Chapter 5.5, Problem 188RP

A building with an internal volume of 400 m³ is to be heated by a 30-kW electric resistance heater placed in the duct inside the building. Initially, the air in the building is at 14°C, and the local atmospheric pressure is 95 kPa. The building is losing heat to the surroundings at a steady rate of 450 kJ/min. Air is forced to flow through the duct and the heater steadily by a 250-W fan, and it experiences a temperature rise of 5°C each time it passes through the duct, which may be assumed to be adiabatic.

1. (a)   How long will it take for the air inside the building to reach an average temperature of 24°C?
2. (b)   Determine the average mass flow rate of air through the duct.

FIGURE P5–173

To determine

The time taken to attain the building’s average temperature of $24°\text{C}$.

Answer to Problem 188RP

The time taken to attain the building’s average temperature of $24°\text{C}$ is $146\text{s}$.

Explanation of Solution

Consider the entire building as system and the air circulates the in the building 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 building $\left({\dot{Q}}_2\right)$. 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 building to the surrounding. The mass of air circulates in the building 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.

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

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

$m=\frac{P_1\nu }{R{T}_1}$ (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 $95\text{kPa}$ for $P_1$, $400{\text{m}}^3$ for $\nu$, $0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$ and $14°\text{C}$ for $T_1$ in Equation (V).

$\begin{array}{c}m=\frac{\left(95\text{kPa}\right)\left(400{\text{m}}^3\right)}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(14°\text{C}\right)}\\ =\frac{38000\text{kPa}\cdot {\text{m}}^3}{\left(0.287\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(14+273\right)\text{K}}\\ =\frac{38000\text{kPa}\cdot {\text{m}}^3}{82.369\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =461.3386\text{kg}\end{array}$

$\simeq 142.3\text{kg}$

Substitute $461.3386\text{kg}$ for $m$ in Equation (VI).

$\begin{array}{c}\dot{m}=\frac{461.3386\text{kg}}{\Delta t}\\ \simeq \frac{461.3\text{kg}}{\Delta t}\end{array}$

Substitute $\frac{461.3\text{kg}}{\Delta t}$ for $\dot{m}$, $0.718\text{kJ/kg}\cdot \text{K}$ for $c_v$, $24°\text{C}$ for $T_2$, $14\text{°C}$ for $T_1$, $30\text{kW}$ for ${\dot{W}}_{\text{e,in}}$, $250\text{W}$ for ${\dot{W}}_{\text{fan,in}}$ and $450\text{kJ/min}$ for ${\dot{Q}}_{\text{out}}$ in Equation (IV).

$\begin{array}{l}30\text{kW}+250\text{W}-450\text{kJ/min}=\left(\frac{461.3\text{kg}}{\Delta t}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)\left(24°\text{C}-14°\text{C}\right)\\ \left\{\begin{array}{l}30\text{kW}+\left(250\text{W}\times \frac{1\text{kW}}{1000\text{W}}\right)\\ -\left(450\text{kJ/min}\times \frac{1\text{min}}{60\text{s}}\times \frac{1\text{kW}}{1\text{kJ/s}}\right)\end{array}\right\}=\left\{\begin{array}{l}\left(\frac{461.3\text{kg}}{\Delta t}\right)\left(0.718\text{kJ/kg}\cdot \text{K}\right)\\ \left[\left(24+273\right)\text{K}-\left(14+273\right)\text{K}\right]\end{array}\right\}\\ 30\text{kW}+0.25\text{kW}-7.5\text{kW}=\left(\frac{461.3\text{kg}}{\Delta t}\right)7.18\text{kJ/kg}\\ 22.75\text{kW}=\frac{1}{\Delta t}\left(3312.134\text{kJ}\right)\end{array}$

$\begin{array}{c}\Delta t=\frac{3312.134\text{kJ}}{22.75\text{kW}\times \frac{1\text{kJ/s}}{1\text{kW}}}\\ \Delta t=\frac{3312.134\text{kJ}}{22.75\text{kJ/s}}\\ =145.588\text{s}\\ \simeq 14\text{6}\text{s}\end{array}$

Thus, the time taken to attain the building’s average temperature of $24°\text{C}$ is $146\text{s}$.

To determine

The average mass flow rate of air through the duct.

Answer to Problem 188RP

The average mass flow rate of air through the duct is $\text{6}\text{.02}\text{kg/s}$.

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 $\dot{m}$.

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

It is given that the temperature rise is $5°\text{C}$.

$T_2-T_1=5°\text{C}$

Substitute $\text{30}\text{kW}$ for ${\dot{W}}_{\text{e,in}}$, $250\text{W}$ for ${\dot{W}}_{\text{fan,in}}$, $1.005\text{kJ/kg}\cdot \text{°C}$ for $c_p$, and $5\text{°C}$ for $T_2-T_1$ in Equation (IX).

$\begin{array}{c}\dot{m}=\frac{\text{30}\text{kW}+250\text{W}}{\left(1.005\text{kJ/kg}\cdot \text{°C}\right)\left(5°\text{C}\right)}\\ =\frac{\left(\text{30}\text{kW}\times \frac{1\text{kJ/s}}{1\text{kW}}\right)+\left(250\text{W}\times \frac{1\text{kJ/s}}{1000\text{W}}\right)}{\left(40\text{kg/min}\times \frac{1\text{min}}{60\text{s}}\right)\left(1.005\text{kJ/kg}\cdot \text{°C}\right)}\\ =\frac{30.25\text{kJ/s}}{5.025\text{kJ/kg}}\\ =6.0199\text{kg/s}\end{array}$

$\simeq \text{6}\text{.02}\text{kg/s}$

Thus, The average mass flow rate of air through the duct is $\text{6}\text{.02}\text{kg/s}$.