}/\text{kg}$ and $\text{6}\text{.3658}\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Since, the entropy at state 5 is equal to state 6s, so the entropy value of $s_{6s}$ is $\text{6}\text{.3658}\text{kJ}/\text{kg}\cdot \text{K}$.

Interpret the value of $s_f$ and $s_{fg}$ for pressure of 0.6 MPa as $\text{1}\text{.9308}\text{kJ}/\text{kg}\cdot \text{K}$ and $4.8285\text{kJ}/\text{kg}\cdot \text{K}$ .

Substitute $\text{6}\text{.3658}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{6s}$, $\text{1}\text{.9308}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, and $\text{4}\text{.8285}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in equation (IX).

$\begin{array}{c}{x}_{6s}=\frac{6.3658-1.9308}{4.8285}\\ =0.9185\end{array}$

Refer to steam tables, and interpret the value of $h_f$ and $h_{fg}$ for a pressure of $0.6\text{MPa}$ as $670.38\text{kJ}/\text{kg}$ and $2,085.9\text{kJ}/\text{kg}$.

Substitute $670.38\text{kJ}/\text{kg}$ for $h_f$, $2,085.9\text{kJ}/\text{kg}$ for $h_{fg}$, and 0.9185 for $x_{6s}$ in equation (X).

$\begin{array}{c}{h}_{6s}=670.38\text{kJ}/\text{kg}+0.9185\left(2,085.9\text{kJ}/\text{kg}\right)\\ =2,586.28\text{kJ}/\text{kg}\end{array}$

Substitute 0.86 for ${\eta }_T$, $\text{3,139}\text{.4}\text{kJ}/\text{kg}$ for $h_5$, and $\text{2,586}\text{.28}\text{kJ}/\text{kg}$ for $h_{6s}$ in equation (XI).

$\begin{array}{l}0.86=\frac{3,139.4-h_6}{3,139.4-2,586.28}\\ 3,139.4-h_6=389.68\\ h_6=2,749.7\text{kJ}/\text{kg}\end{array}$

Since, the entropy at state 5 is equal to the entropy at state 7, the entropy value of $s_7$ is $\text{6}\text{.3658}\text{kJ}/\text{kg}\cdot \text{K}$.

Refer to steam tables A-5 “saturated water-Pressure table”, obtain the value of $s_f$ and $s_{fg}$ for a pressure of $20\text{kPa}$ as $\text{0}\text{.8320}\text{kJ}/\text{kg}\cdot \text{K}$ and $7.0752\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Substitute $\text{6}\text{.3658}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_7$, $\text{0}\text{.8320}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, and $\text{7}\text{.0752}\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in equation (XII).

$\begin{array}{c}{x}_{6s}=\frac{6.3658-0.8320}{7.0752}\\ =0.7821\end{array}$

Refer to steam tables A-5 “saturated water-Pressure table”, obtain the value of $h_f$ and $h_{fg}$ for a pressure of $20\text{kPa}$ as $251.42\text{kJ}/\text{kg}$ and $2357.5\text{kJ}/\text{kg}$ respectively.

Substitute $251.42\text{kJ}/\text{kg}$ for $h_f$, $2,357.5\text{kJ}/\text{kg}$ for $h_{fg}$, and 0.7821 for $x_{7s}$ in equation (XIII)

$\begin{array}{c}{h}_{7s}=251.42\text{kJ}/\text{kg}+0.7821\left(2,357.5\text{kJ}/\text{kg}\right)\\ =2,095.2\text{kJ}/\text{kg}\end{array}$

Substitute 0.86 for ${\eta }_T$, $\text{3,139}\text{.4}\text{kJ}/\text{kg}$ for $h_5$, and $\text{2,095}\text{.2}\text{kJ}/\text{kg}$ for $h_{7s}$ in equation (XIV).

$\begin{array}{l}0.86=\frac{3,139.4-h_6}{3,139.4-2,095.2}\\ 3,139.4-h_6=898.012\\ h_6=2,241.4\text{kJ}/\text{kg}\end{array}$

Refer to steam tables A-17 “Ideal gas properties of air”, obtain the value of $h_{12}$ temperature of 460 K as $\text{462}\text{.02}\text{kJ}/\text{kg}$.

Substitute $3,139.4\text{kJ}/\text{kg}$ for $h_5$, $678.52\text{kJ}/\text{kg}$ for $h_4$, and $844.95\text{kJ}/\text{kg}$ for $h_{11}$, $462.02\text{kJ}/\text{kg}$ for $h_{12}$ in Equation (XVI).

$\begin{array}{c}\frac{{\dot{m}}_{\text{air}}}{{\dot{m}}_s}=\frac{\left(3,139.4-678.52\right)}{\left(844.95-462.02\right)}\\ =6.426\text{kg air}/\text{kg steam}\end{array}$

Thus, the ratio of mass flow rate of air to mass flow rate of steam in the combined power cycle is $\underset{\_}{6.426\text{kg air}/\text{kg steam}}$.

b)

To determine

The rate of heat input in the combustion chamber.

Answer to Problem 80P

The rate of heat input in the combustion chamber is $\underset{\_}{933,879\text{kW}}$.

Explanation of Solution

Write the expression for the energy balance equation for the open feed water heater.

${\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}$ (XVII)

Rewrite Equation (3) and rearrange the terms with mass and enthalpy terms.

${\dot{m}}_2{h}_2+{\dot{m}}_6{h}_6={\dot{m}}_3{h}_3$

$y{h}_6+\left(1-y\right)h_2=h_3$

$y=\frac{h_3-h_2}{h_6-h_2}$ (XVIIII)

Here, fraction of steam extracted is y.

Write the expression for the specific power output of the turbine.

$w_T={\eta }_T\left[h_5-h_6+\left(1-y\right)\left(h_6-h_7\right)\right]$ (XIX)

Write the expression for the specific net work output from the steam.

$w_{\text{net,steam}}=w_T-\left(1-y\right)w_{\text{pI,in}}-w_{\text{pII,in}}$ (XX)

Write the expression for the specific net work output from the gas stream.

$w_{\text{net,gas}}=\left(h_{10}-h_{11}\right)-\left(h_9-h_8\right)$ (XXI)

Write the expression for the net work output per unit mass of gas.

$w_{\text{net}}=w_{\text{net,gas}}+\frac{1}{6.423}{w}_{\text{net,steam}}$ (XXII)

Write the expression for the mass flow rate of air.

${\dot{m}}_{\text{air}}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}$ (XXIII)

Write the expression for the rate of heat input to the cycle.

${\dot{Q}}_{\text{in}}={\dot{m}}_{\text{air}}\left(h_{10}-h_9\right)$ (XXIV)

Conclusion:

Substitute $670.38\text{kJ}/\text{kg}$ for $h_3$, $252.01\text{kJ}/\text{kg}$ for $h_2$, and $2,663.3\text{kJ}/\text{kg}$ for $h_6$ in Equation (XVIIII).

$\begin{array}{c}y=\frac{670.38-252.01}{2,663.3-252.01}\\ =0.1735\end{array}$

Substitute 0.86 for ${\eta }_T$, 0.1735 for y, $3,139.4\text{kJ}/\text{kg}$ for $h_5$, $2,241.3\text{kJ}/\text{kg}$ for $h_7$, and $2,663.3\text{kJ}/\text{kg}$ for $h_6$ in equation (XIX).

$\begin{array}{c}{w}_T=0.86\left[3,139.4-2,663.3+\left(1-0.1735\right)\left(2,663.3-2,241.3\right)\right]\\ =824.5\text{kJ}/\text{kg}\end{array}$

Substitute $824.5\text{kJ}/\text{kg}$ for $w_T$, and 0.1735 for y, $0.59\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$, and $8.15\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$.

$\begin{array}{c}{w}_{\text{net,steam}}=824.5-\left(1-0.1735\right)\left(0.59\right)-8.15\\ =815.9\text{kJ}/\text{kg}\end{array}$

Substitute $1,515.42\text{kJ}/\text{kg}$ for $h_{10}$, and $844.95\text{kJ}/\text{kg}$ for $h_{11}$, $709.1\text{kJ}/\text{kg}$ for $h_9$, and $300.19\text{kJ}/\text{kg}$ for $h_8$.

$\begin{array}{c}{w}_{\text{net,gas}}=1,515.42-844.95-\left(709.1-300.19\right)\\ =261.56\text{kJ}/\text{kg}\end{array}$

Substitute $261.56\text{kJ}/\text{kg}$ for $w_{\text{net,gas}}$, and $815.9\text{kJ}/\text{kg}$ for $w_{\text{net,steam}}$ in equation (XXII)

$\begin{array}{c}{w}_{\text{net}}=261.56+\frac{1}{6.423}\times 815.9\\ =388.58\text{kJ}/\text{kg}\end{array}$

Substitute $388.58\text{kJ}/\text{kg}$ for $w_{\text{net}}$, and $450,000\text{kJ}/\text{s}$ for ${\dot{W}}_{\text{net}}$. In equation (XXIIII)

$\begin{array}{c}{\dot{m}}_{\text{air}}=\frac{450,000\text{kJ}/\text{s}}{388.58\text{kJ}/\text{kg}}\\ =1,158.2\text{kg}/\text{s}\end{array}$

Substitute $1,158.2\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, $1,515.42\text{kJ}/\text{kg}$ for $h_{10}$, and $709.1\text{kJ}/\text{kg}$ for $h_9$.

$\begin{array}{c}{\dot{Q}}_{\text{in}}=1,158.2\left(1,515.42-709.1\right)\\ =933,879\text{kW}\end{array}$

Thus, the rate of heat input in the combustion chamber is $\underset{\_}{933,879\text{kW}}$.

c)

To determine

The thermal efficiency of the combined power cycle

Answer to Problem 80P

The thermal efficiency of the combined power cycle is $\underset{\_}{48.18\%}$.

Explanation of Solution

Write the expression for the thermal efficiency of the combined power cycle.

${\eta }_{\text{th}}=\frac{{\dot{W}}_{\text{net}}}{{\dot{Q}}_{\text{in}}}$ (XXV)

Conclusion:

Substitute $933,879\text{kW}$ for ${\dot{Q}}_{\text{in}}$, and  $450,000\text{kW}$ for ${\dot{W}}_{\text{net}}$ in equation (XXV).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{450,000\text{kW}}{933,879\text{kW}}\\ =0.4818\end{array}$

Thus, the thermal efficiency of the combined power cycle is $\underset{\_}{48.18\%}$.