Making a simple water cooled condenser

[The MG Pony]

How long of copper tube of 1/4 size would I need to make a 3,100BTU water cooled condenser?

I plan on using a 1/2" PEX water pipe as the jacket and cooling the 1/4 inside.

I tried searching the forum but couldn't find any info on the heat rejecting ability of 1/4" copper per foot with a 10K delta-T.

I'm thinking with a 16C inlet temp I'd guestimate 6 feet or so?

Basicly I want to figur out how many BTU per foot of 1/4 at a dT of 10K in a water cooled aplication.

[Peter_1]

This is exactly he same way the evpaorators are working on some Trumpf lasers we're servicing.

In attachment a Kelvinator document from just after WWII. :D Got it some time ago from a RE poster
Hopes this is of any help even if it's ancient.
I have another for cooling but it is 105kB large, so too big to post here.
Anyone has a tool to reduce this size?

[The MG Pony]

Ok cool, so I'm figuring about 100 BTU per 2.4 F^2, but I'll be using it in the revers, but I supos it works all the same only the water is cooling it.

So 3100BTU * 2.4 = 7440 squar feet of tubing? Thats doesn't sound right to me.

[kubae]

Peter_1,Download Winzip or WinRar free off the web or send your email address. You can use any of these to reduce the file.I'm salivating at the thought of it so...

[frank]

How long of copper tube of 1/4 size would I need to make a 3,100BTU water cooled condenser?

I plan on using a 1/2" PEX water pipe as the jacket and cooling the 1/4 inside.

I tried searching the forum but couldn't find any info on the heat rejecting ability of 1/4" copper per foot with a 10K delta-T.

I'm thinking with a 16C inlet temp I'd guestimate 6 feet or so?

Basicly I want to figur out how many BTU per foot of 1/4 at a dT of 10K in a water cooled aplication.
Thats fairly simple but you have a couple of figures missing.
Using the formula Q = U A LMTD for a cross flow heat exchanger you can find A by transposing.

We know that Q = 3100btu's or 908.3W
U (for copper) = 385
A = the area of copper tube ( for the length divide the answer by Pi r^2)
LMTD = (shell & tube) (Q = UAd lmtd)
Shell = (Q = mcd lmtd)
Tube = (Q = mcd lmtd)

So, to complete the formula, we need to know the refrigerant entering and leaving temperature and SHC. From what you state, we assume that you are using water as the cooling fluid with an entering temp of 16 and a leaving temp of 6 (dt = 10K)

[The MG Pony]

Lets assume the average of the discharge of R-22 and of R-134a in a fridge/freezer aplication entering refrigerant and a goal of 30 out or better

Would the average discharge be round 60C?

[frank]

Oh, I forgot - we need the expected mass flow rates as well :)

[The MG Pony]

About 20.84 Kg/H (45.86 pounds/H)

[frank]

I assume that this is the cooling water? Great.

All we need now is the refrigerant side (shell?)

Then we can feed the info into the formula

[The MG Pony]

Oh no thats the refrigerant, sorry.

Water would be 818.3L/h (180 gallons(UK)/H)

[The MG Pony]

*Nickers" Any one? *Trots off*

[The MG Pony]

Help? How do I assemble the formula?

[mohamed khamis]

Help? How do I assemble the formula?

Hi MG Pony

I attached to you a file in which i made for you a rough estimation. Check it and if u need further discussion I'd like to be in ur service. I wish it could help

Best regards:)

[The MG Pony]

Thank you but I don't understand, The pex is holding the water, and inside the pex will be the copper tubing of 1/4. So do you mean I need 6 feet worth of copper tube in a pex jacket?

I'd realy enjoy learning how to assemble then compile the formula, as it is one I feel I'll be oft needing. Math has all ways been my weak point unless it is working on some thing physical, I'm very practical learning slanted.

Or do I need five 6foot tubes in a water jacket?

[frank]

Sorry MG, just read your recent replies and it's been a s**t day today so I'll feed your info into the formula over the weekend.

:)

[The MG Pony]

Ok thank you.

[mohamed khamis]

Thank you but I don't understand, The pex is holding the water, and inside the pex will be the copper tubing of 1/4. So do you mean I need 6 feet worth of copper tube in a pex jacket?

I'd realy enjoy learning how to assemble then compile the formula, as it is one I feel I'll be oft needing. Math has all ways been my weak point unless it is working on some thing physical, I'm very practical learning slanted.

Or do I need five 6foot tubes in a water jacket?

You Welcome, Yes five 6 foot copper tubes in a water Jaket... I attached to u new file to demonstrate further the calculations.

Cheers

[The MG Pony]

Ok I get it now, thanks, I was intending to wind the tube in a spiral to reduce the space that it would take up.

I apreciate all the work you have don, I'll have to re-read it a fiew times in the morning to under stand it better, but I sor tof get it.

[mohamed khamis]

Ok I get it now, thanks, I was intending to wind the tube in a spiral to reduce the space that it would take up.

I apreciate all the work you have don, I'll have to re-read it a fiew times in the morning to under stand it better, but I sor tof get it.

U welcome at any time

Cheers:)

[The MG Pony]

We never did finnish this, I would assum it works in the revers too!