This is not really a model engine build, as usual.
It is more of a competition to see who can match the specific output of a 125 GP two-stroke in a smaller size.
Regarding accuracy- engines are not that consistent from one run to another.
Unless you are at the top of a mountain , in really cold air or really hot conditions, or the air is a bit moist, the conditions won't significantly affect results- maybe a few percent. - In bike racing and drag racing, they do tend to rejet for the conditions- I wouldn't bother.
A single Arduino microcontroller is enough to get good readouts. There is plenty of software available.
It would be nice to emulate an inertial run to get a good output plot, but I also want to do steady rpm power runs.
I will check the torque range, and see if I can use a cheap scales unit as my load cells.
The motor only makes a couple of newton-metres torque, I think , so it is on the low side.
Calcs - 17850 rpm = 1869 rad/s, 30 hp = 22.3 kw, torque = watts/ rad/s = 22/1.8 = 12.4 n-m. if the lever arm is 0.3m, F = 41.4 kg. That seems to be in a measurable range. I won't get the full power, as I am restricting myself to 95 ron, so maybe 12:1 max CR will be my limit.
I don't think there is much to be gained between 12:1 and 18:1 with a spark ignition engine, anyway.
Torque at lower rpms has to be considered as well. The base engine can make 5 hp at 6500 rpm, although the pipe engine won't do that much.
rad/s = 680, pwr = 3.7 kw, torque = 5.4 N-m, force = 18 kg.
For the volume flow needed, a rough estimate is F = m w (sq) r; (force per unit mass thrown out) torque = Fr , m per second = m(dot) = ; F = T/r, m= T/ w(sq)
r(sq) , r= 50mm,
- this force is actually at right angles. - I may have to follow an energy equivalence argument to get back to torque. J = 1/2 m V(sq) and V = wr so J per rev = 1/2 m w(sq) r(sq)
Now power is the rate of doing work, so then I can convert to w= J/t, where t = time for 1 rev. then I can divide w by rad/s to get torque.
time for 1 rev at 6500 rpm = 9 x 10**-3 sec. (t)
substitute back in to find m
T = J/t = 1/2 x m x w(sq)r(sq)/t /rad/s
We will do the first case: running at 6500 rpm.
we know all these, so substitute in:
or kg/s in here as m applies only for one revolution.
m = 2Txtx w/w(sq)r(sq) -[do cancels]=m = 2Txt/wxr(sq)
so m = 2 x 5.4 x 9 x 10**-3 / 680 x (50) (sq) x 10**-6
9.72 x 10**-2 /1.7 = 5.7 x 10**-2 kg by this argument.
Now multiply again be revs / sec = 5.7 x 10**-2 x 108 = 6.2 kg/s
**** can anyone spot any errors in my substitutions here??*** I see that t and rev/s cancel out.
available water = (roughly) pi x (50 (sq)- 20(sq) x 10**-6 x Width x rps rps = 108.3, vol = 2500 - 400, 2100 x 10**-6 x pi = 6.6 x 10**-3 x 50 x 10**-3
= 3.3 x 10**-4; m(dot) = 3.3 x 108.3 x 10**-4 =3.6 x 10**-2 kg/s,
so there is not enough torque at this speed.
Adjust for case 2 at the higher rpm:
Speed multiplier = 1869(sq) - 680(sq) = 3.5 x 10**6 - 0.46 x 10**6 = 3 x 10**6 , so mass flow rate will increase to 1869/ 680x .0356 = 0.26kg/s
and the required flow rate drops to: m = 2T t /wr(sq) where t = 3.4 x 10**-3 sec; and rps , and t cancel out.
So kg/s = 2T/wr(sq) = 2 x 12.4 / 1.869 x 50 x = 24.8/93.45 = 0.27 kg/s
this is just about enough torque- if we can recycle more water around the loop, we should have enough.
At least we are in the right ballpark for 11,000 to 19,500 rpm expected torque.
A better design would have a shallower sawtooth ring instead of the pockets, as the objective is to recycle the water more than once per cycle, and have it drop back on the spinning scoop-wheel, plus remove most of its energy.
A smoother start-up will be obtained by dropping top-up water in at the top, to get the bottom "race" going quicker.
A problem that Alex from "2stroke stuffing" has encountered is bad earthing, and crosstalk between wires, so I will watch out for this.
Another factor is that apparently, engines exist in this capacity which will do 26 hp at the rear wheel, so this may be over 30 hp at the crank.
These people run special race fuel I think which is better than 95 ron.
Kudos applies if I can do it without knowing exactly how they did the job, then post a video on YouTube.
Two areas for attention are: cooling and shaping around the exhaust ports; and trying different tuned pipe layouts and lengths.
The pipe performance depends on the specific speed of sound in each application, which requires a bit of trialling.