Fred B. McGalliard said:
...
But, given the internal pipe plus insulation, adding a layer of external
insulation decreases the net heat loss, but by much less than the internal
layer because the area is larger. Paying more for less improvement, but you
never get to paying more for nothing. That's clear isn't it?
Let's take an 'extreme' example:
You have a wire that is 2mm in diameter. And we'll assume the
free-convective heat-transfer coefficient with air is 20 W/m^2-K, and the
wire surface is 75C above the local air temperature.
For a length (L) of the wire, the surface area is pi*(0.002) * L = 6.28e-3 *
L m^2 so the heat loss per unit length is 20 * 6.28e-3 * L = 1.26e-1 * L
watts.
Now, lets put a 2mm coating of insulation that has a conductivity of 0.035
W/m-K (about the same as fiber-glass). The surface area of the outside of
the insulation is pi*(.006) * L = 1.885e-2 * L m^2. But we don't know the
temperature at the outer surface of the insulation though, so lets just say:
Q = 1.885e-2 * L * 20 *(Tsurface - Tambient) watts
= 3.77e-1 * L * (Tsurface - Tambient) watts.
Similarly, we know the inside temperature of the insulation and the heat
transfer through such a cylinder is
Q = (Twire - Tsurface) * 2*pi* 0.035 * L / ln(.003/.001) (.003 is the outer
radius of the insulation, .001 the inner radius)
= 2.002e-1 * L * (Twire - Tsurface) watts.
The heat loss through the insulation must be equal to the heat lost through
the air film when in the steady-state, so setting the two equations equal to
one another...
3.77e-1 * L *(Tsurface - Tambient) = 2.002e-1 * L * (Twire - Tsurface)
The temperature drop across the insulation is 1.88 times the temperature
drop across the film. If the total is 75C, then the temperature drop across
the insulation is 3.77e-1 / (2.002e-1 + 3.77e-1) * 75 = 49C. Across the air
film (75-49)=26 -or- 49/1.88 = 26
Now we go back to see what the Q is for the insulation = 2.002e-1 * L * 49C
= 9.8 * L watts. Quite a bit more than the bare wire alone (at 0.126 * L
watts). (a check is to calculate the Q for the air film as 3.77e-1 * L *
(75-49) = 9.8 * L watts)
So in this 'extreme' case, putting the insulation on *increased* the heat
loss from 1.26e-1 watts / meter to 9.8 watts / meter. (paying for making
it worse).
The problem doesn't come up in 'ordinary' applications in homes. If the
initial radius of the pipe is larger, then the calculations work out the way
most people expect. But with *tiny* wires and pipes, it can be a surprise
to actually make the heat loss *worse* by adding insulation.
daestrom