Small Delta-T hydronic system

The delta T on a system is a moving number. We choose a ∆ for a design but a system may never operate at the exact number. The delta varies based on the load of the space. In a cold room expect to see a wide delta on start up.

As the room warms the ∆ will tighten up.

The determine the actual operating delta the system needs to get to thermal equilibrium. This is the point where the supply and return temperature have stabilized. 10- 25° is a typical fin tube design ∆.

Fin tube spec sheets usually show BTU output at two different flow rates. Slantfin shows a 1 and 4 gpm.

Here are some graphic of how various ∆s effect the heat output of a strip of fin tube. As flow increased the ∆ closes up. That equates to a higher AWT average water temperature across the heat emitter, higher BTU output. For 3/4" fin tube, 4 gpm is about the max, without getting velocity noise.

What is the temperature at the first fin tube on the loop? Two ways to increase heat output are to raise the supply temperature, or increase the flow rate.

Various brands of residential fin tube will have similar out put. 150 SWT isn't giving you much BTU output. and that temperature drops along the series. Most fin tube designs are, were, based on 180 SWT

You are sure the pump is actually circulating when you take measurements? The tight number you observed could bve the pump has just turned off, no water is moving?

I don't believe that 007E is a ∆T function? Maybe a ∆P function for zone valves?

Bottom line you need temperature differential and flow to transfer heat hydronically.

The ∆P circulators that adjust flow do so based on zone valves opening and closing, so a multi-zoned system, is that what you have?

It will not make any flow adjustments on a single loop circuit.

Here is an example of how delta P circs looks/ works

If the pump is programmed properly you should see, hear it modulate as the valves open and close. Some pumps have a display that shows you the change in W or gpm.

The question I would start with is, has it ever worked satisfactorily? If the answer is yes, let's focus on what has changed since it last worked. If no, that's a little trickier, we're looking at some sort of design issue.

The heat output of a radiator is given by the flow times the temperature drop. If the heat output is less than what is wanted, either the flow or the temperature drop has to increase. Generally, to increase the temperature drop you have to increase the water temperature.

If it worked at one time, my first suspicion would be flow has become diverted or obstructed. I would make sure the system was free of air before checking anything else.

@Dave H_2 : "Flow is not your issue here, if the delta t is low, the flow is high and more than enough."

That's assuming the flow is getting to the radiators. If you have primary-secondary piping, or radiators in parallel, you can have plenty of flow circulating but little of it reaching the radiators. If the flow isn't reaching the radiators there's going to be little heat loss and a small delta t.

Conveniently leaving the "f" out of an example is always suspect :) Sounds like something a member of the alternate reality ∆T club would say:)

No you cannot just raise the SWT and cause the ∆ in a loop to increase.

The ∆ that a system operates under is driven directly by the load at any given point. Further the operating condition of the boiler is driven by the load on the system. You cannot force the delta to be what you want it to be with a basic fixed speed circulator.

Archimedes allegedly said " give me a lever long enough and I can move the earth". Sounds good, could probably even put some numbers to it, but.

While the universal hydronic formula can predict most any possibility, not all will be attainable.

500 X 2 gpm X 5∆= 5000 btu/hr

500 X .01 gpm X 1000∆= 5000 btu/hr

Anyone care to build a working model of the .01 gpm flow example?

A wise man told me

The “danger” with the [ Btu/hr = 500 x gpm x ∆T ] formula is that people think that anything the makes the math work to the desired result can actually physically take place.  For example, if someone just thinks I’ll make the ∆T 40 instead of 20, and cut the flow rate in half, and get the same heat output  (which is true mathematically) but this math absolutely doesn’t imply that the circuit will physically operate under those conditions to give the same heat output.

Isn't the load the room and the heat loss of it, the radiators are the emitters, the the heat exchangers just like the boiler is.

Same with an indirect, the load is the cold gallons of water in the tank.

What you want and what you get. The load of the building, not you or the pump, or you buddies determine the actual operating conditions.

Your change of SWT without any other changes, absolutely will not guarntee the system is operating at the 80∆ you calculated. Granted the increased SWT from 80- 140 is where extra output comes from.

However if this is the 1500 sq ft shop discussed on another site? 120,000 btu ÷1500sq ft. = 80 btu/sq.ft seems like an awful high load you are trying to cover?

For those with open minds tuned in today, more data here

@hot_rod yes it is that building, and no the load is nowhere near 80 BTU/SF although that would be attainable. As was discussed then, the load is 27 BTU/SF but you argued tooth and nail that it was impossible to attain with only 1200LF of tubing because raising the SWT into a high mass radiant floor while maintaining everything else the same would not raise the delta or the BTU output. I showed you two videos as explained above to explain why you were wrong and that raising the SWT absolutely 100% DOES raise the delta and therefore the BTU output, but instead of trying to have an intelligent conversation about it, you got defensive and called me names before leaving the group entirely rather than admitting that you were wrong. As you were told then and still stands, you are welcome to share all the charts and opinionated snippets that you wish but none of them supersede the mathematical equation of DT x GPM x 500 = BTU. I very clearly stated that the delta in the extreme above example would dwindle as the temps reach an equilibrium, but that does not negate the fact that your deflections and opinions do not supersede mathematics and facts. I tried to give you the benefit of the doubt that there was simply a reading comprehension issue on your end, but then you doubled down with the insults and went so far as to remove yourself from the group to avoid being wrong. The reason this is being brought up again, is because it looks as though you're trying to preach the same falsities here and I want to make sure the others know not to trust everything they read- even if it comes from somebody like the almighty Hot Rod.

You said multiple times, and I quote, "Raising SWT and not changing flow rate does not change the delta", among several other avenues saying the same thing to imply that a given BTU output is impossible without a ridiculous amount of flow. That is false. You were/are wrong. Period. Now back to our scheduled program- sorry bauer, did not mean to get off topic.

Unfortunately you have done what many do using 500 for one factor of the equation . That 500 number is basically only a number for 60* water and changes along with the viscosity and temperature of the fluid . Maybe we should use the AWT of each and every example discussed to be dealing with facts . Alpha numbers matter !

477 for 30% PG glycol, and that changes a small amount based on the glycol temperature

temperature of the glycol effects heat transfer and pumping capacity

water 500 X 3 X 20= 30,000 btu/hr

30% PG 477 x 3 X 20= 28,620 btu/hr

The HDS software has a fluid calculator if you want to drill down to exact numbers.

This is why very transient conditions exist until the system reaches equilibrium