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Conduction & Convection
Quiz 8 – 2014.01.24
Determine the heat transfer across the 1 inch-thick slab shown at the right, whose thermal conductivity varies linearly with temperature according to the equation
TIME IS UP!!!
20 0 5k T . T
1 in
350F 250F
2 ft
3 ft
4 ft
o2 o
Btu F
h ft Fk , T
Outline
2. Conduction Heat Transfer
2.1. Series/Parallel Resistances
2.2. Geometric Considerations
Outline
3. Convection Heat Transfer
3.1. Heat Transfer Coefficient
3.2. Dimensionless Groups for HTC
Estimation
Thermal Resistance Circuits
Recall: Fourier’s Law of Heat Conduction
𝑄𝐴
=−𝑘𝑑𝑇𝑑𝑥
ONE-DIMENSIONAL ONLY
For a constant cross-section and isotropic thermal conductivity (e.g. a flat slab).
𝑄=−∆𝑇
(∆ 𝑥𝑘𝐴 )
Driving force*
Thermal Resistanceǂ
* where T1> T2ǂ where x2> x1
Series/Parallel Resistances
For series resistances (flat slab):
𝑅𝐴=∆𝑥𝐴
𝑘𝐴 𝐴𝑅𝐵=
∆ 𝑥𝐵
𝑘𝐵 𝐴
Driving Force at A:
Driving Force at B:
Overall driving force:
(−∆𝑇 𝐴 )=𝑄𝑡𝑜𝑡𝑅𝐴
(−∆𝑇 𝐵 )=𝑄𝑡𝑜𝑡𝑅𝐵
(−∆𝑇 𝑡𝑜𝑡 )=𝑄𝑡𝑜𝑡 (𝑅𝐴+𝑅𝐵 )
Series/Parallel Resistances
For series resistances (flat slab):
𝑄𝑡𝑜𝑡=(−∆𝑇 𝑡𝑜𝑡 )∆ 𝑥𝐴
𝑘𝐴 𝐴+∆ 𝑥𝐵
𝑘𝐵 𝐴
𝑅𝑡𝑜𝑡=𝑅𝐴+𝑅𝐵
𝑄𝑡𝑜𝑡=𝑄𝐴=𝑄𝐵
(−∆𝑇 𝑡𝑜𝑡 )=(−∆𝑇 𝐴 )+(−∆𝑇 𝐵 )𝑅𝐴=
∆𝑥𝐴
𝑘𝐴 𝐴𝑅𝐵=
∆ 𝑥𝐵
𝑘𝐵 𝐴
Series/Parallel Resistances
For parallel resistances (flat slab):
𝑅𝐴=∆𝑥𝐴
𝑘𝐴 𝐴𝑅𝐵=
∆ 𝑥𝐵
𝑘𝐵 𝐴
Driving Force at A:
Driving Force at B:
Overall driving force:
(−∆𝑇 𝑡𝑜𝑡 )=𝑄𝐴𝑅𝐴
(−∆𝑇 𝑡𝑜𝑡 )=𝑄𝐵𝑅𝐵
(−∆𝑇 𝑡𝑜𝑡 )𝑅𝑡𝑜𝑡
=𝑄𝐴+𝑄𝐵
Series/Parallel Resistances
For parallel resistances (flat slab):
1𝑅𝑡𝑜𝑡
=1𝑅𝐴
+1𝑅𝐵
𝑄𝑡𝑜𝑡=𝑄𝐴+𝑄𝐵
(−∆𝑇 𝑡𝑜𝑡 )=(−∆𝑇 𝐴 )=(−∆𝑇 𝐵 )𝑅𝐴=
∆𝑥𝐴
𝑘𝐴 𝐴𝑅𝐵=
∆ 𝑥𝐵
𝑘𝐵 𝐴
(−∆𝑇 𝑡𝑜𝑡 )𝑅𝑡𝑜𝑡
=(−∆𝑇 𝑡𝑜𝑡 )
𝑅𝐴
+(−∆𝑇 𝑡𝑜𝑡 )
𝑅𝐵
Substituting the individual Q:
Series/Parallel Resistances
Exercise! (PIChE Quiz Bowl Nationals 2009, Easy Round, 2 min)
A composite wall consists of 2-in. corkboard (inner), 6-in. concrete, and 3-in. wood (outer). The thermal conductivities of the materials are 0.025, 0.8, and 0.065 Btu/hr/ft/°F, respectively. The temperature of the inner surface of the wall is 55°F while the outer surface is at 90°F. What are the temperatures in °F:
a) Between the cork and concrete?b) Between the concrete and wood?
Series/Parallel Resistances
Exercise! (Combined series & parallel)Four different materials were joined together as a block of constant width shown below. The cross-sectional area of B is equal to that of C. Materials A, B, C, and D have k = 0.1, 0.5, 0.4, and 0.1 W/mK, respectively. The thickness of blocks A and D is 1” and the thickness of blocks B and C is 6”. If the left of A is exposed to 110°C and the right of D is exposed to 60°C, calculate (a) the temperature right after material A and (b) the temperature right before material D.
Geometric Considerations
Heat Conduction Through Concentric Cylinders
𝐴=2𝜋𝑟𝐿
𝑄𝐴
=−𝑘𝑑𝑇𝑑𝑟
∫ 𝑄2𝜋𝑟𝐿
𝑑𝑟=−𝑘∫𝑑𝑇
Integrating both sides:
ln ( 𝑟2
𝑟1)=−2𝜋 𝑘𝐿
𝑄 (𝑇 2−𝑇 1 )
Geometric Considerations
Heat Conduction Through Concentric Cylinders
𝐴=2𝜋𝑟𝐿
𝑄𝐴
=−𝑘𝑑𝑇𝑑𝑟
Rearranging: 𝑄=−𝑘2𝜋 𝐿
ln ¿¿¿¿¿
∫ 𝑄2𝜋𝑟𝐿
𝑑𝑟=−𝑘∫𝑑𝑇
Geometric Considerations
Heat Conduction Through Concentric Cylinders
Define a logarithmic mean area:
𝐴𝐿𝑀=2𝜋 𝐿(𝑟 2−𝑟1)ln ¿¿ ¿¿¿
…and a logarithmic mean radius:
𝑟 𝐿𝑀=(𝑟2−𝑟1)ln¿ ¿¿¿¿
𝑄=−𝑘 𝐴𝐿𝑀 (𝑇 2−𝑇1
𝑟2−𝑟1)
𝐴𝐿𝑀=2𝜋 𝐿𝑟 𝐿𝑀 *Final form
𝑄=−𝑘2𝜋 𝐿
ln ¿¿¿¿¿
Geometric Considerations
Heat Conduction Through Concentric Cylinders
When to use logarithmic mean area and when to use arithmetic mean area?
𝐴𝐿𝑀=2𝜋 𝐿(𝑟 2−𝑟1)ln ¿¿ ¿¿¿
𝐴𝑀=2𝜋 𝐿(𝑟 2+𝑟 1)
2
𝒓 𝑳𝑴
𝒓𝑴
r2/r1 rLM rM
1 #DIV/0! 11.05 1.024797 1.0251.1 1.049206 1.05
1.15 1.073254 1.0751.2 1.096963 1.1
1.25 1.120355 1.1251.3 1.143448 1.15
1.35 1.16626 1.1751.4 1.188805 1.2
Geometric Considerations
Heat Conduction Through Concentric Cylinders
A tube of 60-mm (2.36-in.) outer diameter is insulated with a 50-mm (1.97-in.) layer of silica foam, for which k = 0.032 Btu/hr-ft-°F, followed by a 40-mm (1.57-in.) layer of cork with k = 0.03 Btu/hr/ft/°F. If the temperature of the outer surface of the pipe is 150°C, and the temperature of the outer surface of the cork is 30°C, calculate the heat loss in W/(m of pipe).