Question

A fan is to be selected to cool a computer case whose dimensions are 12 cm
3 40 cm x 40 cm (Fig. below). Half of the volume in the case is expected to
be filled with components and the other half to be air space. A 5-cmdiameter hole is available at the back of the case for the installation of the
fan that is to replace the air in the void spaces of the case once every
second. Small low-power fan–motor combined units are available in the
market and their efficiency is estimated to be 30 percent. Determine (a) the
wattage of the fan–motor unit to be purchased and (b) the pressure
difference across the fan. Take the air density to be 1.20 kg/m3.

Answer 1

Explanations & Calculations

• Total volume of the case = $\small 12mc\times 34cm\times 40cm = 16320cm^3 = 0.01632m^3$
• Available volume ($\small V$ ) to be replaced by the fan $=\large \frac{1}{2}\small (16320) = 8160cm^3= 0.00816m^3$

• Area of the hole ($\small A$ ) = $\large \pi\frac{d^2}{4} = \pi\frac{(5cm)^2}{4} = \small 19.635cm^2=0.0019635m^2$

• That volume of air should be pumped in by the fan in a minute through the 5cm diameter inlet hence the speed ($\small v$) of the incoming air column is

$\qquad \begin{aligned} \small v&=\small \frac{8160cm^3s^{-1}}{19.635cm^2} \\ &= \small 415.584cms^{-1}=4.15584ms^{-1} \end{aligned}$

• Therefore, if the power of the pump is $\small P$ then,

$\qquad\qquad \begin{aligned} \small P\times 30\% &= \small \frac{\frac{1}{2}mv^2}{t} = \frac{1}{2}\big(\frac{m}{t}\big)v^2=\frac{1}{2}(Av\rho)v^2\\ \small P &= \small \frac{100}{30}\times \frac{1}{2}A\rho v^3\\ &= \small \frac{5}{3} \times (0.0019635m^2)\times(1.2kgm^{-3}) \times{(4.15584ms^{-1})^3}\\ &= \small \bold{0.282W} \end{aligned}$

• Then the pressure difference ($\small \Delta P$ ) across the pump,

$\qquad\qquad \begin{aligned} \small \Delta P\times V &= \small (0.282W\times 30\%)\times 1s\\ \small \Delta P &= \small \frac{0.282J\times 0.3}{0.00816m^3}\\ &= \small \bold{10.368Pa} \end{aligned}$