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Delta T / GPM question
Frank_3
Member Posts: 112
The input temperature to the fan coil and output capacity of the boiler are irrelevant. You already have the information you're looking for, unless I'm misunderstanding your question.
You say the fan coil is rated at 2.8" @ 8 gpm, with a 20 degree Delta T. Well, there's your temperature drop, regardless of the input temperature. As long as that flow rate is maintained you should see that delta T.
As for change in delta T based on GPM, that takes a little bit of math. Let's start by calculating the output capacity of the fan coil at its listed flow rate. The formula is:
btuH = GPM x deltaT x 500
So, according to your specs the fan coil will produce
btuH = 8 x 20 x 500 = 80000 btuH
To answer the question about what delta T would result from a change in flow rate, flip the formula around a bit and you get:
deltaT = btuH / (GPM x 500)
Using the 10 gpm figure you mention, that gives us
deltaT = 80000 / (10 x 500) = 16 degrees
However, what you really need to determine is what flow rate the 15-42 pump can give you when it's only looking at a 2.8" pressure drop. My guess is it'd be a lot faster than 10gpm, so that will change the temperature drop as well.
And that's assuming the fan coil is the only thing presenting resistance to the pump. You should include some additional pressure drop for piping and fittings, but that's a whole different book. :-)
You say the fan coil is rated at 2.8" @ 8 gpm, with a 20 degree Delta T. Well, there's your temperature drop, regardless of the input temperature. As long as that flow rate is maintained you should see that delta T.
As for change in delta T based on GPM, that takes a little bit of math. Let's start by calculating the output capacity of the fan coil at its listed flow rate. The formula is:
btuH = GPM x deltaT x 500
So, according to your specs the fan coil will produce
btuH = 8 x 20 x 500 = 80000 btuH
To answer the question about what delta T would result from a change in flow rate, flip the formula around a bit and you get:
deltaT = btuH / (GPM x 500)
Using the 10 gpm figure you mention, that gives us
deltaT = 80000 / (10 x 500) = 16 degrees
However, what you really need to determine is what flow rate the 15-42 pump can give you when it's only looking at a 2.8" pressure drop. My guess is it'd be a lot faster than 10gpm, so that will change the temperature drop as well.
And that's assuming the fan coil is the only thing presenting resistance to the pump. You should include some additional pressure drop for piping and fittings, but that's a whole different book. :-)
0
Comments
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Delta T / GPM question
I just finished Dan's book about Primary / Secondary pumping. Why can't all textbooks be written that way? Today's kids would have higher test scores & stay in school!
Anyway, I'm trying to figure out the DT / GPM relationship so here's the deal:
Using Monitor MZ40C (135,000 net BTUH) condensing boiler to heat a fan coil. Contractor wants to use 15-42 boiler pump for entire system. Two scenarios - 140F to fan coil and 180F to coil. Fan coil rated at 2.8" @ 8gpm for 20F DT. The MZ HX can go as high as 7' @ 10 gpm. 15-42 pump does 10' @ 10 gpm. For those temps, what DT spreads am I looking at? Remeber, no low temp return limit to boiler, and homeowner is an efficiency nut.
Please show all calculations so I can do them later.
Thanks!
Matt Connolly
0 -
very nice
thank you frank
chris smith
paradise porter maine0 -
Thank you - but...
If I used a lower gpm, won't more heat leave the fancoil(given a constant airflow) since it has more time to shed the heat; and a greater delta T would result? That's the heart of my question.
Basically, if the system flows 6 gpm after all losses, what will the delta T be, and will it affect net heat to the house?
Matt0 -
A common misunderstanding
regarding flow rate and the ability of heat to "jump off" If you look at the specs for hw coils or baseboards you will see the greater GPMs result in higher heat output. This article explains it much better than I can. The formula posted in the response above proves that result.
http://www.pmmag.com/CDA/ArticleInformation/features/BNP__Features__Item/0,2379,3760,00.html
hot rod
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Thanks HR, now I see..
the picture. I'm from the racing / automotive world and sometimes when a car overheats at high rpm we put on a bigger water pump pulley to slow the flow thru the engine and radiator down so heat transfer actually occurs. This seems to work - but why? What the hell's going on here!?!? -I don't think it's pump cavitation since the hose size is large and it's under 18 psi system pressure. Your thoughts?
Matt0 -
There are a bunch of things to keep in mind:
1) In any heating/cooling system the exchange occurs from the hotter medium to the cooler medium. The baseboard/fan coil/engine block don't know or care whether heat will be transferred from the water to the surrounding surface or vice-versa. Mother nature and the laws of physics take care of that part, and the simple answer is that heat moves to cold. (You could actually use the "cooling" system of an engine to pre-heat the block by running warm water through it while the engine is cold. Of course, you'd have to pull the thermostat to do that but that's a different story.)
2) When we talk about deltaT it's not always a temperature drop -- it's really just a change in temperature from one end of the heat exchanger to the other, whether that exchanger is a baseboard or engine block.
3) In a heating system we're moving hot water through some heat exchanger which has a cooler surface temperature. The result is that heat is transferred away from the water -- towards the cooler medium -- and the deltaT (temperature change) is presented as a lowering of the water temperature and an increase in the temperature of the surrounding surface.
4) In a cooling system we're moving cold water through some heat exchanger which has a *higher* surface temperature. The result is that is heat is transferred away from the surrounding surface and into the water. The deltaT is presented as an increase in the water temperature and a decrease in the temperature of the surrounding surface.
5) So, how do we increase the heat transfer of that engine's cooling system? We want the deltaT (temperature change) to be at its highest and therefore the water has to be slowed down.
The caveat being -- as the article that HR posted mentions -- there is a law of diminishing returns. The water can only carry so much heat based on volume, and there is a velocity at which the peak heat transfer occurs and anything slower than that will screw you up royally. So don't try running the engine without a pump and expecting gravity to do the work. :-)0 -
Good points Frank
I would suspect the high rpms could really lead to cavatation in an engine cooling pump. As you know all pumps are designed to optimize at a certain rpm (pump curves) I don't know much about the types of pumps used on race engines. Generally regular automotive engines run a centrifical pump of some sort, as I recall from auto shop! I expect at a certain rpm the impeller speed and high engine temperatures would REALLY open the door to cavatation.
I suspect designing a pump that could provide an adequate gpm over such a wide rpm range seen in racing applications would be a challange. Too bad we can't "Ask Smokey" any more
http://www.pmmag.com/CDA/ArticleInformation/features/BNP__Features__Item/0,2379,4270,00.html
hot rod
High head pumps run on that jagged edge.
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Let it Flow
Another article that Larry Drake wrote in last months Fuel Oil News.
http://www.petroretail.net/fon/2002/0211/0211rey.asp
cheese0
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