How to calculate thermal exchange rate of water inside Pex piping.

[Equinefencer]

How can I figure out how to calculate the heat/cooling exchange rate of water inside pex tubing? I want to build a ground source cooling system simular to a chiller system but I want a closed loop system. My game plan is to pump ground water into a holding tank that is constantly flowing ground water into the tank and dumping back out. Inside I'd like to use pex tubing with the cooling water going to my shop inside pex tubing in a closed loop. I have a brand new air handler rated at 12 Ton for the building end. It's a bit overkill I know, but for $250 it's what I have to work with. I'm wondering how much pex tubing I need to use for my loops inside my holding tank to get at least 3-5 tons of cooling. My ground water is 60-62* F. My air temp inside the shop will be 80*F or better.This is my first post, so I'm hoping someone can lead me in the right direction without getting tossed. Thanks, Floyd.

 

[pa5tabear]

What is a "ton" of cooling?

I think I know how to calculate this, but you will need thermal conductivity of tubing and the thickness of the tubing

 

[Equinefencer]

PEX tubing thermal conductivity: 2.43 BTU in/hr-ft^2-F per DIN52612

3.2 (Btu in /h ft2 oF)

2.4 BTU/(hr)(ft2)(°F/in) with ASTM TEST METHOD C117, I'd figure it at the low side of 2.4 rating. I have the vapor barrier type in my concrete floor noe because of the iron in the heating system I use. The 2.4 rated is cheaper. It seems the Thermal Conductivity varies a bit depending on what brand for the searches I've done. I cannot post links as of yet, drop me an e-mail if you like and I'll send you some if this does not suffice alliedpex(DOT)com/LITERATURES/Material_Specification_Sheet_O2_EVOH_Barrier_PEX.pdf

 

[haruspex]

I think you're asking for how the total heat transfer depends on the length of tube etc.
Let the conductivity of the Pex be k, wall thickness w, internal radius r, fluid velocity v, fluid s.h. per unit volume s. Treating w as small compared with r, and pretending heat conducts very rapidly across the fluid stream (as in turbulent flow) the temperature difference between the fluid and the outside at distance x from the start is given by:
T = T₀exp(-2kx/vwsr)
where T₀ is the initial difference.
If w/r is not that small, a more accurate formula would be to replace w in the above formula with r ln(1+w/r)

 

[Equinefencer]

I want to figure out how much 3/4 inch Pex pipe/tubing I need to use to get 4-5 Tons of cooling.


Air conditioners are rated in three ways, by B.T.U.'s, Tons of Refrigeration, or by Horsepower. One Ton of Refrigeration removes the amount of heat needed to melt one ton of ice in 24 hours. One Ton of Refrigeration can remove 12,000 B.T.U.'s of heat in one hour. The B.T.U. is the amount of heat needed to raise 1 lb. of pure water 1 deg F.

 

[SteveThing]

Pardon my ignorance, but a quick look at wiki show PEX as being a thermoplastic flexible pipe (replacement for copper for carrying water). Are you using any type of heat exchanger in your loop? Or are you trying to use geothermal energy (you mentioned ground temps)? Or is the holding tank going to be the point of heat exchange?

I may be misunderstanding. The way I read your original post is that the PEX pipe is going to do all the cooling (i.e. replacement for a radiator). If this is the case, I'm intrigued. I'm going to keep reading up on this stuff. If you have time, can you spell it out for us non HVAC/plumbing types?

 

[haruspex]

Using my earlier notation,
k = 2.43 BTU in/hr-ft^2-F = 0.346 W/(m^oK)
Note that this is a property of the material. You still need to specify the internal and external diameters.
s = 4.2*10⁶ J/(m^3oK)
k/s = 0.8 * 10^-6 m²/s
T = T₀exp(-2(k/s)x/(v.ln(1+w/r)r²)
E.g. with v = 1 m/s, r = 0.01 m, w = 0.005 m, x = 100 m:
T/T₀ = 0.02
I.e. the temperature difference between inside and outside will drop to only 2% of the initial value.
Presumably you're interested in maximising the rate of extraction of heat. For that you want to look at (T₀-T)*v*r². However, this will increase beyond any limit as v increases. The problem is that this model assumes the ambient temperature is fixed. In practice, the stream can only supply heat at a certain rate. Without going into modelling the flow and temperature gradients in the stream, I suggest just calculating the available heat flow (stream rate in m³/sec * T₀ * s) and only attempting to extract say 10% of that.

 

[Equinefencer]

Pardon my ignorance, but a quick look at wiki show PEX as being a thermoplastic flexible pipe (replacement for copper for carrying water). Are you using any type of heat exchanger in your loop? Or are you trying to use geothermal energy (you mentioned ground temps)? Or is the holding tank going to be the point of heat exchange?

I may be misunderstanding. The way I read your original post is that the PEX pipe is going to do all the cooling (i.e. replacement for a radiator). If this is the case, I'm intrigued. I'm going to keep reading up on this stuff. If you have time, can you spell it out for us non HVAC/plumbing types?

The 3/4 Pex is going to be submerged in a tank of water. The ground source water is going to be3 constantly turned over and dumped. The water inside the 3/4 pex is going to be circulated through a cooling coil in my air handler and retured back through the 3/4 Pex to be cooled again in a closed loop. The water inside the cooling coil in the air handler will take the heat out of the air and transfer it to water inside the coil to return to the Pex in the holding tank. Then the heated water inside the 3/4 Pex inside the holding tank will transfer to the holding tank water and be lost through "new" cooled water that is turned over. That way I can control the Ph of the water so it does not eat the coiling coil with untreated ground water.

3/4 Pex 0.875 +/- .04, wall thickness .097 +/- .01, ID 0.677

 

[Equinefencer]

Using my earlier notation,
k = 2.43 BTU in/hr-ft^2-F = 0.346 W/(m^oK)
Note that this is a property of the material. You still need to specify the internal and external diameters.
s = 4.2*10⁶ J/(m^3oK)
k/s = 0.8 * 10^-6 m²/s
T = T₀exp(-2(k/s)x/(v.ln(1+w/r)r²)
E.g. with v = 1 m/s, r = 0.01 m, w = 0.005 m, x = 100 m:
T/T₀ = 0.02
I.e. the temperature difference between inside and outside will drop to only 2% of the initial value.
Presumably you're interested in maximising the rate of extraction of heat. For that you want to look at (T₀-T)*v*r². However, this will increase beyond any limit as v increases. The problem is that this model assumes the ambient temperature is fixed. In practice, the stream can only supply heat at a certain rate. Without going into modelling the flow and temperature gradients in the stream, I suggest just calculating the available heat flow (stream rate in m³/sec * T₀ * s) and only attempting to extract say 10% of that.

I just reread your post. This system will only be used when the ambient air temp is 80*F or higher. The hottest I've ever seen here is 106*; at worst it'll get to maybe 100* once in a while in the summer. It's never been over 101*F in my shop when I'm working. The ground water temp is always 60*-62*F. From what I've been reading I need at least a 20*F temp difference from my air temp to my water temp for it to work for cooling. Another benifit I've found out, is in the winter time, it will heat my shop if it's really cold out side.

 

[haruspex]

With T₀ = 20F, x = 50m (165') I calculate you get from 5 to 10 tons of cooling as the flow rate rises from 2 m/s to 10 m/s.
I can attach the spreadsheet so that you can play with the flow rate and tube length. But I would only be able to attach one with just a few rows (file size limitation of the forum), so you'd have to replicate the last row to get a nice table. Are you ok with spreadsheets and formulas?
For extracting warmth in winter, the snag might be if the groundwater is in danger of freezing around the coldest part of the pipe. That would seriously reduce the heat flow.

 

[Equinefencer]

If you would, e-mail it to me and I'll look at it. To be quite honest, I'm not the sharpest knife in the drawer. I seem to have a knack for figuring things out one way or the other. It's like the heating system for my shop. It's radiant heated concrete, one of the best ways to heat areas like mine. I asked around enough in the right places, I gained enough knowledge to figure out the cheapest way to do it and still have it work right and be efficient. My neighbor keeps telling me I should go back to school to be a mechanical engineer, she's a NCSU Engineering grad. She keeps seeing the way I do things and figure ways around obstacles. Her husband, ex-Marine and semi-retired doctor, calls me MacGyver, I've done so much with so little for so long, I can do almost anything with nothing. I bought a 3Ph Mill and lathe and figured out how to build a rotary phase convertor to run them. I ran up on a broken VFD and took it apart and fixed it, now I can change speeds on either the lathe or mill with the touch of a button. But math gives me a hard time. Once I get into the swing of doing it on a regular basis, I get good at it, until I stop doing it. I think that's why I've not gone back to school. I see something I need and then I build it. If I see someone that needs help, I help them. I'm a believer in what goes around comes around. Sorry for getting off topic, just giving you a little background in case I sound a little lame.

 

[haruspex]

I don't see a way to attach a file to the private message system provided by the forum, and I don't have your email address, so I'll post the spreadsheet here.
No need to apologise for your lack of academic background. I'm in awe of your practical skills.

 

[Equinefencer]

What do the below stand for as units from your spreadsheet? Is k/s Kelvin/seconds? How do I apply your spreadshet to find how many feet of Pex I need to get the required cooling? I expanded your sheet up to the 12T limit of my air handler.
v w r k/s x T/To To

 

[Equinefencer]

Ignore the last post, I just read what they are, sorry. I guess it's too late for me tonight.

 

[haruspex]

What do the below stand for as units from your spreadsheet? Is k/s Kelvin/seconds? How do I apply your spreadshet to find how many feet of Pex I need to get the required cooling? I expanded your sheet up to the 12T limit of my air handler.
v w r k/s x T/To To

These are the same as in my earlier posts:

conductivity of the Pex is k, wall thickness w, internal radius r, fluid velocity v, fluid s.h. per unit volume s
T₀ is the initial difference in temperature, T the difference after distance x.
k = 2.43 BTU in/(hr ft^{2 o}F) = 0.346 W/(m ^oK)
s = 4.2*10⁶ J/(m^{3 o}K)
So k/s = 0.8 * 10^-6 m²/s
Fill in the hose length as x in metres in cell E5, and a (low) flow rate in metres/second as v in cell A5. Subsequent rows keep all the same parameters except they steadily increase the flow rate.

 

[Equinefencer]

Thanks, I'll work on it a little while longer tonight. I have to get up in 5 hours to go to work.

 

[Coolthoughts]

Equine:
So whatever happened with your install? I'm considering something similar.
Upon review, I notice you seem to miss that you have 2 delta-Ts: stream to water tank (say 70F-60F) and water tank to air (say 80F-70F). Since heat transfer is related to deltaT, the heat transfer may be less than you think (10 vs 20F).
In addition, you have a need to stir the water tank to promote transfer.
All in all, you may do better to simply blow a fan past coils of PEX containing stream water to get a 20 degree delta T. It also only needs one circulation pump, or none if you dam the stream & siphon the water to exit lower.

 

[Chestermiller]

Are you saying that you are trying to cool the air in the shop by blowing the hotter air over the pex tubing that is carrying cooled water? Have you considered how much tubing you need in the shop to achieve the cooling you need?

Chet

 

[Equinefencer]

A lot has changed since I posted this. First off, the shop burned to the ground, Major set back. The up side to it, is I now have a newer all steel building better insulated to work in when I finish it. I also enlarged it slightly, it was 60Lx40Wx16T, Now it's 62Lx44Wx16T. The old building was post steel and I had Radiant heat in the flooring; concrete. Eveyone that wanted to put up a steel building wanted to to cut into the old pad for new piers for a steel buiding so we had it enginnered for a 2ft continuous footer around the old pad. I was able to salvage to old heating piping in the floor, it was undamaged inside the floor. While digging the new footer I found a useable 44ft deep well I can use for the cooling system. I recased the well and it's just inside the building wall, what luck huh?
Now back to my original plan. I have an idea to use ground water to cool my shop when the inside temp is above 80*F. Here's what I'm thinking of doing, Using a large water tank, if feasable, to make a closed loop system for the inside water. The inside water will closed loop from a cooling coil I have;it's rated at 12 ton, to the tank. The lines from the cooling coil will be submerged inside the tank with ground water around them. Ground water will be pumped into the tank for cooling the water that's running closed loop inside the 3/4 inch Pex pipe. The Red 3/4 Pex pipe will be the heat exchanger inside the cooling tank. I'll blow warm shop air over the coil inside the shop, the closed loop water will absorb the heat and then be pumped to the cooling tank where the Red 3/4 pex will transfer it to the ground water that's inside the water tank. The water tank will have water pumped into it and free flowing back out so that water stays cool. The water inside the Red 3/4 Pex will stay inside that loop and not mix with the cooling water inside the cooling tank. From what I've been reading, I only need 3-4 Tons of A/C to cool my shop and that was with the old building heat load calcs. What I'm trying to figure out is how much Pex pipe I need in a tank of 55*F-60*F cold water to get 3-4 Tons of cooling out of it? I cannot bury the closed loop lines on the property, we have horses and it takes forever to fill the trenches. She's the boss and I live by her rules as it's really all her stuff, I just do the work and play in the shop.