HEAT TRANSFER 851

dharm
\M-therm\Th15-4.pm5

or

TT TT 14 − (^34) = 34 − 24
27.33 25.25
or TT 14 −= 34 27.33TT 34 − 24
25.25
()
= 1.08 ()TT 34 − 24 = 1.08 T 34 – 1.08 T 24
or 2.08 TT 34 = 14 + 1.08 T 24
or T 34 =^1
2.08
(T 14 + 1.08 T 24 ) = 0.48 (T 14 + 1.08 T 24 ) ...(i)
Q 12 (heat flow without shield)

σ
εε
()()()TT 14 24 σσTT TT
12
1
4
2
4
1
4
2
4
(^1111)
−
+−

−
+−

−
1
0.3
1
0.8
3.58
...(ii)
Q 13 (heat flow with shield)

σ
εε
()()()TT 14 34 σσTT TT
13
1
4
3
4
1
4
3
4
(^1111)
−
+−
= −
+−
1 = −
0.3
1
0.4
27.33 ...(iii)
∴ Percentage reduction in heat flow due to shield

QQ
Q
12 13
12
−
= 1 –
Q
Q
TT
TT
13
12
1
4
3
4
1
4
2
=− (^14)
−
−
σ
σ
()/
()/
27.33
3.58
= 1 – 3.58
27.33
TT
TT
1
4
3
4
1
4
2
4
−
−
L
N
M
M
O
Q
P
P
= 1 – 0.131
TTT
TT
1
4
1
4
2
4
1
4
2
4
−+ 0
−
L
N
M
M
O
Q
P
P
.48()1.08
= 1 – 0.131
TTT
TT
14 14 24
14 24
−+ 0
−
L
N
M
M
O
Q
P
P
.48 0.52
= 1 – 0.131
0 14 24
14 24
.52 (
()
TT
TT
−
−
L
N
M
M
O
Q
P
P
= 1 – 0.131 × 0.52 = 0.932 or 93.2%. (Ans.)

Highlights

1. Heat transfer may be defined as the transmission of energy from one region to another as a result of
   temperature gradient and it takes place by three modes : conduction, convection and radiation.


2. Fourier’s law of conduction : Q = – kA
   dt
   dx
   where, Q = Heat flow through a body per unit time,
   A = Surface area of heat flow (perpendicular to the direction of flow),
   dt = Temperature difference,