Chapter 3 Part 1
One-Dimensional, Steady-State
Conduction
Thermal Resistance
T1
h1
T
T2
h2
T3
k
x
Hot air
q1
L
q2
T4
Cold air
q3
Accumulation of Energy = Energy In - Energy Out + Energy Generated
Therefore:
q1 = q2 =q3
q1  h1 A1 T1  T2 
q1
 T1  T2
h1 A1
q2  kA2
q2 L
 T2  T3
kA2
T3  T2 
L
q3  h2 A3 T3  T4 
+
q3
 T3  T4
h2 A3
q3
q1
q2 L


 T1  T4
h1 A1 kA2 h2 A3
Since: q1 = q2 =q3 =q
and
A1 = A2 =A3 =A
 1

L
1
  T1  T4
q


 h1 A kA h2 A 


we have
T1  T4
q
1
L
1


h1 A kA h2 A
Solving for q
This is analogous to electrical resistance placed in series,
V
V
I
 i n
R1  R2  ...  Rn
 Ri
i 1
T1
h1
T2
h2
T4
T3
k
Hot air
L
Cold air
R2
R3
q
R1
So
1
,
R1 
h1 A
Therefore
L
R2 
kA
1
R3 
h2 A
and
T1  T4
T
q
 i 3
1
L
1


Ri

h1 A kA h2 A i 1
General form Cartesian Coordinate
L
kA
1
R
hA
R
Thermal resistance (conduction)
Thermal resistance (convection)
Heat transfer rate
q
Equation 3.1
Equation 3.2
T
R
i 1
i
Equation 3.3
Composite Wall (series)
T1
h1
T3
T2
A
B
C
Cold air
T4
Hot air
kA
kB
LA
LB
h2
T6
T5
kC
LC
q
R1
R2
R3
R4
R5
So
1
LA
LB
,
,
,
R1 
R2 
R3 
h1 A
kA A
kB A
and
Therefore
LC
R4 
kC A
1
R5 
h2 A
T1  T6
T
q
 i 5
L
1
L
L
1
 A  B  C 
Ri

h1 A k A A k B A kC A h2 A i 1
Composite Wall (series-parallel)
F
kF
T1
A2
H
A1
E
kE
LA
G
kG
A3
LF = LG
kH
T2
LH
RF
q
RE
RH
RG
So
LG
LE
LF
, RF 
, RG 
RE 
k E A1
k F A2
kG A3
R1
Recall
=
and R  LH
H
k H A1
R3
R2
Where
Therefore
1
1
1
 
R3 R1 R2
q
so
R3 
1
1
1

R1 R2
T1  T2


LE 
1

k E A1  k F A2  kG A3
 L
LG
 F


  LH
 k H A1


Equation 3.4
T
T


 R RTotal
Overall Heat Transfer Coefficient
For conveniences we can define the overall heat transfer
coefficient as
1
U
RTotal A
Which yields an expression analogous to Newton’s law
q  UAT
Example 3.1
A 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.
Tw = 400oC
hr= 25 Wm-2k-1
Composite
Window
LA = LB
LA
Ts,i
LB
Ts,o <50oC
Hot air
Ta = 400oC
hi = 25 Wm-2k-1
B, kB =0.08 Wm-1k-1
Cold air
Oven cavity
T∞ = 25oC
ho = 25 Wm-2k-1
A, kA = 0.15 Wm-1k-1
Solution
Assumptions:
1.
2.
3.
4.
Steady-state
Conduction through the window is one dimensional
Contact resistance is negligible
Radiation exchange between window outer surface
and surrounding is negligible
Energy balance
Accumulation of Energy = Energy In - Energy Out + Energy Generated
Solution
Energy In = qB = qA = qr +qi
Energy Out = qo
so
qB = qA = qr +qi = qo
qo  ho ATs,o  T 
qr  qi  hr ATw  Ts,i   hi ATa  Ts,i 
kb A
Ts,o  TA,B 
qB  
LB
kA A
TA,B  Ts,i 
qA  
2 LB
Solution
ho ATs,o  T   hr ATw  Ts,i   hi ATa  Ts,i 
T
s ,o
 T   2Ta  Ts,i 
kb A
Ts,o  TA,B 
ho ATs ,o  T   
LB
kA A
TA,B  Ts,i 
ho ATs ,o  T   
2LB
Solution
Rewrite as
T
s ,o
 T 
2
2LB ho Ts ,o  T 
kA
LB ho Ts ,o  T 
kb
T
s ,o
 Ta  Ts ,i 
 Ts ,i  TA, B 
 TA, B  Ts ,o 
 T  2LB ho Ts ,o  T  LB ho Ts ,o  T 


 Ta  Ts ,o 
2
kA
kb
Solution
Replace with values
50 C  25 C   2L 25
o
o
B
2
W
50 C  25 C   L 25
o
2
m K
o
B
0.15W mK
12.5o C  8333LB
o
C
16146LB
W
m K
C
o
o
0.08W mK
m  7812.5LB
o
50 C  25 C   400 C  50 C 
o
2
o
C
o

350
C
m
o

337
.
5
C
m
337.5o C
LB 
 0.0209m
o
C
16146 m
L  3LB  0.0627m  62.7mm
o
Solution
Thermal Circuit
Tw
1
hr A
Ts ,i
Ta
LA
kA A
LB
kB A
1
hi A
Energy In =
1
ho A
Ts ,o
T
a
 Ts ,o 
R
Energy Out = ho ATs,o T 
T
Solution
Thermal Circuit
ho ATs ,o  T

T

 1 
 1 

 R    h A    h A 
 r 
 i 
1
 Ts ,o 
a
R
1 1
T

  LA  LB

k A A kB A

 Ts ,o 
ho ATs ,o  T  
hi A  hr A1  LA  LB
k A A kB A
a
T
 Ts ,o 
ho Ts ,o  T  
1
2L
L
 B B
hi  hr
k A kB
a
Solution
Thermal Circuit

Ta  Ts ,o 
 1
2 LB LB 


  
 hi  hr k A k B  ho Ts ,o  T 

Ta  Ts ,o 
 2 1
1
LB    

 k A k B  ho Ts ,o  T  hi  hr
 Ta  Ts ,o 
1 

 h T  T   h  h 
o s ,o

i
r 

LB 
 2
1
  
 k A kB 
Solution
Thermal Circuit


o
o


400
C

50
C
1


 25W 2 50o C  25o C  25W 2  25W 2 
m K
m K
m K 
LB  


2
1



W
W
 0.15 mK 0.08 mK 




o
o


400
C

50
C
1


 25W 2 50o C  25o C  25W 2  25W 2 
m K
m K
m K 
LB  


2
1



W
W
 0.15 mK 0.08 mK 


LB  0.0209m
Direct Application of Fourier’s Law
Conditions:
Steady-state
no heat source or sink
one 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 function
of T.
Hence we may use Fourier’s law in the integral form
without knowing the temperature distribution
T1
dx
qx 
   k T dT
x0 A x 
T0
Furthermore if the area and the conductivity are constant
x1
x
qx
  k T
A
Example
Calculate 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.
T2 =600K
x2 = 0.25m
T1 = 400K
x1 = 0.05m
x
Solution
Assumptions:
Steady State
One-dimensional conduction in x direction
No energy source or sink
Constant properties
Based on the assumptions the heat transfer rate is constant along the
x direction. Hence from table A.2, pyroceram(500k):
k = 3.46 W m-1 K-1
From equation 1.1


q
  kT
A
q x   Ak
dT
dx
Where A the cross-section
area is
A
D 2
4
q x  k

a 2 x 2
4
a 2 x 2 dT
4
dx
dx
a 2
q x 2  k
dT
x
4
2
dx
a 2
x qx x 2  T k 4 dT
1
1
x2
T
1 1
a 2
T2  T1 
qx     k
4
 x2 x1 
qx 
k
a 2
T2  T1 
4
1 1
  
 x2 x1 
W  0.252
600K  400K 
 3.46
mK 4
qx 
 2.12W
1 
 1



 0.25m 0.05m 
Thermal Resistance
The Cylinder
Knowing the inside surface temperature Ti
and the outside surface temperature To of a
cylinder and assuming that:
conductivity k is constant
steady-state
no heat source or sink
L is much larger than ro
heat transfer solely radial


than we can use equation 1.1 q  kAT  to
determine the heat transfer rate across the
cylinder, so:
qr  kA
dT
dT
 k 2rL
dr
dr
Ti
L
ri
ro
To
By separation of variables we find
qr
dr   k 2LdT
r
ro
T
o
qr
r r dr   T k 2LdT
i
i
qr lnro   lnri   2kLTi  To 
qr 
2kLTi  To 
 ro 
ln 
 ri 
T Ti  To 
qr 

R
r 
ln o 
 ri 
2kL
 ro 
ln 
ri 

R
2kL
R1 
1
For a cylinder with several layer
2h1r1 L
r 
ln 2 
r1 

R2 
2k AL
r 
ln 3 
r2 

R3 
2k BL
r 
ln 4 
r
R4   3 
2 k C L
1
R5 
2h1r4 L
C
B
A
r1
r2
r3
T∞,1
r4
T1
T2
T3
T4
qr
R1
R2
R3
R4
T∞,4
R5
qr 
T,1  T, 4
 r4 
 r2 
 r3 
ln  ln  ln 
r3 
r1 
r2 
1
1







2h1r1 L 2k AL 2k BL 2k BL 2h4r4 L
qr 
T,1  T, 4
R
 UAT,1  T, 4 
So for A1 which is define as
U1 
A1  2r1L
1
1 r1  r2  r1  r3  r1  r4  r1
 ln   ln   ln  
h1 k A  r1  k B  r2  k B  r3  h4 r4
Similarly for A2 which is define as
A2  2r2 L
1
U2 
r2
r2  r2  r2  r3  r2  r4  r2
 ln   ln   ln  
h1r1 k A  r1  k B  r2  k B  r3  h4 r4
And
U1 A1  U 2 A2  U 3 A3  U 4 A4 
 R 
1
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∞).
r
ri
k
T∞
h
Cold air
Ti
Assumptions:
Steady state
One dimensional temperature gradient (radial)
Negligible tube wall thermal resistance
Constant properties for insulation
Negligible radiation
r
ln 
 ri 
2kiL
q
1
2hL
We can define q’ as the heat transfer rate per unit length and R’t as the
total thermal resistance therefore
r
ln 
 ri 
2ki
And
Optimum at
q’
1
2hr
r
ln 
ri 
1

R 't 

2ki
2 h r
T  Ti
q' 
R't
 r

 ln 



dR't d   ri 
1 
1
1
 



0

2
dr dr  2ki
2hr  2ki 2hr




Hence, we have an optimum at
ki
r
h
we determine whether it is a minimum or maximum using the second
derivative
d 2 R't
1
1


2
2
dr
2kir
hr 3
at
r
ki
h
1
d 2 R't
1
1 
1





0
2 
2

1
dr
ki   ki 2ki  2k 3

 h 
i
2
h


Therefore, we a minimum, which is usually called the critical radius
using the typical values ki  0.055W mK
the critical radius is
and
h  5 W m2 K
ki 0.055W mK
rc  
 0.011m
W
h
5 m2 K
So if, rc is larger than ri, heat transfer will increase with the addition of
insulation up to a thickness of rc  ri . Therefore for ri = 0.005 m
7
R’t
R mk/W
6
5
R’conduction
4
3
R’convection
2
1
0
0
0.005
0.01
rc  ri
0.015
0.02
Thermal Resistance
The Sphere
Knowing the inside surface temperature Ti
and the outside surface temperature To of a
sphere and assuming that:
conductivity k is constant
steady-state
no heat source or sink
heat transfer solely radial


than we can use equation 1.1 q  kAT  to
determine the heat transfer rate across the
sphere, so:
qr   kA
dT
dT
 k 4r 2
dr
dr
ro
ri
k
Ti
To
ro
T
o
dr
qr  2   k 4  dT
r
ri
Ti
 1   1 
qr           k 4 To  Ti 
  ro   ri  
k 4 Ti  To 
qr 
1 1

ri ro
1
r
k 4
1 1
  
 ri ro 
Example
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?
mhfg
Vent
r2=0.275 m
r1=0.25 m T∞,2= 300K
h=20W m-2 K-1
k
Liquid nitrogen
T ∞,1= 77K
ρ = 804 kg m-3
hfg = 2 x 105 J kg-1
Cold air
Assumptions:
Steady state
One dimensional temperature gradient (radial)
Negligible sphere wall thermal resistance
Constant properties for insulation
Negligible radiation
Thermal circuit
q
1
k 4
1 1
  
 r1 r2 
qr 
1
2hr22
T
, 2
1
k 4
 T,1 
1 1
1
   
2
r
r
h
4

r
2
 1 2
qr 
300 77K
1
0.0017Wm 1 K 1 4
1 
1
 1




2
2
1
0
.
25
m
0
.
275
m

 20Wm K 4 0.275m 
223 K
qr 
 13.06W
1
1
17.02W K  0.05W K
What is mass of liquid nitrogen boiling-off?
Performing an energy balance for a control surface about the
nitrogen we have
Ein  Eout  0
Ein  q
 hfg
Eout  m
 h fg  0
qm
q
13.06Js1
5
1

m


6
.
53
x
10
kgs
h fg 2 x105 Jkg1
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
Heat
equation
Temperature
distribution
Heat flux
Spherical Wall
d 2T
0
dx2
1 d  dT 
r
0
r dr  dr 
1 d  2 dT 
r
0
r 2 dr  dr 
x
Ts ,1  T
L
ln r 
r
Ts , 2  T  2 
r
ln 1 
 r2 
  r1  
 1   r  
Ts ,1  T 

r


1
1  

  r2  
T
k
L
Heat rate
T
kA
L
Thermal
resistance
Cylindrical Wall
L
kA
k
T
r
r ln 2 
 r1 
2kL
T
r
ln 2 
 r1 
ln r2 
 r1 
2kL
k
T


r 2  1    1 
 r1   r2 
4k
T
 1  1 
 r  r 
 1  2 
 1  1 
 r  r 
 1  2 
4k