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Chapter 3 Part 1
One-Dimensional, Steady-State Conduction
Thermal Resistance
x
T
Hot air Cold air
T1
T2 T3
T4
L
q1 q2 q3
h1
k
h2
Accumulation of Energy = Energy In - Energy Out + Energy Generated
Therefore: q1 = q2 =q3
21111 TTAhq
L
TTkAq 23
22
43323 TTAhq
2111
1 TTAh
q
322
2 TTkA
Lq
4332
3 TTAh
q+
4132
3
2
2
11
1 TTAh
q
kA
Lq
Ah
q
4121
11TT
AhkA
L
Ahq
Since: q1 = q2 =q3 =q and A1 = A2 =A3 =A we have
Solving for q
AhkAL
Ah
TTq
21
41
11
This is analogous to electrical resistance placed in series,
ni
ii
n R
V
RRR
VI
1
21 ...
Hot air Cold air
T1
T2 T3
T4
L
qR2 R3
h1
k
h2
R1
SoAh
R1
1
1
kA
LR 2
, andAh
R2
3
1
Therefore
3
121
41
11 i
iiR
T
AhkAL
Ah
TTq
1iiR
Tq
Equation 3.3
General form Cartesian Coordinate
Heat transfer rate
Thermal resistance (conduction)
Thermal resistance (convection)hA
R1
kA
LR Equation 3.1
Equation 3.2
Composite Wall (series)
Hot air
Cold air
T1
T2
T5
T6
LA
qR2 R3
h1
h2
R1 R5R4
T4
T3
LB LC
A B C
kA kB kC
SoAh
R1
1
1
Ak
LR
A
A2,
Therefore
5
121
61
11 i
ii
C
C
B
B
A
A R
T
AhAk
L
AkL
AkL
Ah
TTq
,Ak
LR
B
B3
AhR
25
1
Ak
LR
C
C4,
and
Composite Wall (series-parallel)
T2
T1
LA
qRE
RF
LF = LG LH
E
F
H
kE
kF
kG
GkH
RG
RH
A1
A2
A3
So1Ak
LR
E
EE
ThereforeTotal
H
H
G
G
F
FE
E
R
T
R
T
AkL
L
Ak
LAkAk
L
TTq
1321
21
1
, , and2Ak
LR
F
FF
3Ak
LR
G
GG
1Ak
LR
H
HH
R1
R2
=RecallR3
Where213
111
RRR so
21
3 111
RR
R
Equation 3.4
Overall Heat Transfer Coefficient
For conveniences we can define the overall heat transfer coefficient as
ARU
Total
1
Which yields an expression analogous to Newton’s law
TUAq
Example 3.1A leading manufacturer of household appliances is proposing a self-cleaning oven design that involves use of a composite window separating the oven cavity from the room air. The composite is to consist of two high temperature plastic (A and B) of thicknesses LA = 2 LB and thermal conductivities kA = 0.15 Wm-1k-1 and kB = 0.08 Wm-1k-1. During the self cleaning process the oven wall and air temperatures, Tw and Ta are 400oC, while room temperature T∞ is 25oC. The inside convection and radiation heat transfer coefficients hi and hr, as well as the outside convection coefficient ho, are each approximately 25 Wm-2k-1. What is the minimum window thickness L = LA + LB, needed to ensure a temperature that is 50oC or less at the outer surface of the window? This temperature must not be exceeded for safety reasons.
Oven cavity
A, kA = 0.15 Wm-1k-1
B, kB =0.08 Wm-1k-1
LA LB
Ts,o <50oCTs,i
Hot air
Cold air
Composite WindowLA = LB
Tw = 400oChr= 25 Wm-2k-1
Ta = 400oChi = 25 Wm-2k-1
T∞ = 25oCho = 25 Wm-2k-1
Solution
Assumptions:1. Steady-state2. Conduction through the window is one dimensional3. Contact resistance is negligible4. Radiation exchange between window outer surface
and surrounding is negligible
Energy balance
Accumulation of Energy = Energy In - Energy Out + Energy Generated
SolutionEnergy In = qB = qA = qr +qi
isaiiswrir TTAhTTAhqq ,,
Energy Out = qo so qB = qA = qr +qi = qo
isBAB
AA TT
L
Akq ,,2
TTAhq osoo ,
BAosB
bB TT
L
Akq ,,
Solution
isaiiswroso TTAhTTAhTTAh ,,,
isBAB
Aoso TT
L
AkTTAh ,,, 2
BAosB
boso TT
L
AkTTAh ,,,
isaos TTTT ,, 2
Solution
Rewrite as isa
os TTTT
,,
2
BAisA
osoB TTk
TThL,,
,2
osBAb
osoB TTk
TThL,,
,
osab
osoB
A
osoBos TTk
TThL
k
TThLTT,
,,, 2
2
Solution
Replace with values
CCCCLCCLCC oo
mKW
ooKm
WB
mKW
ooKm
WB
oo
5040008.0
255025
15.0
2550252
2
2550 22
CLLC om
CBm
CB
o oo 3505.781283335.12
CL om
CB
o 5.33716146
mC
Lm
C
o
B o0209.0
16146
5.337
mmmLL B 7.620627.03
SolutionThermal Circuit
Ahi
1
Ahr
1
Aho
1
TaT
wT
Ak
L
B
B
Ak
L
A
A
isT ,
osT ,
Energy In =
Energy Out = TTAh oso ,
R
TT osa ,
SolutionThermal Circuit
R
TTTTAh osa
oso,
,
Ak
L
Ak
L
AhAhR
B
B
A
A
ri
111
11
AkL
AkL
AhAh
TTTTAh
B
B
A
Ari
osaoso
1
,,
B
B
A
B
ri
osaoso
kL
kL
hh
TTTTh
21
,,
SolutionThermal Circuit
TTh
TT
k
L
k
L
hh oso
osa
B
B
A
B
ri ,
,21
rioso
osa
BAB hhTTh
TT
kkL
112
,
,
BA
rioso
osa
B
kk
hhTTh
TT
L12
1
,
,
SolutionThermal Circuit
mKW
mKW
KmW
KmWoo
KmW
oo
B
CCCC
L
08.01
15.02
25251
25502550400
222
mKW
mKW
KmW
KmWoo
KmW
oo
B
CCCC
L
08.01
15.02
25251
25502550400
222
mLB 0209.0
Direct Application of Fourier’s Law
Conditions: Steady-stateno heat source or sinkone dimensional system
Consequences: For any differential element dx, qx = qx+dx. Even if the
area is a function of x and the conductivity is a functionof T.Hence we may use Fourier’s law in the integral form without knowing the temperature distribution
1
0
1
0
T
T
x
xx dTTkxA
dxq
Furthermore if the area and the conductivity are constant
TkA
xqx
ExampleCalculate the heat rate through a pyroceram cone of circular cross section with a diameter D = ax = 0.25x. The small end is located at x1 = 0.50 mm and the large end at x2 = 250 mm. The end temperatures are T1 = 400 K and T2 = 600 K. The lateral surfaces are well insulated.
xx2 = 0.25m x1 = 0.05m
T1 = 400KT2 =600K
SolutionAssumptions:
Steady StateOne-dimensional conduction in x directionNo energy source or sinkConstant properties
Based on the assumptions the heat transfer rate is constant along thex direction. Hence from table A.2, pyroceram(500k):
k = 3.46 W m-1 K-1
q
Ak T From equation 1.1
dx
dTAkqx
Where A the cross-section area is
44
222 xaDA
dx
dTxakqx 4
22
dTa
kx
dxqx 4
2
2
2
1
2
14
2
2
T
T
x
x
x dTa
kx
dxq
12
2
12 4
11TT
ak
xxqx
12
12
2
114
xx
TTa
kqx
W
mm
KKmKW
qx 12.2
05.01
25.01
400600425.0
46.32
Thermal ResistanceThe Cylinder
Knowing the inside surface temperature Ti and the outside surface temperature To of a cylinder and assuming that:
conductivity k is constantsteady-stateno heat source or sinkL is much larger than ro
heat transfer solely radialthan we can use equation 1.1 to determine the heat transfer rate across the cylinder, so:
ro
ri
Ti
To
L
TkAq
dr
dTrLk
dr
dTkAqr 2
By separation of variables we find
LdTkdrr
qr 2
o
i
o
i
T
T
r
r
r LdTkdrr
q 2
oiior TTLkrrq 2lnln
i
o
oir
r
r
TTLkq
ln
2
Lk
rr
TT
R
Tq
i
o
oir
2
ln
Lk
rr
R i
o
2
ln
r2r1
r3
r4
For a cylinder with several layer
qr
R2 R3R1 R5R4
Lk
rr
RC2
ln3
4
4
LrhR
111 2
1
Lk
rr
RA2
ln1
2
2
Lk
rr
RB2
ln2
3
3
LrhR
415 2
1
T∞,1
T1
T2
T3T4
T∞,4
AB
C
LrhLk
rr
Lk
rr
Lk
rr
Lrh
TTq
BBA
r
44
3
4
2
3
1
2
11
4,1,
21
2
ln
2
ln
2
ln
21
4,1,4,1,
TTUA
R
TTqr
44
1
3
41
2
31
1
21
1
1
lnlnln1
1
rhr
rr
kr
rr
kr
rr
kr
h
U
BBA
So for A1 which is define as LrA 11 2
Similarly for A2 which is define as LrA 22 2
44
2
3
42
2
32
1
22
11
2
2
lnlnln
1
rhr
rr
kr
rr
kr
rr
kr
rhr
U
BBA
And 1
44332211
RAUAUAUAU
Example
The possible existence of an optimum insulation thickness for radial systems is suggested by the presence of competing effects associated with an increase in this thickness. In particular, although the conduction resistance increases with the addition of insulation, the convection resistance decreases due to increasing outer surface area. Hence there may exist an insulation thickness that minimizes heat loss by maximizing the total resistance to heat transfer.
a) Is there an optimum thickness associated with the application of insulation to a thin-walled copper tube of radius ri used in the transport of a refrigerant. The temperature of the refrigerant (Ti) is less than the temperature of the ambient air (T∞).
Cold air
ri
r
kTi
T∞
h
Assumptions:Steady stateOne dimensional temperature gradient (radial)Negligible tube wall thermal resistanceConstant properties for insulationNegligible radiation
q
Lk
rr
i
i
2
ln
Lh2
1
q’
ii
k
rr
2
ln
rh2
1
We can define q’ as the heat transfer rate per unit length and R’t as the total thermal resistance therefore
Andrhk
rr
Ri
it 2
1
2
ln
'
t
i
R
TTq
''
Optimum at 02
1
2
1
2
1
2
ln'
2
rhkrhk
rr
dr
d
dr
dR
ii
it
Hence, we have an optimum at
at
322
2 1
2
1'
rhrkdr
Rd
i
t
h
kr i
we determine whether it is a minimum or maximum using the second derivative
01
2
1
2
111'
23
22
2
hkkk
hkdr
Rd
iiii
t
h
kr i
Therefore, we a minimum, which is usually called the critical radius
0
1
2
3
4
5
6
7
0 0.005 0.01 0.015 0.02
using the typical values and the critical radius is
mKW
ik 055.0Km
Wh 25
mh
kr
KmW
mKW
ic 011.0
5
055.0
2
So if, rc is larger than ri, heat transfer will increase with the addition of insulation up to a thickness of . Therefore for ri = 0.005 mic rr
R’t
R’convection
R’conduction
ic rr
R m
k/W
Thermal ResistanceThe Sphere
Knowing the inside surface temperature Ti and the outside surface temperature To of a sphere and assuming that:
conductivity k is constantsteady-stateno heat source or sinkheat transfer solely radial
than we can use equation 1.1 to determine the heat transfer rate across the sphere, so:
TkAq
dr
dTrk
dr
dTkAqr
24
ri
ro
kTi
To
o
i
o
i
T
T
r
r
r dTkr
drq 4
2
ioio
r TTkrr
q
4
11
oi
oir
rr
TTkq
114
oi rrkr
11
4
1
A spherical, thin walled metallic container is used to store liquid nitrogen at 77 K. The container has a diameter of 0.5 m and is covered with an evacuated, reflective insulation composed of silica powder. The insulation is 25 mm thick, and its outer surface is exposed to ambient air at 300K. The convection coefficient is known to be 20 W m-2 K-1. The latent heat of vaporization and the density of liquid nitrogen are 2x105 J kg-1 and 804 kg m-3, respectively.
What is the rate of heat transfer to the liquid nitrogen?
Example
Cold air
r1=0.25 mr2=0.275 m
k
Liquid nitrogenT ∞,1= 77K
ρ = 804 kg m-3
hfg = 2 x 105 J kg-1
T∞,2= 300K
h=20W m-2 K-1
mhfg Vent
Assumptions:Steady stateOne dimensional temperature gradient (radial)Negligible sphere wall thermal resistanceConstant properties for insulationNegligible radiation
Thermal circuit
q
222
1
rh
21
11
4
1
rrk
2221
1,2,
4111
41
rhrrk
TTqr
21211 275.0420
1275.01
25.01
40017.01
77300
mKWmmmKWm
Kqr
WKWKW
Kqr 06.13
05.002.17
22311
What is mass of liquid nitrogen boiling-off?
Performing an energy balance for a control surface about the nitrogen we have
0 outin EE
qEin
fgout hmE
0 fghmq
1515
1
1053.6102
06.13
kgsxJkgx
Js
h
qm
fg
Which represents 5.64 kg per day or 7 liters per day
One-dimensional, steady-state solutions to the heat equation with no generation Plane Wall Cylindrical Wall Spherical Wall
Heat equation
Temperature distribution
Heat flux
Heat rate
Thermal resistance
02
2
dx
Td
L
xTTs 1,
L
Tk
L
TkA
kA
L
2
1
1
1,
1
1
rr
rr
TTs
01
dr
dTr
dr
d
r0
1 22
dr
dTr
dr
d
r
2
1
22,
ln
ln
rr
rr
TTs
1
2ln rrr
Tk
21
2 11rrr
Tk
1
2ln2
rrT
kL
21
114
rr
Tk
kL
rr
2
ln1
2
k
rr
4
1121