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chapchap70_2
Posts: **147**Member

I guess you would be looking for the optimum delta T if I understand your question correctly. If the water is flowing too fast, it will not give off its heat. Maybe others know better than me.

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## Comments

540MemberAir handlers. Hypothetically speaking, 4 gpm @ 170* produces 40000 btu output. If gpm is increased, will btu output be increased? If so, what is the calculation for solving this? Thanks.

40MemberOne of the paradoxes of heat transfer. All that effort and not much to show for it; the quietessential law of diminishing returns.

The coil output is a proportional function of the difference between the entering water temperature and the entering air.

You start at 40,000 BTUH and you have a 20 degree delta-T.

Let's assume that the 170 degree water you have is the average water temperature, the point halfway through the coil. A median if you will. Water is entering at 180 and leaving at 160.

As an extreme example, say you doubled the flow rate. You still enter the coil at 180 but now leave at 170, a 10 degree delta-T. What have you accomplished? You have raised the average water temperature by a whopping 5 degrees. Practically bupkes. Nothing to get excited about.

What else have you done? By doubling the flow rate you have quadrupled the coil pressure drop (approximate square function) and the horsepower to do the work increased as an approximate cube function. Velocity doubled with resultant erosion issues.

Conversely the throttling of flow to reduce output, even if a linear valve, will not yield a significant reduction in capacity. Often you can use half the flow and still have 90 percent of initial capacity. Does not seem fair, somehow

The best way to vary output is to vary the incoming water temperature outside the coil by mixing or modulation. Temperature difference is very closely related to output, a good tool to use.

As another tangible example, take a look at baseboard radiation capacity tables. You will see two rating categories: 1 GPM and 4 GPM flow rates. (You use the 1 GPM if the flow rate is unknown, the 4 GPM if it can be confirmed. Alternate rating points are expressed as 500 lbs. per hour and 2000 lbs. per hour respectively.)

If a given 1 GPM rating is, say, 520 BTUH per Linear Foot, the 4 GPM rating might be, oh, 550 BTUH per Linear Foot. Not much reward for effort. So you can see that going the other way can yield reasonable outputs for a lot less transport cost.

7MemberNot many men can explain it that well!

1Member578MemberIf I have an air handler moving 1000 CFM, BTU's are calculated CFM X Delta T X 1.08. The BTU's that the coil can produce versus what it can deliver might be somewhat different. Have to check the Delta T of the air to determine if more heat is actually recovered.

To get 40,000 BTU's delivered we would need a Delta T of 37*

which is giving us a 20 degree drop in water temperature.

If the gpm is increased and the water temperature is only dropping 10* that would mean the air has to be absorbing less and the Delta T would have be less. The BTU's delivered would also have to drop. Explain if I am doing something wrong here?

231MemberI am glad you added the discussion about baseboards, it gives me a clearer understanding of my system.

2,440MemberFair question, Jim. It is not entirely linear because you are dealing with two fluids of different densities and specific heats. Ever changing variables during the process.

And there is also counter-flow, the notion that entering air will see cooler leaving water and leaving air will see the hotter entering water.

Let's say we have a mean coil temperature of 175 degrees (180 in and 170 out; higher by five degrees than the 170 degrees median temperature we started with). And let's assume your 1,000 CFM entering the coil at the same temperature. The air will absorb more heat, not less, because the temperature difference is higher. Not by much, but higher.

For discussion purposes the center of the coil will be a median point for both fluids.

If your first scenario of 40 MBH, 1,000 cfm rising call it 37 degrees. (Say we start at 70 degrees and leave at 107 F.) Water enters the last row of the coil at 180 and leaves the first row of the coil at 160.

The initial entering temperature difference is 90 degrees (entering air to leaving water, 70F and 160F) . The leaving water/air temperature difference is 73 degrees (180-107F).

The swing of the delta-T's is 90 to 73 degrees in this case.

Now double the water flow so that we halve the water TD. Let's assume (because they are moving variables, we must assume) that the new leaving air temperature from the now hotter coil will go up a couple of degrees to 109F. Just a guess, less than half of the mean water temperature difference.

The initial entering temperature difference becomes 100 degrees (entering air to leaving water, 70F and 170F) . The leaving water/air temperature difference is 61 degrees (170-109F).

The swing of the delta-T's is 100 to 61 degrees in this case.

It starts higher (good) but leaves at a lesser delta-T (less good). Almost but not totally cancels out.

You can see how you will get some increased capacity, just not much. The temperature approaches keep changing with every variable.

Granted the above is hypothetical but the principles remain. A finite coil selection program will give you some more accurate figures but order of magnitude these are not too far off.

578MemberLittle more thought and I don't necessarily see a decrease but I still have problem with an increase. Why isn't 10 degrees recovered from 8gpm exactly the same as 20 degrees recovered from 4gpm?

2,440Memberselected at specific parameters (rows, fin spacing, etc.IOW: it is fixed) when you change one variable there is a cascading effect on all of the others. Changing flow is the least of them. Changing temperatures entering the coil has a much greater effect. Issues if turbulence on an increase in flow versus laminar flow on a decrease; the mass flow versus the mass of metal in the coils, any number of factors.

In our coil discussion earlier, with the variable Delta-T's you know definitely that your average water temperature is higher. That begets higher inlet delta-T's and lower leaving delta-T's. The median point shifts a bit, as does the density of the air at a given point. Practically splitting molecules.

But take the earlier comparitive coils: Fact is you will see a slight increase on the air side because the coil is five degrees hotter which means the water side will droop a bit (your original point) maybe a half degree? So instead of a bona-fide 10 degree drop it might be 10.5 degrees.

All that each of the variables are seeking to do is find a steady-state balance.

You do raise a good point, Jim. It is a little paradoxical at first. But as the Harvard Rule of Arts and Sciences says, "Given the most controlled states of temperature, pressure, humidity and other variables, the organism will do as it damned well pleases."

Hope this helps-

Brad

578MemberGood explanations, just not as perfect a world as we would like. I have proven many times in the field when airflow is increased across a heat exchanger or coil there is a point where the heat transfer is substantially decreased. Also do this when setting the gpm on boilers. There is a peak performance for everything and on either side, fast or slow transfer can diminish.

2,440MemberWell said. I am glad you raised the points. See how our collective experience defies the laws we think we know.

"Reality Engineering". Kind of makes you wonder how often data is tailored to meet expectations, does it not?

Thanks.

Brad