heat loss in ducts

[j]

I have a prototype central solar heater which I have been tweaking and
am interested in calculating heat loss in the ducts.

Here is the way I think this would work.

Insulated ducts have an R value. I think this R value relates to the
total inside area. So, Inside Diameter * pi * length. So the heat loss
is proportional to:

ti = temperature inside duct
to = temperature outside duct

(ID * pi * l) * (ti - to) / R

units in feet.

Is that about right?

Obviously if there is no flow through the duct, there is no heat loss.
So the heat loss in BTUs must be proportional to the mass of air flowing
through the duct.

Now, I'm getting confused (at least more confused). That should be the
mass of air per hour * the specific heat of air.

The mass of air/hour = the density of ~ .075 * CFM * 60 minutes

or:
CFM * 4.5

specific heat (dry air) = .24

what the actual specific heat of 50% RH air is beyond me, but I think it
is 5 or 10% higher.

So loss in BTU/hr:

(ID * pi * l) * (ti - to) * (CFM * 4.5) * .24 * 1.1 / R

with ID and l in feet

Is that about right? It seems like I missed something, somewhere...

Jeff

 

[Curbie]

I have a prototype central solar heater which I have been tweaking and
am interested in calculating heat loss in the ducts.

Here is the way I think this would work.

Insulated ducts have an R value. I think this R value relates to the
total inside area.

Not unless you are accounting for the thickness and heat capacity of
the duct itself, otherwise the diameter or OD is just fine.

So, Inside Diameter * pi * length. So the heat loss
is proportional to:

ti = temperature inside duct
to = temperature outside duct

(ID * pi * l) * (ti - to) / R

units in feet.

Is that about right?

Obviously if there is no flow through the duct, there is no heat loss.
So the heat loss in BTUs must be proportional to the mass of air flowing
through the duct.

Now, I'm getting confused (at least more confused). That should be the
mass of air per hour * the specific heat of air.

The mass of air/hour = the density of ~ .075 * CFM * 60 minutes

or:
CFM * 4.5

specific heat (dry air) = .24

what the actual specific heat of 50% RH air is beyond me, but I think it
is 5 or 10% higher.

So loss in BTU/hr:

(ID * pi * l) * (ti - to) * (CFM * 4.5) * .24 * 1.1 / R

with ID and l in feet

Is that about right? It seems like I missed something, somewhere...

Yes, maybe, depends on if the ducts are all in unconditioned space and
you may want to consider duct pressure loss.

Jeff

Curbie