Need help designing chiller system from scratch!

[The MG Pony]

Ok here is the situation, be for I'd all ways make a chiller system based off the max it could move, IE a Brute force type system, naturally it worked, but may be bigger then needed.

So now I want a task specific system to be made.
I don't want it don for me, but I'd like to be shown the formula's and helped through if any are willing, I hope to learn about all the required formula's via this project!

Parameters:

Ambient design temp: 35C
Target fluid temp---: 20C
Load --------------: 195W
Refrigerant---------: R-134a
Expansion device---: Thermal-static
Secondary coolant--: Water/Antifreeze 75%/25%
Evaporator--------: Packless Coaxial Hx Rated for 6,000BTU

So I've determined the end temp desired (21c) I've determined the constant load (195W) and the ambient max temp, How do I enter this data to determine flow rate of my secondary coolant (Water/Antifreeze 75%/25%) and how to determine the actual refrigeration system!

Again, I just am asking for a helping hand as I want to do all the number crunching! (After all I won't learn it if I don't use it!)

Thanks in advance for your time and help!

[mohamed khamis]

Hi MG Pony

there are some things got me confused.

1- could u draw a sketch and point the temperatures inlet and outlet for water.

2- what do u mean by constant load (195W)

3- Why do u want to use antifreeze solution although ur target temperature is 20°C and the ambient is 35C. elaborate pls more data to help u God Willing


Cheers

[The MG Pony]

Antifreeze is just a percuation and to help lubricate the pump along with to discurage life from moving into my cooling circuit.

195Watts will be dumped into the coolant. (Semiconducters running at full load)

I have no idea of the deltas! in fact I have not much idea what the exact coolent temp is at now other then onboard sensores, and they are not all that reliable due to where they are placed!!

Perhaps I'm going about this wrong, this is obviously going to be a very small system, perhaps the Hx should be custom made, and well call system load, 300W
and design for -10 evap.

So what formula's do I need to make such an evap? Assuming 1.5gpm flow

[mohamed khamis]

If u cooling load as u mentioned 300 W and as i understood the load is onboard circuit. So why do u go to thermoelectric cooler with heat sink or heat pipe with heat sink. the load is very light and difficult to find out evaporator size suits this load. ok regarding to chiller

u mean the water comes from river at 35C ambinet condition and u need to drop it to 20 and if it is correct u don not need TEv to be -10C it is about 7.5 or 10C. further data will help much

cheers

[The MG Pony]

Thermal electric is insane waste of energy it is so inefficient compared to a phase system.

Heat pipes suffer from the same problem as any other heat sink, it just moves the heat around in the enclosure.

Nearest river is over 4Km away, piping would be pricey and the bylaws officer would have words

System is a closed loop water cooling circuit cooling at current a 195Watt load, how ever it will be increased as other systems are brought on line.

[frank]

To be quite honest MG, I would build a small chilled water system using a tank and then have a secondary water loop for the cooling.

What I mean is, because the load is so small it would be better to have the compressor cooling a tank full of water with either a small refrig water heat exchanger or a trombone type refrig coil in the tank, say 25 litres, down to say, 6C and then run a pipework loop with circulator/pump to the CPU.

This way, as it's a small load, you will get better cooling of the CPU with a constant supply of chilled water and less cycling of the compressor.

Use the formula Q = m c dt
where
Q = the size of your cooling equipment in kw
m = the mass flow of your water circuit (primary)
c = constant (specific heat capacity of the fluid) for water it's 4.19
dt = the temp difference in your fluid circuit (primary) normally 12 entering and 6 leaving (C)

If you have a CPU load of 195w then you could design your tank to be rated at 1.5 or 2 x that, say 400w.

Using the formula we can now see that Q = 0.4 so,

0.4 = m x 4.19 x 6
Transposing, we get
m = 0.4/(4.19 x 6) = 0.016 kgs/s (or l/s)

If you want your secondary circuit to be 6C leaving and 12C returning then you would need a pump/circulator rated at 0.195/(4.19x6) = 0.0077l/s

Have a play with the numbers in the formula to see what you can do with the kit you have got.

[The MG Pony]

Ah K thanks, Normaly I all ways just built them 3,100BTU using a 6,000BTU Coaxial Hx with a 1/4 TXV and that all ways seemed to do a good job on most mid sized aplications.

I was hoping to be able to realy cut it down to save money on parts.

[monkey spanners]

Why not make a small icebuilder system?

Jon

[The MG Pony]

Too cold, I need to keep the temp up due to humidity issues.

[The MG Pony]

Q= 750*25%= 938W Total+safety factor
M= Veriable
C= Specific heat of water 4.19
dT= Delta Temp 12 return, 6 exit 12-6C= 6CdT

Q=M*C*dT But we need M for coolant flow
M=.938/(4.19*6dT) = .038L/s *60*60 (To get Lph) 136.8Lph = .603gpm

computer cooling loop + exchanger must have .603gpm flowing through it to maintain its proper delta.

For the condensor side same formula
Q= 1.75*25% =2.2Kw
M= x
C= 4.19
dT= 10 is average for condencing

M= 2.2/(4.19*10) = .053Lps .053*60*60= 190.8Lph (.84gpm)

So does that look correct?

[mohamed khamis]

Q= 750*25%= 938W Total+safety factor
M= Veriable
C= Specific heat of water 4.19
dT= Delta Temp 12 return, 6 exit 12-6C= 6CdT

Q=M*C*dT But we need M for coolant flow
M=.938/(4.19*6dT) = .038L/s *60*60 (To get Lph) 136.8Lph = .603gpm

computer cooling loop + exchanger must have .603gpm flowing through it to maintain its proper delta.

For the condensor side same formula
Q= 1.75*25% =2.2Kw
M= x
C= 4.19
dT= 10 is average for condencing

M= 2.2/(4.19*10) = .053Lps .053*60*60= 190.8Lph (.84gpm)

So does that look correct?

Hi MG Pony

Yes it is correct but if u want to add antifreeze material as u mentioned early the water specific heat in the cooler will be less than 4.19 according to percentage of antifreeze material and this will enlarge the calculated mass flow for the chilled water above (.84gpm).

Regards

[The MG Pony]

Ok thanks, I'm going to play with this formula some more

[frank]

Q= 750*25%= 938W Total+safety factor
M= Veriable
C= Specific heat of water 4.19
dT= Delta Temp 12 return, 6 exit 12-6C= 6CdT

Q=M*C*dT But we need M for coolant flow
M=.938/(4.19*6dT) = .038L/s *60*60 (To get Lph) 136.8Lph = .603gpm

computer cooling loop + exchanger must have .603gpm flowing through it to maintain its proper delta.

For the condensor side same formula
Q= 1.75*25% =2.2Kw
M= x
C= 4.19
dT= 10 is average for condencing

M= 2.2/(4.19*10) = .053Lps .053*60*60= 190.8Lph (.84gpm)

So does that look correct?

Well done Pony - you are starting to use the formula
And I note that you used the correct conversion for L/s to G/m i.e. US gallon = 3.8L as opposed to the UK conversion UK gallon = 4.8L

[The MG Pony]

Cool Thanks, it is fun learning new things I greatly appreciate all your help!

You and mohamed khamis have helped greatly, along with others, Many thanks

[mohamed khamis]

Cool Thanks, it is fun learning new things I greatly appreciate all your help!

You and mohamed khamis have helped greatly, along with others, Many thanks

u'welcome at any time MG Pony

Good luck

[star882]

Thermal electric is insane waste of energy it is so inefficient compared to a phase system.

There was a high end Dell workstation that used a thermoelectric cooler to cool a quad core CPU. I was under the impression that phase change is much cheaper than thermoelectrics for a quad core but Dell seems to disagree.
Can't beat the price and simplicity of some surplus modules, though.

[The MG Pony]

There was a high end Dell workstation that used a thermoelectric cooler to cool a quad core CPU. I was under the impression that phase change is much cheaper than thermoelectrics for a quad core but Dell seems to disagree.
Can't beat the price and simplicity of some surplus modules, though.

The initial purchase and set up is some what cheap but operating costs are not, they suck up far more energy then needed, so naturally due to cheap purchase cost of course dell would like them, but the customer ends up with a part of the power company far up their rear! and dell gets a good laugh from their customer being the sucker!