### Document

```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

= −

Driving force*
−∆
=
∆

Thermal
Resistanceǂ
For a constant cross-section and isotropic
thermal conductivity (e.g. a flat slab).
*∆ = 2 − 1 where T1> T2
ǂ∆ =  −  where x > x
2
1
2
1
Series/Parallel Resistances
For series resistances (flat slab):
Driving Force at A:
−∆ =
Driving Force at B:
−∆ =
Overall driving force:
∆
=

∆
=

−∆ =   +
Series/Parallel Resistances
For series resistances (flat slab):

−∆
=
∆ ∆
+

=  =
=  +
∆
=

∆
=

−∆ = −∆ + −∆
Series/Parallel Resistances
For parallel resistances (flat slab):
Driving Force at A:
−∆ =
Driving Force at B:
−∆ =
∆
=

∆
=

Overall driving force:
−∆
=  +

Series/Parallel Resistances
For parallel resistances (flat slab):
Substituting the individual Q:
−∆
−∆
−∆
=
+

=  +
∆
=

∆
=

1
1
1
=
+

−∆ = −∆ = −∆
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

= −

= 2

= −
2
Integrating
both sides:
2
−2
ln
=
2 − 1
1

Geometric Considerations
Heat Conduction Through Concentric Cylinders

= −

= 2

= −
2
Rearranging:
2
= −
2 − 1
ln (2 1 )

Geometric Considerations
Heat Conduction Through Concentric Cylinders
Define a logarithmic mean area:

(2 − 1 )
= 2
ln (2 1 )
2
= −
2 − 1
ln (2 1 )
…and a logarithmic mean radius:

(2 − 1 )
=
ln (2 1 )
= 2
= −
*Final form
2 − 1
2 − 1
Geometric Considerations
Heat Conduction Through Concentric Cylinders
When to use logarithmic mean area and
when to use arithmetic mean area?
r2/r1

(2 − 1 )
= 2
ln (2 1 )

(2 + 1 )
= 2
2

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).
```