INTRODUCTION :
Emissivity of a surface is defined as the ratio of emissive power of a surface (e) to emissive
power of a black surface (eb)
e
e = emissivity = -----
eb
THEORY
Emissivity of a perfect black surface is 1 and Emissivity of a perfect surface is 0. All
the practical surface are having emissivity between 0 to 1.
The knowledge of emissivity is necessary when one want to find out heat exchange by
Thermal radiation. In the design of appliances based on sciar Energy, the knowledge of
emissivities of collector surface and reflecting surfaces is a must for the concerned engineer.
At a particular wavelength, over the spectrum of wavelengths the monochromatic
EMISSIVITY is defined as
e
= --------- Where e is monochromatic emissive power
eb and eb is monochromatic emissive power of black surface
Table given below gives the values of Emissivities of some concerned material surfaces
for ready reference.
Sr.No Surface Temperature Emissivity
1. Metals-polished copper 20-5000C 0 -1.0
Steel,- stinless- steel, nickel
2. Oxidised-copper, steel, Upto 10000C 0.6 -1.0
Stainless- steel, nickel
3. Non Metals : 20-1000C 0.8 – 1.0
Brick, wood, mable, water
OBSERVATION :
To be recorded under study state conditions.
1. Diameter of the dise (d) = 0.15 m
2. Thickness of disc assembly (t) = 0.12 m
3. Ares of disc surface (A)Exposed. = ( d2 / 4 ) m2
4. Heater input to black dise (w1) = Watts
V2 = Volts, A2 Amps.
W2 = V2 x A2
0
5. Temperatures of back dise (W2) = C + 273 = K
V2 = Volts, A2 = Amps.
W2 = V2 x A2
6. Heater input non-back dise (w1) = Watts
V1 = Volts, A1 = Amps.
W1 = V1 x A1
7. Temperatures of non-black disc (T1) 0C = 0
C + 273 = K
8. T4 =(T2 + T1) / 2 = K
0 0
9. Temperatures of the enclosure (T3) C = C + 273 = K
CALCULATION :
Under steady condition, heater input to black disc.
W2 = E
b oA (T44 – T34 ) + conduction loss through base
+ Natural convection loss form black disc.
W1 = E
nb oA (T44 – T34 ) + conduction loss through base
+ Natural convection loss form black disc.
Since the disc assemblies are identical in shape and size and she /are placed in a
symmetrical manner inside the enclosure, Conduction and convection loss must be same for
both the discs when their temperatures are identifical
CONCLUSIONS :
1. Commet on the value of emissivity obtained.
2. Conduct the experiment by changing the values of heater inputs and obtain
emissivity values at different temperatures to study its variation with temperature
REFERENCES-
1) A text book on heat transfer by –Dr. S.P. Sukhatme
2) Fundamental of heat transfer by-M.Mikheyev.
3) Heat transfer by- V.P. Isachenko, V.A.Osipova, A.S.Sukomel
QUESTIONS:
Q.1 Define black body, emissivity & its significance
Q.2 comment on the value of emissivity obtained
Q.3 Explain radiation with suitable example & application.