cooling capacity requirements for a water tank. 4

[somacast]

Hello,

It is desired to cool down the water of a tank using a chiller (simply an air conditioning unit which evap is placed in the water, a common practice over here) this water is used to cool down compressed air from a compressor, so I understand that BTUs required for initial cooling down from 50 C to 25 C can be simply found from M*Cp*(T2-T1), but I want to know the cooling capacity requirement for continuous cooling (Btu/ Hr) ?? I want this to size the unit which I have to buy for the purpose, a schematic of the system being used is attached FYI.

Thanks a lot.

 

[LittleInch]

Surely the same formula but instead input the data for the air that is being cooled?

Not sure why you're not using the cooling coils to directly cool the air instead, or do some of the cooling using an air/air radiator, but hey, if you want to make life complicated then feel free.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[somacast]

hi LittleInch

can you please explain more ? I want the rate , to compare to the suitable chilling unit (1 Ton , 1 1/2 tons , etc), so how do I use the air data , sorry did not get you

and for the reason behind using this is that originally the unit is a package unit of an air compressor, reciprocating, with inter-stage cooler (air to water), so the water comes from a tank, during last years ambient temp goes really high in summer that the water isn't cold enough to exchange heat with air, so we intend to cool the water, the same system was used long ago to cool water in tanks of other purposes.

 

[LittleInch]

For air your know the mass of air per second you are trying to cool(M), you can find the Cp of air at that pressure (google), you know T1 and T2 of the air.

Then convert that to whatever heat flow unit you want to use.

Ton units are usually tons of water per hour, but there is a conversion - just search for it.

I still think there is a more efficient way of cooling air than using an AC machine, but if it's only for occasional use then maybe Ok.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[somacast]

thnx again , but please let me ask more and sorry if it sounds stupid..

for the water inside the tank q = m*cp*dt .... like that iam going to find out Btus required t cool down the water from T1 to T2 ..

then I need to know the rate btu/hr , if I use m*cp*dt again for the air what do I gain from that ?? this equation does not give the rate (per hr or so) right?? isn't it supposed to be the combined heat transfer from water through tank wall to atmosphere ? and using water temp and atmosphere temp ??

sorry again if the question isn't too good but its been really 10 years since I came across heat transfer issue, its not part of our daily projects supervision so I really forgot most of it , and once in a while they throw such a request on me .. so no solution other than asking for help ..

 

[LittleInch]

I kind of assumed that the heat required to cool the air down was much larger than any heat transfer from the tank into the atmosphere and hence this is the heat load being applied to the water that you need to remove from the water to keep the water tank at a steady temperature

You could also use mass of water going into and out of the cooler and the Tin and T out of that if you know what that is. Then you have the real heat load going into the water tank.

If your tank volume is big compared to he heat load then maybe it doesn't matter so much?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[somacast]

Thanks again and again for your help,

its just a small tank of volume of 1.57 m3 , not insulated , and your assumption about air in direct contact with water is good, but I still doubt that convection by atmosphere through the tank body to the water inside is important ... I just need to estimate the heat transfer rate after all and the source should be considered.

thnx

 

[lilliput1]

There are packaged refrigerated air dryers for cooling the compressed air available. The condensed water in the tank is then drained out by float trap to the floor drain. Using refrigerated dryer is preferred if compressed air dew point is required to be between 35 and 40F. Chilled water typically at 44F can not cool down to this level. also chilled water may not be available year round. If lower dew point to -40F is required there are desiccant dryers available. Filter the compressed air downstream to make sure it is dry.

 

[IRstuff]

Here's my simplified approach. Note the fact that it's uninsulated is costing you a lot of heat gain, as the surface temperature is barely distinguishable from the internal temperature. Insulation will increase the surface temperature and reduce the heat gain.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[somacast]

Thank you all for the pointers.

 

[somacast]

what happened overhere ???

my reply was changed by someone !!! and all my questions are gone, and I got 3 emails notifying 3 new replies, however none are here ?? how could this happen ???????????????????????????????????

 

[LittleInch]

If a post gets red flagged, the moderators have a look and can delete posts or modify responses.

try listing your questions again.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[somacast]

I see , although I don't see why would it be red flagged , I will re-post my questions again hoping to get some further answers,

thnx

 

[somacast]

ok , reposting my reply to @IRstuff (I don't know how to quote a post in this forum , sorry),

First I would like to thank you, however I have a few question to ask please (kindly also refer to my attached sketch):

1- I understand that you have not considered the convection between water and tank inner surface as temperature wont really be much different between them, but is that really good practice to do so ? mainly if I have a pump downstream the tank which circulates the chilled water?

2- how do I get the values of HTC for air & water? you have taken 7.5 & 50 , aren't they temperature related?

3- you have assumed a Tsurf = Tin , but later when you calculated power down you calculated Tsurf again, because if Tsurf=Tin then power of conduction in the shell would be zero ?

4- can we use q = A*(Tout-Tin) / (1/HTC1 water) + (thk/kshell) + (1/HTC2 air) ??

5- in the (find power) line , the result iam getting is 1.527 ? kindly correct me if iam wrong

thnx again

 

[IRstuff]

1. if you have circulation, then you have less of a need for a transfer coefficient. The transfer coefficient is essentially modeling the thermal equilibrating of the fluid. If the water has a high circulation, then that interface with the wall will behave like a perfect temperature source.

2. thermal conductivity you can look up in tables or literature for your material. convection coefficient is dependent on air circulation, and there are numerous possible values, depending on the geometry and the author. Both are indeed temperature dependent, but unless you are doing something that's radically different, like putting this tank on Venus, or running molten sodium in the tank, it's not that important, particularly as your geometry impact on these coefficients has a larger variance.

3. as notated in the sheet the initial assignment of T.surf is for a guess value to start the solver

4. you can explicitly solve for the temperature, but not that way. you can algebraically manipulate the simultaneous equation above for T.surf to explcitly get an answer. you can see that the end result does not look like your equation

5. based on what? that amount of heat flow is impossible with an uninsulated system.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[somacast]

@IRstuff , thnx again and again , the flow of the pumps is 26 m3/hr , so I guess I will have to consider forced convection in the pipes downstream the pump , but this again confuses me if the AC(chiller) capacity should cover all of this or just the tank with some margin of increment will do .

 

[IRstuff]

If there is substantial heat gain in those pipes and the water returns, the it should be accounted for, but it might already be accounted for in your 50C to 25C cooling calculation.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[somacast]

@IRstuff cant thank you enough really , thanks a lot , and thanks to everyone who contributed as well.