Equation for Temp of Water OUT of copper water jacket.

[~~~^^^larry^^^~~~]

the bottom line question: WHAT IS THE EQUATION FOR CALCULATING HEAT GAIN (not HEAT LOSS) FROM COPPER TUBING?

my application is a wood stove with 10 feet of 3/4" copper tubing coiled/looped inside an insulated panel on each side of the wood stove. the interest is to calculate the Tout based on expected inputs for the range for the Tair = 500 to 700 deg F, approximately, and the Tin = 50 to 200 deg F, approximately? (these are large ranges, and i can study a variety of ranges for Tair and Tin, so, this is not an inquiry about appropriate ranges for Tair and Tin).

if the resultant elevated Temp of the water exiting the water jacket (Tout) can be determined based on the range of expected air Temp around the copper pipe (Tair) and the expected Temp of the water as it enters the water jacket after being cooled during piping through the heat exchanger of the hot water storage tank (Tin), then the Universal Hydronics Formula can be used to estimate the BTUHs the water jackets are capable of generating.
Universal Hydronics Formula: GPM = BTUH ÷
T and multiplied by 500, or GPM = BTUH/(delta-t * 500).
rearranging the terms gives: BTUH = GPM * (delta-t * 500).

bottom line question: with a given GPM and an estimated range for Tin, is there an equation for solving to get Tout. it seems like there should be a mathematical solution for this. if so, the Universal Hydronics Formula should then give us BTUH from the water jackets.


it may help if i explain that i worked through an equation that i tracked down that works the other way (HEAT LOSS FROM COPPER TUBING), that is, for a baseboard heater, where the water in the copper pipe is HOTTER than the air which is being heated. the goal in the case under study is the other way around. the air temp inside the insulated shields on the sides of the wood stove where the 10' of copper pipe is enclosed is MUCH HOTTER than the water entering the cool end of the copper pipe. the water flowing through the copper pipes is picking up heat, not dumping it off as in the baseboard heater situation. i located a web-based tech article with an equation that works in the baseboard-heater direction, but as soon as the air temp surrounding the pipes inside the insulated panels is entered as HIGHER than the water temp when entering the water jacket, the spreadsheet results appear to blow up, mathematically.
all the best and thanks for your considerations,
~~~^^^larry^^^~~~

 

[stee6043]

The problem in solving for Tout instead of GPM based on a target Tout is that Tout should not exceed boiling in this type of application. You might as well plug in Tout=195 or 200 and work from there to keep the system operating with liquid water and not steam.

I guess I'd start a few steps back from the heat transfer calc you're trying to figure out. 10 feet of 3/4" copper pipe will hold roughly 0.23 gallons of water. Tin of 50 and Tout of 200 provides you with 150 degrees of change inside the coil. With utopian like levels of heat transfer efficiency you'll be adding 300 btu's to every .23 gallons of water you send through the coil if you're able to maintain 50 deg in and 200 deg out, ignoring any other part of the system and also ingnoring the question of whether or not you can actually transfer this amount of heat through the coil at a certain flow rate.

With this you may be able to ballpark some type of flow but realize that in a real world application this small of a volume of water is going to have a very hard time transfering useful amounts of heat.

 

[Bob Rohr]

The problem in solving for Tout instead of GPM based on a target Tout is that Tout should not exceed boiling in this type of application. You might as well plug in Tout=195 or 200 and work from there to keep the system operating with liquid water and not steam.

I guess I'd start a few steps back from the heat transfer calc you're trying to figure out. 10 feet of 3/4" copper pipe will hold roughly 0.23 gallons of water. Tin of 50 and Tout of 200 provides you with 150 degrees of change inside the coil. With utopian like levels of heat transfer efficiency you'll be adding 300 btu's to every .23 gallons of water you send through the coil if you're able to maintain 50 deg in and 200 deg out, ignoring any other part of the system and also ingnoring the question of whether or not you can actually transfer this amount of heat through the coil at a certain flow rate.

With this you may be able to ballpark some type of flow but realize that in a real world application this small of a volume of water is going to have a very hard time transfering useful amounts of heat.

Is the copper coil in contact with the stove jacket? if so you have conductive energy transfer, if it is in the space around the jacket, the transfer is radiant and convective, not as powerful. If the air has a means to move you have better convective heat transfer, like a forced convector, or a convection oven.

From what I understand the calculation is a bit complex as the fluid temperature changes from on end to the other, so you need to average the temperature. Also as the fluid temperature increases the heat transfer slows.

High ∆T will exchange more energy, as the fluid warms the transfer slows. This is why the last few degrees rise in a buffer tank is slower then the start up with cold water.

Probably the best method would be to connect a data logger to the device, log the temperature in and out and the flow rate to give you the answer, via the BTU formula.

Solar controllers have the ability to do this logging and calculation. They "watch" S&R temperature, you input the flow rate, and it spits out a number in KWh. It will even correct if you run the pump in a variable speed mode.

 

[~~~^^^larry^^^~~~]

Is the copper coil in contact with the stove jacket? if so you have conductive energy transfer, if it is in the space around the jacket, the transfer is radiant and convective, not as powerful. If the air has a means to move you have better convective heat transfer, like a forced convector, or a convection oven.

From what I understand the calculation is a bit complex as the fluid temperature changes from on end to the other, so you need to average the temperature. Also as the fluid temperature increases the heat transfer slows.

High ∆T will exchange more energy, as the fluid warms the transfer slows. This is why the last few degrees rise in a buffer tank is slower then the start up with cold water.

Probably the best method would be to connect a data logger to the device, log the temperature in and out and the flow rate to give you the answer, via the BTU formula.

Solar controllers have the ability to do this logging and calculation. They "watch" S&R temperature, you input the flow rate, and it spits out a number in KWh. It will even correct if you run the pump in a variable speed mode.

the 10' of 3/4" copper pipe is sandwiched between the exterior side of the wood stove and a thin metal shield that has its edges turned down. the distance between the exterior side of the wood stove and the interior surface of the thin metal shield is barely more than the thickness of the copper pipe, not more than 1/2" more than the pipe. the copper pipe touches neither the side of the wood stove nor the surface of the thin metal shield. there is a thin layer of insulation between the copper pipe and the thin metal shield, but no insulation between the copper pipe and the side of the wood stove.

plans are in place to monitor the btus being transferred to the water flowing through the water jackets, as suggested. when installed, there will be temp sensors on both the cold and hot sides. the current interest is to get an valid approximation of the btus that the water going through the water jackets will be carrying away from the wood stove, and thus, out of the sauna and into the hot water storage tank. it would be helpful for some of the design considerations if we could get a reasonable estimate for those btus from the water jackets. those copper pipes are receiving both radiant heat and convective heat. a mathematical solution would, i suspect, be quite involved. unless someone here or elsewhere has some kind of similar situation that they have data logged, that might be helpful.

as for Tin and Tout, that is a good suggestion Stee6043 to plug in Tout as 195 or 200 degrees, since we sure don't want to deal with anything that is boiling. and, we know the Tin should be something close to the desired max temp in the hot water storage tank, which we can for now think of as about 140 deg F. that gives us something to play with in terms of the Universal Hydronics Formula, but it does not help us understand the radiant and convective thermodyamics going on inside the water jackets.

~~~^^^larry^^^~~~

 

[maple1]

If I'm picturing it right, I don't think you're getting much convective heat as compared to radiant. Convective requires air movement, and between the insulation on one side of the pipe, and the air restriction that would be caused by the pipe being on top of itself (if I said that right - if you know what I mean), I'm not picturing a lot of convective air currents flowing around the surface of the pipe.

 

[~~~^^^larry^^^~~~]

If I'm picturing it right, I don't think you're getting much convective heat as compared to radiant. Convective requires air movement, and between the insulation on one side of the pipe, and the air restriction that would be caused by the pipe being on top of itself (if I said that right - if you know what I mean), I'm not picturing a lot of convective air currents flowing around the surface of the pipe.

maple1 - i think you are picturing the water jackets correctly. each side of the wood stove has a sheet metal panel bolted to it with spacers keeping it from squeezing the copper tubing against the stove, so there is some air space. but, three of the edges of the sheet metal panel are turned in towards the stove with only the vertical edge along the back of the stove open for the copper pipes to enter and exit this thin space. so, air flow is NOT specifically encouraged within these sheet metal panels and would in fact be somewhat minimal. as my concept of how these water jackets actually work is evolving, i am coming to the realization that the radiant portion of the heat transfer is driving the equation, so to speak.

one additional detail about the arrangement of the copper tubing within the water jacket: starting at the bottom of the sheet metal panel the copper runs from the rear to the front, then has a 180-U turn up a few inches, then returns to the rear of the water jacket, then turns up a few inches again and runs back to the front, turns up, runs back to the rear, turns up, etc., for about 6 lengths of the water jacket where it exits at the top. remember, the edges of the sheet metal are turned in, towards the exterior wall of the stove. thus, air movement within the sheet metal panel is minimal.

thanks, ~~~^^^larry^^^~~~

 

[Bob Rohr]

maple1 - i think you are picturing the water jackets correctly. each side of the wood stove has a sheet metal panel bolted to it with spacers keeping it from squeezing the copper tubing against the stove, so there is some air space. but, three of the edges of the sheet metal panel are turned in towards the stove with only the vertical edge along the back of the stove open for the copper pipes to enter and exit this thin space. so, air flow is NOT specifically encouraged within these sheet metal panels and would in fact be somewhat minimal. as my concept of how these water jackets actually work is evolving, i am coming to the realization that the radiant portion of the heat transfer is driving the equation, so to speak.

one additional detail about the arrangement of the copper tubing within the water jacket: starting at the bottom of the sheet metal panel the copper runs from the rear to the front, then has a 180-U turn up a few inches, then returns to the rear of the water jacket, then turns up a few inches again and runs back to the front, turns up, runs back to the rear, turns up, etc., for about 6 lengths of the water jacket where it exits at the top. remember, the edges of the sheet metal are turned in, towards the exterior wall of the stove. thus, air movement within the sheet metal panel is minimal.

thanks, ~~~^^^larry^^^~~~

critical to this type of HX is over-heat protection. In the event of a power outage or pump failure. I like to see a good quality relief valve at the HX. Some of the brands that were available years ago, had T&P type that protected temperature (210F) as well as pressure. Although 125 or 150 psi is quite high pressure if it is tied into hydronics.

Pipe the relief discharge to a drain, or a bucket if it is glycol, of course.

 

[BoiledOver]

Hi Larry:

Have you considered using sand within your side panels rather than air? I experimented with sand and copper coil on an old stove. In this case the sand and copper were on the upper section of the firebox which was built on an agle inwards at the top. Inside the friebox, down low, was a good amount of firebrick. This tended 80% of the heat load for one winter. The flaw in this partricular design was the stove was not airtigh and firebox temps were often above 1,000 degrees fahrenheit. The fatigue to the steel allowed the weight of the sand to severly droop the 3/16 steel into the firebox. Maybe vertical sides and stays would keep the sagging to a minimum. On the upside, the sand and firebrick was still releasing heat 4 hours after a burn was complete.