Net Power Transfer by Radiation

Click For Summary
SUMMARY

The net power transfer by radiation for a person with a skin temperature of 33°C and an emissivity of 0.9 in a room at 24°C is calculated using the Stefan-Boltzmann Law. The temperature difference is 9°C, leading to an irradiance calculation of Irad = (0.9)(5.67 x 10-8 Wm-2K-4)(306 K4) = 3.34 x 10-4 W/m2. Multiplying by the body area of 1.8 m2 results in a net power transfer of 6.03 x 10-4 Watts. The calculations confirm that temperature differences in Celsius and Kelvin yield the same results for power transfer.

PREREQUISITES
  • Understanding of the Stefan-Boltzmann Law
  • Knowledge of emissivity and its implications in thermal radiation
  • Basic principles of thermodynamics and heat transfer
  • Familiarity with units of measurement in physics, particularly Watts and Kelvin
NEXT STEPS
  • Study the Stefan-Boltzmann Law in detail to understand its applications
  • Explore the concept of emissivity and its role in thermal radiation
  • Learn about heat transfer mechanisms in thermodynamics
  • Investigate the conversion of temperature units and their impact on calculations
USEFUL FOR

Students in physics or engineering, thermal engineers, and anyone interested in understanding heat transfer and radiation principles.

ethanabaker1
Messages
4
Reaction score
0

Homework Statement



A person with a skin temperature of 33 C and an emissivity of .9 is in a room at 24 C. What is the net power transfer by radiation. Let the area of the body be 1.8 m2.

Homework Equations


I=\frac{P}{A}
\DeltaIrad=\epsilon\sigmaT4 (Stefan-Boltzmann Law)


The Attempt at a Solution


Temperature difference is 33-24=9 C
Irad=(.9)(5.67x10-8)(94)
Irad=3.34x10-4
IradA=Power
(3.34x10-4)(1.8)=6.03x10-4
 
Physics news on Phys.org
It probably would be nice to show a few more units in your calculations.
 
Well, power should have units of Watts. Emissivity has no units, and I know the Boltzmann constant is Wm-2K-4, so I wasn't sure if that means I should convert the temperature difference to Kelvin or not. I left it because both Kelvin and Celsius should have the same temperature difference. But I don't even know if that solution is reasonable.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K