Chapter 7.13, Problem 194RP

a)

To determine

The net power delivered to the generator by the turbine.

Answer to Problem 194RP

The net power delivered to the generator by the turbine is 20,448 kW.

Explanation of Solution

Write the expression for the final entropy, $s_2$.

$s_4=s_f+x_4{s}_{fg}$ (I)

Here, final entropy is $s_4$$s_1$, final dry fraction is $x_4$, entropy of saturated liquid is $s_f$ and entropy of evaporation is $s_{fg}$.

Write the expression to calculate the final enthalpy $h_2$.

$h_4=h_f+x_4{h}_{fg}$ (II)

Here, final enthalpy is $s_4$$s_1$, final dry fraction is $x_4$, enthalpy of saturated liquid is $s_f$ and enthalpy of evaporation is $s_{fg}$.

Express the energy balance equation for closed system.

$E_{in}-E_{out}=\Delta E_{system}$ (III)

Here, energy transfer in to the system is $E_{in}$, energy transfer from the system is $E_{out}$ and change in internal energy of system is $\Delta E_{system}$.

Write the energy balance for the compressor using equation (III).

${\dot{W}}_{comp,in}+{\dot{m}}_{air}{h}_1={\dot{m}}_{air}{h}_2$

${\dot{W}}_{comp,in}={\dot{m}}_{air}\left(h_2-h_1\right)$ (IV)

Write the energy balance for the compressor using equation (III).

${\dot{m}}_{steam}{h}_3={\dot{W}}_{turb,out}+{\dot{m}}_{steam}{h}_4$

${\dot{W}}_{turb,out}={\dot{m}}_{steam}\left(h_3-h_4\right)$ (V)

Write the expression for the net work done.

${\dot{W}}_{net,out}={\dot{W}}_{turb,out}-W_{comp,in}$ (VI)

Conclusion:

From the Table A-1, “Molar mass, gas constant, and critical-point properties”, select the gas constant $\left(R\right)$ for air gas as $0.287\text{kJ}/\text{kg}\cdot \text{K}$.

Refer to Table A-17, “ideal gas properties of air”, obtain the following properties of air at a temperature of 295 K.

$\begin{array}{l}{h}_1=295.17\text{kJ}/\text{kg}\\ s_1^{°}=1.68515\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Refer to Table A-17, “ideal gas properties of air”, obtain the following properties of air at a temperature of 620 K.

$\begin{array}{l}{h}_2=628.07\text{kJ}/\text{kg}\\ s_2^{°}=2.44356\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From Table A-6E, “Superheated water”, select the enthalpy $\left(h_3\right)$ and entropy $\left(s_3\right)$ at the pressure of $12.5\text{ MPa}$ and temperature of $500°\text{C}$ as $3343.6\text{kJ}/\text{kg}$ and $6.4651\text{kJ}/\text{kg}\cdot \text{K}$ respectively.

Refer to Table A-3, “saturated water-pressure table”, obtain the following properties of water at pressure of 10 kPa.

$\begin{array}{l}{h}_f=191.81\text{kJ}/\text{kg}\\ h_{fg}=2392.1\text{kJ}/\text{kg}\\ s_f=0.6492\text{kJ}/\text{kg}\cdot \text{K}\\ h_{fg}=7.4996\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $0.6492\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$,$0.92$ for $x_4$, and $7.4996\text{kJ}/\text{kg}\cdot \text{K}$ for $s_{fg}$ in equation (I)

$\begin{array}{c}6.5432\text{kJ}/\text{kg}\cdot \text{K}=1.3028\text{kJ}/\text{kg}\cdot \text{K}+x_2\left(6.0562\text{kJ}/\text{kg}\cdot \text{K}\right)\\ x_2=\frac{\left(6.5432-1.3028\right)\text{kJ}/\text{kg}\cdot \text{K}}{6.0562\text{kJ}/\text{kg}\cdot \text{K}}\\ =0.8653\end{array}$

Substitute $191.8\text{kJ}/\text{kg}$ for $h_f$,$0.92$ for $x_4$ and $2392.1\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (II).

$\begin{array}{c}{h}_2=191.81\text{kJ}/\text{kg}+0.92\left(2392.1\text{kJ}/\text{kg}\right)\\ =2392.5\text{kJ}/\text{kg}\end{array}$

Substitute $\text{10 }\text{kg}/\text{s}$ for $m_{air}$, $628.07\text{kJ}/\text{kg}$ for $h_2$,and $295.17\text{kJ}/\text{kg}$ for $h_1$ in equation (IV).

$\begin{array}{c}{\dot{W}}_{comp,in}=\left(\text{10 }\text{kg}/\text{s}\right)\left(628.07\text{kJ}/\text{kg}-295.17\text{kJ}/\text{kg}\right)\\ =3329\text{ kW}\end{array}$

Substitute $\text{25 }\text{kg}/\text{s}$ for $m_{steam}$, $3343.6\text{kJ}/\text{kg}$ for $h_3$,and $2392.5\text{kJ}/\text{kg}$ for $h_4$ in equation (V).

$\begin{array}{c}{\dot{W}}_{turb,out}=\left(\text{25 }\text{kg}/\text{s}\right)\left(3343.6\text{kJ}/\text{kg}-2392.5\text{kJ}/\text{kg}\right)\\ =23,777\text{ kW}\end{array}$

Substitute $23,777\text{ kW}$ for ${\dot{W}}_{turb,out}$ and $3329\text{ kW}$ for ${\dot{W}}_{comp,in}$ in equation (VI).

$\begin{array}{c}{\dot{W}}_{net,out}=23,777\text{ kW}-3329\text{ kW}\\ =20,448\text{ kW}\end{array}$

Thus, the net power delivered to the generator by the turbine is 20,448 kW.

b)

To determine

The rate of entropy generation with in the turbine and compressor during the process

Answer to Problem 194RP

The rate of entropy generation with in the turbine and compressor during the process is $\underset{\_}{28.02\text{kW}/\text{K}}$.

Explanation of Solution

Write the formula for the entropy change of compressor.

${\dot{S}}_{gen,comp}={\dot{m}}_{air}\left(s_2^{°}-s_1^{°}-R\ln \left(\frac{P_2}{P_1}\right)\right)$ (VII)

Here, pressure at state point 2 is $P_2$, pressure at state point 1 is $P_1$, and mass flow rate of the air is ${\dot{m}}_{air}$.

Write the formula for the entropy change of turbine.

${\dot{S}}_{gen,Turb}={\dot{m}}_{steam}\left(s_4-s_3\right)$ (VIII)

Write the expression for the total entropy generation.

${\dot{S}}_{gen,total}={\dot{S}}_{gen,comp}+{\dot{S}}_{gen,turb}$ (IX)

Conclusion:

Substitute $10\text{kg}/\text{s}$ for ${\dot{m}}_{air}$,$2.44356\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2^{°}$,$1.68515\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1^{°}$,$0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, $1000\text{kPa}$ for $P_2$, and $98\text{ kPa}$ for $P_1$ equation (VII).

$\begin{array}{c}{\dot{S}}_{gen,comp}=10\text{kg}/\text{s}\left(2.44356\text{kJ}/\text{kg}\cdot \text{K}-1.68515\text{kJ}/\text{kg}\cdot \text{K}-0.287\text{kJ}/\text{kg}\cdot \text{K}\times \text{ln}\left(\frac{1000\text{kPa}}{98\text{ kPa}}\right)\right)\\ =0.92\text{kW}/\text{K}\end{array}$

Substitute $25\text{kg}/\text{s}$ for ${\dot{m}}_{steam}$,$7.5489\text{kJ}/\text{kg}\cdot \text{K}$ for $s_4$, and $6.4651\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$ equation (VIII).

$\begin{array}{c}{\dot{S}}_{gen,turb}=25\text{kg}/\text{s}\left(\text{7}\text{.5489 }\text{kJ}/\text{kg}\cdot \text{K}-6.4651\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =27.1\text{kW}/\text{K}\end{array}$

Substitute $0.92\text{kW}/\text{K}$ for ${\dot{S}}_{gen,comp}$ and $27.1\text{kW}/\text{K}$ for ${\dot{S}}_{gen,turb}$ in equation (IX).

$\begin{array}{c}{\dot{S}}_{gen,total}=0.92\text{kW}/\text{K}+27.1\text{kW}/\text{K}\\ =28.02\text{kW}/\text{K}\end{array}$

Thus, the rate of entropy generation with in the turbine and compressor during the process is $\underset{\_}{28.02\text{kW}/\text{K}}$.