Thermodynamics for a geothermal heat pump

[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 anyone time.

So to make things conceptually simple (I think it does?) let's 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

 

[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 into 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.

 

[CWatters]

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

 

[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

 

[sardar]

Hello Steve,

Thank you for your inquiry on thermodynamics for a geothermal heat pump. It is great to see individuals taking an interest in experimental projects, however, it is important to have a thorough understanding of the principles and calculations involved in order to ensure a successful and safe installation.

Based on your description, you are looking to determine the heat transfer rate for a 100' length of 3/4" copper pipe, with water flowing at a rate of 1.887 gallons per minute, submerged in a larger body of water. To calculate the heat transfer rate, we will use the formula Q = m x Cp x ΔT, where Q is the heat transfer rate, m is the mass flow rate of water, Cp is the specific heat of water, and ΔT is the change in temperature.

First, we need to convert the flow rate of 1.887 gallons per minute to mass flow rate in kg/s. Using the conversion factor of 1 gallon = 3.78541 kg, the mass flow rate would be 1.887 gallons per minute x 3.78541 kg/gallon x 1 minute/60 seconds = 0.1189 kg/s.

Next, we need to determine the specific heat of water. At 40 degrees Fahrenheit, the specific heat of water is 1.005 kJ/kg·K, and at 50 degrees Fahrenheit, it is 1.007 kJ/kg·K. We will use the average of these values, which is 1.006 kJ/kg·K.

Now, we can plug in these values in the formula Q = m x Cp x ΔT. The mass flow rate of 0.1189 kg/s, the specific heat of water of 1.006 kJ/kg·K, and the temperature difference of 50 degrees Fahrenheit (10 degrees Celsius) will give us a heat transfer rate of 0.1189 kg/s x 1.006 kJ/kg·K x 10°C = 1.199 kJ/s.

To calculate the total temperature change in the pipe over 1 minute, we need to consider the heat capacity of the pipe as well. Assuming a density of 8.96 g/cm3 and a specific heat of 0.385 J/g·K for copper, the heat capacity of the pipe would be 0.035" x 100' x 12 in/ft x 2