Heat Transfer - finding rate of heat transfer

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Homework Statement


A 0.8m high and 1.5m wide double pane window consisting of two 3mm thick layers of glass(k=0.78W/mK) is separated by a 10mm wide stagnant air space(k=0.026 W/mK). The convection heat transfer coefficients on the inner and outer surfaces of the window are 10W/m^2K and 35W/m^2K respectively. For a day when the room is maintained at 25oC while the temperature outdoors is 10oC, determine:
i) the steady rate of heat transfer through this double pane window, and
ii) the temperature of its inner surface.

Homework Equations


Q=-k*A*ΔT/ΔX
Rth=ΔX/(k*A)

The Attempt at a Solution


i)
Rth,glass=0.003/(0.78*0.8*1.5)*2
Rth,air=0.01/(0.026*0.8*1.5)
Rth,total=2.549 W/m^2.K

What should I do after this?
And is it correct to calculate Rth,air like this?

Thank you very much.
 

Answers and Replies

  • #2
BvU
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Looks fine. Now what about this convective heat transfer on the outer and inner sufaces ?
 
  • #3
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Looks fine. Now what about this convective heat transfer on the outer and inner sufaces ?
Rth,inner=1/(10*0.8*1.5)=0.0833W/m^2.K
Rth,outer=1/(35*0.8*1.5)=0.0231W/m^2.K

Like this?
 
  • #4
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Good. Now write and equation for the temperature difference across each of the layers (and for the inside and outside boundary layers) in terms of Q (which is the same for all the layers). Then add the equations together, and see what you get.

Chet
 
  • #5
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Good. Now write and equation for the temperature difference across each of the layers (and for the inside and outside boundary layers) in terms of Q (which is the same for all the layers). Then add the equations together, and see what you get.

Chet
Q = (25-Tinner wall)/(Rth,inner) = (Tinner wall-TOuter wall)/(2.549) = (TOuter wall-10)/(Rth,outer)
Q = (25-Tinner wall)/(0.0833) = (Tinner wall-TOuter wall)/(2.549) = (TOuter wall-10)/(0.0231)

Is this what you mean?
 
  • #6
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Coming back to post #1:

I forgot to ask about the dimensions. If Q=-k*A*ΔT/ΔX and R is ΔX/(k*A) then what is the dimension of R ?
The about the calculation:

How do you get Rth,total from Rth,glass and Rth,air

Aren't they simply in series ? No suspicion about the values ? The biggest R is the air one: 0.01 divided by approx 0.03 can't yield approx 2.5 !?


Post #3:

Why do you ask "Like this ?" ?
(Slow typist: Chet comes in and gives you the next step -- which in fact you already used when you added the R)
 
  • #7
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Q = (25-Tinner wall)/(Rth,inner) = (Tinner wall-TOuter wall)/(2.549) = (TOuter wall-10)/(Rth,outer)
Q = (25-Tinner wall)/(0.0833) = (Tinner wall-TOuter wall)/(2.549) = (TOuter wall-10)/(0.0231)

Is this what you mean?
Not exactly. Express each of the temperature differences explicitly in terms of Q (just multiply both sides of your equation by R). Do this for each layer. Then add all the temperature differences together, and you will be left with only the inner and outer temperature values.

Chet
 

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