Chapter 9.12, Problem 143P

To determine

The minimum work that must be supplied to the compressor and turbine due to irreversibilities.

Answer to Problem 143P

The minimum work that must be supplied to the compressor due to irreversibilities is $40.1\text{kJ}/\text{kg}$.

The minimum work that is developed by the turbine due to irreversibilities is $34.8\text{kJ}/\text{kg}$.

It can be noted that the compressor is more sensitive to irreversibilities than the turbine.

Explanation of Solution

Show the simple Brayton cycle, with air as the working fluid on $T-s$ diagram.

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

Write the expression of temperature and pressure relation ratio for the compression process 1-2.

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

Here, specific heat ratio is k.

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

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

${\eta }_C=\frac{c_p\left(T_{2s}-T_1\right)}{c_p\left(T_2-T_1\right)}$ (II)

Here, specific heat at constant pressure is $c_p$.

Write the expression of temperature and pressure relation ratio for the expansion process 3-4.

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

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

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

${\eta }_T=\frac{c_p\left(T_3-T_4\right)}{c_p\left(T_3-T_{4s}\right)}$ (IV)

For compression processes

Write the expression for the entropy change for the isentropic process 1-2s $\left(s_{2s}-s_1\right)$.

$s_{2s}-s_1=c_p\ln \frac{T_{2s}}{T_1}-R\ln \frac{P_2}{P_1}$ (V)

Here, gas constant of air is R.

Write the expression for reversible work for the isentropic process 1-2s $\left(w_{\text{rev,1-2s}}\right)$.

$w_{\text{rev,1-2s}}=c_p\left(T_{2s}-T_1\right)-T_0\left(s_{2s}-s_1\right)$ (VI)

Here, temperature of the surroundings is $T_0$.

Write the expression for the entropy change for the process 1-2 $\left(s_2-s_1\right)$.

$s_2-s_1=c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}$ (VII)

Write the expression for the reversible work for the process 1-2 $\left(w_{\text{rev,1-2}}\right)$.

$w_{\text{rev,1-2}}=c_p\left(T_2-T_1\right)-T_0\left(s_2-s_1\right)$ (VIII)

Write the expression for the minimum work that must be supplied to the compressor due to irreversibilities $\left(\Delta w_{\text{rev,C}}\right)$.

$\Delta w_{\text{rev,C}}=w_{\text{rev,1-2}}-w_{\text{rev,1-2s}}$ (IX)

For expansion processes

Write the expression for the entropy change for the isentropic process 3-4s $\left(s_3-s_{4s}\right)$.

$s_3-s_{4s}=c_p\ln \frac{T_3}{T_{4s}}-R\ln \frac{P_3}{P_4}$ (X)

Write the expression for the reversible work for the isentropic process 3-4s $\left(w_{\text{rev,3-4s}}\right)$.

$w_{\text{rev,3-4s}}=c_p\left(T_3-T_{4s}\right)-T_0\left(s_3-s_{4s}\right)$ (XI)

Write the expression for the entropy change for the process 3-4 $\left(s_3-s_4\right)$.

$s_3-s_4=c_p\ln \frac{T_3}{T_4}-R\ln \frac{P_3}{P_4}$ (XII)

Write the expression for the reversible work for the process 3-4 $\left(w_{\text{rev,3-4}}\right)$.

$w_{\text{rev,3-4}}=c_p\left(T_3-T_4\right)-T_0\left(s_3-s_4\right)$ (XIII)

Write the expression for the minimum work that is developed by the turbine due to irreversibilities $\left(\Delta w_{\text{rev,T}}\right)$.

$\Delta w_{\text{rev,T}}=w_{\text{rev,3-4s}}-w_{\text{rev,3-4}}$ (XIV)

Conclusion:

Substitute 288 K for $T_1$, 12 for $\frac{P_2}{P_1}$, and 1.4 for k in Equation (I).

$\begin{array}{c}{T}_{2s}=\left(288\text{K}\right){\left(12\right)}^{1.4-1/1.4}\\ =585.8\text{K}\end{array}$

Rearrange Equation (II), and solve for $T_2$. Substitute 288 K for $T_1$, 585.8 K for $T_{2s}$, and 0.80 for ${\eta }_C$ in Equation (II).

$\begin{array}{c}{T}_2=T_1+\frac{T_{2s}-T_1}{{\eta }_C}\\ =288\text{ K}+\frac{\left(585.8-288\right)\text{K}}{0.80}\\ =660.2\text{}\text{K}\end{array}$

Substitute 873 K for $T_3$, $\frac{1}{12}$ for $\frac{P_4}{P_3}$, and 1.4 for k in Equation (III).

$\begin{array}{c}{T}_{4s}=\left(873\text{K}\right){\left(\frac{1}{12}\right)}^{1.4-1/1.4}\\ =429.2\text{K}\end{array}$

Rearrange Equation (IV), and substitute 873 K for $T_3$, 429.2 K for $T_{4s}$, and 0.80 for ${\eta }_T$.

$\begin{array}{c}{T}_4=T_3-{\eta }_T\left(T_3-T_{4s}\right)\\ =873\text{ K}-\left(0.80\right)\left(873\text{ K}-429.2\text{}\text{ K}\right)\\ =518\text{K}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 585.8 K for $T_{2s}$, 288 K for $T_1$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, and 12 for $\frac{P_2}{P_1}$ in Equation (V).

$\begin{array}{c}{s}_{2s}-s_1=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{585.8\text{K}}{288\text{K}}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(12\right)\\ =0.0003998\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 585.8 K for $T_{2s}$, 288 K for $T_1$, 288 K for $T_0$, and $0.0003998\text{kJ}/\text{kg}\cdot \text{K}$ for $\left(s_{2s}-s_1\right)$ in Equation (VI).

$\begin{array}{c}{w}_{\text{rev,1-2s}}=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(585.8-288\right)\text{K}-\left(288\text{K}\right)\left(0.0003998\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =299.2\text{kJ}/\text{kg}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 660.2 K for $T_2$, 288 K for $T_1$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, and 12 for $\frac{P_2}{P_1}$ in Equation (VII).

$\begin{array}{c}{s}_2-s_1=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{660.2\text{K}}{288\text{K}}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(12\right)\\ =0.1206\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 660.2 K for $T_2$, 288 K for $T_1$, 288 K for $T_0$, and $0.1206\text{kJ}/\text{kg}\cdot \text{K}$ for $\left(s_2-s_1\right)$ in Equation (VIII).

$\begin{array}{c}{w}_{\text{rev,1-2}}=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(660.2-288\right)\text{K}-\left(288\text{K}\right)\left(0.1206\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =339.3\text{kJ}/\text{kg}\end{array}$

Substitute $339.3\text{kJ}/\text{kg}$ for $w_{\text{rev,1-2}}$, and $299.2\text{kJ}/\text{kg}$ for $w_{\text{rev,1-2s}}$ in Equation (IX).

$\begin{array}{c}\Delta w_{\text{rev,C}}=\left(339.3-299.2\right)\text{kJ}/\text{kg}\\ =40.1\text{kJ}/\text{kg}\end{array}$

Thus, the minimum work that must be supplied to the compressor due to irreversibilities is $40.1\text{kJ}/\text{kg}$.

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 873 K for $T_3$, 429.2 K for $T_{4s}$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, and 12 for $\frac{P_3}{P_4}$ in Equation (X).

$\begin{array}{c}{s}_3-s_{4s}=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{873\text{K}}{429.2\text{K}}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(12\right)\\ =0.0003944\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 873 K for $T_3$, 429.2 K for $T_{4s}$, 288 K for $T_0$, and $0.0003944\text{kJ}/\text{kg}\cdot \text{K}$ for $\left(s_3-s_{4s}\right)$ in Equation (XI).

$\begin{array}{c}{w}_{\text{rev,3-4s}}=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(873-429.2\right)\text{K}-\left(288\text{K}\right)\left(0.0003944\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =445.9\text{kJ}/\text{kg}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 873 K for $T_3$, 518.0 K for $T_4$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$, and 12 for $\frac{P_3}{P_4}$ in Equation (XII).

$\begin{array}{c}{s}_3-s_4=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{873\text{K}}{518.0\text{K}}-\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \left(12\right)\\ =-0.1885\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, 873 K for $T_3$, 518.0 K for $T_4$, 288 K for $T_0$, and $-0.1885\text{kJ}/\text{kg}\cdot \text{K}$ for $\left(s_3-s_4\right)$ in Equation (XIII).

$\begin{array}{c}{w}_{\text{rev,3-4}}=\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(873-518.0\right)\text{K}-\left(288\text{K}\right)\left(-0.1885\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =411.1\text{kJ}/\text{kg}\end{array}$

Substitute $445.9\text{kJ}/\text{kg}$ for $w_{\text{rev,3-4s}}$, and $411.1\text{kJ}/\text{kg}$ for $w_{\text{rev,3-4}}$ in Equation (XIV).

$\begin{array}{c}\Delta w_{\text{rev,T}}=\left(445.9-411.1\right)\text{kJ}/\text{kg}\\ =34.8\text{kJ}/\text{kg}\end{array}$

Thus, the minimum work that is developed by the turbine due to irreversibilities is $34.8\text{kJ}/\text{kg}$.

It can be noted that the compressor is more sensitive to irreversibilities than the turbine.