Chapter 1
Conduction through a hollow cylinders
((Simple & composite)) – Problems
Conduction in a Cylindrical system
For composite cylindrical system made up of different layers of materials
Q = (∆Toverall / Roverall)
∆T = Ta – Tb
1 ⎛⎜ 1
1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞ 1 ⎞⎟
R=
+ ln⎜⎜ ⎟⎟ + ln⎜⎜ ⎟⎟ +
⎜
2πL ⎝ ha r1 k1 ⎝ r1 ⎠ k1 ⎝ r2 ⎠ hb r3 ⎟⎠
Electrical analogy:
Ta
Tb
1/r1ha
ln(r2/r1) ln(r3/r2) 1/r3hb
/ k2
/ k1
1/2πL term is common for all resistances
Unit I - Conduction - Problems
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CONDUCTION - Problems
1.
A hollow cylinder
y
5 cm inner radius and 10 cm outer radius has inner
surface temperature of 200o C and outer surface temperature of 100o C. If
the thermal conductivity is 70 W/m K, find heat transfer per unit length.
Given:
Inner radius, r1 = 5 cm
= 0.05 m
Outer radius, r2 = 10 cm
= 0.1 m
Inner surface temperature, T1 = 200 + 273 = 473 K
Outer surface temperature, T2 = 100 + 273 = 373 K
Thermal conductivity, k
= 70 W/m K
To find:
Heat flow per unit length
Solution:
Heat transfer through hollow cylinder is given by
Q = ∆Toverallll / R [From HMT Data book
book, Pg No: 43 and 44]
where
∆T = T1 – T2
and
R = (1/2πLk) ln [r2/r1]
[(1/2πLk)) ln [r
[ 2/r1]]
⇒Q = [[T1 – T2 ] / [(
Unit I - Conduction - Problems
⇒Q/L = 63453.04 W/m
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CONDUCTION - Problems
2
2.
Determine
D
t
i thermal
th
l conductivity
d ti it off asbestos
b t powder
d packed
k d in
i between
b t
t
two
concentric copper pipes 25 mm and 36 mm diameter length. The inner pipe
housing has a heating coil to which 120 W power is supplied. The average
temperature
p
of inner and outer p
pipes
p
are 42.4o C and 27.9o C respectively
p
y
and the length of the pipe is 1 m.
Given:
Inner diameter,, D1
Inner radius, r1
Outer diameter, D2
Outer radius, r2
Length of the pipe
pipe, L
Inner temperature, T1
Outer temperature, T2
Heat transfer, Q
= 25 mm
= 12.5 mm
= 36 mm
= 18 mm
= 0.0125 m
= 0.018 m
=1m
= 42.4o C + 273 = 315.4 K
= 27.9o C + 273 = 300.9 K
= 120 W
To find: Thermal conductivity, k
Unit I - Conduction - Problems
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CONDUCTION - Problems
Solution:
Heat transfer through hollow cylinder is given by
Q = ∆Toverall / R
where
[F
[From
HMT D
Data
t b
book,
k P
Pg N
No: 43 and
d 44]
∆T = T1 – T2
R = (1/2πLk) ln [r2/r1]
⇒Q = [T1 – T2 ] / [(1/2πLk) ln [r2/r1]]
k = 0.48 W/m K
Thermal conductivity, k = 0.48 W/mK
Unit I - Conduction - Problems
5
CONDUCTION - Problems
3
3.
An insulated steel pipe is carrying a hot liquid.
liquid Inner diameter of the pipe
is 25 cm, wall thickness is 2 cm, thickness of insulation surrounding the
steel pipe is 5 cm. Temperature of hot liquid is 100o C and temperature of
surrounding is 20o C. Inside heat transfer co-efficient is 730 W/m2K and
outside heat transfer co-efficient is 12 W/m2K. Calculate the heat loss per
meter length of the pipe.
Take ksteel = 55 W/m K, kinsulating material = 0.22 W/m K
Given:
Inner diameter, d1
= 25 cm
Inner radius, r1
= 12.5 cm
= 0.125 + 0.02
radius, r2 = r1 + thickness of wall
radius, r3 = r2 + thickness of insulation
= 0.145 + 0.05
Temperature of hot liquid, Ta = 100o C + 273
= 373 K
Temperature of surrounding, Tb = 20o C + 273 = 293 K
Inside heat transfer co-efficient, ha
= 730 W/m2K
= 12 W/m2K
Outside heat transfer co-efficient, hb
ksteel = 55 W/
W/m K
kinsulation = 0.22
0 22 W/
W/m K
Unit I - Conduction - Problems
= 0.125 m
= 0.145 m
= 0.195 m
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CONDUCTION - Problems
To find:
Heat loss per meter length
Solution:
Heat transfer through composite cylinder is given by
Q = ∆Toverall / R [From HMT Data book, Pg No: 43 and 45]
where
∆T = Ta – Tb
R = (1/2πL) [(1/ha r1) + (ln (r2 / r1) / k1)
+ (ln (r3 / r2) / k2) + (1/hb r3)]
Q = [Ta – Tb ] /
[(1/2πL) {(1/ha r1) + (ln (r2 / r1) / k1) + (ln (r3 / r2) / k2) + (1/hb r3)}]
Q/L = 281.178 W/m
Unit I - Conduction - Problems
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CONDUCTION - Problems
4.
Air at 90o C flows in a copper
pp tube ((k=380 W/m K)) of 5 cm inner diameter,,
0.7 cm thick wall which is heated from the outside by water at 120o C. A
scale of 0.4 cm thick is deposited on the outer surface of the tube whose
thermal conductivity is 1.82 W/m K. The air and water side unit surface
conductance are 220 W/m2K and 3650 W/m2K respectively.
respectively Calculate
1. Heat transfer rate per unit length of the pipe
2. Temperature drop across the scale deposit
Given:
Inner air temperature, Ta = 90o C + 273 = 363 K
Inner diameter of the copper
copper, d1 = 5 cm
radius, r1 = 2.5 cm = 0.025 m
Thermal conductivity, k1 = 380 W/m K
Outer radius of the copper, r2 = Inner radius + thickness of wall = 0.032 m
Scale deposit radius, r3 = r2 + thickness of scale = 0.036 m
Outside temperature of water, Tb = 120o C + 273 = 393 K
Thermal conductivity, k2 = 1.82 W/m K
Surface conductance of air
air, ha = 220 W/m2K
Surface conductance of water, hb = 3650 W/m2K
Unit I - Conduction - Problems
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CONDUCTION - Problems
To find:
1. Heat transfer rate per unit length of the pipe, (Q/L)
2. Temperature drop across the scale deposit, (T3 – T2)
Solution:
Heat transfer through composite cylinder is given by
Q = ∆Toverall / R [From HMT Data book
book, Pg No: 43 and 45]
Where ∆T = Ta – Tb
R = (1/2πL) [(1/ha r1) + (ln (r2 / r1) / k1)
+ (ln (r3 / r2) / k2) + (1/hb r3)]
Q = [Ta – Tb ] /
[(1/2πL) [(1/ha r1) + (ln (r2 / r1) / k1) + (ln (r3 / r2) / k2) + (1/hb r3)]]
Q/L = - 739.79 W/m
Unit I - Conduction - Problems
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CONDUCTION - Problems
Interface temperatures
Q = ∆T/R = (Ta – Tb)/R = (Ta – T1)/Ra = (T1 – T2)/R1 = (T2 – T3)/R2
= (T3 – Tb)/Rb Æ (1)
(1)
=>
Q = (T2 – T3)/R2
where R2 = [(1/2πL) (ln (r3 / r2) / k2)]
=>
Q = (T2 – T3)/ [(1/2πL) (ln (r3 / r2) / k2)]
=>
=>
T2 – T3 =
T3 – T2 =
-7.6 K
7.6 K
Temperature across the scale deposit,
T3 – T2 = 7.6 K
Unit I - Conduction - Problems
10
Home Work
5.
A composite
p
cylinder
y
consists of 10 cm radius steel p
pipe
p of 25 mm
thickness over which two layers of insulation 30mm and 35 mm are laid.
The conductivities are 25 W/m K, 0.25 W/m K and 0.65 W/m K. The inside
is exposed to convection at 300 oC with h = 65 W/m2K. The outside
temperature is exposed to air at 30 oC with h = 15 W/m2K.
K Determine the
heat loss per unit length of the pipe. Also find the interface temperatures.
Unit I - Conduction - Problems
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