# Convection and Radiation in an Electrically Heated Cylinder

1. Aug 4, 2013

### Liammaloney

1. The problem statement, all variables and given/known data

Hi everyone,
I'm working on an assignment and am totally stumped by this question:

A horizontal cylinder is heated electrically. The cylinder is supplied with a current at a DC Voltage. Using the data supplied and allowing for both natural convection and radiation, estimate the steady state temperature of the cylinder surface.

Hint- You may have to solve the heat balance equation numerically.

A fan is used to blow air over the heated cylinder in a cross flow at an average velocity of 50m/s. Estimate the surface temperature now.

The heating is stopped and the cylinder is left in the airflow of 50m/x velocity. Estimate the time for the cylinder temperature to drop by 10K.

You will need to find the relevant fluid properties, but may assume a surface emissivity of ε=0.9.

2. Relevant equations

Cylinder Diameter = 87mm
Cylinder Length = 254mm
Surrounding air temperature = 20°C
Current = 1.032A
Voltage = 20V
Specific heat of cylinder = 450J/kgK
Density of Cylinder = 7850km/m^3

3. The attempt at a solution

I'm not sure where to start but I have had a go trying to find out where I'm meant to be going.
I first went about finding the resistance in the cylinder using:

R = V/I (20/1.03=19.42Ω)

With this I could then find the heat generated per unit volume in the cylinder using:

q(dot)g = I^2*R/AL ((1.03^2*19.42)/(5.9*10^-3*0.254) = 13.64kW

After this though I'm not sure where I'm meant to be going, and would really appreciate any help!

2. Aug 9, 2013

### CWatters

I'm not an expert but under steady state the power going into the cylinder = power emitted by the cylinder. Presumably the power emitted is comprised of the power lost by convection and power lost by radiation.

So I'd be looking to write an equation of the form...

Power In = Power convected + Power radiated

...in terms of the surface temperature of the cylinder.

I don't think I've ever known the equations for the right hand side but perhaps you can look them up.

Then solve for the surface temperature.

You have to do this twice. Once for still air and once for 50m/s airflow.

As for the last part. If the temperature with 50m/s air flow is quite hot then a 10C fall might not be very much as a percentage. You might be able to assume the power loss is constant over that 10C range(?) or perhaps use an average figure for the power loss(?). Then it's a simple matter of working out how much heat energy must be removed to reduce the temperature by 10C (eg using the SHC and the mass). Then remember Power = energy/time.