Water immersion Heater selection 1

[chem55]

Hi:

I have about 376 gallon(1421 lt) of water in a open cylindrical tank. I would like to use 6kw of heater and I want to increase 20 deg cel temperature. As per the formula,

Kw = [gallons * T(rise)] / [325*Time(hrs)]


I have calculated that it will take about 13hrs to temperature rise of 20 deg cel.

Now my question is, after reaching the setpoint (Temp rise of 20deg cel) if there will be temperature reduction of 1 or 2 deg cel, how much time it will require to reach again the set point? If there will be reduction of 2 deg cel will occur, I want to reach the set point just within 5 to 10 minutes.

For minimize the time, I'm ready to use more than one heaters of 6Kw.

Looking forward your kind reply.

Thanks,
Chem55

 

[ko99]

Chem55,

I don't have a ready answer. Heating time will vary with air temperature, wind, container material, sensor location, etc. Also, the water temp will not be uniform (multiple heaters would help).

Regardless of the specifics, an excellent way to minimize time is to use a PID temperature controller. There are several good suppliers with helpful support staffs.

http://www.omega.com/vh/subsectionVH.html?subsection=A03&book=vhPC

http://www.watlow.com/products/controllers/

Good luck

ko (

http://www.ecooling.biz)

 

[TD2K]

What's the basis for your formula? Is it supposed to take into account heat losses from your tank?

If I go back to basics:

6 kW = 20500 BTU/hr

376 gal (US?) = 3136 lbs.

A change of 20 deg C = a change of 36 deg C

Water has a specific heat of 1 BTU/lbF

The amount of heat required = 112900 BTU and your heater should be able to supply this in 5.5 hours.

Now, if you are raising the water temperature from 5C to 25C, there is going to have much less heat loss due to evaporation from your open tank than if you raise it from 60C to 80C. In addition, you have heat losses from the tank to consider which the heater must supply in addition to the warmup and evaporation losses.

Basically, you haven't provided enough information. Chromolox, just one example, has tables in their catalog showing heat losses from various tanks for various cases including open tanks.

To your second question, if it's going to take 13 hours (just for arguement's sake, I'll use your numbers) to raise the water temperature 20C, a change of 2C (10% of that figure) is going to take about 10% of the time or 1.3 hours.

 

[chem55]

Hi TD2K & Ko99:

I got the correct calculation from Chromalox technical article.!

Thanks for your kind reply.

Chem55

 

[mb44]

Rather than heating all the tank and maintaining the temperature would an in-line heater provide a better solution?

This would only heat the water you are using.

This may not be suitable for your application but worth a mention

MB

 

[25362]

All the comments above are correct and useful. As long as an electric heater is considered a constant heat flux device, the build up of water hardness on the external surface of the heating element wouldn't much affect the time needed for heating. It would, of course, rise the temperature of the element's metal cover and may endanger its integrity.