Lower cost design for a water brake for a dyno.
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Owen_N
Well-Known Member
I want to go to 20-60 hp as my upper limit.
This is a rather large generator, and they generally don't want to spin at 20,000 rpm.
Possibly a DC motor out of a trollybus or train would do it. - not readily available.
Someone was suggesting an eddy current brake, but I think the water brake is more compact and cheaper.
A big disc brake has an unfriendly rpm-torque characteristic, and would possibly need water-mist cooling.
The initial engine of 20 hp at 17500 rpm is only the start- I would like to go to a water-cooled 125 , 55 hp at 13,000 rpm, eventually.
I may not get beyond the 49cc version yet. We will see.
Owen_N
Well-Known Member
New design for a water brake- set rotor and stator segments radially. That way, the outer housing can be a lot of half-pipes, with flat plate sides, and a removable ring to get the rotor out. the rotor blades can be made from pipe sections back-to-back as well, with a machined centre core to pick up the shaft.
This really knocks down material costs and precise machining needs. Possibly a 120mm inner rotor, 50 mm wide? 6 inner segments. I will draw this up
and work out the pipe size needed. I also need to design bearing cups. possibly 5mm aluminium sheet would do for side walls.
I wonder if there if any way to work out in advance what the resistance torque will be at various rpms? how much side clearance? Volume of lower sump?
A small sump would be good. then I need a fluid bleed in and bleed out arrangement. It would be good to incorporate servo-driven valves, and some form of feedback automation, to try to emulate an inertia system for fast graphing of torque-rpm , as well as allowing steady runs at several rpms.
Would bottom water inlet and extraction be a good thing? the rotor will distribute the water right round the outer stator fairly quickly, I imagine.
What are the best type of valves? do I need needle valves, ball valves, butterfly valves? Robotic-type continuous-rotation servos with PWM input?
This really knocks down material costs and precise machining needs. Possibly a 120mm inner rotor, 50 mm wide? 6 inner segments. I will draw this up
and work out the pipe size needed. I also need to design bearing cups. possibly 5mm aluminium sheet would do for side walls.
I wonder if there if any way to work out in advance what the resistance torque will be at various rpms? how much side clearance? Volume of lower sump?
A small sump would be good. then I need a fluid bleed in and bleed out arrangement. It would be good to incorporate servo-driven valves, and some form of feedback automation, to try to emulate an inertia system for fast graphing of torque-rpm , as well as allowing steady runs at several rpms.
Would bottom water inlet and extraction be a good thing? the rotor will distribute the water right round the outer stator fairly quickly, I imagine.
What are the best type of valves? do I need needle valves, ball valves, butterfly valves? Robotic-type continuous-rotation servos with PWM input?
Owen_N
Well-Known Member
Because I want a rising torque with rpm. This allows stable engine operation. with a linear brake, the engine will suddenly stop when lowering rpm, and torque drops off.
I have drawn a 100 mm diameter rotor with 50mm wide buckets, of 30mm tube sections. I think this should provide enough resistance for a 50cc engine.
-image-
This is quite easy to make from cut tube and plate.
the objective of the bottom race is to smooth out surging. It would probably work with half the number of rotor buckets.
The principle of operation is that water will flow into the scoop, then reverse, and be flung out tangentially.
A bit of trial and error will be needed to get it to run smoothly. It probably needs a vertical component to the lower reservoir.
Once it starts to circulate, the water level in the bottom sling chamber will drop, so there needs to be more water storage capacity, and a head to help water flow into the "ring" section.
I would like to work out how much water is in each bucket, and the energy consumed in a complete cycle.
A difficulty is the water velocity is likely to be slower than the tangential velocity, so I don't know how many revs it will take water to get right around the loop.
I think it will give a rising torque characteristic with rotational speed.
Continually piling water in at the start of each tangential pocket will tend to synchronise water speed with the rotor.
Water entering from pocket 1 may totally skip pocket 2 when it gets slung out, so I need to trial to establish the best pocket layout.
The pocket flow may end up going in totally the reverse direction, or circulate into and out of the pockets quite differently.
Attachments
Realistically, do you REALLY need this precision on a model engine.
After all, it's just an indicator or do you intend incorporating atmospheric conditions in your microcontroller programming (I can think of at least 3 required) to be anywhere near accurate.
After all, it's just an indicator or do you intend incorporating atmospheric conditions in your microcontroller programming (I can think of at least 3 required) to be anywhere near accurate.
Owen_N
Well-Known Member
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.
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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.
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.
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Owen_N
Well-Known Member
New post- image of revised outer ring, only 6 rotor cups.
The 5mm sheet seems to be very expensive at $45 nz per kg for cut sheet from RS components- ex Australia- which I think is a rip-off. The base material is worth about $4.50 nz/kg
AliExpress is even worse, at about $55 per kg. I will have a go with 3mm aluminium sheet. It should be stiff enough.
-image-
It looks like a similar hexagonal spacing seems to work out with the tangential inlets.
These smaller steps should be enough to extract the energy and deflect water back into the rotor.
I have ordered a heap of stuff that should turn up towards the end of June.
I got the cheap water-cooled 50cc , but it is a centrifugal clutch motor with no gearbox, and rotates backwards.
I will convert to forwards rotation, and chuck out the centrifugal clutch, and adapt to electric start.
- this will need changes to the left side compression/oil seal on the bearing.
I think forward rotation is easier on the piston ,which tends to catch in the exhaust port, if it is the full 75 degrees wide.
This makes the "cold side" the main thrust face.
The idea is to use this just as a "Bench Engine".
The 65cc KTM pre 2008 is an actual gearbox engine, I think, with the same stroke, but they are hard to find.
Otherwise, the next engine is a Minarelli AM6 50cc brand new, but $1000nz plus freight, which at 20 kg won't be cheap ex Europe.
This has a different stroke and bore by about a mm.
The 5mm sheet seems to be very expensive at $45 nz per kg for cut sheet from RS components- ex Australia- which I think is a rip-off. The base material is worth about $4.50 nz/kg
AliExpress is even worse, at about $55 per kg. I will have a go with 3mm aluminium sheet. It should be stiff enough.
-image-
It looks like a similar hexagonal spacing seems to work out with the tangential inlets.
These smaller steps should be enough to extract the energy and deflect water back into the rotor.
I have ordered a heap of stuff that should turn up towards the end of June.
I got the cheap water-cooled 50cc , but it is a centrifugal clutch motor with no gearbox, and rotates backwards.
I will convert to forwards rotation, and chuck out the centrifugal clutch, and adapt to electric start.
- this will need changes to the left side compression/oil seal on the bearing.
I think forward rotation is easier on the piston ,which tends to catch in the exhaust port, if it is the full 75 degrees wide.
This makes the "cold side" the main thrust face.
The idea is to use this just as a "Bench Engine".
The 65cc KTM pre 2008 is an actual gearbox engine, I think, with the same stroke, but they are hard to find.
Otherwise, the next engine is a Minarelli AM6 50cc brand new, but $1000nz plus freight, which at 20 kg won't be cheap ex Europe.
This has a different stroke and bore by about a mm.
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I was recently looking at and discussing a 'full size' (200 bhp) water brake dyno which is owned by a friend.
Cavitation is a significant issue, especially with higher revving engines. It is necessary to have a substantial head of water at the inlet to ensure that the pump does not pull vacuum (which would allow the engine to over-rev). The example I was looking at had a feed tank approx 6 feet above the impeller. The pump returns water to the same tank, so there is no pressure delta when stationary. Back pressure is generated by a simple adjustable restrictor at the pump outlet.
Cavitation is a significant issue, especially with higher revving engines. It is necessary to have a substantial head of water at the inlet to ensure that the pump does not pull vacuum (which would allow the engine to over-rev). The example I was looking at had a feed tank approx 6 feet above the impeller. The pump returns water to the same tank, so there is no pressure delta when stationary. Back pressure is generated by a simple adjustable restrictor at the pump outlet.
Owen_N
Well-Known Member
This sounds like a large centrifugal water pump. Mine is more similar to a Pelton wheel, with some modifications. There is no actual suction, the main limiting factor I can see is that it will scoop water out of the bottom race faster than it can flow in from the sides and replenish it.
It may be better to have a fully closed circuit, with a water bleed out for cooling, matched by water bleed in as a top up.
It would be nice to automate the valves. variable-speed feedback motors and gear-type water pumps would do it.
I have some fuel pumps that use gears, but they are tiny. the gears are only about 5mm across, and use 2mm "D" shafts as a drive, and brush motors.
Maybe I can drive right through the brush motor to preserve the seal, but I don't know how fast I have to rev them. I have brushless motors capable of 21,000 or so rpms. these motors are not so precise at rpm control, though - the controllers use PWM input, but that is just an indication of desired speed.
giel
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why don't you use 2 brake discs one with magnets glued onto it and a way of bringing them closer together so you can control the braking, this way you get eddy currents that are getting stronger the higher the rpm is and no blocking..
it is used in carnival rides and rollercoasters a metal strip that is in the middle of the cart that when brakes are needed there are magnets on the track. used as a fail safe system as it requires no mechanics to work just the magnets and rhe strip between it.
they use this system in landspeed cars that gives brakes that will not fade and will never lock up. the discs on these machines are made from aluminium. normal when not braking the distance is about 5 cm and when braking it goes to 0.5cm of the disc
other system is hooking up an alternator to the motor and varying the load with resistors or charging batteries/generating power for your house ( it is used in big dyno shops) to not waste energy into heat.
it is used in carnival rides and rollercoasters a metal strip that is in the middle of the cart that when brakes are needed there are magnets on the track. used as a fail safe system as it requires no mechanics to work just the magnets and rhe strip between it.
they use this system in landspeed cars that gives brakes that will not fade and will never lock up. the discs on these machines are made from aluminium. normal when not braking the distance is about 5 cm and when braking it goes to 0.5cm of the disc
other system is hooking up an alternator to the motor and varying the load with resistors or charging batteries/generating power for your house ( it is used in big dyno shops) to not waste energy into heat.
giel
Member
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or you can use a rotary vane pump..
fir cheap version ( some old air tool grinder or drill that is direct drive!) and the inlet+ air out in the handle so you can make a closed loop in the system with a tank in the middle.
replace the bearings with stainless steel versions or ceramic . or use an oil pump from a car only drawback is having to make a mount for it and fill the system with oil. the vane pump acts as a closed system so cavitation is not a problem. and as you are water instead of air you can control the resistance with a valve at the end..
fir cheap version ( some old air tool grinder or drill that is direct drive!) and the inlet+ air out in the handle so you can make a closed loop in the system with a tank in the middle.
replace the bearings with stainless steel versions or ceramic . or use an oil pump from a car only drawback is having to make a mount for it and fill the system with oil. the vane pump acts as a closed system so cavitation is not a problem. and as you are water instead of air you can control the resistance with a valve at the end..
Bentwings
Well-Known Member
at the start a water brake sounds simple and easy . A model of an early hydra magic fluid coupling transmission sounds simple enough , but actual fabrication is not quite as easy as it sounds unless you have some nice shop equipment. . Making toruses is not as simple as pictures show it. Several have made these with limited success. I think the magnetic clutch brake is easier or just a variable load generator. A long time ago some model plane and car guys used various props to provide loads. Props reach a speed where they essentially stall then they cal take a lot of power to turn much faster. So not easy to measure. Various torque arm scales were used but very cumbersome .
Owen_N
Well-Known Member
I think my idea of a scoopng pelton-wheel derivative is quite easy to make.
I can temporarily assemble the rotor with side plates and long bolts.
I have a lot of those.
I may get away with a 10mm shaft.
Getting a good key on the centre hub requires a little thought. There is one on you-tube which used a light knurl and a press fit.
I have some countersunk M5 screws on order.
These would be good for holding on the outer access lid.
I wonder if I can get away with 1.5mm aluminium for the outer deflector scoops?
I will also try joining parts up with aluminium solder.
I can temporarily hold parts in place with CA hobby glue.
It evaporates when heated. You need good ventilation, as it is hard on the eyes.
It doesn't seem to be toxic- less so than ammonia.
I have ordered 1/2" ball valves , and will see if they meter OK.
Also a 70kg scales beam, and a amplifier circuit I can hood up to my Arduino Zero.
I also need an rpm adapter as well-next job.
I just now need some Arduino code, and some software for the notebook to calculate, display, and store plot runs.
There is free software available for an inertial dyno.
I can temporarily assemble the rotor with side plates and long bolts.
I have a lot of those.
I may get away with a 10mm shaft.
Getting a good key on the centre hub requires a little thought. There is one on you-tube which used a light knurl and a press fit.
I have some countersunk M5 screws on order.
These would be good for holding on the outer access lid.
I wonder if I can get away with 1.5mm aluminium for the outer deflector scoops?
I will also try joining parts up with aluminium solder.
I can temporarily hold parts in place with CA hobby glue.
It evaporates when heated. You need good ventilation, as it is hard on the eyes.
It doesn't seem to be toxic- less so than ammonia.
I have ordered 1/2" ball valves , and will see if they meter OK.
Also a 70kg scales beam, and a amplifier circuit I can hood up to my Arduino Zero.
I also need an rpm adapter as well-next job.
I just now need some Arduino code, and some software for the notebook to calculate, display, and store plot runs.
There is free software available for an inertial dyno.
Bentwings
Well-Known Member
id forget the aluminum solder
Check project farm you tube for his tests not are worth the time of day.
Why not make a wood mold then metal form a round torus. You can cut slots for fins then silver solder them in place.
You can search the web for how liquid flows in a fluid coupling.. I gets hot in a hurry if there is much power or mall size ratt and Whitney developed big ones to test the big P & W radial engines .
Owen_N
Well-Known Member
I will check out that video. The outer buckets can be retained and sealed with this J-B weld. It is quite good for surface holding power-several
megapascals in shear and tension. I stuck stuff all over an engine piston with it, and it holds up to around 11 m/s avg piston speed very well.
I can extend and hold the buckets in place with screws, the side plates can be screwed to the hub, and the sealant has very little actual load on it.
This is a different principle to the torus flow couplings,- closer to a centrifugal pump in performance, but uses plain slinging of water to gain kinetic energy, then stop and deflect the water back to the scoops. It is easier to make than the torus layout.
I need to check that it is sufficient in theory to work starting around 9000 rpm. it is insufficient at 6500 rpm, but energy goes up at the square of the rotational speed. I need to arrange so all water drains into the bottom race section . Any remaining slung water from the buckets is supposed to drive the race, which cycles the water back into the impeller buckets as a jet. You can tip the whole thing on its side, with plain radial impeller blades, and a fairly high water level, and it should also work quite well. There is an example on you-tube, where there is an impeller just swirling water around a rough-surfaced can. I think my layout has better effectiveness for radius and volume. You have to splash the water around enough, so that it flies back into the impeller.
<edit>
I watched the Project Farm video, and some aluminium solders held up very well, particularly the Hobart brand.
One brand was very poor, with too high melting point. I brought mine from **** Smith (Kogan) australia, so they should have a reputable brand. The material does not seem to "wick" at all! - fairly lumpy looking weld bead.
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Bentwings
Well-Known Member
I use plenty of JBweld as well as others epoxies in Rc aircraft. It works for it’s purpose. I also TIG weld about anything wieldable. Doing my own version of physical therapy to recover some lost vision. Liken it to muscle memory . It’s incredibly difficult to weld when you see two of things. Analyzing the direction you have to go then recalling proper movements to achieve results I’m not up to old standards but I can make passable welds now.
Project farms also tested all the ocular epoxies. JB is a good one.
Owen_N
Well-Known Member
I found an error in my torque calculations.
I will expand:
Energy to move the water = tip speed V(sq) x m /2 for one revolution.
now tip speed = w x r
power = J/t where t = s/rev = 2 pi/w
substitute in:
J = w(sq) r(sq) m/2 ; J/t = m/t w(sq) r(sq)/2
divide through by t and substitute:
power(P) = w(sq) r(sq) m/2t x w/2 pi = (w**3) (r**2) x m/t / 4pi ------(1)
*Note - multiplier becomes rev/s on the right hand side*
another way to express m/t is volume x density(1) /t where t = 2pi/w
volume = pi(r2(sq) - r1(sq)) x L where L = width.
m/t = w x L x (r2(sq) - r1*(sq)) --------(2)
Substitute (2) in (1)
Power = w(**4) . r2(**2) . (r2(sq) - r1(sq)) . L / 4.pi ( . same as x)
now T (torque) = P/w
we are back to T = w(**3).r2(sq).( r2(sq) - r1(sq)) .L /4pi
This is looking good with a w(cu) multiplier!
substitute in actual values for 9000 rpm, w = 942
r2 = 50 x 10**-3, r1 = 20 x 10**-3, L = 50 x 10**-3
T = (9.42 x 10**2)**3 x 2.5 x 10**-3 x 2.1 x 10**-3 x 50 x 10**-3 /4x 3.14
=(8.36 x 10**8) x 21 x 10**-6
1.76 x 10**4 N-m
We can multiply this up by number of water recycles which can be up to 4
this makes 7.1x 10** 4 N-m --- which is plenty!
the engine itself can make 30 hp at 17800 rpm, or 22.35 kw at 1864 rad/s
Torque = 2.24 x 10**4/1.864 x 10**3 = 12 N-m ** actual expected engine torque.
The energy result sounds suspiciously high as power now becomes Tw = 7.1 x 10**4 x 942 = 47 kw at 9000 rpm
** theoretical brake power.
This could also indicate that braking force at the buckets could be quite high.
The actual stopping force depend on the time it takes to stop a mass at a velocity. if time is known, "a" can be calculated, and f= ma
now v = at and t = 1/150 and m = all the mass in the torus = 2.1 x 10**-3 x pi x 20 x 10**-3 = 132 x 10**-6 cu mt, also mass as well for water.
so V = wr = 942 x 50 x10**-3 = 47 m/s, and a = v/t = 7050 m/s(sq) and f = mA = 132 x 10**-6 x 7050 x rev/s (150)
= 140N, or 14 kgf per bucket.
Multiply by radius to get T where T = Fr ; 140 x 50 x 10^-3 = 7 n-m , x recycle rate (4) = 28N-m.
While this is satisfactory, it doesn't agree with the energy analysis. Why is this?
20mm should be enough for the bucket width, though.
I will expand:
Energy to move the water = tip speed V(sq) x m /2 for one revolution.
now tip speed = w x r
power = J/t where t = s/rev = 2 pi/w
substitute in:
J = w(sq) r(sq) m/2 ; J/t = m/t w(sq) r(sq)/2
divide through by t and substitute:
power(P) = w(sq) r(sq) m/2t x w/2 pi = (w**3) (r**2) x m/t / 4pi ------(1)
*Note - multiplier becomes rev/s on the right hand side*
another way to express m/t is volume x density(1) /t where t = 2pi/w
volume = pi(r2(sq) - r1(sq)) x L where L = width.
m/t = w x L x (r2(sq) - r1*(sq)) --------(2)
Substitute (2) in (1)
Power = w(**4) . r2(**2) . (r2(sq) - r1(sq)) . L / 4.pi ( . same as x)
now T (torque) = P/w
we are back to T = w(**3).r2(sq).( r2(sq) - r1(sq)) .L /4pi
This is looking good with a w(cu) multiplier!
substitute in actual values for 9000 rpm, w = 942
r2 = 50 x 10**-3, r1 = 20 x 10**-3, L = 50 x 10**-3
T = (9.42 x 10**2)**3 x 2.5 x 10**-3 x 2.1 x 10**-3 x 50 x 10**-3 /4x 3.14
=(8.36 x 10**8) x 21 x 10**-6
1.76 x 10**4 N-m
We can multiply this up by number of water recycles which can be up to 4
this makes 7.1x 10** 4 N-m --- which is plenty!
the engine itself can make 30 hp at 17800 rpm, or 22.35 kw at 1864 rad/s
Torque = 2.24 x 10**4/1.864 x 10**3 = 12 N-m ** actual expected engine torque.
The energy result sounds suspiciously high as power now becomes Tw = 7.1 x 10**4 x 942 = 47 kw at 9000 rpm
** theoretical brake power.
This could also indicate that braking force at the buckets could be quite high.
The actual stopping force depend on the time it takes to stop a mass at a velocity. if time is known, "a" can be calculated, and f= ma
now v = at and t = 1/150 and m = all the mass in the torus = 2.1 x 10**-3 x pi x 20 x 10**-3 = 132 x 10**-6 cu mt, also mass as well for water.
so V = wr = 942 x 50 x10**-3 = 47 m/s, and a = v/t = 7050 m/s(sq) and f = mA = 132 x 10**-6 x 7050 x rev/s (150)
= 140N, or 14 kgf per bucket.
Multiply by radius to get T where T = Fr ; 140 x 50 x 10^-3 = 7 n-m , x recycle rate (4) = 28N-m.
While this is satisfactory, it doesn't agree with the energy analysis. Why is this?
20mm should be enough for the bucket width, though.
Last edited:
harryedward
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Aluminum is ideal for the impeller and housing. You can use 4-6 straight radial blades, with overhung bearings for easy mounting. Design the housing to contain water with durable ceramic seals, and use a stepup drive system for proper gear ratios and protection at high speeds. Mechanical estimating services can help calculate material costs and fabrication time.
Bentwings
Well-Known Member
Tooth belt drive in metric 8 mm or 20 mm race car stuff. These could handle any reasonable load. Aluminum pulleys available from “ good vibrations” racing “
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