equivalent pipe length method issues

[LeadByExample]

Hi kirkgas,

I understand, it may not make a lot of difference. I had already assumed main circuits of 22mm and 15mm or 10mm to rads.

I thought I asked a few simple questions (where are these values and how are they applied), hence my extremely simplistic examples as when answered it would get me out of the woods (or cold as it is for heating our home). But it seems a few of the 'professionals' either don't know or unwilling to divulge that information, both are fine with me, just don't be smug about it (I don't mean you, or actually most other responders).

Because it doesn't have mathematically correct, the equivalent pipe length method seemed like a simple approach to a more complex problem and it would mean I wouldn't have to go through a whole bunch of much more complex equations etc. However, I would like it to be relatively accurate.

Anyway, ......

..... as you asked, here is the 'extended' version of what I am trying to achieve.

We need to replace the existing piping, some rads (even adding a rad or two) and the combi boiler (never worked since we moved in, repossesion property, everything shut down, drained down etc, etc. etc.).

I calculated I require a boiler able to provide 12 liter per minute (assuming a 12°C drop, if this is not a suitable assumption, let me know). I saw a suitable one able to provide 14 liter per minute, but I wanted to make sure the speed of the water wouldn't fall below about 0.5 meter per second and also wouldn't exceed 2 meters a second by too much in the system.

Just replacing the existing one with a similar is not desireable (not to mention 'lazy') as those appear to have far too much overcapacity and thus ineffective and inefficient and above all too expensive (not only in purchase).

I'm not sure how this will help, but there you go. You may feel you need to know how the property is laid out,number of rads etc... but that I have already calculated the heat requirements.

And while I am explaining, the only other thing I need to take into account is the system will go up and down over three floors (one rad down two rads up and if the difference in elevation will have a significant impact.

Of course if you know of other issues significantly impacting a domestic heating system, feel free to contribute.

Regards

PS.

When I have solved my prob, I will come back and let you (and all know).

 

[LeadByExample]

Hi benny_headland,

I'll have a look at it later, could be helpful, although it is for copper, I believe.

Regards

 

[doitmyself]

I calculated I require a boiler able to provide 12 liter per minute (assuming a 12°C drop, if this is not a suitable assumption, let me know).

Are you talking about Domestic Hot Water or Central Heating???

If CH, why are you quoting a flow rate - the heating requirement (kW) is the relevant figure.

CH systems should now be designed for a 20°C drop.

 

[BLOD]

as above, you should design system for 20 degree c drop for a condensing boiler.

The Domestic heating design guide is quite good and has some worked examples of heating designs
HHIC » Domestic Heating Design Guide

 

[kirkgas]

LeadBy
i have done loads of systems and dont know the answer to your question, but i honestly believe i have designed and fitted very good systems, some of which are now being pulled out after nearly 20 yrs good service, i think there are loads of guys who know how to fit top systems but dont know the level of info you are asking for, in a domestic scenario it simply is over complicating quite a straight forward process,
re your reply to me, you are confusing me with a couple of bits of info, (flow of 12L/min and 12 degree drop) when i discuss flow it is to a tap, i appreciate there is info on flow of heating pipes but we never bother with it in houses ( i remember vaguely one of the sections on the mears calculator having info on it) also the 12degree drop, you need to expand where that is for, as condensing boilers are designed to have a 20 degree drop between flow and return to maximise boiler condensing (i think)
so although some may be holding info, i think you will find most dont know, and that is no slight on them/me as it isnt info we need on a daily basis, and lets be honest we have enough info to keep to the front of the brain as it is, but as i said, sometimes a wee project keeps the mind active and keeps us away from Corrie, and im on tablets watchin my footy team play at the minute so that doesnt help either, keep us posted

 

[tamz]

You are over complicating things and looking for things that are not there. I would be interested on how this plays out. It is quite facinating to me the extent some people will go to find the obvious.
Your question has already been answered if you can get over your aversion to internet pages you don't trust.
Keep counting but i recon you should go back to base on this or phone some plumber /heating eng who may not understand the theory but through experience knows what works.
I await your comments.

 

[LeadByExample]

Hi Kirkgas,

I understand the depth of what I am asking in practice may not matter too much as a decent margin is taken into account. And as said, I don't mind if people don't know, it's just if some are being condescending to me about saying 'I just know'. I get a bit narky as it's neither constructive or helpful and only meant to make them feel superior (which they are not, I am, hahah ---- nah, just kidding). You and some others seem genuinely interested willing to provide help and I appreciate that.

Anyway back to the matter at hand.

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.

It is this drop that makes it important how much a boiler can pump per minute.

Let me explain (bare with me and figures and values are representative).

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)

This means for a temperature drop of 12°C:
Energy Released: 49047.5 J/liter (or 12*4.0873*1000)

OK that were the figures, now to a practical example.

Lets assume only one (yes 1) radiator with a heat output of 1788W and ignore all other losses.

This 1788W is delivered by the energy released by the temperature drop of the water flowing across the radiator (I know I'm telling you nothing you don't know, but keep reading).

The amount of water (in liters) per second required to deliver this can then be calculated by dividing the 1788W by 49047.5 J/liter.

1788/49047.5 = 0.03645 liter per second.

To know how much water we need per minute just multiply by 60 → 2.187 liter per minute.

So in other words in order to have the radiator to operate at full capacity we need to supply it with 2.187 liter per minute (anything less and the radiator would not emit the full 1788W).

Now assume I have 8 radiators as above (other losses ignored), what flow rate do I require minimally? Well 8*2.187 = 17.5 liter per minute at least! But for good measure you should go for 20 liters per minute.

To Doitmyself
Of course, the boiler also needs to be able to heat this volume to the required operational temperature. Therefore to increase the water by 20°C at the boiler it would need to transfer at least 27.25kW to the water, but again best to go for a 36kW (on average 10% is lost, so this would be an effective output of 32kW).

All in all, the boiler not only needs to be able to heat the water (the kilowatts), it also needs to be able to deliver that heated water to the radiator (the liters per minute).

But the calculated values above are base on 12°C drop across the radiator. If it would be a 11°C drop across a radiator, then per radiator 2.386 liter per minute is required (and for 8 similar ones it would require a flow rate of 19.1 liter per minute and thus probably a boiler of 22 liters per minute).

If it would be a 13°C drop then 2.019 liter per minute per radiator is required and for 8 a flow rate of 16.2 liter per minute and thus a boiler of 'only' 18 liters per minute would suffice.

Now back to the my situation.

I know there are other losses and that's why I needed to know how the Equivalent Pipe Length would work. And to estimate what the expected flow speed through my system would be.

Regards

PS.
That was an hour of my life, I could use a beer right now, and it's only 1 pm!!!!

 

[LeadByExample]

Hi Doitmyself,

In order to prevent to have all the info spread all over, I answered you in my reply to kirkgas.

Regards

 

[LeadByExample]

Hi Blod,

I'll have a look at it later, thank you.

Regards

 

[LeadByExample]

Hi tamz,

Welcome to the discussion.

I have been given a short cut (in the form of a spreadsheet) however that did not explain why I need to apply it where it says (or enter in the spreadsheet in this case). I can make spreadsheets (or design entire IT systems) to do similar things, however, that wouldn't explain to users why they need to enter what where.

Don't get me wrong I appreciate what you are saying, however, my original question has actually not been answered. It seems to me you think about the destination (the spreadsheet), I like to know how to get there.

Regards

 

[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.

​

 

[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.