Chapter 9, Problem 177RQ

a)

To determine

The thermal efficiency and the required mass flow rate of helium when the isentropic efficiency is 100%.

Explanation of Solution

Given:

Pressure ratio $\left(r_p\right)$ is $8$.

Net power output $\left({\dot{W}}_{\text{net}}\right)$ is $\text{60}\text{MW}$.

Effectiveness $\left(\epsilon \right)$ is $0.75$.

Compressor inlet temperature $\left(T_1\right)$ is $\text{300}\text{K}$.

Turbine inlet temperature $\left(T_3\right)$ is $\text{1800}\text{K}$.

Calculation:

Draw the $T-s$ diagram of the cycle as in Figure (1).

Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties of the helium.

  $\begin{array}{l}{c}_p=5.1926\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.667\end{array}$

Calculate the temperature at state 2s$\left(T_{2s}\right)$.

  $\begin{array}{c}{T}_{2s}=T_1{\left(\frac{P_1}{P_2}\right)}^{\frac{k-1}{k}}\\ =T_1{\left(r_P\right)}^{\frac{k-1}{k}}\\ =\left(300\text{K}\right){\left(8\right)}^{\frac{1.667-1}{1.667}}\\ =689.4\text{K}\end{array}$

Calculate the temperature at state 4s$\left(T_{4s}\right)$.

  $\begin{array}{c}{T}_{4s}=T_3{\left(\frac{P_4}{P_{32}}\right)}^{\frac{k-1}{k}}\\ =T_3{\left(\frac{1}{r_P}\right)}^{\frac{k-1}{k}}\\ =\left(1800\text{K}\right){\left(\frac{1}{8}\right)}^{\frac{1.667-1}{1.667}}\\ =783.3\text{K}\end{array}$

Calculate the temperature at state 5$\left(T_5\right)$.

  $\begin{array}{c}{T}_5=T_2+\epsilon \left(T_4-T_2\right)\\ =689.4\text{K}+\left(0.75\right)\left(783.3\text{K}-689.4\text{K}\right)\\ =759.8\text{K}\end{array}$

Calculate the mass flow rate of the helium $\left(\dot{m}\right)$.

  $\begin{array}{c}\dot{m}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}\\ =\frac{{\dot{W}}_{\text{net}}}{c_p\left[\left(T_3-T_4\right)-\left(T_2-T_1\right)\right]}\\ =\frac{60,000\text{kJ}/\text{s}}{\left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\begin{array}{l}\left(1800\text{K}-783.3\text{K}\right)-\\ \left(689.4\text{K}-300\text{K}\right)\end{array}\right]}\\ =18.42\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the helium is $18.42\text{kg}/\text{s}$.

Calculate the thermal efficiency of the cycle $\left({\eta }_{\text{th}}\right)$.

  $\begin{array}{c}{\eta }_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}\\ =\frac{c_p\left[\left(T_3-T_4\right)-\left(T_2-T_1\right)\right]}{c_p\left[\left(T_3-T_5\right)\right]}\end{array}$

  $\begin{array}{c}=\frac{\left(1800\text{K}-783.3\text{K}\right)-\left(689.4\text{K}-300\text{K}\right)}{1800\text{K}-759.8\text{K}}\\ =0.603\\ =60.3\%\end{array}$

Thus, the thermal efficiency of the cycle is $60.3\%$.

b)

To determine

The thermal efficiency and the required mass flow rate of helium when the isentropic efficiency is 80%.

Explanation of Solution

Calculation:

Calculate the temperature at state 2s$\left(T_{2s}\right)$.

  $\begin{array}{c}{T}_{2s}=T_1{\left(\frac{P_1}{P_2}\right)}^{\frac{k-1}{k}}\\ =T_1{\left(r_P\right)}^{\frac{k-1}{k}}\\ =\left(300\text{K}\right){\left(8\right)}^{\frac{1.667-1}{1.667}}\\ =689.4\text{K}\end{array}$

Calculate the temperature at state 2$\left(T_2\right)$.

  $\begin{array}{c}{T}_2=T_1+\frac{\left(T_{2s}-T_1\right)}{{\eta }_C}\\ =300\text{K}+\frac{\left(689.4\text{K}-300\text{K}\right)}{0.80}\\ =786.8\text{K}\end{array}$

Calculate the temperature at state 4s$\left(T_{4s}\right)$.

  $\begin{array}{c}{T}_{4s}=T_3{\left(\frac{P_4}{P_{32}}\right)}^{\frac{k-1}{k}}\\ =T_3{\left(\frac{1}{r_P}\right)}^{\frac{k-1}{k}}\\ =\left(1800\text{K}\right){\left(\frac{1}{8}\right)}^{\frac{1.667-1}{1.667}}\\ =783.3\text{K}\end{array}$

Calculate the temperature at state 4$\left(T_4\right)$.

  $\begin{array}{c}{T}_4=T_3-{\eta }_T\left(T_{4s}-T_3\right)\\ =1800\text{K}-\left(0.80\right)\left(1800\text{K}-783.3\text{K}\right)\\ =986.6\text{K}\end{array}$

Calculate the temperature at state 5$\left(T_5\right)$.

  $\begin{array}{c}{T}_5=T_2+\epsilon \left(T_4-T_2\right)\\ =786.8\text{K}+\left(0.75\right)\left(986.6\text{K}-786.8\text{K}\right)\\ =936.7\text{K}\end{array}$

Calculate the mass flow rate of the helium $\left(\dot{m}\right)$.

  $\begin{array}{c}\dot{m}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}\\ =\frac{{\dot{W}}_{\text{net}}}{c_p\left[\left(T_3-T_4\right)-\left(T_2-T_1\right)\right]}\\ =\frac{60,000\text{kJ}/\text{s}}{\left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\begin{array}{l}\left(1800\text{K}-986.6\text{K}\right)-\\ \left(786.8\text{K}-300\text{K}\right)\end{array}\right]}\\ =35.4\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of the helium is $35.4\text{kg}/\text{s}$.

Calculate the thermal efficiency of the cycle $\left({\eta }_{\text{th}}\right)$.

  $\begin{array}{c}{\eta }_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}\\ =\frac{c_p\left[\left(T_3-T_4\right)-\left(T_2-T_1\right)\right]}{c_p\left[\left(T_3-T_5\right)\right]}\end{array}$

  $\begin{array}{c}=\frac{\left(1800\text{K}-986.6\text{K}\right)-\left(786.8\text{K}-300\text{K}\right)}{1800\text{K}-936.7\text{K}}\\ =0.378\\ =37.8\%\end{array}$

Thus, the thermal efficiency of the cycle is $37.8\%$.