Convection Heat Transfer Problem

[AWOC]

Hi all,

I have a unique problem that I need some direction on. I have a .5x.5x.75 meter adiabatic chamber. No heat loss from chamber walls to the atmosphere.

Inside the chamber there is an electrical flat heater plate .5x.05 meter with constant 100W. A fan blows air over the heater bar at 0.25 kg/s (V =3m/s).

The air temp in the chamber rises from 20°C to a controlled 40°C. It never goes over 40° and it takes 60 minutes to reach a temp of 40° from 20°.

Inside this chamber is a 2L soda bottle. How long would it take to warm up the soda bottle to a temp of 40°C from 20°? The bottle is placed in the middle of the chamber not touching the chamber ( no Conduction).

I need to graph Temp vs. time of the soda bottle.

I am having problems with a lot of aspects:
-First the heat transfer from the heater bar to the air is difficult to do because the air temp is constantly rising. My forced convection over a plate calculations don’t make for a constant convection coefficient this way.

-The air volume in the chamber is constant, but a lot of heat transfer problems I have come across don’t account for this. They account for the heater bar volume and the time it takes to cool off, but my heater bar is a constant 100W output.

-How can I relate the air heat transfer from the heater bar to the soda bottle? The soda bottle will take longer to heat up than the air inside the chamber.

The problems sounds easier than it is, but there are a lot of things to consider. I would appreciate any information .

 

[AWOC]

Anyone?

 

[Mech_Engineer]

It's a difficult problem there's no doubt about it. I think you have your work cut out for you!

I'll try to address a couple of the issues you brought up:

-First the heat transfer from the heater bar to the air is difficult to do because the air temp is constantly rising. My forced convection over a plate calculations don’t make for a constant convection coefficient this way.

Well the heat output of the bar is constant in the problem statement, and no volume or material properties are given for the heating plate, so its temperature will rise to whatever is necessary to dissipate 100W of energy. This tells me that if you ignore other heat transfer modes (radiation and conduction) you can just consider it to be putting a constant 100W of power into the system and just ignore the extraneous data regarding convection across it.

My guess the data regarding the air blowing across the heater plate is only there to distract you, or possibly to help you estimate a forced convection coefficient around the bottle.

-The air volume in the chamber is constant, but a lot of heat transfer problems I have come across don’t account for this. They account for the heater bar volume and the time it takes to cool off, but my heater bar is a constant 100W output.

The thermal mass of the air is important because it controls how fast the air heats up around the bottle. Since you have been directly given the power input, you don't need to worry about convection (except perhaps to have an estimate of the forced convection about the bottle) and instead just calculate how fast the mass of air heats up with 100W of power input. Since they spec 60minutes, maybe you just say it's a linear rise since power input is constant too...

If the air only heats up to 40C and no farther, I would assume the mass of air heats up similarly with or without the bottle in there, and then use that T(t) function as the boundary forcing function for the bottle heating up. Basically, you might be able to ignore the effect of the bottle on the bulk air temp.

I'd also guess the air will heat up relatively quickly compared to the bottle, and you may be able to approximate its heating time as small compared to the bottle's heating time. If this is the case, you might also be able to approximate the temperature around the bottle as constant, simplifying your solve of the bottle's T(t).

-How can I relate the air heat transfer from the heater bar to the soda bottle? The soda bottle will take longer to heat up than the air inside the chamber.

If you utilize how fast the air heats up, and then use that temperature vs. time function, you can approximate the temp on the outside of the bottle and calculate its heating next. It could be this temperature looks nearly constant compared to the T(t) of the bottle.

 

[AWOC]

Sorry I forgot to mention that the heater bar material is pure aluminum.

I will definitely assume the constant temp around the bottle. Thanks for tip. Here is where I am stuck now (Heater bar heating air).

Doing a lumped capacitance for the convection from the bar to air. The numbers don’t make sense. I get results for the time it takes the heater bar material to cool off. This doesn’t help at all because the heater bar doesn’t cool off in the short time I get. I will take your advise and look into just generalizing the constant 100W energy coming out of the bar, but I am not sure how the surface area of the bar ties into this problem.

Did a forced convection over a hot plate calculation and got the Re#, Nu#, convection coefficient h. When I plug these numbers in q=hL(ΔT), my q which should be 100W is not even close.

Since I am given time it takes the air to warm up, what equation can I use to determine to right convection coefficient?

Thanx for the heading me in the right track btw. You are making some key points for me to consider.

 

[Mech_Engineer]

I think the point is you don't even have to calculate the convection over the heater bar, because you're given the power it's putting into the system and for that matter you're given how long the air takes to heat up too. Since power input is constant, if you assume air's peoperties are constant within the temperature range you're looking at (probably a good assumption) you can just assume the temperature rises linearly between 20C and 40C in a 60 minute time window. No need for any calculations w.r.t. the air heating.

 

[AWOC]

Sorry for not telling you right off the bet about the second part. In the second part of the problem I have to analyze how a smaller surface area heater bar changes Temp vs. time of the soda bottle. Same 100W going into the heater bar with 25% reduction in surface area. This is why I was going in the route relating the heater bar in the problem.