Chapter 8.8, Problem 120RP

Repeat Prob. 8–114 if heat were supplied to the pressure cooker from a heat source at 180°C instead of the electrical heating unit?

8–114 A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid water and the other half by water vapor. The cooker is now placed on top of a 750-W electrical heating unit that is kept on for 20 min. Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the amount of water that remained in the cooker and (b) the exergy destruction associated with the entire process.

FIGURE P8–114

To determine

The final mass of water that remained in the cooker.

Answer to Problem 120RP

The final mass of water that remained in the cooker is $1.507\text{kg}$.

Explanation of Solution

Express the mass balance for a pressure cooker which acts as a system.

$\begin{array}{l}{m}_{\text{in}}-m_{\text{out}}=\Delta m_{\text{system}}\\ m_e=m_1-m_2\end{array}$ (I)

Here, initial mass is $m_{\text{in}}\left(m_1\right)$, difference in the mass of a states is $m_e$, change in mass of a system is $\Delta m_{\text{system}}$, and final mass is $m_{\text{out}}\left(m_2\right)$.

Write an energy balance for a system.

$\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ W_{e,\text{in}}+m_e{h}_e=m_2{u}_2-m_1{u}_1\end{array}$ (II)

Here, internal energy at state 1 and 2 is $u_1\text{and}{u}_2$, enthalpy at state e is $h_e$, energy transfer at inlet and outlet condition is $E_{\text{in}}\text{and}{E}_{\text{out}}$ respectively, and change in energy transfer of a system is $\Delta E_{\text{system}}$, and amount of electrical energy supplied during the process is $W_{e,\text{in}}$.

Calculate the initial mass in the tank $\left(m_1\right)$.

$\begin{array}{c}{m}_1=m_f+m_g\\ =\frac{V_f}{v_f}+\frac{V_g}{v_g}\end{array}$ (III)

Here, saturated liquid specific volume is $v_f$, saturated vapor specific volume is $v_g$, volume at saturated liquid and vapor is $V_f\text{and}{V}_g$ respectively.

Calculate the initial internal energy $\left(U_1\right)$ in the tank.

$\begin{array}{c}{U}_1=m_1{u}_1\\ =m_f{u}_f+m_g{u}_g\end{array}$ (IV)

Here, specific internal energy of saturated fluid is $u_f$ and the specific internal energy of saturated vapor is $u_g$.

Calculate the initial entropy $\left(S_1\right)$ in the tank.

$\begin{array}{c}{S}_1=m_1{s}_1\\ =m_f{s}_f+m_g{s}_g\end{array}$ (V)

Here, specific entropy of saturated fluid is $s_f$ and the specific entropy of saturated vapor is $s_g$.

Write the internal energy at state 2$\left(u_2\right)$.

$u_2=u_f+x_2\left(u_g-u_f\right)$ (VI)

Here, dryness fraction at state 2 is $x_2$.

Write the specific volume at state 2$\left(v_2\right)$.

$v_2=v_f+x_2\left(v_g-v_f\right)$ (VII)

Write the formula to calculate the mass of the water remained in tank $\left(m_2\right)$.

$m_2=\frac{V}{v_2}$ (VIII)

Here, volume of the cooker is $V$ and the final specific volume of water is $v_2$.

Conclusion:

From the Table A-5 of “Saturated water: Pressure”, obtain the saturated liquid specific volume $\left(v_f\right)$, saturated vapor specific volume $\left(v_g\right)$, saturated liquid enthalpy $\left(u_f\right)$, saturated vapor enthalpy $\left(u_g\right)$, saturated liquid entropy $\left(s_f\right)$, and saturated vapor entropy $\left(s_g\right)$ at initial pressure $\left(P_1=175\text{kPa}\right)$ as

$\begin{array}{l}{v}_f=0.001057{\text{m}}^{\text{3}}/\text{kg}\\ v_g=1.0037{\text{m}}^{\text{3}}/\text{kg}\\ u_f=486.82\text{kJ}/\text{kg}\\ u_g=2524.5\text{kJ}/\text{kg}\end{array}$

$\begin{array}{l}{s}_f=1.4850\text{kJ}/\text{kg}\cdot \text{K}\\ {\text{s}}_g=7.1716\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From the Table A-5 of “Saturated water: Pressure”, obtain the enthalpy $\left(h_e\right)$ and entropy $\left(s_e\right)$ at the microscopic energy of flowing at pressure $\left(P_e=175\text{kPa}\right)$ as,

$\begin{array}{l}{h}_e=h_g\\ h_e=2700.2\text{kJ}/\text{kg}\\ s_e=s_g\\ s_e=7.1716\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Since one half of the volume is filled with liquid water and other half by water vapor, calculate the volume of tank $\left(V_f\right)$.

$\begin{array}{c}{V}_f=V_g\\ =2\text{L}\\ =2\text{L}\times \frac{\text{0}\text{.001}{\text{m}}^{\text{3}}}{\text{L}}\\ =0.002{\text{m}}^{\text{3}}\end{array}$

Substitute $0.002{\text{m}}^{\text{3}}$ for $V_f$, $0.002{\text{m}}^{\text{3}}$ for $V_g$, $0.001057{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, and $1.0037{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ in Equation (III).

$\begin{array}{c}{m}_1=\frac{0.002{\text{m}}^{\text{3}}}{0.001057{\text{m}}^{\text{3}}/\text{kg}}+\frac{0.002{\text{m}}^{\text{3}}}{1.0037{\text{m}}^{\text{3}}/\text{kg}}\\ =1.893+0.002\\ =1.8945\text{kg}\end{array}$

Substitute 1.893 kg for $m_f$, $486.82\text{kJ}/\text{kg}$ for $u_f$, 0.002 kg for $m_g$, and $2524.5\text{kJ}/\text{kg}$ for $u_g$ in Equation (IV).

$\begin{array}{c}{U}_1=\left(\text{1}\text{.893}\text{kg}\right)\left(486.82\text{kJ}/\text{kg}\right)+\left(0.002\text{kg}\right)\left(2524.5\text{kJ}/\text{kg}\right)\\ =926.6\text{kJ}\end{array}$

Substitute 1.893 kg for $m_f$, $1.4850\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$, 0.002 kg for $m_g$, and $7.1716\text{kJ}/\text{kg}\cdot \text{K}$ for $s_g$ in Equation (V).

$\begin{array}{c}{S}_1=\left(\text{1}\text{.893}\text{kg}\right)\left(1.4850\text{kJ}/\text{kg}\cdot \text{K}\right)+\left(0.002\text{kg}\right)\left(7.1716\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =2.8239\text{kJ}/\text{K}\end{array}$

Substitute $486.82\text{kJ}/\text{kg}$ for $u_f$ and $2524.5\text{kJ}/\text{kg}$ for $u_g$ in Equation (VI).

$u_2=486.82\text{kJ}/\text{kg}+x_2\left(2524.5\text{kJ}/\text{kg}-486.82\text{kJ}/\text{kg}\right)$

$u_2=486.82+x_2\left(2037.68\right)$

Substitute $0.001057{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$ and $1.0037{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ Equation (VII).

$v_2=0.001057{\text{m}}^{\text{3}}/\text{kg}+x_2\left(1.0037{\text{m}}^{\text{3}}/\text{kg}-0.001057{\text{m}}^{\text{3}}/\text{kg}\right)$

$v_2=0.001057+x_2\left(1.002643\right)$ (IX)

Re-write the Equation (II) using the Equation (I),

$W_{e,\text{in}}=\left(m_1-m_2\right)h_e+m_2{u}_2-m_1{u}_1$ (X)

Substitute $900\text{kJ}$ for $W_{e,\text{in}}$, $1.8945\text{kg}$ for $m_1$, $2700.2\text{kJ}/\text{kg}$ for $h_e$, $\left[486.82+x_2\left(2037.68\right)\right]$ for $u_2$, $\frac{0.004}{v_2}$ for $m_2$, and $926.6\text{kJ}$ for $m_1{u}_1$ in Equation (X).

$900\text{kJ}=\left\{\begin{array}{l}\left(1.8945\text{kg}-\frac{0.004}{v_2}\right)2700.2\text{kJ}/\text{kg}\\ +\left(\frac{0.004}{v_2}\right)\left(\left[486.82+x_2\left(2037.68\right)\right]\right)-926.6\text{kJ}\end{array}\right\}$

Substitute $\left[0.001057+x_2\left(1.002643\right)\right]$ for $v_2$.

$\begin{array}{l}900\text{kJ}=\left\{\begin{array}{l}\left(1.8945\text{kg}-\frac{0.004}{\left[0.001057+x_2\left(1.002643\right)\right]}\right)2700.2\text{kJ}/\text{kg}\\ +\frac{0.004}{0.001057+x_2\left(1.002643\right)}\left[486.82+x_2\left(2037.68\right)\right]\\ -926.6\text{kJ}\end{array}\right\}\\ x_2=0.001918\end{array}$

Substitute 0.001918 for $x_2$ in Equation (IX).

$\begin{array}{c}{v}_2=0.001057+0.001918\left(1.002643\right)\\ =0.002654{\text{m}}^3/\text{kg}\end{array}$

Substitute $0.004{\text{m}}^3$ for $V$ and $0.002654{\text{m}}^3/\text{kg}$ for $v_2$

$\begin{array}{c}{m}_2=\frac{0.004{\text{m}}^3}{0.002654{\text{m}}^3/\text{kg}}\\ =1.507\text{kg}\end{array}$

Thus, the final mass of water that remained in the cooker is $1.507\text{kg}$.

To determine

The amount of exergy destructed during the heat supplied to pressure cooker from heat source.

Answer to Problem 120RP

The amount of exergy destructed during the heat supplied to pressure cooker from heat source is $96.8\text{kJ}$.

Explanation of Solution

For an extended system, write the simplification rate form of the entropy balance.

$\begin{array}{l}{S}_{\text{in}}-S_{\text{out}}+S_{\text{gen}}=\Delta S_{\text{system}}\\ \frac{Q_{\text{in}}}{T_{\text{b,in}}}-m_e{s}_e+S_{\text{gen}}=m_2{s}_2-m_1{s}_1\end{array}$

$S_{\text{gen}}=m_e{s}_e+m_2{s}_2-m_1{s}_1-\frac{Q_{\text{in}}}{T_{\text{b,in}}}$ (XI)

Here, entropy generation is $S_{\text{gen}}$, change of entropy is $\Delta S_{\text{system}}$, entropy at inlet condition is $S_{\text{in}}$, entropy at outlet condition is $S_{\text{out}}$, heat input is $Q_{\text{in}}$, and the surrounding’s temperature is $T_{\text{b,in}}$.

Calculate the exergy destroyed during the process $\left(X_{\text{destroyed}}\right)$.

$X_{\text{destroyed}}=T_0{S}_{\text{gen}}$ (XII)

Here, dead state temperature is $T_0$.

Substitute Equation (XI) in Equation (XII).

$X_{\text{destroyed}}=T_0\left(m_e{s}_e+m_2{s}_2-m_1{s}_1-\frac{Q_{\text{in}}}{T_{\text{b,in}}}\right)$ (XIII)

Calculate the entropy at state 2$\left(s_2\right)$.

$s_2=s_f+x_2\left(s_g-s_f\right)$ (XIV)

Conclusion:

Substitute 0.001918 for $x_2$, $1.4850\text{kJ}/\text{kg}\cdot \text{K}$ for $s_f$ and $7.1716\text{kJ}/\text{kg}\cdot \text{K}$ for $s_g$ in Equation (XIV).

$\begin{array}{c}{s}_2=1.4850\text{kJ}/\text{kg}\cdot \text{K}+0.001918\left(7.1716\text{kJ}/\text{kg}\cdot \text{K}-1.4850\text{kJ}/\text{kg}\cdot \text{K}\right)\\ =1.5642\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Substitute 1.8945 kg for $m_1$ and 1.507 kg for $m_2$ in Equation (I).

$\begin{array}{c}{m}_e=1.8945\text{kg}-1.507\text{kg}\\ =0.3875\text{kg}\end{array}$

Substitute 298 K for $T_0$, 1.507 kg for $m_2$, $1.5642\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$,$2.8239\text{kJ}/\text{K}$ for $m_1{s}_1$, 0.3875 kg for $m_e$, $7.1716\text{kJ}/\text{kg}\cdot \text{K}$ for $s_e$, $900\text{kJ}$ for $Q_{\text{in}}$, and $180°\text{C}$ for $T_{\text{b,in}}$ in Equation (XIII)

$\begin{array}{c}{X}_{\text{destroyed}}=298\text{K}\left(\begin{array}{l}1.507\text{kg}\left(1.5642\text{kJ}/\text{kg}\cdot \text{K}\right)-\left(2.8239\text{kJ}/\text{K}\right)+\\ 0.3785\text{kg}\left(7.1716\text{kJ}/\text{kg}\cdot \text{K}\right)-\frac{900\text{ kJ}}{\left(180+273\right)\text{K}}\end{array}\right)\\ =298\left(2.3572-2.8239+2.7144-\frac{900}{453}\right)\\ =96.8\text{kJ}\end{array}$

Thus, the amount of exergy destructed during the heat supplied to pressure cooker from heat source is $96.8\text{kJ}$.