CHENG 314 Heat Transfer
Dr. Zakir Hossain
E.mail: zhossain@uob.edu.bh
Office: 15-312
Tel: 1787-6374
Department of Chemical Engineering
University of Bahrain
1
2
Part 1
1-Dimensional, Steady-State
Conduction without
Heat Generation
3
The Plane wall
Consider a plane wall between two fluids of different temperature.
Heat flux and heat rate:
dT k
q  k
 (T  T )
dx L
"
x
s ,1
s,2
dT k A
q  k A

(T  T )
dx
L
x
s ,1
s,2
4
Wall
g
KAT2)
=
R
-
Plain
wall
En
=
cylindrical
qu
=
2πkT2)
>
-
cylindrical wall
1/2)
-e &
& ↑
I
P
Inue
=
o
o
resistance
solid
Cylinder
spherican
qV
=
4 πK(T1
-
T2)
(i i)
TT
R
R
=
(I
-
-2) /YπK
&
Uz
-
V
-
4πkv , Vz
R
=
R =
>
-
Plain wall
MV)
>
-
Cylindrical
2πkL
R
=
r-r
-
>
-
spherical
ukrivz
-
The Plane wall
Thermal Resistances and Thermal Circuits
qx  k A
ΔT
q 
R
T
T

L
 L 


k
A


x
t , cond .
R t ,cond .  thermal resistance 
L
kA
ohn' s law, i 
T
q  h A T 
 1 


 h A
R
conv

t ,conv .
V
R
1
hA
Radiation resistance
R
t , rad .

1
R
h A
rad
h    T  T
r
s
surr
"
t , rad .
1

h
T  T 
2
s
rad
2
surr
5
The Plane wall
Thermal circuit for plane
wall with adjoining fluids:
R
t , total
1
L
1



h A kA h A
1
2
(T  T )
T
q 

R
R
 ,1
,2
x
t , total
t , total
6
6
Thermal resistance
for unit surface area:
R
"
t , cond .
L

k
R
"
t , conv .
1

h
7
Composite Wall
k A
k A
k A
T  T  
T  T  
T  T 
q
x
x
x
A
B
1
C
2
2
A
3
3
B
4
C
Solving each equation for ΔT
x
T T  q
k A
x
T T  q
k A
A
1
B
2
2
A
3
B
T T q
3
4
x
k A
C
C
Adding the equations for T1 – T2, T2 – T3,
and T3 – T4, the final rearranged equation is
q
cond

T T
x
x
x


k A k A k A
1
4
A
A
q
cond

B
T T
R  R R
1
A
C
B
4
B
C
T T
q 
R
1
C
4
8
Composite Wall
T T T T T T T T T T
q 




R
R
R
R
R
 ,1
s,1
s,1
2
2
3
3
s,4
 ,4
s,4
x
1
q
2
T T

R R R
s ,1
cond
2
q 
T
4
5
s,4
3
4
T
R
 ,1
x
3
 ,4
total
R
total
11 L L L
1
      
Ah k k k h 
1
A
B
C
A
B
C
4
9
Series – Parallel Composite Wall
Writing Fourier's equation
for each solid and summing;
ΔT
T1  T2
q

 Rt RE  RF RG  RH
RF  RG
L
R 
kA
i
i
i
i
10
Overall Heat Transfer Coefficient
q x  h1 A(T,1  Ts,1)  kA / L(Ts,1  Ts,2 )  h2 A(Ts,2  T,2 )
Overall HT:
q 
Toverall
     (1)
1
L
1


h1 A kA h2 A
The overall HT by combined convection and conduction is expressed in
terms of overall HT coefficient, U (analogus to newton’s law of cooling)
q  UAToverall    (2)
From eq 1 & 2
1
U
1 / h1  L / k  1 / h2
11
Radial System- Cylinder
Fourier's law for radially form
q
dT
 k
A
dr
The cross-sectional area normal to the heat flow is : A = 2 π r L
q dr
  k  dT

2πL r
q
r2
Ts, 2
r1
Ts, 1
2π L
 k
(T  T )
r 
ln 

r


s, 1
s, 2
2
1
q

(T  T ) (T  T )

R
 r
 ln 
 r
2π k L
s, 2
s, 1
s, 2
s, 1
2
1
 r2 
 ln 
r1 
where, R t, cond  
2πk L
12
Radial System- Cylinder
q

(T,1  T,2 )
R
13
Composite Cylindrical Wall
qt 
(T,1  T,4 )
 UA(T,1  T,4 )
R
Note that
UA  Rtot 1
is a constant independent of radius,
14
Radial System-Hollow Sphere
Fourier's law for radially form
q
dT
 k
A
dr
The cross-sectional area
normal to the heat flow is:
A = 4 π r2
q
t

4π k
T  T 
 1   1 
 r    r 
   
s,1
1
2
s,2
15
Spherical Shell
q

t
T  T 
s,1
s,2
R
t
R
t, cond
 1  1
  
r r


4π k
1
2
r2  r1
R t, cond 
4 π kr2 r1
16
Contact Resistance
Handhetrue
RI
RC
Values depend on: Materials A and B, surface finishes, interstitial
conditions, and contact pressure (Tables 3.1 and 3.2)
q
q


TA  TB
T

(x / kA)
R
TA  TB T

(x / k ) Rt,c

Rt,c
Rc

=
x
(m 2 .k / W )
k
·
KA
17
Contact Resistance
T1
T2
Tc1
k1
k2
Tc2
LA
q

LB
T1  T2
LA
LB


 Rt ,c 
k1
k2
18
Contact Resistance
19
Critical Thickness of Insulation
insulation
Consider a metallic cylinder pipe with
high thermal conductivity, radius r1 and
a fixed length of L. the cylinder is
covered with a layer of insulation. The
inside wall temperature is T1 and is
exposed to an environment at To.
K
r2
To
r1
ho
q
T1
T2
Does adding more insulation with a
thermal conductivity of k will
decrease/increase the heat transfer
rate?
20
Critical Thickness of Insulation
To determine this, we write the
equation of q using the two
resistance, assuming the drop in
temperature within the metallic wall
is negligible (i,e . Twi = Two = T1)
q
2 π L (T1  To )
 r2 
ln  
 r1   1
k
r2 h o
21
Critical Thickness of Insulation
To determine the effect of thickness of insulation on q, we take the
derivative of q with respect to r2, equate this result to zero, and obtain the
following maximum heat flow
dq

dr2
- 2 π L (T1  To ) (
1
1
 2 )
r2 k r2 h o
  r2 

 ln  

r
1
1





 k
r2 h o 




2
0
solving
r   k
2
crit.
h
(r2)crit is the critical radius when q is maximum
o
22
Critical thickness of insulation
Ronv
q
dominant
>
-
Rins
R cond
dominant
Uz
(a) If the outer radius r2 is less than rcrit then adding more insulation
will actually increase q.
(b) If the outer radius r2 is greater than rcrit then adding more
insulation will decrease q.
23
What happens when adding more insulation to
wall, cylinder and spherical shell
Wall:
We know that adding more insulation to a wall always decreases heat
transfer. The thicker the insulation, the lower the heat transfer rate. This
is expected, since the heat transfer area A is constant, and adding
insulation always increases the thermal resistance (due to increases of
L) of the wall without increasing the convection resistance.
Cylinder & Sphere
Adding insulation to a cylindrical pipe or a spherical shell, however, is a
different matter. The additional insulation increases the conduction
resistance of the insulation layer but decreases the convection
resistance of the surface because of the increase in the outer surface
area for convection. The heat transfer from the pipe may increase or
decrease, depending on which effect dominates.
24
•
Part 2
General Heat Equation with and
Without Heat source
(Chap 2 +Chap 3)
25
Multidimensional Conduction
• Scalar quantity vs Vector quantity
-scalar: measure magnitude
- Vector: measure both magnitude and direction
•
• Heat flux is a vector quantity
• General statement of conduction rate equation:
Where
is the 3D del operator and T(x, y, z) is the scalar temperature field
26
Heat Flux Components
T  x, y , z 

T  T  T 
q    k
i k
jk
k
x
y
z
• Cartesian Coordinates:
q x
(2.3)
q z
q y
• Cylindrical Coordinates: T  r ,  , z 

T 
T  T 
q   k
i k
jk
k
r
r 
z
qr
q
(2.24)
q z
• Spherical Coordinates: T  r ,  ,  


T 
T 
T
q    k
i k
jk
k
r
r 
r sin  
qr
q
(2.27)
q
27
General Heat Equation or Heat Diffusion Equation
• Wish to know the T distribution, which represents how T varies with position in
the medium.
• A differential equation whose solution provides the temperature distribution in a
stationary medium.
• Cartesian Coordinates:
E in  E out  E g  E st
28
General Heat Equation
• Cartesian Coordinates:
one dimensional
constant property
↑
My
Cancels
if
/
  T    T    T 
T
 k    k    k   q  c p
x  x  y  y  z  z 
t
Net transfer of thermal energy into the
control volume (inflow-outflow)
Thermal energy
Ex()
generation
=
no
heat
generation
steady
state
(2.19)
Change in thermal
energy storage
0
Where
29
>
-
find
distribution
T
Assumptions
Steady-state
1
En(tt) 2
E
/GT)
+
=
=
B C
.
.
1
.
2
T
:
Tz
:
Substract
T
-
=
dimensional
one
.
constant
3.
It
end Integration
.
2
property
-
st integration
B2
0
.
-q x
T
=
4
-
2
=
=
=
-
+
xt
+
+
4x
a
2
4, L +
egl -e92
T2
4
=
0
24 , L
-
T
T
TE
=
2L
-
T
- TET
-
=
c =
T +
TT
+
e
.
12
-
TET 222
+
x =
-
T= TI
L
L
>
-
&
- C I
B
2k
·
O
L
B
-
C
.
2
X = L
TITZ
>
-
When T1
T
=
E
2k
>
-
find
To
=
T2
=
TS
11 -22) s
+
delta To
=
E
+
s
• Cylindrical Coordinates:
1   T  1   T    T 
T
kr

k

k

q


c
p




r r  r  r 2     z  z 
t
(2.26)
where
30
• Spherical Coordinates:
1   2 T 
1
  T 
1
 
T 
T
kr

k

k
sin


q


c
p




r  r 2 sin 2      r 2 sin   
 
t
r 2 r 
(2.29)
where
31
Assume
:
state
steady
>
>
-
one
>
-
no
·
dimensional
T
heat generation
property
T2
constant
>
-
-
one
⑳o
1
/kr22)
E(22
dimensional
0
=
1st
Integration
:
0
~ ST
C
=
-
da
# =
and Integration
T
-1
=
+
(2
V
V
T+ C
:
= I
z
>
-
general
Solution
r2
B. C 1
:
.
T
=
=
-4
+
,
T
=
C2
T
-
-
2/
B C
.
.
2
V = V2
:
T2
T -T2
=
T=
,
Tz
ET
V2
heatouration
steady
State
22
=
vo
subtract
T
-
T2
c
=
4)-tc)
=
Substitute
substitute
i T
C
in
, and
TAt
Equation
c
in
+
I
or
general
2
solution
The Heat Equation-Special cases
• One-Dimensional Conduction in a Planar Medium with Constant Properties
and No Generation
  T 
T
k
  c p
x  x 
t
becomes
 2T
x


2
1 T
 t
k
 thermal diffusivity of the medium  m 2 /s 


c p
↑
32
Boundary and Initial Conditions
• For transient conduction, heat equation is first order in time, requiring
specification of an initial temperature distribution: T  x,t t=0 = T  x,0 
• Since heat equation is second order in space, two boundary conditions
must be specified. Some common cases:
Constant Surface Temperature:
T  0,t  = Ts
Constant Heat Flux:
Applied Flux
-k
T
|x=0= qs
x
33
Boundary and Initial Conditions
Convection:
-k
T
|x=0= h T - T  0,t  
x
Insulated Surface
T
|x=0= 0
x
34
1
2
3
One dimensional
.
steady state
Uniform properties
.
.
Y
No
.
T
generation
heat
g
TX
N
/kuz)
G(r(t)
Integration
0
=
=
0
C
:
=
ST
2nd
I
=
Tr
Integration
J
Plnr
T=
:
B C
. I
<2
+
-
.
V= V
Econd
,
conv
=
kd
=
h/T
+)
-
di
k2
c
=
=
h(TX T)
-
hv
Th
x
,
-
k
4
=
4
B. C
.
i
TS
cylinder
Is t
V
=
h (To -[414 (2)
c (n
r
,
n)
,
T
-
2
T
V = Vz
=
Ts
(, /
=
C2
=
Ts
TS
U2 + (2
-clnUz
+
2
general Solution
4
h
=
4k
(T1-414
TX
=
4 lur
-
-
-
T+
TS +
c
,
/NU
<Inrz
hu
4k
4
4 /InVz -Invi) + R
=
4lV2)
-
4) K
-
=
T=
=
T
122)
as
Tx-TS
inve
Inv
-In
T
=
U
+
is
i
TI
InrlInv
It
4r /
T
=
Un
TATSnr
-
UV
In
-FGsInin
e
-Inv is a
/ E)
+
Ts
T distribution e
Boundary and Initial Conditions
Radiation:
Interface B.C:
35
Temperature Distribution in ID, SS Plane wall with no Heat
generation
•
Consider a plane wall between two fluids of different temperature:
Heat Equation:
•
d  dT 
k
0
dx  dx 
(3.1)
General solution:
T(x)=C1x + C2
X=0
Boundary Conditions:
•
•
x=L
T  0   Ts ,1, T  L   Ts ,2
Temperature Distribution for Constant
k K :
x
T  x   Ts ,1  Ts ,2  Ts ,1 
L
(3.3) 36
Temperature Distribution in ID, SS Plane wall with no Heat
generation
T  x   Ts ,1  Ts ,2  Ts ,1 
x
L
Heat flux  qx  is independent of x.
Heat rate  qx  is independent of x.
37
Temperature Distribution in ID, SS Plane wall
with Heat generation
Heat flux, q and heat rate, q
x
x
is independent of x or not ? Prove it
38
General Heat conduction equation:
  T    T    T 
T
k

k

k

q


c




p


x  x  y  y  z  z 
t
Assumption:
SS, 1D, constant properties
.
d  dT  q

 0
dx  dx  k
Integrating once with respect to x yield
.
q
 dT 
    x  C1
k
 dx 
Integration once again, we obtain
.
T(x )  
q 2
x  C1 x  C2
2k
Where C1 and C2 are an arbitrary constants
39
If B. C. 1
1) x =0; T = Ts1 then C2=Ts1
2) x=L; T = Ts2 then
Ts 2  
.

q
L2  C1 L  Ts1
2k
.



q

2 
C1   Ts 2  Ts1 
L / L
2k




The temperature profile is:
.
T(x )  
.
q 2 Ts 2  Ts1 q
x (

L) x  Ts1
2k
L
2k
Non linear equation
Note:
Heat flux, q and heat rate, q
x
x
are dependent on x
40
If B. C. 2
1) T(x =0) = T0 (center temperature) and,
2) T(x = -L) = Ts1 and T(x = L) = Ts2 , Then;
T distribution
If
then T distribution
.
q 2
x2 
T( x )
L  1  2   Ts
2k 
L 
The center temperature
.
q L2
T  T 
o
s
2k
41
Radial Systems
Cylindrical (Tube) Wall
Solid Cylinder (Circular Rod)
• Heat Equations:
Cylindrical
1 d  dT 
 kr
q0
r dr  dr 
Spherical Wall (Shell)
Solid Sphere
Spherical
1 d  2 dT 
 kr
q0
r 2 dr 
dr 
Temperature Distribution in SS cylindrical
system
Heat flux, q and heat rate, q
x
x
is independent of r or not ? Prove it
43
Radial Systems; heat generation in a hollow
cylinder
When the cylinder is hollow, the boundary conditions
are: at r = r1, T = T1 and at r = r2, T = T2 , and
.
q r 2
T (r )  
 C1 ln( r )  C2
4k
which reduce to
T1
r1
T2
r
r2
  r 
.
.
 ln   
2
2
2
2
q (r1  r )
q (r2  r1 )   r1  
T (r )  T1 
 [(T1  T2 ) 
]
  r1  
4k
4k
 ln   
  r2   44
Hollow cylinder with outside surface insulated (adiabatic)
Te general solution for temperature distribution
.
q r 2
T (r )  
 C1 ln( r )  C2
4k
(1)
.
dT
q r
C1


dr
2k
r
(2)
The constants of the integration are determined from the relevant boundary
conditions which are
T = T1 at
r = r1
and at r = r2 , the conduction region is perfectly insulated and hence heat
flow, q=0
dT
From Fourier’s law q   k A
dr
and accordingly
dT
 0 at r = r2
dr
45
Applying the boundary conditions to expressions (1) and (2) ,
the general solution for temperature distribution becomes
.
 r 
q  (r12  r 2 )
2
T (r )  T1 
 r2 ln  

2k 
2
 r1 
Apparently the temperature distribution is parabolic
46
Electrical to Thermal Energy
Involves a local (volumetric) source of thermal energy due to conversion
from another form of energy in a conducting medium.
The source may be uniformly distributed, as in the conversion from
electrical to thermal energy (Ohmic heating):
I2 R
q
q 

V
V
.
q
Is the volumetric generation heat transfer rate = W/m 3
Generation affects the temperature distribution in the medium and causes
the heat rate to vary with location, thereby precluding inclusion of the
medium in a thermal circuit.
47
Radial Systems; heat generation in a solid cylinder
.
d (r dT ) q r

dr dr
k
 0
Solid Cylinder (Circular Rod)
Upon integration
.
r dT q r 2

 C1
dr
2k
(1)
.
and
dT q r C1


dr
2k
r
.
q r 2
T (r )  
 C1 ln( r )  C2
4k
B.C.
1) r = 0, dT/dr = 0 ; then C1= 0
2) r=ro, T=Ts
Solid
Cylinder
48
Radial Systems; heat generation in a solid cylinder
.
q ro2
Ts  
 C2
4k
(2)
.
and
 ro2
q
C 2  Ts 
4k
Now the temperature distribution is:
.
q ( ro2  r 2 )
T ( r )  Ts 
4k
49
Radial Systems; heat generation in a solid cylinder
The temperature at the center of the solid cylinder (r= 0)
will be
.
2
o
r
q
To  Ts 
4k
50
Heat conduction equation in a sphere
Consider a sphere with cylinder ρ,
specific heat is C, and outer radius
R, the area of the sphere normal to
the direction of heat transfer at any
0
location is A= 4 π r2 where r is the
value of the radius at that location.
R
r
Note that the heat transfer area A
depends on r in this case, and thus
varies with location.
Volume element
Fig. 3
51
Heat conduction equation in a sphere
One dimensional heat conduction equation for sphere (constant k) is
determined to be
1   T 
r
0
r r  r 
2
2
After simplification
T (r )  
r1 r2
r r2  r1 
T1  T2  
r2 T2  r1T1
r2  r1 
T1  T2
q sphere  (4 π k r1 r2 )
r2  r1
52
Example 1
Find the heat transfer through the composite wall sketched.
>
-
>
-
find
Teand Ts
Draw
T-Profile
....
53
RA
1
=
Ka A
=
0
1 67 x
.
=
RB
10-3
k/W
0 05
=
.
ox
=
o
=
PD
/W
0
/2
=
Rc
1
.
0
-
k/w
01
.
0
=
ot
0
=
Reg
K/W
0214
-
RRP
=
RD
0
=
.
05/0 0214)
-
0 05 +
0
.
ER
Ra +
=
.
0
=
q
=
.
0214
0
-
0149 +
0 01
-
02657
o
=
:
Reg + Ra
1 67y 103 +
=
:
0
=
/1400
=
370
0
W
.
-
66
02657
0149
k/W
>
-
finding
q
Tz
T
=
RA
11400
30To -3
=
T2 =
>
-
350
finding
E
.
°
9
C
TB
T3T4
=
RC
11400
T3
=
=
To e
180
°
L
Example 2
54
Example
(a)
2
v
525
=
22
V3
683
=
3
=
King
E
ER
65 X10
m
3015x10
=
m
.
015 + 2
0
=
5
.
015x10-2
Wim k
43
=
=
T
.
.
150
=
T3
04
m
=
%
C
25 °
#
=
Assume
ER
R
=
.
R2 + 2
=
kpipe
2
=
Pipe
RPipe
C= 1 m
Rins
+
In
=
2πk
3015x104/2 65X10 )
in
=
.
=
4
26 x10
.
2π x 43
Rins
Instea
=
In
I
015x10 /3 015x10-2)
(5.
.
2πX 0 06
.
q=
13 5 +
I
92 58
.
4 776 x 10
.
W/m
-
Y
=
1
.
35
(b)
<
TT
=
91 34
.
10 x10
=
Tz
pe
=
149
.
50 C
Example 3
Steam at T∞,1 = 320°C flows in a cast iron
pipe (k = 80 W/m.K) whose inner and outer
diameters are D1 = 5 cm and D2 = 5.5 cm,
respectively. The pipe is covered with 3cm-thick glass wool insulation with k =
0.05 W/m.K. Heat is lost to the
surroundings at T∞, = 5°C by natural
convection and radiation with a combined
heat transfer coefficient of h2 = 18
W/m2.K. Taking the heat transfer
coefficient inside the pipe to be
h1 = 60W/m2.K,
1) Determine the rate of heat loss from the
steam per unit length of the pipe.
2) Also determine the temperature drops
across the pipe shell and the insulation.
55
Example 3
1
TI
TI
T3
Fentantmum
Ronr
Tal
2
5/2
=
5/2
5
~
22 +
=
.
Kins
5
0
=
Pipe
q
.
=
.
75
Assume
(
cr
thickness
-
=
2
=
RConvA2
<M
.
2 75 +
=
Rins
2 5
=
W2 =
3
Rpipe
,
3
75 Cm
.
05
Tai
So
Th2
=
320 °
5
=
1
=
2R
R Conv
I
=
,
>
-
Sconr
he Al
I
I
60 x 2π X 2 5 x 10:2
=
host
.
0
=
&
Pipe
.
R
104
2πkL
=
12 75/2 5)
In
.
.
-
2TTX 80 X
=
1 84 x
.
t
ha
tri
I
=
=
18
/
-Y
i
R
=
1
m
Rins
Inrscr
l
=
in
=
(5 <5/2 75)
.
.
+ 1
x 0 05
2
2
I
R Conve
.
.
35
L
=
H2Az
↳
I
18 X 2π U3L
Fix 5C5x10-2
=
0
I
E R
q
=
.
154
2 61
.
Tar-Tai
=
-
ER
I
5220
2
=
-
120
61
Wim
2)
find
9
>
-
T2 and T3
Tai-Te
=
-
R CONVI
120
32
=
0
T1
>
-
9
.
104
307 28
.
=
·
T
pe
=
120
=
e
308
-
>
-
T2
=
q
=
307
26
.
Y
°
C
TEB
Rins
120
=
B
307
. 35
2
Tz
=
25 26
.
·
L
Example 4: Problem 3.26
Stainless Steel (AISI): k= 25.4 W/m.K)
Beryllium oxide: k= 21.5 W/m.K
56
898
Liquich
Tx
=
↓ =
2600 °
50
I
Taz
w/m"
.
K
h2
ki
d
k2
=
K
Ta
21
TH
m" K/W
0 05
-
.
=
4
Wlmik
5
.
22
m
=
25
.
4
TU
T3
TI
20X10-3
=
K2
RBO
RONUI
T22
Rss
Rc
R CONVz
Do
-
I
2600
-
100
+
1
4/ A
K, A
+
RC"
+
A
+
50
=
IA
1
naA
2500
-100-3
-
34,600
Wlmk
+
-entkat
-Me
&
=
m
the
M
E
wim 2. K
SS
10x10
=
1000
=
R
BO
H
100 °
=
.
9
Wimt
0 05 +
.
200
I
1000
find
>
-
>
-
T2 , T3 , Th
bubon = Ther
e
150
T
>
-
1907 48 %
.
=
34600 4
.
IT2
=
34600 9
190
e
.
Tz
>
-
=
1891 9
=
°
.
C
34600 9 =
* T3
I
34600 9
1891 9
.
RC
.
=
:
T3
-
-
0
T
3
>
-
161
=
34600 9
.
=
.
86
:
05
oc
TO0
R conv2
34600. 4
Ty
=
=
o
Th
134 6
.
%
C
Example 5:
Calculate heat loss per unit length of pipe using:
a) Overall Ui
b) Overall Uo
c) Are the heat loss values same?
57
K
kz
=
=
Btu/h ft of
26
0
.
.
037
Al
-
=
=
V2
-i
no
=
Btu/hift of
2πVL
πDL
2
A ITTV L
=
vo
W
steam
Thi
261 of
=
m
vi
1000
-
=
22
Vo
0
ui
Ai
=
=
=
-
=
.
216
f72
ITTVoL
2 iTx 0
1
:
.
.
062
169
f+ 2
n
=
/p fin
=
(1 5/12)
:
F
Taz
+MKM+M
Pipe
ins
00343
At
12 :N
169 ft
CONVI
2πx 0 0343
.
0438 +
+
2πTViL
0
°
.
MK
Y ERAi
=
=
Ao
*
269
.
0
=
=
1 05
=
I
V2 + t
=
=
UASA
0824 if /
=
Conve
00438
+t
↑
Conv ,
1
=
hi A ;
I
I
=
4 63 x 10-3
.
1000 X 0 216
.
R pipe
(n/22/vi)
=
2πKL
In
=
100438/00343)
1 49
.
=
x10-3
2πx 24
Rins
Inoes
=
2πKL
0438)
1069/0
-
.
5
=
.
81
2π X 0 037
.
R
conv2
1
=
no to
1
=
=
2 x 2π x 0
ER
169
6 284
=
ni
.
.
=
6 28 x 0 216
.
.
=
Yo
=
=
0
.
737
.
YER Ho
Y 6 28
.
=
BtU/n +2 of
0 15
.
x
1062
Btu/lft2 of
0
.
471
hi f/Btu
Example 6
An electrical wire having a outer radius of
insulation
r1 = 1.5 mm and covered with a plastic
insulation (thickness = 2.5 mm) is exposed
to air at 300 K and ho
= 20 W/m2.K. The
insulation has a k of 0.4 W/m.k . It is
assumed that wire surface temperature(T1)
is constant at 400 K and is not affected by
K
r2
q
To= 300 K
r1
ho= 20 W/m2K
T1
T2
covering.
(a) Calculate the value of critical insulation.
(b) Calculate the heat loss per m of wire length with no insulation
(c ) Repeat (b) for the insulation present.
58
VI
1 5
Ve
=
(a) Ycrit
mm
.
=
=
1
5 + 2 5
.
4
=
.
mm
1 ins
L
0
I
.
04
-
20
0 02
I
q
(b)
u
.
~
AT
=
convection
and
0
I
mm
only
>
-
Rconv
q
20
no
insulation
don't har
we
2
r's
Rond
calculating
/A
E
=
I
HOXITTU /
q
o
I
T
L
Y20x2πX 1 5x 10
=
-
Wim
18 8
.
3
.
400
(c)
-
Rins
300
+
You ne
=
Ronv
e
100
it
>
-
Not
+
I
42 02
.
=
20X2πX Ux10-3
good design
because
2
z
in
umm
f
comm
f
<
U
crit
wim
for
Example 7
Calculate the critical radius of insulation for asbestos [k = 0.17 W/m.K]
surrounding a pipe and exposed to a room air at 20oC with h = 3 W/m2K.
Calculate the heat loss from a 200oC, 5 cm diameter pipe when:
1) covered with the critical radius of insulation and
2) without insulation.
59
Ucrit
Assume
I
=
3
=
2
I
=
1)E
.
056
5/2
n =
V2
0
.
5
m
10 2
x
Vcrit =
m
0 056
m
.
#
=
R Conr +
Rins
Peo
-
is
+
no 2TTU2h
+
2πkL
-oooo
105
=
2)
=
=
.
7
Win
Extric
200
-
20
3 x 2T X 2
=
84 8
.
.
5 xc0-2
wIm
trixos
e
Im
L
Example 8
Use the following schematic process , calculate the heat loss per tube
length and the outer temperature of the steel pipe:
a) with no insulation and
b) with insulation if the heat loss to the surrounding is reduced by 95% .
T(i) ?
Insulation
k = 0.1 W/mK
Steam
T = 250oC
h = 500 W/m2K
SS pipe
> ri
ID = 60 mm >
U2
OD = 75 mm k = 16 W/mK
T (ii) ?
Air, 20oC, 25 W/m2K
60
T2 =?
Ti ?
TXI
+
4)
Ronv ,
q
TA
:
ER
R convi
Ronvz
RPipe
1T
=
TX2
+
m
M
Im
2
&
RPipe
Conr +
RonV2
+
HA
=
YuzTV L
=
,
1
-
-
500XITTX30X10-3
Ppipe-Auris
os
=
0
.
0104
2 22x10-3
=
.
2 TX /P
R
conv2
-
=
h2//
-I L
[sxTx3 5x10-3
.
q
25020
=
+ 2 22 x 103 x 0
1
.
1260
=
outer
WIm
temperature
T22T
=
16
0
-
1260
T2
=
=
IiG
234
°
C
6
=
0
.
1698
T =
T = 234
250
T
=?
T
=
20
TMFt
Ronv ,
b)
1260x0 05
q
WIm
It
=
↳
63
=
.
RonV2
Rins
RPipe
ERO
-
RPipe + Rins + Ron an
63
250
=
-
20
-
Inuri)
+
K/A /
63
In
+
I
heAz
230
=
0
.
2
0106 +
22x103 +
5 o
Inr13
.
using solver
Us
0
=
6273
.
627 3
=
Outer
o
2πkzL
.
temperature
63
=
=
z
63
=
/25 2πx
x
T
=
20 64
.
°
C
0
.
623
m
mm
-
25 x 2π x V3
x 10
3
Example 9: Exercise 2.3
Temperature Unit: in K
61
>
-
-
&
q(x
T
=
ET
qx
= 1
2
9 + bx +
=
q
0
cx
b+
=
q
=
-
KALI
Ex
=
Elx
=
O
-
=
=
KA(b + 22x)
kAb
-
-
9/x
=
=
=
2
.
-
1
=
40X10X-300
120 KW
=
-
ka/b + 22)
YOX101-300
-
100)
160 kW
Ein-Eont + Eg
=
Est
120kW -160kw
+
1000 40xmB
,
Est
=
-30 kw
Kw
te
=
Est
.
3
=
SCP
*
(10X1)
It
-
30kW
I
Ot
=
1600 x 4 x 10
=
-
4
:
67 x 10" K/s
x
It
Ot
m
=
Example 10: heat loss through a steam pipe
Consider a steam pipe of length L = 20 m, inner radius r1 = 6 cm, outer
radius r2 = 8 cm, and thermal conductivity k = 20 W/m.K (as in Fig. 2) . The
inner and outer surface of the pipes are maintained at average
temperatures of T1 = 150oC and T2 = 60oC, respectively. Obtain a general
relation for the temperature distribution inside the pipe under steady
state conditions, and determine the rate of heat loss from the steam
through the pipe.
Solution: the mathematical formulation of this problem can be expressed as
1   T 
r
0
r r  r 
62
VI = 6 Cm
-2 =
8 Cm
(v)
(v)
1st
=
↑
0
W
-
7
T=
°
20
steam
&
T=
n
60
°
m
C
T
Integration
C
rt
=
and Integration
B- C
.
+
Cl
J
(vc + C2
=
General solution
-
1
2 = V
T
= 150
°
I
150 =
/NV1
150
=
In
=
150
C2
41
+ C2
106) 4 + C2
In 10 06) CI
.
-
. 2
B. C
V = 22
T=
,
60
°
C
60 =
In U2C1 + C2
60 =
In 10.08) < , +
[150
-
In 10063 (17
solver
using
3/2 85
.
c
22 =
=
T=
-
312
.
-
=
150
-
85/Mr
In 10 6) (
-
730
-
.
.
-
3/2
.
85)
2
130
·
2
↑ distribution
.
K
1
i
150
0
=
ZowIm
k=
Example 11
A hollow cylinder (k = 30 W/mK) of inner radius 3 cm and outside radius 4.5
cm has a heat generation rate of 5×106 W/m3. the inner and outer surfaces
are maintained at temperatures of 380oC and 360oC respectively. Determine
the temperature at the mid radius.
  r 
.
.
 ln   
2
2
2
2
q (r1  r )
q (r2  r1 )   r1  
T (r )  T1 
 [(T1  T2 ) 
]
  r1  
4k
4k
 ln   
  r2  
>
-
temp
distribution
equation
  0.0375  
ln 

5  10 6 (0.03  0.03752 )
5  10 6 (0.0452  0.032 )   0.03  

T (r  3.75 cm)  380 
 [(380  360) 
]
4  30
4  30
  0.03  
 ln  0.045  


r = (3+4.5)/2 = 3.75
.
2
.
T(r = 3.75 cm) = 373.69oC
63
Example 12
A hollow cylindrical conductor with r1 = 0.6 cm and r2 = 0.75 cm is made
of a metal of thermal conductivity 17.5 W/m.K and electrical resistance
of 2.5×10-2 Ω/m. Find the maximum allowable current if the temperature Tmax
is not exceeding 48.5oC anywhere in the conductor. The cooling fluid at
the inside is at 37.5oC.
insulated
.
 r 
q  (r12  r 2 )
2
 
T (r )  T1 

r
ln

2
2k 
2
 r1 
Tmax occurs at the insulated surface, i.e. at the outside radius and is equal
.
 r2 
q  (r12  r22 )
2
   48.5
Tmax  T1 

r
ln

2
2k 
2
 r1 
I
.
q  1584105
W
m3
q.V
R
64
Insulated
B.C
.
hollow
Cylinder
C
I
~ = VI
Maximum
TITI
,
9
B. C
.
82
Tnax/rere
.
FR
M
=
.
+
+
solver
q
1584 x105
=
1564 x
=
/(2)
+
v
In/v) is
/1000600052)
=T
634 9
.
A
+
6
-
0075
n
Z
WIm3
Volume-Tr, -To
=
= 40-2
=
R/volume
I
Ct
,
37 5
=
using
I
=
2
~ =
48 5
case
= 0
=
π(/v22
-
TX//0 00152
-
-
42)
Example 13
An electric current of 200 A is passed through a stainless steel wire having
a radius ro of 0.001268 m. the wire is L = 0.91 m long and has a resistance R
of 0.126 Ω. The outer surface temperature Ts is held at 422.1 K. the average
thermal conductivity is k = 22.5 W/m.K. calculate the center temperature.
.
q ro2
T0 
T
s
4k
Since power = I2 R = q (Watt)
.
.
2
= q  V  q    r  L
 = 1.096 ×109 W/m3
q
.
q ro2
T0 
T
s
4k
To = 441.7 K
65
Example 14
Consider a spherical container of inner radius r1 = 8 cm, outer radius r2 = 10
cm, and thermal conductivity k = 45 W/m.K (as in Fig. 3) . The inner and outer
surface of the pipes are maintained at average temperatures of T 1 = 200oC
and T2 = 80oC, respectively. Obtain a general relation for the temperature
distribution inside the shell under steady state conditions, and determine the
rate of heat loss from the steam through the container.
Hints: the mathematical formulation of this problem can be expressed as
1   T 
r
0
r r  r 
2
2
66
-(v)
(2)
1st
=
0
=
Integration
2 I
:
21
=
Of
o
and
Integration
.
B
T=
:
C
=
-1
+
C2
-
I
.
C
V=V,
T=
,
200
T H
=
0
(1
.
08
200 +
=
C2
+
=
4
0
.
08
.C 2
B
-
V=
-
T= 1 2
U2
80
=
-
4 +
0
80
-
1
-2
=
using
22
+
/200
C08]
+
solver
c
=
12 =
-
48
200
-
U8
-
0 08
.
=
T
=
48-400
y
-
400
T
distribution
general
solution
Acknowledgement
67
Cartesian
Coordinates
/167)
9
+
Site)
st
Integration
-
=
&
-Exc
=
and Integration
T
-
=
-
<2
o
+ <, K +
12
-
general
Solution o
. (
B C
.
T
=
T
T
B C
.
.
C2
=
2
T= T2
T2
=
=
-
E
+
T2
ci
T
+
/
T2
=
It
+
=
=
-[22
T
L
x=
,
-
=
4
q"
x = 0
,
1
+
942)
2
+
kd
dx
=
-
k)-Zx
+
TET
+
Ite
T
T distribution
o
coordinates
cylindrical
(kv , )
q
+
=
0
(v) =
1st Integration
UI
-
:
Du
T
by
and Integration
Solid
.C
B
.
c
*
I
r +
I
.
c
.
E
+
Cylinder
,
U= 0
B c
T
Er
=
0
0
=
2
V =
Vo
Ts
T=
,
Ts
-So
=
e
(2
Ts +
=
T=
-qu
T
Ts + = (V0
=
+
Uk
ro
is
+
-
&e
22)
↑ distribution
Cylinder
Hollow
Insulated
. I
B C
.
V =
T
B
.
C
.
VI
T
=
,
-Ev
=
Ti
4 Ini
+
2
~= 22
o
-
=
c
I
,
=
o
q
Insulated
-
0
=
KAST
=
O
=
0
-
If
I
zr
T
2
En e
=
·FF- art & I
e
2k
2
e
F
=
-gr2
+
q
Inr
+
i +
invi
Ev -gr
n e
FFtCrrtEr/Mrndistribution
Spherical
coordinates
3 =
-(kv2z)
E(v2() =
+
=
1st
Integration
=
-
Af
A
and Integration
T
<3
+
Cl
3k
t
Er
=
-
22
3k
I
=
-
qV
-
6K
-1
↓
+
<