Chapter 9.12, Problem 126P

Repeat Prob. 9–125, assuming an efficiency of 86 percent for each compressor stage and an efficiency of 90 percent for each turbine stage.

To determine

The back work ratio and thermal efficiency of the gas-turbine without regenerator.

Answer to Problem 126P

The back work ratio for the ideal gas-turbine cycle without regenerator is $43.3\%$.

The thermal efficiency of the gas-turbine without regenerator is $29.5\%$.

Explanation of Solution

Draw the $T-s$ diagram for an ideal gas turbine cycle with two stages of compression and expansion as shown in Figure (1).

Write the pressure ratio relation for the process 1-2.

$P_{r_2}=\frac{P_2}{P_1}{P}_{r_1}$ (I)

Here, relative pressure at state 1 is $P_{r_1}$ and relative pressure at state 2 is $P_{r_2}$.

Write the pressure ratio relation for the process 5-6.

$P_{r_6}=\frac{P_6}{P_5}{P}_{r_5}$ (II)

Here, pressure at state 6 is $P_6$ , pressure at state 5 is $P_5$ , relative pressure at state 6 is $P_{r_6}$ and relative pressure at state 5 is $P_{r_5}$.

Write the expression to calculate the work input per kg to the compressors $\left(w_{\text{C,in}}\right)$.

$w_{\text{C,in}}=2\left(h_2-h_1\right)$ (III)

Here, enthalpy at state 2 is $h_2$  and enthalpy at state 1 is $h_1$ .

Write the expression to calculate the work done per kg by the turbines $\left(w_{\text{T,out}}\right)$.

$w_{\text{T,out}}=2\left(h_5-h_6\right)$ (IV)

Here, enthalpy at state 5 is $h_5$  and enthalpy at state 6 is $h_6$.

Write the expression to calculate the back work ratio $\left(r_{bw}\right)$.

$r_{bw}=\frac{w_{\text{C,in}}}{w_{\text{T,out}}}$ (V)

Write the expression to calculate the heat input for ideal gas-turbine cycle

$\left(q_{\text{in}}\right)$.

$q_{\text{in}}=\left(h_5-h_4\right)+\left(h_7-h_6\right)$ (VI)

Here, enthalpy at state 4 is $h_4$  and enthalpy at state 7 is $h_7$.

Write the expression to calculate the net work output per kg by the gas-turbine cycle

$\left(w_{\text{net}}\right)$.

$w_{\text{net}}=w_{\text{T,out}}-w_{\text{C,in}}$ (VII)

Write the expression to calculate the thermal efficiency of the gas-turbine cycle $\left({\eta }_{\text{th}}\right)$.

${\eta }_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}$ (VIII)

Write the expression for the efficiency of the compressor $\left({\eta }_C\right)$

$\begin{array}{l}{\eta }_C=\frac{h_{2s}-h_1}{h_2-h_1}\\ h_2=h_1+\left(h_{2s}-h_1\right)/{\eta }_C\end{array}$ (IX)

Here, the specific heat at constant pressure is $c_p$.

Write the expression for the efficiency of the turbine $\left({\eta }_T\right)$.+

$\begin{array}{l}{\eta }_T=\frac{h_5-h_6}{h_5-h_{6s}}\\ h_6=h_5-{\eta }_T\left(h_5-h_{6s}\right)\end{array}$ (X)

Conclusion:

From Table A-17, “Ideal-gas properties of air”, obtain the following properties at the temperature of $300\text{K}$.

$\begin{array}{l}{h}_1=300.19\text{kJ}/\text{kg}\\ P_{r_1}=1.386\end{array}$

Substitute 3 for $\frac{P_2}{P_1}$, and 1.386 for $P_{r_1}$ in Equation (I).

$\begin{array}{c}{P}_{r_2}=\left(3\right)\left(1.386\right)\\ =4.158\end{array}$

From the Table A-17, “Ideal-gas properties of air”.

Obtain the value of enthalpy on isentropic state $\left(h_{2s}\right)$ at the relative pressure $\left(P_{r_2}\right)$ of $4.158$ by using interpolation method.

Write the formula of interpolation method of two variables.

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

Here, the variables denoted by x and y are relative pressure and enthalpy on isentropic state.

Show relative pressure and enthalpy on isentropic state values from the Table A-17.

Relative pressure $\left(P_{r_2}\right)$	Enthalpy $\left(h_{2s}\right)$, in $\text{kJ}/\text{kg}$
4.153	411.12
4.158	?
4.522	421.26

Substitute $4.153$ for $x_1$, $4.158$ for $x_2$, $4.522$ for $x_3$, $411.12$ for $y_1$, and $421.26$ for $y_3$ in Equation (XI).

$\begin{array}{c}{y}_2=\frac{\left(4.158-4.153\right)\left(421.26-411.12\right)}{\left(4.522-4.153\right)}+411.12\\ =411.257\\ =411.26\end{array}$

The enthalpy on isentropic state $\left(h_{2s}\right)$ at the relative pressure $\left(P_{r_2}\right)$ of $4.158$ by using interpolation method is $411.26\text{kJ}/\text{kg}$.

Substitute $300.19\text{kJ}/\text{kg}$ for $h_1$, $411.26\text{kJ}/\text{kg}$ for $h_{2s}$, and $0.86$ for ${\eta }_C$ in Equation (IX).

$\begin{array}{c}{h}_2=300.19\text{kJ}/\text{kg}+\left(411.26\text{kJ}/\text{kg}-300.19\text{kJ}/\text{kg}\right)/0.86\\ h_2=h_4=429.34\text{kJ}/\text{kg}\end{array}$

From Table A-17, “Ideal-gas properties of air”, obtain the following properties at the temperature of $1200\text{K}$.

$\begin{array}{l}{h}_5=h_7=1277.79\text{kJ}/\text{kg}\\ P_{r_5}=238\end{array}$

Substitute $\frac{1}{3}$ for $\frac{P_6}{P_5}$, and 238 for $P_{r_5}$ in Equation (II).

$\begin{array}{c}{P}_{r_6}=\left(\frac{1}{3}\right)\left(238\right)\\ =79.33\end{array}$

From the Table A-17, “Ideal-gas properties of air” obtain the values of enthalpy on isentropic states $h_{6s}$ and $h_{8s}$.at the relative pressure $\left(P_{r_4}\right)$ of $79.33$ as $946.36\text{kJ}/\text{kg}$ by interpolation method as shown in Equation (XI).

Substitute $1277.79\text{kJ}/\text{kg}$ for $h_5$, $946.36\text{kJ}/\text{kg}$ for $h_{6s}$, and 0.90 for ${\eta }_T$ in Equation (X).

$\begin{array}{c}{h}_6=1277.79\text{kJ}/\text{kg}-0.90\left(1277.79\text{kJ}/\text{kg}-946.36\text{kJ}/\text{kg}\right)\\ h_6=h_8=979.50\text{}\text{kJ}/\text{kg}\end{array}$

Substitute $429.34\text{kJ}/\text{kg}$ for $h_2$, and $300.19\text{kJ}/\text{kg}$ for $h_1$ in Equation (III).

$\begin{array}{c}{w}_{\text{C,in}}=2\left(429.34\text{kJ}/\text{kg}-300.19\text{kJ}/\text{kg}\right)\\ =258.3\text{kJ}/\text{kg}\end{array}$

Substitute $1277.79\text{kJ}/\text{kg}$ for $h_5$, and $258.3\text{kJ}/\text{kg}$ for $h_6$ in Equation (IV).

$\begin{array}{c}{w}_{\text{T,out}}=2\left(1277.79\text{kJ}/\text{kg}-979.50\text{}\text{kJ}/\text{kg}\right)\\ =596.6\text{kJ}/\text{kg}\end{array}$

Substitute $258.3\text{kJ}/\text{kg}$ for $w_{\text{C,in}}$, and $596.6\text{kJ}/\text{kg}$ for $w_{\text{T,out}}$ in Equation (V).

$\begin{array}{c}{r}_{bw}=\frac{258.3\text{kJ}/\text{kg}}{596.6\text{kJ}/\text{kg}}\\ =0.433\times 100\%\\ =43.3\%\end{array}$

Thus, the back work ratio for the ideal gas-turbine cycle without regenerator is $43.3\%$.

Substitute $1277.79\text{kJ}/\text{kg}$ for $h_5$, $429.34\text{kJ}/\text{kg}$ for $h_4$, $1277.79\text{kJ}/\text{kg}$ for $h_7$, and $979.50\text{}\text{kJ}/\text{kg}$ for $h_6$ in Equation (VI).

$\begin{array}{c}{q}_{\text{in}}=\left(1277.79-429.34\right)\text{kJ}/\text{kg}+\left(1277.79-979.50\right)\text{kJ}/\text{kg}\\ =1146.7\text{kJ}/\text{kg}\end{array}$

Substitute $596.6\text{kJ}/\text{kg}$ for $w_{\text{T,out}}$, and $258.3\text{kJ}/\text{kg}$ for $w_{\text{C,in}}$ in Equation (VII).

$\begin{array}{c}{w}_{\text{net}}=\left(596.6\text{kJ}/\text{kg}-258.3\text{kJ}/\text{kg}\right)\\ =338.3\text{kJ}/\text{kg}\end{array}$

Substitute $338.3\text{kJ}/\text{kg}$ for $w_{\text{net}}$, and $1146.7\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (VIII).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{338.3\text{kJ}/\text{kg}}{1146.7\text{kJ}/\text{kg}}\\ =0.295\times 100\%\\ =29.5\%\end{array}$

Thus, the thermal efficiency of the gas-turbine without regenerator is $29.5\%$.

To determine

The thermal efficiency of the gas turbine with regenerator.

Answer to Problem 126P

The thermal efficiency of the gas turbine with regenerator is $46.1\%$.

Explanation of Solution

Write the expression to calculate the heat used for the regeneration process $\left(q_{\text{regen}}\right)$.

$q_{\text{regen}}=\epsilon \left(h_8-h_4\right)$ (X)

Here, the effectiveness of the regenerator is $\epsilon$ and enthalpy at state 8 is $h_8$.

Write the expression to calculate the new heat input to the gas-turbine cycle $\left(q_{\text{in}}\right)$.

$q_{\text{in}}=q_{\text{in,old}}-q_{\text{regen}}$ (XI)

Write the expression to calculate the thermal efficiency of the gas-turbine with regenerator $\left({\eta }_{\text{th}}\right)$.

${\eta }_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}$ (XII)

Conclusion:

Substitute 0.75 for $\epsilon$, $979.50\text{kJ}/\text{kg}$ for $h_8$, and $429.34\text{kJ}/\text{kg}$ for $h_4$ in Equation (X).

$\begin{array}{c}{q}_{\text{regen}}=\left(0.75\right)\left(946.36\text{kJ}/\text{kg}-429.34\text{kJ}/\text{kg}\right)\\ =412.6\text{kJ}/\text{kg}\end{array}$

Substitute $1146.7\text{kJ}/\text{kg}$ for $q_{\text{in,old}}$, and $412.6\text{kJ}/\text{kg}$ for $q_{\text{regen}}$ in Equation (XI).

$\begin{array}{c}{q}_{\text{in}}=\left(1146.7\text{kJ}/\text{kg}-412.6\text{kJ}/\text{kg}\right)\\ =734.1\text{kJ}/\text{kg}\end{array}$

Substitute $338.3\text{kJ}/\text{kg}$ for $w_{\text{net}}$, and $734.1\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (XII).

$\begin{array}{c}{\eta }_{\text{th}}=\frac{338.3\text{kJ}/\text{kg}}{734.1\text{kJ}/\text{kg}}\\ =0.461\times 100\%\\ =46.1\%\end{array}$

Thus, the thermal efficiency of the gas turbine with regenerator is $46.1\%$.