rfr1tz
asked on
Basic Thermosdynamics Problem
I have a 100 watt light bulb inside a steel cube. The outside air temperature is 75 degrees. How how is the air inside the cube?
I actually want the general solutionm - I've just concocted a typical problem. Approximations are definitely appropriate - I'm looking for a method of quick estimation.
Also, I'm not I sure even asked the right question, I may have left out some needed info (which you can add, but don't make is too hard.)
I actually want the general solutionm - I've just concocted a typical problem. Approximations are definitely appropriate - I'm looking for a method of quick estimation.
Also, I'm not I sure even asked the right question, I may have left out some needed info (which you can add, but don't make is too hard.)
ozo
How large is the cube? Is the outside air free to circulate around the cube (this could make the problem complicated)? Is the cube in thermal contact with anything besides the air?
You can find reference data you need for this sort of calculation at the Watlow website:
http://www.watlow.com/reference/
Look in particular at the first few topics in the reference section:
Heat Loss Factors & Graphs
Quick Estimates of Wattage Requirements (PDF)
Properties of Gases & Air (PDF)
The best way to solve this problem is by experiment. Actually put a light bulb in a steel box.
But you can do a good job modeling thermal behavior with an electrical circuit.
Power ==> Current Temperature ==> Voltage
The light bulb is a current source of 100 V
The air is the box is a thermal resistance.
The steel box is a series resistance and a large thermal capacitor.
Convection between the box and the ambient as resistance.
The ambient temperature is a voltage source of 75 deg.
The air in the box will still be hottest near the bulb (where the power is coming in)
and coolest near the wall (where the power is going out).
And the top of the box will be hotter than the bottom.
But the Watlow tables give you all the infor you need to start to solve this problem.
http://www.watlow.com/reference/
Look in particular at the first few topics in the reference section:
Heat Loss Factors & Graphs
Quick Estimates of Wattage Requirements (PDF)
Properties of Gases & Air (PDF)
The best way to solve this problem is by experiment. Actually put a light bulb in a steel box.
But you can do a good job modeling thermal behavior with an electrical circuit.
Power ==> Current Temperature ==> Voltage
The light bulb is a current source of 100 V
The air is the box is a thermal resistance.
The steel box is a series resistance and a large thermal capacitor.
Convection between the box and the ambient as resistance.
The ambient temperature is a voltage source of 75 deg.
The air in the box will still be hottest near the bulb (where the power is coming in)
and coolest near the wall (where the power is going out).
And the top of the box will be hotter than the bottom.
But the Watlow tables give you all the infor you need to start to solve this problem.
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There are a wide variety of assumptions one can make. I can't strongly defend any of the assumptions I made. I assumed that heat transfer between faces would be insignificant.
In this case, I believe convection and conduction inside the box are mcuh more important than radiation.
Although the bottom surface fo the box certainly has the highest thermal resistance due to convection,
I expect it will be the coolest.
Convection inside the box delivers far more heat to the top than the sides or bottom. Hot air really does rise
Although the bottom surface fo the box certainly has the highest thermal resistance due to convection,
I expect it will be the coolest.
Convection inside the box delivers far more heat to the top than the sides or bottom. Hot air really does rise
ASKER
That was pretty good everyone. I now feel more comfortable baking a pie in my easy bake oven.
If the heat transfer between faces is insignificant, the temperature difference between the faces would also be insignificant.
According to the graphs, radiation transfers more heat outside than convection.
Radiation would be even more dominant inside, where the ambient air temperature will be much closer to the temperature of each surface.
According to the graphs, radiation transfers more heat outside than convection.
Radiation would be even more dominant inside, where the ambient air temperature will be much closer to the temperature of each surface.
I believe the air in an easy bake oven is able to flow between the outside and the inside.
Radiation is probably more important than convection in transfering heat from the bulb to the pie.
Radiation is probably more important than convection in transfering heat from the bulb to the pie.
The 0.63 factor is only for convective heat loss, you shouldn't divide the radiation loss for the bottom surface.