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

## 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?

Mapes
Homework Helper
Gold Member
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.)

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.

Mapes