Convection and Radiation in an Electrically Heated Cylinder

In summary, the task is to estimate the steady state temperature of a horizontally heated cylinder using data such as cylinder dimensions, surrounding air temperature, and electrical current and voltage. Natural convection and radiation must also be taken into account. A fan blowing air at 50m/s is then introduced, and the surface temperature is estimated again. For the final part, the task is to estimate the time it takes for the cylinder temperature to drop by 10K once the heating is stopped. The equations for power balance and relevant fluid properties must be used in order to solve for the surface temperature. The process must be repeated for both still air and airflow conditions. The power loss over the temperature range of 10K must also be taken into consideration for
  • #1
Liammaloney
1
0

Homework Statement



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.



Homework 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



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!
 
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  • #2
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.
 

Related to Convection and Radiation in an Electrically Heated Cylinder

1. What is the difference between convection and radiation in an electrically heated cylinder?

Convection is the transfer of heat through the movement of fluids or gases, while radiation is the transfer of heat through electromagnetic waves. In an electrically heated cylinder, convection occurs when the heated air inside the cylinder rises and is replaced by cooler air. Radiation occurs when the heat from the cylinder is transferred through electromagnetic waves to the surrounding environment.

2. How does the shape of the electrically heated cylinder affect convection and radiation?

The shape of the cylinder can affect the rate of convection and radiation. A cylindrical shape provides a larger surface area for convection and radiation to occur, allowing for more efficient heat transfer. However, if the cylinder is too long, it may take longer for the heat to reach the end of the cylinder, resulting in slower heat transfer.

3. What factors influence the rate of convection and radiation in an electrically heated cylinder?

The rate of convection and radiation in an electrically heated cylinder is influenced by several factors, including the temperature difference between the cylinder and the surrounding environment, the properties of the fluid or gas in contact with the cylinder, and the surface area and shape of the cylinder.

4. How does the material of the electrically heated cylinder affect convection and radiation?

The material of the cylinder can also impact the rate of convection and radiation. Materials with high thermal conductivity, such as metals, allow for faster heat transfer through convection and radiation. On the other hand, materials with low thermal conductivity, such as plastics, may result in slower heat transfer.

5. Can convection and radiation be controlled in an electrically heated cylinder?

Yes, there are ways to control the rate of convection and radiation in an electrically heated cylinder. For example, using insulation around the cylinder can reduce heat loss through convection and radiation. Additionally, adjusting the temperature of the cylinder and the surrounding environment can also affect the rate of heat transfer through convection and radiation.

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