Chapter 8.8, Problem 85P

a)

To determine

The temperature of the saturated steam $\left(T_2\right)$ entering the chamber

Answer to Problem 85P

The temperature of the saturated steam $\left(T_2\right)$ entering the chamber is $\underset{\_}{129.2\text{°C}}$.

Explanation of Solution

Write expression for the mass balance on the chamber.

$\begin{array}{l}{\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2={\dot{m}}_3{h}_3\\ {\dot{m}}_1{h}_1+{\dot{m}}_2{h}_2=\left({\dot{m}}_1+{\dot{m}}_2\right)h_3\\ h_2=\frac{\left[\left({\dot{m}}_1+{\dot{m}}_2\right)h_3\right]-{\dot{m}}_1{h}_1}{{\dot{m}}_2}\end{array}$ (I)

Here, mass rate of liquid water at state 1 is ${\dot{m}}_1$, mass flow rate of saturated steam at state 2 is ${\dot{m}}_2$, mixture mass flow rate of liquid water and saturated steam at state 3 is ${\dot{m}}_3$, enthalpy at state $\text{1,2}\text{and}\text{3}$ is $h_1,h_2\text{and}{h}_3$ respectively.

Conclusion:

Refer to Table A-4, “Saturated water-temperature table”, obtain the following properties at the temperature of $20°\text{C}$.

$\begin{array}{c}{h}_{1,h_f@20\text{°C}}=h_0\\ =83.91\text{kJ}/\text{kg}\\ s_{1,s_f@20\text{°C}}=s_0\\ =0.29649\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, enthalpy of saturated liquid is $h_f$, entropy of saturated liquid is $s_f$, dead-state enthalpy is $h_0$ and dead-state entropy is $s_0$.

Refer to Table A-4, “Saturated water-temperature table”, obtain the following properties at the temperature of $45°\text{C}$ and state 1 quality $\left(x_1\right)$ of $0$.

$\begin{array}{c}{h}_{3,h_f@45\text{°C}}=188.44\text{kJ}/\text{kg}\\ s_{3,s_f@45\text{°C}}=0.63862\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $\text{4}\text{.6}\text{kg}/\text{s}$ for ${\dot{m}}_1$, $\text{83}\text{.91}\text{kJ}/\text{kg}$ for $h_1$, $\text{0}\text{.19}\text{kg}/\text{s}$ for ${\dot{m}}_2$, $\text{188}\text{.44}\text{kJ}/\text{kg}$ for $h_3$ in Equation (I).

$\begin{array}{c}{h}_2=\frac{\left[\left(\text{4}\text{.6}\text{kg}/\text{s}+\text{0}\text{.19}\text{kg}/\text{s}\right)\text{188}\text{.44}\text{kJ}/\text{kg}\right]-\left(\text{4}\text{.6}\text{kg}/\text{s}\right)\left(\text{83}\text{.91}\text{kJ}/\text{kg}\right)}{\text{0}\text{.19}\text{kg}/\text{s}}\\ =\frac{\left[\left(\text{4}\text{.79}\text{kg}/\text{s}\right)\text{188}\text{.44}\text{kJ}/\text{kg}\right]-\text{385}\text{.986}\text{kJ}/\text{s}}{\text{0}\text{.19}\text{kg}/\text{s}}\\ =\frac{902.627\text{kJ}/\text{s}-\text{385}\text{.986}\text{kJ}/\text{s}}{\text{0}\text{.19}\text{kg}/\text{s}}\\ =2,719\text{kJ}/\text{kg}\end{array}$

Refer to Table A-4, “saturated water–temperature table”, for the enthalpy of $2719\text{kJ}/\text{kg}$ obtain the following properties using interpolation method.

Enthalpy $\text{kJ}/\text{kg}$	Temperature, $°\text{C}$
2713.1	125
2719	?
2720.1	130

Here, temperature of the saturated steam at state 2 is $T_2$ and enthalpy of saturated vapor is $h_g$.

Substitute $2713.1\text{kJ}/\text{kg}$ for $x_1$, $2719\text{kJ}/\text{kg}$ for $x_2$, $2720.1\text{kJ}/\text{kg}$ for $x_3$, $125°\text{C}$ for $y_1$, and $130°\text{C}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{T}_2=\frac{\left(2719-2713.1\right)\left(130-125\right)}{\left(2720.1-2713.1\right)}+125\\ =4.2+125\\ =129.2\text{°C}\end{array}$

Thus, the temperature of the saturated steam $\left(T_2\right)$ entering the chamber is $\underset{\_}{129.2\text{°C}}$.

b)

To determine

The exergy destruction $\left({\dot{X}}_{\text{dest}}\right)$ during the mixing process.

Answer to Problem 85P

The exergy destruction $\left({\dot{X}}_{\text{dest}}\right)$ during the mixing process is $\underset{\_}{105.05\text{kW}}$.

Explanation of Solution

Calculate the specific exergy of state 2 $\left({\psi }_2\right)$.

${\psi }_2=h_2-h_0-T_0\left(s_2-s_0\right)$ (II)

Here, dead state temperature is $T_0$.

Calculate the specific exergy of state 3 $\left({\psi }_3\right)$.

${\psi }_3=h_3-h_0-T_0\left(s_3-s_0\right)$ (III)

Calculate the specific exergy of state 1 $\left({\psi }_1\right)$.

${\psi }_1=h_1-h_0-T_0\left(s_1-s_0\right)$ (IV)

Calculate the exergy destruction $\left({\dot{X}}_{\text{dest}}\right)$ during the mixing process.

$\begin{array}{c}{\dot{X}}_{\text{dest}}={\dot{m}}_1{\psi }_1+{\dot{m}}_2{\psi }_2-{\dot{m}}_3{\psi }_3\\ ={\dot{m}}_1{\psi }_1+{\dot{m}}_2{\psi }_2-\left({\dot{m}}_1+{\dot{m}}_2\right){\psi }_3\end{array}$ (V)

Conclusion:

Refer to Table A-4, “saturated water–temperature table”, for the enthalpy of $2719\text{kJ}/\text{kg}$ obtain the following properties using interpolation method.

Temperature, $°\text{C}$	Entropy $\text{kJ}/\text{kg}\cdot \text{K}$
125	7.0771
129.2	?
130	7.0265

Substitute $125°\text{C}$ for $x_1$, $129.2°\text{C}$ for $x_2$, $130°\text{C}$ for $x_3$, $7.0771\text{kJ}/\text{kg}\cdot \text{K}$ for $y_1$, and $7.0265\text{kJ}/\text{kg}\cdot \text{K}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{s}_{2@129.2\text{°C}}=\frac{\left(129.2-125\right)\left(7.0265-7.0771\right)}{\left(130-125\right)}+7.0771\\ =-0.0425+7.0771\\ =7.0348\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute $2,719\text{kJ}/\text{kg}$ for $h_2$, $83.91\text{kJ}/\text{kg}$ for $h_0$, $\text{20°C}$ for $T_0$, $7.0348\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ and $0.29649\text{kJ}/\text{kg}\cdot \text{K}$ for $s_0$ in equation (II).

$\begin{array}{c}{\psi }_2=\left[\left(2,719-83.91\right)\text{kJ}/\text{kg}\right]-\left(\text{20°C}\right)\left[\left(7.0348-0.29649\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\\ =\left(2,635.09\text{kJ}/\text{kg}\right)-\left\{\left[\left(20+273\right)\text{K}\right]\left(6.73831\text{kJ}/\text{kg}\cdot \text{K}\right)\right\}\\ =2,635.09\text{kJ}/\text{kg}-\left[\left(293\text{K}\right)\left(6.73831\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ =660.76\text{kJ}/\text{kg}\end{array}$

Substitute $188.44\text{kJ}/\text{kg}$ for $h_3$, $83.91\text{kJ}/\text{kg}$ for $h_0$, $\text{20°C}$ for $T_0$, $0.63862\text{kJ}/\text{kg}\cdot \text{K}$ for $s_3$ and $0.29649\text{kJ}/\text{kg}\cdot \text{K}$ for $s_0$ in equation (III).

$\begin{array}{c}{\psi }_2=\left[\left(188.44-83.91\right)\text{kJ}/\text{kg}\right]-\left(\text{20°C}\right)\left[\left(0.63862-0.29649\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\\ =104.53\text{kJ}/\text{kg}-\left[\left(293\text{K}\right)\left(0.34213\text{kJ}/\text{kg}\cdot \text{K}\right)\right]\\ =4.276\text{kJ}/\text{kg}\end{array}$

Substitute $83.91\text{kJ}/\text{kg}$ for $h_1$, $83.91\text{kJ}/\text{kg}$ for $h_0$, $\text{20°C}$ for $T_0$, $0.29649\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ and $0.29649\text{kJ}/\text{kg}\cdot \text{K}$ for $s_0$ in equation (IV).

$\begin{array}{c}{\psi }_1=\left[\left(83.91-83.91\right)\text{kJ}/\text{kg}\right]-\left(\text{20°C}\right)\left[\left(0.29649-0.29649\right)\text{kJ}/\text{kg}\cdot \text{K}\right]\\ =0\end{array}$

Substitute $\text{4}\text{.6}\text{kg}/\text{s}$ for ${\dot{m}}_1$, $\text{0}$ for ${\psi }_1$, $\text{0}\text{.19}\text{kg}/\text{s}$ for ${\dot{m}}_2$, $\text{660}\text{.7}\text{kJ}/\text{kg}$ for ${\psi }_2$ and $\text{4}\text{.276}\text{kJ}/\text{kg}$ for ${\psi }_3$ in Equation (V).

$\begin{array}{c}{\dot{X}}_{\text{dest}}=\left\{\begin{array}{l}\left[\left(\text{4}\text{.6}\text{kg}/\text{s}\right)\left(0\right)\right]+\left[\left(\text{0}\text{.19}\text{kg}/\text{s}\right)\left(\text{660}\text{.7}\text{kJ}/\text{kg}\right)\right]\\ -\left[\left(\text{4}\text{.6}\text{kg}/\text{s}+\text{0}\text{.19}\text{kg}/\text{s}\right)\left(\text{4}\text{.276}\text{kJ}/\text{kg}\right)\right]\end{array}\right\}\\ =0+125.533\text{kJ}/\text{s}-20.482\text{kJ}/\text{s}\\ =105.05\text{kJ}/\text{s}\left(\frac{\text{kW}}{\text{kJ}/\text{s}}\right)\\ =105.05\text{kW}\end{array}$

Thus, the exergy destruction $\left({\dot{X}}_{\text{dest}}\right)$ during the mixing process is $\underset{\_}{105.05\text{kW}}$.

c)

To determine

The second law efficiency $\left(n_{{\rm I}{\rm I}}\right)$ of the mixing chamber

Answer to Problem 85P

The second law efficiency $\left(n_{{\rm I}{\rm I}}\right)$ of the mixing chamber is $\underset{\_}{15.8\%}$.

Explanation of Solution

Write the expression for the second law efficiency $\left(n_{{\rm I}{\rm I}}\right)$ of the mixing chamber.

$\begin{array}{c}{n}_{{\rm I}{\rm I}}=\frac{{\dot{X}}_{\text{recoverd}}}{{\dot{X}}_{\text{expended}}}\\ =\frac{{\dot{m}}_1\left({\psi }_3-{\psi }_1\right)}{{\dot{m}}_2\left({\psi }_2-{\psi }_3\right)}\end{array}$ (VI)

Here, rate of exergy recovered is ${\dot{X}}_{\text{recovered}}$ and rate of exergy expended is ${\dot{X}}_{\text{expended}}$.

Conclusion:

Substitute $\text{4}\text{.6}\text{kg}/\text{s}$ for ${\dot{m}}_1$, $\text{0}$ for ${\psi }_1$, $\text{0}\text{.19}\text{kg}/\text{s}$ for ${\dot{m}}_2$, $\text{660}\text{.7}\text{kJ}/\text{kg}$ for ${\psi }_2$ and $\text{4}\text{.276}\text{kJ}/\text{kg}$ for ${\psi }_3$ in Equation (VI).

$\begin{array}{c}{n}_{{\rm I}{\rm I}}=\frac{\text{4}\text{.6}\text{kg}/\text{s}\left(\text{4}\text{.276}\text{kJ}/\text{kg}-0\right)}{\text{0}\text{.19}\text{kg}/\text{s}\left(\text{660}\text{.7}\text{kJ}/\text{kg}-\text{4}\text{.276}\text{kJ}/\text{kg}\right)}\\ =\frac{19.66\text{kJ}/\text{s}}{\text{0}\text{.19}\text{kg}/\text{s}\left(\text{656}\text{.424}\text{kJ}/\text{kg}\right)}\\ =\frac{19.66\text{kJ}/\text{s}}{\text{124}\text{.720}\text{kJ}/\text{s}}\\ =0.158\end{array}$

$=15.8\%$

Thus, the second law efficiency $\left(n_{{\rm I}{\rm I}}\right)$ of the mixing chamber is $\underset{\_}{15.8\%}$.