Chapter 6, Problem 90P

(a)

To determine

The power rating of the electric heater.

Explanation of Solution

Given:

The length of room $\left(l\right)$ is $\text{4}\text{m}$.

The width of the room $\left(b\right)$ is $\text{5}\text{m}$.

The height of the room $\left(h\right)$ is $\text{6}\text{m}$.

The initial pressure of air $\left(P_1\right)$ is $\text{98}\text{kPa}$.

The initial temperature of air $\left(T_1\right)$ is $\text{15°C}$.

The electrical work input of fan $\left({\dot{W}}_{\text{fan,in}}\right)$ is $\text{200 W }\left(\text{or}\right)\text{ 0}\text{.2}\text{kJ}/\text{s}$.

The heat dissipated out of the room $\left({\dot{Q}}_{\text{out}}\right)$ is $\text{150}\text{kJ}/\text{min}$.

The mass flow rate of water $\left(\dot{m}\right)$ is $40\text{ kg/min}$.

The time required by air to evacuate the room $\left(\Delta t\right)$ is $20\min$.

The exit temperature of air $\left(T_2\right)$ is $\text{25°C}$.

Calculation:

Refer Table A-1, “Gas constant of common gases”, obtain the gas constant of air as $\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$.

Calculate the volume of room $\left(V\right)$.

  $V=lbh$

  $\begin{array}{c}V=4\text{m}\times 5\text{m}\times 6\text{m}\\ =120{\text{m}}^{\text{3}}\end{array}$

Calculate the total mass of air in the room $\left(m\right)$ using the ideal gas equation.

  $m=\frac{P_{1}V}{R{T}_1}$

  $\begin{array}{c}m=\frac{\left(\text{98}\text{kPa}\right)\left(120{\text{m}}^{\text{3}}\right)}{\left(\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{15°C}\right)}\\ =\frac{11,760\text{kPa}\cdot {\text{m}}^{\text{3}}}{\left(\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{15}+273\right)\text{K}}\\ =142.3\text{kg}\end{array}$

Consider the entire room as the steady-flow system that is a control volume as mass traverses the boundary.

Write the energy balance for system in the rate form.

  ${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=\Delta {\dot{E}}_{\text{system}}$

Here, rate of net energy transfer into the control volume is ${\dot{E}}_{\text{in}}$, rate of net energy transfer’s exit from the control volume is ${\dot{E}}_{\text{out}}$ and rate of change in internal energy of system is $\Delta {\dot{E}}_{\text{system}}$.

At steady state, rate of change in internal energy of the system is zero. Thus rewrite the energy balance equation for the system.

  $\begin{array}{l}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\dot{W}}_{\text{e,in}}+{\dot{W}}_{\text{fan,in}}-{\dot{Q}}_{\text{out}}=\Delta U\\ \Delta t\left({\dot{W}}_{\text{e,in}}+{\dot{W}}_{\text{fan,in}}-{\dot{Q}}_{\text{out}}\right)=m{c}_{v,\text{avg}}\left(T_2-T_1\right)\end{array}$

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

Refer Table A-2, “Ideal – gas specific heats of common gases”, obtain the constant volume specific heat of air as $\text{0}\text{.718}\text{kJ}/\text{kg}\cdot °\text{C}$.

Calculate the power rating of the electric heater $\left({\dot{W}}_{\text{e,in}}\right)$ using the equation (III).

  $\begin{array}{c}{\dot{W}}_{\text{e,in}}=\left(\text{150}\text{kJ}/\text{min}\right)-\text{0}\text{.2}\text{kJ}/\text{s}+\frac{\left(\text{142}\text{.3}\text{kg}\right)\left(\text{0}\text{.718}\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{25°C}-1\text{5°C}\right)}{\left(20\min \right)}\\ =\left[\begin{array}{c}\left(\text{150}\text{kJ}/\text{min}\right)\left(\frac{1\text{kJ}/\text{min}}{60\text{kJ}/\text{s}}\right)-\text{0}\text{.2}\text{kJ}/\text{s}\\ +\frac{\left(\text{142}\text{.3}\text{kg}\right)\left(\text{0}\text{.718}\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(\text{25°C}-1\text{5°C}\right)}{\left(20\min \right)\left(\frac{60\text{s}}{1\min }\right)}\end{array}\right]\\ =3.151\text{kJ}/\text{s}\end{array}$

  $\begin{array}{c}=3.151\text{kJ}/\text{s}\left(\frac{\text{1}\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =3.151\text{kW}\end{array}$

Thus, the power rating of the electric heater is $\text{3}\text{.151}\text{kW}$.

(b)

To determine

The temperature rise of air as it passes through the heating duct.

Explanation of Solution

Refer Table A-2,“Ideal – gas specific heats of common gases”, obtain the constant pressure specific heat of air as $\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}$.

Write the mass balance equation for the flow of air.

  $\begin{array}{c}{\dot{m}}_2={\dot{m}}_1\\ =\dot{m}\end{array}$

Here, mass flow rate of air at the inlet is ${\dot{m}}_1$, and mass flow rate of air at the outlet is ${\dot{m}}_2$.

Assume the heating duct as the steady-flow system that controls the volume as mass traverses the boundary.

Write the energy balance for system in the rate form as follows:

  ${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$

Re-write the energy balance equation for the system as follows:

  $\begin{array}{l}{\dot{E}}_{\text{in}}={\dot{E}}_{\text{out}}\\ {\dot{W}}_{\text{e,in}}+{\dot{W}}_{\text{fan,in}}+\dot{m}{h}_1=\dot{m}{h}_2\\ {\dot{W}}_{\text{e,in}}+{\dot{W}}_{\text{fan,in}}=\dot{m}\left(h_2-h_1\right)\end{array}$

  $\begin{array}{l}{\dot{W}}_{\text{e,in}}+{\dot{W}}_{\text{fan,in}}=\dot{m}{c}_p\left(T_2-T_1\right)\\ T_2-T_1=\frac{{\dot{W}}_{\text{e,in}}+{\dot{W}}_{\text{fan,in}}}{\dot{m}{c}_p}\end{array}$

Here, initial specific enthalpy of air is $h_1$ and final specific enthalpy of air $h_2$ and specific heat at constant pressure for air at room temperature is $c_p$.

  $\begin{array}{c}{T}_2-T_1=\frac{\text{3}\text{.151}\text{kW}+\left(\text{0}\text{.2}\text{kJ}/\text{s}\right)}{\left(\text{40}\text{kJ}/\min \right)\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =\frac{\text{3}\text{.151}\text{kW}\left(\frac{\text{1}\text{kJ}/\text{s}}{\text{1}\text{kW}}\right)+\left(\text{0}\text{.2}\text{kJ}/\text{s}\right)}{\left[\left(\frac{\text{40}\text{kJ}/\min }{60\text{kJ}/\text{s}}\right)\left(1\text{kg}/\text{s}\right)\right]\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =\frac{\text{3}\text{.151}\text{kJ}/\text{s}+\text{0}\text{.2}\text{kJ}/\text{s}}{0.666\text{kg}/\text{s}\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}\right)}\end{array}$

  $\begin{array}{c}=\frac{\text{3}\text{.351}\text{kJ}/\text{s}}{0.666\text{kg}/\text{s}\left(\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ =5°\text{C}\end{array}$

Thus, the temperature rise of air as it passes through the heating duct is $\text{5°C}$.