Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 10 (Q.No. 17)
17.
The radiation heat flux from a heating element at a temperature of 800°C, in a furnace maintained at 300°C is 8 kW/m2. The flux, when the element temperature is increased to 1000°C for the same furnace temperature is
Discussion:
5 comments Page 1 of 1.
Vicky said:
3 years ago
Right @Raj Pandit.
RAJ PANDIT said:
4 years ago
q1= Q/A = σ (T1^4-T2^4) ------------> 1
q2 = Q/A = σ (t1^4-T2^4) -----------> 2
DIVIDING 2 BY 1.
q2/q1 = (t1^4-T2^4)/(T1^4-T2^4).
q2/8 = (t1^4-T2^4)/(T1^4-T2^4).
Also, don't forget to convert the temperatures to Kelvin.
You will get 16.54 kW/m2.
q2 = Q/A = σ (t1^4-T2^4) -----------> 2
DIVIDING 2 BY 1.
q2/q1 = (t1^4-T2^4)/(T1^4-T2^4).
q2/8 = (t1^4-T2^4)/(T1^4-T2^4).
Also, don't forget to convert the temperatures to Kelvin.
You will get 16.54 kW/m2.
(1)
Mr.g said:
7 years ago
It is a radiation mode of heat transfer.
not the conduction mode.
So that formula will be Q = σ A (T1^4-T2^4).
not the conduction mode.
So that formula will be Q = σ A (T1^4-T2^4).
Shafiq said:
9 years ago
THE CORRECT ANS IS A.
q1= Q/A===>-KAdT/x------------> 1
q2 = Q/A====>-KAdt/x-----------> 2
DIVIDING 2 BY 1
q2/q1= dt/DT.
q2/8 = (t1-T2)/T1-T2.
q2 = 8(1000-300)/800-300 = 11.2.
q1= Q/A===>-KAdT/x------------> 1
q2 = Q/A====>-KAdt/x-----------> 2
DIVIDING 2 BY 1
q2/q1= dt/DT.
q2/8 = (t1-T2)/T1-T2.
q2 = 8(1000-300)/800-300 = 11.2.
MFQUY said:
9 years ago
Please, I want the solution for this. Anyone explain it.
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