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Water Cooling Heat Transfer Calculations

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franksnsd
#1
Jan19-11, 05:30 PM
P: 4
I am currently working on a project that cools water moving through a chilled aluminum block. I am not familiar with the math/physics involved and would like some direction or recommendations. My project is currently setup so that I suction 32oz of 70 degree F water out of tank1 and pump it through a chilled aluminum water block into tank2. I am aiming to get the water coming out of the chilled water block to 33 degrees F. I am trying to figure out what calculation I would use to calculate how many BTUs need to be removed from the water to chill it to my desired temperature for the whole 32oz. The water is flowing at a rate of 8oz every 42 seconds. If there is anyone that may be able to help me with some calculations I would appreciate it.

Frank
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Alvasin
#2
Jan22-11, 03:41 AM
P: 10
UsuallyThere are two primary classifications of heat exchangers according to their flow arrangement. In parallel-flow heat exchangers, the two fluids enter the exchanger at the same end, and travel in parallel to one another to the other side. In counter-flow heat exchangers the fluids enter the exchanger from opposite ends. The counter current design is most efficient, in that it can transfer the most heat from the heat (transfer) medium.For efficiency, heat exchangers are designed to maximize the surface area of the wall between the two fluids, while minimizing resistance to fluid flow through the exchanger. The exchanger's performance can also be affected by the addition of fins or corrugations in one or both directions, which increase surface area and may channel fluid flow or induce turbulence.
franksnsd
#3
Jan23-11, 11:31 AM
P: 4
Thanks for the information. I am using a Thermo Electric Cooling module with the cold side attched to an aluminum water block and the hot side attched to a finned heat sink with and fan. I want to see if there is a calculation or formula to figure out exactly what I need to have the TEC cold temperature set to in order to meet my requirements as stated above. I have been experimenting with different voltages and temps but I cant seem to figure out a formula to predict what will happen when I change the variables.

thopsy
#4
Feb7-11, 05:28 PM
P: 25
Water Cooling Heat Transfer Calculations

Hey Frank,

The calculation to determine the amount of heat you need to remove is quite simple. I am going to use units of degrees Celsius and Joules instead of F and BTUs, but you can convert after. Also, please check my math, because I will probably make a mistake, but the methodology is correct.

The energy flow into the chiller is:

Qin = flowrate*heat_capacity*Tin

The energy flow out of the chiller is:

Qout = flowrate*heat_capacity*Tout

The amount of heat you need to remove is:

Q = Qin - Qout = flowrate*heat_capacity*(Tout - Tin)

For water: heat_capacity = 4.18 kJ/kg-C
8 oz = 0.24 L => 0.24 kg
The flowrate is 0.24 kg/42 s = 0.0057 kg/s
Tout = 0.5 C <-- will be hard to get this low (freezing), maybe aim for a few degrees higher
Tin = 70 C

Now we can calculate Q:

Q = 0.0057*4.18*(70-0.5) = 1.65 kW
This is about 94 BTU/min

Therefore you will need to extract 94 BTU/min from your device. This is the easy part. The more difficult part is to determine the temperature of the aluminum block required to get this heat removal rate. If you are still interested...I can also help you with this.
franksnsd
#5
Feb7-11, 09:06 PM
P: 4
Hey thopsy,

Thanks for the information, that is what I was looking for. I will also take you up on your offer to help me with the temperature of the Aluminum Block. If you have any input as to what size the block should be and how much water should be in the block or flowing through the block at any given time I would appreciate the help.

Send me a PM if you want talk via email.

Thanks,

Frank


Quote Quote by thopsy View Post
Hey Frank,

The calculation to determine the amount of heat you need to remove is quite simple. I am going to use units of degrees Celsius and Joules instead of F and BTUs, but you can convert after. Also, please check my math, because I will probably make a mistake, but the methodology is correct.

The energy flow into the chiller is:

Qin = flowrate*heat_capacity*Tin

The energy flow out of the chiller is:

Qout = flowrate*heat_capacity*Tout

The amount of heat you need to remove is:

Q = Qin - Qout = flowrate*heat_capacity*(Tout - Tin)

For water: heat_capacity = 4.18 kJ/kg-C
8 oz = 0.24 L => 0.24 kg
The flowrate is 0.24 kg/42 s = 0.0057 kg/s
Tout = 0.5 C <-- will be hard to get this low (freezing), maybe aim for a few degrees higher
Tin = 70 C

Now we can calculate Q:

Q = 0.0057*4.18*(70-0.5) = 1.65 kW
This is about 94 BTU/min

Therefore you will need to extract 94 BTU/min from your device. This is the easy part. The more difficult part is to determine the temperature of the aluminum block required to get this heat removal rate. If you are still interested...I can also help you with this.
thopsy
#6
Feb8-11, 03:26 AM
P: 25
Hey Frank,

Now that you know the thermodynamics, i.e. how to calculate Q, the next step is the heat transfer, i.e. how to calculate how big a block you need, and at what temperature. For this, it would be helpful to know a little more about how you will feed the water through the aluminum block: i.e. how many passages, the shape of the passages (i.e. circular), any dimensions you have.
franksnsd
#7
Feb8-11, 09:45 PM
P: 4
Hey thopsy,

I have made a couple differnt blocks. I am able to cast aluminum so I can make any size. I was just shooting in the dark trying to figure out what might work. I also have several different TEC modules with ratings of 70watts, 168 watts, 226 watts and 400 watts.

The first block I made was a 3" square, 1" thick. The channels were drilled out using 1/2" bit. The bottom and top are about 1/8" thick. I attached a rough sketch of what it looked like.

The second block I made consists of a 1/8" stainless steel tubing coiled inside a 2 3/8" square, 1 1/2" high aluminum block. I figured the stainless steel tubing would make the water flow faster and smoother to help the heat transfer.

If you have any recomendations for a block please let know as I can fabricate anything.

Thanks,

Frank





Quote Quote by thopsy View Post
Hey Frank,

Now that you know the thermodynamics, i.e. how to calculate Q, the next step is the heat transfer, i.e. how to calculate how big a block you need, and at what temperature. For this, it would be helpful to know a little more about how you will feed the water through the aluminum block: i.e. how many passages, the shape of the passages (i.e. circular), any dimensions you have.
Attached Thumbnails
Block.jpg  
thopsy
#8
Feb9-11, 09:42 AM
P: 25
Thanks for the sketch. It's quite difficult to calculate the heat transfer for a complex geometry like this, but you can do a simplified calculation.

In general smooth flow has worse heat transfer. The rough profile caused from the drill holes will help to create turbulence which increases the heat transfer. Also the metal-to-metal contact from steel-to-aluminum will reduce the heat transfer a lot. I think the drilled aluminum is the best idea.

I'm swamped at the moment, so will have to get back to you in a day or two.
Arshak
#9
Feb1-12, 06:37 AM
P: 2
Quote Quote by thopsy View Post
Hey Frank,

The calculation to determine the amount of heat you need to remove is quite simple. I am going to use units of degrees Celsius and Joules instead of F and BTUs, but you can convert after. Also, please check my math, because I will probably make a mistake, but the methodology is correct.

The energy flow into the chiller is:

Qin = flowrate*heat_capacity*Tin

The energy flow out of the chiller is:

Qout = flowrate*heat_capacity*Tout

The amount of heat you need to remove is:

Q = Qin - Qout = flowrate*heat_capacity*(Tout - Tin)

For water: heat_capacity = 4.18 kJ/kg-C
8 oz = 0.24 L => 0.24 kg
The flowrate is 0.24 kg/42 s = 0.0057 kg/s
Tout = 0.5 C <-- will be hard to get this low (freezing), maybe aim for a few degrees higher
Tin = 70 C

Now we can calculate Q:

Q = 0.0057*4.18*(70-0.5) = 1.65 kW
This is about 94 BTU/min

Therefore you will need to extract 94 BTU/min from your device. This is the easy part. The more difficult part is to determine the temperature of the aluminum block required to get this heat removal rate. If you are still interested...I can also help you with this.
Hellow thopsy.

My Name is Arshak and Iam working on Photovoltaic concentrationg Thermal hybrid system. And now I need some one helps to calculate heat transfare from plate collector to whater which flows under the plate. As I understand it is a dinamic process and calculation quit dificulte. Now I need to now How can I flow whater that GaAs photovoltaic solar cell will have some stable temerature. for example 25C. If you can help it will be very usefull for me, as I am always two month cannot calculate it. Thank you in advance
onkarj
#10
Jan4-14, 12:26 AM
P: 1
Hello Thopsy
If I can figure out the surface area and mass of Aluminium block then how shud I approch to calculate the time required to cool the water , can u explain by taking any example.

Thanks in Advance


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