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First law of thermodynamics & steady flow devices

  • Thread starter 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.
 

Answers and Replies

  • #2
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This looks about right, except that you have to decide whether the 90 W is heat or work. There really isn't any what I would call shaft work, so I would consider the 90 W as Q. The heat being given off by the CPU transfers to the gas stream, so the "heater" inside the gas stream is not considered part of the system. That's why I call the 90W to be Q rather than -W. The air mass flow rate looks like it is about 3 gm/sec, which doesn't seem outrageous to me. Check what volume flow rate this would entail by dividing by the density.

Chet
 
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  • #3
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m*= -90W/33.23kJ/kg
Clean up your units. Do you get g/s?

And yes, delta T is the same in K or oC.
 
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  • #4
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To elaborate a little further on my answer in post #2, there is a bit of flexibility on the choice of the "system" in this problem, and each choice involves a different Q and W.

1. If one chooses only the gas within the case as the system, then the CPU is not part of the system, and Q is the 90 W given off by the CPU, with W being equal to zero.

2. If one chooses the entire case as the system, then the CPU is part of the system, and Q = 0; in this case there is electrical work being done on the system (since electrons enter the case through the wires at one potential and exit the case through the wires at a different potential). This work -W = 90 W is not associated with the work required to push the air into and out of the case, and thus counts as "shaft work."

For each of these two choices for the "open system," the change in enthalpy of the air between entrance and exit is the same.

Chet
 
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  • #5
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I'm really bad when it comes to different units - I'm afraid I don't know how I would clean that up.

There is a second part to the question which asks : "calculate the average flow speed (units: m/sec) of air through the side port." Would that basically be the volume flow rate through the cross sectional area of the side port? so (3mm*10mm) * 10 = 300mm^2?
 
  • #6
SteamKing
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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 :\
I'm really bad when it comes to different units - I'm afraid I don't know how I would clean that up.
Well, you should at least try.

Units aren't something which are just randomly tacked onto the end of numbers. Understanding the units is key to understanding the calculation.

What's the definition of a Watt?
 
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  • #7
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Joule per second..so it would work out to be kg/s?
 
  • #8
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Joule per second..so it would work out to be kg/s?
You've got watts divided by (kiloJoules / kilogram). Manipulate these units according to the arithmetic and see what's left over, after making the substitution for the units which make up watts.
 
  • #9
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Joule per second..so it would work out to be kg/s?
Thats what I did haha
 
  • #10
SteamKing
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Thats what I did haha
No, you haven't done the unit cancellations correctly.

Remember 1 kJ = 1000 J, just like 1 kg = 1000 g. 1 W = 1 J/s.

You must account for any numerical factors along with the units.
 
  • #11
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I doubled checked it in wolfram..I get 0.001 Kg/s

That was simply cancelling the units..I didn't put in the 90W or the m*

Putting the values in however I got 0.0027 kg/s...90/33.23*0.001 so 2.7 grams/s
 
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  • #12
SteamKing
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I doubled checked it in wolfram..I get 0.001 Kg/s
0.001 kg /s = 0.001 × 1000 g /s = g/s

Always get the units to the simplest form.
 
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  • #13
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Thank you SteamKing
Would I be correct to assume the following?
There is a second part to the question which asks : "calculate the average flow speed (units: m/sec) of air through the side port." Would that basically be the volume flow rate through the cross sectional area of the side port? so (3mm*10mm) * 10 = 300mm^2?
 
  • #14
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Thank you SteamKing
Would I be correct to assume the following?
So it would seem.
 
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  • #15
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Thank you SteamKing
Would I be correct to assume the following?
You need to divide the volume flow rate by the area of the slots to get the speed.
 
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  • #16
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Hi Chester, sorry to revive an old thread again..although I'm having trouble with attaining the flow speed (mainly with the units)

To get v*=m*/p ...so 2.7g/s /1.176kg/m^3 = 0.0023m^3/s

and like you said " divide the volume flow rate by the area of the slots to get the speed."

(0.0023m^3/s)/300mm^2) ..when I put that into wolfram it comes out as 7.6m/s...but when I work it out I get 0.0076
I convert the 300mm to m so I end up with (0.0023m/s)/0.3=0.0076...can't figure out how wolfram gets its answer :\
 
  • #17
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(0.0023m^3/s)/300mm^2) ..when I put that into wolfram it comes out as 7.6m/s...but when I work it out I get 0.0076
I convert the 300mm to m so I end up with (0.0023m/s)/0.3=0.0076...can't figure out how wolfram gets its answer :\
Unlike you, wolfram can keep units straight...........
 
  • #18
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Yes, thank you for that 'insightful' advice. That's why I'm here, for help. Not insults.
 
  • #19
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Yes, thank you for that 'insightful' advice. That's why I'm here, for help. Not insults.
Even before you responded, insightful had already been given some warning points for this sarcasm. Here in Physics Forums, we try to maintain a pleasant and supportive environment.

Chet
 
  • #20
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Sorry to keep reviving this problem....
but I was looking at the question again and noticed " a single (motherboard powered) 10W electric fan is mounted inside the casing."

So if the fan is taking 10W away from the 90W..and we then applied that to the formula: Q*-W*=m*[h2-h1+(V22-V12)/2+g(z2-z1]

Would I still put Q=90 and W=10 or W=80 and W=10 Seeing as that power is subtracted straight away from the fan?
 
  • #21
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Sorry to keep reviving this problem....
but I was looking at the question again and noticed " a single (motherboard powered) 10W electric fan is mounted inside the casing."

So if the fan is taking 10W away from the 90W..and we then applied that to the formula: Q*-W*=m*[h2-h1+(V22-V12)/2+g(z2-z1]

Would I still put Q=90 and W=10 or W=80 and W=10 Seeing as that power is subtracted straight away from the fan?
If the contents of the insulated case is taken as the system, then Q=0 and W=-90 watts.
 
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  • #22
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That makes sense..was just getting confused in thinking only 80W is really doing any heating. Thank you for your patience Chester. You've helped a great deal!
 
  • #23
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Hi, I have a feeling Luchekv and I are both Thermo students at the same uni cause I've got the exact same question in my assignment. I get everything except why we don't subtract the 10W of the fan from the initial 90W, since that is being converted to mechanical energy, and not heating energy? So you'd have Q*=80W and W*=10? Is it like when you stir a cup of water it increases the temperature? So the fan contributes to total energy of the system?
My Equation I used reduces to:

Q*(in) = W(out)* + m*(Cp * dT)
80-10 = m*(Cp * dT)
m* = 80/(M*(Cp * dT))

Though looking at it now, I guess the fan might be W*(in) as opposed to W*(out), leaving it -ve on the RHS, and making it:
Q*(in) - [-W*(out)] = m*(Cp * dT) = 90; anyways.

Then again, my approach might be completely wrong, so feel free to correct me where I am, Please ha
 
  • #24
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Hi, I have a feeling Luchekv and I are both Thermo students at the same uni cause I've got the exact same question in my assignment. I get everything except why we don't subtract the 10W of the fan from the initial 90W, since that is being converted to mechanical energy, and not heating energy? So you'd have Q*=80W and W*=10? Is it like when you stir a cup of water it increases the temperature? So the fan contributes to total energy of the system?
My Equation I used reduces to:

Q*(in) = W(out)* + m*(Cp * dT)
80-10 = m*(Cp * dT)
m* = 80/(M*(Cp * dT))

Though looking at it now, I guess the fan might be W*(in) as opposed to W*(out), leaving it -ve on the RHS, and making it:
Q*(in) - [-W*(out)] = m*(Cp * dT) = 90; anyways.

Then again, my approach might be completely wrong, so feel free to correct me where I am, Please ha
It all depends on what you call your "system." If your system is just the gas inside the case, then your system is receiving heat at the rate of 80 W from the CPU and is receiving work from the fan at the rate of 10 W, so the net is 80 - (-10) = 90 W. If you system consists of the entire contents of the (adiabatic) computer case, then your system is receiving no heat and is receiving electrical work at the rate of 90 W, so the net is 0 - (-90) = 90 W. Either way, you get the same answer.

Chet
 
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  • #25
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Thanks Chet, that makes sense, thank you for the concise explanation
 

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