HEAT TRANSFER
The heating and cooling of liquid and solids, condensation of vapours and the removal of heat liberated by chemical reactions are
common examples which involve heat transfer. It is important to an industrial chemist to understand the practical aspects of heat
transfer as well as the basic laws governing this operation.
The unit operation of heat transfer is usually one component, a part of the overall process and the inter-relations among the
different operations involved must be recognized.
The driving force for heat flow is the temperature difference between the points, where heat is received and where heat originates.
Heat tend to flow from a point of high temperature to a point of low temperature.
There are several methods of heat transfer:
• conduction in solids, transfer through vibrations
• convection in fluids through physical movement
• radiation through emission of radiant energy
HEAT TRANSFER BY CONDUCTION: FOURIER’S LAW
The instantaneous rate of heat transfer through a homogenous body by conduction is directly proportional to the temperature
difference driving force across the body, to the cross sectional area of the body through which the heat flows and inversely
proportional to the thickness of the body.
dQ
dT
= −kA
dt
dx
where dQ/dt = instantaneous rate of heat transfer, A = cross sectional area of body perpendicular to the direction of heat flow,
dT/dx = rate of change of temperature, T with respect to the length of the heat transfer flow path, x and k = thermal conductivity
of the material through which the heat is flowing (thermal conductivity of a homogenous substance depends on its temperature
and varies linearly).
STEADY FLOW OF HEAT IN HOMOGENOUS BODIES
In most industrial processes, heat is transferred from one point to another under steady conditions of temperature difference,
length of heat flow path and cross sectional area. In such cases, the Fourier’s equation is modified to:
Q
∆T
= −kA
=𝑞
t
x
where Q = total amount of heat transferred in time, t.
EXAMPLE
Assume one wall of a furnace is 8 in. thick and is made of brick having an average thermal conductivity of 2.2 Btu/hr ft °F over
the temperature range involved. The inner side of the furnace has a temperature of 1700°F and the temperature of the outer
side is 700°F. Assuming a wall area of 80 ft2, calculate the amount of heat lost per hour through the wall.
SOLUTION
𝑞=
𝑘𝐴
2.2 𝐵𝑡𝑢 ℎ−1 𝑓𝑡 −1 ℉−1 × 80 𝑓𝑡 2
∆𝑇 =
× 1000 ℉ = 262687 𝐵𝑡𝑢 ℎ−1
𝑥
0.67 𝑓𝑡
RESISTANCE TO FLOW OF HEAT
When heat flows through a solid by conduction, there is a certain resistance to the flow determined by the thermal conductivity,
cross sectional area and the thickness of that particular substance.
To obtain any given rate of heat flow, the resistance must be overcome by setting up a certain temperature difference driving
force.
𝑥
𝑅=
𝑘𝐴
RESISTANCE IN SERIES
If one solid is placed in series with another solid, the resistance to the flow of heat will be greater than it would be for one of the
solids alone.
If three solids of equal cross sectional area are placed in series when steady flow of heat exists, exactly the same amount of heat
per unit time is transferred through each of the solids.
HOT SIDE
R3
R2
R1
Direction of heat flow
Amount of heat per unit time that passes through each of the solids is the same, i.e.
𝑄
𝑡
=
1
𝑄
𝑡
=
2
𝑄
𝑡
=
3
𝑄
𝑡
𝑇𝑂𝑇𝐴𝐿
Rate of transfer through solid 1:
𝑄
𝑡
=
𝑘𝐴
1
∆𝑇
𝑥
𝑘𝐴
𝑥
=
1
∆𝑇1
1
Similarly, for solids 2 and 3
∴ ∆𝑇𝑇𝑂𝑇𝐴𝐿
𝑄
𝑡 1
=
+
𝑘𝐴
𝑥 1
𝑄
𝑡 2
+ …..
𝑘𝐴
𝑥 2
Rearranging, we get
𝑄𝑇𝑂𝑇𝐴𝐿 =
∆𝑇
1
𝑘𝐴
𝑥
+
1
1
𝑘𝐴
𝑥
+ …
2
EXAMPLE
The temperature at the inside surface of an oven is 460°F. The inside wall of the oven is constructed of brick and is 8 in. thick.
The thermal conductivity of the brick is 2.2 Btu/hr ft °F. The outside of the oven is covered with a 3 in. layer of asbestos which
has a thermal conductivity of 0.11 Btu/hr ft °F. If the outside surface of the insulation has a temperature of 100°F, calculate the
amount of heat lost through 2 ft2 of wall in 3 hours.
SOLUTION
𝑘𝐴
𝑥
𝑘𝐴
𝑥
𝑏𝑟𝑖𝑐𝑘
𝑎𝑠𝑏𝑒𝑠𝑡𝑜𝑠
2.2 𝐵𝑡𝑢/ℎ𝑟𝑓𝑡 ℉ × 2 𝑓𝑡 2
=
= 6.6 𝐵𝑡𝑢/ℎ𝑟℉
0.67 𝑓𝑡
0.11 𝐵𝑡𝑢/ℎ𝑟𝑓𝑡 ℉ × 2 𝑓𝑡 2
=
= 0.88 𝐵𝑡𝑢/ℎ𝑟℉
0.25 𝑓𝑡
𝑄𝑇𝑂𝑇𝐴𝐿 =
360
1
1
+
6.6
0.88
× 3 ℎ𝑟 = 839 𝐵𝑡𝑢
RESISTANCE IN PARALLEL
The total heat transferred through these resistances in a given time is the sum of the heat transferred through each one
individually.
Thus, for steady flow of heat through homogenous bodies
R1
HOT SIDE
R2
R3
Direction of heat flow
𝑄
𝑡
=
𝑇𝑂𝑇𝐴𝐿
𝑄
𝑡
+
1
𝑄
𝑡
+
2
𝑄
𝑡
3

𝑄
𝑡
=
𝑘𝐴
𝑇𝑂𝑇𝐴𝐿
∆𝑇
𝑥
+ 𝑘𝐴
1
∆𝑇
𝑥
+ 𝑘𝐴
2
∆𝑇
𝑥
Assuming that ΔT is the same for each of the three solids
𝑄
𝑡
=
𝑇𝑂𝑇𝐴𝐿
Substituting
𝑅=
𝑘𝐴
𝑥
+
1
𝑘𝐴
𝑥
+
2
𝑘𝐴
𝑥
∆𝑇
3
𝑥
𝑘𝐴
have
𝑞=
𝑄
𝑡
=
𝑇𝑂𝑇𝐴𝐿
1
1
1
+
+
∆𝑇
𝑅1 𝑅2 𝑅3
It is also possible to determine the temperature drop across any one of the solids:
∆𝑇1 = ∆𝑇𝑇𝑂𝑇𝐴𝐿
𝑅1
𝑅1 + 𝑅2 + 𝑅3
3
EXAMPLE
A glass window with an area of 6 ft2 is installed in a wooden wall of a room. The dimensions of this wall are 8 by 10 ft. The
wood is 1 in. thick and has a thermal conductivity of 0.087 Btu/hr ft °F. The glass is 1/8 in. thick and has a thermal conductivity
of 0.40 Btu/hr ft °F. If the inside of the wall and the glass temperature is 90 °F and the outside wall and glass is 30 °F,
calculate the total amount of heat conducted through the wall and glass per hour.
SOLUTION
𝑄𝑇𝑂𝑇𝐴𝐿 =
𝑘𝐴
𝑥
𝑔𝑙𝑎𝑠𝑠
𝑘𝐴
𝑥
𝑘𝐴
𝑥
+
𝑔𝑙𝑎𝑠𝑠
𝑘𝐴
𝑥
∆𝑇
𝑤𝑜𝑜𝑑
0.4 𝐵𝑡𝑢 ℎ−1 𝑓𝑡 −1 ℉−1 × 6 𝑓𝑡 2
=
= 230.3 𝐵𝑡𝑢 ℎ−1 ℉−1
0.01042 𝑓𝑡
𝑤𝑜𝑜𝑑
0.087 𝐵𝑡𝑢 ℎ−1 𝑓𝑡 −1 ℉−1 × 74 𝑓𝑡 2
=
= 77.3 𝐵𝑡𝑢 ℎ−1 ℉−1
0.0833 𝑓𝑡
∴ 𝑄𝑇𝑂𝑇𝐴𝐿 = 230.3 + 77.3 𝐵𝑡𝑢 ℎ−1 ℉−1 × 60 ℉ = 18456 𝐵𝑡𝑢 ℎ−1
USE OF MEAN AREA AND MEAN TEMPERATURE DIFFERENCE
In many types of heat transfer equipment, the cross sectional area varies along the heat flow path and the temperature
difference changes from one point to the another.
When water flows through straight pipes and is heated by steam condensing on the pipes, the inside area of the pipe wall is
smaller than the outside area of the pipe. If cold water comes into the heater, the overall temperature difference between cold
water and the steam at the entrance of the heater will be greater than the overall temperature difference between the hot water
and the stream at the heater exit.
Therefore, to apply the basic heat transfer equations, some type of mean or average values for the area and temperature
differences must be employed.
Arithmetic mean values:
𝐴𝑚𝑒𝑎𝑛 =
𝐴1 + 𝐴2
𝑇1 + 𝑇2
𝑎𝑛𝑑 𝑇𝑚𝑒𝑎𝑛 =
2
2
Logarithmic mean values: applicable to closed cylindrical bodies of circular sections.
𝐴𝑚𝑒𝑎𝑛 =
𝐴2 − 𝐴1
𝑇2 − 𝑇1
𝑎𝑛𝑑 𝑇𝑚𝑒𝑎𝑛 =
2.3𝑙𝑜𝑔 𝐴2 − 𝐴1
2.3𝑙𝑜𝑔 𝑇2 − 𝑇1
As long as A1 and A2 or T1 or T2 ratios are less than 2, difference between arithmetic and logarithmic mean is less than 5%.
EXAMPLE
A furnace wall is constructed of fire brick, 10 cm thick. The temperature at the inside of the wall is 800°C and the temperature of
the outside of the wall is 40°C. If the mean thermal conductivity of the brick under these conditions is 0.28 W m-1 °C-1, determine
the rate of heat loss through 10 m2 of wall surface.
SOLUTION
𝑞=
𝑘𝐴
0.28 𝑊 𝑚−1 ℃−1 × 10 𝑚2
∆𝑇 =
× 760 ℃ = 21280 𝑊
𝑥
0.10 𝑚
If the coefficient of heat transfer between the outside furnace wall and air is 14 W m-2 °C-1 and the air temperature is 20°C when
the temperature inside the furnace is 800°C. What is the rate of heat loss through 10 m2 of wall surface? (Air acts as an
insulator and should reduce the outward heat flow. Consider air as a layer). What is the temperature of the outside furnace wall?
SOLUTION
FURNACE
WALL
800°C
AIR LAYER
R1
20°C
R2
ΔTwall
ΔTair
𝑅𝑤𝑎𝑙𝑙 =
𝑅𝑎𝑖𝑟 =
𝑞=
𝑄
𝑡
=
𝑇𝑂𝑇𝐴𝐿
𝑥
0.1 𝑚
=
= 0.0357 ℃ 𝑊 −1
−1
−1
2
𝑘𝐴
0.28 𝑊 𝑚 ℃ × 10 𝑚
𝑥
1
=
= 0.00714 ℃ 𝑊 −1
−1
−1
2
𝑘𝐴
14 𝑊 𝑚 ℃ × 10 𝑚
1
𝑅𝑤𝑎𝑙𝑙
∆𝑇𝑎𝑖𝑟 = ∆𝑇𝑇𝑂𝑇𝐴𝐿
+
1
∆𝑇 =
𝑅𝑎𝑖𝑟
1
1
+
0.0357 0.00714
× 780 ℃ = 131 𝑘𝑊
𝑅𝑎𝑖𝑟
0.00714
= 800 ℃
= 133 ℃
𝑅𝑤𝑎𝑙𝑙 + 𝑅𝑎𝑖𝑟
0.0357 + 0.00714
If the furnace wall was instead constructed of an inside layer of 8 cm thick firebrick and an outside layer of chrome brick, 2 cm
thick, what is the rate of heat loss through 10 m2 of surface, the temperature of the outside wall under these conditions and the
temperature between the two layers of brick? (kchrome = 1.38 W m-1 °C-1)
Assume that the coefficient of heat transfer between the outside furnace wall and the air does not change.
SOLUTION
𝑞=
𝑅𝑏𝑟𝑖𝑐𝑘 =
𝑄
𝑡
=
𝑇𝑂𝑇𝐴𝐿
1
𝑅𝑤𝑎𝑙𝑙
+
1
𝑅𝑐ℎ𝑟𝑜𝑚𝑒
+
1
∆𝑇
𝑅𝑎𝑖𝑟
𝑥
0.08 𝑚
=
= 0.0286 ℃ 𝑊 −1
−1
−1
2
𝑘𝐴
0.28 𝑊 𝑚 ℃ × 10 𝑚
Similarly,
Rchrome and Rair = 0.00145 and 0.00714 °C W-1, respectively.
𝑞=
𝑄
𝑡
=
𝑇𝑂𝑇𝐴𝐿
1
𝑅𝑤𝑎𝑙𝑙
∆𝑇𝑏𝑟𝑖𝑐𝑘 = ∆𝑇𝑇𝑂𝑇𝐴𝐿
∆𝑇𝑐ℎ𝑟𝑜𝑚𝑒 = ∆𝑇𝑇𝑂𝑇𝐴𝐿
+
1
𝑅𝑐ℎ𝑟𝑜𝑚𝑒
+
1
∆𝑇 =
𝑅𝑎𝑖𝑟
1
1
1
+
+
0.0286 0.00145 0.00714
× 780 ℃ = 674.7 𝑘𝑊
𝑅𝑏𝑟𝑖𝑐𝑘
0.0286
= 800 ℃
= 615 ℃
𝑅𝑏𝑟𝑖𝑐𝑘 + 𝑅𝑐ℎ𝑟𝑜𝑚𝑒 + 𝑅𝑎𝑖𝑟
0.0286 + 0.00145 + 0.00714
𝑅𝑐ℎ𝑟𝑜𝑚𝑒
0.00145
= 800 ℃
= 31.2 ℃
𝑅𝑏𝑟𝑖𝑐𝑘 + 𝑅𝑐ℎ𝑟𝑜𝑚𝑒 + 𝑅𝑎𝑖𝑟
0.0286 + 0.00145 + 0.00714