Chapter 6, Problem 169RQ

(a)

To determine

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

Explanation of Solution

Given:

The volume of the building $\left(v\right)$ is $400{\text{ m}}^3$.

The local atmospheric pressure $\left(P_1\right)$ is $95\text{ kPa}$.

The initial temperature of the $\left(T_1\right)$ building is $14°\text{C}$.

The heat lost to the surroundings $\left({\dot{Q}}_{\text{out}}\right)$ is $450\text{ kJ/min}$.

The work input to the fan $\left(W_{\text{fan,in}}\right)$ is $250\text{W}$.

The temperature rise $\left(T_2-T_1\right)$ is $5°\text{C}$

Calculation:

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

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

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

  $m=\frac{P_1\nu }{R{T}_1}$

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

Calculate the mass flow rate $\left(\dot{m}\right)$.

  $\dot{m}=\frac{m}{\Delta t}$

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

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

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)

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 $0.718\text{kJ/kg}\cdot \text{K}$.

Substitute $\dot{m}=\frac{461.3\text{kg}}{\Delta t}$, $c_v=0.718\text{kJ/kg}\cdot \text{K}$, $T_2=24°\text{C}$, $T_1=14\text{°C}$, ${\dot{W}}_{\text{e,in}}=30\text{kW}$, ${\dot{W}}_{\text{fan,in}}=250\text{W}$ and ${\dot{Q}}_{\text{out}}=450\text{kJ/min}$ 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 $146\text{s}$.

(b)

To determine

The average mass flow rate of air through the duct.

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.

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

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

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

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 $\dot{m}$.

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

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 $1.005\text{kJ/kg}\cdot \text{°C}$.

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

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

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

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