PM3125: Lectures 4 to 6
Content of Lectures 1 to 6:
Heat transfer:
•
•
•
•
•
•
R. Shanthini
17 May 2010
Source of heat
Heat transfer
Steam and electricity as heating media
Determination of requirement of amount of
steam/electrical energy
Steam pressure
Mathematical problems on heat transfer
1
Heat Transfer
is the means by which
energy moves from
a hotter object to
a colder object
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2
Mechanisms of Heat Transfer
Conduction
is the flow of heat by direct contact between a
warmer and a cooler body.
Convection
is the flow of heat carried by moving gas or liquid.
(warm air rises, gives up heat, cools, then falls)
Radiation
is the flow of heat without need of an intervening
medium.
(by infrared radiation, or light)
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3
Mechanisms of Heat Transfer
Latent heat
Conduction
Convection
Radiation
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4
Conduction
HOT
(lots of vibration)
COLD
(not much vibration)
Heat travels
along the rod
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5
Conduction
Conduction is the process whereby heat is transferred
directly through a material, any bulk motion of the
material playing no role in the transfer.
Those materials that conduct heat well are called
thermal conductors, while those that conduct heat poorly
are known as thermal insulators.
Most metals are excellent thermal conductors, while
wood, glass, and most plastics are common thermal
insulators.
The free electrons in metals are responsible for the
excellent thermal conductivity of metals.
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6
Conduction: Fourier’s Law
Cross-sectional area A
L
Q =kA
ΔT
L
( )t
What is the unit of k?
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17 May 2010
Q = heat transferred
k = thermal conductivity
A = cross sectional area
DT = temperature difference
between two ends
L = length
7
t = duration of heat transfer
Thermal Conductivities
Substance
Thermal
Conductivity
k [W/m.K]
Substance
Thermal
Conductivity
k [W/m.K]
Syrofoam
0.010
Glass
0.80
Air
0.026
Concrete
1.1
Wool
0.040
Iron
79
Wood
0.15
Aluminum
240
Body fat
0.20
Silver
420
Water
0.60
Diamond
2450
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8
Conduction through Single Wall
Use Fourier’s Law:
T1
Q =kA
.
.
Q
( )t
Q
Δx
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ΔT
L
T2  T1
x
.
Q =
k A (T1 – T2)
Δx
9
Conduction through Single Wall
.
T1
Q =
.
Δx
.
Q
Q
Δx
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k A (T1 – T2)
T2  T1
x
=
T1 – T2
Δx/(kA)
Thermal resistance (in k/W)
(opposing heat flow) 10 10
Conduction through Composite Wall
T1
B
C
A
.
.
T2
Q
Q
T3
.
Q =
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17 May 2010
kA
kB
kC
ΔxA
ΔxB
ΔxC
T1 – T2
(Δx/kA)A
=
T4
T2 – T3
(Δx/kA)B
x
=
T3 – T4
(Δx/kA)C
11
11
Conduction through Composite Wall
.
Q =
.
T1 – T2
(Δx/kA)A
=
T2 – T3
(Δx/kA)B
=
T3 – T4
(Δx/kA)C
[
Q (Δx/kA)A + (Δx/kA)B + (Δx/kA)C
]
= T1 – T2 + T2 – T3 + T3 – T4
.
Q =
T1 – T4
(Δx/kA)A + (Δx/kA)B + (Δx/kA)C
12
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12
Example 1
An industrial furnace wall is constructed of 21 cm thick
fireclay brick having k = 1.04 W/m.K. This is covered on
the outer surface with 3 cm layer of insulating material
having k = 0.07 W/m.K. The innermost surface is at 1000oC
and the outermost surface is at 40oC. Calculate the steady
state heat transfer per area.
Solution: We start with the equation
.
Q =
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17 May 2010
Tin – Tout
(Δx/kA)fireclay + (Δx/kA)insulation
13
Example 1 continued
.
(1000 – 40) A
Q =
(0.21/1.04) + (0.03/0.07)
.
Q
A
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17 May 2010
= 1522.6 W/m2
14
Example 2
We want to reduce the heat loss in Example 1 to 960 W/m2.
What should be the insulation thickness?
Solution: We start with the equation
.
Q =
.
Tin – Tout
(Δx/kA)fireclay + (Δx/kA)insulation
Q
(1000 – 40)
2
W/m
=
960
=
A
(0.21/1.04) + (Δx)insulation /0.07)
(Δx)insulation = 5.6 cm
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15
Conduction through hollow-cylinder
ro
Ti
ri
To
L
.
Q =
Ti – To
[ln(ro/ri)] / 2πkL
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16
Conduction through the composite
wall
in
a
hollow-cylinder
r3
r2
Ti
.
Q =
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To
Material A
r1
Material B
Ti – To
[ln(r2/r1)] / 2πkAL+ [ln(r3/r2)] / 2πkBL
17
Example 3
A thick walled tube of stainless steel ( k = 19 W/m.K) with
2-cm inner diameter and 4-cm outer diameter is covered
with a 3-cm layer of asbestos insulation (k = 0.2 W/m.K).
If the inside-wall temperature of the pipe is maintained at
600oC and the outside of the insulation at 100oC, calculate
the heat loss per meter of length.
Solution: We start with the equation
.
Q =
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Ti – To
[ln(r2/r1)] / 2πkAL+ [ln(r3/r2)] / 2πkBL
18
Example 3 continued
2 π L ( 600 – 100)
.
Q =
[ln(2/1)] / 19 + [ln(5/2)] / 0.2
.
Q
L
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= 680 W/m
19
Mechanisms of Heat Transfer

Conduction
is the flow of heat by direct contact between a
warmer and a cooler body.
Convection
is the flow of heat carried by moving gas or liquid.
(warm air rises, gives up heat, cools, then falls)
Radiation
is the flow of heat without need of an intervening
medium.
(by infrared radiation, or light)
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20
Convection
Convection is the process in which heat is carried from
place to place by the bulk movement of a fluid (gas or
liquid).
Convection currents are set up when
a pan of water is heated.
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21
Convection
It explains why breezes come from the ocean in the day
and from the land at night
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22
Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid
Heated surface at Tsurface
.
Qconv. = h A (Tsurface – Tfluid)
Area exposed
Heat transfer coefficient (in W/m2.K)
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23
Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid
Heated surface at Tsurface
.
Qconv. =
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Tsurface – Tfluid
1/(hA)
Convective heat resistance (in k/W)
24
Example 4
The convection heat transfer coefficient between a surface at 50oC and
ambient air at 30oC is 20 W/m2.K. Calculate the heat flux leaving the
surface by convection.
Solution:
Use Newton’s Law of cooling :
.
Flowing fluid at Tfluid = 30oC
Q
= h A (Tsurface – Tfluid)
conv.
= (20 W/m2.K) x A x (50-30)oC
Heated surface at Tsurface = 50oC
h = 20 W/m2.K
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Heat flux leaving the surface:
.
Q
conv.
= 20 x 20 = 400 W/m2
A
25
Example 5
Air at 300°C flows over a flat plate of dimensions 0.50 m by 0.25 m.
If the convection heat transfer coefficient is 250 W/m2.K, determine
the heat transfer rate from the air to one side of the plate when the
plate is maintained at 40°C.
Solution:
Flowing fluid at Tfluid = 300oC
Heated surface at Tsurface = 40oC
Use Newton’s Law of cooling :
.
Q
= h A (Tsurface – Tfluid)
conv.
= 250 W/m2.K x 0.125 m2
x (40 - 300)oC
= - 8125 W/m2
h = 250 W/m2.K
A = 0.50x0.25 m2
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Heat is transferred from
the air to the plate.
26
Forced Convection
In forced convection, a fluid is forced by external forces
such as fans.
In forced convection over external surface:
Tfluid = the free stream temperature (T∞), or a
temperature far removed from the surface
In forced convection through a tube or channel:
Tfluid = the bulk temperature
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27
Free Convection
In free convection, a fluid is circulated due to buoyancy
effects, in which less dense fluid near the heated surface rises
and thereby setting up convection.
In free (or partially forced) convection over
external surface:
Tfluid = (Tsurface + Tfree stream) / 2
In free or forced convection through a tube or
channel:
Tfluid = (Tinlet + Toutlet) / 2
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28
Change of Phase Convection
Change-of-phase convection is observed with
boiling or condensation
.
It is a very complicated mechanism and
therefore will not be covered in this course.
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29
Overall Heat Transfer through a Plane Wall
Fluid A
at TA > T1
T1
.
.
Q
Q
T2
Δx
.
Q =
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17 May 2010
TA – T1
1/(hAA)
=
x
T1 – T2
Δx/(kA)
Fluid B
at TB < T2
=
T2– TB
1/(hBA)
30
Overall Heat Transfer through a Plane Wall
.
Q =
TA – T1
1/(hAA)
=
Δx/(kA)
=
T2– TB
1/(hBA)
TA – TB
.
Q =
T1 – T2
1/(hAA) + Δx/(kA) + 1/(hBA)
.
Q = U A (TA – TB)
where U is the overall heat transfer coefficient given by
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1/U = 1/hA + Δx/k + 1/hB
31
Overall heat transfer through hollow-cylinder
Fluid A is inside the pipe
Fluid B is outside the pipe
TA > TB
ro
Ti r
i
To
.
L
Q = U A (TA – TB)
where
1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + 1/(hBAo)
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32
Example 6
Steam at 120oC flows in an insulated pipe. The pipe is
mild steel (k = 45 W/m K) and has an inside radius of 5
cm and an outside radius of 5.5 cm. The pipe is covered
with a 2.5 cm layer of 85% magnesia (k = 0.07 W/m K).
The inside heat transfer coefficient (hi) is 85 W/m2 K, and
the outside coefficient (ho) is 12.5 W/m2 K. Determine the
heat transfer rate from the steam per m of pipe length, if
the surrounding air is at 35oC.
Solution: Start with
.
Q = U A (TA – TB) = U A (120 – 35)
What is UA?
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33
Example 6 continued
1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + … + 1/(hBAo)
1/UA = 1/(85Ain) + ln(5.5/5) / 2π(45)L
+ ln(8/5.5) / 2π(0.07)L
+ 1/(12.5Aout)
Ain = 2π(0.05)L and Aout = 2π(0.08)L
1/UA = (0.235 + 0.0021 +5.35 + 1) / 2πL
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34
Example 6 continued
UA = 2πL / (0.235 + 0.0021 +5.35 + 1)
.
Q = U A (120 – 35)
air
steel
= 2πL (120 – 35) / (0.235 + 0.0021 +5.35 + 1)
steam
= 81 L
insulation
.
Q / L = 81 W/m
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35
Mechanisms of Heat Transfer

Conduction
is the flow of heat by direct contact between a
warmer and a cooler body.

Convection
is the flow of heat carried by moving gas or liquid.
(warm air rises, gives up heat, cools, then falls)
Radiation
is the flow of heat without need of an intervening
medium.
(by infrared radiation, or light)
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36
Radiation
Radiation is the process in which energy is transferred by
means of electromagnetic waves of wavelength band between
0.1 and 100 micrometers solely as a result of the temperature
of a surface.
Heat transfer by radiation
can take place through
vacuum. This is because
electromagnetic waves
can propagate through
empty space.
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37
The Stefan–Boltzmann Law of Radiation
Q
t
= ε σ A T4
ε = emissivity, which takes a value between 0 (for
an ideal reflector) and 1 (for a black body).
σ = 5.668 x 10-8 W/m2.K4 is the Stefan-Boltzmann
constant
A = surface area of the radiator
T = temperature of the radiator in Kelvin.
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38
Why is the
mother shielding
her cub?
Ratio of the surface
area of a cub to its
volume is much larger
than for its mother.
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39
What is the Sun’s surface temperature?
The sun provides about 1000 W/m2 at the Earth's surface.
Assume the Sun's emissivity ε = 1
Distance from Sun to Earth = R = 1.5 x 1011 m
Radius of the Sun = r = 6.9 x 108 m
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40
What is the Sun’s surface temperature?
Q
t
= ε σ A T4
(4 π 1.52 x 1022 m2)(1000 W/m2)
(4 π 6.92 x 1016 m2)
= 2.83 x 1026 W
T4 =
ε
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= 5.98 x 1018 m2
2.83 x 1026 W
(1) (5.67 x 10-8 W/m2.K4) (5.98 x 1018 m2)
σ
T = 5375 K
41
If object at temperature T is surrounded by
an environment at temperature T0, the net
radioactive heat flow is:
Q
t
= ε σ A (T4 - To4 )
Temperature of the radiating surface
Temperature of the environment
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42
Example 7
What is the rate at which radiation is emitted by a surface
of area 0.5 m2, emissivity 0.8, and temperature 150°C?
Solution:
Q
t
0.8
Q
t
= ε σ A T4
[(273+150) K]4
0.5 m2
5.67 x 10-8 W/m2.K4
= (0.8) (5.67 x 10-8 W/m2.K4) (0.5 m2) (423 K)4
= 726 W
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43
Example 8
If the surface of Example 7 is placed in a large, evacuated chamber
whose walls are maintained at 25°C, what is the net rate at which
radiation is exchanged between the surface and the chamber walls?
Solution:
Q
t
= ε σ A (T4 - To4 )
[(273+25) K]4
[(273+150) K]4
Q
t
= (0.8) x (5.67 x 10-8 W/m2.K4) x (0.5 m2)
x [(423 K)4 -(298 K)4 ]
= 547 W
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44
Example 8 continued
Note that 547 W of heat loss from the surface occurs at
the instant the surface is placed in the chamber. That
is, when the surface is at 150oC and the chamber wall
is at 25oC.
With increasing time, the surface would cool due to
the heat loss. Therefore its temperature, as well as the
heat loss, would decrease with increasing time.
Steady-state conditions would eventually be achieved
when the temperature of the surface reached that of the
surroundings.
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45
Example 9
Under steady state operation, a 50 W incandescent light bulb has a
surface temperature of 135°C when the room air is at a temperature
of 25°C. If the bulb may be approximated as a 60 mm diameter
sphere with a diffuse, gray surface of emissivity 0.8, what is the
radiant heat transfer from the bulb surface to its surroundings?
Q
t
Solution:
= ε σ A (T4 - To4 )
[(273+135) K]4
Q
t
[(273+25) K]4
= (0.8) x (5.67 x 10-8 J/s.m2.K4) x [π x (0.06) m2]
x [(408 K)4 -(298 K)4 ]
= 10.2
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W
(about 20% of the power is dissipated by radiation)
46
Mathematical Problems on Heat Exchanger
Tc,in
Th,out
Th,in
Tc,out
.
.
.
Q = mc cc (Tc,out – Tc,in) = mh ch (Th,in – Th,out)
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47
Mathematical Problems on Heat Exchanger
Tc,in
Parallel-flow heat exchanger
Th,out
Th,in
Th,in
Tc,out
high heat
transfer
Tc,in
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17 May 2010
low heat
transfer
Th,out
Tc,out
48
Mathematical Problems on Heat Exchanger
Parallel-flow heat exchanger
Th,in
Th,out
ΔTb T
c,out
ΔTa
Tc,in
a
b
.
Q = U A ΔT
where ΔT =
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ΔTa - ΔTb
ln(ΔTa / ΔTb)
is the log mean
temperature
49
difference (LMTD)
Mathematical Problems on Heat Exchanger
Tc,out
Counter-flow heat exchanger
Th,out
Th,in
Tc,in
.
.
.
Q = mc cc (Tc,out – Tc,in) = mh ch (Th,in – Th,out)
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50
Mathematical Problems on Heat Exchanger
Tc,out
Th,in
Th,in
Tc,out
Counter-flow heat exchanger
Th,out
Tc,in
Th,out
Tc,in
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51
Mathematical Problems on Heat Exchanger
Counter-flow heat exchanger
Th,in
ΔTa
Tc,out
ΔTb
a
Th,out
Tc,in
b
.
Q = U A ΔT
where ΔT =
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17 May 2010
ΔTa - ΔTb
ln(ΔTa / ΔTb)
is the log mean
temperature
52
difference (LMTD)
Example in heat Exchanger Design
An exhaust pipe, 75 mm outside diameter, is cooled by
surrounding it by an annular space containing water.
The hot gases enters the exhaust pipe at 350oC, gas flow
rate being 200 kg/h, mean specific heat capacity at
constant pressure 1.13 kJ/kg K, and comes out at 100oC.
Water enters from the mains at 25oC, flow rate 1400 kg/h,
mean specific heat capacity 4.19 kJ/kg K.
The heat transfer coefficient for gases and water may be
taken as 0.3 and 1.5 kW/m2 K and pipe thickness may be
taken as negligible.
Calculate the required pipe length for (i) parallel flow, and
for (ii) counter flow.
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53
Example in heat Exchanger Design
Solution:
.
.
.
Q = mc cc (Tc,out – Tc,in) = mh ch (Th,in – Th,out)
(1400 kg/hr) (4.19 kJ/kg K) (Tc,out – 25)oC
= (200 kg/hr) (1.13 kJ/kg K) (350 – 100)oC
The temperature of water at the outlet = Tc,out = 34.63oC.
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54
Example in heat Exchanger Design
Solution continued:
(i) Parallel flow:
ΔTa = 350 – 25 = 325oC
ΔTb = 100 – 34.63 = 65.37oC
ΔT =
ΔTa - ΔTb
ln(ΔTa / ΔTb)
=
325 – 65.37
= 162oC
ln(325 / 65.37)
.
Q = U A ΔT = (UA) 162oC
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17 May 2010
What is UA?
55
Example in heat Exchanger Design
Solution continued:
1/U = 1/hwater + 1/hgases
= 1/1.5 + 1/0.3 = 4 (kW/m2 K)-1
Therefore, U = 0.25 kW/m2 K
A = π (outer diameter) (L) = π (0.075 m) (L m)
.
Q = (UA) 162oC
= (0.25) π (0.075) L (162) kW
.
What is Q?
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56
Example in heat Exchanger Design
Solution continued:
.
.
.
Q = mc cc (Tc,out – Tc,in) = mh ch (Th,in – Th,out)
= (200 kg/h) (1.13 kJ/kg K) (350 – 100)oC
= 15.69 kW
Substituting the above in
.
Q = (UA) 162oC
= (0.25) π (0.075) L (162) kW
we get
L = 1.64 m
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57
Example in heat Exchanger Design
Solution continued:
(ii) Counter flow:
ΔTa = 350 – 34.63 = 315.37oC
ΔTb = 100 – 25 = 75oC
ΔT =
ΔTa - ΔTb
ln(ΔTa / ΔTb)
=
315.37 – 75
= 167.35oC
ln(315.37 / 75)
.
.
Q = U A ΔT = (UA) 167.35oC
Q = 15.69 kW;
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17 May 2010
U = 0.25 kW/m2 K ;
A = π (0.075) L m2
58
Therefore, L = 1.59 m
Other Heat Exchanger Types
Cross-flow heat exchanger
with both fluids unmixed
The direction of fluids are perpendicular to each other.
The required surface area for this heat exchanger is
usually calculated by using tables.
It is between the required surface area for counter-flow
and parallel-flow heat exchangers.
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59
Other Heat Exchanger Types
One shell pass and
two tube passes
Th,in
Tc,in
Tc,out
Th,out
The required surface area for this heat
exchanger is calculated using tables.
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60
Other Heat Exchanger Types
Two shell passes and
two tube passes
Tc,in
Tc,out
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Th,in
Th,out
The required surface area for this heat
exchanger is calculated using tables.
61
Batch Sterilization (method of heating):
Electrical
heating
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Direct steam
sparging
Steam
heating
62
For batch heating with constant rate heat flow:
Total heat lost by the coil to the medium
= heat gained by the medium
M
- mass of the medium
T0 - initial temperature of the medium
T
- final temperature of the medium
c
- specific heat of the medium
q
- rate of heat transfer from the
electrical coil to the medium
t
- duration of electrical heating
.
Electrical
heating
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.
q t = M c (T - T0)
63
For batch heating by direct steam sparging:
M - initial mass of the raw medium
T0 - initial temperature of the raw medium
.
ms - steam mass flow rate
t
H
- duration of steam sparging
- enthalpy of steam relative to the
enthalpy at the initial temperature
of the raw medium (T0)
T
c
- final temperature of the mixture
- specific heat of medium and water
.
. st) c T
(ms t) (H + cT0) + M c T0 = (M + m
Direct steam
sparging
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.
.
ms t H = (M + ms t) c (T – T0)
64
For batch heating with isothermal heat source:
M - mass of the medium
T0 - initial temperature of the medium
TH - temperature of heat source (steam)
Steam
heating
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17 May 2010
T
- final temperature of the medium
c
- specific heat of the medium
t
- duration of steam heating
U
- overall heat transfer coefficient
A
- heat transfer area
U A t = M c ln
(
T0 - TH
T - TH
)
65
Could you prove the above?
For batch heating with isothermal heat source:
U A t = M c ln
(
T0 - TH
T - TH
)
(
T = TH + (T0 - TH) exp - U A t
cM
Steam
heating
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66
)
Example of batch heating by direct steam sparging:
A fermentor containing 40 m3 medium at 25oC is going to be
sterilized by direct injection of saturated steam. The steam at 350
kPa absolute pressure is injected with a flow rate of 5000 kg/hr,
which will be stopped when the medium temperature reaches 122oC.
Determine the time taken to heat the medium.
Additional data required:
Enthalpy of saturated steam at 350 kPa = ??
Enthalpy of water at 25oC = ??
The heat capacity of the medium 4.187 kJ/kg.K
The density of the medium are 4.187 kJ/kg.K and 1000 kg/m3,
respectively.)
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67
Example of batch heating by direct steam sparging:
A fermentor containing 40 m3 medium at 25oC is going to be
sterilized by direct injection of saturated steam. The steam at 350
kPa absolute pressure is injected with a flow rate of 5000 kg/hr,
which will be stopped when the medium temperature reaches 122oC.
Determine the time taken to heat the medium.
Additional data: The enthalpy of saturated steam at 350 kPa and water at
25oC are 2732 and 105 kJ/kg, respectively. The heat capacity and density
of the medium are 4.187 kJ/kg.K and 1000 kg/m3, respectively.
Solution:
Use the equation below:
.
.
ms t H = (M + ms t) c (T – T0)
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.
.
ms t H = (M + ms t) c (T – T0)
(5000 kg/hr) (th) (2732-105) kJ/kg
= [(40 m3)(1000 kg/m3) + (5000 kg/hr)(th)](4.187 kJ/kg.K)(122-25)K
Taking the heating time (th) to be in hr, we get
(5000 th) (2627) kJ = [40000 + 5000 t](4.187)(97)kJ
(5000 th) [2627 – 4.187 x 97] = 40000 x 4.187 x 97
th = 1.463 hr
Therefore, the time taken to heat the medium is 1.463 hours.
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Example of batch heating with isothermal heat source:
A fermentor containing 40 m3 medium at 25oC is going to be
sterilized by an isothermal heat source, which is saturated steam at
350 kPa absolute pressure. Heating will be stopped when the
medium temperature reaches 122oC. Determine the time taken to
heat the medium.
Additional data: The saturated temperature of steam at 350 kPa is 138.9oC.
The heat capacity and density of the medium are 4.187 kJ/kg.K and 1000
kg/m3, respectively.
Solution:
Use the equation below:
U A t = M c ln
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(
T0 - TH
T - TH
)
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U A t = M c ln
(
T0 - TH
T - TH
)
(2500 kJ/hr.m2.K) (40 m2) (tc)
= (40 m3) (1000 kg/m3) (4.187 kJ/kg.K) ln[(25-138.9)/(122-138.9)]
Taking the heating time (th) to be in hr, we get
(2500 kJ/K) (40) (th) = (40) (1000) (4.187 kJ/K) ln[113.9/16.9]
(2500 kJ/K) (40) (th) = (40) (1000) (4.187 kJ/K) (1.908)
th = 3.1955 hr
Therefore, the time taken to heat the medium is 3.1955 hours.
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Explain why heating with isothermal heat source takes
twice the time taken by heating with steam sparging,
even though we used the same steam.
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Question from PM3125 / Jan 2010 past paper
A steel pipeline (inside diameter = 52.50 mm; outside diameter =
60.32 mm) contains saturated steam at 121.1oC. The line is insulated
with 25.4 mm of asbestos. Assume that the inside surface temperature
of the metal wall is at 121.1oC and the outer surface of the insulation is
at 26.7oC. Taking the average value of ksteel as 45 W/m.K and that of
kasbestos as 0.182 W/m.K, calculate the following:
(a) Heat loss for 30.5 m of pipe length.
[10 marks]
(b) Mass (in kg) of steam condensed per hour in the pipe due to the
heat loss.
[10 marks]
Additional data given on the next slide:
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Question from PM3125 / Jan 2010 past paper
Additional Data:
i) Heat transfer rate through the pipe wall is given by,
Q 
2  L ( T1  T 2 )
 ln( r2 / r1 ) ln( r3 / r2 ) 



k asbestos 
 k steel
where L is the length of pipe, T1 and T2 are the respective temperatures
at the inner and outer surfaces of the insulated pipe, r1 and r2 are the
respective inner and outer radius of the steel pipe, and r3 is the outer
radius of the insulated pipe.
ii) Latent heat of vapourization of steam could be taken as 2200 kJ/kg.
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Group Assignment will be uploaded at
http://www.rshanthini.com/PM3125.htm
(keep track of the site)
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End of slides for the heat transfer lecture
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Additional material not used in the lectures.
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Critical Radius of Insulation
To
r
Insulation
ro
Ti
ri
Ti – To
.
Q =
Pipe
[ln(ro/ri)] /2πkPL + [ln(r/ro)] /2πkIL + 1/hairA
Pipe resistance could be neglected
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A=2πrL
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Critical Radius of Insulation
Ti – To
.
Q =
[ln(r/ro)] /2πkIL + 1/(hair 2πrL)
=
Insulation
resistance
2π L ( Ti – To)
[ln(r/ro)] /kI + 1/(hair r)
Convective
resistance
Increasing r increases insulation resistance and decreases
heat transfer.
Increasing r decreases convective resistance and increases
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17 May 2010
heat transfer.
Critical Radius of Insulation
.
dQ /dr = 0 at the critical radius of insulation,
which leads to rcr = kI / hair
If the outer radius of the pipe (ro) < rcr and if
insulation is added to the pipe, heat losses will first
increase and go through a maximum at the
insulation radius of rcr and then decrease.
If the outer radius of the pipe (ro) > rcr and if
insulation is added to the pipe, heat losses will
continue to decrease.
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