equivalent pipe length method issues

[ecowarm]

I think this may answer your question PIPE FITTING LOSSES

 

[doitmyself]

Yes, I know the drop is 20°C between flow and return at the boiler, however the 12°C drop was referring to across the radiator. I should have made that clear.

You can't do that! If you have a 20°C drop across the boiler you must also have a 20°C drop across the radiators.

BS EN442 says that rad output has to be measured with a 10°C drop, so that is what you will find in manufacturers' catalogues. If a 20°C drop is used the radiator output will be reduced. You can see this if you compare the pages headed 50Δt and 40Δt in the Stelrad Elite Catalogue.

50Δt = (75+65)/2 - 20; [75-65=10]
40Δt = (70+50)/2 -20; [70-50=20]

First some values for water at 75 °C: Density 0.97486 g/cm³
Heat Capacity (per unit weight) 4.1927 J/g
Heat Capacity (per unit volume) 4.0873 J/cm³ (Density x Heat Capacity per unit weight)

It's not really necessary to go into such detail. Most heating engineers (if they bothered to do the calculation) would use the standard number of 4.18 Joules per cc.

In any case you are starting at the wrong end!

First you calculate the required boiler size using the Whole House Boiler Size Calculator. You then deduct 2kW (Domestic Hot Water Allowance) from the result to give the central heating requirement.

Let's say this is 20kW. If we assume a 20C differential, the flow rate will be: (20 x 60)/(4.18 x 20) = 14.35 litres/min.

I'm not sure where you are including in the 10% loss.

 

[LeadByExample]

Hi Doitmyself,

I would like to say beforehand I appreciate your input and I get the impression you are genuinely trying to help and OK guy (I could have said person, but I'm not trying to be pc, but I can't tell gender by just reading your posts. Though I am not psychic, I don't think your a women, apologies if you are.)

Back to replying to your post, lets start with the last bit before correcting a few errors, misconceptions and contractible statements.
​

In any case you are starting at the wrong end!

First you calculate the required boiler size using the Whole House Boiler Size Calculator. You then deduct 2kW (Domestic Hot Water Allowance) from the result to give the central heating requirement.

No, I started at the right end, I have calculated my heat requirements and included 2kW. I found radiators closest to what I need and I think I have found a suitable boiler, I only needed to find out a few more things. That site (and some others) I've done and they confirm what I have calculated.

I'm not sure where you are including in the 10% loss.

​

What it says on the box of a boiler is its maximum conversion of fuel (usually gas) to heat, not usable heat (hence the 90% performance combi boilers etc....).


Now, don't be offended, but I will now show you why you are actually very incorrect in your last post.

You can't do that! If you have a 20°C drop across the boiler you must also have a 20°C drop across the radiators.

BS EN442 says that rad output has to be measured with a 10°C drop (...)

​

A misconception and contradiction in one.

Which one is it? A 20°C drop or 10°C across a radiator?

As well, only ideal (read theoretical) situations could a boiler deliver heat with no losses to a radiator and thus have a drop of 20°C at the boiler be a drop of 20°C across a radiator. Or in other words, only an ideal system would have the flow from the boiler have no losses (thus no temperature drop), then have the drop only across the radiator and the return have no losses either. Doesn't happen in practice.

If a 20°C drop is used the radiator output will be reduced. You can see this if you compare the pages headed 50Δt and 40Δt in the Stelrad Elite Catalogue.

50Δt = (75+65)/2 - 20; [75-65=10]
40Δt = (70+50)/2 -20; [70-50=20]

Δt is comprised of Δ and t; t is temperature; Δ is the (ancient) Greek letter for delta and is used in maths as a symbol meaning difference. Thus 50Δt means a temperature difference of 50°C. With heating this is to indicate the difference between the temperature of the water in the radiator and air temperature. Rule of thumb air temperature will be room temperature of 20°C.

As you see, the 20°C and 20°C drop at the boiler are only the same in they are expressed in 20 degrees Celcius, it is purely coincidental they are the same, but they describe two completely different things.

This makes your equation 50Δt =(75+65)/2 – 20 nonsense.
(75+65)/2 – 20 = 50, not a difference of 50, what you have there is an equation with randomly selected numbers, because they 'fit'.


In order to prevent confusion I'll express the following as follows:
boilΔt : temperature difference at the boiler (drop at the boiler between flow and return)
wtaΔt : temperature difference water to air (water in radiator and temperature in room)
radΔt : temperature difference across a radiator

Hence it is better expressing 20°C drop at the boiler as 20 boilΔt. Take note, this does not say anything about the start and end temperature, only the difference! Or in other words the boilΔt of 70-50 is the same as 90-70, both are 20 boilΔt!


The speed of transfer of energy from one medium to another (in this case water to air, the radiator is the membrane) is directly relational (hyperbolic in this case) to the difference of the water to air temperature. The higher the wtaΔt the faster the transfer or in heating the higher the output of the radiator. That's the reason why the output of a radiator decreases with a lower wtaΔt.

This means that the output of a drop across a radiator is dependent on the wtaΔt. Or in other words, a similar radΔt has a lower output with a lower wtaΔt. This is expressed in the Elite catalogue. However a 20radΔt will always have a higher output than a 10radΔt at any wtaΔt!

Heat Capacity (per unit volume) 4.0873 J/cm³ (…)

It's not really necessary to go into such detail. Most heating engineers (if they bothered to do the calculation) would use the standard number of 4.18 Joules per cc.

It appears to me, given so many misconceptions it is important to go into detail! Two things are incorrect about this. It is not in cc, but in grams and more importantly, this is for water at 20°.

A difference of 0.11J on 4.18 and 4.09 is a 2.25% overestimation on the heat delivered! Although small, a significant difference (just go to you bank and ask them to drop their mortgages by that).


Anyway, all this still does not answer my question, I'll reiterate as people may have forgotten!

Lets assume a straight pipe of one meter with a diameter 22mm connecting to a one meter straight pipe with a diameter of 10mm with a reduction fitting. Now we assume the fitting adds one meter of pipe.

Which diameter? Or in other words, do you have 'two' meters of 22mm and one 10mm or one meter of 22mm and 'two' meters of 10mm?

Thank you 🙂
​

 

[doitmyself]

No, I started at the right end, I have calculated my heat requirements and included 2kW. I found radiators closest to what I need and I think I have found a suitable boiler, I only needed to find out a few more things. That site (and some others) I've done and they confirm what I have calculated.

So you did a heat loss calculation for each room and then compared the result with that obtained by using the whole house method? That's a "belt and braces" approach, but you gave the impression that you were just adding up the theoretical flows though each rad to obtain the boiler flow.

What it says on the box of a boiler is its maximum conversion of fuel (usually gas) to heat, not usable heat (hence the 90% performance combi boilers etc....).

Don't understand what you are getting at. If your calculations show that you need a 20kW boiler, then you need one with an output of 20kW. Manufacturer's literature always show the output, so there's no need to assume a 10%loss.

I agree that it is only in theoretical situations that you will get exactly the same drop across the boiler as across each radiator, due to heat losses in the pipework. However trying to obtain a 20C drop across the boiler when the drop across each rad is only 10C will be virtually impossible.

Δt is comprised of Δ and t; t is temperature; Δ is the (ancient) Greek letter for delta and is used in maths as a symbol meaning difference. Thus 50Δt means a temperature difference of 50°C. With heating this is to indicate the difference between the temperature of the water in the radiator and air temperature. Rule of thumb air temperature will be room temperature of 20°C.

Grandmothers and eggs comes to mind!

This makes your equation 50Δt =(75+65)/2 – 20 nonsense.(75+65)/2 – 20 = 50, not a difference of 50, what you have there is an equation with randomly selected numbers, because they 'fit'.

It is not nonsense! They are the temperatures used when testing radiators to BS EN442. This states that the flow temperature must be 75C, the return temperature 65C and the room temperature 20C (so that's not just "rule of thumb"). The mean water temperature is therefore 70C or (65+75)/2 and the mean difference between the radiator and the room is 70-20 or 50C, generally referred to as a 50C delta.

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[AlexGas]

Have you got the "Domestic Heating design guide"?

 

[LeadByExample]

Hi Doitmyself,

Again all with due respect, but you don't understand what the catalogue actually describes. And everything you say just seems to confirm you don't know the theory behind it. This is OK, but this is a tangent on my original question. I just don't want to write another 'essay' explaining why you are wrong here.

I suggest we start a new topic, discussing this. How does that sound?

Regards

 

[LeadByExample]

Hi AlexGas

No, would you recommend it?

 

[AlexGas]

Yes, I would recommend it, it's a good read.

 

[AlexGas]

He's gone! :smile5:

 

[doitmyself]

Hi Doitmyself,
Again all with due respect, but you don't understand what the catalogue actually describes. And everything you say just seems to confirm you don't know the theory behind it.

Which catalogue are you referring to?

I just don't want to write another 'essay' explaining why you are wrong here.

I didn't know you were an expert on this topic!

 

[LeadByExample]

Hi Doitmyself,

As said, I don't mind discussing this, but I strongly suggest we start a new topic to cover this.

Regards

 

[LeadByExample]

Addendum to previous reply.


PS
If not then there is no point continuing and I will no longer reply unless it is on topic.

 

[ryanlee]

Do the calcs, fit, rads, balance system to achieve required results. Done.