Chapter 12, Problem 54P

a)

To determine

The wattage of the fan-motor unit to be purchased.

Explanation of Solution

Given:

The dimension of the bathroom is $2\text{m}\times 3\text{m}\times 3\text{m}$.

The combine efficiency of the fan-motor unit is 50%.

The entire volume of air in the bathroom is replace by the ventilator for every 10 minutes.

Calculation:

Draw the free body diagram of the system as in Figure 1.

Assume the frictional losses along the flow is negligible and the flow is steady and incompressible.

The effect of kinetic energy correction factor is negligible $\alpha =1$.

Take the density of air $\rho =1.25{\text{kg/m}}^{\text{3}}$ .

Consider the acceleration due to gravity (g) as $9.81{\text{m/s}}^2$.

Calculate the total volume of air $\left(V\right)$ in the bathroom;

  $\begin{array}{c}V=2\text{m}\times 3\text{m}\times 3\text{m}\\ =18{\text{m}}^{\text{3}}\end{array}$

Calculate the volume flow rate $\dot{V}$ using the time it takes to replace the entire volume of air.

  $\dot{V}=\frac{V}{t}$

Here, t is time.

Substitute $18{\text{m}}^{\text{3}}$ for V and 10 min for t.

  $\begin{array}{c}\dot{V}=\frac{18{\text{m}}^{\text{3}}}{10\min }\times \frac{1\min }{\text{60s}}\\ =0.03{\text{m}}^{\text{3}}\text{/s}\end{array}$

Calculate the mass flow rate $\dot{m}$ using the relation;

  $\dot{m}=\rho \dot{\text{V}}$

Substitute $1.25{\text{kg/m}}^{\text{3}}$ for $\rho$ and $0.03{\text{m}}^{\text{3}}\text{/s}$ for $\dot{\text{V}}$.

  $\begin{array}{c}\dot{m}=1.25\times 0.03\\ =0.0375\text{kg/s}\end{array}$

Refer to figure 1,

The pressure at the point 1 and point 2 as same as the atmospheric pressure $P_1=P_2=P_{atm}$ .

The elevation of the point 1 and 2 are same $z_1=z_2$ .

Consider the point 1 is far enough from the fan. Thus the flow velocity at the point 1 $V_1=0$ .

Write the energy equation for a steady incompressible flow that relates the pressure $\left(P\right)$ , velocity $\left(V\right)$ and height $\left(z\right)$ for between the point 1 and point 2 as below;

  $\dot{m}\left(\frac{P_1}{\rho }+{\alpha }_1\frac{V_1{}^2}{2}+z_{1}g\right)+{\dot{W}}_{pump}=\dot{m}\left(\frac{P_2}{\rho }+{\alpha }_2\frac{V_2{}^2}{2}+z_{2}g\right)+{\dot{W}}_{turbine}+{\dot{E}}_{mech,loss}$

Substitute $z_1$ for $z_2$ , 0 for $V_1$ , $P_1$ for $P_2$ and 0 for ${\dot{W}}_{pump}$.

  $\begin{array}{c}\dot{m}\left(\frac{P_1}{\rho }+0+z_{1}g\right)+{\dot{W}}_{pump}=\dot{m}\left(\frac{P_1}{\rho }+{\alpha }_2\frac{V_2{}^2}{2}+z_{1}g\right)+0+{\dot{E}}_{mech,loss}\\ {\dot{W}}_{pump}-{\dot{E}}_{mech,loss}=\dot{m}{\alpha }_2\frac{V_2{}^2}{2}\end{array}$

Substitute ${\dot{W}}_{fan,elect}$ for ${\dot{W}}_{pump}-{\dot{E}}_{mech,loss}$  .

  ${\dot{W}}_{fan,elect}=\dot{m}{\alpha }_2\frac{V_2{}^2}{2}$

Recall that the electric power rating of fan is equal to the effective power divided by the power rating.

  ${\dot{W}}_{fan,elect}=\frac{1}{\eta }\dot{m}{\alpha }_2\frac{V_2{}^2}{2}$

Substitute 0.5 for $\eta$, $0.0375\text{kg/s}$ for $\dot{m}$, 1 for ${\alpha }_2$ and 8 m/s for $V_2$ .

  $\begin{array}{c}{\dot{W}}_{fan,elect}=\frac{1}{0.5}\times 0.0375\times 1\times \frac{{\left(8\right)}^2}{2}\\ =2.4\text{W}\end{array}$

Thus, the wattage of the fan-motor unit to be purchased is $\underset{\_}{2.4\text{W}}$.

b)

To determine

The diameter of fan casing.

Explanation of Solution

Given:

The dimension of the bathroom is $2\text{m}\times 3\text{m}\times 3\text{m}$.

The combine efficiency of the fan-motor unit is 50%.

The entire volume of air in the bathroom is replace by the ventilator for every 10 minutes.

Calculation:

Calculate the volume flow rate $\dot{V}$ using the diameter of fan and velocity of flow.

  $\dot{V}=A_2{V}_2$

Here, $A_2$ is the area of the fan.

Substitute $\frac{\pi D^2}{4}$ for $A_2$, $0.03{\text{m}}^{\text{3}}\text{/s}$ for $\dot{V}$ and 8 m/s for $V_2$ .

  $\begin{array}{c}0.03=\frac{\pi D^2}{4}\times 8\\ D=\sqrt{\frac{4\left(0.03\right)}{8\pi }}\\ =6.9\text{cm}\end{array}$

Thus, the diameter of the casing is $\underset{\_}{6.9\text{cm}}$.

c)

To determine

The pressure difference across the fan.

Explanation of Solution

Given:

The dimension of the bathroom is $2\text{m}\times 3\text{m}\times 3\text{m}$.

The combine efficiency of the fan-motor unit is 50%.

The entire volume of air in the bathroom is replace by the ventilator for every 10 minutes.

Calculation:

Take that the point 3 and 4 to be on the two sides of the fan on the horizontal line.

The elevation of the point 3 and 4 are same $z_3=z_4$ .

The velocity of the point 3 and 4 are same $V_3=V_4$ .

Write the energy equation for a steady incompressible flow that relates the pressure $\left(P\right)$ , velocity $\left(V\right)$ and height $\left(z\right)$ for between the point 3 and point 4 as below;

  $\dot{m}\left(\frac{P_3}{\rho }+{\alpha }_3\frac{V_3{}^2}{2}+z_{3}g\right)+{\dot{W}}_{pump}=\dot{m}\left(\frac{P_4}{\rho }+{\alpha }_4\frac{V_4{}^2}{2}+z_{4}g\right)+{\dot{W}}_{turbine}+{\dot{E}}_{mech,loss}$

Substitute $z_3$ for $z_4$ , $V_3$  for $V_4$ and 0 for  ${\dot{W}}_{turbine}$.

  $\begin{array}{c}\dot{m}\left(\frac{P_3}{\rho }+\alpha \frac{V_3{}^2}{2}+z_{3}g\right)+{\dot{W}}_{pump}=\dot{m}\left(\frac{P_4}{\rho }+\alpha \frac{V_3{}^2}{2}+z_{3}g\right)+0+{\dot{E}}_{mech,loss}\\ \dot{m}\frac{P_3}{\rho }+{\dot{W}}_{pump}-{\dot{E}}_{mech,loss}=\dot{m}\frac{P_4}{\rho }\\ {\dot{W}}_{pump}-{\dot{E}}_{mech,loss}=\dot{m}\frac{P_4}{\rho }-\dot{m}\frac{P_3}{\rho }\end{array}$

Substitute ${\dot{W}}_{fan,elect}$ for ${\dot{W}}_{pump}-{\dot{E}}_{mech,loss}$  .

  $\begin{array}{c}{\dot{W}}_{fan,elect}=\dot{m}\frac{P_4}{\rho }-\dot{m}\frac{P_3}{\rho }\\ {\dot{W}}_{fan,elect}=\frac{\dot{m}\left(P_4-P_3\right)}{\rho }\\ P_4-P_3=\frac{{\dot{W}}_{fan,elect}}{\dot{m}/\rho }\end{array}$

Substitute $\dot{V}$ for $\dot{m}/\rho$.

  $P_4-P_3=\frac{{\dot{W}}_{fan,elect}}{\dot{V}}$

Substitute 1.2 W for ${\dot{W}}_{fan,elect}$ and $\text{0}\text{.03}{\text{m}}^{\text{3}}\text{/s}$ for $\dot{V}$.

  $\begin{array}{c}{P}_4-P_3=\frac{\text{1}\text{.2W}}{\text{0}\text{.03}{\text{m}}^{\text{3}}\text{/s}}\\ =40\text{Pa}\end{array}$

Thus, the pressure difference across the fan is $\underset{\_}{40\text{Pa}}$.