1. Conduction &
Convection

2. TIME IS UP!!! Quiz 8 – 2014.01.24 1 in 350F 250F 2 ft 3 ft 4 ft
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
Source: McCabe 6th Ed. Example 10.3
Answer: 29.1 W/m or 30.3 Btu/ft/hr
TIME IS UP!!!

3. Outline Conduction Heat Transfer 2.1. Series/Parallel Resistances
2.2. Geometric Considerations

4. Outline Convection Heat Transfer 3.1. Heat Transfer Coefficient
3.2. Dimensionless Groups for HTC Estimation

5. Thermal Resistance Circuits
Recall: Fourier’s Law of Heat Conduction
ONE-DIMENSIONAL ONLY
Driving force*
𝑄= −∆𝑇 ∆𝑥 𝑘𝐴
𝑄 𝐴 =−𝑘 𝑑𝑇 𝑑𝑥
Thermal Resistanceǂ
The differential in dT/dx becomes simple delta T/delta x because of a constant cross section and material isotropy (constant k).
All the delta signs in our convention is point 2 – point 1, where the direction of heat flow is from point 1 to point 2.
For a constant cross-section and isotropic thermal conductivity (e.g. a flat slab).
*∆𝑇= 𝑇 2 − 𝑇 1 where T1> T2
ǂ∆𝑥= 𝑥 2 − 𝑥 1 where x2> x1

6. Series/Parallel Resistances
For series resistances (flat slab):
Driving Force at A:
Driving Force at B:
Overall driving force:
−∆ 𝑇 𝐴 = 𝑄 𝑡𝑜𝑡 𝑅 𝐴
−∆ 𝑇 𝐵 = 𝑄 𝑡𝑜𝑡 𝑅 𝐵
Consider 2 flat slabs of constant cross-section but different k and thickness deltaX.
Note that (-deltaTA) + (-deltaTB) = -(T2 – T1). Temperature drops are additive.
Meanwhile, heat flow (current) is constant and resistances are additive also. Compare these with ohm’s law.
𝑅 𝐴 = ∆ 𝑥 𝐴 𝑘 𝐴 𝐴
𝑅 𝐵 = ∆ 𝑥 𝐵 𝑘 𝐵 𝐴
−∆ 𝑇 𝑡𝑜𝑡 = 𝑄 𝑡𝑜𝑡 𝑅 𝐴 + 𝑅 𝐵

7. Series/Parallel Resistances
For series resistances (flat slab):
𝑄 𝑡𝑜𝑡 = −∆ 𝑇 𝑡𝑜𝑡 ∆ 𝑥 𝐴 𝑘 𝐴 𝐴 + ∆ 𝑥 𝐵 𝑘 𝐵 𝐴
𝑄 𝑡𝑜𝑡 = 𝑄 𝐴 = 𝑄 𝐵
This is a summary of the equations for Series Conduction
𝑅 𝑡𝑜𝑡 =𝑅 𝐴 + 𝑅 𝐵
𝑅 𝐴 = ∆ 𝑥 𝐴 𝑘 𝐴 𝐴
𝑅 𝐵 = ∆ 𝑥 𝐵 𝑘 𝐵 𝐴
−∆ 𝑇 𝑡𝑜𝑡 = −∆ 𝑇 𝐴 + −∆ 𝑇 𝐵

8. Series/Parallel Resistances
For parallel resistances (flat slab):
Driving Force at A:
Driving Force at B:
Overall driving force:
−∆ 𝑇 𝑡𝑜𝑡 = 𝑄 𝐴 𝑅 𝐴
−∆ 𝑇 𝑡𝑜𝑡 = 𝑄 𝐵 𝑅 𝐵
Actually, thickness is constant, so deltaXA = deltaXB and (-deltaTA) = (-deltaTB) = -(T2 – T1).
Heat flow rate is additive and resistances are reciprocal-additive. Compare these with ohm’s law.
𝑅 𝐴 = ∆ 𝑥 𝐴 𝑘 𝐴 𝐴
𝑅 𝐵 = ∆ 𝑥 𝐵 𝑘 𝐵 𝐴
−∆ 𝑇 𝑡𝑜𝑡 𝑅 𝑡𝑜𝑡 = 𝑄 𝐴 + 𝑄 𝐵

9. Series/Parallel Resistances
For parallel resistances (flat slab):
Substituting the individual Q:
−∆ 𝑇 𝑡𝑜𝑡 𝑅 𝑡𝑜𝑡 = −∆ 𝑇 𝑡𝑜𝑡 𝑅 𝐴 + −∆ 𝑇 𝑡𝑜𝑡 𝑅 𝐵
𝑄 𝑡𝑜𝑡 = 𝑄 𝐴 + 𝑄 𝐵
1 𝑅 𝑡𝑜𝑡 = 1 𝑅 𝐴 𝑅 𝐵
This is a summary of the equations for Parallel Conduction
𝑅 𝐴 = ∆ 𝑥 𝐴 𝑘 𝐴 𝐴
𝑅 𝐵 = ∆ 𝑥 𝐵 𝑘 𝐵 𝐴
−∆ 𝑇 𝑡𝑜𝑡 = −∆ 𝑇 𝐴 = −∆ 𝑇 𝐵

10. 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 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:
Between the cork and concrete?
Between the concrete and wood?
Answers: degF and degF (Source: PIChE Quiz Bowl 2009)
The time limit is just there to intimidate hehe. Try to draw temperature profile in the diagram. It’s just linearly dropping.
Explain why concrete had the least temperature drop because of low thermal resistance (deltax/kA).

11. 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.
Answer: 95 degC and 75 degC (Source: Gawa gawa ko lang)
We can compute that the temperature drop in A and D are both 15 degC. This is because materials A and D have the same thermal resistance.
Also, you can arrive at the same answer even if the thicknesses were reported in meters instead of inches. This is because the relative magnitudes of
the thermal resistances remain the same.

12. Geometric Considerations
Heat Conduction Through Concentric Cylinders
𝑄 𝐴 =−𝑘 𝑑𝑇 𝑑𝑟
𝐴=2𝜋𝑟𝐿
𝑄 2𝜋𝑟𝐿 𝑑𝑟=−𝑘 𝑑𝑇
Emphasize why dx became dr in cylindrical coordinates. The direction of heat flow is now radial.
Make the students guess the equation for area (A = 2*pi*r*L) before flashing the equation. This should be
the area perpendicular to the flow. You may also opt the students to integrate before flashing the final equation.
Integrating both sides:
ln 𝑟 2 𝑟 1 = −2𝜋𝑘𝐿 𝑄 𝑇 2 − 𝑇 1

13. Geometric Considerations
Heat Conduction Through Concentric Cylinders
𝑄 𝐴 =−𝑘 𝑑𝑇 𝑑𝑟
𝐴=2𝜋𝑟𝐿
𝑄 2𝜋𝑟𝐿 𝑑𝑟=−𝑘 𝑑𝑇
We arranged it like this so that we can introduce a new concept called the logarithmic mean area (next slide).
𝑄=−𝑘 2𝜋𝐿 ln (𝑟 2 𝑟 1 ) 𝑇 2 − 𝑇 1
Rearranging:

14. Geometric Considerations
Heat Conduction Through Concentric Cylinders
Define a logarithmic mean area:
𝑄=−𝑘 2𝜋𝐿 ln (𝑟 2 𝑟 1 ) 𝑇 2 − 𝑇 1
𝐴 𝐿𝑀 =2𝜋𝐿 ( 𝑟 2 − 𝑟 1 ) ln (𝑟 2 𝑟 1 )
…and a logarithmic mean radius:
The final form is made analogous to the form of Fourier’s Law for a flat slab.
The thermal resistance is now deltaR/kA(LM), analogous to deltaX/kA in flat slabs.
Good question: What if we want to solve for heat transfer starting from r1 = 0 to r2 > 0?
𝑄=−𝑘 𝐴 𝐿𝑀 𝑇 2 − 𝑇 1 𝑟 2 − 𝑟 1
𝑟 𝐿𝑀 = ( 𝑟 2 − 𝑟 1 ) ln (𝑟 2 𝑟 1 )
𝐴 𝐿𝑀 =2𝜋𝐿 𝑟 𝐿𝑀
*Final form

15. Geometric Considerations
Heat Conduction Through Concentric Cylinders
When to use logarithmic mean area and when to use arithmetic mean area?
r2/r1
rLM rM 1
#DIV/0!
1.05
1.025
1.1
1.15
1.075
1.2
1.25
1.125
1.3
1.35
1.175
1.4
𝐴 𝐿𝑀 =2𝜋𝐿 ( 𝑟 2 − 𝑟 1 ) ln (𝑟 2 𝑟 1 )
𝒓 𝑳𝑴
Sometimes we use arithmetic mean instead of logarithmic mean for ease of computation.
The table shows how close the values differ for each. For BWG pipes, the usual r2/r1 is 1.1 to 1.5.
In Geankoplis, he says that if A2/A1 < 1.5, then the arithmetic mean is within 1.5% error from logarithmic mean.
Therefore, we can use this rule of thumb in engineering practice. (But I prefer logarithmic mean all the time)
𝐴 𝑀 =2𝜋𝐿 ( 𝑟 2 + 𝑟 1 ) 2
𝒓 𝑴

16. 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 = 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).
Source: McCabe 6th Ed. Example 10.3
Answer: 29.1 W/m or 30.3 Btu/ft/hr