Delta question..For interest sake..

[wipeoutdude]

So in recent weeks I have been learning quite a bit and getting my new heating system setup right thanks to the helpful experts on this forum.. I have a question of the "Delta" and its effect on the heating ability of the system..

So as I understand it the Delta is calculated as..

(flow temp + return temp)
----------------------------- - 20
2

So if we have two situations..

Flow 70, return 50 = Delta 40
or
Flow 65, return 55 = Delta 40

So the delta is the same..

Does that means the heat output (and ability to heat the house) of the system the same in both cases?

Is one likely to be more efficient than the other?

I guess I am trying to understand the heat output characteristics as it related to the Delta value..

 

[Steelmasons]

So if we have two situations..

Flow 70, return 50 = Delta 40

Delta would be 20c

Flow 65, return 55 = Delta 40

Delta 10c

Don't confuse MRT with deltaT.

 

[Steelmasons]

Twice.

 

[Dan Robinson]

Coconuts at the ready?

 

[wipeoutdude]

So if we have two situations..

Flow 70, return 50 = Delta 40

Delta would be 20c

Flow 65, return 55 = Delta 40

Delta 10c

Don't confuse MRT with deltaT.

Not according to many sources on the internet that I have looked at..

Delta is calculated by determining the mean/average radiator temperature so in my two examples this would be..

(70C + 50C) / 2 = 60C average radiator temperature
or
(65C + 55C) / 2 = 60C average radiator temperature

Then you subtract the standard 20C room temperature to get the Delta..
So.
60C - 20C = 40C Delta in both cases..

So assuming that is correct, back to the original questions..
Would both these scenarios have the same heat output from the radiator?

Would there be much difference in efficiency from the boiler since both scenarios have a return temperature of 55C or less??

 

[Biggles..]

You are correct in that the difference between radiator temp and air temp is termed the Delta T.

If both calculations have the same Delta T as the answer, then there is only one correction factor, that being 0.59 so you will get the same output from both MWT calculations.

MWT= (F+R)/2

DT= MWT-AT

 

[D_Hailsham]

If both calculations have the same Delta T as the answer, then there is only one correction factor, that being 0.59 so you will get the same output from both MWT calculations.

MWT= (F+R)/2

DT= MWT-AT

Where do you get 0.59 from? According to Stelrad the correction factor for a DT of 40C is 0.748.

As for the factor being the same provided the MWT is the same, this is not strictly true but the difference is very small. For example:

70/50/20 has a correction factor of 0.733;

65/55/20 has a correction factor of 0.748.

 

[Biggles..]

Don't get to caught up by what stelrad quote or use, lots of different manufacturers have a different n value for their radiators.

Correction factor being (ΔT/50) to the power n.

n is found to be anywhere from 1 to 3 depending on manufacturer.

Edited to correct a mistake.

Can sometimes be read is cf=(dΔt/bΔt)^n

 

[Steelmasons]

Correction factor is a simple calculation.

Double panel radiators of the same output as single panels would yield differing correction factors.

 

[D_Hailsham]

Don't get to caught up by what stelrad quote or use, lots of different manufacturers have a different n value for their radiators.

Correction factor being (ΔT/50) to the power n.

That's a very simplified version of the equation which is:

View media item 89432
Tf = Flow temp; Tr = Return temp; Ta = Room Temp

The constant (49.83) depends on the default temperatures, i.e 75/65/20.

n is found to be anywhere from 1 to 3 depending on manufacturer.

Never seen a value of n as high as 3. The value does vary with make and type of radiator, but most rads have a value about 1.3; Stelrads table is based on an n of 1.29. The highest value I can find is 1.35.

 

[Biggles..]

Jaga will be around 1.35
Mhs 1.30
Myson 1.3

We fitted some cast columns that were 2.6