# Thermodynamics for a geothermal heat pump

1. Aug 2, 2012

### sday

I'm going a bit off the reservation to try something experimental when I attempt my own HVAC installation and I'm having some trouble determining a few things. I want to transfer heat from water flowing in a 3/4" copper tube that is submersed in a larger body of water.

The copper pipe has a wall thickness of .035" so that makes an inside diameter of .68". I'd like to determine how much heat I can transfer in a 100' length of pipe for every gallon per minute flow. Being that I know just enough to be dangerous, I'm not comfortable with many of the various units, so don't laugh at my selection of units. :) My calculations show 1.887 gallons of water are in that pipe at any one time.

So to make things conceptually simple (I think it does?) lets assume the water is moving at 1.887 gallons per minute. I would think from there you could just calculate it as if the pipe was a long cylinder with no flow rate Ignoring friction heat, etc. although I would assume as the heat transfers you would start to get turbulence inside the pipe and in the tank as convection begins moving the water around. In the tank, if the pipe was suspended in the middle (not laying on the floor of the tank), the cooler water would sink being replaced with the more plentiful warmer water. In the pipe the top would be warmer than the bottom, but with flowing water I would suspect more turbulence and a better temperature distribution within the pipe? I would like to calculate the total temperature change in the pipe over 1 minute.

Water in pipe = 40 degrees F
Water in larger tank = 50 degrees F

Water in pipe after one minute = ? degrees F (and I would actually like to know how to do it myself. :) Like they always told me in school... show your work. lol I want to come away from this a little smarter than going in.

Any suggestion, corrections, etc. are very welcome. College physics was 20 years ago and to my detriment I had very little interest in any class let alone physics. :(

Thanks in advance for any input...
-Steve

2. Aug 4, 2012

### CWatters

Sorry but I work in metric/SI units..

Power = Thermal conductivity x Temperature gradient across pipe wall

The thermal conductivity depends on the bulk material, it's surface area and thickness. Look up the thermal conductivity of copper which will be specified for a unit thickness and area (eg a 1m cube). Multiply by the area of pipe and divide by the thickness. Plug it in to the above.

However note that this applies to the average temperature of water in the pipe. So if it goes in at 40 and comes out at say 45 use an average of 42.5. Then the temperature gradient across the pipe wall is an average of 50-42.5 = 7.5 degrees

If all the temperatures are constant/stable then the power going into the tank equals the power coming out. You can calculate the power coming out using the temperature gain (45-42.5=2.5), the flow rate in Kg/S and the specific heat capacity of water.

Power out = SHC x flow rate x temperature gain.

If the pipe work can scale up that can seriously effect the thermal properties.

3. Aug 4, 2012

### CWatters

You might also be able to find an online calculator that's intended for tank heater coil design.

4. Aug 7, 2012

### sday

Thanks for the feedback. I'm pretty sure I have a handle on the formula now. I broke it down to square feet of pipe surface. BTU/hr=(A(Th-Tc))/(R * thickness in inches)

Hopefully I'm doing it right, it mirrored an example in my old college physics book with two water tanks separated by a 4" concrete barrier. I replaced the concrete with copper and the .035 inch pipe wall thickness and the K value of 401. The R value appears to be .14/k and A = 1.

I'm not clear on the pipe gradient. why do you use an average? Is it because of the convective flow that builds up and starts moving around the fluid in the pipe as the temperature changes? ...and the average happens to be a close approximation of all the secondary effects that occur?

Thanks
-Steve