Question about calculating BTU

[ysh1317]

Hi, I do have a couple of question, let's say I have a pond with 1000 liter of water and I'd like to bring the temperature down to 4 degree Celsius, how much BTU do I need? And how do I calculate the diameter and length of the caliper pipe?

Thank you in advance!

[NoNickName]

It depends what temperature the pond is before starting cooling

1 BTU is the energy necessary for cooling 1 pound of water by 1°F.

[Toosh]

Hi, I do have a couple of question, let's say I have a pond with 1000 liter of water and I'd like to bring the temperature down to 4 degree Celsius, how much BTU do I need? And how do I calculate the diameter and length of the caliper pipe?

Thank you in advance!


My question is why are you quoting metric, then trying to work out in BTU`s which is an imperial measurement

[NoNickName]

That's easy. Split systems are sold by the BTU/h, which is an awful habit and should be deemed as unlawful in EU

[james10]

That's easy. Split systems are sold by the BTU/h, which is an awful habit and should be deemed as unlawful in EU
We buy ours in Kw's.

[dougheret0]

One btu will change the temperature of one liter of water by 1.23 degrees C. So to cool 1000 liters from 15 to 4C will require removing about 9000 btu from the water. Unfortunately, I have no idea what a caliper pipe is.

[Brian_UK]

Caliper Pipe probably equals a capillary tube, use Dancap.

http://www.danfoss.com/BusinessAreas/RefrigerationAndAirConditioning/Product+Selection+Tools+Details/DanCap.htm

[mad fridgie]

One btu will change the temperature of one liter of water by 1.23 degrees C. So to cool 1000 liters from 15 to 4C will require removing about 9000 btu from the water. Unfortunately, I have no idea what a caliper pipe is.
I do not think so.
some clues
1 litre of water weighs more than a pound of water!
1F is a smaller energy value than 1C
look at Nonicknames description above
Best to use either imperial or metric to stop confussion
Have ago again?!

[NoNickName]

Let's do it in metric.
1000 Kg x 1 Kcal/KgK x 11K = 11000Kcal = 43651 BTU.
If you want to cool it in one hour then you need a chiller by the capacity of 43651 BTU/h or 12.4 kW.

[ysh1317]

Let's do it in metric.
1000 Kg x 1 Kcal/KgK x 11K = 11000Kcal = 43651 BTU.
If you want to cool it in one hour then you need a chiller by the capacity of 43651 BTU/h or 12.4 kW.

Thanks for all the replies so far, so helpful.

I am from Indonesia btw, nice to see all of you :)

[ysh1317]

Another question, is it possible to bring the temperature of 1000 liter of water down around 4-8°C with 9000 BTU chiller? Ambient temperature is around 30-31°C.

If it is, any tips and trick to do so? Or is it too much for a 9000 BTU chiller? Thanks and sorry for the basic question :)

[NoNickName]

I think you mean BTU/H e not BTU.
Yes, it is possible, but instead of taking one hour, it will take 43651/9000= slightly more than 5 hours

[ysh1317]

I think you mean BTU/H e not BTU.
Yes, it is possible, but instead of taking one hour, it will take 43651/9000= slightly more than 5 hours

The reason I am asking that is because I tried do it with a 9000BTU/H HBP rotary compressor, but I wasnt able to get the water down even to just 10°C. I wonder what I did wrong? Perhaps the sizing of the pipe is incorrect?

[NoNickName]

It means that the insulation is not enough or that the unit is not performing as declared.

[chilliwilly]

The reason I am asking that is because I tried do it with a 9000BTU/H HBP rotary compressor, but I wasnt able to get the water down even to just 10°C. I wonder what I did wrong? Perhaps the sizing of the pipe is incorrect?

What are the dimensions of the pond? is it long and shallow having a large contact area with the air, or short and deep with a small contact area with the air, and what is the RH of the air? And what method of heat exchanger are you using? Ie a direct pump system thats capable of moving the total volume of water very quickly via a cooling tower then through a double walled evaporator, or an evaporator directly submerged in the pond, or a glycol/brine coil/loop submerged directly in the pond?

[ysh1317]

What are the dimensions of the pond? is it long and shallow having a large contact area with the air, or short and deep with a small contact area with the air, and what is the RH of the air? And what method of heat exchanger are you using? Ie a direct pump system thats capable of moving the total volume of water very quickly via a cooling tower then through a double walled evaporator, or an evaporator directly submerged in the pond, or a glycol/brine coil/loop submerged directly in the pond?

The dimension of the tank is 100cm wide and 60 cm deep, I've tried both with direct pump and submerged evaporator with bad result... I wonder what can be done to fix it. Thanks for the response :)

[NoNickName]

It also depends at what conditions the cooling capacity is declared.

[dougheret0]

Nonickname's calculation is correct. My unfamiliarity with SI led me to an error. Glad it was picked up.