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

[Sentience]

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]

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.)

 

[Sentience]

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]

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.

 

[DeeNos]

I would first check the given information and equations to make sure they are accurate and relevant to the problem. In this case, the given equation for convection heat transfer coefficient (h) is incorrect as it does not include any temperature difference (delta T) term. The correct equation for convection heat transfer coefficient is h = Q/A(delta T), where Q is the heat transfer rate, A is the surface area, and delta T is the temperature difference between the surface and the surrounding fluid.

To answer the first question, it is possible that the given radiation heat loss is a typo and should actually be convection heat loss. However, if the problem truly intends to include radiation heat loss, then it is important to note that radiation heat transfer is typically much smaller than convection heat transfer in air at low temperatures. Therefore, it can be assumed that the majority of heat loss from the wire is due to convection.

As for the second question, the power consumption information is not necessary to solve for the convection heat transfer coefficient. However, it can be used to check the accuracy of the calculated coefficient by comparing it to the expected power consumption based on the calculated coefficient and the given temperature difference.

In conclusion, the correct approach to solving this problem would be to use the equation h = Q/A(delta T) and calculate the heat transfer rate Q by subtracting the given radiation heat loss from the total power consumption. Then, the calculated coefficient can be compared to the expected power consumption to check for accuracy.