Engine torque or power

[trimix]

Hi,

I was wondering which parameter is important for a marine engine which drives a propeller.

For a given boat when do I consider the torque of the Engine and when do I consider the power ?

(for a car I know that the torque allows the car to accelerate and the power to maintain a high speed)

Thanks for your help

[rasorinc]

You need torque to push a boat through water. High power low torque is of less value then high torque less power

In our Marine Diesel engine world, you may simplify,
high speed, high rpm, small prop diam. high power
low speed, low rpm, larger prop, high torque

Regards
Richard

[FAST FRED]

For a high speed boat HP is the game.

For most of us that pay for our own fuel the ability to match and not UNDERLOAD the engine to the prop at the speeds we cruise is the name of the game.

Most cruisers are overengined by 50% to 300% , at the cost of fuel economy and reduced engine life.

Shoot to use 70% 80% of the engines 24/7 rating at your favorite speed , hopefully near the torque peak on the Mfg chart.

FF

[Guest625101138]

Originally Posted by trimix

Hi,

I was wondering which parameter is important for a marine engine which drives a propeller.

For a given boat when do I consider the torque of the Engine and when do I consider the power ?

(for a car I know that the torque allows the car to accelerate and the power to maintain a high speed)

Thanks for your help

The relationship between power and torque is:
Power = Torque X Rotational Speed.

Rated power for any engine is the peak power it can achieve. Normally is it close to the peak rpm.

If a motor has a flat torque curve it is easier to match to the torque demand from a prop over a range of operating conditions. With most boats using typical IC engine there is little to no advantage in having variable gearing. If you have the gearing and prop best suited to the boat and motor then it usually works well through the range of speed.

Designing the propulsion system for a boat and given motor gets down to what is more important or what compromise you want to make. You can usually gain better fuel economy by overpropping the motor, which may not then deliver full power before it is torque limited at the top end. Having two gears would enable both conditions to be achieved but then there is extra weight and complexity so the benefit is likely to be very small.

If you give more idea on the type of boat and motor along with what objective you hope to achieve you might get more useful input.

Also your statement about the car could be misleading. The more important aspect for acceleration is for the motor to develop high torque at low rpm. You could have different motors with identical peak torque and peak power that perform differently. The motor that has the wider torque band will out accelerate the motor with a narrow torque band.

Rick W

[Ad Hoc]

trimix

"..I was wondering which parameter is important for a marine engine which drives a propeller..."

It really should be said the other way around.

Without knowing a single bit of info regarding your "problem", one can summarise as thus:

The RPM, delivered power, diameter and the speed of advance all dictate the design of the prop and hence its performance. In addition, you need to check the cavitation no., so you'll need to know, or, estimate the static pressure and dynamic pressure.

So given that all you have is an engine with "delivered power", and possibly a gear box to obtain the some kind of RPM, the prop still has many other variables that affect the overall performance.

That's before getting in BAR ,number of blades and so on...It is all a balancing act.

However a quick rough and ready guide has been posted by apex1 above.

[Mark Emaus]

Originally Posted by Ad Hoc

trimix

"..I was wondering which parameter is important for a marine engine which drives a propeller..."

It really should be said the other way around.

Without knowing a single bit of info regarding your "problem", one can summarise as thus:

The RPM, delivered power, diameter and the speed of advance all dictate the design of the prop and hence its performance. In addition, you need to check the cavitation no., so you'll need to know, or, estimate the static pressure and dynamic pressure.

So given that all you have is an engine with "delivered power", and possibly a gear box to obtain the some kind of RPM, the prop still has many other variables that affect the overall performance.

That's before getting in BAR ,number of blades and so on...It is all a balancing act.

However a quick rough and ready guide has been posted by apex1 above.

What does BAR stand for? What does it mean?

Mark

[Ad Hoc]

Mark

BAR = Blade Area Ratio.

It is the total, projected or developed area of the blades on the prop. Where the developed area is the sum of the face area of all the blades, and projected area is the projection of the blades (like normal dwg projection one may have done at school or uni) onto a plane normal to the prop axis.

The thrust developed by the prop varies directly with the surface area.

[alan white]

A thing that is pushed around in a circle with one pound of pressure at one foot from the axis covers 6.2918 feet per revolution (2 x radius x pi). If it takes a minute to go around, the force required to push it is said to be one ft lb at one RPM.
This can be used to figure what that means in horsepower.
Really, the circle can be straightened, so you can also say that raising a pound 6.2918 feet vertically is the same thing, as the pressure exerted is in addition to gravity.
A horsepower is the power required to raise 33,000 lbs to a height of one foot in one minute.
If we are raising one lb to 6.2918 feet in one minute, we need much less than a horsepower. We can also say we can raise 6.2918 lbs one foot in one minute. 33,000 divided by 6.2918 is 5,252.
One horsepower is then equal to one ft lb @ 1 RPM x 5,252.
Also, since 1 RPM x 1 ft lb x 5,252 = 1 hp, and the square root of 5,252 is 72.5, then 72.5 ft lbs @ 72.5 RPM = 1 hp. So if the RPM is 50 times as much, at 3625 RPM, at 72.5/50 lbs (1.45 lbs) you still have 1 hp. A diesel boat engine might run at 3625 RPM, and if it had 100 hp at that speed, it would have 72.5 ft lb of torque. Or, it could have 72.5 HP and 100 ft lbs of torque @ 3625 RPM.
I think.

[Mark Emaus]

lol....That is a lot of info....but it now gives me a place to start to figure what I need to know. One of the ideas we ave is to make the mudmotor a hydraulic system. I can see where this will help. I saw a BAR spread sheet and I was wondering what it stood for, (not any more).

Mark

[alan white]

A lot of the time, engine power is expressed in kilowatts, since both measurements describe actual power over time (They could be expressed as kilowatt hours (KWH), or horsepower hours when applied to continuous output like engines..

Deriving HP from torque and RPM also gives Kilowatts. The simple way to instantly remember how many KW are equal to a horsepower is to recall that two HP are equal to 1492 watts, the date Columbus "discovered" America
(1.492 KW). So one HP is half that, 746 watts (.746 KW).
One hundred HP is the same as 74.6 KW.

[kroberts]

At what size/scale of boat, either power level or length of boat, does it become practical to use a motor-generator as a transmission, the way a modern train does?

Sorry if this sounds like an off-topic post, but the whole idea in this case is to more accurately match the torque and power generated by the engine to the needs of the propeller in the water.

[Guest625101138]

Originally Posted by kroberts

At what size/scale of boat, either power level or length of boat, does it become practical to use a motor-generator as a transmission, the way a modern train does?

Sorry if this sounds like an off-topic post, but the whole idea in this case is to more accurately match the torque and power generated by the engine to the needs of the propeller in the water.

The power demand of the propeller is well under the power produced by an IC engine right up to the design point. There is no need for gearing unless you want to aim for optimised fuel efficiency. There is a size where the fuel efficiency gained by using an electric transmission would offset the losses in the electrical system plus the higher capital costs involved.

There are other factors that might come into the equation such as the boat may have a high electric demand because of the function it serves and propulsion is only one of the main loads. There could be space restrictions to mount the IC engines. This is not uncommon in SWATH type vessels so electric is a good choice for the smaller vessels here. There could be need for distributed thrust with a number of side thrusters locate around a vessel. These could be electric.

For a simple cruiser I would expect you would be looking at 1000kW or more before the economics of diesel electric come into play. There have been businesses set up doing smaller systems but I think they have withered through lack of interest.

Rick W

[kroberts]

That's exactly what I wanted to know, Rick. Thanks.

[alan white]

Good question, and while I could maybe guess what efficiencies are served in each scenario, I'd like to hear what folks have to say on the subject. My feeling is there are many many factors at play, maybe the least of which would be a slight difference in fuel consumption.