Heat transfer, single and double paned window. Stove and pot.

AI Thread Summary
The discussion focuses on calculating heat loss through single and double-paned windows, highlighting the significant difference in thermal resistance between the two configurations. For the single pane, the heat loss is calculated at 19,600W, while the double-pane setup results in a much lower heat loss of 70.75W due to the added resistance from the air gap. Participants note the importance of considering factors like radiation and gas motion between panes, which can further influence heat transfer in real-life scenarios. Additionally, calculations for heat transfer in a pot made of copper and aluminum show minimal temperature differences at the bottom surface in contact with the stove. The conversation emphasizes the complexities of thermal conductivity and the impact of material properties on heat transfer efficiency.
Ulyaoth
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A glass window 1m x 2m and .005m thick has thermal conductivity of 1.4W/m K. Inner surface temp of 15C and outside of -20C, what is the heat loss through the glass.

double paned construction two panes separated by .01m, if glass surfaces are 10C and -15C, with air thermal conductivity of .024W/m K

q''= -k dT/L
q’’= 1.4W/m•K((288K-253K)/(.005m)), q’’=9800W/m2, q=9800W/m2 • 2m2 = 19600W

Double Paned: Rtot = L/kg+L/ka+L/kg,
Rtot = (.005m)/(1.4W/(m•K))+(.010m)/(.024W/(m•K))+(.005m)/(1.4W/m•K)
Rtot=0.424K/W,
q=ΔT/Rtot, q=(283K-253K)/(.0424K/W), q=70.75W


It just seemed like too big of a difference to be the correct answer. Just one more I wasn't sure of.

The 5mm thick bottom of a 200mm diameter pan may be copper(k = 390W/m K) or aluminum (240W/m K). When used to boil water the surface of the bottom exposed to the water is nominally 110C. IF the heat transferred from the stove to the pot is 600W, what is the temperature of the surface in contact with the stove for the two materials
A=0.031416m2, q''=19100W
q=k(Tbottom-Tin)/L (q•L)/k+Tin=Tbottom,alum
Tbottom,alum=(19100W/m^2(.005m))/(240W/m•K)+383K = 383.398K
Tbottom,copper=(19100W/m^2(.005m))/(390W/m•K)+383K = 383.245K


Again, just wasn't sure, it seemed the numbers were too close.
 
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q=ΔT/Rtot, q=(283K-253K)/(.0424K/W), q=70.75W

You have a slight error in the above. It should be 82.6. The .0424 should be .424.

The huge difference comes from the additional resistance of the air. Look at the ratios of k/dx for air gap and for a pane of glass. The difference is large.

In a real life problem, you also have the radiation factor where the hotter pane loses heat to the cooler pane. It has an effect but is not considered in this problem. To complicate the issue further, there is motion of the trapped gas between the panes that further enhances the heat transfer.
 
Second part dealing with the pot on the stove is correct.
 
Thank you, just wanted to make sure I had the right idea for the problem.
 
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