Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 3 (Q.No. 43)
43.
Out of 100 kcal/second of incident radiant energy on the surface of a thermally transparent body, 300 kcal/second is reflected back. If the transmissivity of the body is 0.25, the emissivity of the surface will be
Discussion:
11 comments Page 1 of 2.
Zahraa said:
3 years ago
I think it's wrong in question because the incident radiant energy must be more than the reflected energy so it must be 1000 kcal/s not 100 kcal/s.
(1)
VINOD said:
7 years ago
Very good, Thanks @Monika.
Nik said:
7 years ago
Thanks @Monika.
Shriraj said:
7 years ago
Thanks @Monica.
Ritesh said:
7 years ago
Thanks @Monika.
Monika said:
7 years ago
A+t+r=1.
R=300/1000=0.3,
t=0.25,
A=1-0.55=0.45,
Emmisivity=A of sample/A of black.
O.45/1=0.45.
R=300/1000=0.3,
t=0.25,
A=1-0.55=0.45,
Emmisivity=A of sample/A of black.
O.45/1=0.45.
(5)
Swathi said:
8 years ago
Please explain it briefly.
(1)
Philip said:
8 years ago
@Nizam Ali.
Thats true, or maybe the incident energy should be 1000 kcal/s.
Thats true, or maybe the incident energy should be 1000 kcal/s.
Danish said:
8 years ago
How? explain me.
Nizam ali said:
8 years ago
I think it should be 30kcal/sec not 300kcal/sec.
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