Chapter 9.12, Problem 146P

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 146P

The exergy destruction associated with process 1-2 of the given Brayton cycle is $38.42\text{kJ}/\text{kg}$.

The exergy destruction associated with process 3-4 of the given Brayton cycle is $24.23\text{kJ}/\text{kg}$.

The exergy destruction associated with regeneration process of the given Brayton cycle is $4.47\text{kJ}/\text{kg}$.

The exergy destruction associated with process 5-3 of the given Brayton cycle is $24.16\text{kJ}/\text{kg}$.

The exergy destruction associated with process 6-1 of the given Brayton cycle is $148.7\text{kJ}/\text{kg}$.

The exergy of the exhaust gases at the exit of the regenerator is $148.9\text{kJ}/\text{kg}$.

Explanation of Solution

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

For the given regenerative Brayton cycle with air as the working fluid, let $T_i$, $P_i$, $h_i$ and $P_{r_i}$ be the temperature, pressure, specific enthalpy and relative pressure at $i^{th}$ state respectively.

Write the expression of pressure ratio for the regenerative Brayton cycle $\left(r_p\right)$.

$r_p=P_2/P_1$ (I)

Write the pressure ratio and 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 turbine $\left({\eta }_T\right)$.

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

Write the expression of heat added due to regeneration $\left(q_{\text{regen}}\right)$.

$q_{\text{regen}}=\epsilon \left(h_4-h_2\right)$ (IV)

Here, the effectiveness of the regenerator is $\epsilon$.

Write the expression of net work output of the regenerative Brayton cycle $\left(w_{\text{net}}\right)$.

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

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

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

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

$q_{\text{in}}=\left(h_3-h_2\right)-q_{\text{regen}}$ (VI)

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

$q_{\text{out}}=q_{\text{in}}-w_{\text{net}}$ (VII)

Write the expression of specific enthalpy at state 6 $\left(h_6\right)$, using the heat rejection relation.

$h_6=h_1+q_{\text{out}}$ (VIII)

Write the specific enthalpy relation for the regenerator.

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

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

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

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

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 $\left(x_{\text{destroyed,34}}\right)$.

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

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

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 $\left(x_{\text{destroyed,regen}}\right)$.

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

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

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 $\left(x_{\text{destroyed,53}}\right)$.

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

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

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 $\left(x_{\text{destroyed,61}}\right)$.

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

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

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

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

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

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

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

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

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 310 K $\left(T_1\right)$, 650K $\left(T_2\right)$ and 1400 K $\left(T_3\right)$, which gives $310.24\text{kJ}/\text{kg}$ for $h_1$, $659.84\text{kJ}/\text{kg}$ for $h_2$, $1,515.42\text{kJ}/\text{kg}$ for $h_3$ and 450.5 for $P_{r_3}$ respectively.

Substitute 900 kPa for $P_2$, and 100 kPa for $P_1$ in Equation (I).

$\begin{array}{c}{r}_p=900\text{kPa}/100\text{kPa}\\ =9\end{array}$

Substitute $\frac{1}{9}$ for $\frac{P_4}{P_3}$, and 450.5 for $P_{r_3}$ in Equation (II).

$\begin{array}{c}{P}_{r_4}=\left(\frac{1}{9}\right)\left(450.5\right)\\ =50.06\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at 50.06 $\left(P_{r_4}\right)$, which gives $832.44\text{kJ}/\text{kg}$ for $h_{4s}$.

Rearrange Equation (III) and substitute $1,515.42\text{kJ}/\text{kg}$ for $h_3$, $832.44\text{kJ}/\text{kg}$ for $h_{4s}$, and 0.90 for ${\eta }_T$.

$\begin{array}{c}{h}_4=h_3-{\eta }_T\left(h_3-h_{4s}\right)\\ =1,515.42\text{kJ}/\text{kg}+\left(0.90\right)\left(1,515.42-832.44\right)\text{kJ}/\text{kg}\\ =900.74\text{}\text{kJ}/\text{kg}\end{array}$

Substitute 0.80 for $\epsilon$, $900.74\text{}\text{kJ}/\text{kg}$ for $h_4$, and $659.84\text{}\text{kJ}/\text{kg}$ for $h_2$ in Equation (IV).

$\begin{array}{c}{q}_{\text{regen}}=\left(0.80\right)\left(900.74-659.84\right)\text{kJ}/\text{kg}\\ =192.7\text{kJ}/\text{kg}\end{array}$

Substitute $1,515.42$ $\text{kJ}/\text{kg}$ for $h_3$, $900.74$ $\text{kJ}/\text{kg}$ for $h_4$, 659.84 $\text{kJ}/\text{kg}$ for $h_2$, and 310.24 $\text{kJ}/\text{kg}$ for $h_1$ in Equation (V).

$\begin{array}{c}{w}_{\text{net}}=\left(1,515.42-900.74\right)\text{kJ}/\text{kg}-\left(659.84-310.24\right)\text{kJ}/\text{kg}\\ =265.08\text{kJ}/\text{kg}\end{array}$

Substitute $192.7\text{kJ}/\text{kg}$ for $q_{\text{regen}}$, $1,515.42\text{kJ}/\text{kg}$ for $h_3$, and $659.84\text{kJ}/\text{kg}$ for $h_2$ in Equation (VI).

$\begin{array}{c}{q}_{\text{in}}=\left[\left(1,515.42-659.84\right)-192.7\right]\text{kJ}/\text{kg}\\ =662.88\text{kJ}/\text{kg}\end{array}$

Substitute $662.88\text{kJ}/\text{kg}$ for $q_{\text{in}}$, and $265.08\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (VII).

$\begin{array}{c}{q}_{\text{out}}=\left(662.88-265.08\right)\text{kJ}/\text{kg}\\ =397.80\text{kJ}/\text{kg}\end{array}$

Substitute 310.24 $\text{kJ}/\text{kg}$ for $h_1$, and 397.80 $\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (VIII).

$\begin{array}{c}{h}_6=\left(310.24+397.80\right)\text{kJ}/\text{kg}\\ =708.04\text{kJ}/\text{kg}\end{array}$

Rearrange Equation (IX), and substitute 659.84 $\text{kJ}/\text{kg}$ for $h_2$, 900.74 $\text{kJ}/\text{kg}$ for $h_4$, and 708.04 $\text{kJ}/\text{kg}$ for $h_6$.

$\begin{array}{c}{h}_5=\left(h_4-h_6\right)+h_2\\ =\left(900.74-708.04\right)\text{kJ}/\text{kg}+\text{659}\text{.84}\text{kJ}/\text{kg}\\ =852.54\text{kJ}/\text{kg}\end{array}$

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at 310 K $\left(T_1\right)$, $659.84\text{kJ}/\text{kg}$ $\left(h_2\right)$, 1400 K $\left(T_3\right)$, $900.74\text{kJ}/\text{kg}$ $\left(h_4\right)$, $852.54\text{kJ}/\text{kg}$ $\left(h_5\right)$ and $708.04\text{kJ}/\text{kg}$ $\left(h_6\right)$ which gives $1.73498\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1^0$, $2.49364\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2^0$, $3.36200\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3^0$, $2.81216\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4^0$,$2.75537\text{kJ}/\text{kg}\cdot \text{K}$ for $s_5^0$, and $2.56525\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6^0$.

Substitute 300 K for $T_0$, $1.73498\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1^{\circ }$, $2.49364\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, and 9 for $\frac{P_2}{P_1}$ in Equation (X).

$\begin{array}{c}{x}_{\text{destroyed,12}}=\left(300\text{K}\right)\left[\left(2.49364-1.73498\right)\text{kJ}/\text{kg}\cdot \text{K}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(9\right)\right]\\ =38.42\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 1-2 of the given Brayton cycle is $38.42\text{kJ}/\text{kg}$.

Substitute 300 K for $T_0$, $3.36200\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3^{\circ }$, $2.81216\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, and $\frac{1}{9}$ for $\frac{P_4}{P_3}$ in Equation (XI).

$\begin{array}{c}{x}_{\text{destroyed,34}}=\left(300\text{K}\right)\left[\left(2.81216-3.36200\right)\text{kJ}/\text{kg}\cdot \text{K}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(\frac{1}{9}\right)\right]\\ =24.23\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 3-4 of the given Brayton cycle is $24.23\text{kJ}/\text{kg}$.

Substitute 300 K for $T_0$, $2.75537\text{kJ}/\text{kg}\cdot \text{K}$ for $s_5^{\circ }$, $2.81216\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4^{\circ }$, $2.49364\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2^{\circ }$, and $2.56525\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6^{\circ }$ in Equation (XII).

$\begin{array}{c}{x}_{\text{destroyed,regen}}=\left(300\text{K}\right)\left[\begin{array}{l}\left(2.75537-2.49364\right)\text{kJ}/\text{kg}\cdot \text{K}\\ +\left(2.56525-2.81216\right)\text{kJ}/\text{kg}\cdot \text{K}\end{array}\right]\\ =4.47\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with regeneration process of the given Brayton cycle is $4.47\text{kJ}/\text{kg}$.

Substitute 300 K for $T_0$, $3.36200\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3^{\circ }$, $2.75537\text{kJ}/\text{kg}\cdot \text{K}$ for $s_5^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, 1260 K for $T_H$, and $662.88\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (XIII). For the process 5-3, pressure remains constant, hence substitute 0 for $\ln \frac{P_3}{P_5}$.

$\begin{array}{c}{x}_{\text{destroyed,53}}=\left(300\text{K}\right)\left[\begin{array}{l}\left(3.36200-2.75537\right)\text{kJ}/\text{kg}\cdot \text{K}\\ -\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0\right)-\frac{662.88\text{kJ}/\text{kg}}{1260\text{K}}\end{array}\right]\\ =24.16\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 5-3 of the given Brayton cycle is $24.16\text{kJ}/\text{kg}$.

Substitute 300 K for $T_0$, $1.73498\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1^{\circ }$, $2.56525\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6^{\circ }$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, 300 K for $T_L$, and $397.80\text{kJ}/\text{kg}$ for $q_{\text{out}}$ in Equation (XIV). For the process 6-1, pressure remains constant, hence substitute 0 for $\ln \frac{P_1}{P_6}$.

$\begin{array}{c}{x}_{\text{destroyed,61}}=\left(300\text{K}\right)\left[\left(1.73498-2.56525\right)\text{kJ}/\text{kg}\cdot \text{K}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0\right)+\frac{397.80\text{kJ}/\text{kg}}{300\text{K}}\right]\\ =148.7\text{kJ}/\text{kg}\end{array}$

Thus, the exergy destruction associated with process 6-1 of the given Brayton cycle is $148.7\text{kJ}/\text{kg}$.

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at 300 K $\left(T_0\right)$ which gives $1.70203\text{kJ}/\text{kg}\cdot \text{K}$ for $s_0^0$ and $300.19\text{kJ}/\text{kg}$ for $h_0$.

Substitute $2.56525\text{kJ}/\text{kg}\cdot \text{K}$ for $s_6^{\circ }$, and $1.70203\text{kJ}/\text{kg}\cdot \text{K}$ for $s_0^{\circ }$ in Equation (XVI). Also, at the exit of the regenerator, pressure remains constant, hence substitute 0 for $\ln \frac{P_6}{P_1}$.

$\begin{array}{c}{s}_6-s_0=\left(2.56525-1.70203\right)\text{kJ}/\text{kg}\cdot \text{K}\\ =0.86322\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $708.04\text{kJ}/\text{kg}$ for $h_6$, $300.19\text{kJ}/\text{kg}$ for $h_0$, 300 K for $T_0$, and $0.86322\text{kJ}/\text{kg}\cdot \text{K}$ for $\left(s_6-s_0\right)$ in Equation (XV).

$\begin{array}{c}{\varphi }_6=\left(h_6-h_0\right)-T_0\left(s_6-s_0\right)\\ =\left(708.04-300.19\right)\text{kJ}/\text{kg}-\left(300\text{K}\right)\left(0.86322\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =148.9\text{kJ}/\text{kg}\end{array}$

Thus, the exergy of the exhaust gases at the exit of the regenerator is $148.9\text{kJ}/\text{kg}$.