Conduction & Convection Quiz 8 – 2014.01.24 Determine the heat transfer across the 1...

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).