Question

The fan on a personal computer draws 8.6 L/s of air at 100 kPa and 20 deg C through the box containing the CPU and other components. Air leaves at 100 kPa and 27 deg C. Calculate the electrical power, in kW, dissipated by the PC components.

answer: 0.0719 KW

Answer #1

▪︎ Givens :

V_{1} = 8.6 L/s = 0.3 ft^{3}/s

P_{1} = 100 kPa = 14.5 psia

T_{1} = 20°C = 528°R = 80.6°F

T_{2}= 27°C = 68°F

Energy balance

E_{input} - E_{output} =
= 0

Therefore,

We have,

From ideal gas equation,

= 13.49 ft^{3}/lbm

Therefore,

= 0.02224 lbm/s

We have,

W_{output} = mc_{p}(T_{2} -
T_{1})

= (0.02224 lbm/s)(0.240 Btu/lbm.°R)(80.6-68)Btu/lbm

**W _{output} = 0.0709 kW**

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