Chapter 5.5, Problem 100P

A 4-m × 5-m × 6-m room is to be heated by an electric resistance heater placed in a short duct in the room. Initially, the room is at 15°C, and the local atmospheric pressure is 98 kPa. The room is losing heat steadily to the outside at a rate of 150 kJ/min. A 200-W fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of 40 kg/min. The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 25 min for the room air to reach an average temperature of 25°C, find (a) the power rating of the electric heater and (b) the temperature rise that the air experiences each time it passes through the heater.

To determine

The power rating of the electric heater.

Answer to Problem 100P

The power rating of the electric heater is $\text{2}\text{.981}\text{kW}$.

Explanation of Solution

Consider the entire room as system and the air circulates the in the room itself. There is no leakage to the surrounding.

The air flows at steady state through one inlet and one exit system (pipe and duct flow). Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the energy balance equation.

$\begin{array}{l}{E}_{in}-E_{out}=\Delta E_{system}\\ \left\{\begin{array}{l}\left[Q_{\text{in}}+W_{\text{in}}+m_{\text{in}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{in}}\right]\\ -\left[Q_{\text{out}}+W_{\text{out}}+m_{\text{out}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{out}}\right]\end{array}\right\}={\left[\begin{array}{l}{m}_2{\left(u+\text{ke}+\text{pe}\right)}_2\\ -m_1{\left(u+\text{ke}+\text{pe}\right)}_1\end{array}\right]}_{\text{system}}\end{array}$ (I)

Here, the heat transfer is $Q$, the work transfer is $W$, the enthalpy is $h$, the internal energy is $u$, the kinetic energy is $\text{ke}$, the potential energy is $\text{pe}$ and the change in net energy of the system is $\Delta E_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

In this system two work inputs are involved namely, the work input to the electric heater $\left(W_{\text{e},.\text{in}}\right)$- used to heat the air, the work input to the fan $\left(W_{\text{fan,in}}\right)$- used to circulate the air. There is an heat loss from the room $\left({\dot{Q}}_2\right)$. Neglect work transfer at the outlet, kinetic and potential energies.

The Equations (I) reduced as follows.

$\begin{array}{l}\left\{\begin{array}{l}\left[0+\left(W_{\text{e,in}}+W_{\text{fan,in}}\right)+m\left(h_{\text{in}}+0+0\right)\right]\\ -\left[Q_{\text{out}}+0+m\left(h_{\text{out}}+0+0\right)\right]\end{array}\right\}=m{u}_2-m{u}_1\\ W_{\text{e,in}}+W_{\text{fan,in}}+m{h}_{\text{in}}-Q_{\text{out}}-m{h}_{\text{out}}=m\left(u_2-u_1\right)\end{array}$ (II)

Here, there is no mass leakage from the room to the surrounding. The mass of air circulates in the room itself. Hence, inlet and exit enthalpies are neglected.

The change in internal energy is expresses as follow.

$u_2-u_1=c_v\left(T_2-T_1\right)$

Here, the specific heat at constant volume is $c_v$, the exit temperature is $T_2$ and the inlet temperature is $T_1$.

Neglect the inlet and exit enthalpies and substitute $c_v\left(T_2-T_1\right)$ for $u_2-u_1$ in

Equation (II).

$\begin{array}{l}{W}_{\text{e,in}}+W_{\text{fan,in}}+m\left(0\right)-Q_{\text{out}}-m\left(0\right)=m{c}_v\left(T_2-T_1\right)\\ W_{\text{e,in}}+W_{\text{fan,in}}-Q_{\text{out}}=m{c}_v\left(T_2-T_1\right)\end{array}$ (III)

Express the Equation (III) with respect to change of time and rearrange it to obtain ${\dot{W}}_{\text{e,in}}$ as follows.

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

Write the formula for mass of air $\left(m\right)$ that circulates inside the room.

$\begin{array}{c}m=\frac{P_1\nu }{R{T}_1}\\ =\frac{P_1\left(4\times 5\times 6\right){\text{m}}^3}{R{T}_1}\\ =\frac{P_1\left(120{\text{m}}^3\right)}{R{T}_1}\end{array}$ (V)

The mass flow rate $\left(\dot{m}\right)$ is expressed as follows.

$\dot{m}=\frac{m}{\Delta t}$ (VI)

Here, the change in time or time interval is $\Delta t$.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant of air $\left(R\right)$ is $0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$.

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant volume $\left(c_v\right)$ of air is $0.718\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $98\text{kPa}$ for $P_1$, $0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$ and $15°\text{C}$ for $T_1$ in Equation (V).

$\begin{array}{c}m=\frac{\left(98\text{kPa}\right)\left(120{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(15°\text{C}\right)}\\ =\frac{11760\text{kPa}\cdot {\text{m}}^3}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(15+273\right)\text{K}}\\ =\frac{11760\text{kPa}\cdot {\text{m}}^3}{82.656\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =142.2764\text{kg}\end{array}$

$\simeq 142.3\text{kg}$

Substitute $142.3\text{kg}$ for $m$ and $25\text{min}$ for $\Delta t$ in Equation (VI).

$\begin{array}{c}\dot{m}=\frac{142.3\text{kg}}{25\text{min}}\\ =\frac{142.3\text{kg}}{25\text{min}\times \frac{60\text{s}}{1\text{min}}}\\ =0.09485\text{kg/s}\end{array}$

Substitute $0.09485\text{kg/s}$ for $\dot{m}$, $0.718\text{kJ/kg}\cdot \text{K}$ for $c_v$, $25°\text{C}$ for $T_2$, $15\text{°C}$ for $T_1$, $200\text{W}$ for ${\dot{W}}_{\text{fan,in}}$ and $150\text{kJ/min}$ for ${\dot{Q}}_{\text{out}}$ in Equation (IV).

$\begin{array}{c}{\dot{W}}_{\text{e,in}}=\left[\begin{array}{l}\left(0.09485\text{kg/s}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)\left(25°\text{C}-15°\text{C}\right)\\ -200\text{W}+150\text{kJ/min}\end{array}\right]\\ =\left[\begin{array}{l}\left(0.09485\text{kg/s}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)\left[\left(25+273\right)\text{K}-\left(15+273\right)\text{K}\right]\\ -\left(200\text{W}\times \frac{1\text{kJ/s}}{{10}^3\text{W}}\right)+\left(150\text{kJ/min}\times \frac{1\text{min}}{60\text{s}}\right)\end{array}\right]\\ =0.681\text{kJ/s}-0.200\text{kJ/s}+2.5\text{kJ/s}\\ =2.981\text{kJ/s}\times \frac{1\text{kW}}{1\text{}\text{kJ/s}}\end{array}$

$=\text{2}\text{.981}\text{kW}$

Thus, the power rating of the electric heater is $\text{2}\text{.981}\text{kW}$.

To determine

The temperature rise that the air experiences each time it passes through the heater.

Answer to Problem 100P

The temperature rise that the air experiences each time it passes through the heater is $4.75°\text{C}$.

Explanation of Solution

Consider the heating duct with fan and heater only as the system. The air passes through in it steadily.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

The heating duct is an adiabatic duct. Hence, there is no heat loss.

The Equations (II) reduced as follows.

$\begin{array}{l}\left[0+\left(W_{\text{e,in}}+W_{\text{fan,in}}\right)+m\left(h_{\text{in}}+0+0\right)\right]-\left[0+0+m\left(h_{\text{out}}+0+0\right)\right]=0\\ W_{\text{e,in}}+W_{\text{fan,in}}+m{h}_{\text{in}}-m{h}_{\text{out}}=0\\ W_{\text{e,in}}+W_{\text{fan,in}}=m{h}_{\text{out}}-m{h}_{\text{in}}\\ W_{\text{e,in}}+W_{\text{fan,in}}=m\left(h_{\text{out}}-h_{\text{in}}\right)\end{array}$ (VII)

Express the Equation (VII) with respect to change of time as follows.

${\dot{W}}_{\text{e,in}}+{\dot{W}}_{\text{fan,in}}=\dot{m}\left(h_{\text{out}}-h_{\text{in}}\right)$ (VIII)

The change in enthalpy is expresses as follow.

$\begin{array}{c}{h}_{\text{out}}-h_{\text{in}}=c_p\left(T_{\text{out}}-T_{\text{in}}\right)\\ =c_p\left(T_{\text{2}}-T_1\right)\end{array}$

Here, the specific heat at constant pressure is $c_p$, the outlet temperature is $T_2$ and the inlet temperature is $T_1$.

Substitute $c_p\left(T_2-T_1\right)$ for $h_{\text{out}}-h_{\text{in}}$ in Equation (VIII) and rearrange it to obtain $\left(T_2-T_1\right)$.

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

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_p\right)$ of air is $1.005\text{kJ/kg}\cdot \text{°C}$.

Conclusion:

Substitute $\text{2}\text{.981}\text{kW}$ for ${\dot{W}}_{\text{e,in}}$, $200\text{W}$ for ${\dot{W}}_{\text{fan,in}}$, $40\text{kg/min}$ for $\dot{m}$ and $1.005\text{kJ/kg}\cdot \text{°C}$ for $c_p$,  in Equation (IX).

$\begin{array}{c}{T}_2-T_1=\frac{\text{2}\text{.981}\text{kW}+200\text{W}}{\left(40\text{kg/min}\right)\left(1.005\text{kJ/kg}\cdot \text{°C}\right)}\\ =\frac{\left(\text{2}\text{.981}\text{kW}\times \frac{1\text{kJ/s}}{1\text{kW}}\right)+\left(200\text{W}\times \frac{1\text{kJ/s}}{1000\text{W}}\right)}{\left(40\text{kg/min}\times \frac{1\text{min}}{60\text{s}}\right)\left(1.005\text{kJ/kg}\cdot \text{°C}\right)}\\ =\frac{3.181\text{kJ/s}}{\left(0.6667\text{kg/s}\right)\left(1.005\text{kJ/kg}\cdot \text{°C}\right)}\\ =4.7478\text{°C}\end{array}$

$\simeq 4.75\text{°C}$

Thus, the temperature rise that the air experiences each time it passes through the heater is $4.75°\text{C}$.