Heat transfer-SS Tank 1

[harrinshotmail]

Stainless steel tank 27cu. feet volume, is filled with water. The top has a lid with a few small holes for the water vapours.
Water has to be heated from 20 deg celsius to 95 deg celsius in 25 minutes(not critical). Once the water is hot, the water is pumped from the tank using a 2 gpm pump and is dumped back into the same tank accept that the returning water is 20 degree celsius.

The question is what is the wattage and number of screw plug heaters that will be required to keep the water around 95 deg celsius? What will be rate of heat loss from the surface of water?

 

[desertfox]

Hi harrinshotmail

I haven't enough information to calculate the heat through the tank wall or lid ie:- wall thickness, vessel diameter etc or whether they have insulation and what insulation material it is but very approximatly I estimated the heat required based purely on the mass of water as follows:-

E= mass of water * specific heat capacity(water) * temp diff


coverting to metric units:-

E = 765kg * 4.187kj/kgK * (368K-293K)= 240229 Joules of heat

now 2gallons = about 9.1kg based on uk gallons

now remove 9.1kg of water and return it at 20 deg C

heat gained by the 9.1kg must equal the heat lost by the 755.9kg left in the tank:-

755.9*4.187*(368-T)= 9.1*4.187*(T-293)

0= 1164702.814-3164.9533T= 38.1T-11163.7981

the T works out at 367.1K

therefore additional heat req to maintain 368K or 95 deg C

755.9*4.187*(368-367.1)= 2823.01J

total heat req = 240229+2823=243052J

now power is joules/second

so 243052/(25min*60)= 162 watts.

You need to consider heat losses from tank and water surface and put a safety margin on the power I have calculated to get a better idea of the heat and power req.

regards

desertfox

 

[harrinshotmail]

The material used is 316 stainless steel. the wall thickness is 0.187".

How do u calculate the heat losses from the surface and the wall?

The container dimensions are 36" x 36" x 36". The water will be filled 5" below the surface.

 

[desertfox]

You need to know the outside temp surrounding the vessel,insulation if any and if its subject to any other elements.
Then you need to refer to a heat transfer book and look up heat conduction and convection from which you should get some idea of your heat required.

 

[harrinshotmail]

Please assume no insulation used. The outside temp is 25 degree celsius.

 

[desertfox]

Hi again

Your heat loss will be massive by conduction through the tank wall you need to insulate it, also I can't find the coefficient for convection from top of the tank etc.
but if you don't insulate this tank the projects a non starter

 

[harrinshotmail]

OK! Thanks. Can we use some insulation materiald similar to what are in the dry walls etc?

 

[harrinshotmail]

If in this tank,I dump a piece of steel for 10s how much heat it will take away from the water. Imagine the piece of steel is AISI 1010 and its weight is 1 pound.

What will be the difference if I dump 50 such pieces for 10seconds?

 

[desertfox]

Hi harrinshotmail

Firstly I made a mistake in my first post where it says watts read kilowatts.
I can't help with your last question placing steel in water for 10 seconds becomes a transient heat transfer problem all my calculations above are based on steady state.

regards

desertfox

 

[harrinshotmail]

Currently we are using 10KW for a larger tank to maintain temperature between 50 and 60 celsius? There are 2 element heaters 5KW each. 160KW sounds a lot practically!

I think it is 1.6KW??????????

 

[desertfox]

hi harrinshotmail

Well maintaiing a temperature is one thing heating a liquid
from 20 degrees C to 95 degrees C over 25 minutes which is what I have calculated takes more power.
The figure of 162 watts is needed to heat 765kg from 20C to 95C. Once you reach the temp you want, if it is to be maintained between say 90C and 95C you need to put in this amount of energy:-

Then:- E= m*Cp*(T1-T2) = 765*4.187*(368K-363K)=16015.275kJ

over what time period you put this energy in determines your power so lets say 25minutes :-


16015.275kj/(25mins*60)= 10.6 KW of power

 

[dvd]

Download this manual:

http://www.chromalox.com/content/training-manuals/TrainingTankHeating.pdf

Call your local Chromalox rep. and have them go over your application. You will end up wasting time posting back and forth here.

For your transient steel heating question, you might gain insight by understanding Biot number and lumped-capacitance. If it is really important, place a thermocouple in the center of one of the pieces of steel and record the transient response.