Unit 42: Heat Transfer and
Combustion
Lesson 3: Convection
Aim
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• LO1: Understanding Heat Transfer Rates for
Composite Systems.
Convection
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• In order to consider heat transfer from fluids
to solids and visa versa, rather than just
between solids, we need to understand how
heat is transferred by convection
Nature of Convection
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• Heat transfer by convection consists of two
mechanisms.
• In addition to energy transfer by random molecular
motion (diffusion) there is also energy being
transferred by the bulk motion of the fluid.
• Thus in the presence of a temperature gradient, large
numbers of molecules are moving together in bulk at
the same time as the random motion off the individual
molecules takes place.
• The cumulative effect of both these energy transfer
methods is referred to as heat transfer by convection.
Heat Transfer by
Convection
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Hot Particles
Cold Particles
Heat flow to
Particles
Hot Surface
Nature of Convection
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• In engineering we are especially interested in
heat transfer by convection that occurs between
a fluid in motion and a solid boundary surface.
• At the fluid-surface interface there is a region
where the velocity of the fluid varies from zero at
the surface to some infinite value u associated
with the flow velocity.
• This region of fluid is known as the velocity
boundary layer
Heat Transfer by
Convection
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Variation of velocity with distance from a solid boundary
Y
y
Sold
Boundary
Velocity, u
Nature of Convection
Velocity boundary layer
between fluid 1 and
solid
Tf1
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Velocity boundary layer
between solid and fluid
2
Ts1
Solid
Material
Bulk fluid at
temperature
T1
Bulk fluid at
temperature
T2
Ts2
Tf2
Temperatures
at solid
surface
B1
Thermal boundary
layer between fluid
1 and solid
B2
Thermal boundary
layer between solid
and fluid 2
B1 & B2 are the upper
boundaries of the
velocity boundary
layers
Nature of Convection
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• If the surface of the fluid flow temperatures
differ there is also a region of the fluid
through which the temperature varies from Ts
at the solid surface to Tf in the flow, as the
heat energy is transferred.
• This region is called the thermal boundary
layer
Nature of Convection
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• It is important to note that at the surface
where the bulk velocity is zero, all heat energy
is transferred by random molecular motion i.e.
by conduction.
• As we move further away from the surface the
boundary layer grows and heat transfer is
more and more dependent on the
contribution made by the bulk fluid motion.
Nature of Convection
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• Convection heat transfer may be classified
according to the nature of the flow.
– When the fluid flow is caused by external means
we refer to heat transfer as forced convection e.g.
consider the use of a fan to provide forced
convection air cooling of electrical components in
a PC.
– Natural convection is induced by forces that result
from density differences caused by temperature
variations within a fluid.
Nature of Convection
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• Consider an electrical circuit board: air makes
contact with the hot electrical components
and experiences an increase in temperature.
This causes a reduction in density. In this
lighter air buoyancy forces causes vertical
motion and the warm air rises. As this
happens so it is replaced by an inflow of
cooler ambient air – these convection currents
now act as the heat transfer mechanism.
Nature of Convection
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• Regardless of the heat transfer process, the
rate of heat transfer that is of particular
interest to engineers may be determined by
using Newton’s law of cooling given by the
equation…
.
Q = hA(Ts – Tf)
Where h is called the surface heat transfer
coefficient (W/m2K)
Heat Transfer with Convection through
Composite Walls
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• If the materials of the composite wall involve
heat transfer from fluids to solids and visa
versa rather than just between solids, then
there is a similar electrical analogy to that for
heat transfer by conduction between solids.
• The surface heat transfer coefficient h
depends on both the type of fluid and the
fluid velocity.
Heat Transfer with Convection through
Composite Walls
NDGTA
• The thermal resistance of a fluid at the interface
with a solid can be shown to be…
Fluid thermal resistance, R = 1/hA
Note the similarity to thermal resistance for
conduction R = x/kA, both have the same units K/W.
• The following example shows how we combine
heat transfer by conduction though solids and
heat transfer by conduction at the solid-fluid
interface…
Heat Transfer with Convection through
Composite Walls
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• The wall of a house consists of an inner thermal block
125mm wide and a 125mm house brick separated by
an air gap. The inner surface wall is at a temperature
of 25oC and the outside air film temperature is 5oC. For
simplicity no additional insulating material has been
included in the construction.
• Calculate the heat flux (the rate at which heat is lost
per m2) of the wall surface when, the surface heat
transfer coefficient from the outside wall to air is 5
W/m2.K and the resistance to heat flow of the air gap is
0.15 k/W and the thermal conductivities of thermal
block and outer brick are 0.2 and 0.5 W/m.K
respectively.
Composite Walls
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T0=25oC
T1
T2
T3
Thermal
Brick
.
Heat
flow Q
Air Gap
House
Brick
T4= 5oC
Outside
air layer
.
.
Q
Q
R1
R2
R3
R4
Heat Transfer with Convection through
Composite Walls
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R = x/kA
Considering unit Area R =x/k
RTB = 0.125/0.2 = 0.625 K/W
RHB = 0.125/0.5 = 0.25 K/W
Rair gap = 0.15 K/W
Rwall/air = 1/hA = 1/5 = 0.2 K/W
RT = RTH + RAG + RHB + Rwall/Air
= 0.625 + 0.15 + 0.25 + 0.2 = 1.225 K/W
Heat Transfer with Convection through
Composite Walls
NDGTA
• Using
. equation…
Q = (T0 – T4)/ RT
We find the heat flux (heat loss per square metre to
be…
q = (25 – 5) / 1.225 = 16.3 W
Suppose we want to find the interface temperature
t1, then
16.3 = (25 – t1)/0.625
thus t1 = 14.8oC
Heat Transfer through
Cylindrical Walls
Q
dr
r2
.
t
r
2
1
t1
.
Unit length l = 1
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.
Heat Transfer through
Cylindrical Walls
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Q = -kAdt/dr
In this
case
A
=
2πr
(this
assumes
length
l
=
1)
.
So Q = -k2πrdt/dr
Note: for a plane wall the area perpendicular to
the heat flow is constant, but this is not the case
for a cylindrical wall
Heat Transfer with Convection through
Composite Walls
NDGTA
• For a plane wall the area perpendicular to the
heat flow is constant, but this is not the case
for a cylinder wall.
• At any radius within the cylinder, assuming
negligible axial and circumferential heat
transfer…
.
Q = -k2πrldt/dr (or –k2πrdt/dr for unit length)
Heat Transfer with Convection through
Composite Walls
NDGTA
• Thus…
.
Qdr/k2πr = -dt
• Integrating between r2 and r1 gives…
.
t2
r2
Q/(2πk) dr/r = - dt
.
r1
t1
Thus Q ln r2 = - (t2 – t1)
2πk r1
Heat Transfer with Convection through
Composite Walls
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• Thus the heat transfer per unit length radially
outwards
is given by…
.
Q = 2πk(t1 – t2)/ln(r2/r1)
Again considering the electrical analogue (I=V/r and
t1 – t2 = V then the thermal resistance per unit
length for a cylinder wall is…
R = ln(r2/r1)/2πk
(Verify this relationship!)
Heat Transfer with Convection through
Composite Walls
NDGTA
A steel pipe carrying steam at 250oC has an internal diameter of
100mm and an outside diameter of 120mm. The pipe is
insulated using an inner layer of specially treated plastic based
composite 40mm thick and an outer layer of felt 50mm thick.
The ambient temperature surrounding the pipe is 20oC.
Calculate the overall thermal resistance of the assembly and so
find the heat loss per unit length of pipe. Also calculate the
temperature on the outer surface of the pipe assembly. Assume
that the heat transfer rates for internal and external surfaces of
the pipe assembly are 500 and 15 W/m2K and the thermal
conductivities of the steel, plastic composite and felt are 45, 0.2
and 0.05 W/mK respectively
Heat Transfer with Convection through
Composite Walls
NDGTA
150mm
Plastic
composite
100mm
Steel pipe
. .
T3
Felt
T2
. . steam
60mm
T1 T0
50mm
Heat Transfer with Convection through
Composite Walls
NDGTA
Considering unit length of pipe…then the thermal
resistance of the steam film…
R = 1/hA = 1/(500)(2π x 0.05) = 0.00637 K.W
The thermal resistance of the steel pipe…
R = ln(r2/r1)/2πk = Ln (60/50)/(2π)(45) = 0.000645K/W
The thermal resistance of the plastic composite..
R = ln(r2/r1)/2πk = Ln (100/60)/(2π)(0.2) = 0.406K/W
The thermal resistance of the felt…
R = ln(r2/r1)/2πk = Ln (150/100)/(2π)(0.05) = 1.29K/W
The thermal resistance of air…
R = 1/hA = 1/(15)(2π x 0.15) = 0.0707 K.W
Heat Transfer with Convection through
Composite Walls
NDGTA
Thus total thermal resistance is…
RT = 0.00637 + 0.000645 + 0.406 + 1.29 + 0.0707
Thus RT = 1.774 K/W
Thus the heat loss per unit length…
Q = (t0 – t4) /1.774 = (250-20)/1.774 = 1.297 W
The temperature on the outer surface of the
pipe…
129.7 = (t3 – t4)/0.071 = (t3 – 20)/ 0.071 = 29.2oC
Heat Transfer with Convection through
Composite Walls
NDGTA
• Note the insignificant amount of thermal
resistance provided by the steel pipe. For this
reason, to a first approximation, the thermal
resistance of metal pipes is often ignored in
these calculations.
• Imagine the energy losses if the pipe were not
lagged!