Chapter 9, Problem 168RQ

a)

To determine

The thermal efficiency of the cycle using the constant specific heats at room temperature.

Explanation of Solution

Given:

Compression ratio $\left(r\right)$ is $20$.

Temperature of air at state 1$\left(T_1\right)$ is $45°\text{F}$.

Temperature of air at state 3$\left(T_3\right)$ is $1800°\text{F}$.

Pressure of air at state 1$\left(P_1\right)$ is $\text{13 psia}$.

Calculation:

Draw the $P-v$ diagram of the cycle as in Figure (1).

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

$\begin{array}{l}{c}_p=0.240\text{Btu}/\text{lbm}\cdot \text{R}\\ c_v=0.171\text{Btu}/\text{lbm}\cdot \text{R}\\ k=1.4\end{array}$

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

$\begin{array}{c}{T}_2=T_1{\left(r\right)}^{k-1}\\ =\left(45°\text{F}\right){\left(20\right)}^{1.4-1}\\ =\left(45+460\text{R}\right){\left(20\right)}^{1.4-1}\\ =1673.8\text{R}\end{array}$

Calculate the ratio between the volumes at state 3 and state 2$\left(\frac{V_3}{V_2}\right)$.

$\begin{array}{c}\frac{V_3}{V_2}=\frac{T_3}{T_2}\\ =\frac{2260\text{R}}{1673.8\text{R}}\\ =1.350\end{array}$

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

$\begin{array}{c}{T}_4=T_3{\left(\frac{V_3}{V_4}\right)}^{k-1}\\ =T_3{\left(\frac{1.350V_2}{V_4}\right)}^{k-1}\end{array}$

$\begin{array}{c}=T_3{\left(\frac{1.350}{r}\right)}^{k-1}\\ =\left(2260\text{R}\right){\left(\frac{1.350}{20}\right)}^{1.4-1}\\ =768.8\text{R}\end{array}$

Calculate the net specific work produced by the cycle $\left(w_{\text{net}}\right)$.

$\begin{array}{c}{w}_{\text{net}}=q_{\text{in}}-q_{\text{out}}\\ =c_p\left(T_3-T_2\right)-c_v\left(T_4-T_1\right)\\ =\left\{\begin{array}{c}\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2260\text{R}-1673.8\text{R}\right)-\\ \left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(768.8\text{R}-505\text{R}\right)\end{array}\right\}\\ =95.59\text{Btu}/\text{lbm}\end{array}$

$\begin{array}{c}=\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(1460\text{R}-717.8\text{R}+480\text{R}-976.3\text{R}\right)\\ =59\text{Btu}/\text{lbm}\end{array}$

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

$\begin{array}{c}{\eta }_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}\\ =\frac{w_{\text{net}}}{c_p\left(T_3-T_2\right)}\\ =\frac{95.59\text{Btu}/\text{lbm}}{\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2260\text{R}-1673.8\text{R}\right)}\end{array}$

$\begin{array}{c}=\frac{95.59\text{Btu}/\text{lbm}}{140.7\text{Btu}/\text{lbm}}\\ =0.6794\\ =67.9\%\end{array}$

Thus, the thermal efficiency of the cycle using the constant specific heats at room temperature is $67.9\%$.

b)

To determine

The thermal efficiency of the cycle using the variable specific heats.

Explanation of Solution

Calculation:

Refer table A-21E, “Ideal gas properties of the air”, obtain the specific internal energy and relative specific volume of air at the temperature of $505\text{R}$.

$\begin{array}{l}{u}_1=\text{86}\text{.06}\text{Btu}/\text{lbm}\\ v_{r_1}=\text{170}\text{.52}\end{array}$

Calculate the relative specific volume at state 2$\left(v_{r_2}\right)$.

$\begin{array}{c}{v}_{r_2}=\frac{v_2}{v_1}{v}_{r_1}\\ =\left(\frac{1}{r}\right)v_{r_1}\\ =\left(\frac{1}{20}\right)\left(170.82\right)\\ =8.541\end{array}$

Refer table A-21E, “Ideal gas properties of the air”, obtain the specific enthalpy and temperature of air at the relative pressure of $8.541$.

$\begin{array}{l}{h}_2=\text{391}\text{.01}\text{Btu}/\text{lbm}\\ T_2=\text{1582}\text{.3}\text{R}\end{array}$

Refer table A-21E, “Ideal gas properties of the air”, obtain the specific enthalpy and relative specific volume of air at the temperature of $2260\text{R}$.

$\begin{array}{l}{h}_3=\text{577}\text{.52}\text{Btu}/\text{lbm}\\ v_{r_3}=\text{2}\text{.922}\end{array}$

Calculate the ratio between the specific volumes at state 3 and state 2$\left(\frac{v_3}{v_2}\right)$.

$\begin{array}{c}\frac{v_3}{v_2}=\frac{T_3}{T_2}\\ =\frac{2260\text{R}}{1582.3\text{R}}\\ =1.428\end{array}$

Calculate the relative specific volume at state 4$\left(v_{r_4}\right)$.

$\begin{array}{c}{v}_{r_4}=\frac{v_4}{v_3}{v}_{r_3}\\ =\left(\frac{v_4}{1.428v_2}\right)v_{r_3}\\ =\left(\frac{r}{1.428}\right)v_{r_3}\end{array}$

$\begin{array}{c}=\left(\frac{20}{1.428}\right)\left(2.922\right)\\ =40.92\end{array}$

Refer table A-21E, “Ideal gas properties of the air”, obtain the specific internal energy of air at the relative specific volume of $40.92$.

$u_4=\text{152}\text{.62}\text{Btu}/\text{lbm}$

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

$\begin{array}{c}{\eta }_{\text{th}}=1-\frac{q_{\text{out}}}{q_{\text{in}}}\\ =1-\frac{u_4-u_1}{h_3-h_2}\end{array}$

$\begin{array}{c}=1-\frac{\left(152.65\text{Btu}/\text{lbm}-86.06\text{Btu}/\text{lbm}\right)}{\left(577.52\text{Btu}/\text{lbm}-391.01\text{Btu}/\text{lbm}\right)}\\ =0.643\\ =64.3\%\end{array}$

Thus, the thermal efficiency of the cycle using the variable specific heats is $64.3\%$.