Could some kind soul tell me how to calculate the steam requirements of a steam engine - let us say a twin cylinder, double acting, 2 in. bore x 2 in. stroke operating at 80 psi and 500 rpm- in lbs./hour? Thanks in advance for any knowledge offered. Mike in BC
On or around Tue, 03 Apr 2007 03:29:29 GMT, michael gray enlightened us thusly:
should be able to get a ball-park figure...
lessee... 2xpi² for cylinder volume... 19.7 ci. x2 again 'cos it's double-acting, x500 and x60, 1,184,352 ci of displacement per hour. That's the maximum you can get through it in terms of volume... /12/12/12 will give it in cu ft... 685, more or less.
now, if you know how much a cu ft of steam @80psi weighs, that should be your maximum consumption; however, on most engines, you'll be using part live steam and part expansion (there's a proper term for that which I forget).
aha.
formatting link
says 4.67 cu ft per lb at 80 psig, so that 685 / 4.67 gives us about 147 lb per hour, max comsumption.
now watch someone come and say that's all bollox.
The figures sound high, but then again, try working out the airflow through an IC engine... which is why forced induction on same takes a noticeable amount of power.
Woo hoo, I got something right, as opposed to the frack to bunt clack valve I spent 3 hours on today, maybe I'd better stick to the sums, nah, swarfs much more fun!
Now who's going to take it further and deal with percentage water content and superheating!
Don't get too excited my comment only applied to your arithmetic re the cylinder! Long since forgotten what little I understood about steam calculations.
I'd endorse the calculation derived from Tubal Cain's book - and the thought that it's well worthwhile getting a copy.
Note that the answer will relate to the engine operating continuously under load. The same engine will use much less steam if the load is removed and the steam supply throttled back to maintain the same rpm.
On or around Tue, 03 Apr 2007 19:08:00 GMT, "Dragon" enlightened us thusly:
yeah, you're right.
what was it, 2" bore by 2" stroke, double acting?
working swept volume is thus 2* stroke * area, area = pi-r-squared,
but in this case r=1 so area = pi, thus swept volume = 4xpi. I think I can claim a draw: Andy P has an extra factor of 2 in his...
12.56 ci per revloution, x500 x60 = 376991 ci per hour, /12³ = 218 cu ft per hour, /4.67 = 46.7 lb per hour max. if it's not expanding.
what IS the proper word for that - there's a big lever on steam engines (well, on the ones I've played with) which adjusts where in the cycle the inlet valves are closed - you start off using live steam for most-or-all of the stroke, then when it's up and running you shift the big lever a notch or several and use less steam by letting it expand in the cylinder(s).
I think that's called expansive working. When I was playing on 0-4-0s in the sixties I was told it was the gear lever but it only ever had 2 positions for us - all forward and all reverse! teenagers eh! Come to think of it the regulator was a bit like that too!
Tubal Cain estimates that the actual steam consumption figures will be around 50% of those calculated for model size engines.
Of course this really is all book knowledge 'cos I've built nowt yet!
It's all to do with the valve timing gear,which controls both reversing and "cut-off". The power stroke is initially at constant pressure, inlet valve closes, then (hopefully) adiabatic expansion until exhaust valve opens and constant pressure exhaust. Incredibly inefficient to just use steam pressure to push the piston like an hydraulic motor. There's a concept of "expansion ratio" and efficiency (what little there is) varies with expansion ratio. Cut off varies with just about every other parameter, expect between 10 and 50%.
There's an assumption in the original post: that if fed with steam at
80psig then it will spin at 500rpm. It will spin up until the power absorbed by the load equals that generated by the engine. To run at a specific speed you need a governor to control the steam input. How much steam therefore depends on the load (no surprise here), not on swept volume*speed. If you know the hp of the engine then the approx steam consumption is available from tabulated data. For a non-condensing engine at 70psig, no superheat, about 26 lb per ihp-hour.
Actually, it should be isentropic expansion. Adiabatic expansion is where there is no external energy input, like an ideal throttle valve (enthalpy stays the same). Isentropic expansion is where all of the energy that can be extracted is extracted (entropy stays the same).
I don't know the arithmetic, just worked around people that are obsessed with it.
On or around Thu, 05 Apr 2007 18:34:57 +0100, David Powell enlightened us thusly:
nevertheless, there's a maximum amount of steam that can go through the engine under the given conditions. The actual steam requirement will be less, depending on the power required, unless the power required is the maximum possible for the combination of engine and steam pressure. Not that I'm a steam expert.
similar calculations apply to IC engines and forced air feed - for a given erpm there's a certain volume of air going through it whether or not it's actually running, and to increase the inlet pressure above ambient your blower has to supply more than that volume of air.
The swept volume on one side of a cylinder is 2pi - each cylinder is double acting, so the swept volume per cylinder is 2 x 2pi - there are two cylinders, so the total volume is twice that, ie 2 x 2 x 2pi or 8 pi cubic inches.
If the steam is x% wet you will need 100+x % as much steam.
As a first approximation, if the steam is superheated then you will need Tc/Ts times less. Ts is the temperature of the steam in kelvin, Tc is the temperature where the steam and the water are in equilibrium at the pressure used, eg 429 K at 80 psi.
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.