Reverse engineering prop RPMs

[bdgWesternMass]

Hello All,

I'm doing basic boat research and I'm tryin to figure out the actual RPMs of the propshaft.

For instance if I was running a Volvo D2-75 drive shaft at 1800 RPM it should produce around 212 Nm at 40KWs or 75 HP.

If it's connected to a 2.8:1 torque converter. I believe the prop shaft will be rotating at around 650 RPMs the torque would go up to around 600 Nm and the power would drop to about 7 KWs to 12 HP.

Thank you for any clarifications,

Ben

[StuM]

Originally Posted by bdgWesternMass

Hello All,

I'm doing basic boat research and I'm tryin to figure out the actual RPMs of the propshaft.

For instance if I was running a Volvo D2-75 drive shaft at 1800 RPM it should produce around 212 Nm at 40KWs or 75 HP.

If it's connected to a 2.8:1 torque converter. I believe the prop shaft will be rotating at around 650 RPMs the torque would go up to around 600 Nm and the power would drop to about 7 KWs to 12 HP.

Thank you for any clarifications,

Ben

40KW is about 56HP, not 75!


Assuming you have a suitable propeller: Prop RPM and Torque will change proportionally.

Why would you expect the engine power to change? TANSTAAFL!

[bdgWesternMass]

Originally Posted by StuM

40KW is about 56HP, not 75!


Assuming you have a suitable propeller: Prop RPM and Torque will change proportionally.

Why would you expect the engine power to change? TANSTAAFL!

Whoops not only did I screw up the KW to HP but I meant reduction gear not torque converter.

I was expecting that reducing the RPMs would reduce the horsepower but increase the torque.

[StuM]

Originally Posted by bdgWesternMass

Whoops not only did I screw up the KW to HP but I meant reduction gear not torque converter.

I did wonder about that, but assumed the key part of your question ws concerning the reduction ratio.

I was expecting that reducing the RPMs would reduce the horsepower but increase the torque.

HP at RPM is dependent on load.
Reducing the engine RPM will reduce the HP. And so will reducing the load on the engine. (An engine in neutral at 1800 RPM will use much less fuel and generate much less HP than one under load).


Reducing the prop RPM with a reduction gear will reduce the load on the engine assuming you keep the same prop since you will driving the boat a lot lower. That will reduce the HP produced and the fuel burnt.

But if you fit a prop (larger diameter and/or greater pitch) that is designed to work efficiently and get the same boat speed at 600 RPM then fuel/HP required will be the same at 1800 engine RPM .

[Pauls]

And, just to be clear, the reduction gear does not in itself change horsepower. It reduces prop shaft RPM and increases prop shaft torque. Horsepower in = horsepower out, minus friction losses in the reduction gear which are probably in the vicinity of 15%, this varies with the gear.

[DougR]

The formula to calculate horsepower is: HP = torque x rpm / 5252 ( torque in FtLb )

So you can see that if hp stays the same and rpm goes down, then torque must go up, and visa versa.

When we hang a reduction gear on the back of the engine with, for example, a 2.8:1 reduction ratio, the rpm of the crankshaft will be reduced by a factor of 2.8 and the prop shaft rpm will be that much slower. ( divide crankshaft rpm by reduction ratio) However, the torque on the prop shaft will be 2.8 times greater than the torque on the engine crankshaft. Thus, the numbers in the formula change, but the horsepower on the prop shaft should remain the same as on the crankshaft.

I say the hp should remain the same, but that ignores power losses within the reduction gear. These losses can include gear friction, oil churning losses, and sometimes oil pump losses. The amount of loss depends on the type and configuration of the transmission, but typically the total gear losses amount to about 3% or 4% of the crankshaft power. This power loss disappears in the form of heat which radiates from the gear case or is rejected in the gear cooler.
Typically, total power loss from the engine to the propeller is about 7%. Three percent loss in the transmission and four percent loss in the stuffing box and cutless bearing.

[bdgWesternMass]

Originally Posted by Pauls

And, just to be clear, the reduction gear does not in itself change horsepower. It reduces prop shaft RPM and increases prop shaft torque. Horsepower in = horsepower out, minus friction losses in the reduction gear which are probably in the vicinity of 15%, this varies with the gear.

Great this tidbit was really helps me understand, although essentially is identical to StuM's notion that Horsepower is only present when available and used.

Also since I have higher torque at lower RPM = the same amount of HP.

I believe I truly understand the system up to the prop-shaft

My follow question would be regarding feathering props.

How does one design around a prop that changes it's properties?
Does one establish a few points along the load curves likely to be encountered and plan off those points?

This has been really helpful and appreciate everyone's contribution.

Ben

[DougR]

Normally engine manufacturers recommend (require) that an engine be propped so that it is able to obtain 100% of its rated rpm when operated at full throttle in the vessels “normal” load condition. Sometimes a propping rpm range is given, and this takes into consideration that load conditions may vary widely.

Because a feathering prop doesn’t really change its properties, (it’s either 100% or 0%), feathering props are treated the same as fixed blade props. Even a true prop that can change properties, such as a controllable pitch prop, is sized to allow the engine to obtain rated rpm at full throttle.

Typically a naval architect or designer has a good feel for the propeller requirements for a given vessel, but the process goes something like this......

The designer takes the physical attributes of the boat, like waterline length, beam waterline, displacement, windage, available hp, location of LCG, etc. and he inputs that into his propeller software program. The program will calculate the max speed that the vessel is capable of reaching, and if the program is any good it will also produce a hull resistance curve for the vessel.

He also inputs the engine hp, rpm and a trial guess at gearbox reduction ratio, and the program calculates the prop thrust available at various speeds, and then the program lays this curve on the same graph as the hull resistance curve. Now he does the same thing again with a different reduction ratio, and he comes up with another thrust line in the graph. And so on....

This process allows the designer to test out multiple cases of engine power and reduction ratio in a short amount of time, and lets him optimize the entire drive train from engine to prop in a fairly short period of time.

[bdgWesternMass]

Thank you Doug for that explanation. Again my green-ness in terminology stumbled my question. I really did mean a controllable pitch prop not feathering

I would have used the only brand names I know like Auto-prop or max-prop but I have been lurking for a awhile and wanted to open the discussion to all I believe controllable pitch props.

It sounds like you plan on the extremes and let the middle take care of itself.

One thing I did get right is the idea of reverse engineering starting with a known values and figure out the hull shape...

I'm a slow learner and tend to do things backwards.

Thank you,

Ben

[Cadence]

Originally Posted by Pauls

And, just to be clear, the reduction gear does not in itself change horsepower. It reduces prop shaft RPM and increases prop shaft torque. Horsepower in = horsepower out, minus friction losses in the reduction gear which are probably in the vicinity of 15%, this varies with the gear.

Paul, thanks for beating me to a sensible explanation.

[a64pilot]

Only propellor that adjusts pitch when underway for smaller boats that I know of is an Autoprop, and you can’t control it. It self adjusts pitch based on centripetal force that tends to drive the prop to a flatter pitch, and water flow that tends to drive the prop to a higher pitch.
The idea is if you increase the load or resistance on a boat by stiff head winds etc, the pitch will decrease and not load the engine more than it should, but if you Decrease the load like by motor sailing, then the prop will increase pitch keeping the motor loaded.

Any fixed pitch prop, whether it be a folder or feathering, will be matched correctly to the engines output at one RPM and hull speed, that being flat water and max RPM.

Look at the attached photo, you will see that the only time the prop matches engine output is at max continuous power RPM, on this engine that’s 3500 RPM, at any other RPM the engine is very much under loaded.
But this chart also assumes wave and wind resistance to not be a variable, which it always is of course.
This is why many overprop, by overproppimg you move that propellor curve up closer to the engines curve, but as you do so, at higher RPM the curve will go higher than the engine curve. It can’t because the engine is power limited, so what happens is the engine is overloaded, so have enough careful if your overpropped to not overload.

A variable pitch propellor would allow the engine and prop curve to be the same. That is what an auto’s transmission does, it matches the engine power output to the power required and reduces fuel consumption, noise and vibration, and engine wear.

The Autoprop does to some extend match them, but I believe it’s conservative in order to ensure the engine is never overloaded.
Larger boats a controllable prop is available, and that with ideally a torque meter and a pyrometer would allow the operator to keep engine demand closely matched to power available.

[bdgWesternMass]

Well you there you go I was trying to figure out how one reasons out the loading of an Autoprop that made me post this question.

Thanks for that chart.