None! I would lift those pipes up onto the warm side of the insulation.

Insulation will not stop hear loss only reduce and if you need a heating system all the heat that it generates needs to go into your house not escape.

Posted By: Peter_in_Hungarywhat I did whith my heating pipework unavoidably outside the heated envelope was to put 2 layers of insulation on. I found that by choosing the sizes it was possible to get another layer on the pipes.

I'm just working on that. A thinner layer to start, followed by a thicker layer that better covers the compression fittings is the aim.

However, the current pipe insulation sizes do not easily lend themselves to this approach :(

Also, where do you hit the law of diminishing returns? What thickness is the sweetspot?

how far below the joists are the pipes? I think the best performance for suspended timber floors is with insulation between the joists and under them if poss to remove the cold bridge through each joist. Would it be possible to insulate under the joists (mineral wool rolls or slabs) and enclose the pipework at the same time?

The thickness that is most cost effective will, I suspect, depend upon the temperatures concerned. If you are running traditional rads at 80 deg or UHF at less than 50 then your target losses /savings will need a different thickness of insulation. Or you take the view that the cost is not so much and its a one time expense for a forever saving and go for lots.

For DHW pipes, there is also the issue that heat may move into cold insinuation faster then cold air. So insulation that helps with pipes that are hot for a long time (CH) may not with hot water.

Posted By: barneyAdding significantly more insulation will increase the heat loss as the surface are increases faster than the insulation surface temperature decreases

Are you sure about that? Sources on the intertubes give conflicting information. There is definitely a phenomenon known as the critical radius of pipe insulation and some sources describe what you say. But others explain that starting from an uninsulated pipe, adding insulation causes the heat loss to increase for the reason you state, until it reaches the critical radius, but then adding more causes the heat loss to decrease in the usual way. I'm inclined to believe the latter explanation. The implication is that it's worse than useless putting a small amount of insulation on a pipe (for the appropriate definition of 'small' which is quite small).

Posted By: ringiWhat if the insulation is covered in foil?

If I remember correctly, one of the assumptions of the critical radius theory is that there is negligible radiative transfer.

r = radius of the pipe (m)
λ = conductivity of the insulation material (W/m·K)
t = thickness of the insulation material (m)
s = surface resistance of the outer face of the insulation (m²·K/W)

For a 1 metre length of pipe:

Resistance of the insulation material: i(t) = ∫(r,r+t) 2πx/λ dx = (2π/λ)·(rt + t²/2)
Resistance of the surface: S(t) = s/(2Ï€(r + t))

Setting:
λ = 0.030 W/m·K
s = 0.1 m²·K/W
r = 0.0075m (15 mm pipe)
t = 0..0.1m (no insulation to 100 mm)

gnuplot> l = 0.030
gnuplot> s = 0.1
gnuplot> r = 0.0075
gnuplot> tp = 2*3.14159
gnuplot> i(t) = tp/l * (r*t + t*t/2)
gnuplot> S(t) = s/(tp * (r + t))
gnuplot> R(t) = i(t) + S(t)
gnuplot> plot [0:0.1] R(x)

That seems to say that up to 30 mm you're incrementally doing harm and you have to put a lot more on to beat having no insulation at all. I don't believe that. What have I got wrong?

Update: I think I got this wrong: i(t) = ∫(r,r+t) 2πx/λ dx. Perhaps it should be i(t) = ∫(r,r+t) 1/(λ2πx) dx or something. More thought later.

Yep, I'm fairly sure that the insulation resistance should be:

i(t) = ∫(r,r+t) 1/(λ2πx) dx

Less confidently, I think the definite integral should be: 1/(2πλ)[ln(r+t) - ln(r)].

With the previous assumed values the resistance increases continuously with thickness. Doubling the surface resistance to 0.2 m²·K/W (maybe by adding Ringi's foil) and increasing the conductivity of the insulation to 0.050 W/m²·K gives a little dip a few mm wide before more insulation starts to increase resistance.

Red line: resistance due to insulation.
Green line: resistance due to outer surface.
Blue line: total resistance.

So, unless I've made a further upcock, this indicates that this effect is real, works the way DJH remembers (reduction is in the first part of the insulation increase) but for practical purposes is irrelevant. Add mhor insultation.

pipe.plt;

#!/usr/bin/gnuplot -p

set terminal png size 450,300
set output 'pipe.png'

l = 0.050
s = 0.2
r = 0.0075
tp = 2*3.14159
i(t) = 1/(tp*l) * (log(r+t)-log(r))
S(t) = s/(tp * (r + t))
R(t) = i(t) + S(t)
plot [0:0.1] i(x), S(x), R(x)

I had a go independantly on paper then plotted with excel - I think I ended up with exactly the same formula as you Ed -

1/Loss = 1/(2.pi.L)*[(log(R/r)/lambda + 1/R.Kair]
Where R is outer radius, r is pipe radius, Kair is 10W/m^2.degC, lamda = 0.04W/m.degC

As the pipe gets thinner the wierd effect gets much more noticeable - that is, an 8mm diameter pipe in still air needs 30mm OD of insulation or more to be better than bare pipe, while a 2mm hot object is virtually impossible to insulate usefully. A 15mm diameter pipe just shows the effect- any thicker and the effect is gone.

It's maybe relevent for solar thermal installations, as they tend to have super thin pipes. I suppose the insulation there serves 2 purposes though - whilst the insulation on our indoor ST pipes may be useless from a heat loss perspective, we don't want exposed potentially 100degC pipework.

If it's right, what about thin electrical cables running in free air versus in insulation? Conventional wisdom says you should be cautious of covering them, in case they overheat. Maybe that's a safe oversimplification.

Hopefully there's a pic attached...

Thanks Rob. Yes, our formulas look the same. It took my a while to work out how your log(R/r) is the same as my ln(r+t) - ln(r). Doh! That's what logarithms were invented for. It's getting late. Nice generalization to look at the different pipe sizes and something worth bearing in mind.

Posted By: RobLI had a go independantly on paper then plotted with excel

Could you post the Excel file?

Posted By: djhIt's because the heat flow isn't uniform throughout the volume of the insulation; the flow lines aren't parallel.

We're assuming an ideal insulator (conductivity doesn't change with temperature, heat flux or direction as it would in a real insulator because of convection in the little cells) but we are taking the basic geometry of the higher heat flux (W/m²) in the inner bit into account.

There's a lesson for installation here as well. I've got a hot water recirculation loop. It's all within the insulated envelope but of course is wasted heat in summer. I'd planned to heavily insulate the pipes but of course when I came to do it I found the outbound and return had been installed close together, and in many places attached to the wall with pipe clips which meant it was impossible to use insulation thicker than the standard 10mm polyfoam.

If I was doing it again I'd ask for wider spacing and set off surfaces so thicker insulation could be used.

Posted By: tonyNone! I would lift those pipes up onto the warm side of the insulation.

Insulation will not stop hear loss only reduce and if you need a heating system all the heat that it generates needs to go into your house not escape.

Unfortunately it is not possible for me to put the pipes onto the warm side of the insulation. I have to insulate between the joists, there is only about five inches of space beneath the joists and that must be kept clear. I cannot put the pipes between the joists or through the joists.

I am installing ThermaSkirt which I have found has the advantage of shortening the pipes under the floor in my situation.