Chapter 9, Problem 57P

To determine

The cycle’s net specific work, the specific heat addition and the thermal efficiency.

Explanation of Solution

Given:

Compression ratio $\left(r\right)$ is $15$.

Cut off ratio $\left(r_c\right)$ is $1.4$.

Pressure ratio $\left(r_p\right)$ is $1.1$.

Temperature of air at state 1$\left(T_1\right)$ is $75°\text{F}$.

Pressure of air at state 1$\left(P_1\right)$ is $\text{14}\text{.2}\text{psia}$.

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.

  $\begin{array}{l}R=0.37047\text{psia}\cdot {\text{ft}}^3/\text{lbm}\cdot \text{R}\\ c_v=0.240\text{Btu}/\text{lbm}\cdot \text{R}\\ c_p=0.171\text{Btu}/\text{lbm}\cdot \text{R}\\ k=1.4\end{array}$

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

  $\begin{array}{c}{T}_2=T_1{\left(\frac{v_1}{v_2}\right)}^{k-1}\\ =T_1{\left(r\right)}^{k-1}\\ =\left(535\text{R}\right){\left(15\right)}^{1.4-1}\\ =1580\text{R}\end{array}$

Calculate the pressure at state 2$\left(P_2\right)$.

  $\begin{array}{c}{P}_2=P_1{\left(\frac{v_1}{v_2}\right)}^k\\ =P_1{\left(r\right)}^k\\ =\left(14.2\text{psia}\right){\left(15\right)}^{1.4}\\ =629.2\text{psia}\end{array}$

Calculate the pressure at state x$\left(P_x\right)$.

  $\begin{array}{c}{P}_x=P_3\\ =r_p{P}_2\\ =\left(1.1\right)\left(629.2\text{psia}\right)\\ =692.1\text{psia}\end{array}$

Calculate the temperature at state x$\left(T_x\right)$.

  $\begin{array}{c}{T}_x=T_2\left(\frac{P_x}{P_2}\right)\\ =\left(1580\text{R}\right)\left(\frac{692.1\text{psia}}{629.2\text{psia}}\right)\\ =1738\text{R}\end{array}$

Calculate the temperature at state 3$\left(T_3\right)$.

  $\begin{array}{c}{T}_3=T_x\left(\frac{v_3}{v_x}\right)\\ =T_x\left(r_c\right)\\ =\left(1738\text{R}\right)\left(1.4\right)\\ =2433\text{R}\end{array}$

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

  $\begin{array}{c}{T}_4=T_3{\left(\frac{v_3}{v_4}\right)}^{k-1}\\ =T_3{\left(\frac{r_c}{r}\right)}^{k-1}\\ =\left(2433\text{R}\right){\left(\frac{1.4}{15}\right)}^{1.4-1}\\ =942.2\text{R}\end{array}$

Calculate the amount of work during the process 1-2$\left(w_{1-2}\right)$.

  $\begin{array}{c}{w}_{1-2}=c_v\left(T_2-T_1\right)\\ =\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(1580\text{R}-535\text{R}\right)\\ =178.7\text{Btu}/\text{lbm}\end{array}$

Calculate the amount of heat during the process 2-x$\left(q_{2-x}\right)$.

  $\begin{array}{c}{q}_{2-x}=c_v\left(T_x-T_2\right)\\ =\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(1738\text{R}-1580\text{R}\right)\\ =27.02\text{Btu}/\text{lbm}\end{array}$

Calculate the amount of heat during the process x-3$\left(q_{x-3}\right)$.

  $\begin{array}{c}{q}_{x-3}=c_p\left(T_3-T_x\right)\\ =\left(0.240\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2433\text{R}-1738\text{R}\right)\\ =166.8\text{Btu}/\text{lbm}\end{array}$

Calculate the amount of work during the process x-3$\left(w_{x-3}\right)$.

  $\begin{array}{c}{w}_{x-3}=q_{x-3}-c_v\left(T_3-T_x\right)\\ =166.8\text{Btu}/\text{lbm}-\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2433\text{R}-1738\text{R}\right)\\ =47.96\text{Btu}/\text{lbm}\end{array}$

Calculate the amount of work during the process 3-4$\left(w_{3-4}\right)$.

  $\begin{array}{c}{w}_{3-4}=c_v\left(T_3-T_4\right)\\ =\left(0.171\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(2433\text{R}-942.2\text{R}\right)\\ =254.9\text{Btu}/\text{lbm}\end{array}$

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

  $\begin{array}{c}{w}_{\text{net}}=w_{3-4}+w_{x-3}-w_{1-2}\\ =254.9\text{Btu}/\text{lbm}+47.96\text{Btu}/\text{lbm}-178.7\text{Btu}/\text{lbm}\\ =124.2\text{Btu}/\text{lbm}\end{array}$

Thus, the cycle’s net specific work is $124.2\text{Btu}/\text{lbm}$.

Calculate the specific heat addition $\left(q_{\text{in}}\right)$.

  $\begin{array}{c}{q}_{\text{in}}=q_{2-x}+q_{x-3}\\ =27.02\text{Btu}/\text{lbm}+166.8\text{Btu}/\text{lbm}\\ =193.8\text{Btu}/\text{lbm}\end{array}$

Thus, the specific heat addition is $193.8\text{Btu}/\text{lbm}$.

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

  $\begin{array}{c}{\eta }_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}\\ =\frac{124.2\text{Btu}/\text{lbm}}{193.8\text{Btu}/\text{lbm}}\\ =0.641\\ =64.1\%\end{array}$

Thus, the thermal efficiency of the cycle is $64.1\%$.