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Frank

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- Thread starter franksnsd
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- #1

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Frank

- #2

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- #3

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- #4

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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.

- #5

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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

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.

- #6

- 25

- 0

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.

- #7

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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

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

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.

- #8

- 25

- 0

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.

- #9

- 2

- 0

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 25°C. If you can help it will be very usefull for me, as I am always two month cannot calculate it. Thank you in advance

- #10

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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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