Heat Transfer:
Thermal Resistance
windmill pumping water for cows – west Texas
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Class Problem: The properties of the composite material shown is being testing in a lab. If a test section of the
material is 0.75m by 0.5m, then find the rate of heat transfer through material for the given conditions.
Assume no heat transfer due to convection is occurring, and the urethane and polystyrene are equal widths.
rubber
glass fiber
k = 0.13
k = 0.043
𝑊
𝑚∙ ℃
k = 0.026
𝑊
𝑚∙ ℃
𝑊
𝑚∙ ℃
urethane
𝟑𝟖℃
k = 0.17
0.75m
𝟏𝟗℃
𝑊
𝑚∙ ℃
polystyrene
4cm
5cm
7cm
0.5m
How will you draw the “circuit” diagram for this composite material?
The thermal resistance of the middle section will be treated as a parallel section.
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Class Problem: The properties of the composite material shown is being testing in a lab. If a test section of the
material is 0.75m by 0.5m, then find the rate of heat transfer through material for the given conditions.
Assume no heat transfer due to convection is occurring.
1
R t,eq = R t,rubber +
1
R t,𝑢𝑟𝑒𝑡ℎ𝑎𝑛𝑒
∆x𝑟𝑢𝑏𝑏𝑒𝑟
k 𝑟𝑢𝑏𝑏𝑒𝑟 A
𝐑 𝐭,𝐫𝐮𝐛𝐛𝐞𝐫
𝐪
𝟑𝟖℃
R t,eq =
∆x𝑢𝑟𝑒𝑡ℎ𝑎𝑛𝑒
𝐴
k 𝑢𝑟𝑒𝑡ℎ𝑎𝑛𝑒
2
𝐑 𝐭,𝐮𝐫𝐞𝐭𝐡𝐚𝐧𝐞
∆xglass fiber
k𝑔𝑙𝑎𝑠𝑠 𝑓𝑖𝑏𝑒𝑟 A
∆x𝑟𝑢𝑏𝑏𝑒𝑟
+
k 𝑟𝑢𝑏𝑏𝑒𝑟 A
+
+ R t,glass fiber
1
R t,polysytrene
1
1
∆x𝑢𝑟𝑒𝑡ℎ𝑎𝑛𝑒
𝐴
k 𝑢𝑟𝑒𝑡ℎ𝑎𝑛𝑒
2
+ ∆x
+
1
∆xglass fiber
k𝑔𝑙𝑎𝑠𝑠 𝑓𝑖𝑏𝑒𝑟 A
𝑝olysytren𝑒
k 𝑝𝑜𝑙𝑦𝑠𝑦𝑡𝑟𝑒𝑛𝑒
𝐴
2
𝐑 𝐭,𝐠𝐥𝐚𝐬𝐬 𝐟𝐢𝐛𝐞𝐫
𝟏𝟗℃
𝐑 𝐭,𝐩𝐨𝐥𝐲𝐬𝐲𝐭𝐫𝐞𝐧𝐞
∆x𝑝olysytren𝑒
𝐴
k 𝑝𝑜𝑙𝑦𝑠𝑦𝑡𝑟𝑒𝑛𝑒
2
Why is the area divided by 2 for the middle section layers
(polystyrene and urethane)?
Looking at the area perpendicular to the flow of heat
transfer, theses material each make up only half the area.
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Class Problem (continued): Find the rate of heat transfer through material for the given conditions.
𝐑 𝐭,𝐮𝐫𝐞𝐭𝐡𝐚𝐧𝐞
𝐑 𝐭,𝐠𝐥𝐚𝐬𝐬 𝐟𝐢𝐛𝐞𝐫
𝐑 𝐭,𝐫𝐮𝐛𝐛𝐞𝐫
𝐪
𝟑𝟖℃
𝟏𝟗℃
𝐑 𝐭,𝐩𝐨𝐥𝐲𝐬𝐲𝐭𝐫𝐞𝐧𝐞
∆x𝑟𝑢𝑏𝑏𝑒𝑟
R t,eq =
+
k 𝑟𝑢𝑏𝑏𝑒𝑟 A
R t,eq =
0.04𝑚
𝑊
0.13 𝑚 ∙ ℃ (0.75m ∙ 0.5𝑚)
+
∆xglass fiber
+
k𝑔𝑙𝑎𝑠𝑠 𝑓𝑖𝑏𝑒𝑟 A
1
1
∆x𝑢𝑟𝑒𝑡ℎ𝑎𝑛𝑒
𝐴
k 𝑢𝑟𝑒𝑡ℎ𝑎𝑛𝑒
2
+ ∆x
1
𝑝olysytren𝑒
k 𝑝𝑜𝑙𝑦𝑠𝑦𝑡𝑟𝑒𝑛𝑒
𝐴
2
1
1
1
+
0.05𝑚
0.05𝑚
𝑊
0.75m ∙ 0.5𝑚
𝑊
0.75m ∙ 0.5𝑚
0.026 𝑚 ∙ ℃
0.17 𝑚 ∙ ℃
2
2
R t,eq = 6.522
℃
𝑊
+
0.07m
𝑊
0.043 𝑚 ∙ ℃
0.75m ∙ 0.5𝑚
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Class Problem (continued): Find the rate of heat transfer through material for the given conditions.
𝐑 𝐭,𝐮𝐫𝐞𝐭𝐡𝐚𝐧𝐞
𝐑 𝐭,𝐫𝐮𝐛𝐛𝐞𝐫
𝐪
𝟑𝟖℃
𝐑 𝐭,𝐠𝐥𝐚𝐬𝐬 𝐟𝐢𝐛𝐞𝐫
𝟏𝟗℃
𝐑 𝐭,𝐩𝐨𝐥𝐲𝐬𝐲𝐭𝐫𝐞𝐧𝐞
R t,eq = 6.522
Does it make sense that the rate of
heat transfer is so low? Why?
q=
℃
𝑊
(38℃ −19℃)
℃
6.522𝑊
q = 2.913𝑊
Extra Challenge: determine the inside
temperature at the boundaries:
• rubber and urethane boundary
• rubber and polystyrene boundary
• urethane and glass fiber boundary
• polystyrene and glass fiber
boundary
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