TITLE.PM5

850 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-4.pm5

Now substituting the values in the above equation, we get

Q 12 =

0 283 5 67^91

100

303

100

1003

003

1 1003

003

0 4444

44

..

.

.

.

.

.

× F

HG

I

KJ

−F

HG

I

KJ

L

N

M

M

O

Q

P

P

F −

HG

I

KJ++

F −

HG

I

KJ×

=

0.283 5.67(0.686 84.289)

32.33 1 14.37

×−

++

= – 2.81 W

• ve sign shows heat flows from outside to inside. (Ans.)
  Example 15.32 (Radiation shield). The large parallel planes with emissivities 0.3 and 0.8
  exchange heat. Find the percentage reduction when a polished aluminium shield of emissivity 0.04 is
  placed between them. Use the method of electrical analogy.
  Solution. Given : ε 1 = 0.3 ; ε 2 = 0.8 ; ε 3 = 0.04
  Consider all resistances (surface resistances and space resistances) per unit surface area.
  For steady state heat flow,
  EEbb 13
  1 1 1 1
  1

3

3

−

F −

HG

I

KJ

++−

F

HG

I

KJ

ε

ε

ε

ε

=

EEbb 32

1 3 1 1

3

2

2

−

F −

HG

I

KJ

++−

F

HG

I

KJ

ε

ε

ε

ε

[aQ AAA12 3=== 11 m^2 ndFF13 32−−,=]

ε 1 ε 3 ε 3 ε 2

Radiation

shield

Eb 1 J 1 J 3 Eb3 J′ 3 J 2 Eb 2

1 –

A

ε

ε

1

11

1

AF 1 1–3

1 –

A

ε

ε

3

13

1 –

A

ε

ε

3

33

1

AF 3 3–2

1 –

A

ε

ε

2

22

Fig. 15.57

or

σ

εε

σ

εε

()()TT 14 34 TT

13

3

4

2

4

32

(^111111)
−
+−
= −
+−
or
TT 14 34 TT 34 24
1
004
−
+−
= −
(^1) +−
0.3
1
0.04
1 1
0.8
1
.