;D
I agree, the engine is much nicer to look at than a load of numbers, however, since you asked!!!
here are a few: -   1, 17, 5, 36, 19, 45....... can't give you any big ones, Bog's has used them all up adding at least 4 zero's to the National debt ;D ;D ;D... just kidding John.... good to see you are getting back on track.
Ok, try this for size: -
For boiler shell/flu calculations the applied safety factor is generally accepted a being 8, which results in an accepted tensile strength of: -
25,000/8 = 3,125psi = Tensile strength (t)
1. The formula for plate/wall thickness is given as: -
T = P x D / 2t
2. And for Pressure as: -
P = 2T x t / D
Where D = Exposed dia
t = Tensile strength
T = Thickness
And P = Working pressure
As an example let’s take the max working pressure for the boiler to be 90psi
Minimum wall thickness required.
Using formula 1. (Above)
NOTE…. D is expressed as the EXPOSED dia….. this means the dia actually exposed directly to the steam pressure.
In the case of an outer barrel, this would normally be the INSIDE DIA.
In the case where the ID is unknown, which is the case when the wall thickness required is an unknown, it is quite satisfactory to use the known OD to make a first order calculation. The resulting value/figure will be slightly higher than actually necessary, however, since the thickness required increases with dia for any given pressure, the small error will be in favour of safety.
This can be verified in a subsequent pressure calculation.
To return to our calculations: -
Boiler outer shell thickness required.
Given an OD of 4” this becomes: -
T = 90 x 4  / 2 x 3125  = 360 / 6250  = 0.0576”  
       
Which is just a thousandth of an inch or so over 17swg.
To prove the case for the small error described above: -
Correction to D for calculated thickness.
This gives the correct ID for the calculation.
ID = OD – 2 x calculated thickness  = 4” – 2 x 0.0576”  = 4” – 0.1152” = 3.8848”
Using formula 2.
P = 2T x t / D  = 2 x 0.0576” x 3125 / 3.8848”
= 0.1152” x 3125 / 3.8848” = 92.668psi.
Which, as you can see, is a little higher than required, however, it is, as stated above, on the safety side.
Ok, as calculated, the gauge of material required is just a little over 17swg.
This is not an easy thickness to source; therefore, it would be more usual to select/use a barrel material of 16swg (0.064”
since this would be the more readily available material.
NOTE: - In such cases, always go to a higher, more readily available, thickness. Never go lower.
Using this thicker gauge will result in the following pressure capability: -
D becomes 4” – 2 x 0.064”  = 4” – 0.128”  = 3.872”
P = 2T x t / D  = 2 x 0.064” x 3125 / 3.872”
= 0.128 x 3125 / 3.872”  = 103.3psi.
Which gives an even greater safety margin.
The portable engine design you are contemplating appears to be a 2 cylinder compound type which would probably be operating at around 120psi – 150psi.
Taking this at its highest level (150psi) and using a barrel dia of 5” OD, then a first order calculation would yield the following: -
Thickness required (T)
T = P x D / 2 x tensile strength  = 150 x 5” / 2 x 3125
= 750 / 6250  = 0.120”
This lies between 11swg (0.116”
and 10swg(0.128”
So choose 10swg or thicker.
Ok Rich, I think this should be sufficient for you to see how the numbers fit with the formulae, however, if anything is still not clear then just ask.
Best regards to all who are still awake..... ;D ;D ;D ;D ;D ;D ;D
Sandy
