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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*[h

_{2}-h

_{1}+(V

_{2}

^{2}-

**V**

_{1}^{2})/2+g(z_{2}-z_{1}]Δh=Cp(T

_{2}-T

_{1})[/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*[h

_{2}-h

_{1}]

So then using Δh=Cp(T

_{2}-T

_{1}) 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.