Author Topic: 12/11/2015 Thermo  (Read 803 times)

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12/11/2015 Thermo
« on: December 11, 2015, 09:05:17 AM »
Question:

Two identical bodies radiate heat to each other.  One body is at 30 ºC and the other is at 250 ºC. The emissivity of both is .7; Find the net heat transfer per square meter:

(A) 3693 Watts

(B) 37 Watts
 
(C) 6237 Watts

(D) 2637 Watts

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Re: 12/11/2015 Thermo
« Reply #1 on: December 16, 2015, 07:53:03 AM »
The answer is (D) 2637 Watts.


Refer to page 116 of the FE Handbook: Heat Transfer, Radiation:


The radiation emitted by a body is given by:

Q = ε• σ •A•T4

Given:
ε = .7
σ = 5.67 × 10-8 W / (m2•K4)    (the Stefan-Boltzmann constant)
A = 1 m2 (not given, however, the problem statement says per square meter, so just use 1 m2)

T = Temperatures must be in absolute temperatures (ºR or K)

Using page 1 of the FE Handbook : Temperature Conversion:

Since the problem statement gives us the temperature in ºC, it’s easy to convert to K.


K = ºC + 273.15


TL:
K = 30 + 273.15
K = 303.15

TH:
K = 250 + 273.15
K = 523.15


Therefore:

Q = ε• σ •A•T4
Q = (.7) • (5.67 × 10-8 W / (m2•K4)) • 1 m2 • (523.154 K4 – 303.154 K4)
Q = (.7) • (5.67 × 10-8 W / (m2•K4)) • 1 m2 • (66,458,388,419.2K4)
Q = 2637 Watts


The answer is (D) 2637 Watts.