Chapter 9.12, Problem 173RP

Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor inlet temperature is 300 K, and the turbine inlet temperature is 1800 K. The effectiveness of the regenerator is 75 percent. Determine the thermal efficiency and the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 80 percent.

To determine

The thermal efficiency and the required mass flow rate of helium for a net power

output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of  100 percent.

Answer to Problem 173RP

The thermal efficiency of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $100\%$ is $60.3\%$.

The required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $100\%$ is $18.42\text{kg}/\text{s}$.

Explanation of Solution

Draw the $T-s$ diagram for regenerative Brayton cycle as shown 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.

Consider ${\eta }_{\text{T}}={\eta }_{\text{C}}=100\%$

Write the expression to calculate the temperature and pressure relation ratio for the isentropic compression process 1-2s.

$T_{2s}=T_1{\left(\frac{P_2}{P_1}\right)}^{\left(k-1\right)/k}$ (I)

Here, the specific heat ratio is k.

Write the expression to calculate the temperature and pressure relation ratio for the isentropic expansion process 3-4s.

$T_{4s}=T_3{\left(\frac{P_4}{P_3}\right)}^{\left(k-1\right)/k}$ (II)

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

$\begin{array}{l}\epsilon =\frac{T_5-T_2}{T_4-T_2}\\ T_5=T_2+\epsilon \left(T_4-T_2\right)\end{array}$ (III)

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

$\begin{array}{c}{w}_{\text{net}}=w_{\text{T,out}}-w_{\text{C,in}}\\ =\left(h_3-h_4\right)-\left(h_2-h_1\right)\\ =c_p\left[\left(T_3-T_4\right)-\left(T_2-T_1\right)\right]\end{array}$ (IV)

Here, the specific heat of helium at constant pressure is $c_p$, the 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 to calculate the heat input for the regenerative Brayton cycle $\left(q_{\text{in}}\right)$.

$\begin{array}{c}{q}_{\text{in}}=h_3-h_5\\ =c_p\left(T_3-T_5\right)\end{array}$ (V)

Write the expression to calculate the thermal efficiency of the given regenerative Brayton cycle $\left({\eta }_{th}\right)$.

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

Write the expression to calculate the mass flow rate of helium flowing through the given regenerative Brayton cycle $\left({\dot{m}}_{\text{air}}\right)$.

$\dot{m}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}$ (VII)

Here, the net power output produced by the given regenerative Brayton cycleis ${\dot{W}}_{\text{net}}$.

Conclusion:

From Table A-2, “Ideal-gas specific heats of various common gases”, obtain the following values for helium gas.

$\begin{array}{l}{c}_p=5.196\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.667\end{array}$

Substitute 300 K for $T_1$, 8 for $\frac{P_2}{P_1}$, and 1.667 for k in Equation (I).

$\begin{array}{c}{T}_{2s}=\left(300\text{K}\right){\left(8\right)}^{1.667-1/1.667}\\ =689.4\text{K}\end{array}$

Substitute 1800 K for $T_3$, $\frac{1}{8}$ for $\frac{P_4}{P_3}$, and 1.667 for k in Equation (II).

$\begin{array}{c}{T}_{4s}=\left(1800\text{K}\right){\left(\frac{1}{8}\right)}^{1.667-1/1.667}\\ =783.3\text{K}\end{array}$

Substitute 0.75 for $\epsilon$, 689.4 K for $T_2$, and 783.3 K for $T_4$ to find $T_5$ in Equation (III).

$\begin{array}{c}{T}_5=689.4\text{K+}\left(0.75\right)\left(783.3-689.4\right)\text{K}\\ =759.8\text{K}\end{array}$

Substitute $5.1926\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1800 K for $T_3$, 783.3 K for $T_4$, 689.4 K for $T_2$, and 300 K for $T_1$ in Equation (IV).

$\begin{array}{c}{w}_{\text{net}}=\left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(1800-783.3\right)\text{K-}\left(689.4-300\right)\text{K}\right]\\ =3257.3\text{kJ}/\text{kg}\end{array}$

Substitute $5.1926\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1,800 K for $T_3$, 759.8 K for $T_5$ to find $q_{\text{in}}$ in

Equation (V).

$\begin{array}{c}{q}_{\text{in}}=\left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1,800-759.8\right)\text{K}\\ =5,401.3\text{kJ}/\text{kg}\end{array}$

Substitute $3257.3\text{kJ}/\text{kg}$ for $w_{\text{net}}$, and $5401.3\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (VI).

$\begin{array}{c}{\eta }_{th}=\frac{3257.3\text{kJ}/\text{kg}}{5401.3\text{kJ}/\text{kg}}\\ =0.603\times 100\%\\ =60.3\%\end{array}$

Thus, the thermal efficiency of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $100\%$ is $60.3\%$.

Substitute $\text{60,000}\text{kJ}/\text{s}$ for ${\dot{W}}_{\text{net}}$, and $\text{3257}\text{.3}\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (VII).

$\begin{array}{c}\dot{m}=\frac{\text{60,000}\text{kJ}/\text{s}}{\text{3257}\text{.3}\text{kJ}/\text{kg}}\\ =18.42\text{kg}/\text{s}\end{array}$

Thus, the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $100\%$ is $18.42\text{kg}/\text{s}$.

To determine

The thermal efficiency and the required mass flow rate of helium for a net power

output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of  80 percent.

Answer to Problem 173RP

The required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $80\%$ is $35.4\text{kg}/\text{s}$.

The thermal efficiency of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $80\%$ is $37.8\%$.

Explanation of Solution

Consider ${\eta }_{\text{T}}={\eta }_{\text{C}}=80\%$

Write the expression to calculate the temperature and pressure relation for the isentropic compression process 1-2.

$T_{2s}=T_1{\left(\frac{P_2}{P_1}\right)}^{\left(k-1\right)/k}$ (VIII)

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

$\begin{array}{c}{\eta }_C=\frac{h_{2s}-h_1}{h_2-h_1}\\ {\eta }_C=\frac{c_p\left(T_{2s}-T_1\right)}{c_p\left(T_2-T_1\right)}\\ T_2=T_1+\frac{T_{2s}-T_1}{{\eta }_C}\end{array}$ (IX)

Write the expression to calculate the temperature and pressure relation for the isentropic expansion process 3-4.

$T_{4s}=T_3{\left(\frac{P_4}{P_3}\right)}^{\left(k-1\right)/k}$ (X)

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

$\begin{array}{c}{\eta }_T=\frac{h_3-h_4}{h_3-h_{4s}}\\ {\eta }_T=\frac{c_p\left(T_3-T_4\right)}{c_p\left(T_3-T_{4s}\right)}\\ T_4=T_3-{\eta }_T\left(T_3-T_{4s}\right)\end{array}$ (XI)

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

$\begin{array}{l}\epsilon =\frac{T_5-T_2}{T_4-T_2}\\ T_5=T_2+\epsilon \left(T_4-T_2\right)\end{array}$ (XII)

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

$\begin{array}{c}{w}_{\text{net}}=w_{\text{T,out}}-w_{\text{C,in}}\\ =\left(h_3-h_4\right)-\left(h_2-h_1\right)\\ =c_p\left[\left(T_3-T_4\right)-\left(T_2-T_1\right)\right]\end{array}$ (XIII)

Here, the specific heat of helium at constant pressure is $c_p$, the 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 to calculate the heat input for the regenerative Brayton cycle $\left(q_{\text{in}}\right)$.

$\begin{array}{c}{q}_{\text{in}}=h_3-h_5\\ =c_p\left(T_3-T_5\right)\end{array}$ (XIV)

Write the expression to calculate the thermal efficiency of the given regenerative Brayton cycle $\left({\eta }_{th}\right)$.

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

Write the expression to calculate the mass flow rate of helium flowing through the given regenerative Brayton cycle $\left({\dot{m}}_{\text{air}}\right)$.

$\dot{m}=\frac{{\dot{W}}_{\text{net}}}{w_{\text{net}}}$ (XVI)

Here, the net power output produced by the given regenerative Brayton cycle is ${\dot{W}}_{\text{net}}$.

Conclusion:

Substitute 300 K for $T_1$, 8 for $\frac{P_2}{P_1}$, and 1.667 for k in Equation (VIII).

$\begin{array}{c}{T}_{2s}=\left(300\text{K}\right){\left(8\right)}^{1.667-1/1.667}\\ =689.4\text{K}\end{array}$

Substitute 300 K for $T_1$, 689.4 K for $T_{2s}$, and 0.80 for ${\eta }_C$ in Equation (IX).

$\begin{array}{c}{T}_2=300\text{K}+\frac{\left(689.4-300\right)\text{K}}{0.80}\\ =786.8\text{}\text{K}\end{array}$

Substitute 1800 K for $T_3$, $\frac{1}{8}$ for $\frac{P_4}{P_3}$, and 1.667 for k in Equation (X).

$\begin{array}{c}{T}_{4s}=\left(1800\text{K}\right){\left(\frac{1}{8}\right)}^{1.667-1/1.667}\\ =783.3\text{K}\end{array}$

Substitute 1800 K for $T_3$, 783.3 K for $T_{4s}$, and 0.80 for ${\eta }_T$ in Equation (XI).

$\begin{array}{c}{T}_4=1800\text{K}-\left(0.80\right)\left(1,800-783.3\right)\text{K}\\ =986.6\text{K}\end{array}$

Substitute 0.75 for $\epsilon$, 786.8 K for $T_2$, and 986.6 K for $T_4$ in Equation (XII).

$\begin{array}{c}{T}_5=786.8\text{K+}\left(0.75\right)\left(986.6-786.8\right)\text{K}\\ =936.7\text{K}\end{array}$

Substitute $5.1926\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1,800 K for $T_3$, 986.6 K for $T_4$, 786.8 K for $T_2$, and 300 K for $T_1$ in Equation (XIII).

$\begin{array}{c}{w}_{\text{net}}=\left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(1,800-986.6\right)\text{K}-\left(786.8-300\right)\text{K}\right]\\ =1695.9\text{kJ}/\text{kg}\end{array}$

Substitute $5.1926\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 1800 K for $T_3$, 936.7 K for $T_5$ to find $q_{\text{in}}$ in Equation (XIV).

$\begin{array}{c}{q}_{\text{in}}=\left(5.1926\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1800-936.7\right)\text{K}\\ =4482.\text{8}\text{kJ}/\text{kg}\end{array}$

Substitute $\text{60,000}\text{kJ}/\text{s}$ for ${\dot{W}}_{\text{net}}$, and $\text{1695}\text{.9}\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (XVI).

$\begin{array}{c}\dot{m}=\frac{\text{60,000}\text{kJ}/\text{s}}{\text{1695}\text{.9}\text{kJ}/\text{kg}}\\ =35.4\text{kg}/\text{s}\end{array}$

Thus, the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $80\%$ is $35.4\text{kg}/\text{s}$.

Substitute $1695.9\text{kJ}/\text{kg}$ for $w_{\text{net}}$, and $4482.8\text{kJ}/\text{kg}$ for $q_{\text{in}}$ in Equation (XV).

$\begin{array}{c}{\eta }_{th}=\frac{1695.9\text{kJ}/\text{kg}}{4482.8\text{kJ}/\text{kg}}\\ =0.378\times 100\%\\ =37.8\%\end{array}$

Thus, the thermal efficiency of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of $80\%$ is $37.8\%$.