steam requirements for an engine

Mike, this culled from Model Engineers Handbook by Tubal Cain which I use every day!

swept volume = 2pi * 4 per rev = 25.136 cu in

at 500 rpm = 12568 in cu or 7.27 cu ft per min

which is 436 cu ft/hour

steam tables give Volume of dry steam at 80psi as 4.66 cu ft/lb

so 436/4.66 is about 94 lbs/hour.

This is a bit simplistic - ignores wetness of steam, superheating, any leakage etc but it's a start!

At risk of repeating myself get the book - it's brilliant.

Cheers, Andy

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.

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.

I wouldn't be so rude, but try that cylinder volume again! You've 'squared' the wrong bit. Andy's post above got it right.

Henry

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!

Andy

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.

Henry

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.

Wilfrid Underwood

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).

In article , Austin Shackles writes

IIRC, it's called a reverser. Often a screw-driven device, at least on later steam locos.

David

Not so fast.....

Yes the area is pi. So the swept volume above one piston is the stroke times the area, so equals 2xpi.

So a double acting piston has twice that, and two of them has twice again. So the total volume is (2xpi) x4

So Andy was right, after all.

Wilfrid Underwood

PS and the reversing lever is often called the "pole"

Thank you all for the help - I really do need to get Tubal Cain's book. Mike in BC

In article , Wilfrid Underwood writes

Never heard of that one. Where are you based, Wilfrid - it could be a locational difference.

David

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!

Cheers, Andy

Notching up :-)

Mark Rand RTFM

South of England.

I must have picked it up from the Model Engineering club - or perhaps ME magazine.

I googled "pole reverser" and got a link to StationRoadSteam in Lincoln and one to an article on Terriers - Southern again!

So perhaps I should have said "sometimes" instead of "often".

Wilf

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.

Regards,

David P.

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.

Mark Rand RTFM

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.

On or around 4 Apr 2007 09:17:49 -0700, "Wilfrid Underwood" enlightened us thusly:

twice 2 x pi is 4xpi, not 2 x pi x 4. 2 x pi x 2...

and four times 2 x pi is 8 pi.

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.