Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Explanation of Solution

Given:

The volume of the saturated mixture $\left(V_1\right)$ is $0.01{\text{m}}^3$.

The quality of the mixture is $0.2$.

The initial temperature of the mixture $\left(T_1\right)$ is $120°\text{C}$.

The heat of fusion is $333.7\text{ kJ/kg}$.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as $2.11\text{kJ}/\text{kg}°\text{C}$ and $4.18\text{kJ}/\text{kg}°\text{C}$.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as $120°\text{C}$.

  The specific volume of ice for insulated cylinder device $\left(v_f\right)$ is $0.001060$

  The specific volume of ice for insulated cylinder device $\left(v_g\right)$ is $0.89133{\text{m}}^{\text{3}}/\text{kg}$

  The specific enthalpy of ice $\left(h_f\right)$ is $503.81\text{kJ}/\text{kg}$.

  The specific enthalpy of the mixture $\left(h_{fg}\right)$ is $2202.1\text{kJ}/\text{kg}$

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

  $\begin{array}{c}{v}_1=v_f+x{v}_{fg}\\ =v_f+x\left(v_g-v_f\right)\end{array}$

  $\begin{array}{c}{v}_1=\left(0.001060\right)+\left(0.2\right)\times \left(0.89133{\text{m}}^{\text{3}}/\text{kg}-0.001060\right)\\ =\left(0.001060\right)+\left(0.2\right)\times \left(0.89027{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.179114{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

  $h_1=h_f+x{h}_{fg}$

  $\begin{array}{c}{h}_1=\left(503.81\right)+\left(0.2\right)\times \left(2202.1\text{kJ}/\text{kg}\right)\\ =944.23\text{kJ}/\text{kg}\end{array}$

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as $503.81\text{kJ}/\text{kg}$.

Determine the initial specific volume of the refrigerant.

  $\begin{array}{l}{v}_1=\frac{V_1}{m}\\ m=\frac{V_1}{v_1}\end{array}$

  $\begin{array}{c}m=\frac{0.01{\text{m}}^{\text{3}}}{0.179114{\text{m}}^{\text{3}}/\text{kg}}\\ =0.05583\text{kg}\end{array}$

Write the expression for the energy balance equation.

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

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

  $\begin{array}{l}{W}_{\text{b,in}}=\Delta U\\ W_{\text{b,in}}-\Delta U=0\end{array}$        (II)

Here, the boundary work to be done into the system is $W_{\text{b,in}}$ and the change in the internal energy is $\Delta U$.

Substitute $\Delta H$ for $W_{\text{b,in}}-\Delta U$ in Equation (II).

  $\begin{array}{l}\Delta H=0\\ \Delta H_{\text{ice}}+\Delta H_{\text{water}}=0\\ {\left[mc{\left(0°\text{C}-T_1\right)}_{\text{solid}}+m{h}_{if}+mc{\left(T_2-0°\text{C}\right)}_{\text{liquid}}\right]}_{\text{ice}}+{\left[m\left(h_2-h_1\right)\right]}_{\text{water}}=0\end{array}$        (III)

Here, the mass of the ice is $m$, the heat of fusion of ice is $h_{if}$, the specific heat of ice at liquid state is $c$, the initial temperature of ice is $T_1$, the final temperature of ice is $T_2$, the initial specific enthalpy of ice is $h_1$, and the final specific enthalpy of ice is $h_2$.

Substitute $0°\text{C}$ for $T_{\text{1,solid}}$, $2.11\text{kJ}/\text{kg}°\text{C}$ for $c_{\text{solid}}$, $333.7\text{kJ}/\text{kg}$ for $h_{if}$, $4.18\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_{\text{liquid}}$, $120\text{°C}$ for $T_2{}_{\text{,liquid}}$, $0.05583\text{kg}$ for $m_{\text{water}}$, $503.81\text{kJ}/\text{kg}$ for $h_2$, and $944.24\text{kJ}/\text{kg}$ for $h_1$ in Equation (III).

  $\begin{array}{l}\left[\begin{array}{l}{\left[\begin{array}{l}m\left(2.11\text{kJ}/\text{kg}°\text{C}\right){\left(0°\text{C}-0°\text{C}\right)}_{\text{solid}}+m\left(333.7\text{kJ}/\text{kg}\right)\\ +m\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right){\left(\left(120\text{°C}\right)-0°\text{C}\right)}_{\text{liquid}}\end{array}\right]}_{\text{ice}}\\ +{\left[\left(0.05583\text{kg}\right)\left(\left(503.81\text{kJ}/\text{kg}\right)-\left(944.24\text{kJ}/\text{kg}\right)\right)\right]}_{\text{water}}\end{array}\right]=0\\ m{\left[0+\left(333.7\text{kJ}/\text{kg}\right)+\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(120\text{°C}\right)\right]}_{\text{ice}}+\left[\left(0.05583\text{kg}\right)\left(-440.43\text{kJ}/\text{kg}\right)\right]=0\\ m=0.0294\text{kg}\times \left(\frac{1000\text{g}}{1\text{kg}}\right)\\ m=29.4\text{g}\end{array}$

Thus, the amount of ice that needs to be added in an insulated cylinder is $\underset{\_}{29.4\text{g}}$.