Chapter 9, Problem 91P

a)

To determine

The power delivered by the plant assuming constant specific heats for air.

Explanation of Solution

Given:

Temperature of the air at the compressor inlet $\left(T_1\right)$ is $\text{290}\text{K}$.

Pressure of the air at the compressor inlet $\left(P_1\right)$ is $\text{95}\text{kPa}$.

Temperature of the air at the turbine inlet $\left(T_2\right)$ is $\text{1100}\text{K}$.

Pressure of the air at the turbine inlet $\left(P_2\right)$ is $\text{880}\text{kPa}$.

Rate of heat transferred to air $\left({\dot{Q}}_{\text{in}}\right)$ is $30,000\text{kJ}/\text{s}$.

Effectiveness for the regenerator $\left(\epsilon \right)$ is $0.100$.

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 nitrogen.

  $\begin{array}{l}R=0.2968\text{kJ}/\text{kg}\cdot \text{K}\\ c_v=0.743\text{kJ}/\text{kg}\cdot \text{K}\\ c_p=1.039\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.4\end{array}$

Calculate the pressure ratio $\left(r_P\right)$.

  $\begin{array}{c}{r}_P=\frac{P_2}{P_1}\\ =\frac{\text{880}\text{kPa}}{\text{95}\text{kPa}}\\ =9.263\end{array}$

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

  $\begin{array}{c}{T}_2=T_1{\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}\\ =T_1{\left(r_P\right)}^{\frac{k-1}{k}}\\ =\left(290\text{K}\right){\left(9.263\right)}^{\frac{1.4-1}{1.4}}\\ =547.8\text{K}\end{array}$

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

  $\begin{array}{c}{T}_4=T_3{\left(\frac{P_4}{P_3}\right)}^{\frac{k-1}{k}}\\ =T_3{\left(\frac{1}{r_P}\right)}^{\frac{k-1}{k}}\\ =\left(1100\text{K}\right){\left(\frac{1}{9.263}\right)}^{\frac{1.4-1}{1.4}}\\ =582.3\text{K}\end{array}$

For 100% effectiveness of the regenerator, the temperatures at states 5 and 4 are equal and the temperatures at states 6 and 2 are equal.

  $\begin{array}{c}{T}_5=T_4\\ =582.3\text{ K}\\ T_6=T_2\\ =547.8\text{ K}\end{array}$

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

  $\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{c_P\left(T_6-T_1\right)}{c_P\left(T_3-T_5\right)}\\ =1-\frac{T_6-T_1}{T_3-T_5}\end{array}$

  $\begin{array}{c}=1-\frac{547.8\text{K}-290\text{K}}{1100\text{K}-582.3\text{K}}\\ =0.5020\\ =50.2\%\end{array}$

Calculate the power developed by the plant $\left({\dot{W}}_{\text{net}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{net}}={\eta }_{\text{th}}\left({\dot{Q}}_{\text{in}}\right)\\ =\left(0.5020\right)\left(30,000\text{kJ}/\text{s}\right)\\ =\left(0.5020\right)\left(30,000\text{kW}\right)\\ =15,060\text{kW}\end{array}$

Thus, the power delivered by the plant assuming constant specific heats for air is $\text{15,060}\text{kW}$.

b)

To determine

The power delivered by the plant accounting for the variation of specific heats with temperature.

Explanation of Solution

Calculation:

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative pressure of air at the temperature of $\text{290}\text{K}$.

  $\begin{array}{l}{h}_1=\text{290}\text{.16}\text{kJ}/\text{kg}\\ P_{r_1}=1.2311\end{array}$

Calculate the relative pressure at state 2$\left(P_{r_2}\right)$.

  $P_{r_2}=\frac{P_2}{P_1}{P}_{r_1}$

  $\begin{array}{c}{P}_{r_2}=\left(r_P\right)P_{r_1}\\ =\left(9.263\right)\left(1.2311\right)\\ =11.40\end{array}$

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy of air at the relative pressure of $11.40$.

  $h_2=548.85\text{kJ}/\text{kg}$

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy and relative pressure of air at the temperature of $\text{1100}\text{K}$.

  $\begin{array}{l}{h}_3=\text{1161}\text{.07}\text{kJ}/\text{kg}\\ P_{r_3}=167.1\end{array}$

Calculate the relative pressure at state 4$\left(P_{r_4}\right)$.

  $P_{r_4}=\frac{P_4}{P_3}{P}_{r_3}$

  $\begin{array}{c}{P}_{r_4}=\left(\frac{1}{r_P}\right)P_{r_3}\\ =\left(\frac{1}{9.263}\right)\left(167.1\right)\\ =18.04\end{array}$

Refer table A-21, “Ideal gas properties of the air”, obtain the enthalpy of air at the relative pressure of $18.04$.

  $h_4=624.89\text{kJ}/\text{kg}$

For 100% effectiveness of the regenerator, the enthalpies at states 5 and 4 are equal and enthalpies at states 6 and 2 are equal.

  $\begin{array}{c}{h}_5=h_4\\ =624.89\text{kJ}/\text{kg}\\ h_6=h_2\\ =548.85\text{kJ}/\text{kg}\end{array}$

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

  $\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{h_6-h_1}{h_3-h_5}\end{array}$

  $\begin{array}{c}=1-\frac{548.85\text{kJ}/\text{kg}-290.16\text{kJ}/\text{kg}}{1161.07\text{kJ}/\text{kg}-624.89\text{kJ}/\text{kg}}\\ =0.5175\\ =51.75\%\end{array}$

Calculate the power developed by the plant $\left({\dot{W}}_{\text{net}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{net}}={\eta }_{\text{th}}\left({\dot{Q}}_{\text{in}}\right)\\ =\left(0.5175\right)\left(30,000\text{kJ}/\text{s}\right)\\ =\left(0.5175\right)\left(30,000\text{kW}\right)\\ =15,525\text{kW}\end{array}$

Thus, the power delivered by the plant accounting for the variation of specific heats with temperature is $\text{15,525}\text{kW}$.