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Heat Transfer from stainless steel tank of waters

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luktek

Electrical
Nov 25, 2009
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I am using Liquid Cooled Variable Frequency Drives and am wondering if it would be possible to cool the drives sufficiently using a large stainless steel tank full of water and then a pump to cycle water through the VFDs. The ambient temperature will be a constant 30C. I am not really familiar with this type of application, being a power electronics guy, and am wondering what formulas I need to consider in sizing the tank for heat transfer to the ambient air.
 
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How big a tank and what is the duty cycle of your drives?

Heat has to get out of the tank faster than your drives put heat into the tank. If not the water in the tank gets hotter and hotter.

If your drives are off for a long time each day then maybe the water will be cool by morning. Or maybe not.

And by the way, lots of nasty things like Legionella like to live in big tanks of warm water.

All in all, you'll be much better off with a properly sized water-to-air heat exchanger or cooling tower.
 
Hi There,

We would determine the size of the tank based on how large it needs to be to reject 36,230 Btu/hr.

The environment we are working in is an underground mine where water to air heat exchangers seem to not stand up to the dust.

I am wondering if this might be a simpler solution.ie, a big tank recirculating water as opposed to a fluid cooler.
 
For example, lets say we were to look at a single VFD with the following characteristics:

losses = 7.9kw
flow rate = minimum= 20L/60s, maximum 40L/60s
nominal flow rate = 25L/60s
pressure drop across drive = 1.96bar @ 40L/60s, 0.43bar @ 20L/min
ambient temperature 30C
how big would the tank need to be, assuming it was constructed of stainless steel, and filled with water, in order to never allow the rejected heat by the drives to raise the temperature of the tank above 40C.
 
Just as a back of the envelope thing.

If the water in the tank is at 40C you'll get about 6 watts per square foot of tank surface.

7900 W/6W/ft^2 = 1316 square feet of tank surface needed.

For a cube that's almost 15 feet each edge.

Probably bigger than you want?
 
Yes, we can deal with large, but we have height issues. Basically, the tank can be 15 feet long but no larger in diameter than 48".
 
and that's for a single VFD.

but how long would it take for the water to heat up to 40deg ?

Assume 10' cube of water, 5 faces radiating, 1 VFD ...
After 1hr you've spat out 8 kW.hrs, this'll increase the water temperature, so now the water can radiate heat to the outside (30deg air).

In the second hour the energy input to the water is (8-radiation heat calc'd above) kW.hrs ...

Assumes a recirculation fan inside the water, to distribute the heat.

hours 7 and 8 of the cycle radiate heat only, dropping the temp of the water. back to 30deg ??
 
Well, 15 feet long and 4 feet diameter gives you 213 square feet of surface. So you'll need 6 of them per drive.

I didn't count pumping heat, so make it 7 just to be safe.

I'm thinking this is a non-starter. But hey, it's your hole in the ground.
 
Another thought: What if we were to mount a large fan blowing on the tank to circulate air on it and create a wind-chill? Would that give us significant gains in cooling and allow for a size of tank that will work?
 
Air flow would definitely improve the performance, as would pin fins on the outside and plenty of conduction pipes within the tank.

However, tank size is a bit misleading, since the critical component is the heat transfer coefficient, which is proportional to the air cooling efficiency divided by the area. We've used liquid to air heat exchangers that dump 500W on a 5ºC delta at 1 gpm liquid flow and 90 CFM air flow in about 1/2 cubic ft volume. Assuming 10kW load, then you'd need something like 2000 CFM fresh air flow. In a mine, that's probably a nontrivial task.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
I think what you've described is what led to the development of the basic liquid-air heat exchanger.

Would a geothermal cooling design be an option? (Of course that all depends on mine depth)
 
quote, "...Assuming 10kW load, then you'd need something like 2000 CFM fresh air flow. In a mine, that's probably a nontrivial task."

In undeground coal mines MSHA(Mine Sasfety and Health Act) requires minimum of 3,000cfm airflow sweepinig the mine face in front of the miner to deal with methane evolving and respirable dust. So maybe the wind chill is available?

 
My understanding, based on Big Branch and others, is that air flow is often not up to code, and keeping the flow rates up is always a challenge. Additionally, since the outlet air would be hot, it wouldn't be suitable for human use, so this would be an additional load to the air handlers, and additional air handling in the exchanger itself and its compartment.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
Look at liquid cooled computer modules for some inspiration. You'll want to use a water/coolant mix to improve the BTU efficiency, and you basically need to setup some sort of heat sink to effectively cool the areas you need cooled. Doing it as an ambient cooling isn't the way to go. Making a finned liquid cooling mount to go on a chip or device with high heat output would be way better. There's all sorts of thermal conductive material you can use to move the heat where you need it. Again, look CPU/GPU heat sinks/fans, north/south bridge heat sinks/fas, etc in computers.

This is also a good candidate for running a thermal FEA/CFD analysis to optimize your design and reduce the total flow of coolant, etc to reach your target temperature.

James Spisich
Design Engineer, CSWP
 
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