Copper Coil Delta T Formula?

[afblue]

I have searched here and on the engineers tool box, and I have yet to find a calculation give me a solid answer on how long I need to make my DHW coil in my storage tank to reliably get useable hot water out on my tank. I have concluded that 3/4" copper @ 4ft/sec flow is 6.5GPM. So what I am looking for is the formula that shows me how long a coil of copper I need. I have used nofossil's spread sheet which has given me a idea, but I really am looking for the formula that shows me a solid answer, then I can TLAR from there.

 

[afblue]

Someone has to have a formula......
I know Copper water to water is 60-80BTU/f^2 hr F. But what does that mean?

I looking to heat water from 50F to 120F at a design flow of 4-5GPM with a storage tank of best case 185F and worse case 140F.

I know that 3/4 coil will support the water flowing at less than 4ft/sec, but i need to know how long the coil needs to be.

 

[DaveBP]

Any formula you might find for that sort of thing is going to have many variables.

That heat transfer formula you quote means that for every square foot of surface area, 60 to 80 BTUs will move for every degree Fahrenheit in temperature difference between inside the pipe and the water around it every hour.

So you need to figure how many square feet of area you have for each foot of your pipe size, what the temp difference is (you'll hear this referred to as delta T around here a lot)
and you'll be on your way to getting an answer.

Of course, in the real world there are a lot more variables going on but that would get you some idea of how many BTUs you'll exchange under your chosen conditions. Notice that there is a pretty hefty fudge factor even in this simplified formula (60-80).

Hope that helps.

 

[Chris Hoskin]

DaveBP sounds like he is getting you on the right track as usual. I would just add that your design parameters are pretty demanding and will likely result in requiring a huge heat exchanger. I think more realistic would be 50 degrees to 110 degrees rise (60 degree delta T) at a flow rate of 2.5 gallons per minute (one shower running) and a tank temp of 140 degrees.

let us know what you come up with! Chris

 

[afblue]

DaveBP sounds like he is getting you on the right track as usual. I would just add that your design parameters are pretty demanding and will likely result in requiring a huge heat exchanger. I think more realistic would be 50 degrees to 110 degrees rise (60 degree delta T) at a flow rate of 2.5 gallons per minute (one shower running) and a tank temp of 140 degrees.

let us know what you come up with! Chris

Ok guys I think I am getting on the right track here, I did some math while I was burning holes over the skies of Afghanistan today. Applying this formula in theory so I could have a more educated TLAR (that looks about right).

So this I why I was thinking I was crazy, becuase my #s seemed out of whack like I had the wrong formula.

raising water 50-120F with 180F water the average delta T is 95.

5GPM=42lbs water X 70deg=2940BTU/min x60min=176400BTU/hr
176K BTU / 70BTU / 95 delta T = 26.4sq ft Copper.

3/4" copper is .196sqft/liner foot. 26.4/.196=135 feet of 3/4" copper

So now using 2.5 GPM 50f-110f with 180f tank=11.3sqft =58ft of 3/4"
Worse case 2.5 Gpm 50-110f with 140F tank=92ft of 3/4"

This is all based on instantanious flow.

So if I was to account for a static water heated to tank temp when a faucet is first turned on and mixed down to 110F.

If I was to use 100ft of 3/4 tubing, thats 1.23cu ft X 7.48gal/cu ft=9.2gal@180F mixing with 50F water to 110F output=21gals of stored hot water.

Something tells me sinking a 30gal pressurized stainless tank in my thermal storage tank and letting it heat up in between hot water uses is going to give me a better outcome, vs trying to heat it as it comes rushing by in a copper tube.
This makes sense why Varmebaronen Aqualux CU tanks have a DHW tank inside. http://smokelessheat.com/Media/FurnaceDocuments/Varm Aqualux/AqualuxBrochure.pdf

 

[Bad Wolf]

Creativity is fine but plagiarism is faster.

www.stsscoinc.com

Based on what STSS does I made two 150' coils out of 3/4 tubing for a 1200 gallon tank. I'm able to heat it up to the upper 170's. Even with the TARM Excel 2000 firing flat out I get return water equal to the tank temp at the bottom. That's telling me that its able to shed all of its heat to the tank and reach equilibrium. This is using a grunfos 15-58 pump set at high.
The TARM is rated at 102,000 btu/hr so at 80% efficient I should get 80,000 BTU/hr. That works about right, if the house is not calling for heat I get about an 8 degree rise per hour on a 10,000 lb tank.
For the DHW I used one 150' coil and use a tempering valve to get a constant 115 degrees, so I heat 58 degree water up to what ever the tank is at then temper it back down to 115.
I have never run out of hot water even when the tank is down to 125 degrees.

 

[Lukas060606]

What about those furnaces that have DHW off a coil in the water jacket (like mine). I know they run out of hot water if you use the hot at more than one location, but I could take a three hour shower if I wanted to and there'd be no temperature drop. There can't be anywhere near 150' of copper coil in there. I'd be surprised if that coil was more than 20'. I guess I would think if you had a coil three or four times longer than the one in a furnace like mine, you'd never run out of hot water. Maybe I'm missing something.

 

[PassionForFire&Water]

What about those furnaces that have DHW off a coil in the water jacket (like mine). I know they run out of hot water if you use the hot at more than one location, but I could take a three hour shower if I wanted to and there'd be no temperature drop. There can't be anywhere near 150' of copper coil in there. I'd be surprised if that coil was more than 20'. I guess I would think if you had a coil three or four times longer than the one in a furnace like mine, you'd never run out of hot water. Maybe I'm missing something.

These have cupro-nickel finned coils in them.
Better conductivity and much much much more square feet because of the fins per unit length.

 

[mole]

I'd be interested to see a formula that calculates pipe length as well. But I suspect that the calculation is actually pretty complicated, since the Delta T between the inside of the pipe and the tank changes each foot as the temperature in the pipe rises. Then there's other factors' such as wall thickness, velocity, turbulence, etc that play into it. I think the easiest path, like Greg H says, is to see the STSS site or you could poll some of the people here to see what they have and then compare with flow/ volume requirements. There's certainly enough people here who have built 'em!

 

[afblue]

I'd be interested to see a formula that calculates pipe length as well. But I suspect that the calculation is actually pretty complicated, since the Delta T between the inside of the pipe and the tank changes each foot as the temperature in the pipe rises. Then there's other factors' such as wall thickness, velocity, turbulence, etc that play into it. I think the easiest path, like Greg H says, is to see the STSS site or you could poll some of the people here to see what they have and then compare with flow/ volume requirements. There's certainly enough people here who have built 'em!

After looking through the STSS website, it supports all the math that I have done. There is a certain amount of variables such as what you have listed above, but outside factors I would think only fudge the theoretical #s by less than +-5%.

It all starts with the flow in GPM that you want to move, that determines the pipe diameter, or how many of a certain diameter you want to use. The key is using a diameter or multiple coils to meet your flow requirements, and maintain the optimal heat transfer of copper which is 2-4ft/sec. which translates to 3-6gpm for 3/4 NOM copper. Nom means the inside diameter is 3/4". (refridgeration copper is OD)

So for domestic coils you want to support the usually at least 2.5-3gpm. Where 1/2" copper would support this, but it would be on the higher end of the velocity, or head pressure on the hot side.

So all of STSS's coils support that copper will transfer 60 BTU per hour per every sqft of copper, per degree of degree of delta T of water change.

So using one of there coils part no. 425 120ft of 3/4 NOM type L soft copper (My #s will be in bold)

Temperature rise from 47 Deg. F. to 105 Deg F delta T=58F. @ 2 GPM. 2GPM X 60 gal /min X 8.3 X delta t 58=57,768BTU/hr pretty close to below

Average tank temperature 120 Deg. F. The way to calculate delta T use the difference between the tank temp to average DHW temp you want. In this case 120-average DHW temp of 76=44F delta T

Transfer rate 57,976 BTU’s per hour. 57,976 BTUs/60BTU/hr divided by 44F= 21.96 sqft of copper. 3/4" is .185 sqft per linear ft divide them= 118.4ft of copper pretty darn close if I say so myself


So let me see if can simplify the equation to length of ft. This is going to be for 3/4 NOM copper. Example: 2.5gpm of DHW from 50-110F with a 140deg tank

1) design GPM x 8.3lbs per gal x delta T you are trying to achieve in DHW = BTU/min you are trying to heat. 2.5gpm x 8.3 x 60F= 1245 BTU/min

2) find average DHW temp and subtract from tank temp= design delta T average of 50 to 110 is 80F. 140F-80F= Design Delta T of 60F

3) take design BTU/min and divide by design delta T = sq ft of copper in the coil. 1245 BTU/min divided by 60= 20.75sqft

4) .185sqft/linear ft of 3/4" copper. 20.75 sqft divided by .185 = 112.2linear ft of copper

5) from there I would add 5-10% for the wife kids, and Fluffy the dog so that gives me 117-123ft of 3/4" copper


6) now what you can also count on is if the water sits in the coil for a long period of time its going to equalize with the tank and you will have a volume of that will be 140F that needs to be mixed down to 110F. I will spare you the math, but it equates to roughly 2 minutes of "free heat" at 2.5gpm before you have to rely on the heat transfer of the copper to heat to instantaneously heat your hot water.

I hope this helps someone along the way.......or I really like to hear myself talk?

 

[DaveBP]

I don't know if you are single, AF, but there will come a time (perhaps quite often in the middle of coldest winter) when after a long, cold night the heat in the tanks is low first thing in the morning. And your housemate will go to take a shower before rushing off to work and the water coming out of the shower nozzle will be heated by 117F tank water, not 140F.

Do the math on that.

 

[afblue]

I don't know if you are single, AF, but there will come a time (perhaps quite often in the middle of coldest winter) when after a long, cold night the heat in the tanks is low first thing in the morning. And your housemate will go to take a shower before rushing off to work and the water coming out of the shower nozzle will be heated by 117F tank water, not 140F.

Do the math on that.

I understand you completely, You start working to where the tank heat is close to the DHW temp, you start talking about a coil thats 200+ft long, or the time when the tank is lower than the desired DHW then you are completely SOL. I wouldnt dare try to run my house solely on this DHW coil, there needs to be a back up to just the coil. Something like an instant gas or electric hot water heater. Doing the math and to find BTU requirements vs the price of copper coil, and adding a backup heating source, I start seeing the $$$$ add to the point where I should be considering an dual coil indirect water tank with an electric back up. By the time I add a instant hot water heater, and a solar coil, I am at the point a indirect water heater starts to sound alot more reasonable heating my hot water.
I am very much married and my wife gets up before me for work, so getting up at 3am to charge the thermal storage tank for her to have hot water, and yes she does like it hot, then I need to design a hot water system that will support that. I know how quickly a cold shower, can ruin anyone's day, and I will sure hear it if I ruin hers!
To add more to this, when I am deployed with the military, like you said on the coldest day of the year, and she has to be at work early, there is no way she is going to go into the basement load the boiler, and wait for hot water.

Its like our solar electric system, we are still tied to the grid as our backup, we arent that hardcore that we cant watch TV because the sun didnt shine that day.. haha

 

[mole]

[/quote]

After looking through the STSS website, it supports all the math that I have done. There is a certain amount of variables such as what you have listed above, but outside factors I would think only fudge the theoretical #s by less than +-5%.

It all starts with the flow in GPM that you want to move, that determines the pipe diameter, or how many of a certain diameter you want to use. The key is using a diameter or multiple coils to meet your flow requirements, and maintain the optimal heat transfer of copper which is 2-4ft/sec. which translates to 3-6gpm for 3/4 NOM copper. Nom means the inside diameter is 3/4". (refridgeration copper is OD)

So for domestic coils you want to support the usually at least 2.5-3gpm. Where 1/2" copper would support this, but it would be on the higher end of the velocity, or head pressure on the hot side.

So all of STSS's coils support that copper will transfer 60 BTU per hour per every sqft of copper, per degree of degree of delta T of water change.

So using one of there coils part no. 425 120ft of 3/4 NOM type L soft copper (My #s will be in bold)

Temperature rise from 47 Deg. F. to 105 Deg F delta T=58F. @ 2 GPM. 2GPM X 60 gal /min X 8.3 X delta t 58=57,768BTU/hr pretty close to below

Average tank temperature 120 Deg. F. The way to calculate delta T use the difference between the tank temp to average DHW temp you want. In this case 120-average DHW temp of 76=44F delta T

Transfer rate 57,976 BTU’s per hour. 57,976 BTUs/60BTU/hr divided by 44F= 21.96 sqft of copper. 3/4" is .185 sqft per linear ft divide them= 118.4ft of copper pretty darn close if I say so myself


So let me see if can simplify the equation to length of ft. This is going to be for 3/4 NOM copper. Example: 2.5gpm of DHW from 50-110F with a 140deg tank

1) design GPM x 8.3lbs per gal x delta T you are trying to achieve in DHW = BTU/min you are trying to heat. 2.5gpm x 8.3 x 60F= 1245 BTU/min

2) find average DHW temp and subtract from tank temp= design delta T average of 50 to 110 is 80F. 140F-80F= Design Delta T of 60F

3) take design BTU/min and divide by design delta T = sq ft of copper in the coil. 1245 BTU/min divided by 60= 20.75sqft

4) .185sqft/linear ft of 3/4" copper. 20.75 sqft divided by .185 = 112.2linear ft of copper

5) from there I would add 5-10% for the wife kids, and Fluffy the dog so that gives me 117-123ft of 3/4" copper


6) now what you can also count on is if the water sits in the coil for a long period of time its going to equalize with the tank and you will have a volume of that will be 140F that needs to be mixed down to 110F. I will spare you the math, but it equates to roughly 2 minutes of "free heat" at 2.5gpm before you have to rely on the heat transfer of the copper to heat to instantaneously heat your hot water.

I hope this helps someone along the way.......or I really like to hear myself talk?[/quote]

I'll agree with the end conclusion of 120ft. The math appears to works in this particular case , but I think it may be overly simplified for other cases. You're averaging the low and the high temperature of the incoming water (50) and the desired outgoing temp (110) to get an average deltaT over the pipe (50+ 110)/2 = 80. Using this average dT may work fine when when the temperature difference between the tank and the desired DHW temp is fairly large, like it is in this example, but if you're trying to figure out a system where you want the outcoming DHW to be fairly close to the tank temp, I think the error in averaging the delta T becomes significant.

For instance, for your last example above, if you had a desired DHW temp of 120F and wanted to be able to operate your tank down close to 120, (for your solar heat in the summer, your weekly summer fire, or the cold tank morning Dave referes to), then you'll get a calculated length of 225ft. In acutality you would probably need a lot more because the heat flow rate becomes asymptotic approaching zero as the coil temperature approaches the tank temperature. The error of averaging the temperature becomes even more obvious if you want a DHW temp of 120 and you want to run the tank down to 115F. Common sense says it aint never gonna happen, but the calcualtion says...262ft.

Please don't read me wrong. I'm not trying to be righteous or argumentative. I'm just searching for the answer. I had some closer desired approach temperatures on my project. I ended up guestimating and everything worked out fine. But it still irks me that I don't know how to calculate it!! And we want to make sure that Fluffy is happy.

JR

 

[Bad Wolf]

Going with real life experience, as I said earlier I've never run out of hot water with a 150' coil. There have been days in the summer when I've had the tank down to 124 at the top so I'm sure the bottom was even colder. A little extra doesn't hurt where as a little under means cold water.
This summer should be interesting; I have another 150' coil that I'm going to tie into 6 solar panels. My calculations say I should get around 90k btu's/day and my load should be 45k btu's. With standby loses and cloudy days I hope to hold even. I'm going to plumb an electric water heater inline so I can run either wholly from the storage tank, or wholly from the heater or use the tank as a pre-heater for the electric.

The electric water heater I got for free on craigs list. Even came with a timer.

 

[afblue]

I am starting to think a better heat transfer to DHW is to have a separate external hot water tank. I must be missing something but why is a solar or indirect hot water heater so expensive? they are $800+ where a standard electric hot water tank is as cheap as $220? I am wondering if there is some way to coil a copper tube inside of a standard tank? I am thinking of you take the pressure relief valve and feed the end of a 3/8 or 1/2" copper coil in until you could fish it out the drain port, then could use bring the end of the coil out through a reducer fitting like on a side arm?
This would allow the use of a far smaller coil of copper to heat the water, and a reliable electric water backup.

 

[DaveBP]

Have you considered an ordinary hot water heater as you seem to be coming around to, but heated via a sidearm heat exchanger?

This puts everything outside where you can get at it and doesn't void any warranty you have on the water heater. It can even thermosiphon if you want to avoid electric pumps.

Searching the archives here will keep you busy awhile.

 

[afblue]

Have you considered an ordinary hot water heater as you seem to be coming around to, but heated via a sidearm heat exchanger?

This puts everything outside where you can get at it and doesn't void any warranty you have on the water heater. It can even thermosiphon if you want to avoid electric pumps.

Searching the archives here will keep you busy awhile.

I will search the archives about sidearms some more but I was thinking that it wouldnt provide the nesesary amount of surface area to properly heat the DHW with a lower storage tank temp. I was thinking it would need to be a high delta for them to work well?

 

[DaveBP]

I will search the archives about sidearms some more but I was thinking that it wouldnt provide the nesesary amount of surface area to properly heat the DHW with a lower storage tank temp. I was thinking it would need to be a high delta for them to work well?

Good question and I don't know much about sidearms. After browsing past threads here you could use a new thread to get a response from people who are actually experienced with them and could answer your further detailed questions.