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

*> 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

> 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.