Boiler Flue Calculations

[arnoldb]

I'm trying to get my head around all the basic calculations that goes into designing boilers, with a specific interest in all-copper boilers silver brazed together.

A good amount of my hobby time have of late gone into trying to get together and understand as much information in this regard as possible, and I think I have a fair understanding of what is needed in terms of calculating pressures / material thicknesses and how safety factors and material strengths and temperatures are incorporated in those when doing calculations related to boiler shells and stays.

So far, my reference material has included KN Harris's book on model boilers and construction and information I could find here on HMEM, as well as some Google finds - unfortunately I have nearly no information on this subject in my own "library".

One aspect of calculating that I can't get a clear understanding of, and find little information relating to, is formulas needed to do calculations related to the "external" pressure limits / thicknesses required for flue tubes, and that is where I would like some clarification or help if anybody is willing.

One good thread started by Rich I found here on HMEM - with some formulas provided by SandyC.

For reference in the formulas that follow:
P: Pressure in psi
S: Maximum allowable stress value of the design material at design temperature
T: Tube wall thickness in Inches
D: Outer diameter of tube in Inches

Sandy's formula for calculating the maximum operating pressure for a flue is:
a) P = S [2xT –0.01 x D / D - (T - 0.005 x D)]
I'm not entirely sure how this formula is to be interpreted as written (No offense to Sandy !) - I _think_ it could be:
b) P = S (2T - 0.01D) / (D - T - 0.005D) ?

If I use the formula specified in the same thread for T:
c) T = (PxD / 2xS + P) + 0.005 x D
Which I think means :
d) T = (PD / (2S+P)) + 0.005D
And try to work it back to find P, I always end up with:
e) P = S(2T-0.01D) / (D + 0.005D - T)
Which is close to the original one for P as in b) , except that I get a positive instead of negative 0.005D in the "Divisor" term on the Right hand of the equation.
Given, I may have made a repetitive mistake, as my algebra is a bit rusty, or I could be misunderstand either or both the original formulas as specified for a) and c)

Another simpler formula I found on the Internet ( :-[ I didn't jot down the reference site!) is
f) P = 2ST/D
with the source specifying that this is an "Ideal" (Hypothetical and in the Ideal World) formula for "Thin Wall tube" usage, where the internal stresses/strengths and deformations between the difference of the outer diameter and inner diameter of the cylinder being pressurized is not taken into account.

The formula in f) is visibly close to the one provided in a) and I suspect the "complications" in the formula in a) compared to f) are due to incorporating factors for deformations and inherent material properties. However, when calculating P from formulas e) and f), the result is very close.

So, am I barking up the wrong tree ? - Did I miss something obvious ? - I'd really like to get a better understanding of this particular facet of boiler design.

Any input, suggestions and help will be most welcome - even if somebody wants to bash my head in for being a nitwit ;D

Kind regards, Arnold
(And apologies If this lot gave anybody a headache!)

 

[Maryak]

Arnold,

Starting with Sandy's formula

T = (PxD / 2xS + P) + 0.005 x D

Then

T = {PD / (2S + P)} + 0.005D

Using 1/2" copper pipe as an example and a boiler pressure of 125psig - 1/2" pipe has an OD of 5/8"

T = {125x0.625 / (2*893 + 125)} + 0.005 x 0.625

T = { 78.125 / (1786 + 125)} + 0.003

T = { 78.125 / 1911} + 0.003

T = 0.041 + 0.003

T = 0.044................ Schedule K has a wall thickness of 0.049 and L 0.040 so K it is.

Transposing for P.......... I get

P = [(2T - 0.01D) x S] / D

P = [(0.098 - 0.006) x 893]/0.625

P = 0.093 x 893 / 0.625

P = 83 / 0.625

P = 133 psi

Which tallies with assumed pressure of 125 psi and wall thickness of 0.049" for "K" pipe.

Hope this helps (my algebra is more than a little rusty, it's almost jammed solid, I'm going to sit in the oven for a while)

Best Regards
Bob

 

[arnoldb]

Thanks Bob ;D

Yes, it helps; it seems the formulas come out pretty much the same, and that any minor differences in final calculations could be disregarded - erring on the side of safety of course.
I'll do some more research though - I'm finding this all very interesting - makes me wish I'd studied mechanical engineering...

At least I saw you got out of the oven; there's just a lamb shoulder with red wine roasting off in there now!

Kind regards, Arnold

 

[Dan Rowe]

Arnold,
Good question and and thanks for the link to the thread with SandyC's flue formula. Can anyone supply further reference to that formula?

I checked all the model boiler books in my library.
K.N. Harris Chapter 8 pg. 135 flue tube recomendations.

Martin Evans states that the shell formula should not be used to determine the flue size because they need to be made thicker because of abrasive effect of combustion gas. He gives a table of recommended thickness for flue tubes for service between 80 and 120 PSI.

Tubal Cain in "Model Engineers Handbook" summarizes Martin Evans and gives suggestions for large fire tubes as found in Cornish and Lancashire boilers.

Kozo Hiraoka in "Safety of Copper Boilers" does not mention flue tubes.

The AMBSC Code Part 1 has tables of minimum thickness for flue tubes and water wall tubes, 3.10.1 and 3.10.2 respectively.

The only other model boiler code I know about is the code for Maryland, and no mention of the flue sizes or calculations are listed in that code.

As for the simpler formula you listed - that is the formula for tangential stress on a thin walled tube. The same formula can be used for internal pressure and external pressure. The difference to the formula and the normal one used for boiler calculations includes factors for tubes with joints and corrosion and temperature factors.

The simpler formula is found on a lot of engineering sites including this one:
http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/thin-walled-pressure-vessels
Note that the tangential stress is twice the longitudinal stress.

Dan

 

[doubletop]

T = {125x0.625 / (2*893 + 125)} + 0.005 x 0.625

Bob

I think your calc is for 1/4" tube. I can't find a copy of the SandyC's calcs but isn't that the cross tubes he was calculating?

EDIT - this is nonsense

Arnold

If you can take a look at Tubal Cains Model Engineers Handbook

Section 10 has a sub section on flue tubes. The issue is that tubes under compressive pressure act differently to those under expansive pressure. So whereas a safety factor of 8 (which is what I think SandyC used giving an 'S' of 893) may be appropriate for a boiler barrel it isn't for a flue tube.

From the book the fail pressure for different tube sizes is given as

3/8" 1600Lbf/sqin
1/2" 1150Lbf/sqin
5/8" 900Lbf/sqin
3/4" 650Lbf/sqin
7/8" 460Lbf/sqin
1" 330Lbf/sqin

Although this is for long tubes it serves to show it is non-linear relationship and the reason for putting in cross tubes in the flue is not only to increase the thermal efficiency.

Pete

 

[Maryak]

Bob

I think your calc is for 1/4" tube. I can't find a copy of the SandyC's calcs but isn't that the cross tubes he was calculating?

As far as I recall 0.625 = 5/8" which is the OD of 1/2" pipe.

Copper has a nominal tensile strength of 25000psi with a FOS of 8 TS = 3125psi the fiddle factor for tubes/flues is 3.5 giving near enough to 893psi.

Hope this helps.

Best Regards
Bob

 

[doubletop]

You are right what am I thinking??? (1/4" = 6.35mm not .625" ???)

Back into my box.........

Pete

 

[Dan Rowe]

In the thread linked by the first message SandyC said that the formula for flues was a simplified formula but adequate for model boiler work. I am searching for the complete formula and what boiler code it is from.

The additional factor of adding .005D is similar to the ASME factor for corrosion allowance as that is also added.

It in not the collapse pressure that we are concerned with as that figure is much higher than the safety margin boilers should be built to. Here is a link to online collapse pressures for a reality check.
http://www.copper.org/applications/plumbing/techref/cth/cth_3design_gencon.html
About halfway down under Collapse Pressure of Copper Tube click on "Figure 2".

Dan

 

[Maryak]

Dan,

Found this on the net

Pc = (2*E /(1-M^2))*(t/D)^3
Where E= Elasticity [Young] modulus, M= Poisson ratio, t= wall thickness, D= external pipe diameter

Pc being the collapse pressure.

Now all I have to find is Youngs Modulus for annealed copper and its Poisson ratio. : :

Best Regards
Bob

Edit: Young's Modulus for Copper is 15.7 x 106 psi

Poisson Ratio is 0.34

So Pc = [2 x 15700000 / (1 - 0.342)] x (0.49/0.625)3

Pc = some 3456 psi - A little above model boiler working or test pressures.


Table 9.3 Code Formulas for Calculation of Vessel Component Thickness
Cylindrical shell t = PR/SE¡0:6P
Hemispherical head or spherical shell t = PR/2SE¡0:2P
2:1 ellipsoidal head t = PD/2SE¡0:2P
Flanged and dished head t = 1:77PL/2SE¡0:2P
Flat head t = dpCP=SE
where
t = Minimum required thickness (in.)
P = Design pressure (psi)
R = Inside radius (in.)
S = Allowable stress (psi)
D = Inside diameter (in.)
L = Inside spherical crown radius (in.)
E = Weld joint efficiency factor, determined by joint location and degree of examination
C = Factor depending upon method of head-to-shell attachment

 

[arnoldb]

Bob, Dan & Pete - thank you very much for your input :bow: :bow:

Pete, I saw Tubal Cain's graph, but found it lacking in additional information (tube thicknesses etc) - that's part of the reason I posted this thread Please get out of that box - it's reserved for the dog

Dan & Bob - :bow: :bow: :bow: - thank you for going the extra mile on the research!

Things are clear for me now; and there is now satisfactory information available for me to plonk formulas and values in a spreadsheet to "play" with design parameters ;D

Bob, the last research and formula you posted clinched things for me. The Pc calculated is the theoretical one without safety factors, and if I factor in the safety margins (first a margin of 8 for normal internal pressure and then another 3.5 on top of that for flues - resulting in a total safety factor of 28 for flues), I get 3456/28 = 123psi. Given, this is a linear conversion, but still very close to what was derived from the original formulas ;D

Given the high safety factor involved, I'd say these are safe and completely usable for boiler design - or to see what would be a safe operating pressure working back from available material, as is the case with me currently ;D

Once again, thanks very much guys!

Kind regards, Arnold

 

[Maryak]

Arnold,

Your very welcome. It was for me very enlightening to delve a little deeper into the whys and wherefores of the formulae. Not that I doubted Sandys work, just a small quest for a little more understanding.

Thanks for posing the questions. :bow:

Best Regards
Bob

 

[Dan Rowe]

Arnold,
I agree with Bob this was a very interesting topic and I need to think about this a bit more.

If we take Bob's example of a copper tube 0.625 OD and 0.049 wall thickness and check that on the copper.org charts for collapse pressures we get about 5,500 psi for drawn copper and about 630 psi or so for annealed copper. This gives us a safety factor of 5 with a 125psi working pressure and annealed copper.

SandyC's formula is more conservative than the AMBSC code. The AMBSC code is nearly the same as Martin Evans flue size recommendations.

I missed it in Harris he gives recommendations in the flue layout chapter and he gives a range of gauges for flue tubes.

Dan

Edit:
Bob I was going to ask what boiler code you gave the formulas from and I googled the title phase and found this link:
http://faculty.washington.edu/vkumar/me356/pv_rules.pdf
The boiler code is ASME 1994 edition.

I googled my ASME reference and found this:
http://www.hrsgdesign.com/asmecalc.htm

 

[ianjkirby]

Hi Arnold,
Your maths is mostly ok, with the exception of the following;

Sandy's formula for calculating the maximum operating pressure for a flue is:
a) P = S [2xT –0.01 x D / D - (T - 0.005 x D)]
I'm not entirely sure how this formula is to be interpreted as written (No offense to Sandy !) - I _think_ it could be:
b) P = S (2T - 0.01D) / (D - T - 0.005D) ?

In (b), the divisor should be D - T + 0.005D.

While it has been mentioned that the same expression may be used for external pressure as well as internal pressure, and I agree with this, no-one has mentioned one important factor.

The tubes used for flues must be hard drawn, for the reason that they MUST be perfectly round and straight in order to develop their full strength. Internal pressure tends to force the tube to straighten and become round, essentially becoming stronger as the pressure rises. Any variation from round and straight will weaken the tube to external pressure, leading to potential collapse. I forget now which is the thickest tube (of type K, L, or M), but use the thickest you can get.

Regards, Ian.

 

[Maryak]

Bob I was going to ask what boiler code you gave the formulas from and I googled the title phase and found this link:
http://faculty.washington.edu/vkumar/me356/pv_rules.pdf
The boiler code is ASME 1994 edition.

Dan,

Yep that's the one I found. ;D

Best Regards
Bob

 

[GWRdriver]

Hi Arnold, . . . . I forget now which is the thickest tube (of type K, L, or M), but use the thickest you can get. Regards, Ian.

Ian,
You have it right, and in order. K = thicker, M = thinner.

. . . and the reason for putting in cross tubes in the flue is not only to increase the thermal efficiency. - Pete

I have to disagree with this statement, at least with the implications of it as I read it. Certainly cross tubes will increase the resistance to collapse in a flue but I've never heard it referred to as The reason. The primary reason is thermal efficiency and generally any staying aginst collpase they might is never mentioned . . . well, until now. In boiler designs with cross tubes, which are invariably small, first regard is typically given to the number of tubes and balancing maximum heat absorption against minimal obstruction of gas flow, and second is the relationship of the cross tubes to the heat source so as to minimize the potential for overheating and burnout. The usual practice is to move the cross tube "nest" forward in the flue, away from the fire. There is no benefit to having the cross tubes flues in or very near the fire, at least when balanced against the potential problems. (Firebox thermic siphons are a different matter.) In order for there to be any benefit the cross tubes would have to be equally spaced axially and radially along the entire length of the flue subject to pressure, otherwise the flue would be only partially stayed which is tantamount to having no staying. The pressure would find the path of least resistance.

 

[Dan Rowe]

Here is one of the ASME formulas for thin wall tubes from:
http://www.copper.org/applications/plumbing/overview/cu_alloy_tube_pipe.html
http://www.maksal.com/Tubes6.aspx

Barlow formula for thin walled, hollow cylinders:
P = 2Stm/(D-.08tm)
where
P = allowable pressure
S = allowable stress
tm= wall thickness
D = outside diameter

The value of S is the allowable design strength for continuous long-term service of the tube, as determined by the ASME Boiler and Pressure Vessel Code, Section

Table 3. Allowable Stresses for Copper Tube
as a Function of Temperature

Temperature (ºF) Allowable Stress (psi)
Annealed Drawn
100 6000 10300
150 5100 10300
200 4900 10300
250 4800 10300
300 4700 10000
350 4000 9700
400 3000 9400

I have not found how the ASME code handles external pressure on a copper tube, but I have found nothing to suggest that the allowable stress should be divided by 3.5 as SandyC did in his calculations.

Here is a page with flue gauge recomentations which uses tubes a little thicker than Martin Evans recomendations.
http://www.alanstepney.info/page18.html

Bob I found your formula but only here, most of what I have found on this subject is very complicated with tests graphs and a whole lot of greek letters.
http://wiki.answers.com/Q/What_is_t...pe_using_diameter_wall_thickness_and_material

My copy of Marks led me to "A STUDY OF THE COLLAPSING PRESSURE OF THIN-WALLED CYLINDERS" by R.G Sturm 1941. This is free on the web as a PDF.

Dan

 

[arnoldb]

Once again gentleman, thanks for all the responses!

Ian, thanks - at least I got the formula right when I worked it back :big:. As to only hard drawn tubes for flues, I agree that flue tubes should be drawn and as round and straight as possible. However IMHO, the process of silver brazing them will anneal them taking the "hard" part out of "hard drawn". I'd go as far as to say that for a silver brazed boiler, all calculations should be done based on annealed values for all parts. The only exception to this I can think of would be in the case of rolled flues - e.g steel boiler shell with copper flues rolled in.

Dan, I think SandyC use the additional 3.5 safety factor for flues to compensate for any out-of-roundness or defects that could be present in the flue tubes, as well as to compensate for possible corrosion from firing through the flues. I'm not sure if it is a requirement of the UK Code perhaps; but I do know that SandyC made/makes UK certified boilers, so I'm happy to add that factor.

To quote Bob:

Not that I doubted Sandys work, just a small quest for a little more understanding

That says it in a nutshell for me; I didn't doubt Sandy either; but I have an inquisitive mind, and like to know why things are done - not just how

Kind regards, Arnold

 

[ianjkirby]

Hi Arnold,
Yes, you are correct in that the soldering will soften the tubes, but you should select hard-drawn tubes to ensure you are inserting straight and round tubes. You might be surprised how many people ask "Why can't I just straighten out some coiled tube?" which is easily obtainable from the local air conditioning tradesman or scrap metal dealer.
Regards, Ian.

 

[Dan Rowe]

The strength of the tubes was covered in this thread but not the diameter to length ratio. The only thing I have found on the L/D ratio is in Martin Evans "Model Locomotive Boilers". He states the ratio of suscesfull boilers have a ratio of 50 to 70 times the square of the tube inner diameter. He chose 65 for his calculations.

For further information see:
http://www.model-engineer.co.uk/forums/postings.asp?th=34610

Dan

Edit: Harris also covers this with a very similar formula in Chapter 8 in the tube layout section.