week_5_heat_transfer_lecture.pdf
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Lecture Notes for CO2 (Part 3) 1-D STEADY STATE HEAT
CONDUCTION
Wan Azmi bin Wan Hamzah Universiti Malaysia Pahang
Week – 5
Course Outcome 2 (CO2)
Students should be able to understand and
evaluate one-dimensional heat flow and in
different geometries
2
Lesson Outcomes from CO2 (Part 3)
To derive the equation for temperature
distribution in various geometries
Thermal Resistance concept – to derive
expression for various geometries
To evaluate the heat transfer using thermal
resistance in various geometries
To evaluate the critical radius of insulation
To evaluate heat transfer from the rectangular fins
3
CRITICAL RADIUS OF INSULATION
Adding more insulation to a wall or
to the attic results:
a) decrease heat transfer and,
b) increase the thermal resistance
of the wall.
In a cylindrical pipe or a spherical
shell, the additional insulation
results :
a) increase the conduction
resistance but,
b) decreases the convection
resistance because of the surface
for convection become larger .
An insulated cylindrical pipe exposed to
convection from the outer surface and
the thermal resistance network
associated with it.
• Cylindrical pipe of outer radius r1 whose outer
surface temperature T1 is maintained constant.
• The pipe is covered with an insulator (k and r2).
• Convection heat transfer at T∞ and h.
• The rate of heat transfer from the insulated
pipe to the surrounding air can be expressed
as
1 1
2 1
2
ln / 1
2 2
ins conv
T T T TQ
r rR R
Lk h r L
5
The critical radius of insulation
for a cylindrical body:
The critical radius of insulation
for a spherical shell:
The variation of heat transfer
rate with the outer radius of the
insulation r2 when r1 < rcr.
• The rate of heat transfer from the cylinder increases with the addition of insulation for r2 < rcr
• Reaches maximum, when r2 = rcr
• Decreases when r2 > rcr
• Thus, insulating the pipe may actually increase the rate of heat transfer instead of decreasing it when r2 < rcr.
A 3 mm diameter and 5 m long electrical wire is tightly wrapped with a 2 mm
thick plastic cover whose thermal conductivity is k = 0.15 W/m°C. Electrical
measurements indicate that a current of 10A passes through the wire and there
is a voltage drop of 8V along the wire.
If the insulating wire is exposed to a medium at T∞ = 30°C with a heat transfer
coefficient of h = 24 W/m2 °C, find the temperature at the interface of the wire
and the plastic cover in steady state operation. Also, determine the effect of
doubling the thickness of the plastic cover on the interface temperature.
6
Problem
1. Heat transfer is steady state.
2. Heat transfer is 1-D,
3. Thermal properties are constant.
4. The thermal contact resistance at the interface is negligible.
5. Heat transfer coefficient accounts for the radiation effects, if any.
6. Heat generation is uniform
Properties:
k= 0.15 W/m0C, h= 12 W/m2 0C
Analysis:
,
7
W80108 VIWQ e
Solution
• Outer surface Area
• Thermal resistances
• The interface temperature can be determined
• Critical radius
• Doubling thickness?
8
2
2 m 11.050035.022 LrA
C/W 0.940180.76R
C/W 18.0515.02
5.15.3ln
2
ln
C/W 76.011.012
11
0
plasticconvtotal
012plastic
0
2
conv
RR
kL
rrR
hAR
C 103T94.0
03-T80
T-T 0
11
total
1
RQ
mm 12.5m 0012512
15.0rcr
h
k
cr
crnew2,
r reaches radiusouter theuntilfer heat trans theincreasing Thus
r than less is which mm, 6r
A 5-mm-diameter spherical ball at 50 0C is covered by a 1-mm-thick
plastic insulation (k=0.13 W/m ·0C). The ball is exposed to a medium
at 15 0C, with a combined convection and radiation heat transfer
coefficient of 20 W/m2·0C.
Determine if the plastic insulation on the ball will help or hurt heat
transfer from the ball.
9
Problem
Assumptions : 1. steady state, 2. one-dimensional , 3. Thermal properties are constant.
4 The thermal contact resistance at the interface is negligible.
Properties: k = 0.13 W/m⋅°C.
Analysis: The critical radius of plastic insulation is
10
2
cr
ran greater th iswhich
mm 13m 013.020
13.022
h
kr
Solution
11
HEAT TRANSFER FROM FINNED SURFACES
When Ts and T are fixed, there are two
ways to increase the rate of heat transfer:
• Increase the convection heat transfer
coefficient h. This may require the
installation of a pump or fan or add larger
size of fan. Not adequate.
• Increase the surface area As by attaching
to the surface extended surfaces called
fins made of highly conductive materials
such as aluminum.
Newton’s law of cooling: The rate of heat transfer
from a surface to the surrounding medium
12
The thin plate fins of a car
radiator greatly increase the
rate of heat transfer to the air.
13
Fin Equation
Volume element of a fin at location x
having a length of x, cross-sectional
area of Ac, and perimeter of p.
14
Fin Equation
Volume element of a fin at location x
having a length of x, cross-sectional
area of Ac, and perimeter of p.
where
Temperature
excess
and
15
The general solution of the
differential equation
Boundary condition at fin base
Boundary conditions at the fin
base and the fin tip.
16
Under steady conditions, heat
transfer from the exposed surfaces
of the fin is equal to heat conduction
to the fin at the base.
1 Infinitely Long Fin
(Tfin tip = T)
Boundary condition at fin tip
The variation of temperature along the fin
The steady rate of heat transfer from the entire fin
17
2 Negligible Heat Loss from the Fin Tip
(Adiabatic fin tip, Qfin tip = 0)
Boundary condition at fin tip
The variation of temperature along the fin
Heat transfer from the entire fin
Fins are not likely to be so long that their temperature approaches the
surrounding temperature at the tip. A more realistic assumption is for
heat transfer from the fin tip to be negligible since the surface area of
the fin tip is usually a negligible fraction of the total fin area.
18
3 Specified Temperature (Tfin,tip = TL)
In this case the temperature at the end of the fin (the fin tip) is
fixed at a specified temperature TL.
This case could be considered as a generalization of the case of
Infinitely Long Fin where the fin tip temperature was fixed at T.
19
4 Convection from Fin Tip
The fin tips, in practice, are exposed to the surroundings, and thus the proper
boundary condition for the fin tip is convection that may also include the effects
of radiation. Consider the case of convection only at the tip. The condition
at the fin tip can be obtained from an energy balance at the fin tip.
20
Replace the fin length L in the relation for
the insulated tip (slide 49) by a corrected
length defined as
Corrected fin length Lc is defined such
that heat transfer from a fin of length Lc
with insulated tip is equal to heat transfer
from the actual fin of length L with
convection at the fin tip.
t is the thickness of the rectangular fins.
D is the diameter of the cylindrical fins.
4 Convection from Fin Tip - cont
21
Fin Efficiency
Zero thermal resistance or infinite
thermal conductivity (Tfin = Tb)
22
mL
23
Efficiency of straight fins of rectangular, triangular, and parabolic profiles.
Triangular and parabolic are more efficient than rectangular, contain less
material and more suitable for applications requiring less weight and less
space
24
25
• The fin efficiency decreases with increasing fin length because
of decrease in fin temperature with length.
• Fin lengths that cause the fin efficiency to drop below 60 percent
usually cannot be justified economically.
• The efficiency of most fins used in practice is above 90 percent.
26
Fin
Effectiveness
Effectiveness
of a fin
• The thermal conductivity k of the fin
material should be as high as possible.
Use aluminum, copper, iron.
• The ratio of the perimeter to the cross-
sectional area of the fin p/Ac should be
as high as possible. Use slender or thin
pin fins.
• Low convection heat transfer coefficient
h. Place fins on the gas (air) side.
27
Proper Length of a Fin
Because of the gradual temperature drop
along the fin, the region near the fin tip makes
little or no contribution to heat transfer.
mL = 5 an infinitely long fin
mL = 1 offers a good compromise
between heat transfer
performance and the fin size.
28
• Heat sinks: Specially
designed finned surfaces
which are commonly used in
the cooling of electronic
equipment, and involve one-
of-a-kind complex
geometries.
• The heat transfer
performance of heat sinks is
usually expressed in terms of
their thermal resistances R.
• A small value of thermal
resistance indicates a small
temperature drop across the
heat sink, and thus a high fin
efficiency.
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