Chapter 4.5, Problem 130RP

A well-insulated 3-m × 4–m × 6-m room initially at 7°C is heated by the radiator of a steam heating system. The radiator has a volume of 15 L and is filled with superheated vapor at 200 kPa and 200°C. At this moment both the inlet and the exit valves to the radiator are closed. A 120-W fan is used to distribute the air in the room. The pressure of the steam is observed to drop to 100 kPa after 45 min as a result of heat transfer to the room. Assuming constant specific heats for air at room temperature, determine the average temperature of air in 45 min. Assume the air pressure in the room remains constant at 100 kPa.

FIGURE P4–136

To determine

The average temperature of air in 45 min.

Answer to Problem 130RP

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

Explanation of Solution

Show the free body diagram of the well-insulated room.

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).

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

Here, the mass is $m$, the initial specific internal energy is $u_1$, and the final specific internal energy is $u_2$.

From 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 $200°\text{C}$.

$\begin{array}{l}{v}_1=1.08049{\text{m}}^{\text{3}}/\text{kg}\\ u_1=2654.6\text{kJ}/\text{kg}\end{array}$

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

$\begin{array}{l}{v}_f=0.01043\\ u_f=417.40\\ v_g=1.6941{\text{m}}^{\text{3}}/\text{kg}\\ u_{fg}=2088.2\text{kJ}/\text{kg}\end{array}$

Write the expression for final quality at the final state.

$x_2=\frac{v_1-v_f}{v_{fg}}$ (III)

Here, the specific volume of saturated liquid is $v_f$, the specific volume of saturated vapour is $v_g$, and the specific volume change upon vaporization is $v_{fg}$.

Write the expression for final specific internal energy of a well-insulated room.

$u_2=u_f+x_2{u}_{fg}$ (IV)

Here, the specific internal energy of saturated liquid is $u_f$ and the specific internal energy change upon vaporization is $u_{fg}$.

Write the expression for total mass of a well-insulated room.

$m=\frac{V_1}{v_1}$ (V)

Here, the initial specific volume is $v_1$ and the volume of tank is $V_1$.

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

$V=lbh$ (VI)

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

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

Here, the initial pressure is $P_1$, the initial volume is $V_1$, the initial temperature is $T_1$, and gas constant is $R$.

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

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

Conclusion:

Here, the specific volume $\left(v_2=v_1\right)$.

Substitute $1.08049{\text{m}}^{\text{3}}/\text{kg}$ for $v_2$, $0.01043{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $1.693057{\text{m}}^{\text{3}}/\text{kg}$ for $v_{fg}$ in Equation (III).

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

Substitute $417.40$ for $u_f$, $0.6376$ for $x_2$, and $2088\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (IV).

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

Substitute $0.015{\text{m}}^{\text{3}}$ for $V_1$ and $1.08049{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (V).

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

Substitute $1748.7\text{kJ}/\text{kg}$ for $u_2$, $2654.6\text{kJ}/\text{kg}$ for $u_1$, and $0.01388\text{kg}$ for $m$ in Equation (II).

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

Substitute $3\text{m}$ for $l$, $4\text{m}$ for $b$ and $6\text{m}$ for $h$ in Equation (VI).

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

Substitute $100\text{kPa}$ for $P_1$, $72{\text{m}}^3$ for $V_1$, $200\text{K}$ for $T_1$, and $0.2870\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $l$ in Equation (VII).

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

Substitute $0.120\text{kJ}/\text{s}$ for ${\dot{W}}_{\text{fan,in}}$ and $45\min$ for $\Delta t$ in Equation (VIII).

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

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).

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

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

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

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

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

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 $\underset{\_}{10.7°\text{C}}$.