I want to experiment with geothermal cooling, and have an r-10 insulated workshop which I'd like to cool during the worst of the summer heat this year. The workshop is about 400 square feet (37 m2), and 11' (3.3 m) ceilings, with a concrete floor.

To keep costs down, I'm not interested in investing in a heat pump. I'd like to try laying long horizontal loops of pipe in the nearby ground 4–6' down (1.2–1.8 m), run water through them, and then through a radiator in the shop I can blow air through.

The ground around here is the typical 55 °F (13 °C) average temperature, and in this area is mostly heavy dense clay. The water table will be many feet below the loops in the area I plan to lay them.

While I could do this experimentally, lay an arbitrary amount of pipe, and take measurements (inflow/outflow temp difference and flow rate), I'm wondering what equations and engineering knowledge I should be looking at to give me rough back-of-the-napkin numbers on how much pipe I'm looking at to produce a certain temperature differential in the shop.

  • $\begingroup$ Related: How to design an house to be cooled passively? $\endgroup$ May 12, 2015 at 12:53
  • 3
    $\begingroup$ I have a feeling this question may be a bit too broad for this site. I did something very similar (design of a ice rink cooling system) for the final project in a senior/graduate level thermodynamics class. The issue is that you can't say "this much pipe gets this much cooling," because you have to consider flow rate, which considers pumping power, which considers losses (i.e. how much pipe you lay.) I'm not sure there's any non-iterative way to solve this, and that definitely increases the complexity. $\endgroup$ May 12, 2015 at 13:49
  • $\begingroup$ @TrevorArchibald That's a very good point. $\endgroup$
    – Adam Davis
    May 12, 2015 at 13:53
  • 2
    $\begingroup$ @TrevorArchibald A method for developing a rough estimate doesn't need to be iterative or lengthy, especially given the context. $\endgroup$
    – Air
    May 12, 2015 at 16:22
  • $\begingroup$ There's several questions here. One, how much thermal power does the op need to get into the ground. Two, how does the ground conduct heat in these circumstances. Ground is a good insolator. $\endgroup$
    – mart
    May 12, 2015 at 19:54

1 Answer 1


The numerical answer depends a lot on the other numerical data that is missing.

  1. What sort of a summer is this? I mean, length of days (geolocation of the house), solar radiation vs time (is it 1000KW per square meter?)
  2. Are there already heat isolators in place? Is the building covered by a roof and is there air or space in between the ceiling and the roof-tiles? or is the roof cover just touching the internal air?
  3. how much power could you invest into the pump? If it is infinite, than with enough mass flow rate, you could use a simple 2 meter pipe, and as long as you can heat the water in the pipes, you would be giving away the heat to the ground. Like trying to heat the Earth.
  4. how much heat conduction is there, between the pipe and the soil? This could be best tested.
  5. how much heat can you subtract from the radiator? You cannot heat the ground more than you can cool the air. So a generous radiator is a must (which is probably simpler than piping).

As the numbers would follow the above clarifications, experimentation is the only valid method for determining the required piping and the radiator.

Another point is, would you be trying to cool the workshop as a whole, or is it only the working point (one person standing) that you would like to cool. In the second case, using a spray-ventilator could work as well (if it is safe and no side-effects for the type of work you are doing). For the first case, cooling as a whole, isolating the sunrays from the building is the number 1 approach to take. A dark coated surface can rise to 80degrees Celcius under the sun, which essentially becomes a wonderful heater, and if you can put shadow-panels over your building, you would get rid of the sun's radiation and only deal with the hot air.

One last point: why not employ some solar panels, and use their power to power your cooling solution?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.