Propeller shaft torque?

[Arnot]

I am looking for a few words from the wise here. I am trying to get a rough idea of what the rotational torque on my propeller shaft might be (and I realise that it is going to have to be approximate).

Why? well I want to convert the existing mechanical gearbox and dry clutch system to a full hydrosatic drive system. I want to do this partially to allow me to lower the cabin floor which is presently raised to clear the propshaft and partially to allow me constantly variable drive ratio to better match the engine speed to the power requirement. A variable prop is not an option on my boat.

Hydrostatic motors have a fairly fixed upper limit to the shaft torque they can provide and the pump choice is dictated by the motor selected.

My boat is a narrow boat, it is 72' long, 7' wide has a flat bottom and draws about 20". It operates on the English canal network and this is very narrow and shallow, often only 16' wide and 3'6" deep (so no clear water). There is a speed limit of 4mph on canals but on rivers there may be a current or tidal flow so I would like to work on about 8 knots on clear water.

The propeller is a bronze three bladed version with a 22" diameter and a 16" pitch but this may be changed for one with a 22" pitch.

The engine is a Ruston and Hornsby 3VRO built in 1937 and rated at 33hp at 1000 rpm, they didn't have torque curves in those days but I suspect that peak torque is between 400 and 500 RPM. The typical cruising engine speed is about 300 RPM. This is why torque is such a consideration.

There is often a lot of drag and occasionally I have to grind up shopping trolleys or other tough debris. Although I think that this is where pressure relief valves come into play... I think I need to be looking at "bollard" conditions for the calculations.

Any ideas please?

Regards

Arnot

[erik818]

Arnot,
First calculate what's easy, the torque at rated power and speed:
30 hp at 1000 rpm equals 22 kW at 105 radians/second.
Torque = 22 kW/105 rad/s = 210 Nm.

Next thing is to guess the shape of the torque curve. My guess is that it's reasonably flat from 300 rpm to 1000 rpm (it is on modern diesels), and that anyway maximum torque was excerted at maximum speed.

You can do the calculations in imperial units instead, if you really enjoy conversion factors.

Erik

[Arnot]

Originally Posted by erik818

Arnot,
First calculate what's easy, the torque at rated power and speed:
30 hp at 1000 rpm equals 22 kW at 105 radians/second.
Torque = 22 kW/105 rad/s = 210 Nm.

Next thing is to guess the shape of the torque curve. My guess is that it's reasonably flat from 300 rpm to 1000 rpm (it is on modern diesels), and that anyway maximum torque was excerted at maximum speed.

Erik

Thanks for your input, that's what I first thought but...

Unlike modern engines the torque curve is anything but flat! Its more the power curve that's flat. Isn't Sweden the home of Bolinders?

The engine probably still produces nearly 20 hp at 500 rpm but it's difficult to prove. I suspect that the peak torque lies between 500 and 600 rpm and is about 300 Nm.

More importantly, as I see it, the engine's possible output is irrelevant as the propeller will never be able to transmit it to the water. If you try to rev the engine quickly, the prop just slips, drags air in and makes the canal into an imitation of a cappuchino! So, in effect, the propeller slippage and inapropriate stern design acts as a shaft torque limiter.

Bear in mind that when this boat was made in 1884 it was designed to be propelled by one horse power - on the end of a piece of rope. Sometime after engines were invented, it was converted to a motor drive and with very little modification a propeller shaft was poked through the stern and a pair of anti cavitation plates welded on just below the water line. the result is that it actually slips through the water very easily and it takes very little power to get it up to speed eventually, it just won't accellerate.

As well as this it has all the stopping ability of an oil tanker. From 4 mph it will stop in about 200m on a good day! An anchor isn't workable because there is nowhere within reach to stow it and if deployed would probably bring up something vile from the festering bed of the canal.

One other reason I want to use hydraulic drive is to be able to slide the power in smoothly without changing the engine speed to accellerate and then reduce the power again without altering the engine speed when cruising speed has been reached. Not quite warp factor... A boat - but not as we know it!

Regards

Arnot

[Guest625101138]

The peak torque that the 16" pitch prop will absorb is 330Nm. That is under heavy towing condition at 3.6kts with shaft at 1000rpm. So clearly the engine will not do this as it requires more power than it has.

At 900rpm the peak torque absorbed is 270Nm at 3.2kts - might be possible if the torque curve drops rapidly.

At 800rpm the peak torque is 210Nm at 3kts - around engine torque at rated rpm.

At 700rpm the peak torque is 160Nm at 2.5kts.

The peak torque absorbed drops off considerably after that.

If you increase to the 22" pitch then the peak torque absorbed will go up to 560Nm at 1000rpm - well beyond the engine capability.

At 800rpm the peak torque absorbed is 360Nm - requires 30kW so beyond the engine rated power.

At 700rpm the peak torque absorbed is 275Nm - this would require 20kW and could be possible if the curve is peaky.

At 500rpm this prop will absorb a peak torque of 140Nm.

So using either prop there is no possibility that they will absorb more than 300Nm with the power you have available irrespective of the engine torque curve.

Rick W

[Ad Hoc]

Arnot

Do you know what the reduction ratio is in the gear box?

Sounds like a lovely boat

[peter radclyffe]

Originally Posted by Ad Hoc

Arnot

Do you know what the reduction ratio is in the gear box?

Sounds like a lovely boat

one of the most helpful & highly trained designers on this forum might design you a kort nozzle to increase efficiency

[Ad Hoc]

Peter

Don't understand, care to explain?

[peter radclyffe]

Originally Posted by Ad Hoc

Peter

Don't understand, care to explain?

dont you think a nozzle would be good in this instance

[Ad Hoc]

Peter

You must have me confused with someone else!

The guy is asking about torque on his boat. The only way to ascertain the actual toque of the engine, for converting to hydraulic drive, is to establish what torque he is actually getting from his very old engine at the prop; with some reverse engineering. A few basics are required to calculate it. Knowing how fast the prop is rotating, is one of the parameters required.

Don't understand how this relates to a kort nozzle?..as i said, you must have me confused with somone else!

[peter radclyffe]

Originally Posted by Ad Hoc

Peter

You must have me confused with someone else!

The guy is asking about torque on his boat. The only way to ascertain the actual toque of the engine, for converting to hydraulic drive, is to establish what torque he is actually getting from his very old engine at the prop; with some reverse engineering. A few basics are required to calculate it. Knowing how fast the prop is rotating, is one of the parameters required.

Don't understand how this relates to a kort nozzle?..as i said, you must have me confused with somone else!

i am thinking to supplement a new drive system with something that protects his prop, a nozzle with a cage , fwd of the prop will protect, & increase torque, as he has restricted draft, there is only so much you can increase pitch

[Ad Hoc]

How does that answer his question, viz: "....I am trying to get a rough idea of what the rotational torque on my propeller shaft might be.."..?

The key word here is, "torque", still fail to see what this has to do with a kort nozzle and increasing the pitch?

[Arnot]

Thanks Rick, this is the sort of thing I was hoping for and is very helpful.

The figures you have laid out make sense to me intuitively but I am interested in where they came from.

Starting with the statement “The peak torque that the 16" pitch prop will absorb is 330Nm. That is under heavy towing condition at 3.6kts with shaft at 1000rpm”, is this calculated or based on measurement? By the way the prop diameter is 22”. If it is based on calculation, how can I replicate this? And following this, what would be the calculated shaft torque be if the diameter is kept the same and the pitch increased to 22”?

Again intuitively, I guess that the shaft torque under heavy towing conditions (that probably approximate to acceleration from rest) will fall dramatically as the boat reaches a steady cruising speed.

It seems to me that there is a cubic relationship between propeller speed and torque under any given set of conditions and that makes sense to me (more or less) but I am struggling to see how to get a starting figure. So, for example, how would I calculate the torque that my present propeller (22x16) would absorb at 500rpm under heavy towing conditions?

The two hydraulic motors I am looking at purchasing for this are quoted as being able to deliver either 0.62 or 0.94 Nm per bar of pressure and it looks as though suitable pumps will have about 400 bar available. Either motor has a maximum speed rating of 1000 rpm that should be OK.

The flexible couplings that connect the engine to the pump and the motor to the propeller shaft are rated at 560Nm so that should be OK. The pipe work can be arranged to operate reliably at 500 bar so that should be OK.

The conditions that a narrow boat operates under are a bit out of the norm, there are lots of locks to contend with and sometimes they are very close together so being able to accelerate and slow down in a controlled and timely manner are helpful. Also there is often a lot of traffic on canals, low blind bridges and they are far from straight so it is common to find a boat coming towards you with very little notice and no room to pass. This and this give some idea of the sort of thing…

They are both taken on smaller recreational narrow boats and mine is a full length full width boat that is quite a lot more difficult to navigate round the obstacles.

Regards

Arnot

[Ad Hoc]

Arnot

If all you want is "...how would I calculate the torque that my present propeller (22x16) would absorb at 500rpm under heavy towing conditions.."

is a simple formula.

Power (delivered) = 2*Pi*N*Q
N= shaft/prop Revolutions
Q= torque

[Arnot]

Originally Posted by Ad Hoc

Arnot

If all you want is "...how would I calculate the torque that my present propeller (22x16) would absorb at 500rpm under heavy towing conditions.."

is a simple formula.

Power (delivered) = 2*Pi*N*Q
N= shaft/prop Revolutions
Q= torque

Yes this is the right question but I dont see that the formula supplies the answer unless I am missing something...

It seems to be for calculating power at a given speed once you already know the torque, in my case it is the torque I want to know.

Regards

Arnot

[Guest625101138]

Arnot
I have some idea of what you have to contend with as it is aptly describe in a book I happen to be reading right now about an Australian who sailed/rowed a Mirror dinghy from North Wales to the Black Sea through the lock system. He had some happy nights on boats similar to yours when owners took pity on him.

The numbers I provided came from calculations. They will be close or higher than your prop is capable of. If you want to do the same calculations you can use the JavaProp applet that can be started from this link:
http://www.mh-aerotools.de/airfoils/.../javaprop.jnlp
This is an analytical tool that is based on the fundamental physics not a series of test data. Hence it can be used over a wide range and is suitable for air or water providing the "Options" are set to match the medium.

I have already given you the data for the 22 X 22 prop. I did not give it for 900rpm so the attached provides that data. This is a straight screen image from JavaProp. The peak torque is 398N and is achieved around 5kts. However note that to get this torque the motor would need to deliver 37Kw to the prop - not possible with a 25kW motor.

The second image is the 22X22 prop at 800rpm. You can see here that 25kW will give just over 300Nm.

The third image is 22X22 at 700rpm. At this shaft speed the prop can never absorb more than 17.7kW so peak torque is 240Nm.

Bearing in mind that your hydrostatic transmission is likely to lose 10 to 15% of engine power if it is variable speed, the numbers above are conservative.

In any event you can protect the system by providing pressure relief as noted.

You will find you will get the best braking by gradually slowing the prop so it does not slip too much. Once you get high slip there is a lot of turbulence and the prop will probably suck air.

If you want to play with JavaProp I can get you started. It helps if you have an understanding of foils and propellers but it will start to make sense after a little use.

Rick W