Heat Loss Through a Box with a hole in it

Click For Summary
SUMMARY

The discussion focuses on calculating heat loss from a rectangular styrofoam box with a 10 W lightbulb inside and a hole at the top for heat dissipation. Participants emphasize the importance of considering both conduction through the walls and convection through the hole to accurately determine heat loss. The complexity of the problem arises from the interplay of conduction, convection, and radiation, with suggestions that an analytical solution may not be feasible without iterative methods. Key insights include the lower thermal conductivity of air compared to wood and the significance of mass transfer in heat loss dynamics.

PREREQUISITES
  • Understanding of heat transfer principles, including conduction and convection.
  • Familiarity with thermal conductivity concepts and their implications.
  • Knowledge of steady-state heat transfer analysis.
  • Basic grasp of thermal radiation effects in heat loss scenarios.
NEXT STEPS
  • Study the equations governing heat conduction, specifically Fourier's law.
  • Learn about convection heat transfer, focusing on Newton's law of cooling.
  • Explore the concept of thermal resistance and its application in heat transfer problems.
  • Investigate the role of mass transfer in heat loss, particularly in open systems.
USEFUL FOR

Engineers, physicists, and students studying thermodynamics or heat transfer, particularly those interested in practical applications of thermal analysis in insulated systems.

dchau503
Messages
13
Reaction score
0
This is mainly just a theoretical question: say you have a rectangular box made of styrofoam (one of those ice cooler things) and it is levitating in air at room temperature conditions. It also has one hole at the top of the container so that heat can get out of it. Inside the box is a lightbulb that provides heat of 10 W. What equations can I use to find out how much heat is dissipated out the container?

I know how to use the equations for conduction to calculate the heat loss from the wall but I'm confused on whether to use just convection equations for the hole or if I should also combine convection equations to the conduction of the wall. Any help would be appreciated. I'm also thrown off by the fact that air has a lower thermal conductivity than wood, which doesn't make intuitive sense because that would make it seem like you would be more insulated in the outside than if you were in a house.
 
Engineering news on Phys.org
dchau503 said:
This is mainly just a theoretical question: say you have a rectangular box made of styrofoam (one of those ice cooler things) and it is levitating in air at room temperature conditions. It also has one hole at the top of the container so that heat can get out of it. Inside the box is a lightbulb that provides heat of 10 W. What equations can I use to find out how much heat is dissipated out the container?

At steady state 10 Watts are dissipated.

How much is lost out the walls vs out the hole is a very difficult problem to solve.
There will be radiation to the walls and out the hole, filament temp. is high so it's probably significant. Conduction to the air in the box. Convection to the walls and out the hole. convection and radiation (probably insignificant) at the outer walls etc etc. I don't think this could be solved analytically, if it could it'd take an iterative solution.

[/QUOTE]I'm also thrown off by the fact that air has a lower thermal conductivity than wood, which doesn't make intuitive sense because that would make it seem like you would be more insulated in the outside than if you were in a house.[/QUOTE]
I can't parse the italics.
What's surprising about air having lower thermal conductivity than wood? Generally, solids are much better conductors of heat than gasses. that's why you can put your hand very close to a hot surface without getting burnt..
 
How about if we just make this problem simpler: i.e. we can ignore radiation, there's just one rectangular slab of styrofoam with a hole in it. If you're just armed with convection and conduction, how would you calculate the heat loss if 10 W is applied equally throughout one side of the slab?

Also, it's confusing that air has a lower thermal conductivity because say I live in a wooden house that is at 70 degrees and the outside weather is -20 Celsius. If wood has a higher thermal conductivity, then that would mean that the temperature of the house would drop down much more quickly than if I were to have a house made of air (so no house) at the same initial conditions. I'm sure I'm ignoring something pretty important in this case though and it's driving me nuts.
 
In the winter try opening all of the windows. That gets you closer to a house "made of air". Why does it get cold inside?

BoB
 
It gets cold inside because of how heat goes from hot to cold. Because the thermal conductivity of air is lower than that of wood, I assume that the heat loss to the outside occurs at a slower pace when I open the windows than when I keep the (wooden) windows shut though. What am I missing?
 
dchau503 said:
It gets cold inside because of how heat goes from hot to cold. Because the thermal conductivity of air is lower than that of wood, I assume that the heat loss to the outside occurs at a slower pace when I open the windows than when I keep the (wooden) windows shut though. What am I missing?

I take it you've never opened a window on a cold day?

The difference is that closed windows prevent mass transfer, that's why we have windows! to let light in without mass transfer. Thermal conductivity doesn't matter one bit if cold air is blowing straight in through open windows.
Even if the air outside is completely still there will be mass transfer via convection currents. That is, the warm air in your house will float up right out your windows, and cool air will be drawn in, no conduction necessary.
 
I think a circuit diagram (nodal analysis) is the way to go.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K