- #1
Luchekv
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Homework Statement
The motherboard in a laptop is run by a PSU of 20V/4.5A. This motherboard is inside a plastic case where there is not heat transfer through the casing. To stop the motherboard from over heating a single (motherboard powered) 10W electric fan is mounted inside the casing.
The fan sucks in cool air through a vent at the bottom, it passes over the motherboard and leaves through a side port. The side port has 10 openings, each measuring about 3mm (wide) by 10mm (high)
- For steady operating conditions the air inside the laptop should not exceed 60 degrees Celsius
- Ambient temperature of air sucked into the laptop is 27 degrees Celsius.
- Pressure inside the laptop is 101.325 kPa absolute
- Air behaves as an ideal gas with constant specific heats of Cv=0.721 kJ/kg.K and Cp=1.007kJ/kg.K.
- W = 20*4.5 = 90W
It asks for:
A.) Apply the first law of thermodynamics and calculate the mass flow rate of air (units: g/sec)
-- Flow rates are noted with a '*' as the dot on top of the letter
Homework Equations
Q*-W*=m*[h2-h1+(V22-V12)/2+g(z2-z1]
Δh=Cp(T2-T1)[/B]
The Attempt at a Solution
Looking at that first equation..I don't have any heat, PE or KE..so I put them to zero. What I'm left with is:
-W*=m*[h2-h1]
So then using Δh=Cp(T2-T1) Δh = 1.007kJ/kg.K (60-27) = 33.231 (wasn't sure if it needed to be Kelvin)
∴
m*= -90W/33.23kJ/kg
and that just doesn't seem right :\
Thank you in advance.