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BTU condutivity of copper pipe
Oregonbythesea
Member Posts: 2
I have a heat exchanger and need to know the capacity. There are 24 runs of 5/8" copper refrigerant tubing in a 6" barrel. The tube length is 10 feet. The water flow is 100 GPM over the tubes and the temperature is 60F. The gas inside the tubes is 0 F. How many BTU's per hour can this heat exchanger absorb?
0
Comments

Need the flow rate
of the gas and what the specific heat of the gas is."If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0 
System Syzer Calculator, B&G
Let's take a look at your water: That seems to be a fact from you.
[email protected] gpm/ton= 460K/hr or 38.33 ton with a 9.5 "Delta T" This will work, but not very efficient.
Lower your gpm to [email protected]/ton= 500K at a 20* "Delta T" accross the barrel with out changing anything except flow. Aprox tonnage= 45.83 ton/hr.
Since I am not sure what you are doing with this application, the Above tool from B&G costs about $25.00 and lets you solve gpm, volume, speed of fluid, head,pressure drop, noise senerios etc...
With a chiller barrel at 60*f and diff at 0*f, technically you could acheive 60 ton since 1 ton at 2.5gpm=12,000@ 10*f "Delta T". but your water would move so slow it would be non usable.
As far as your rated capacity, it should be stamped somewhere on the barrel or compressor. Since you stated 100 GPM,...right there it's telling me that its at or above 35 ton.
Side note, Eugene S will probably chime in and there are alot of questions before an accurate representation can be figured.
I hope this was of some kind of help for you.......
Peace:
Mike T.0 
Almost sounds like an exam question,
but first a few disclaimers :
1. I'm not a mechanical engineer,
2. I only had one course in thermodynamics (it was in 1985, and I got a C) . . .
so I took a more basic approach to the problem. I used a linear heat transfer equation, computed a few volumes and surface areas, looked up some constants, and made a few (hopefully not too incorrect) assumptions. I was hoping to come up with a reasonable ballpark answer, not a completely analytical solution.
I started by computing the surface area that the heat flows across. This is found by multiplying the tubing diameter (5/8 = .625") x pi x the tube length (120") x the number of tubes (24) which equals 5654.9 square inches, which, when divided by 144 equals 39.3 square ft.
I then computed the volume of water the heat exchanger holds. This was found by first computing the volume of the six inch barrel (3" x 3" x pi x 120") = 3392.9 cubic inches and subtracting the volume of the tubes (.3125" x .3125" x 120" x 24 tubes) = 883.6 cubic inches leaving a water volume of 2509.3 cubic inches, or 1.45 cubic ft. or 10.85 gallons.
Next, I researched rates of heat transfer, and found equation 2.1 in "Modern Hydronic Heating" by John Siegenthaler. This equation states that the rate of heat transfer via conduction through a material (Btu/hr) is equal to the area of heat transfer (square ft.) x the temperature differential (delta T, degrees F) x the ratio of the thermal conductivity of the material (k, Btu/degree F hr ft) to the thickness of the material (delta X, ft.).
The disadvantage of this equation is that it is designed for a steady state condition, such as computing heat loss of a fixed size wall, of a given R value, with constant indoor and outdoor temperatures. This is not what happens in the heat exchanger. As the refrigerant and water travel through it, the temperature of the two fluids is constantly changing as heat is transferred, which causes the delta T part of the equation to become dependent on other variables, which makes it much harder (impossible) for me to solve.
The advantave of this equation is that I can solve it as is, so I went ahead and pretended that the fluids were stationary, just to get started. I looked up the wall thickness of 5/8 in. tubing (.040" or .00333 ft.) and the thermal conductivity of copper (231.2 Btu/ft hr degree F). Using the surface area of 39.3 square ft. and a delta T of 60 degrees F, I calculated an exchange rate of 163.7 million Btu/hr. This result is wrong, of course, because the fluids are not just sitting around waiting for every last bit of energy to be transferred, they are constantly in motion. Their temperature also insn’t constant, and we don’t know (as Brad White said) anything about the capacity of the refrigerant to supply energy. The result also ignores the phase change that would happen to the water at 32 degrees F.
So I tried to account for the fact that the fluids are in motion my calculating how long the water takes to pass through the exchanger. I assumed (probably incorrectly) that the heat exchanger capacity can be more or less linearly downrated by multiplying the static capacity by the fraction of time that the water spends in the exchanger. I took the 10.85 gallon volume and divided by the 6000 gallon per hour flow and found out that there is an exchange of water every .00181 hours. When the 163.7 million Btu/hr static fluid result is multiplied by .00181, I came up with 296,300 Btu/hr.
Is that anywhere close to what a heat exchanger of this size should be rated at, or should I have gone to bed hours ago ?Chris Bors
Land Mobile Corporation0 
Darn fine job. Chris....Really!
And you said you only got a "C"? I say, "Ringer!"
The thing about waterbased heat exchangers which include a change of state (presuming in this case it sounds like a water chiller evaporator, cooling water in the shell) is that it is nonlinear at the ends. (The transfer rate curve is either a slope or in some cases a bell curve).
Also the turbulence factors (Reynolds Number) plays into this. In a shell and tube setup with water, there are baffles to force the tube water back and forth across the tubes to increase the "wiping action" and avoid stagnation of the chilled media. (Imagine a tube bundle in a size you could carry inserted in a shell the size of the Astrodome. How do you convince the larger volume to interact efficiently?)
The way I would have approached this is similar to what you would have done, but frankly I would have used a selection program to find something comparable (but have not). I also might have taken the leap that the refrigerant was held at zero degrees in the tubes by whatever was necessary. As we have been saying, you need all of the variables to find one operating point. But I think you got close, probably within 510%.
By your numbers, the selection is consistent with a 25ton evaporator and would take that 60 degree water down by about 6 degrees F. to 54 F. Not a real "chiller" range, (too warm for dehumidification), but an able cooling jacket temperature for some processes.
I would disclaim the result for actual application though. We have no idea what the duty is and these bits of advice are informational, not applicationspecific in this case.
I would give you an A."If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0 
An "A" for sure
Thank you for the great work on this problem. It all started out as a way
to prove that a copper tubed chiller would be way more efficient than one
constructed of titanium or stainless steel tubes. The chiller is supposed to be
a nominal 10 tons and I think you have proven that quite well. You are right
about a number of things concerning the operation of the chiller.
The refrigerant gas starts out about 45 degrees and eventually gets down to 10,
depending on how much brine (salt water with salt added) there is to chill. The
water starts at about 60 degrees and will be held at 0. Thanks again for the
figures, it supports our guess.0 
Before you go away,
Why are you checking out a copper chiller barrel w/ salt water ?What is the assumed product?0 
I was thinking
the same thing Cupronickel or Admiralty Metal would be far better for salt water. Also the barrel and baffles would want to be SS, probably 316L."If you do not know the answer, say, "I do not know the answer", and you will be correct!"
Ernie White, my Dad0
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