Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm working on low budget small solar power project.

I want to transport heat (peak heat power about 1kW) over ~10 meter (~30ft) long pipe or hose.

I was thinking about thin hose (6-8mm / about 1/4 inch inner diameter ) to keep small surface area and make water flow faster (and loose less energy) - im right?

Im thinking about some kind of synthetic, transparent hose. I dont know how this material is called in english, it's similar to fuel hoses used in old motocycles.

I want to cover pipe/hose with some kind of thermal insulation (maybe mineral wool) and maybe aluminium foil to keep/reflect IR radiation inside - it's good idea?

1. I need help to decide what pipe/hose diameter shall I use.

2. Any suggestions about "im right" and "it's good idea" questions above?

I don't need exact formulas, just want to know what is proportional to what etc.

share|cite|improve this question
From thermal loss, less surface seems to be the right thing to do. But note that the pressure needed to pump the water through smaller pipes is higher (Hagen-Poisseule law) – Hagen von Eitzen Jan 6 '13 at 22:21
Rock wool (asbestos) is an environmental hazard. You'd need a really good reason to use it over, say, fiberglass (less of a hazard) or closed-cell foam pipe insulation (no hazard). – Dave Tweed Jan 6 '13 at 23:58
@DaveTweed sorry for my bad english. I mean mineral wool used on large scale to insulate buildings etc. Its not asbestos. – Kamil Jan 7 '13 at 3:13

Don't take this as authoritative, but my back-of-the-envelope calculation suggests that the cube-square law applies in this case, meaning that a bigger pipe is better in terms of heat loss than a smaller pipe.

Consider a cylindrical volume of water that only loses heat through the sides of the cylinder, not through the two faces. The rate of heat loss will be proportional to the fraction of the total area that is the sides:

$$rate = \frac{1}{1 + \frac{{\pi}r^3}{Vol}}$$

which basically varies with 1/r3.

That volume of water will be subject to heat loss for the amount of time that it is in the pipe. Assuming you're pumping a constant volume per time, the velocity will vary with 1/r2 and the time duration of the heat loss will be proportional to r2.

Let's assume the rate of heat loss is constant along the pipe (i.e., it doesn't depend on the fact that the water temperature, and therefore the temperature differential, varies), so the total heat loss will be just rate×duration

This means that the total heat lost for a given volume of water will be proportional to 1/r.

If I've gone off the rails at any point in this, I'd appreciate it if someone would point out my error.

share|cite|improve this answer
Thanks Dave, it looks very logical and makes sense for me, but ill wait, maybe someone write another opinion, or some other useful info. – Kamil Jan 7 '13 at 3:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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