Chapter 5.5, Problem 99P

(a)

To determine

The power rating of the electric heater.

Answer to Problem 99P

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

Explanation of Solution

Write the formula to calculate the volume of room $\left(V\right)$.

$V=lbh$ (I)

Here, length of room is $l$, width of the room is $b$ and height of the room is $h$.

Write the ideal gas equation to calculate the total mass of air in the room $\left(m\right)$.

$m=\frac{P_{1}V}{R{T}_1}$ (II)

Here, initial pressure of air is $P_1$, gas constant of air is $R$ and initial temperature of air is $T_1$.

Assume 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)

Here, power rating of the electric heater or electrical work input is ${\dot{W}}_{\text{e,in}}$, electrical work input of fan is ${\dot{W}}_{\text{fan,in}}$, heat dissipated out of the room is ${\dot{Q}}_{\text{out}}$, internal energy change is $\Delta U$, time required by air to evacuate the room is $\Delta t$, constant volume specific heat of air at room temperature is $c_v$ and exit temperature of air is $T_2$.

Conclusion:

Substitute $\text{4}\text{m}$ for $l$, $\text{5}\text{m}$ for $b$, and $\text{6}\text{m}$ for $h$ in Equation (I).

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

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

Substitute $\text{98}\text{kPa}$ for $P_1$, $120{\text{m}}^{\text{3}}$ for $V$, $\text{0}\text{.287}\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$ and $\text{15°C}$ for $T_1$ in Equation (II).

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

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

Substitute $\text{150}\text{kJ}/\text{min}$ for ${\dot{Q}}_{\text{out}}$, $\text{0}\text{.2}\text{kJ}/\text{s}$ for ${\dot{W}}_{\text{fan,in}}$, $\text{142}\text{.3}\text{kg}$ for $m$, $\text{0}\text{.718}\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_v$, $\text{25°C}$ for $T_2$, $\text{15°C}$ for $T_1$ and $20\min$ for $\Delta t$ in 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.

Answer to Problem 99P

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

Explanation of Solution

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:

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

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

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 $c_p$.

Conclusion:

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

Substitute $\text{3}\text{.151}\text{kW}$ for ${\dot{W}}_{\text{e,in}}$, $\text{0}\text{.2}\text{kJ}/\text{s}$ for ${\dot{W}}_{\text{fan,in}}$, $\text{1}\text{.005}\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$ and $\text{40}\text{kJ}/\text{min}$ for $\dot{m}$ in Equation (IV).

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

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

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