Phase Change Heat Transfer and Heat Exchanger Sorted-out Important Problems (Solved) from HMT Databook

HEAT AND MASS
TRANSFER
Solved Problems
By
Mr. P. Raveendiran
Asst. Professor, Mechanical

, Heat and mass Transfer
Unit I
November 2008
1. Calculate the rate of heat loss through the vertical walls of a boiler furnace of size 4
m by 3 m by 3 m high. The walls are constructed from an inner fire brick wall 25 cm
thick of thermal conductivity 0.4 W/mK, a layer of ceramic blanket insulation of
thermal conductivity 0.2 W/mK and 8 cm thick, and a steel protective layer of
thermal conductivity 55 W/mK and 2 mm thick. The inside temperature of the fire
brick layer was measured at 600 o C and the temperature of the outside of the
insulation 600 C. Also find the interface temperature of layers.
Given:
Composite Wall
l= 4m b= 3m h= 3m
Area of rectangular wall lb = 4x3 = 12m2
L1 = 25 cm Fire brick
kı = 0.4 W/mK
L2 =0.002m Steel
k2 = 54 W/mK
L3 = 0.08 m insulation
kı = 0.2 W/mK
T1 = 6000 C
T2 = 600 C
Find
(i) Q (ii) (T3 –T4)
Solution
We know that,
( )
=

Here
(ΔT) overall = T1 – T4
And ΣR th = Rth1 + Rth2 + Rth3
.
Rth1 = = =0.0521K/W
.
.
Rth2 = = =0.0333K/W
.
.
Rth3 = = =0.0000031K/W

, –
=

600 − 60
=
0.0521 + 0.0000031 + 0.0333
Q = 6320.96 W
(i) To find temperature drop across the steel layer (T2 - T3)
–
=

T3 - T 4 = Q Rth2
= 6320.96 0.0000031
T3 - T 4 = 0.0196 K .


2. A spherical container of negligible thickness holding a hot fluid at 1400 and having
an outer diameter of 0.4 m is insulated with three layers of each 50 mm thick
insulation of k 1 = 0.02: k 2 = 0.06 and k3 = 0.16 W/mK. (Starting from inside). The
outside surface temperature is 300C. Determine (i) the heat loss, and (ii) Interface
temperatures of insulating layers.
Given:
OD = 0.4 m
r1 = 0.2 m
r2 = r1 + thickness of 1st insulation
= 0.2+0.05
r2 = 0.25m
r3 = r2 + thickness of 2nd insulation
= 0.25+0.05
r3 = 0.3m
r4 = r3 + thickness of 3rd insulation
= 0.3+0.05
r4 = 0.35m
Thf = 140o C, Tcf = 30 o C,
k1 = 0.02 W/mK
k2 = 0.06 W/mK
k3 = 0.16 W/mK.
Find (i) Q (ii) T2, T 3




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, Solution
( )
=

ΔT = Thf – Tcf
ΣR th = Rth1 + Rth2 + Rth3
( . . )
Rth1 = = =3.978o C/W
. . .
( . . )
Rth2 = = . . .
=0.8842o C/W
( . . )
Rth1 = = =0.23684o C/W
. . .

140 − 30
=
0.0796 + 0.8842 + 0.23684
Q = 21.57 W
To find interface temperature (T2 , T3 )
–
=

T2 = T1 – [Q x ]
= 140 – [91.620.0796]
T2 = 54.17 0C
–
=

T3 = T2 – [Q  ]
= 132.71- [91.620.8842]
T3 = 35.09 o C
3. May 2008
A steel tube with 5 cm ID, 7.6 cm OD and k=15W/m o C is covered with an insulative
covering of thickness 2 cm and k 0.2 W/m oC. A hot gas at 330 o C with h = 400 W/m2oC
flows inside the tube. The outer surface of the insulation is exposed to cooler air at 30oC
with h = 60 W/m2oC. Calculate the heat loss from the tube to the air for 10 m of the tube
and the temperature drops resulting from the thermal resistances of the hot gas flow,
the steel tube, the insulation layer and the outside air.
Given:
Inner diameter of steel, d 1 = 5 cm =0.05 m
Inner radius,r1 = 0.025m
Outer diameter of steel, d2 = 7.6 cm = 0.076m
Outer radius,r2 = 0.025m
Radius, r3 = r2 + thickness of insulation
= 0.038+0.02 m

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