Momentary Flow Rate Confusion

hr

is one number thatwill change, 1 g[m is a bit high for a 300 foot loop of 1/2". Typically 0.5- 0.6 gpm for that loop. The HDS would show you the pressure drop difference between these loop lengths.

hot rod

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Unknown

Ok, so..

I'll probably wake up in the morning and smack myself in the forehead, but this is driving me a little nuts right now.

Say you have a room that is losing 5kBTUs/hr, with radiant heating in the floor.
The basic flow rate calculation (load = 500xGPMxDeltaT) would solve to 1 GPM a minute needed to heat the space with a 10 degree delta T.

Aha! But, look at two scenarios. Scenario one (just for the sake of easy numbers) say you heat it with a single 300' loop, and in Scenario two, you have one 150' loop heating it (and you don't care that half the floor is cold).

Obviously you would need a different flow rate to have the water drop 10 degrees over 150' of pipe than you would over 300' of pipe... so what is the puzzle piece I'm missing here?

yes, sometimes I AM that much of a fool!

Andrew Hagen (ALH)

Delivery

You'll have to run the 150' loop at a higher temperature to achieve the same output as the 300' loop. Same delta-t but different supply and return temps.

-Andrew

Unknown

forgive me, since I posted this, now I have to try and talk it out in public hehe..

Basically, the part I'm disconnecting on is.. If the fluid is travelling at the same rate, for instance 1 GPM, then the Delta T would be different between the two loops (you'll lose twice as much heat in the 300' loop vs the 150' loop, right?), even though you 'designed' for a certain delta T. Or, you could increase flow to make the Delta-T's the same.. but your output would still be more from the 300' then. So (Load = 500xDTxGPM) falls apart here, or it seems to. Your load is constant, but either your delta T or your GPM vary based on the length of the loop you are servicing.

I guess the real question is, if you have a certain load, and a certain size emitter, how do you find the true delta T and GPM... preferably without doing finite element analysis?

Jamie_5

I think ALH is right and you're simply forgetting the different transfer requirements of the different emitters.

Put one way, the 150 foot length of tube needs to transfer 33.3 BTUH per foot of tube to the room while the 300 foot tube has only to transfer half that amount per foot. So, as ALH points out, the water temperature for the 150 foot tube must be higher, resulting in a higher average temperature on the surface of the tube and hence a higher difference between the tube surface temperature and the room temperature and hence a greater rate of transfer from the tube to the room.

Of course, I could be wrong.

Steve Ebels_3

Seems to me

The load and the amount of BTU's the system will supply are two different things. The capacity of the system can change based on water temp and flow rate which determine temp drop. The load varies by the heat loss of the structure or room. A good case for variable water temps obviously.

But then maybe I'm confused........ I'll have to ask my wife.........

Andrew Hagen (ALH)

dT

Will the assumed temperature differential be the case in reality? We're delivering the right number of btu's in the fluid, but are they being transferred to the space?

As long as we're talking about conductive heat transfer it's a little easier. It's another AUdT calculation. The fluid temperature is dropping linearly with respect to the distance travelled along the tube. Therefore the average temperature is in the middle of your dT assumption. Your maximum fluid temperature at design conditions is determined based on the fea-determined relationship between average floor surface temperature and fluid temperature. Fully turbulent flow within the tube is assumed. So you assumed your dT, and know the U value of the tube, and your load. You can calculate the required surface area of contact between the tube and the floor. This is where circular snap-fit channels are important. You can calculate the length of the tube required to achieve that delta-t by backing it out of the surface area of the tube per unit length of the tube. There are quite a few assumptions, most involve perfect contact between the components of the radiant floor. Of course this is all at design conditions.

There's some fea involved, but someone else already did it, or at least they should have if they're selling radiant floor products.

Obviously dT's can be much wider with panels that are not in contact with humans. Flow rates and dT's are continually changing with radiators and trv's.

At least that's how I see it. There are quite a few assumptions involved, but they don't seem to be wildly unconservative. Suspend the tube and it's a lot more complicated.

-Andrew

Mike T., Swampeast MO

Resistance is the Missing Piece in Your Problem

As you use less tube and the distance between runs increases, so does the resistance as seen by the room.

From the wonderfully simple Rad Pad (available here):

Say this is tube-in-slab with the tube 4" deep. With 12" spacing the slab has an effective R-value of 0.5; decrease spacing to 6" and the effective R-value decreases to 0.3

As resistance increases so does the required supply temperature to achieve a given average surface temperature.

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Given your scenario neither delta-t nor flow need change with the length of the tube. What will change however is the supply temperature required to achieve the same average surface temperature given changing resistance. As you said, "You don't care that half the floor is cold".

George Peteya_2

Two more cents from George P.

Rob, let's say the room is 200 square feet. The load is 5000 Btuh, so the output in the first case is 25 Btuh/sf. In the second case, you only have tubing in 1/2 the room, so the output has to be 50 Btuh/sf from the heated area (95° surface temp if the room's at 70° ... ouch). In each case, the DESIGN delta T and therefore flow rate are the same. What has to change is the water temperature. You may need only 100°F. for Case 1, 140° in Case 2. HOWEVER, the circulator is still in its box and does not know that we are having this conversation. So, although we can assume that the "design" flow rate and delta T are the same, the circulator will have other ideas if it has only 150' of (assumed) 1/2" pex to pump through. At 100°F., the head loss through 1/2" pex is 3.6 feet of head per 100' length, or 10.8 in Case 1, and 3.33 feet per 100 feet in Case 2 or 5 feet (Wirsbo CDAM, 3rd ed., Appendix H). Bottom line is that you'll get more flow in Case 2, how much we don't know because of supply and return line losses. Whew.

Unknown

So I did wake up and smack myself in the head... though you guys definitely are making for some interesting conversation and are bringing up some interesting methodologies, thank you! The goal is to find the true flow rate needed here (and to make sure the emitter is large enough).

I can do both, I think, simply from the output charts (which is the fea someone else already did). Obviously I can size the emitter using an output chart, that's pretty easy. But that BTUs/hr/sq ft number also tells me how fast the heat is leaving the pipe (law of conservation of energy, right?), and so with some conversion I can tell what flow each loop would need to acheive a certain Delta T.. or, if I can figure out the flow it will have, I can figure what Delta T it will have as well. *SMACK*

Thanks for the help guys. You helped clear the jam in my head quite nicely!!