Convection/(radiation ?) heat loss problem. (Thermodynamics)

  • Thread starter Sentience
  • Start date
  • #1
78
0

Homework Statement



A 40-cm-long, 0.6-cm-diameter electric resistance wire is used to determine the convection heat transfer coefficient in air at 25 degrees C experimentally. The surface temperature of the wire is measured to be 150 degrees C when the electric power consumption is 90 W. If the radiation heat loss from the wire is calculated to be 30 W, the convection heat transfer coefficient is : ?


Homework Equations



Q (convection) = h * A * (delta T)

h = stefan-boltzmann constant

A = surface area

The Attempt at a Solution



I set the equation equal to h and solve. I get h = 0.318 W/ M^2, which is not one of the given answers (multiple choice). Two things that confuse me :

1 .Why was I given the amount of radiation transfer when all of the other information pertains to convectrion? (I think this might be a typo, I'm going to email my instructor)

2. Why was the amount of power going through the wire ( 90 Watts ) given? Do I need this information?
 

Answers and Replies

  • #2
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
20
Yes, you need all the information. The wire is losing energy due to radiation as well. Try sketching and then writing an energy balance for the wire.

(And check h; it's not the Stefan-Boltzmann constant.)
 
  • #3
78
0
I ended up getting the right answer, in which 30 watts was lost to radiation and then 60 watts to convection, to get the answer of h ~ 63 or so. (Yeah it's the convection coeffiecient, sorry about the typo)

My question is though, is that if you have 90 watts of power flowing through that wire, and all of it ends up being lost to either convection or radiation, does any power end up actually making it through? Seems like a crappy wire to me.
 
  • #4
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
20
My question is though, is that if you have 90 watts of power flowing through that wire, and all of it ends up being lost to either convection or radiation, does any power end up actually making it through? Seems like a crappy wire to me.

Well, for all we know, 9000 W are transmitted and only 90 W are dissipated as heat. It's not mentioned in the problem statement because it's not important in a heat transfer problem. But in practice, such wires (think heating elements in your toaster oven) are designed to have a high resistivity because heat generation is their only purpose.
 

Related Threads on Convection/(radiation ?) heat loss problem. (Thermodynamics)

Replies
1
Views
5K
Replies
5
Views
4K
Replies
1
Views
1K
Replies
1
Views
4K
Replies
12
Views
11K
Replies
1
Views
1K
  • Last Post
Replies
14
Views
2K
Replies
4
Views
2K
Top