Confused

Paul48

The following statement is copied from Caleffi's Idronic Series on Hydronic Balancing<span><span>

• The faster a heated fluid passes through a heat

emitter, the greater the rate of heat transfer,when all

other conditions are equal

This statement seems to directly oppose the Universal Hydronic Formula, whereas to achieve a higher Delta T, you supply less gpm, which implies a slowing of the flow.Can someone clear this up for me?<span></span></span></span><span></span>

Jamie Hall

Deja vu all over again?

seems to me that this came up just recently in another thread...



Anyway.  The amount of heat -- BTU/hour -- given off by an emitter -- a radiator or section of baseboard, let's say -- is dependent only on the temperature of the emitter.  That temperature is, in turn, to a very good first approximation, the average of the incoming temperature of the water and the leaving temperature of the water.  The math gets a little messy, but the outgoing temperature will be equal to the incoming temperature minus 1 degree F per pound of water going through per BTU given off.



So... if you reduce the flow you will have a greater delta T for a given emitter and input temperature.  Or, if you want more heat out of the emitter for a given input temperature, you have to increase the flow so that you have more pounds of water going through and therefore less delta T and a higher average temperature.



The confusion comes, perhaps, from a tendency to equate the delta T with amount of heat delivered, without taking into account the flow.  The actual amount of heat delivered is the delta T times the total flow.



Or you could switch to steam and figure 240 BTU/square foot and be done with it...

Paul48

Conditions

With all other conditions being equal, then can you have a higher DeltaT across an emitter and not have emitted more heat? I am talking design considerations, using two known quantities to calculate the third. Or do you just say, I want more heat over there, so I'll pump it faster?

Jamie Hall

The tradeoff

is between delta T and flow.  If you want the same heat output but more delta T, you reduce the flow -- but to maintain that average temperature, you have to increase the input temperature (it has to go up half as much as the delta T goes up).  If you want more heat, you either have to put more water through at the same input or you have to increase the input temperature -- both of which will increase the average temperature and hence the heat output.



It does seem sort of backwards sometimes, doesn't it?!

hot_rod

some good reading

http://www.pmmag.com/articles/95877-nature-vs-math

remodel

counter intuitive

If the boiler is sized to the load all is good, i.e. gpmx500xDT=boiler output=hopefully load.  Slowing down pump speeds doesn't give you more heat just because it gives you a better DT, depends on emitters, pumps, system deign etc..  The moment the emitters or boiler are not right this gets buggered up and DT sucks  etc...  Its the whole thing ME says:

100 gpm x 1* =  1gpm x 100* 

Whish I would have learned this a little while back.  Great forum for sure.

 

Paul48

Thanks hot rod

After reading the article, it appears there is a kink in the Universal Hydronic Formula. And now for the million dollar question.......Why won't anyone just say it is wrong?

Zman

Broken?

There is  nothing wrong with the formula. There are just conditions where other factors should also be considered.

I think what John is saying that in some applications, the formula is not perfectly linear. When you start getting to the extremes you need to be careful.

Carl

hot_rod

second part of article

http://www.pmmag.com/articles/95958-parsing-a-number

EBEBRATT-Ed

flow

I have had trouble understanding this myself. I do know that turbulent flow is better than laminar or what I call lazy flow when it comes to the heat transfer rate of an emitter. Something to do with the molecules bouncing off each other.

Paul48

It is wrong

It is mathematics. I don't buy into the new mathematics. It is either right or wrong.

Jamie Hall

Sorry...

but, within very broad limits, the universal hydronic formula is correct.  The heat delivered is equal to 500 times the flow in gpm times the delta T in degrees Fahrenheit (the 500 contains a bunch of unit conversions which we don't have to worry about here).  Let's take a look at the formula, bit by bit.



There are two items on the right of the "equals".  Flow and delta T.  Let's consider just one of the two items on the right of the "equals" at a time.



Suppose we change just the flow -- the gpm.  If we increase the flow, the formula tells us that we will increase the heat delivered, assuming that we hold delta t constant.  That stands to reason.



Suppose we change just the delta T.  Again, if we increase just the delta T, holding the flow constant, we increase the heat delivered.  Again, stands to reason.



But... and there is always a but.



Let's suppose, instead, that we hold the heat delivered constant.  Then, if we want to increase the delta T, we have to decrease the flow by the same amount.  Conversely, if we want to increase the flow, we have to decrease the delta T by the same amount.



Where things get a little messy, however, is when we take this and put the water into a real world emitter -- radiator or baseboard or, for that matter, radiant floor.  Then we run into another formula: the heat emitted is equal to another constant times the average temperature of the emitter.  There is nothing we can do about it -- but again, it stands to reason: if you want to get more heat out of a radiator, the radiator has to be hotter (or looked at another way -- if the radiator is hotter, it will give out more heat).



So... if what we are concerned about is how much heat do we get out of the radiator, what we need to do is get the radiator to the proper temperature.  That temperature is simply the temperature of the water coming into the radiator minus, almost exactly, half of the delta T.



Need less heat out of the radiator?  You can either reduce the input temperature and keep the flow constant -- delta T will be less at the same time, as you are taking less heat out.  Or you can decrease the flow -- delta T will be greater.



Delta T can never be greater than the difference between the space temperature of the radiator and the input water temperature, by the way.



I don't know if all this helps... but the formula is correct.

remodel

Got to look at emitter closely

110 water over an emitter just can't output 180 over same emitter. Even though we could assume DT is the same, ie (110-90)=(180-160) DT. Is 4gpm x (180-160) x500 = 4gpm x (110-90) x 500? I think that is where the discussion started talking about 1.057 in the article.

Zman

New Math

Paul,

Just because it is not linear does not mean it is wrong.

The simple hydronic formula you have been using works very well. It is a bit simpler than what is really going on.

If you want o go a little deeper, the article is very interesting. It is not "wrong" or "new math"  it is a non linear formula that is a bit more complicated.

Carl

SWEI

The other ∆T

is that between the emitter temp and the space temp.  That takes care of the performance difference.

remodel

man this little formula

Thanks SWEI.