Post on 26-Dec-2015
BKF2422
Chapter 2
Principles of Steady-state Heat Transfer in
Conduction(Part 1)
Topic Outcomes
It is expected that students will be able to:
• Explain the mechanisms of conduction heat transfer
• Solve problems using steady-state conduction principles
Contents• One Dimensional Conduction Heat Transfer
– Conduction Through a Plane Wall – Conduction Through Solids In Series– Conduction Through Solids In Parallel– Conduction Through a Hollow Cylinder– Conduction Through a Multilayer Cylinders– Conduction Through a Hollow Sphere
• Combined Conduction and Convection and Overall Heat Transfer Coefficient
Conduction• For steady state, the equation can be integrated,
This equation is basically a matter of putting in values to solve.
2112
2
1
2
1
TTxx
k
A
q
dTkdxA
q T
T
x
x
x
Conduction
Conduction Through a Plane Wall
The temperature various linearly with distance.
0 ΔxΔx Distance,x (m)
T1
T2
Temperature, (K)
T2
T1q
R
TT
kAx
TTTT
x
kAq 2121
21
ConductionConduction Through Solids In Series
T1 A B C
q T2 T3
∆xA ∆xB ∆xC T4
• The heat flow, q must be the same in each layer;
C
RRR
TTTT
X
AkTT
X
AkTT
X
Akq
BAC
c
B
B
A
A
41433221
Ak
xR
A
AA
Ak
xR
B
BB
Ak
xR
C
CC
Try This
A cold storage room is constructed of an inner layer of 12.7 mm of pine, a middle layer of 101.6 mm of cork board and an outer layer of 76.2 mm concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of concrete. The conductivites, k for pine: 0.151; cork board: 0.0433; and concrete: 0.762 W/m.K. Calculate
a)the heat loss in W for 1 m2 and b)the temperature at the interface between wood and cork board.
Answer: (-16.48 W, 256.79 K)
ConductionConduction Through Solids In Parallel
• The heat flow, q must be the sum of the heat flow through A and B,
q
A
B
T1T2
T3T4
4321 TTX
AkTT
X
Akqqq
B
B
A
ABA
T1=T3 and T2=T4 21
2121 11
//TT
RRXAk
TT
XAk
TTq
BABBAA
Try This
From figure above:a)Draw the electrical analog. b)Derive the heat flow equation.
A
B
C
D
E
F
G
T1 T2 T3 T4 T5
ConductionConduction Through A Hollow Cylinder T2 L
r2
q T1
r1
• The cross-sectional area normal to the heat flow is, A =2∏rL.
• The rate of heat transfer,
dr
dTk
A
q
Conduction Through A Hollow Cylinder
2
1
2
12
2r
r
T
TdTk
r
dr
L
q
rLAdr
dTk
A
q
12
12
12
12
22ln
22
ln
LrLr
LrLr
AA
AAAlm
12
21
rr
TTkAq lm
or
2112ln
2TT
rr
Lkq
Where:
kL
rrTT
q
2ln 12
21
Try This
A thick wall cylindrical tubing of hard rubber (k=1.15 W/m.K) having an inside radius of 5 mm and an outside radius of 20 mm is being used as a temporary cooling coil in a bath. Ice water is flowing rapidly inside, and the inside wall temperature is 274.9 K. The outside temperature is at 297.1 K. Calculate the heat flow.
(Answer: -15.2 W)
ConductionConduction Through a Multilayer Cylinders
Example, heat is being transferred through the walls of an insulated pipe.
T1T2T3
T4
r1r2
r3
r4q
A
B
C
Conduction• At steady-state, the heat-transfer rate q, be the
same for each layer.
• The rate of heat transfer,
or
Lkrr
TT
LkrrTT
Lkrr
TTq
CBA 2ln
2ln
2ln 34
43
23
32
12
21
CBA Lkrr
Lkrr
Lkrr
TTq
2ln
2ln
2ln
)(
342312
41
ClmCBlmBAlmA Akrr
Akrr
Akrr
TTq
342312
41 )(
Try This
A thick walled tube of stainless steel (A) having a k = 21.63 W/m.K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254 thick layer of insulation (B), k = 0.2423 W/m.K. The inside wall temperature of the pipe is 811 K and the outside is at 310.8 K. For a 0.305 m length pipe, calculate
a)the heat loss and b)the temperature at the interface between the metal and the insulation.
Answer: (331.7 W, 805.5 K)
ConductionConduction Through a Hollow Sphere T2 r2
q r1 T1
• The cross-sectional area normal to the heat flow is, A = 4∏r2.
• The rate of heat transfer,
dr
dTk
A
q
Conduction
2
1
2
124
r
r
T
TdTk
r
drq
k
rr
TTTT
rr
kq
41111
4
21
2121
21
ConductionCombined Conduction and Convection and
Overall CoefficientTemperature Profile for Heat Transfer with convective
boundary: Plane Wall
Metal wall
Warm liquid ACold fluid B
T1
T2
T3
T4
q
hi
ho
Δx
A
ConductionCombined Convection and Conduction and Overall
Coefficients.• Heat flow with convective boundaries: plane wall
41
41
11TTUA
hkxh
TTAq
oi
AhkAxAh
TTq
oi 1141
4303221 TTAhTTx
kATTAhq i
Where: oi hkxh
U11
1
Overall
Coefficients
ConductionTemperature Profile for Heat Transfer with convective boundary: Cylindrical Wall
T1
T2
T3
T4
ro
ri
ho
hi
Aiq
Ao
Conduction
ooiioiii hAAkLrrAh
U
2ln1
1
Since the areas depends on inside diameter and outside diameter
Where,
and
ooioii AhkLrrAh
TTq
12/ln141
Overall Coefficients of inside area
Overall Coefficients
of outside area
4141 TTAUTTAUq ooii
oiooiioo hkLrrAhAA
U12ln
1
Try This
Water flows at 50C inside a 2.5 cm inside diameter tube such that hi=3500 W/m2.C. The 1 m tube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m.C. The outside of tube heat loss by convection in air with ho= 7.6 W/m2.C. Calculate the overall heat transfer coefficient and heat loss to surrounding air at 20C.
(Answers; Uo=7.58 W/m2.C, q=19 W)