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
2 ft
3 ft
4 ft
k  T   20  0 . 5T
k 
350F
B tu
, T  F
o
h  ft  F
2
250F
o
1 in
TIME IS UP!!!
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
ONE-DIMENSIONAL ONLY
𝑄
𝑑𝑇
= −𝑘
𝐴
𝑑𝑥
Driving force*
−∆𝑇
𝑄=
∆𝑥
𝑘𝐴
Thermal
Resistanceǂ
For a constant cross-section and isotropic
thermal conductivity (e.g. a flat slab).
*∆𝑇 = 𝑇2 − 𝑇1 where T1> T2
ǂ∆𝑥 = 𝑥 − 𝑥 where x > x
2
1
2
1
Series/Parallel Resistances
For series resistances (flat slab):
Driving Force at A:
−∆𝑇𝐴 = 𝑄𝑡𝑜𝑡 𝑅𝐴
Driving Force at B:
−∆𝑇𝐵 = 𝑄𝑡𝑜𝑡 𝑅𝐵
Overall driving force:
∆𝑥𝐴
𝑅𝐴 =
𝑘𝐴 𝐴
∆𝑥𝐵
𝑅𝐵 =
𝑘𝐵 𝐴
−∆𝑇𝑡𝑜𝑡 = 𝑄𝑡𝑜𝑡 𝑅𝐴 + 𝑅𝐵
Series/Parallel Resistances
For series resistances (flat slab):
𝑄𝑡𝑜𝑡
−∆𝑇𝑡𝑜𝑡
=
∆𝑥𝐴 ∆𝑥𝐵
+
𝑘𝐴 𝐴 𝑘𝐵 𝐴
𝑄𝑡𝑜𝑡 = 𝑄𝐴 = 𝑄𝐵
𝑅𝑡𝑜𝑡 = 𝑅𝐴 + 𝑅𝐵
∆𝑥𝐴
𝑅𝐴 =
𝑘𝐴 𝐴
∆𝑥𝐵
𝑅𝐵 =
𝑘𝐵 𝐴
−∆𝑇𝑡𝑜𝑡 = −∆𝑇𝐴 + −∆𝑇𝐵
Series/Parallel Resistances
For parallel resistances (flat slab):
Driving Force at A:
−∆𝑇𝑡𝑜𝑡 = 𝑄𝐴 𝑅𝐴
Driving Force at B:
−∆𝑇𝑡𝑜𝑡 = 𝑄𝐵 𝑅𝐵
∆𝑥𝐴
𝑅𝐴 =
𝑘𝐴 𝐴
∆𝑥𝐵
𝑅𝐵 =
𝑘𝐵 𝐴
Overall driving force:
−∆𝑇𝑡𝑜𝑡
= 𝑄𝐴 + 𝑄𝐵
𝑅𝑡𝑜𝑡
Series/Parallel Resistances
For parallel resistances (flat slab):
Substituting the individual Q:
−∆𝑇𝑡𝑜𝑡
−∆𝑇𝑡𝑜𝑡
−∆𝑇𝑡𝑜𝑡
=
+
𝑅𝑡𝑜𝑡
𝑅𝐴
𝑅𝐵
𝑄𝑡𝑜𝑡 = 𝑄𝐴 + 𝑄𝐵
∆𝑥𝐴
𝑅𝐴 =
𝑘𝐴 𝐴
∆𝑥𝐵
𝑅𝐵 =
𝑘𝐵 𝐴
1
1
1
=
+
𝑅𝑡𝑜𝑡 𝑅𝐴 𝑅𝐵
−∆𝑇𝑡𝑜𝑡 = −∆𝑇𝐴 = −∆𝑇𝐵
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
𝑄
𝑑𝑇
= −𝑘
𝐴
𝑑𝑟
𝐴 = 2𝜋𝑟𝐿
𝑄
𝑑𝑟 = −𝑘
2𝜋𝑟𝐿
Integrating
both sides:
𝑟2
−2𝜋𝑘𝐿
ln
=
𝑇2 − 𝑇1
𝑟1
𝑄
𝑑𝑇
Geometric Considerations
Heat Conduction Through Concentric Cylinders
𝑄
𝑑𝑇
= −𝑘
𝐴
𝑑𝑟
𝐴 = 2𝜋𝑟𝐿
𝑄
𝑑𝑟 = −𝑘
2𝜋𝑟𝐿
Rearranging:
2𝜋𝐿
𝑄 = −𝑘
𝑇2 − 𝑇1
ln (𝑟2 𝑟1 )
𝑑𝑇
Geometric Considerations
Heat Conduction Through Concentric Cylinders
Define a logarithmic mean area:
𝐴𝐿𝑀
(𝑟2 − 𝑟1 )
= 2𝜋𝐿
ln (𝑟2 𝑟1 )
2𝜋𝐿
𝑄 = −𝑘
𝑇2 − 𝑇1
ln (𝑟2 𝑟1 )
…and a logarithmic mean radius:
𝑟𝐿𝑀
(𝑟2 − 𝑟1 )
=
ln (𝑟2 𝑟1 )
𝐴𝐿𝑀 = 2𝜋𝐿𝑟𝐿𝑀
𝑄 = −𝑘𝐴𝐿𝑀
*Final form
𝑇2 − 𝑇1
𝑟2 − 𝑟1
Geometric Considerations
Heat Conduction Through Concentric Cylinders
When to use logarithmic mean area and
when to use arithmetic mean area?
r2/r1
𝐴𝐿𝑀
(𝑟2 − 𝑟1 )
= 2𝜋𝐿
ln (𝑟2 𝑟1 ) 𝒓
𝑳𝑴
(𝑟2 + 𝑟1 )
𝐴𝑀 = 2𝜋𝐿
2
𝒓𝑴
rLM
rM
1
#DIV/0!
1
1.05
1.024797
1.025
1.1
1.049206
1.05
1.15
1.073254
1.075
1.2
1.096963
1.1
1.25
1.120355
1.125
1.3
1.143448
1.15
1.35
1.16626
1.175
1.4
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).