846 ENGINEERING THERMODYNAMICS

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\M-therm\Th15-4.pm5

(ii)When the body is not grey :
Absorptivity when source is at 500°C = Emissivity when body is at 500°C
i.e., absorptivity, α = 0.72

Then, energy emitted = εσ T 14 = 0.42 × 5.67 × 10–8 × (1273)^4
and, Energy absorbed = ασT 24 = 0.72 × 5.67 × 10–8 × (773)^4
i.e., q = Energy emitted – Energy absorbed
= 0.42 × 5.67 × 10–8 × (1238)^4 – 0.72 × 5.67 × 10–8 × (773)^4
= 62538 – 14576 = 47962 W
i.e., Heat loss per m^2 by radiation = 47.962 kW. (Ans.)
Example 15.27. A long steel rod, 22 mm in diameter, is to be heated from 420°C to 540°C. It is
placed concentrically in a long cylindrical furnace which has an inside diameter of 180 mm. The
inner surface of the furnace is at a temperature of 1100°C, and has an emissivity of 0.82. If the surface
of the rod has an emissivity of 0.62, find the time required for the heating operation.
Take for steel : c = 0.67 kJ/kg K, ρ= 7845 kg/m^3.
Solution. Refer Fig. 15.53.
Diameter of the steel rod
= 22 mm = 0.022 m
Inside diameter of the furnace
= 180 mm = 0.18 m
Emissivity ε 1 = 0.62
Emissivity ε 2 = 0.82
Specific heat of steel, c = 0.67 kJ/kg K
Density of steel, ρ = 7845 kg/m^3
T 1 = 420 + 273 = 693 K ..... 1st case
and = 540 + 273 = 813 K ..... 2nd case
T 2 = 1100 + 273 = 1373 K
The surface area of the rod, A 1 = π × 0.022 × l m^2
The surface area of the furnace, A 2 = π × 0.18 × l m^2
Time required for the heating operation, th :
Initial rate of heat absorption by radiation, when the rod is at 420°C or 693 K

Qi =

AT T

A

A

11

4

2

4

1

1

22

(^111)
σ
εε
()−
+−F
HG
I
KJ

π
π
π
×××× −

• ××
  ××
  F
  HG
  I
  KJ
  L −
  NM
  O
  QP
  0 022 1 5 67 10− 693 1373
  1
  062
  0 022
  018
  1
  082
  1
  ..(^844 )
  .
  .
  ..
  l
  l
  = −13022.5
  1.64
  = – 7940.5 W/m [Q lm= 1 ...assumed]
  Rate of heat absorption at the end of the heating process, when the rod is at 540°C or 813 K
  Qe =
  AT T
  A
  A
  21
  4
  2
  4
  1
  1
  22
  (^111)
  σ
  εε
  ()−
  +−F
  HG
  I
  KJ
  Fig. 15.53
  T 1
  T 2
  Steel rod
  A 2
  A 1