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Calculate size of low Loss Header

View the thread, titled "Calculate size of low Loss Header" which is posted in UK Plumbers Forums on UK Plumbers Forums.

G

gary j

On the recommendation of my Wolf boiler (CGB-50) supplier, I need to install a LLH, can you please help me calculate the the size of the LLH required



Brgds & thanks Gary
 
Size of the boiler aside, can your supplier not advise of one suitable for the boiler?
 
Forget the LLH for a moment, do you really need a 50kw boiler? If you do, your house must be massive!

The house is 4,000 square feet with under floor heating downstairs, radiators upstairs and hot water tank - 300 LTRS

Brgds Gary
 
As I understand it sizing a LLH will require some site specific information of the system & how you intend to run it. (Delta T, existing pipe circuits sizes & locations to name but a few) In its simple form it is just a large pipe where the velocity of the water flowing through it falls to a low rate, to which the boiler flow & returns are connected & the other circuits are connected to draw away.
If you need a copper one which can be made to suit your site layout rather than a manufactures supplied one which are normally iron & just have two in & two out then it can recommend these people V A Heating Ltd home page
 
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The house is 4,000 square feet with under floor heating downstairs, radiators upstairs and hot water tank - 300 LTRS

Brgds Gary

A LLH in my opinion is no good for you you need a buffer tank if your using a 50kw boiler as I'd say its nor quite big enough for full load
 
i'm impressed gray, without any information on the build, construction or location of this house, you've calculated it's heat loss requirements! :ciappa:
 
If got a spread sheet somewhere for sizing LLH's i'll see if I can dig it out. Never used it myself, but was given to me by a heating consultant who teaches on HETAS courses so must be reasonably accurate, i hope!
 
Tell a lie, it's a word document and here it is copied and pasted

With regards to the sizing of low loss headers the equations below are what I normally use. The low loss state is classed as a flow rate of less than 0,5 m/s and to achieve this the general formula I have used in the past to design for a low loss in the header is:

Flow rate through the header in m³/s = Heating load in kW/((4.2 x delta T) x 1000)
This will now be known as X
Then put this value into the formula:
Dia (m) = Ö((X*1.28)/0.5)
Example
25kW load with a temperature drop around the heating system of 10⁰C
25/((4.2x 10) x 1000) = 0.000592
Diameter = Ö((0.000592 x 1.28)/ 0.5)
= Ö(0.000758 / 0.5)
=Ö 0.00152
= 0.0389m or 39mm
Always go to the next size up or the velocity will rise and potentially give problems.
The length of this will be at least 500mm and there should be at least 400mm between the nearest flow and return tapping.
 
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Most of the boiler manufacturers will be able to size LLH. Check out their websites and you will find some info on them.
 
The house is 4,000 square feet with under floor heating downstairs, radiators upstairs and hot water tank - 300 LTRS
OK, it's big, but the type of heating is irrelevant when calculating the requirement.

It would be interesting to know what result you obtained by using the Whole House Boiler Size Calculator. The HW allowance of 2kW may be on the low side for a 300 litre cylinder, but that will depend on what type of controls you have.
 
What is Ö? You haven't defined it.
Where does the 1.28 come from?

If you look in the part you've quoted, it says Ö = diameter

The 1.28 is predefined in the equation, I have no idea where it is from, sorry. This was just passed on from another source.

Me and Chris Watkins had a look at it last night briefly and actually couldn't quite work out the example above. If I get a chance I will try and get confirmation from the source.
 
If you look in the part you've quoted, it says Ö = diameter
No it doesn't, it says: Dia (m) = Ö((X*1.28)/0.5), which, to anyone with a basic knowledge of mathematics, means you multiply the result of evaluating the data within the brackets by Ö. Hence my question.
The 1.28 is predefined in the equation, I have no idea where it is from, sorry. This was just passed on from another source.
I worked that one out! its 2²/pi

It's from the formula to calculate the cross section area of the pipe= pi x r² = pi x (Ö/2)² = pi x Ö²/4.
 
Yes, your right and that makes sense as to why we couldn't suss it out.

I think you square it from messing around on the calc, seems to make sense
 
No it doesn't, it says: Dia (m) = Ö((X*1.28)/0.5), which, to anyone with a basic knowledge of mathematics, means you multiply the result of evaluating the data within the brackets by Ö. Hence my question.
I worked that one out! its 2²/pi

It's from the formula to calculate the cross section area of the pipe= pi x r² = pi x (Ö/2)² = pi x Ö²/4.

Yes, your right and that makes sense as to why we couldn't suss it out.

I think you square it from messing around on the calc, seems to make sense
Not been following along & being a bit thick does the formula now work ? If so how & does Ö = diameter ?

2kw very very low
Regarding allowances for hot water production I really see no point in allowing any extra kW's for simultaneous heating & HW production. 2 or even 3 kW allowances make little impacted on the modern cylinders coil ability to exchange heat, most S/Steel one are now rated 20 + to achieve the 25min recovers required, so are still going to rob from the heating side. Much better, surely, to use time control to have all boiler power heating HW (for half hour) before then move over to just heating. If WC is used, this will have to be the case anyway & the boiler sizing is not increased for those odd times when you may wish to reheat the cylinder when the heating is on.
 
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Yes Chris, basically as you worked through last night, but then square root it, and you end up with the resulting diameter in mtrs, so divide by 1000 to get pipe size 🙂


Didn't figure out the logic behind it entirely, being honest, couldn't be arsed tonight!
 
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So we don't have to allow for the maximum amount of usage for heating demand and hot water demand all at same time then is that what your saying
And any new house will have all zoned and timed controlled but most just set the hole lot to come on at once imho
 
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So we don't have to allow for the maximum amount of usage for heating demand and hot water demand all at same time then is that what your saying
Well yes, for most domestic system allowing that 2-3 when the coil can exchange 20+ would mean that neither is being heated effectively, surely better to do HW first then over to heating, don't you think ?
 
Yes I do think that
but I as well as you know they don't do this most don't even know how to programme a digital time clock or a stat
 
Me and Chris Watkins had a look at it last night briefly and actually couldn't quite work out the example above. If I get a chance I will try and get confirmation from the source.

Your formula is slightly wrong
The first part is right which gives you your flow rate in 3m/s
If you use 0.5 as your velocity through the header you are sizing for the worst case
The next part should be diameter of header(mm) = (square root of ((flow rate in m3/s x 1.28)/design flow velocity through the header)) x 1000
= 55mm header
You should really always go to the next size up so 67mm copper (in reality 54mm would do as the specific capacity of the water was rounded up when working out the flow rate) or 2" steel as steel (size for size) has a greater volume than copper.

A quicker way to do it is 3 x diameter of the primaries. That may oversize it but it won't be wrong.
 
Low loss headers are sized on the heat input(total boiler kw as more than one boiler could be fitted) the delta T of the boiler(s) (between 11c and 20c depending on design) and the velocity of the header (normally less than 0.5 metres per second). Once you know this you can size the header from the CIBSE pipework charts.
 
If you guys were stood in front of the boiler and knew what was being connected to the boiler from the system, all the different circuits and the boiler needed to be 50kW as a rule of thumb, what length and diameter LLH would you suggest this chap fits, you must have some idea and as Tamz says it would not hurt to be even a few diameter oversized, " would it", shall I have a stab, 100 mm dia MS 750 mm long, am I anywhere near, have to be welded up and would not look good in a kitchen, just a thought rule of thumb.
 
If got a spread sheet somewhere for sizing LLH's i'll see if I can dig it out. Never used it myself, but was given to me by a heating consultant who teaches on HETAS courses so must be reasonably accurate, i hope!

If you guys were stood in front of the boiler and knew what was being connected to the boiler from the system, all the different circuits and the boiler needed to be 50kW as a rule of thumb, what length and diameter LLH would you suggest this chap fits, you must have some idea and as Tamz says it would not hurt to be even a few diameter oversized, " would it", shall I have a stab, 100 mm dia MS 750 mm long, am I anywhere near, have to be welded up and would not look good in a kitchen, just a thought rule of thumb.
Perfect
 
I shall I have a stab, 100 mm dia MS 750 mm long, am I anywhere near, have to be welded up and would not look good in a kitchen, just a thought rule of thumb.

50kw boiler, 1¼" primaries x 3 = 3¾" = 4" :smile:
Near enough :wink:
 

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