Determine temp drop

cheftim

A 150' loop of 3/8" pex is desinged to have an input temp of 100 and a return of 90. What determines this 10F drop? If I increase the input to 130F does the return still see a 10F drop? Do you have to increase the tube size or lower flow to get more output?

Brad White_47

Temperature Drop

Temperature drop is really a design ideal, a balancing point among other variables. It is really based on design steady-state conditions.

For example, when you start with a cold slab, a given flow rate and a given temperature supply water, the temperature drop will be higher than when the slab and the room are at design temperature. Say the room is 58 degrees and so is the slab. Water is at 100 degrees. The first pass may well come back very close to room temperature, call it 60 or be generous, 65 degrees. Your temperature drop is then 35 to 40 degrees. Every pass of the water though, things change as it warms up. But a funny thing happens. As the slab and room warms, the heat output is less. The temperature differences are less you see. But then, we are approaching the balance point.

Now say your slab is at 83 degrees, your design point. The room is coming to or is at temperature. By then your design water temperatures could well be in balance, 100 degrees in and 90 degrees out.

If you raise the water temperature you will raise the heat output better than any other way. Initially you will have the same situation (wider temperature drop) and when it settles at design your temperature drop will be apporximately the same. Think Equilibrium.

If you increase the flow rate at a given temperature, the increase in output will be marginal. If you double the flow rate you will halve the temperature drop is one way to look at it, but will raise your average temperature only five degrees.

If you lower the flow from your 10 degree drop, (100 to 90 =95 average) you will get less output but as with raising it, that too is only marginal. Say you cut the flow in half such that the temperatures settle to 100 in and 80 out (90 average). Again, conversely this time, you will lower your average temperature by five degrees.

Tube size goes to flow rate possible and surface area contact potential with the slab or track or floor. Hence the heat output potential rises.

Remember though- all of this works as a system.

cheftim

So will higher water temp give more btu out?

Thanks Brad.

You said:
"If you raise the water temperature you will raise the heat output better than any other way. Initially you will have the same situation (wider temperature drop) and when it settles at design your temperature drop will be apporximately the same. Think Equilibrium.""

So does this mean you will still end up with the same 10F drop? Will the room be warmer though?

I'm trying to figure out how to build a raidant wall and keep the size of it down. My thinking is that as I raise the temp of the water, I'll get more btu output. How can I calculate the btu output of, let's say, a 4x10 panel made of 3/8 tube at 4" OC with 100F water? Then at 140F? (at least theoretical).

Brad White_47

If the panel surface temperature is warmer

the room will be warmer. You should keep in mind that any surface temperature over 85 degrees should be carefully considered. Too warm and the occupants may feel overheated.

The delta-T (temperature drop) is almost incidental. It is what it will be. Key is prediction of surface temperature and the connection of the tubing to the substrate has far more to do with that than the delta-T.

There are abundant tables available to determine the temperature of the surface, heat output (varies with room temperature!) and other variables.

Best value in my opinion for a great tool? One you will not regret purchasing?

The Rad Pad, available on the "Shop for Books and More" link on this page. That can answer far more than I can given the variables you otherwise control.

hr

I would guess

with a 90 f panel surface (90 isn't too hot if you are not barefooted on the wall , 68 room temperature something like 44 BTU/ square foot times 40 square feet= 1760.

I think you could run that surface temperature up to 100 comfortably. for an output of 2560 BTU/hr on that panel.

Some of the early, primitive, electric resistence celings got so hot you could barly touch tem. Enough to cause tall folks to sweat. Buy they would crank out some serious BTUs/ hr. Think low grade infared heater, like you see in carwashes

hot rod

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Mike T., Swampeast MO

I'm trying to figure out how to build a raidant wall and keep the size of it down. My thinking is that as I raise the temp of the water, I'll get more btu output. How can I calculate the btu output of, let's say, a 4x10 panel made of 3/8 tube at 4" OC with 100F water? Then at 140F? (at least theoretical).

Two ways to "build a radiant wall and keep the size of it down".

1) Increase the average temperature of the panel. As Hot Rod so wisely said, you CAN make the panel too hot...

2) Reduce the heat loss of the space.

40 square feet isn't too much to work with for a radiant panel unless the room is correspondingly small. Say your panel is 90F in a 70F room. 2 btu/hr per square foot per degree of temp difference between panel and room air is a very reasonable output estimate. So, you have 90-70 = 20 degrees of temp difference. 20 * 40 = 800 btu/hr. I hope that the room is quite small and/or climate is very mild and/or insulation & infiltration approaches that of a camping "cooler".