Chapter 9.12, Problem 148P

A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The minimum and maximum temperatures in the cycle are 310 and 1150 K. Take an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator. Determine the total exergy destruction associated with the cycle, assuming a source temperature of 1500 K and a sink temperature of 290 K. Also, determine the exergy of the exhaust gases at the exit of the regenerator. Use variable specific heats for air.

To determine

The exergy destruction associated with each process of the Brayton cycle and the exergy of the exhaust gases at the exit of the regenerator.

Answer to Problem 148P

The exergy destruction associated with process 1-2 of the given Brayton cycle is $38.91\text{kJ}/\text{kg}$.

The exergy destruction associated with process 3-4 of the given Brayton cycle is $35.83\text{kJ}/\text{kg}$.

The exergy destruction associated with regeneration process of the given Brayton cycle is $4.37\text{kJ}/\text{kg}$.

The exergy destruction associated with process 5-3 of the given Brayton cycle is $58.09\text{kJ}/\text{kg}$.

The exergy destruction associated with process 6-1 of the given Brayton cycle is $142.6\text{kJ}/\text{kg}$.

The exergy of the exhaust gases at the exit of the regenerator is $143.2\text{kJ}/\text{kg}$.

Explanation of Solution

Show the regenerative Brayton cycle with air as the working fluid, on $T-s$ diagram as in Figure (1).

Consider, the pressure is $P_i$ , the specific volume is $v_i$, the temperature is $T_i$, the entropy is $s_i$, the enthalpy is $h_i$ corresponding to $i^{th}$ state.

Write the pressure and relative pressure relation for the process 1-2.

$P_{r_2}=\frac{P_2}{P_1}{P}_{r_1}$ (I)

Write the pressure and relative pressure relation for the process 3-4.

$P_{r_4}=\frac{P_4}{P_3}{P}_{r_3}$ (II)

Write the expression of efficiency of the compressor $\left({\eta }_C\right)$.

${\eta }_C=\frac{h_{2s}-h_1}{h_2-h_1}$ (III)

Write the expression of efficiency of the turbine $\left({\eta }_T\right)$.

${\eta }_T=\frac{h_3-h_4}{h_3-h_{4s}}$ (IV)

Write the expression of net work output by the gas turbine $\left(w_{\text{net}}\right)$.

$w_{\text{net}}=w_{\text{T,out}}-w_{\text{C,in}}$

$w_{\text{net}}=\left(h_3-h_4\right)-\left(h_2-h_1\right)$ (V)

Here, work done by the turbine is $w_{\text{T,out}}$, and the work input to the compressor is $w_{\text{C,in}}$.

Write the expression of effectiveness of the regenerator $\left(\epsilon \right)$.

$\epsilon =\frac{h_5-h_2}{h_4-h_2}$ (VI)

Write the expression of heat input to the regenerative Brayton cycle $\left(q_{\text{in}}\right)$.

$q_{\text{in}}=h_3-h_5$ (VII)

Write the expression of heat rejected by the regenerative Brayton cycle $\left(q_{\text{out}}\right)$.

$\begin{array}{l}{q}_{\text{out}}=q_{\text{in}}-w_{\text{net}}\\ q_{\text{out}}=h_6-h_1\end{array}$ (VIII)

Write the expression of thermal efficiency of the given turbine $\left({\eta }_{th}\right)$.

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

Write the energy balance equation on the heat exchanger.

$h_5-h_2=h_4-h_6$ (X)

Write the expression of exergy destruction associated with the process 1-2 of the given Brayton cycle $\left(x_{\text{destroyed,12}}\right)$.

$x_{\text{destroyed,12}}=T_0\left(s_2-s_1\right)$

$x_{\text{destroyed,12}}=T_0\left(s_2^{\circ }-s_1^{\circ }-R\ln \frac{P_2}{P_1}\right)$ (XI)

Here, the temperature of the surroundings is $T_0$, the gas constant of air is R, entropy of air at state 2 as a function of temperature alone is $s_2^{\circ }$, and entropy of air at state 1 as a function of temperature alone is $s_1^{\circ }$.

Write the expression of exergy destruction associated with the process 3-4 of the given Brayton cycle $\left(x_{\text{destroyed,34}}\right)$.

$x_{\text{destroyed,34}}=T_0\left(s_4-s_3\right)$

$x_{\text{destroyed,34}}=T_0\left(s_4^{\circ }-s_3^{\circ }-R\ln \frac{P_4}{P_3}\right)$ (XII)

Here, entropy of air at state 3 as a function of temperature is $s_3^{\circ }$, and entropy of air at state 4 as a function of temperature is $s_4^{\circ }$.

Write the expression of exergy destruction associated with the regeneration process of the given Brayton cycle $\left(x_{\text{destroyed,regen}}\right)$.

$x_{\text{destroyed,regen}}=T_0\left[\left(s_5-s_2\right)+\left(s_6-s_4\right)\right]$

$x_{\text{destroyed,regen}}=T_0\left[\left(s_5^{\circ }-s_2^{\circ }\right)+\left(s_6^{\circ }-s_4^{\circ }\right)\right]$ (XIII)

Here, entropy of air at state 5 as a function of temperature alone is $s_5^{\circ }$, and entropy of air at state 6 as a function of temperature alone is $s_6^{\circ }$.

Write the expression of exergy destruction associated with the process 5-3 of the given Brayton cycle $\left(x_{\text{destroyed,53}}\right)$.

$x_{\text{destroyed,53}}=T_0\left(s_3-s_5+\frac{-q_{\text{in}}}{T_H}\right)$

$x_{\text{destroyed,53}}=T_0\left(s_3^{\circ }-s_5^{\circ }-R\ln \frac{P_3}{P_5}-\frac{q_{\text{in}}}{T_H}\right)$ (XIV)

Here, the temperature of the heat source is $T_H$.

Write the expression of exergy destruction associated with the process 6-1 of the given Brayton cycle $\left(x_{\text{destroyed,61}}\right)$.

$x_{\text{destroyed,61}}=T_0\left(s_1-s_6+\frac{q_{\text{out}}}{T_L}\right)$

$x_{\text{destroyed,61}}=T_0\left(s_1^{\circ }-s_6^{\circ }-R\ln \frac{P_1}{P_6}+\frac{q_{\text{out}}}{T_L}\right)$ (XV)

Here, the temperature of the sink is $T_L$.

Write the expression of stream exergy at the exit of the regenerator (state 6) $\left({\varphi }_6\right)$.

${\varphi }_6=\left(h_6-h_0\right)-T_0\left(s_6-s_0\right)$ (XVI)

Here, the specific enthalpy of the surroundings is $h_0$.

Write the expression of change entropy for the exit of the regenerator $\left(s_6-s_0\right)$.

$s_6-s_0=s_6^{\circ }-s_0^{\circ }-R\ln \frac{P_6}{P_1}$ (XVII)

Here, entropy of air at the surroundings as a function of temperature alone is $s_0^{\circ }$, and the pressure of air at the surroundings is $P_0$.

Conclusion:

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of $310\text{ K}$ $\left(T_1\right)$.

$\begin{array}{l}{h}_1=310.24\text{kJ}/\text{kg}\\ P_{r_1}=1.5546\\ s_1^0=1.73498\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute 7 for $\frac{P_2}{P_1}$, and 1.5546 for $P_{r_1}$ in Equation (I).

$\begin{array}{c}{P}_{r_2}=\left(7\right)\left(1.5546\right)\\ =10.88\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a relative pressure of 10.88 $\left(P_{r_2}\right)$.

$h_{2s}=541.26\text{kJ}/\text{kg}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of $1150\text{ K}$ $\left(T_3\right)$.

$\begin{array}{l}{h}_3=1219.25\text{kJ}/\text{kg}\\ P_{r_3}=200.15\\ s_3^0=3.129\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $\frac{1}{7}$ for $\frac{P_4}{P_3}$, and 200.15 for $P_{r_3}$ in Equation (II).

$\begin{array}{c}{P}_{r_4}=\left(\frac{1}{7}\right)\left(200.15\right)\\ =28.59\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the property of enthalpy $\left(h_{4s}\right)$ at a relative pressure of 28.59 $\left(P_{r_4}\right)$.

$h_{4s}=711.8\text{kJ}/\text{kg}$

Rearrange Equation (III), and substitute $310.24\text{kJ}/\text{kg}$ for $h_1$, $541.26\text{kJ}/\text{kg}$ for $h_{2s}$, and 0.75 for ${\eta }_C$.

$\begin{array}{c}{h}_2=h_1+\frac{h_{2s}-h_1}{{\eta }_C}\\ =310.24\text{kJ}/\text{kg}+\frac{\left(541.26-310.24\right)\text{kJ}/\text{kg}}{0.75}\\ =618.26\text{kJ}/\text{kg}\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the property of entropy $\left(s_2^0\right)$ at a enthalpy of $618.26\text{kJ}/\text{kg}$ $\left(h_2\right)$.

$s_2^0=2.42763\text{kJ}/\text{kg}\cdot \text{K}$

Rearrange Equation (IV), and substitute $1219.25\text{kJ}/\text{kg}$ for $h_3$, $711.8\text{kJ}/\text{kg}$ for $h_{4s}$, and 0.82 for ${\eta }_T$.

$\begin{array}{c}{h}_4=h_3-{\eta }_T\left(h_3-h_{4s}\right)\\ =1219.25\text{kJ}/\text{kg}-\left(0.82\right)\left(1219.25-711.8\right)\text{kJ}/\text{kg}\\ =803.14\text{kJ}/\text{kg}\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the property of entropy $\left(s_4^0\right)$ at a enthalpy of $803.14\text{kJ}/\text{kg}$ $\left(h_4\right)$.

$s_4^0=2.69407\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $1219.25\text{kJ}/\text{kg}$ for $h_3$, $618.26\text{kJ}/\text{kg}$ for $h_2$, $803.14\text{kJ}/\text{kg}$ for $h_4$, and $310.24\text{kJ}/\text{kg}$ for $h_1$ in Equation (V).

$\begin{array}{c}{w}_{\text{net}}=\left(1219.25-803.14\right)\text{kJ}/\text{kg}-\left(618.26-310.24\right)\text{kJ}/\text{kg}\\ =108.09\text{kJ}/\text{kg}\end{array}$

Substitute 0.65 for $\epsilon$, $803.14\text{kJ}/\text{kg}$ for $h_4$, and $618.26\text{kJ}/\text{kg}$ for $h_2$ in Equation (VI).

$\begin{array}{c}0.65=\frac{\left(h_5-618.26\text{kJ}/\text{kg}\right)}{\left(803.14-618.26\right)\text{kJ}/\text{kg}}\\ h_5=738.43\text{kJ}/\text{kg}\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the property of entropy $\left(s_5^0\right)$ at a enthalpy of $738.43\text{kJ}/\text{kg}$ $\left(h_5\right)$.

$s_5^0=2.60815\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $1219.25\text{kJ}/\text{kg}$ for $h_3$, and $738.43\text{kJ}/\text{kg}$ for $h_5$ in Equation (VII).

$\begin{array}{c}{q}_{\text{in}}=\left(1219.25-738.43\right)\text{kJ}/\text{kg}\\ =480.82\text{kJ}/\text{kg}\end{array}$

Substitute $738.43\text{kJ}/\text{kg}$ for $h_5$, $803.14\text{kJ}/\text{kg}$ for $h_4$, and $618.26\text{kJ}/\text{kg}$ for $h_2$ in Equation (X).

$\begin{array}{l}738.43\text{kJ}/\text{kg}-618.26\text{kJ}/\text{kg}=803.14\text{kJ}/\text{kg}-h_6\\ h_6=682.97\text{kJ}/\text{kg}\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a enthalpy of $682.97\text{kJ}/\text{kg}$ $\left(h_6\right)$.

$s_6^0=2.52861\text{kJ}/\text{kg}\cdot \text{K}$

Substitute $682.97\text{kJ}/\text{kg}$ for $h_6$, and $310.24\text{kJ}/\text{kg}$ for $h_1$ in Equation (VIII).

$\begin{array}{c}{q}_{\text{out}}=682.97\text{kJ}/\text{kg}-310.24\text{kJ}/\text{kg}\\ =372.73\text{kJ}/\text{kg}\end{array}$

Substitute $108.09\text{kJ}/\text{kg}$ for $w_{\text{net}}$, and $480.82\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (IX).

$\begin{array}{c}{\eta }_{th}=\frac{108.09\text{kJ}/\text{kg}}{480.82\text{kJ}/\text{kg}}\\ =0.225\\ =22.5\%\end{array}$

Substitute 290 K for $T_0$, $1.73498\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1^{\circ }$, $2.42763\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, and 7 for $\frac{P_2}{P_1}$ in Equation (XI).

$\begin{array}{c}{x}_{\text{destroyed,12}}=\left(290\text{K}\right)\left[\left(2.42763-1.73498\right)\text{kJ}/\text{kg}\cdot \text{K}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(7\right)\right]\\ =38.91\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 1-2 of the given Brayton cycle is $38.91\text{kJ}/\text{kg}$.

Substitute 290 K for $T_0$, $3.129\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3^{\circ }$, $2.69407\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, and $\frac{1}{7}$ for $\frac{P_4}{P_3}$ in Equation (XII).

$\begin{array}{c}{x}_{\text{destroyed,34}}=\left(290\text{K}\right)\left[\left(2.69407-3.129\right)\text{kJ}/\text{kg}\cdot \text{K}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{1}{7}\right)\right]\\ =35.83\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 3-4 of the given Brayton cycle is $35.83\text{kJ}/\text{kg}$.

Substitute 290 K for $T_0$, $2.60815\text{kJ}/\text{kg}\cdot \text{K}$ for $s_5^{\circ }$, $2.69407\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4^{\circ }$, $2.42763\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2^{\circ }$, and $2.52861\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6^{\circ }$ in Equation (XIII).

$\begin{array}{c}{x}_{\text{destroyed,regen}}=\left(290\text{K}\right)\left[\begin{array}{l}\left(2.60815-2.42763\right)\text{kJ}/\text{kg}\cdot \text{K}\\ +\left(2.52861-2.69407\right)\text{kJ}/\text{kg}\cdot \text{K}\end{array}\right]\\ =4.37\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with regeneration process of the given Brayton cycle is $4.37\text{kJ}/\text{kg}$.

Substitute 290 K for $T_0$, $3.129\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3^{\circ }$, $2.60815\text{kJ}/\text{kg}\cdot \text{K}$ for $s_5^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, 1500 K for $T_H$, and $480.82\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (XIV). For the process 5-3, pressure remains constant, hence substitute 0 for $\ln \frac{P_3}{P_5}$.

$\begin{array}{c}{x}_{\text{destroyed,53}}=\left(290\text{K}\right)\left[\begin{array}{l}\left(3.129-2.60815\right)\text{kJ}/\text{kg}\cdot \text{K}\\ -\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0\right)-\frac{480.82\text{kJ}/\text{kg}}{1500\text{K}}\end{array}\right]\\ =58.09\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 5-3 of the given Brayton cycle is $58.09\text{kJ}/\text{kg}$.

Substitute 290 K for $T_0$, $1.73498\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1^{\circ }$, $2.52861\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, 290 K for $T_L$, and $372.73\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XV). For the process 6-1, pressure remains constant, hence substitute 0 for $\ln \frac{P_1}{P_6}$.

$\begin{array}{c}{x}_{\text{destroyed,61}}=\left(290\text{K}\right)\left[\begin{array}{l}\left(1.73498-2.52861\right)\text{kJ}/\text{kg}\cdot \text{K}\\ -\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0\right)+\frac{372.73\text{kJ}/\text{kg}}{290\text{K}}\end{array}\right]\\ =142.6\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 6-1 of the given Brayton cycle is $142.6\text{kJ}/\text{kg}$.

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of $290\text{ K}$ $\left(T_0\right)$.

$\begin{array}{l}{h}_0=290.16\text{kJ}/\text{kg}\\ s_0^0=1.66802\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

At the exit of the regenerator, pressure remains constant, $\left(P_6=P_1\right)$.

Substitute $P_6$ for $P_1$ , $2.52861\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6^{\circ }$, and $1.66802\text{kJ}/\text{kg}\cdot \text{K}$ for $s_0^{\circ }$ in Equation (XVII).

$\begin{array}{c}{s}_6-s_0=\left(2.52861-1.66802\right)\text{kJ}/\text{kg}\cdot \text{K}-R\ln \frac{P_6}{P_6}\\ =\left(2.52861-1.66802\right)\text{kJ}/\text{kg}\cdot \text{K}-R\ln \left(1\right)\\ =0.86059\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $682.97\text{kJ}/\text{kg}$ for $h_6$, $290.16\text{kJ}/\text{kg}$ for $h_0$, 290 K for $T_0$, and $0.86059\text{kJ}/\text{kg}\cdot \text{K}$ for $\left(s_6-s_0\right)$ in Equation (XVI).

$\begin{array}{c}{\varphi }_6=\left(682.97-290.16\right)\text{kJ}/\text{kg}-\left(290\text{K}\right)\left(0.86059\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =143.2\text{kJ}/\text{kg}\end{array}$

Thus, the exergy of the exhaust gases at the exit of the regenerator is $143.2\text{kJ}/\text{kg}$.