Chapter 9.12, Problem 164RP

Consider a simple ideal Brayton cycle with air as the working fluid. The pressure ratio of the cycle is 6, and the minimum and maximum temperatures are 300 and 1300 K, respectively. Now the pressure ratio is doubled without changing the minimum and maximum temperatures in the cycle. Determine the change in (a) the net work output per unit mass and (b) the thermal efficiency of the cycle as a result of this modification. Assume variable specific heats for air.

To determine

The change in net work output per unit mass.

Answer to Problem 164RP

The change in net work output per unit mass is $41.5\text{kJ}/\text{kg}$.

Explanation of Solution

Draw the $T-s$ diagram for the simple Brayton cycle as shown in Figure (1).

Write the expression for the pressure ratio in terms of relative pressure to calculate the relative pressure at state 2 $\left(P_{r2}\right)$.

$\begin{array}{l}\frac{P_2}{P_1}=\frac{P_{r2}}{P_{r1}}\\ r_p=\frac{P_{r2}}{P_{r1}}\end{array}$ (I)

Here, compressor inlet pressure is $P_1$, compressor exit pressure is $P_2$, the relative pressure at state 1 is $P_{r1}$, and relative pressure at state 2 is $P_{r2}$.

Write the expression for the pressure ratio in terms of relative pressure to calculate the relative pressure at state 4 $\left(P_{r4}\right)$.

$\begin{array}{l}\frac{P_4}{P_3}=\frac{P_{r4}}{P_{r3}}\\ \frac{1}{r_p}=\frac{P_{r4}}{P_{r3}}\end{array}$ (II)

Here, compressor inlet pressure is $P_1$, compressor exit pressure is $P_2$, the relative pressure at state 3 is $P_{r3}$, and relative pressure at state 4 is $P_{r4}$.

Write the expression for the heat added $\left(q_{in}\right)$ in the combustion chamber.

$\begin{array}{c}{q}_{in}=h_3-h_2\\ =c_p\left(T_3-T_2\right)\end{array}$ (III)

Here, specific enthalpy of air from the exit of compressor is $h_2$, specific enthalpy of air from the exit of the combustion chamber is $h_3$.

Write the expression for the heat rejected $\left(q_{out}\right)$ to atmosphere.

$\begin{array}{c}{q}_{out}=h_4-h_1\\ =c_p\left(T_4-T_1\right)\end{array}$ (IV)

Here, specific enthalpy of air at the exit of the turbine is $h_4$ and specific enthalpy of air at the inlet to the compressor is $h_1$.

Write the expression for net work done $\left(w_{net}\right)$ per unit mass when the pressure ratio is 6.

$w_{net}=q_{in}-q_{out}$ (V)

Write the expression for net work done $\left(w{\text{'}}_{net}\right)$ per unit mass when the pressure ratio is 12.

$w{\text{'}}_{net}=q_{in}-q_{out}$ (VI)

Write the expression for the thermal efficiency $\left({\eta }_{th}\right)$ of the system when the pressure ratio is 6.

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

Write the expression for the thermal efficiency $\left(\eta {\text{'}}_{th}\right)$ of the system when the pressure ratio is 12.

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

Write the expression for change in the net work $\left(\Delta w_{net}\right)$ output per unit mass of the cycle.

$\Delta w_{net}=w{\text{'}}_{net}-w_{net}$ (IX)

Here, net work done for the initial cycle is $w_{net}$ and net work done for the newer cycle is $w{\text{'}}_{net}$.

Conclusion:

For relative pressure $\left(r_p\right)$ is 6.

From the Table A-17, “Ideal-gas properties of air” obtain the following properties at temperature $\left(T_1\right)$ of 300 K.

$\begin{array}{l}{h}_1=300.19\text{kJ}/\text{kg}\\ P_{r1}=1.386\end{array}$

From the Table A-17, “Ideal-gas properties of air” obtain the following properties at temperature $\left(T_1\right)$ of 1300 K.

$\begin{array}{l}{h}_3=1395.97\text{kJ}/\text{kg}\\ P_{r3}=330.9\end{array}$

Substitute $1.386$ for $P_{r1}$ and $6$ for $r_p$ in Equation (I).

$\begin{array}{l}6=\frac{P_{r2}}{1.386}\\ P_{r2}=8.316\end{array}$

Substitute $330.9$ for $P_{r1}$ and $6$ for $r_p$ in Equation (II).

$\begin{array}{l}\frac{1}{6}=\frac{P_{r4}}{330.9}\\ P_{r4}=55.15\end{array}$

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

Obtain the value of enthalpy at state 2 $\left(h_2\right)$ at the relative pressure at state 2 $\left(P_{r2}\right)$ of $8.316$ by using interpolation method.

Write the formula of interpolation method of two variables.

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

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

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

Relative pressure $\left(P_{r2}\right)$	Enthalpy $\left(h_2\right)$, in $\text{kJ}/\text{kg}$
7.824	492.74
8.316	?
8.411	503.02

Substitute $7.824$ for $x_1$, $8.316$ for $x_2$, $8.411$ for $x_3$, $492.74$ for $y_1$, and $503.02$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left(8.316-7.824\right)\left(503.02-492.74\right)}{\left(8.411-7.824\right)}+492.74\\ =501.4\end{array}$

The value of enthalpy at state 2 $\left(h_2\right)$ at the relative pressure $\left(P_{r2}\right)$ of $8.316$ is $501.4\text{kJ}/\text{kg}$.

Similarly by using interpolation method obtain the value of enthalpy at state 4 $\left(h_4\right)$ at the relative pressure $\left(P_{r2}\right)$ of $55.15$ as $855.3\text{kJ}/\text{kg}$.

Substitute $1395.97\text{kJ}/\text{kg}$ for $h_3$ and $501.4\text{kJ}/\text{kg}$ for $h_2$ in Equation (III).

$\begin{array}{c}{q}_{in}=1395.97\text{kJ}/\text{kg}-501.4\text{kJ}/\text{kg}\\ =894.97\text{kJ}/\text{kg}\end{array}$

Substitute $855.3\text{kJ}/\text{kg}$ for $h_4$ and $300.19\text{kJ}/\text{kg}$ for $h_1$ in Equation (IV).

$\begin{array}{c}{q}_{out}=855.3\text{kJ}/\text{kg}-300.19\text{kJ}/\text{kg}\\ =555.11\text{kJ}/\text{kg}\end{array}$

Substitute $555.11\text{kJ}/\text{kg}$ for $q_{out}$ and $894.97\text{kJ}/\text{kg}$ for $q_{in}$ in Equation (V).

$\begin{array}{c}{w}_{net}=894.97\text{kJ}/\text{kg}-555.11\text{kJ}/\text{kg}\\ =339.46\text{kJ}/\text{kg}\end{array}$

Substitute $339.46\text{kJ}/\text{kg}$ for $w_{net}$ and $894.97\text{kJ}/\text{kg}$ for $q_{in}$ in Equation (VII).

$\begin{array}{c}{\eta }_{th}=\frac{339.46\text{kJ}/\text{kg}}{894.97\text{kJ}/\text{kg}}\\ =0.379\times 100\%\\ =37.9\%\end{array}$

For relative pressure $\left(r_p\right)$ is 12.

Substitute $1.386$ for $P_{r1}$ and $12$ for $r_p$ in Equation (I).

$\begin{array}{l}12=\frac{P_{r2}}{1.386}\\ P_{r2}=16.63\end{array}$

From the Table A-17, “Ideal-gas properties of air” obtain the values of enthalpy $\left(h_2\right)$ at the relative pressure $\left(P_{r2}\right)$ of $16.63$ as $610.6\text{kJ}/\text{kg}$ by interpolation method as shown in Equation (I).

Substitute $330.9$ for $P_{r1}$ and $12$ for $r_p$ in Equation (II).

$\begin{array}{l}\frac{1}{12}=\frac{P_{r4}}{330.9}\\ P_{r4}=27.58\end{array}$

From the Table A-17, “Ideal-gas properties of air” obtain the values of enthalpy $\left(h_4\right)$ at the relative pressure $\left(P_{r4}\right)$ of $25.575$ as $704.6\text{kJ}/\text{kg}$ by interpolation method as shown in Equation (I).

Substitute $1395.97\text{kJ}/\text{kg}$ for $h_3$ and $610.6\text{kJ}/\text{kg}$ for $h_2$ in Equation (III).

$\begin{array}{c}{q}_{in}=1395.97\text{kJ}/\text{kg}-610.6\text{kJ}/\text{kg}\\ =785.37\text{kJ}/\text{kg}\end{array}$

Substitute $704.6\text{kJ}/\text{kg}$ for $h_4$ and $300.19\text{kJ}/\text{kg}$ for $h_1$ in Equation (IV).

$\begin{array}{c}{q}_{out}=704.6\text{kJ}/\text{kg}-300.19\text{kJ}/\text{kg}\\ =404.41\text{kJ}/\text{kg}\end{array}$

Substitute $404.41\text{kJ}/\text{kg}$ for $q_{out}$ and $785.37\text{kJ}/\text{kg}$ for $q_{in}$ in Equation (VI).

$\begin{array}{c}w{\text{'}}_{net}=404.41\text{kJ}/\text{kg}-785.37\text{kJ}/\text{kg}\\ =380.96\text{kJ}/\text{kg}\end{array}$

Substitute $380.96\text{kJ}/\text{kg}$ for $w{\text{'}}_{net}$ and $785.37\text{kJ}/\text{kg}$ for $q_{in}$ in Equation (VIII).

$\begin{array}{c}\eta {\text{'}}_{th}=\frac{339.46\text{kJ}/\text{kg}}{785.37\text{kJ}/\text{kg}}\\ =0.485\times 100\%\\ =48.5\%\end{array}$

Substitute $380.96\text{kJ}/\text{kg}$ for $w{\text{'}}_{net}$ and $339.46\text{kJ}/\text{kg}$ for $w_{net}$ in Equation (IX).

$\begin{array}{c}\Delta w_{net}=380.96\text{kJ}/\text{kg}-339.46\text{kJ}/\text{kg}\\ =41.5\text{kJ}/\text{kg}\end{array}$

Thus, the change in net work output per unit mass is $41.5\text{kJ}/\text{kg}$.

To determine

The change in thermal efficiency of the cycle.

Answer to Problem 164RP

The change in thermal efficiency of the cycle is $10.6\%$.

Explanation of Solution

Write the expression to calculate the change in thermal efficiency $\Delta {\eta }_{th}$.

$\Delta {\eta }_{th}={{\eta }^{\prime }}_{th}-{\eta }_{th}$ (XI)

Conclusion:

Substitute $48.5\%$ for ${{\eta }^{\prime }}_{th}$ and $37.9\%$ for ${\eta }_{th}$ in Equation (XI).

$\begin{array}{c}\Delta {\eta }_{th}=48.5\%-37.9\%\\ =10.6\%\end{array}$

Thus, the change in thermal efficiency of the cycle is $10.6\%$.