ChE 126

SEPARATION PROCESSES

HEAT & MASS TRANSFER

CONDUCTION

Conduction Thru a Hollow Sphere

- another case of one-dimensional conduction

𝑞 𝑘𝑑𝑇
=− ; 𝐴 = 4𝜋𝑟 2
𝐴 𝑑𝑟
𝑞 𝑟2 𝑑𝑟 𝑇2
∫ = −𝑘 ∫ 𝑑𝑇
4𝜋 𝑟1 𝑟 2 𝑇1
𝑞 1 1
[ − ] = 𝑘 (𝑇1 − 𝑇2 )
4𝜋 𝑟1 𝑟2
4𝜋𝑘 ( 𝑇1 − 𝑇2)
𝑞=
1 1
𝑟1 − 𝑟2

𝐷𝐹 ( 𝑇1 − 𝑇2 )
𝑞= =
𝑅 1 1
( − )
𝑟1 𝑟2
4𝜋𝑘

Conduction Thru Solids in Series – in the case where there is a multilayer wall of more
than one material present

T4

- Heat flow q must be the same in each layer, hence the Fourier’s equation for
each layer is written as:

𝑞 𝑘𝐴 (𝑇1 − 𝑇2 ) 𝑘𝐴 𝐴(𝑇1 − 𝑇2 )
𝐴: = ⇒ 𝑞=
𝐴 𝑥2 − 𝑥1 ∆𝑥𝐴

Page 1 of 13
 𝑞 𝑘𝐵 (𝑇1 − 𝑇2 ) 𝑘𝐵 𝐴(𝑇1 − 𝑇2 )
𝐵: = ⇒ 𝑞=
𝐴 𝑥3 − 𝑥2 ∆𝑥𝐵
𝑞 𝑘𝐶 (𝑇1 − 𝑇2) 𝑘𝐶 𝐴(𝑇1 − 𝑇2 )
𝐶: = ⇒ 𝑞=
𝐴 𝑥4 − 𝑥3 ∆𝑥𝐶
𝐷𝐹
𝑟𝑎𝑡𝑒 =
𝑅

𝑇1 − 𝑇2 𝑇2 − 𝑇3 𝑇3 − 𝑇4
𝑞= = =
∆𝑥𝐴 ∆𝑥𝐵 ∆𝑥𝑐
𝑘𝐴 𝐴 𝑘𝐵 𝐴 𝑘𝑐 𝐴

OR solving for ΔT in each:

1 2 3
𝑞∆𝑥𝐴 𝑞∆𝑥𝐵 𝑞∆𝑥𝐶
𝑇1 − 𝑇2 = 𝑇2 − 𝑇3 = 𝑇3 − 𝑇4 =
𝑘𝐴 𝐴 𝑘𝐵 𝐴 𝑘𝐶 𝐴

Combining 1, 2, 3:
𝑞 ∆𝑥𝐴 ∆𝑥𝐵 ∆𝑥𝐶
𝑇1 − 𝑇2 ; 𝑇2 − 𝑇3 ; 𝑎𝑛𝑑 𝑇3 − 𝑇4 = ( + + )
𝐴 𝑘𝐴 𝑘𝑏 𝑘𝐶
∆𝑥𝐴 ∆𝑥𝐵 ∆𝑥𝑐
𝑇1 − 𝑇4 = 𝑞 ( + + )
𝑘𝐴 𝐴 𝑘𝐵 𝐴 𝑘𝑐 𝐴
Solving for q;
𝑇1 − 𝑇4
𝑞=
∆𝑥𝐴 ∆𝑥𝐵 ∆𝑥𝑐
(𝑘 𝐴 + 𝑘 𝐴 + 𝑘 𝐴)
𝐴 𝐵 𝑐

𝑇1 − 𝑇4
𝑞=
𝑅𝐴 + 𝑅𝐵 + 𝑅𝐶

Conduction Thru Multi-Layer Cylinders

- In the process industries, heat transfer often occurs through multilayers of
cylinders, as for example when heat is being transferred through the walls of
an insulated pipe.
- Consider the figure below showing a pipe with two layers of insulation around
it, that is, a total of three concentric hollow cylinders.

q is constant; and at steadystate (ss) process

Page 2 of 13
 𝑘𝐴 𝐴𝐴𝑙𝑚 (𝑇1 − 𝑇2)
𝑞= ⇒ for single layer
𝑟2 − 𝑟1
𝑘𝐵 𝐴𝐵𝑙𝑚 (𝑇2 − 𝑇3 )
=
𝑟3 − 𝑟2
𝑘𝐶 𝐴𝐶𝑙𝑚 (𝑇3 − 𝑇4 )
=
𝑟4 − 𝑟3

(𝐴 −𝐴1 ) (𝐴 −𝐴2 ) (𝐴 −𝐴3 )

Where: 𝐴𝐴𝑙𝑚 = 𝑙𝑛(𝐴2 𝐴𝐵𝑙𝑚 = 𝑙𝑛(𝐴3 𝐴𝐶𝑙𝑚 = 𝑙𝑛(𝐴4
2 /𝐴1 ) 3 /𝐴2 ) 4 /𝐴3 )

Eliminating 𝑇2 𝑎𝑛𝑑 𝑇3 :
𝑇1 − 𝑇4
𝑞=
(𝑟2 − 𝑟1 ) (𝑟3 − 𝑟2 ) (𝑟4 − 𝑟3 )
𝑘𝐴 𝐴𝐴𝑙𝑚 + 𝑘𝐵 𝐴𝐵𝑙𝑚 + 𝑘𝐶 𝐴𝐶𝑙𝑚

𝑇1 − 𝑇4 𝑇1 − 𝑇4
𝑞= =
𝑅𝐴 + 𝑅𝐵 + 𝑅𝐶 ∑𝑅
Example 1:

𝑊
A thick-walled tube of stainless steel (A) having a 𝑘 = 21.62 𝑚−𝐾 with dimensions
of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254-m layer of asbestos (B)
𝑊
insulation, 𝑘 = 0.2423 𝑚−𝐾. The inside wall temperature of the pipe is 811 K and
the outside surface of the insulation is at 310.8 K. For a 0.305-m length of pipe,
calculate the heat loss and also the temperature at the interface between the metal
and the insulation.

Answer: 𝑞 = 331.7 𝑊; 𝑇2 = 805.5 𝐾

Conduction Thru Materials in Parallel

- Suppose that two plane solids A and B are placed side by side in parallel and
the direction heat flow is perpendicular to the plane of the exposed surface of
each solid. Then the total heat flow is the sum of the heat flow through solid A
plus that through B.
- Writing Fourier’s equation for each solid and summing,

𝑘𝐴 𝐴𝐴 𝑘 𝐴 𝑇2 𝑇4
𝑞𝑇 = 𝑞𝐴 + 𝑞𝐵 = (𝑇1 − 𝑇2 ) + 𝐵 𝐵 (𝑇3 − 𝑇4 )
∆𝑥𝐴 ∆𝑥𝐵

𝑇1 = 𝑇3 ; 𝑇2 = 𝑇4

equal rear temperatures 𝑇1 A 𝑇3 B

front temperatures are the same for A and B

𝑇 −𝑇 𝑇 −𝑇 1 1
𝑞𝑇 = ∆𝑥 1/𝑘 2𝐴 + ∆𝑥 1/𝑘 2𝐴 = (𝑅 + 𝑅 )(𝑇1 − 𝑇2 )
𝐴 𝐴 𝐴 𝐵 𝐵 𝐵 𝐴 𝐵

Page 3 of 13
Examples:
1) Insulated wall (A) of a brick oven where steel reinforcing members (B) are in
parallel and penetrate the wall. 𝐴𝐵 < 𝐴𝐴 but 𝑘𝐵 ≫ 𝑘𝐴 ; allows for a large portion of
the heat lost to be conducted by the steel.

2) Increasing heat conduction to accelerate the freeze drying of meat. Spikes of metal
in the frozen meat conduct heat more rapidly into the insides of the meat.

Conduction with Internal Heat Generation

- in certain systems heat is generated inside the conducting medium if a uniform
heat source is present.
Example: Electric resistance heaters and nuclear fuels

- also, if a chemical reaction is occuring uniformly in a medium heat of reaction

is given off.

- in agricultural and sanitation fields, compost heaps and trash heaps in which
biological activity is occurring will have heat given off.

- Food processing, heat of respiration of fresh fruits and vegetables is present

𝑊 𝐵𝑇𝑈
⇒ 0.3 𝑡𝑜 0.6 𝑘𝑔 𝑜𝑟 0.5 𝑡𝑜 1 ℎ−𝑙𝑏
𝑚

Example: Heat Removal of a Cooling Coil

A cooling coil of 1 ft of 304 stainless steel tubing having an inside diameter of 0.25
in (0.0208 ft) and an outside diameter of 0.40 in (0.0333 ft) is being used to remove
heat from a bath. The temperature of the inside surface of the tube is 40℉ & 80℉
on the outside. The conductivity of 304 stainless steel is a function of temperature,
𝑘 = 7.75 + 7.78 × 10−3 𝑇
𝐵𝑇𝑈
where k is in ℎ−𝑓𝑡−℉ and T is in ℉. Calculate the heat removed in BTU/s & Watts.
𝐵𝑇𝑈
Answer: 1.225 ; 1292 𝑊
𝑠
Solution:

Use 𝑘𝑚 instead of 𝑘:
𝑇1 +𝑇2
𝑘𝑚 = 𝑎 + 𝑏 ( ) (4.2-3)
2

40 + 80
𝑘𝑚 = 7.75 + 0.00778 ( )
2
𝐵𝑇𝑈
𝑘𝑚 = 7.75 + 0.00778(60) = 8.2168
ℎ−𝑓𝑡−℉

𝑘𝑚 𝐴𝑙𝑚 (𝑇1 −𝑇2 )

𝑞= 𝑟2 −𝑟1
(4.2-9)

Page 4 of 13
 𝑟2 −𝑟1 (0.0333−0.0208)𝑓𝑡
𝐴𝑙𝑚 = 2𝜋𝐿 𝑟 = 2𝜋(1𝑓𝑡) = 0.1671 𝑓𝑡 2 (4.2-10)
ln( 2) ln(0.0208)
𝑟1

𝐵𝑇𝑈
8.2168 (0.1671 𝑓𝑡 2 )(40℉ − 80℉) 𝐵𝑇𝑈
ℎ − 𝑓𝑡 − ℉
𝑞= = 4,393.81
0.0333 ft − 0.0208 ft ℎ

𝑩𝑻𝑼 𝑱
𝒒 = 𝟏. 𝟐𝟐𝟓 = 𝟏, 𝟐𝟖𝟕 (𝑾)
𝒔 𝒔

Heat Generation in Plane Wall

At steady state 1-D flow (x-direction)

L L

−𝑥 𝑥

𝑞̇ (generated)
Conduction thru x-direction only, other walls are insulated;
𝑊
𝑞̇ = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛
𝑚3

Derivation:
From general property balance equation:

𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 + 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦

= 𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 𝑜𝑢𝑡 + 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦

𝜕𝑇
𝑞𝑥 + 𝑞̇ (∆𝑥 ⋅ 𝐴) = 𝑞𝑥|𝑥+∆𝑥 + 𝜌𝐶𝑝 (∆𝑥 ⋅ 𝐴)
𝜕𝑡
𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 0
𝑞𝑥|𝑥 + 𝑞̇ (∆𝑥 ⋅ 𝐴) = 𝑞𝑥|𝑥+∆𝑥
↑
cross-sectional area of the plate

A 𝑞̇ (∆𝑥 ⋅ 𝐴) = 𝑞𝑥|𝑥+∆𝑥 − 𝑞𝑥|𝑥

𝑞̇ (∆𝑥 ⋅ 𝐴) (𝑞𝑥|𝑥+∆𝑥 − 𝑞𝑥|𝑥 )
=
∆𝑥 ∆𝑥
𝑞𝑥|𝑥+∆𝑥 − 𝑞𝑥|𝑥
𝑞̇ (𝐴) = lim
𝑥→0 ∆𝑥
𝑥 ← ∆𝑥 → 𝑥 + ∆𝑥 𝜕𝑞𝑥
𝑞̇ (𝐴) =
𝜕𝑥
𝜕𝑞𝑥
− + 𝑞̇ (𝐴) = 0
𝜕𝑥
Page 5 of 13
 Recall:
𝑞𝑥 𝑑𝑇 𝑑𝑇
= −𝑘 ; 𝑞𝑥 = −𝑘𝐴
𝐴 𝑑𝑥 𝑑𝑥
𝑑𝑇
𝑑 (−𝑘𝐴 )
𝑑𝑥
− + 𝑞̇ ⋅ 𝐴 = 0
𝑑𝑥
𝑘𝐴𝑑2 𝑇
[ + 𝑞̇ ⋅ 𝐴 = 0] divide both sides by kA
𝑑𝑥 2
𝑑 2 𝑇 𝑞̇
+ =0
𝑑𝑥 2 𝑘
𝑑2𝑇 𝑞̇
∫ 2 = ∫−
𝑑𝑥 𝑘
2
𝑑 𝑇 𝑞̇
∫ 2 == ∫ − + 𝐶1
𝑑𝑥 𝑘
𝑞̇ 𝑥 2
𝑇=− + 𝐶1 𝑥 + 𝐶2
2𝑘

𝐶1 &𝐶2 ⇒ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠; boundary conditions are 𝑥 = 𝐿 𝑜𝑟 − 𝐿, 𝑇 =

𝑇𝑊 , 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 0, 𝑇 = 𝑇0 (𝑐𝑒𝑛𝑡𝑒𝑟 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒)

Then the temperature profile is:

𝑞̇ 𝑥 2 𝑞̇ 𝐿2
𝑇=− + 𝑇0 𝑇0 = 𝑇𝑊 +
2𝑘 2𝑘

𝑞̇ 𝑥 2
𝑇=− + 𝐶1 𝑥 + 𝐶2
2𝑘
At x = 0, T = T0
𝑇0 = 0 + 0 + 𝐶2
𝑞̇ 𝑥 2
𝑇=− + 𝐶1 𝑥 + 𝑇0
2𝑘

At x = L or − L; T = TW
𝑞̇ 𝐿2
𝑇𝑊 = − + 𝐶1 𝐿 + 𝑇0
2𝑘
𝑞̇ 𝐿2
𝑇𝑊 − 𝑇0 = − + 𝐶1 𝐿
2𝑘
𝑞̇ 𝐿2
𝑇𝑊 − 𝑇0 + = 𝐶1
2𝑘
𝐿
𝑇𝑊 − 𝑇0 𝑞̇ 𝐿
𝐶1 = +
𝐿 2𝑘
Thus;

𝑞̇ 𝑥 2 𝑇𝑊 − 𝑇0 𝑞̇ 𝐿 𝑞̇ 𝑥 2 𝑇𝑊 − 𝑇0 𝑞̇ 𝐿
𝑇=− +( + ) 𝐿 + 𝑇0 𝑇=− +( + ) 𝑥 + 𝑇0
2𝑘 𝐿 2𝑘 2𝑘 𝐿 2𝑘
𝑞̇ 𝑥 2 𝑞̇ 𝐿2
𝑇=− + 𝑇𝑊 − 𝑇0 + + 𝑇0
2𝑘 2𝑘

Page 6 of 13
 𝑞̇ 𝑥 2 𝑞̇ 𝐿2
𝑇=− + 𝑇𝑊 +
2𝑘 2𝑘

↑
𝑇0

Total heat lost from the two faces at steady-state is equivalent to the total heat
generated, 𝑞𝑇 , in W:
𝑞̇ 𝑇 = 𝑞̇ (2𝐿𝐴)
where A is the surface area at wall temperature (T W) of the plate.
𝑞̇ 𝑇 = 𝑞̇ (∆𝑥 ⋅ 𝐴)

Heat Generation in Cylinder

- An equation can be derived for a cylinder of radius R with uniformly distributed

heat sources and constant thermal conductivity. The heat is assumed to flow
only radially (the ends are neglected or insulated).

𝑞̇
𝑇= (𝑅2 − 𝑟 2 ) + 𝑇𝑊
4𝑘
R
r
𝑞̇ 𝑅2
𝑇0 = + 𝑇𝑊
4𝑘

Example:

An electric current of 200 A is passed through a stainless steel wire having a radius
R of 0.001268 m. The wire is L= 0.91 m long and has a resistance R of 0.126Ω.
The outer surface temperature 𝑇𝑊 is held at 422.1 K. the average thermal
𝑊
conductivity is 𝑘 = 22.5 𝑚⋅𝐾. Calculate the center temperature.

Solution:

To get 𝑇0, get 𝑞̇ first.

Recall: 𝑃𝑜𝑤𝑒𝑟 (𝑊𝑎𝑡𝑡𝑠) = 𝐼 2 𝑅 = 𝑞̇ (∆𝑥 ⋅ 𝐴)(𝑟𝑎𝑡𝑒 𝑜𝑓 ℎ𝑒𝑎𝑡 ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛)

𝑞̇ (𝑅 ⋅ 2𝜋𝑅𝐿)
← ∆𝑥 → 𝐼2𝑅 =
2

𝐼 2 𝑅 = 𝑞̇ 𝜋𝑅2 𝐿 (4.3-31)

𝐼2 𝑅𝑟𝑠 𝑊
𝑞̇ = 𝜋𝑅2 = 1.096 × 109 𝑚3 in (4.3-30)
𝑤𝑖𝑟𝑒 𝐿

𝑞̇ 𝑅2
𝑇0 = + 𝑇𝑊
4𝑘

Page 7 of 13
 𝑊
1.096 × 109 (0.001268 m2 )
𝑇0 = 𝑚3 + 422.1 K
𝑊
4(22.5 𝑚 ⋅ 𝐾 )

𝑇0 = 441.7 𝐾

Convection
- transfer of heat by bulk transport and mixing of macroscopic elements of
warmer portions with cooler portions of gas or a liquid.
- It also often involves the energy exchange between a solid surface and a liquid

Two Types of Convection

1. Natural or free convection

- a warmer or cooler fluid next to the solid surface causes a circulation
because of a density difference resulting from the temperature
differences in the fluid

2. Forced convection
- a fluid is forced to flow past a solid surface by a pump, fan or other
mechanical means.
- ex. Loss of heat from a car radiator where the air is being circulated by
fan; Cooking of foods in a vessel being stirred and; Cooling a hot cup of
coffee by blowing over the surface

Convective Heat-Transfer Coefficient

Newton’s Law of Cooling

- rate of heat transfers from the solid to fluid, or vice versa whichever is hotter or
cooler.
𝑞 = ℎ𝐴(𝑇𝑊 − 𝑇𝑓 ) ⇒ 𝑞 = ℎ𝐴∆𝑇

rate of heat transfer from the solid to the fluid or

vice-versa whichever is hotter or cooler
where:
𝑞 = ℎ𝑒𝑎𝑡 − 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑟𝑎𝑡𝑒 𝑖𝑛 𝑊
𝐴 = 𝑎𝑟𝑒𝑎 𝑖𝑛 𝑚2
𝑇𝑊 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑖𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑖𝑛 𝐾
𝑇𝑓 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑟 𝑏𝑢𝑙𝑘 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑓𝑙𝑜𝑤𝑖𝑛𝑔 𝑖𝑛 𝐾
𝑊
ℎ = 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 − 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑖𝑛 2
𝑚 ⋅𝐾
𝐵𝑇𝑈
𝑜𝑟
ℎ𝑟 ⋅ 𝑓𝑡 2 ⋅ ℉

ℎ = 𝑓(𝑠𝑦𝑠𝑡𝑒𝑚 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦, 𝑓𝑙𝑢𝑖𝑑 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠, 𝑓𝑙𝑜𝑤 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑡𝑒𝑚𝑝. 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒)

Page 8 of 13
 In many cases, empirical correlations are available to predict the coefficient, since
it cannot be predicted theoretically.

film (thin, almost stationary, offers

resistance to heat transfer)

h→ film coefficient (see table)

Water gives the highest values of h.

English to SI Unit
𝑊
1𝐵𝑇𝑈/(𝑘 ⋅ 𝑓𝑡 2 ⋅ ℉ = 5.5783
𝑚2 ⋅𝐾

Approximate Values/
Magnitude of Some Heat-
Transfer Coefficient
(Table 4.1-2 in Geankoplis)

Combined Convection and Conduction and Over-All Coefficient

Consider:

ℎ𝑖 − 𝑖𝑛𝑠𝑖𝑑𝑒 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

ℎ𝑜 − 𝑜𝑢𝑡𝑠𝑖𝑑𝑒

The heat transfer rate, 𝑞, for

steady-state process:
𝑘𝐴 𝐴
𝑞 = ℎ𝑖 𝐴(𝑇1 − 𝑇2 ) = (𝑇 − 𝑇2)
∆𝑥𝐴 1
𝑞 = ℎ𝑜 𝐴(𝑇1 − 𝑇2 )

𝐷𝑟𝑖𝑣𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒
In terms of 𝑟𝑎𝑡𝑒 = 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

(𝑇1 − 𝑇2 ) (𝑇2 − 𝑇3 ) (𝑇3 − 𝑇4 )

𝑞= = =
1 ∆𝑥𝐴 1
ℎ𝑖 𝐴 𝑘𝐴 𝐴 ℎ0 𝐴
Combining;
(𝑻𝟏 −𝑻𝟒 ) 𝑻𝟏 −𝑻𝟒
𝒒= 𝟏 ∆𝒙 𝟏 =
+ 𝑨+
𝒉𝟏 𝑨 𝒌𝑨 𝑨 𝒉𝟎 𝑨
∑𝑹

Page 9 of 13
 Over-all heat-transfer by combined convection and conduction is often
expressed in terms of an Over-All Heat Transfer Coefficient, U;

𝑞 = 𝑈𝐴∆𝑇
where ∆𝑇:

∆𝑇𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = (𝑇1 − 𝑇4 )

and U is:
1 𝑊 𝐵𝑇𝑈
𝑈= 𝑖𝑛 𝑜𝑟
1 ∆𝑥𝐴 1 𝑚2⋅𝐾 ℎ𝑟 ⋅ 𝑓𝑡 2 ⋅ ℉
ℎ1 𝐴 + 𝑘𝐴 𝐴 + ℎ0 𝐴

Combined Conduction and Convection in Cylinder

Using the same procedure;

𝑘𝐴 𝐴𝑙𝑚𝐴 (𝑇2 − 𝑇3)
𝑞 = ℎ𝑜 𝐴0 (𝑇1 − 𝑇2 ) =
𝑟0 − 𝑟𝑖
𝑞 = ℎ𝑖 𝐴𝑖 𝑇3 − 𝑇4 )
(
where;
𝐴0 = 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 (2𝜋𝐿𝑟0 )
𝐴𝑖 = 𝑖𝑛𝑠𝑖𝑑𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝐴 (2𝜋𝐿𝑟𝑖 )
𝐴𝑙𝑚𝐴 = log 𝑚𝑒𝑎𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝐴
= 2𝜋𝐿𝑟𝑙𝑚
2𝜋𝐿(𝑟0−𝑟𝑖 )
= 𝑟
ln 0
𝑟𝑖

Combining;
(𝑇1 − 𝑇2) (𝑇2 − 𝑇3 ) (𝑇3 − 𝑇4 )
𝑞= + +
1 (𝑟0 − 𝑟𝑖 ) 1
ℎ0 𝐴0 𝑘𝐴 𝐴𝑙𝑚𝐴 ℎ𝑖 𝐴𝑖

(𝑇1 − 𝑇4 )
𝑞=
1 (𝑟0 − 𝑟𝑖 ) 1
ℎ0 𝐴0 + 𝑘𝐴 𝐴𝑙𝑚𝐴 + ℎ𝑖 𝐴𝑖

𝑻𝟏 − 𝑻𝟒
𝒒=
∑𝑹

Page 10 of 13
 U BASED ON 𝑨𝒊 = 𝑼𝒊 U BASED ON 𝑨𝒐 = 𝑼𝟎
𝒒 = 𝑼𝒊 𝑨𝒊 (𝑻𝟏 − 𝑻𝟒 ) 𝑞 = 𝑈0 𝐴0 (𝑇1 − 𝑇4 )
𝟏 𝟏 1 1
= 𝑼𝒊 = = 𝑈0 =
𝑨𝒊 ∑𝑹 𝟏 ( 𝒓 − 𝒓𝒊 )𝑨𝒊 𝑨𝒊 𝐴0 ∑𝑅 𝐴0 ( 𝑟0 − 𝑟𝑖 )𝐴0 1
+ 𝟎 + +
𝒉𝒊 𝒌𝑨 𝑨𝒍𝒎𝑨 𝑨 𝟎 𝒉𝟎 ℎ𝑖 𝐴𝑖 𝑘𝐴 𝐴𝑙𝑚𝐴 + ℎ0

Over-all heat transfer coefficient maybe based on the inside area or outside area
of the tube;
𝑇1 − 𝑇4
𝑞 = 𝑈𝑖 𝐴𝑖 (𝑇1 − 𝑇4 ) = 𝑈0 𝐴0 (𝑇1 − 𝑇4 ) =
∑𝑅

Example:
3
Saturated steam at 267℉ is flowing inside a 4 𝑖𝑛. steel pipe having an ID of 0.824
in. and OD of 1.050 in. The pipe is insulated with 1.5 in of insulation on the outside.
The convective coefficient for the inside steam surface of the pipe is estimated as
𝐵𝑇𝑈
ℎ𝑖 = 1,000 2 , and the convective coefficient on the outside of the lagging is
ℎ𝑟.𝑓𝑡 ⋅℉
𝐵𝑇𝑈 𝑊
estimated as ℎ0 = 2.0 ℎ⋅𝑓𝑡 2 ⋅℉. The mean thermal conductivity of the metal is 45 𝑚⋅𝐾
𝐵𝑇𝑈 𝑊 𝐵𝑇𝑈
or 26 and 0.064 or 0.037 for insulation.
ℎ⋅𝑓𝑡 2 ⋅℉ 𝑚⋅𝐾 ℎ⋅𝑓𝑡 2 ⋅℉

(a) Calculate the heat loss for 1 ft of pipe using resistances if the surrounding
air is at 80℉.
(b) Repeat using the over-all UI based on the inside area of the pipe.

𝐴 − 𝑚𝑒𝑡𝑎𝑙 𝐵 − 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛
𝐵𝑇𝑈
ℎ𝑖 = 1,000
ℎ𝑟 ⋅ 𝑓𝑡 2 ⋅ ℉
𝐵𝑇𝑈
ℎ0 = 2.0
ℎ𝑟 ⋅ 𝑓𝑡 2 ⋅ ℉

𝑊 𝐵𝑇𝑈
𝑘𝐴 = 45 or 26
𝑚⋅𝐾 ℎ ⋅ 𝑓𝑡 2 ⋅ ℉

𝑊 𝐵𝑇𝑈
𝑘𝐵 = 0.064 or 0.037
𝑚⋅𝐾 ℎ ⋅ 𝑓𝑡 2 ⋅ ℉

(𝑇1 − 𝑇4 )
𝑞=
1 (𝑟0 − 𝑟𝑖 )𝐴 (𝑟0 − 𝑟𝑖 )𝐵 1
+
ℎ𝑖 𝐴𝑖 𝑘𝐴 𝐴𝑙𝑚𝐴 + 𝑘𝐵 𝐴𝑙𝑚𝐵 + ℎ0 𝐴0

𝐿 = 1 𝑓𝑡 0.824 1
𝐴𝑖 = 2𝜋𝐿 ( ) ( ) = 0.2157 𝑓𝑡 2
2 12
1.050 1
𝐴𝐴 = 2𝜋𝐿 ( ) ( ) = 0.2750 𝑓𝑡 2
2 12
2.025 1
𝐴𝐵 = 2𝜋𝐿 ( ) ( ) = 1.060 𝑓𝑡 2
2 12

𝐴𝐵 = 𝐴 𝑜

Page 11 of 13
 2𝜋𝐿(𝑟0 − 𝑟𝑖 ) 2𝜋𝐿(0.04375 − 0.03433)
𝐴𝑙𝑚𝐴 = 𝑟 = = 0.245
ln 𝑟0 0.04375
𝑖
ln (0.03433 )

2𝜋𝐿(0.16875 − 0.04375)
𝐴𝑙𝑚 𝐵 = = 0.583
0.16875
ln (0.04375)

𝑻𝟏 − 𝑻𝟒
𝒒= ;
𝑹 𝑨 + 𝑹 𝑩 + 𝑹 𝑪 + 𝑹𝑫
1 1
𝑅𝑖 = = = 0.00464
ℎ𝑖 𝐴𝑖 (1000)(0.2157)

1
(𝑟0 − 𝑟𝑖 )𝐴 (0.525 − 0.412) (12)
𝑅𝐴 = = = 0.00148
𝑘𝐴 𝐴𝑙𝑚𝐴 (26)(0.245)

1
(𝑟0 − 𝑟𝑖 )𝐵 (2.025 − 0.525) (12)
𝑅𝐵 = = = 5.80
𝑘𝐵 𝐴𝑙𝑚𝐵 (0.037)(0.583)

1 1
𝑅0 = = = 0.472
ℎ0 𝐴0 2(1.060)

𝑇𝑖 − 𝑇0
𝑞=
𝑅𝑖 + 𝑅𝐴 + 𝑅𝐵 + 𝑅𝑜

267 − 80 267 − 80
𝑞= =
0.00464 + 0.00148 + 5.80 + 0.472 6.278

𝐵𝑇𝑈
𝑞 = 29.8
ℎ

𝒒 = 𝑼𝒊 𝑨𝒊 (𝑻𝟏 − 𝑻𝟎 );
1 1 𝐵𝑇𝑈
𝑈𝑖 = = = 0.738
𝐴𝑖 ∑𝑅 0.2157(6.278) ℎ𝑟 ⋅ 𝑓𝑡 2 ⋅ ℉

𝒒 = 0.738(0.2157)(267 − 80)

𝐵𝑇𝑈
𝑞 = 29.8 = 8.73 𝑊
ℎ𝑟

Problems to be solved.

𝐵𝑇𝑈
An exterior wall’s insulation is made up of the following materials: 2” cork (𝐾 = 0.025 ℎ𝑟−𝑓𝑡−°𝐹 ; ), 6”
𝐵𝑇𝑈 𝐵𝑇𝑈
concrete (𝐾 = 0.98 ) and 3” wood (𝐾 = 0.065 ). Approximate the heat transferred per unit
ℎ𝑟−𝑓𝑡−°𝐹 ℎ𝑟−𝑓𝑡−°𝐹
area if the inside wall temperature and outside surface temperature are 40°F and 80°F, respectively.
What is the temperature between the cork and concrete (T 2)?

Page 12 of 13
Two concentric stainless steel pipes, with the inner pipe having a diameter of 3” and the outer pipe having
𝑊
a thickness of 1.5” have a 𝑘 = 389.8 𝑚−𝐾. The inside wall temperature of the pipe is 300°C and the outside
surface temperature is at 100°C. For a foot length of pipe, calculate the heat loss.

4.3-3

4.3-8

4.3-10

Page 13 of 13