Question

The fan on a personal computer draws 0.3 ft3/s ofair at 14.7 psia and 708F through the box containing the CPU and other components. Air leaves at 14.7 psia and 838F.Calculate the electrical power, in kW, dissipatedby the PCcomponents

Answer 1

0.12 kW

Step-by-step explanation:

Given that

The flow rate of air (V)=0.3 ft³/s

V=0.008 m³/s

Pressure, P=14.7 psia

P=1.013529 atm=101.325 kPa

Inlet temperature = 70° F=294.261 K

Exit temperature = 83° F=301.483 K

We know that , specific heat capacity of the air

Cp=1.005 kJ/kg.K

The mass flow rate of air is given as

`\dot{m}=(P* V)/(R* T)\\\dot{m}=(101.325* 0.008)/(0.287* 294.261)\\\dot{m}= 0.0095\ kg/s`

By using energy conservation

`Electric\ power =\dot{m}* C_p* (T_2-T_1)\\Electric\ power =0.0095* 1.005* (83-70)=0.12\ kW`

Therefore electric power dissipate by components will be 0.12 kW.