Posted by Gary Wooding on 18/11/2015 08:05:25:
Posted by Andrew Johnston on 16/11/2015 23:03:13:
The HP cylinder diameter is 2", stroke is 3.75" and assuming a cutoff of 80% and a maximum speed of 600rpm we can calculate the volume of steam used per minute. I get:
0.185m³/min
Sorry, can you explain that bit please?
Cylinder diameter is 2", fixed by the design
Stroke is 3.75", fixed by the design
Cutoff is 80%, this is arbitrary, but is what I will be aiming for when I re-design the valve gear
Maximum engine speed is 600rpm, probably a bit high even for a 4" scale engine, but one has to put a peg in the ground somewhere, and it's what I used when re-designing the water pump
We only need to consider the HP cylinder since the LP cylinder steam comes from the HP cylinder, not from the boiler. Of course we're ignoring use of the simpling valve, but that is only used very short term.
The volume of a cylinder is:
V = pi x radius squared x length
So for one stroke (TDC to BDC or vice versa) the swept volume is:
pi x 1² x 3.75 = 11.78 in³
But with a cutoff of 80% steam only enters the cylinder for 80% of the stroke so the volume we're concerned with is:
11.78 x 0.8 = 9.423 in³
The engine is double acting, so steam fills the swept volume twice per revolution, so:
9.423 x 2 = 18.85 in³
Multiply by 600 (rpm) to get the swept volume per minute:
18.85 x 600 = 11309.73 in³/min
I'm too idle to look up the conversion factor from in³ to m³, so I multiply by 25.4³ and divide by 10^9 to get cubic metres:
11309.73 x 25.4³ / 10^9 = 0.185m³/min
Of course there are all sorts of ifs and buts that can be made against my assumptions, but you have to start somewhere, and the calculations are only a marker to size the injector.
Andrew