Chemical Engineering - Heat Transfer - Discussion

Discussion Forum : Heat Transfer - Section 3 (Q.No. 21)
21.
The thermal radiative flux from a surface of emissivity = 0.4 is 22.68 kW/m2. The approximate surface temperature (K) is
(Stefan-Boltzman constant = 5.67xl0-8 W/m2.K4)
1000
727
800
1200
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
6 comments Page 1 of 1.

Kowsal said:   4 years ago
22.68 * 1000 = 0.4 * 5.678 * 10^4 * T^4.
(1)

Ramsajanpaswan said:   2 years ago
E = e*K*T^4 ,
22.68*10^3 = (.4) * 5.67 * 10^-8 * T^4
22.68/2.268 * 10^3 * 10^8 = T^4.
10^12=T^4.
(10^3)^4=T^4,
T =10^3,
T =1000.
(1)

Ccjayesh said:   10 years ago
j = e*(5.67*10^-8)*(T^4).

Muhammad Ilyas said:   5 years ago
Kindly explain it.

Maha said:   4 years ago
22.68 * 1000 = (0.4) * (5.678 * 10^-8) * T^4,
So, T^4 = 9.98 * 10^ 11.
T = 999.6 K.

Athraa said:   3 years ago
Here, E=e* boltz * T4.

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