
Hi PD's
This is fun is it not?
Firstly, Moritz, I am unable to furnish you with much data on the ‘Waverley’s’ paddle wheels other than they are 18ft (5.48metres) dia.
This, and other info quoted was obtained from the ‘Waverley’ website, but they don’t provide much more detail.
Derek’s PS DECOY calculations.
Now that is interesting Moritz, as we now appear to have got the paddle wheel rotation speeds into the generally accepted area, more or less,… i.e. 91 – 166 RPM.
So it would appear that just using the scale square root factor, as in my original calculations, does not provide a satisfactory wheel speed for a model.
It is very clear that much more account must be taken of wheel dia, linear speed/ circumferential speed etc, in order to arrive at a much more accurate figure for our model.
OK, now that we have some more accurate rotation speed figures to work with I thought it might be a good idea to use these to assess the steam usage, and also see the effect of changing something like, say working pressure, so: -
Lets take a look at the requirements for both 6:1 and 8:1 ratios, since these appear to be the preferred ones.
Since my steam tables are imperial, I will take the liberty of using this for the following calculations, hence 10mm bore and stroke become 0.3937†bore and stroke.
And, therefore, volumes etc will be in ‘cu ins’ rather than ‘cc’.
First we need to find out how much steam is required for each cylinder full of steam.
Cylinder swept volume = Bore area x Stroke
= pi x 0.3937â€2 /4 x 0.3937†= 0.048 cu in.
This is the volume of steam required to fill the cylinder when the piston travels from one end to the other (1/2 revolution)
Since we have double acting cylinders (same amount of steam is required to drive the piston back again) so we need to multiply this by 2
= 0.048 x 2 = 0.096 cu in per revolution.
We also have 2 cylinders so again we need to multiply x 2
= 0.096 x 2 = 0.192 cu in per revolution.
So, assuming 100% steam cut-off (worst case) and probably more useful for small model engines anyway.
If the engine is of the oscillating type (or of the piston valve type using port reversal as a means of reversing the engine) this will be the case anyway, since these would have 100% cut-off (as near as) by design.
Slide valve engines can use early cut-off between (40% and 85% typically) they would certainly be operating at the higher end of the range for a good part of their work, so using 100% is not as big a problem for the calculations.
So how much steam is required?
Ok, this is where the paddle wheel gearing, and hence engine speed plays the major role in the calculations.
taking the figures provided/calculated by Moritz...
If we look at the 6:1 ratio: -
Wheel speed = 129rpm
Engine speed = 773 rpm
So for this case we need to multiply the revolution volume by the required rpm.
In this case this becomes = 0.192 cu in x 773 = 148.416 cu in/minute
Which is 8904.96 cu in/hr
For the 8:1 ratio: -
Wheel speed = 149 rpm
Engine speed = 1191 rpm
For this case the steam consumption becomes: -
= 0.192 x 1191 = 228.67 cu in/minute
Which is 13,720.2 cu in/hr
So what do these figures mean for the boiler?
How much water do we need to turn into steam? And at what rate?, in order to satisfy the above requirements.
The first criteria is the WORKING PRESSURE, since this has a big effect on the volume of steam available for each cu ins of boiler feed water evaporated.
With a working pressure of 1.5bar = (approx 22psi)…….. (why so low Derek)?
Each cu in of evaporated water can provide 691 cu in steam.
At 3bar pressure (45psi) this figure changes to 437 cu in steam.
We will look at the effect of this difference a bit later on.
To return to our calculations: -
To satisfy the 6:1 case (148.416 cu in steam/min) at a working pressure of 1.5bar (22psi) the boiler will be required to evaporate: -
148.416/691 cu in water per minute
= 0.214 cu in water/minute.
= 12.84 cu in water/hour.
Which is approx equal to 209.959 grams/hour.
For the 8:1 case (228.67 cu in steam/minute) at the same pressure the boiler will be required to evaporate: -
228.67/691 cu in water per minute.
= 0.330 cu in water/minute.
= 19.8 cu in water/hour.
Which is approx equal to 323.77 grams/hour.
The rate at which a boiler can evaporate water is directly proportional to the area of heated surface available, and for model boilers it is usual to use the figure of between 1 and 3 cu in per minute per 100square in of heated surface.
For a simple pot boiler the lowest figure would be use, whilst for a multi water tube boiler of, e.g. Yarrow type, the higher figure would be more appropriate.
For the purpose of this discussion I will use the figure of 1cu in evaporated per minute per 100sq in Heated Surface. i.e. the worst case.
So: - for the 6:1 case the required heated surface would be: -
Heated surface sq in =100/1 x 0.214 = 21.4 sq in.
And for the 8:1 case: -
Heated surface sq in =100/1 x 0.330 = 33 sq in.
NOTE WELL These figures represent the minimum heated surface requirements the boiler must have in order to meet the steam requirements at the required pressure if the evaporation rate is 1 cu in per minute per100 sq in heated surface.
Lets take a quick look at the effects of increasing the working pressure.
Lets take 3 bar (45psi) as the new working pressure: -
In the 6:1 case the cu in water/minute evaporation rate required would change to: -
148.416/437 cu in water per minute
= 0.339 cu in water/minute.
And the required HEATED SURFACE would increase to: -
100/1 x 0.339 = 33.9 sq in.
For the 8:1 case the cu in water/minute evaporation rate required would change to: -
228.67/437 cu in water per minute
= 0.523 cu in water/minute.
And the required HEATED SURFACE would increase to: -
100/1 x 0.339 = 52.3 sq in.
Derek’s boiler has a HEATED Surface area of approx 45sq in.
Therefore, given the above figures it would be ok for the 6:1 case
but would fall well short requirement of the 8:1 case. OK DEREK , PANIC OVER, you have nothing to worry about, since your boiler has an evaporation factor of approaching 2 cu in water per 100 sq in heated surface.
And anyway, you are running at 1.5 bar (22psi) not 3 bar (45psi), however, the figures do serve to illustrate that any given boiler has a maximum conversion rate and that it is not always obvious, to the un-initiated, why a seemingly small change (like increasing the working pressure) should make such a difference.
WARNING….Never, under any circumstances, change the working pressure (safety valve setting) of your boiler to a setting which is higher than that for which it is designed.
To do so could lead to a boiler failure and possibly a nasty accident.
Similarly, never attempt to use a larger burner in order to get more steam…. This will certainly overload the safety valve, which will not then be able to vent off overpressure steam volume fast enough to stay within the boiler certification requirements.
The burner fitted to your boiler, by the manufacturer, is carefully matched to ensure this requirement is met.
OK: -, to tie this together with Moritz’s figures for the 1.5bar (22psi) case lets adjust for 50% cut-off.
This is a simple matter of halving all the above results: -
Total cylinder Steam required was 0.192 cu in steam/rev.
This now becomes 0.096 cu in/rev.
So for 6:1
@773 rpm total steam volume required becomes 74.208 cu in/minute
Evaporation rate required becomes 74.208/691 = 0.1074 cu in water per minute.
= 6.444 cu in water/hour
= 105.372 grams/hour
and for the 8:1 case: -
@1191 rpm total steam volume required becomes 114.335 cu in/minute
Evaporation rate required becomes 114.355/691 = 0.1655 cu in water per minute.
= 9.93 cu in water/hour
= 162.375 grams/hour
These figures relate to dry saturated steam.
I will let you work out the ‘Heated Surface’ requirements; however, this should not prove to difficult.
Ok enough already

Keep happy.
best regards.
Sandy
