Question re: Cooling Sea Water

[Gastronom]

I need to maintain 400 liters of salt water at 6°C in a closed system. Outside air temperature at peak cooling periods averages between 18 and 22°C, but can be as warm as 30°C. The tank is 12mm molded plexiglass. Commercial cooling units rated for sea water available here range in power output from 350 to 4000 watts. How much power do I need? (Circulation pump achieves a throughput of approx 350L/Hour, if that is indeed relevant).

Thanks for your help.

[Lowrider]

I would gues anywhere between 8 and 70kW. would also depend what else is in the tank.

Most of our machines (trane) will also work with salt water and a capacity up to 1MW.

We just installed two CGAN's on a fishtank for a zoo in Holland!

[Peter_1]

Welcome Gastronorm, we don't have enough figures to help you.
Therefore, I agree Sanderh somewhere between 5 and 70 kW.

[TXiceman]

You need the total volume of the tank and a time period that it has to be cooled in.

Then you the total heat contained in the tank based on the delta T from 30 dC to 6 dC. This will give you the heat in the tank. total heat / time period = refig capacity.

You also need to consider the method that the tank gets to the temp. is this a one tie pull down or is it a batch process or is there a cyclic load being dumped in the tank.

Lots of questions, not many answers.

[The MG Pony]

Sounds like this is a display tank for fish and such, so living load.

[mohamed khamis]

I need to maintain 400 liters of salt water at 6°C in a closed system. Outside air temperature at peak cooling periods averages between 18 and 22°C, but can be as warm as 30°C. The tank is 12mm molded plexiglass. Commercial cooling units rated for sea water available here range in power output from 350 to 4000 watts. How much power do I need? (Circulation pump achieves a throughput of approx 350L/Hour, if that is indeed relevant).

Thanks for your help.

Hi Gastronom

Firstly the cooling load of the chilling sea water is :

= (350/3600)*4.2*(30-6) = 9.8 kW, u should add 15% over to compensate the heat gain from the surrounding to the tank therefore,

the cooling capacity = 9.8*1.15 = 11.27 kW = 12 kW

Thus u need almost a cooling unit with capacity of 12 kW to 15 kW.

Regarding to the available unit is of 4.0 kW, u have to use a pump of capacity of 140.0 L/hour not 350 L/hour.

Best regards:)

[Samarjit Sen]

I remember that I had visited a client who wanted to make ice from sea water. They had three tanks with sea water and was being maintained at a temperature of about 4 Deg. C. These tanks were being used to store crabs. After which they were shipped out alive. The refrigeration plant was about 65 Kw. But the tanks were large.

Compared to that, I feel load calculated by Mohamed Khamis is correct.

[Ravi]

Designing a system based upon the pull down loads brings unnecessary redundancy in cases related to process cooling.

Suppose, if I consider 2hrs pull down time then the system capacity reduces by half to that calculated by Mohamed Khamis. Though the formula is correct, I would agree with other posters about lack of data.

[mohamed khamis]

Designing a system based upon the pull down loads brings unnecessary redundancy in cases related to process cooling.

Suppose, if I consider 2hrs pull down time then the system capacity reduces by half to that calculated by Mohamed Khamis. Though the formula is correct, I would agree with other posters about lack of data.

Hi Ravi

Unfortunately there is a big mixing between a continuous, tubular water chilling and intermittent stored water chilling . when u chill water from let us say 30°C to 6°C by continuous flow rate of 400 L/hr the cooling load calculation is totally different when u chill the same conditions of 400 L stored in a tank at a rest. In other term, if 400 L of water enter steadily a tank of 30C and exit of 6C each hour differs from 400 L of water was if 30C and u want to chill it to 6C with a pull down time of 2 or 3 hours. In the first case the cooling load is = 400/3600*4.2*(30-6) = 11.2 kW while in the second case the product load is = 400/(2*3600)*4.2*(30-6) = 5.6 kW. And this explain to u the difference in cooling load calculation in case of AC and Refrigeration applications which is Why in AC the product load is not involved in the cooling load calculations. I wish it is clear
Cheers:)

[The MG Pony]

So what is the 3600? Is this a constanant for salt water?

What is the constanant for fresh water?

All so where does the 4.2 come from?

I'd enjoy learning more about this formula and how to apply it, so far this thread has been interesting.

Thank you.

[Lowrider]

3600 is 60*60, the number of seconds in an hour.

4,2 kJ/kg*K is specific heat for water

Q=m*c*dT

Q= heat in kJ/s (kW)
m= mass or mass flow (kg or kg/s)
c= specific heat in kJ/kg*K
dT= temperature difference.

[The MG Pony]

Ah Ok, cool, thanks!

Now I tried using the formula a fiew times, Now my aplication is much the same, how ever I know the exact heat in, and the volume of the water that it is going into

400 Watts/H In total (Some times less)
2 L Total system volume
1100L/H
This is for chilling water for cooling a computer system @ 0C to -30C Or for times I want to to figure out how to cool say 64L of water down from say 25C to 4C in 4-5 Hours (For cooling drinks in an average camping cooler) what formula do I use for thees cases? as when I use the above I get massive numbers that are much to big for the intended aplication.

Do I simply take the resualting number and devide it by the number of hours it can take?

[Lowrider]

Yep! But then the time in seconds, not hours!

[mohamed khamis]

So what is the 3600? Is this a constanant for salt water?

What is the constanant for fresh water?

All so where does the 4.2 come from?

I'd enjoy learning more about this formula and how to apply it, so far this thread has been interesting.

Thank you.

Hi The MG Pony

3600 = is the conversion factor from L/hr to L/s

constant for water is named here specific heat 9kJ/(kg.K) and it equals to 4.187 at 25C but for salted water it is slightly larger than that due to the salts concentration and yet i don not have this value and i took it as 4.2 kJ/(kg.K) so i mentioned in the first post here in the froum the cooling load swings between to 12 kW and 15 kw to take into account any diversity factor. and thank u for ur courtesy. I'd like to be in ur service

cheers:)

[mohamed khamis]

Ah Ok, cool, thanks!

Now I tried using the formula a fiew times, Now my aplication is much the same, how ever I know the exact heat in, and the volume of the water that it is going into

400 Watts/H In total (Some times less)
2 L Total system volume
1100L/H
This is for chilling water for cooling a computer system @ 0C to -30C Or for times I want to to figure out how to cool say 64L of water down from say 25C to 4C in 4-5 Hours (For cooling drinks in an average camping cooler) what formula do I use for thees cases? as when I use the above I get massive numbers that are much to big for the intended aplication.

Do I simply take the resualting number and devide it by the number of hours it can take?

for 64 L, the chilling time is 4 hours and factor of safty is of 15%

cooling load = 64/(4*3600)*4.2*(25-4)*1.15 =0.5 kW

therefore u need almost 0.5 kw to 1.5 kw as a system cooling capacity. however i think 4 hours as chilling time is big so what about a hour or two hours i think it reasonable. I wish it could help

Cheers:)

[mohamed khamis]

Ah Ok, cool, thanks!

Now I tried using the formula a fiew times, Now my aplication is much the same, how ever I know the exact heat in, and the volume of the water that it is going into

400 Watts/H In total (Some times less)
2 L Total system volume
1100L/H
This is for chilling water for cooling a computer system @ 0C to -30C

For this case i can not understand some figures as "400 Watts/H" what this "H" it refers to hours and i think no becuase this power not energy then this is very small cooling capcity. what 2 L and 1100 L/H which of both is the water capcity. What about 0C to -30C the water must frist enters at ambinet condition or less than that by 4 - 5 K not at 0C. Pls clarify that to give a light.

[Ravi]

For any coooling system, be it for HVAC or process, we have to consider the end load and this is our objective for deciding a cooling system. Once, the fluid is cooled to 6C from 30C, this 6C fluid is used to cool a process and then temperature rises depending upon the load, for a constant flowrate. Now mass flowrate times specific heat times the new temperature difference is the actual load experienced by the cooling system on a routine basis.

Most of the heat transfer equipment, for cooling, use a temperature differece of 5C for economical heat transfer device sizing and that is why I posted those comments and not to point out anything against your post.

That is why, I once again reiterate for more data from the OP.

[The MG Pony]

I figured it out thanks.

A computer puts in 400watts of energy so 400watts per houre.

System volume is 2L and flow is 1100L per houre.

Average opperating temp will be for the end user -30C.

[wkd]

Be aware of the specific heat of sea water.The salt makes it act like Brine solution which has a much lower specific heat than pure water, so you may need to revise the calc for the difference.