Chapter 5, Problem 80P

A fan is to be selected to ventilate a bathroom whose dimensions are $2\text{m}\times \text{3}\text{m}\times \text{3}$ . The air velocity is not to exceed 7 m/s to minimize vibration and noise. The combined efficiency of the fan-motor unit to be used can be taken to be 50 percent. If the fan is to replace the entire volume of air in 15 mm. determine (a) the vattage of the fan-motor unit to be purchased. (b) the diameter of the fan casing, and (c) the pressure difference across the fan. Take the air density to be 1.25 kg rn3 and disregard the effect of the kinetic energy correction factors.

To determine

(a)

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

Answer to Problem 80P

The wattage of the fan-motor unit to be purchased is $1.225\text{W}$.

Explanation of Solution

Given information:

The dimension of the bathroom is $2\text{m}\times \text{3}\text{m}\times \text{3}\text{m}$, the velocity of the air is $7\text{m}/\text{s}$, the overall efficiency is $50\%$, the time taken by fan to replace the entire volume is $15\min ,$ and the density of the air is $1.25\text{kg}/{\text{m}}^{\text{3}}$.

Write the expression for the total volume of the bathroom.

  $V=l\times b\times h$  .......(I)

Here, the length of the bathroom is $l$, the breath of the bathroom is $b,$ and the height of the bathroom is $h$.

Write the expression for the volume flow rate of the air.

  $\dot{V}=\frac{V}{t}$  .......(II)

Here, the time taken by fan to replace the entire volume is $t$.

Write the expression for the mass flow rate of air.

  $\dot{m}=\rho \dot{V}$  .......(III)

Here, the density of the air is $\rho$.

Write the expression for the electric power rating of the fan.

  ${\dot{W}}_{fan}=\frac{1}{\eta }\dot{m}{\alpha }_2\left(\frac{V_2^2}{2}\right)$  .......(IV)

Here, the velocity of the air is $V_2$, the overall efficiency is $\eta ,$ and the kinetic energy correction factor is ${\alpha }_2$.

Calculation:

Substitute $2\text{m}$ for $l$, $3\text{m}$ for $b$ and $3\text{m}$ for $h$ in Equation (I).

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

Substitute $\text{18}{\text{m}}^3$ for $V$ and $15\min$ for $t$ in Equation (II).

  $\begin{array}{c}\dot{V}=\frac{18{\text{m}}^3}{15\min }\\ =1.2{\text{m}}^3/\min \left(\frac{1\min }{60\sec }\right)\\ =0.02{\text{m}}^3/\sec \end{array}$

Substitute $0.02{\text{m}}^3/\sec$ for $\dot{V}$ and $1.25\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (III).

  $\begin{array}{c}\dot{m}=\left(1.25\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.02{\text{m}}^3/\sec \right)\\ =0.025\text{kg}/\sec \end{array}$

Here, the kinetic energy correction factor is $1$.

Substitute $0.025\text{kg}/\sec$ for $\dot{m}$, $7\text{m}/\sec$ for $V_2$, $0.5$ for $\eta ,$ and $1$ for ${\alpha }_2$ in Equation (IV).

  $\begin{array}{c}{\dot{W}}_{fan}=\frac{1}{0.5}\left(0.025\text{kg}/\sec \right)\left(1\right)\left(\frac{{\left(7\text{m}/\sec \right)}^2}{2}\right)\\ =\frac{1}{0.5}\left(0.025\text{kg}/\sec \right)\left(1\right)\left(24.5{\text{m}}^2/{\sec }^2\right)\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^2/{\sec }^3}\right)\\ =1.225\text{W}\end{array}$

Conclusion:

Thus, the wattage of the fan-motor unit to be purchased is $1.225\text{W}$.

To determine

(b)

The diameter of the fan casing.

Answer to Problem 80P

The diameter of the fan casing is $6.03\text{cm}$.

Explanation of Solution

Given information:

The volume flow rate is $0.02{\text{m}}^3/\sec$.

Write the expression for the diameter of the casing.

  $D=\sqrt{\frac{4\dot{V}}{\pi V_2}}$  .......(V)

Calculation:

Substitute $0.02{\text{m}}^3/\sec$ for $\dot{V}$, $7\text{m}/\text{sec}$ for $V_2$ in Equation (V).

  $\begin{array}{c}D=\sqrt{\frac{4\left(0.02{\text{m}}^3/\sec \right)}{\pi \left(7\text{m}/\text{sec}\right)}}\\ =\sqrt{\frac{\left(0.08{\text{m}}^3/\sec \right)}{\left(21.99\text{m}/\text{sec}\right)}}\\ =0.0603\text{m}\left(\frac{\text{100}\text{cm}}{\text{1}\text{m}}\right)\\ =6.03\text{cm}\end{array}$

Conclusion:

Thus, the diameter of the fan casing is $6.03\text{cm}$.

To determine

(c)

The pressure difference across the fan.

Answer to Problem 80P

The pressure difference across the fan is $30.625\text{Pa}$.

Explanation of Solution

Given information:

The electric power of the fan is $1.225\text{W}$.

Write the expression for the pressure difference of the fan by applying the energy equation.

  $\begin{array}{l}\dot{m}\frac{P_3}{\rho }+{\dot{W}}_{fan}=\dot{m}\frac{P_4}{\rho }\\ P_4-P_3=\frac{{\dot{W}}_{fan}}{\frac{\dot{m}}{\rho }}\\ P_4-P_3=\frac{{\dot{W}}_{fan}}{\dot{V}}\end{array}$  .......(VI)

Write the expression for the effective power of the fan.

  ${\dot{W}}_{eff}={\dot{W}}_{fan}\times \eta$  .......(VII)

Calculation:

Substitute $0.5$ for $\eta$ and $1.225\text{W}$ for ${\dot{W}}_{fan}$ in Equation (VII).

  $\begin{array}{c}{\dot{W}}_{eff}=1.225\text{W}\left(0.5\right)\\ =0.6125\text{W}\end{array}$

Substitute $0.02{\text{m}}^3/\sec$ for $\dot{V}$, $0.6125\text{W}$ for ${\dot{W}}_{\text{eff}}$ in Equation (VI).

  $\begin{array}{c}{P}_4-P_3=\frac{\left(0.6125\text{W}\right)}{\left(0.02{\text{m}}^3/\sec \right)}\\ =30.625\text{Pa}\end{array}$

Conclusion:

Thus, the pressure difference across the fan is $30.625\text{Pa}$.