Hi Guy's,

For the benefit of any of you that may be following this thread here is the main text of the information I sent to Rich (firebird).

He asked the question... did I think a PYRTE type boiler would be suitable for a 3 1/2" gauge locomotive based on Martin Evans 'Conway' narrow gauge locomotive model...

Short answer... NO... not much chance at all since it just does not have sufficient ‘Effective Heated Surface’ this being only approx. 38 - 40 sq ins**. (at best 90sq in with studs fitted)

** approx. Effective Heated Surface area = half circumference of barrel x length of firebox.

PYRTE boiler is 4" in dia. and the firebox is around 6" long and is fired from underneath similar to a 'Mamod' style boiler.

Long answer, to clarify the above... and to answer SOME of Julian’s request for calculation methods... Boiler design is complex I am afraid with a huge number of variables.

'Conway' cylinders are, if I recall, 1 1/4” dia with a stroke of 2”.

Piston area x stroke x 2 for double acting x 2 for 2 cylinders = Total swept volume per revolution.

: - The total swept volume per revolution is 9.8176 cu in per rev.

Assuming a max wheel speed of around 100rpm, since the cylinders are directly coupled, this would amount to 981.76cu in steam required per minute (total swept volume x rpm).

How much steam a boiler can produce is a product of the working pressure, it’s conversion factor (type specific) and the available effective heated surface area (HS).

The volume of steam produced by a given volume of water is determined by the working pressure...

Assuming a working pressure of 80psi (as per ‘Conway’) then 1 cu in of water will produce 289.65 cu in of steam.

(Note... this is calculated from steam tables which for a given working pressure give specific volume of steam in cu ft/lb x 1728 (cu in per cu ft) to give cu ins/lb divided by 27.741 (since 1 lb of water occupies 27.741 cu in) which gives cu in steam per cu in water.)

: - 981.76 / 289.65 = 3.3895 cu ins water need evaporating per minute.

The PYRTE boiler type even with studs fitted would have a maximum conversion factor of x2 (even if fired very hard... and that is being generous) typically it would be more like x1.5.

The conversion factor is the number of cu in water evaporated per 100 sq in HS a particular type of boiler can evaporate per minute.

The required heated surface required would then become: -

HS required = the number of cubic inches of water to be evaporated per minute x 100 / conversion factor...

therefore heated surface required = 3.3895 x 100 / 2 = 169.475 sq in HS required for a x2 conversion factor.

At the more typical x1.5 conversion factor this becomes: -

3.3895 x 100 / 1.5 = 225.966 sq in HS required.

You only have around 40 sq in HS.... and even with studs added it would not exceed 90sq ins at best... so either way you are well short I am afraid.

You could choose to reduce the working pressure which will help a little, however, you will loose tractive power if you go much below 60psi.

At 60 psi 1 cu in water will give you 363.753 cu in steam.

So evaporated water required becomes 981.76 / 363.753 = 2.6989 cu in water per minute.

: – Heated surface becomes 2.6989 x 100 / 2 = 134.945 sq in HS. for a x2 conversion factor.

or 2.6989 x 100 / 1.5 = 179.926 sq in HS. for the more typical x1.5 conversion factor.

So... still well short of HS.

If you consider the actual heated surface available then the actual number of cu in water evaporated per minute becomes... (Available HS in sq in / 100) x conversion factor cu in per minute.

e.g.... for 40sq in heated surface you would only be able to evaporate (40/100) x conversion factor cu in per minute.

hence for a x1.5 conversion rate this becomes (40 / 100) x 1.5 = 0.6 cu in water evaporated per minute rather than the 1.5 cu in per minute per 100sq in HS required... i.e. well under half the required amount.

When considered in terms of the actual conversion rate required at 80 psi being 3.3895 cu ins water per minute then this 0.6 cu in per minute clearly indicates the effect of having too small a heated surface area.

Similarly for the 60psi case being 2.6989 cu in water per minute required then again the 0.6cu in per minute is clearly not going to work.

I hope this is of some help to anyone following.

and to everyone... Have a very Happy Christmas and a peaceful New Year.

Sandy.