Hard info' about model engineering sizes of tube sought
|257 forum posts|
I am interested in finding out what kind of pressure copper tube in the dimensions used in model engineering can withstand. I have looked in the usual suspects (AMBSC Code, Model Engineer's Handbook, & internet) in an attempt to find the maximum safe pressures without success.
My utilisation is as follows - but I am also interested in the general information for future use:
Half-hard copper tube 3/16 O/D for use in locomotive (cold) water boiler feed. Boiler max pressure 100lbs sq. in. Tube wall thicknesses available: 26, 22, 20, 18, 16 swg. There would be a significant advantage to using the thinnest tube possible.
Reference to this table & imaginative extrapolation would seem to indicate that 26 swg will more than adequate. Is this correct... or can you advise otherwise, or even give exact maximum safe pressure information for 'our' sizes of tube??
Edited By Weary on 12/01/2019 15:13:48
Edited By Neil Wyatt on 13/01/2019 14:22:18
|Brian H||12/01/2019 15:17:43|
956 forum posts
Half hard tube would lose some of its properties if silver soldered.
A good book is "Model Boilers & Boilermaking" by K.N.Harris.
|Clive Brown 1||12/01/2019 15:30:11|
|185 forum posts|
!00 psi in 3/16" dia tube with .018" wall thickness gives a hoop stress of ~500psi. Peanuts really.
|Keith Long||12/01/2019 15:43:58|
|776 forum posts|
If you download the publication in this https://www.copper.org/publications/pub_list/pdf/copper_tube_handbook.pdf it will probably give you all the information that you need as well as a load more that you didn't know that you wanted.
|3601 forum posts|
Barlows Formula for the burst pressure in psi of a pipe is:
P = (2 x wallThickness x tensileStrength) / outsideDiameter
The tensile strength of copper varies between 36000psi and 47000psi. Unless you know better use the lower value!
For 3/16" pipe applying the formula gives:
As a check the calculated burst pressure of an ordinary copper water pipe would be about 2400psi. This, I think, agrees fairly well with Phil's link to Table Z which gives a working pressure for that sized pipe of 725psi. ie for a cold water pipe, the safety factor is about x3⅓
With a steam pipe you wouldn't want to get anywhere near burst pressure especially as the higher temperature will reduce the metal's tensile strength slightly. I'd apply a safety factor of at least x10.
Applying Barlow with safety factors, for 3/16 o/d 26swg copper pipe an x10 safety factor is 691psi and x20 is 345psi. On that basis I'd say 26swg 3/16" pipe is plenty good enough for a 100psi boiler and any reasonable hydraulic test.
Although the theory looks reasonable to me prefer the opinion of any boiler expert or mathematician who happens to disagree!
Edited By SillyOldDuffer on 12/01/2019 16:20:33
|Neil Wyatt||12/01/2019 16:21:49|
15199 forum posts
Simplifying Tubal Cain's calculation so it only applies to drawn (seamless) tube:
t = (p * d)/2f
t = thickness (mm or inches)
p = internal pressure (N/sq. mm or lbf/sq. inch)
d = diameter (mm or inches)
f = working stress (N/sq. mm of lbf/sq. in)
For f he suggests 0.15 x UTS for and suggests working stresses for steam at temperature.
Model Engineer's Handbook, 3rd ed. P. 10.2
|Andrew Johnston||12/01/2019 16:50:01|
4308 forum posts
The problem with relying on a formula taken out of context is that one doesn't know where it comes from, or what limitations apply. There are three sorts of stress to take into account; hoop, longditudinal and radial. Radial is only applicable is the tube is considered thick walled, ie, tube diameter is greater than 10 to 20 times the wall thickness. Longditudinal is applicable is the tube is under internal pressure, as I suspect a tube on a steam engine will be. The formulae, some calculators, can be found online.
The biggest problem is the value of tensile strength to use. It can vary rather more widely than suggested by SoD depending upon the exact material and it's state, eg, annealed or work hardened. And the copper losses strength at higher temperatures, even with steam at 100psi the loss needs to be accounted for. Unless one knows for certain what state the material is in take the lowest value for tensile strength.
|257 forum posts|
Thanks to all who replied. Your contributions individually and taken together have been very useful indeed to me.
Thanks to Dave and Neil for giving me the relevant formula, and in Neil's case directing me to where it can be found in The Model Engineer's Handbook. I had not considered that for mathematical modelling purposes a tube could be considered to be a long, thin, seamless boiler. Even though Andrew's response indicates that this may be a somewhat crude analogy and that reality is somewhat more complex I suspect that this 'approximate model' will suit my purposes.
I will probably use a thicker tube than the minimum because, as was pointed out, heating to silver-solder end-fittings will result in the tube being annealed rather than 'half-hard'. Study of the document suggested by Keith indicates that this reduces the burst pressure by around 40%. But then, a number of cycles of pressurising and de-pressurising the tube will (work) harden it again ..... so further research and consideration on my part required.
Anyway, I have the basic vocabulary now for better directed searching, so thanks again one-and-all for your assistance.
|Neil Wyatt||13/01/2019 14:24:03|
15199 forum posts
That's why I gave a reference in my post as TC does discuss issues around his formula
|3601 forum posts|
I plead guilty again, I could have said I got Barlow's Formula from Machinery's Handbook (20th edition) and provided a link to it on Wikipedia. It's also discussed on many other websites and engineering books.
I had a go at working out how the formula is derived but didn't get far. I thought it boiled down to the metal rim area times tensile strength, but the numbers didn't work out. Another mystery.
Comparing Tubal Cain with Barlow, I see TC puts a safety factor in the calculation, whereas Barlow doesn't. So Barlow is a sanity check rather than a design goal! (Unless you want the pipe to go bang.)
Opened up another area of general ignorance for me. I'd imagined that applying pressure to soft copper would work harden it, therefore although a soft pipe might balloon first, its burst pressure would be much the same as a half-hard or fully hardened copper pipe of the same size. Now I doubt it.
Barlow is an interesting chap. He invented the Barlow Lens used in astronomy and the spiky electric motor demo we had at school.
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