Chapter 5, Problem 111RQ

To determine

The average temperature of air in 45 min.

Explanation of Solution

Given:

The size of the room $\left(V\right)$ is (3-m × 4–m × 6-m =$72{\text{m}}^3$).

The air pressure in the room remains constant $\left(p_1\right)$ at 100 kPa.

The initial temperature of the room $\left(T_1\right)$ is 7°C.

The volume of the radiator $\left(V_1\right)$ is 15 L.

The pressure of superheated vapor in radiator ${\left(P_1\right)}_{rad}$  is 200 kPa

The temperature of superheated vapor ${\left(T_1\right)}_{rad}$  is 200°C.

The work input for the fan $\left({\dot{W}}_{\text{fan,in}}\right)$ is 120-W.

The final pressure of the steam $\left(P_2\right)$  is 100 kPa

The change in time $\left(\Delta t\right)$ is 45 min.

Calculation:

Draw the free body diagram of the well-insulated room as shown in figure (1).

Refer the Table (A-4 through A-6), obtain the value of specific volume, specific internal energy at the initial and the final states.

At initial pressure and temperature of $200\text{kPa}$ and $200°\text{C}$.

  The initial specific internal energy $\left(u_1\right)$ is $2654.6\text{kJ}/\text{kg}$.

  The initial specific volume $\left(v_1\right)$ is $1.08049{\text{m}}^{\text{3}}/\text{kg}$,

At final pressure of $100\text{kPa}$.

The specific volume of saturated liquid $\left(v_f\right)$ is $0.01043$.

The specific internal energy of saturated liquid $\left(u_f\right)$ is $417.40$.

The specific volume of saturated vapour $\left(v_g\right)$ is $1.6941{\text{m}}^{\text{3}}/\text{kg}$

The specific internal energy of vapour mixture $\left(u_{fg}\right)$ is $2088.2\text{kJ}/\text{kg}$.

Calculate the final quality at the final state.

  $\begin{array}{c}{x}_2=\frac{v_1-v_f}{v_{fg}}\\ x_2=\frac{\left(1.08049{\text{m}}^{\text{3}}/\text{kg}\right)-\left(0.01043{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(1.693057{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =0.6376\end{array}$

Calculate the final specific internal energy of a well-insulated room.

  $\begin{array}{c}{u}_2=u_f+x_2{u}_{fg}\\ u_2=417.40+0.6376\left(2088\text{kJ}/\text{kg}\right)\\ =1748.7\text{kJ}/\text{kg}\end{array}$

Write the expression for total mass of a water in the radiator.

  $m=\frac{V_1}{v_1}$

  $\begin{array}{c}m=\frac{0.015{\text{m}}^{\text{3}}}{1.08049{\text{m}}^{\text{3}}/\text{kg}}\\ =0.01388\text{kg}\end{array}$

Write expression for the volume of the well-insulated room.

  $V=lbh$        (VI)

  $\begin{array}{c}V=3\text{m}\times \text{4}\text{m}\times \text{6}\text{m}\\ \text{=72}{\text{m}}^3\end{array}$

Write the expression of mass of air in a well-insulated room.

  $m_{\text{air}}=\frac{P_1{V}_1}{R{T}_1}$

  $\begin{array}{c}{m}_{\text{air}}=\frac{\left(100\text{kPa}\right)\left(72{\text{m}}^3\right)}{\left(0.2870\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(200\text{K}\right)}\\ =89.60\text{kg}\end{array}$

Write the expression for amount of fan work done in 45 min.

  $W_{\text{fan,in}}={\dot{W}}_{\text{fan,in}}\Delta t$        (VIII)

  $\begin{array}{c}{W}_{\text{fan,in}}=\left(0.120\text{kJ}/\text{s}\right)\left(40\min \right)\\ =\left(0.120\text{kJ}/\text{s}\right)\left(40\min \times \left(\frac{60\text{s}}{1\min }\right)\right)\\ =324\text{kJ}\end{array}$

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 $\Delta E_{\text{system}}$.

Substitute $0$ for $E_{\text{in}}$, $Q_{\text{out}}$ for $E_{\text{out}}$, and $m\left(u_1-u_2\right)$ for $\Delta E_{\text{system}}$ in Equation (I).

  $\begin{array}{c}-Q_{\text{out}}=\Delta U=m\left(u_2-u_1\right)\\ =m\left(u_1-u_2\right)\end{array}$

  $\begin{array}{c}{Q}_{\text{out}}=\left(0.01388\text{kg}\right)\left(2654.6-1748.7\right)\text{kJ}/\text{kg}\\ =12.58\text{kJ}\end{array}$

For well-insulated room the energy balance equation.\

Substitute $Q_{\text{in}}+W_{\text{fan,in}}$ for $E_{\text{in}}$, $W_{\text{b,out}}$ for $E_{\text{out}}$, and $\Delta U$ for $\Delta E_{\text{system}}$ in Equation (I).

  $\begin{array}{l}{Q}_{\text{in}}+W_{\text{fan,in}}-W_{\text{b,out}}=\Delta U\\ Q_{\text{in}}+W_{\text{fan,in}}=\Delta H\cong m{c}_P\left(T_2-T_1\right)\end{array}$        (II)

Express the change in internal energy and boundary work into the constant pressure expansion and compression in Equation (IX).

  $\left({\dot{Q}}_{\text{in}}+{\dot{W}}_{\text{fan,in}}\right)\Delta t=m{c}_{P,\text{avg}}\left(T_2-T_1\right)$        (III)

Here, the rate of heat transfer entering is ${\dot{Q}}_{\text{in}}$, the rate of work of fan entering is ${\dot{W}}_{\text{fan,in}}$, change in temperature is $\Delta t$, the mass is $m$, the constant average pressure is $c_{P,\text{avg}}$, the initial temperature is $T_1$, and the final temperature is $T_2$.

Substitute ${\dot{Q}}_{\text{in}}=12.58\text{kJ}$, ${\dot{W}}_{\text{fan,in}}=324\text{kJ}$, $m=89.60\text{kg}$, $c_{P,\text{avg}}=1.005\text{kJ}/\text{kg°C}$, and $T_1=7°\text{C}$ in Equation (III).

  $\begin{array}{l}\left(12.58\text{kJ}\right)+\left(324\text{kJ}\right)=\left(89.60\text{kg}\right)\left(1.005\text{kJ}/\text{kg°C}\right)\left(T_2-\text{7°C}\right)\\ \left(336.58\text{kJ}\right)\text{=}\left(90.048\text{kJ°C}\right)\left(T_2-\text{7°C}\right)\\ T_2=10.7°\text{C}\end{array}$

Therefore, the well-insulated room temperature rises from $\text{7°C}\text{to 10}\text{.7}°\text{C}$ in 45 min time.

Thus, the average temperature of air in 45 min is $\underset{\_}{10.7°\text{C}}$.