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Author Topic: Jet pump theory and efficiencies.  (Read 4577 times)
Snikwah
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« on: February 23, 2010, 04:05:25 AM »

Here is some info I've been using to improve my Jetpump designs.

About the best way to tell if you have made a better pump is by measuring the static thrust as well as top boat speed. Though boat speed is also largely related to hull design if you dont design a decent intake your boat could become unstable a nose dive at speed. Thrust can be measured with some big fish scales and boat speed and be measured with a GPS or radar gun.

However, improving jetpump efficiency and matching your motors peak power to the right pitched impeller is the best way gaunrantee performance.

Part 1: Water Power

Potential energy = mass*height*gravity(9.8m/s^2) in joules. eg 10kg on earth at a height of 10m will have 980 J of potetial energy

The potential power of moving water can be calulated by measuring (mass*height*gravity)/s so if 10kg of water is falling through a 5m high hole every second towards a power turbine, it could potentially harness 490.5 J/s or 980 watts.

This can be applied to the water moving out of our jet units too. The mass/second would be flow*the density of water = Jet Velocity * nozzle area* 1000

The height or pressure of the jet stream could be measured by pointing the jet upwards to seee how high it travels or you could use Benoulis equation to calculate the dynamic presure

Dynamic Pressure =1/2 *water density * Jet Velocity^2 (very similar to kenictic enegry 1/2mass*V^2).
But this is measured in pascals not head of water
Pressure is also =water density*gravity*height so rearrage the two equations and you get

Height= 1/2*V^2/9.81

If we put this into the power equation we get
Water power = (Jet Velocity*nozzle area*1000)* (1/2*Jet Velocity^2/9.81) *(9.81)
                  = 1/2*Jet Velocity^3*1/4*pi*nozzle diameter^2*1000
                  = Jet Velocity^3*3.141*nozzle diameter^2 *125

So a 0.028m nozzle diameter pumping water out at a velocity of 20m/s(72KPH) requires 2463 watts of motor power if it were 100% efficient

More to come
« Last Edit: February 24, 2010, 03:00:04 AM by Snikwah » Logged
Snikwah
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« Reply #1 on: February 23, 2010, 04:06:01 AM »

Part 2: Thrust

Now the problem with calculating water power using jet velocity is that its not that easy to measure velocity. However, if you have set of big fish scales or some other kind of force gauge you can work out the work out the velocity from static thrust.

Thrust = massFlow*velocity so 10 liters per sec is going to give you a lot more thrust if its coming out a tiny little hole compare to a large hole becasue the velocity will be a lot higher.
          =density*velocity*nozzle area*velocity
          =1000*velocity^2*(1/4*pi*diameter^2)

so you can work out velocity from the square root of thrust/(250*pi*diameter^2)

The Series II Hydrojet has a static thrust of 210N at 16,500rpm and a 0.028m diameter nozzle. So if you plug those two into the equation and you will get a jet velocity on 18.47m/s or 66.5kph

and if you put 18.47m/s into the water power formulae above you get 1940 watts.
If it were only 36% efficient the motor would need to produce 5.388Kw +transmission losses on top of that




« Last Edit: February 24, 2010, 03:25:52 AM by Snikwah » Logged
Snikwah
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« Reply #2 on: February 23, 2010, 04:06:25 AM »

Part 3: Improving efficiency

So it is quite easy to work out how much water power your pump is producing by measuring thrust with some fish scales, but it is not that easy to work out your efficiency unless you know exactly how much power or torque your motor is inputing into the shaft. You might think your motor is supposed to be producing 5hp but what about mechanical losses and power changes from temperature?

Lets take my boat as an example.
My motor is dynoed to produce 6hp (4.476kw) at 15900rpm (see attachment) but with transmission loses there might only be 4.2kw getting to the impeller.

My pump has a 29mm nozzle diameter and the static thrust is normally around 19kg which gives top speed of 50-55kph but I did get 23kg once and a top speed of 59kph. Maybe the air was cool, the carb was tuned properly and I cleaned out the muffler.

Anyway 19kg of thrust gives a jet velocity of 16.8 m/s (60.4kph) and a water power of 1563 kw
whereas 23kg of thrust gives a jet velocity of 18.4 m/s (66.5kph) and a water power of 2081 kw

So the worst efficiency I could get would be 1563/4476 = 35%
And the best I could get would be 2081/4200= 49%
If the pump was 53% efficient then there might have only been 1563/0.49 = 3.190kw = 4.28 hp at the shaft to give the lower thrusts.

Conclusion It is quite hard to accurately work out true efficiencies unless you can accurately measure shaft horsepower at the time of your experiment and therefore it is hard to know if your improvement in pump thrust was due to a impeller/stator mod or if was due to good weather conditions.

This is where CFD come in handy.
I was lucky enough to have acess to some nice CFD software which allowed me to model the flow through a pump and measure what I want.
My above setup was programed into CFD and gave a result of 27kg thrust at 16000 rpm. From the calcualted torque of 2.38Nm required to turn the shaft I could work out the input power to be 4.0 kw and therefore the pump efficiency of 67%

Now these result show quite a bit of difference between the real thing and the model which is probaly due to a combination of inaccurate modeling as well as inaccurate real life thrust measurement and engine power estimates. However the good thing about CFD is that you can easily change one parameter to see what benefits you get.

For example decreasing the nozzle down to 27mm increased the efficiency to 69%.
Just changing the pitch from 50mm to 55mm increase the efficency to 72%.
changing the rpm had not effect on efficiency.

In a different pump design, going from 6 to 8 stator blades increased the efficiency from 81-83%
but going from 9-10 dropped it back to 82%

The 2nd attacthment shows you a good example where you can optimise the nozzle diameter for a given impeller pitch.
The same effect happens when you change the impeller pitch for the same nozzle diameter. Each impeller pitch has an optimal nozzle diameter to match it.

Matching your motor to your pump
Ideally if you use a motor with lots of power and an effecient pump you will be doing the best you can do. Right? However, this is only the case if you motor power curve matches you pump power curve.
See the 3rd attachment for a good example. The pump curve crosses the motor curve at about 95% of the peak power.
If the impeller was a a bit smaller, the pump curve would be flatter and might not cross until 18,000rpm where the power is only ~5hp
If the impeller was a a bit bigger, the pump curve would be steeper and might cross too early ~14,000rpm where the power is only ~5hp

This is why you want to have a motor without a peaky max power region. The flatter the curve the easier it will be to cross through at a high hp spot. Other wise you will need a bunch of different impellers to change your pump curve.

Power and nozzle chart

Finally I thought I would add my power nozzle chart. See 4th attachment. It uses the previous thrust and velocity formulae to let you see what input power and nozzle diameter would be required to give you a certain thrust and jet velocity.
This chart uses an effiecieny of 50% so you might need to allow for a bit more power if you are using something unknown.

For example, If I wanted to build an electic jetpump that goes about 40-50kph and gives about 5kg of thrust, the chart might recomend a 600w motor with a 20mm nozzle to give 43kph and 4.7kg of thrust. The size and pitch of the impeller are not important so long as you can make it 50% efficient. Though obviously you won't get 50% without having an impeller around the 30mm range but it lets you know where to start.

If I decided to increase the motor power upto 800w later on, I know I'll roughly get 47kph and 5.7kg of thrust.

Anyway, I hope some of this info can be usful and didn't bore everyone to death.

Cheers

Peter

« Last Edit: February 28, 2010, 03:47:03 AM by Snikwah » Logged
Doctorswash
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« Reply #3 on: February 23, 2010, 06:13:38 PM »

G'day Pete

Wow, some impressive calc's, i only ever went to school to "eat my lunch", not math !

Any way here is what Wolfston marine had to say on the Series II Swashdrive Jet.

We recently took delivery of two Swashdrive jets which have now been fitted into our free running model. I would like to take this opportunity to say how impressed we are with the jet construction and performance. The efficiency of your jets is almost twice that of the next best offering from Graupner.

 I thought you might like some information on your jet performance in our model:
Propulsion system - Twin jet drives driven by twin Zenoah 260PUM engines

Model displacement = 28.9kg

Model Length = 2.185m

Deadrise angle at transom = 12 degrees

Model beam = 0.450m

Maximum speed (both engines fully open) = 23.6 knots = 12.13m/sec

Resistance of model @ 23.6 knots = 128N

Effective power = 128 x 12.13 = 1552W

Delivered power per jet = 776W

Delivered engine power = 2.32kW

Assume transmission losses of 5%, delivered engine power = 2.32 x 0.95 = 2.20kW

 Jet Efficiency = 776 / 2200 = 36%

The best efficiency we have achieved to date with other jets (4 different designs to date) is 21%, so this represents a really significant improvement well done!

I hope you find this interesting, Best Regards, Dickon Buckland.

***********************************
Wolfson Unit MTIA
University of Southampton
SO17 1BJ

Link to their websitehttp://www.wumtia.soton.ac.uk/

« Last Edit: February 24, 2010, 06:23:46 PM by Doctorswash » Logged
Doctorswash
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« Reply #4 on: February 24, 2010, 08:47:27 PM »

Pete

Quote
if you put 18.47m/s into the water power formulae above you get 1940 watts.
If it were only 36% efficient the motor would need to produce 5.388Kw +transmission losses on top of that
That works out pretty close !
I have a Full mod 260 (CNC ported) with a 1mm stroker crank which estimated out at approx 6.8 to 7.0Hp @16500rpm, havent had it on a dyno  so couldn't say for sure !
So maybe the jet is a lil more efficient. But boat speed had been GPS'd at 68 to 70 Kph, this could be atributed to the + inlet pressure when at speed over static measurment ?
When it comes to efficiency of a mechanism i have always made the final output measured in COP (Coefficient Of Performance) this takes into account any enviromental input as well as opperator input, add operator input and any enviromental input,subract the  losses to give you devise efficiency in COP, so in the case of the jet it would have a static COP=0.36
Its the same thing, the only difference is COP takes the enviroment into account.
We all know that we cannot make something that has an efficiency of over 100% otherwise known as perpetual motion, not possible !
Most electric motors are not even 86% efficient, only some of the high end coreless motors are only recently now in the 90's.
But you can certainly make a devise that has a COP greater than 1 (over unity) eg solar cell, hydro electric dam, sailing boat etc etc !
Here is a link for an explanation of COP for a heat pump, but can be applied to any devise
http://en.wikipedia.org/wiki/Coefficient_of_performance

some Examples below of how COP is layed out

regards Craig
« Last Edit: March 01, 2010, 01:20:34 AM by Doctorswash » Logged
Doctorswash
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« Reply #5 on: February 25, 2010, 01:21:52 AM »

 A Pump Theory
So when you take COP of a jet into consideration, one might start looking to were there might be an enviromental input ?
I wouldn't say its my discovery, but by adding mixed flow into the jet has aided in jet efficiency, this can be explained by adding a tapered wear ring and taper design impellor adds a centerafugal component to the jet. Centerfugal force is an eviromental effect that happens by turning the shaft. In turn this allows the fluid to expand as its driven over the impellor aiding in static force / pressure of the jet for free !
One other part of the jet were this could happen is in the inlet. This can be explained that when the jet is static the impellor is required to draw or suck the fluid into the inlet and up to the leading edge of the impellor before it can be prosessed, this could be seen as an in-efficiency. Once the boat has velocity, at a  certain point the inlet pressures change from - to a + pressure, from this point on it is aiding the loading of the jet, almost free of charge or very lil cost thus aiding its COP, unfortunatly at a higher point the inlet has to much + pressure and fluid begins the backflow turbulence out of the inlet and unsettles the boat.
So from this observation one could conclude that good efficiency gains could be made from constructing an inlet with a variable volume, ie having the sides of the inlet slide in or out to reduce or increase inlet volume and be servo'd to jey inlet pressure, somehow ? maybe like a picture frame that sits between the jet and the hull with slidable sides reduceing flow into the jet, possibly 2 -3 % gain for free...food for thought.
Regards
Craig
« Last Edit: February 27, 2010, 04:34:10 PM by Doctorswash » Logged
Scott Schneider
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« Reply #6 on: March 02, 2010, 10:18:34 PM »

Peter, Craig
Nice to find an active site again where your collective insights are again shared and able to be learned from  Grin.
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Snikwah
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« Reply #7 on: March 04, 2010, 03:24:42 AM »

Hi Craig, I think they are basically working out the moving pump efficeincy rather than the static thrust efficiency. Thats why the thrust/velocity numbers don't look right.

If they had a light weight boat with low drag, it would be going a lot faster and the thrust would be a lot lower since the jet velocity minus the boat velocity would be a lot lower.

I think my way and their way are ok to work out efficeincies so long as you use them to compare another design in the same way.

Peter

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