This is the first time I see that Nm is quoted for an e-motor

I came across information on an e-motor with the following detail concerning the torque on the axle. If we take the maximum continuous power of 12 Kw and their quoted current of 165 Ampere which is at 72 Volt = 11880 watt. Are we able to compare this with an diesel/gas - petrol engine and see whether the quoted is substantial better then what Ocean Volt and Torqeedo trying to sell to their customers.

POWER: Output Power of 12 kW Continuous, 30 kW Peak
These patented axial flux motors are designed to work with sinusoidal (sine wave) controllers such as the Sevcon Gen4.
Features: Quieter Operation
Improved durability
High 92% Efficiency
Smoother acceleration
Lighter weight (only 35 lbs.)
Low torque ripple at low speed
NEW Fully enclosed design
NEW Water cooled stators

NEW Sine/cosine speed sensor
NEW Anodized aluminum
NEW Protective phase terminals cover
Specifications
- 4 pole motor (8 magnets).
- Phase to Phase winding resistance is 0.065 Ohms with 20 turns per phase.
- Maximum recommended rotor speed: 6500 RPM
- Voltage: 0 to 72 VDC input to the control
- Torque constant of 0.12 Nm per Amp
- The Inductance Phase to Phase is 0.05 Milli-Henry
- Armature Inertia: 45 Kg Cm Squared
- Current: 165 Amps AC continuous (200 Amps DC into the motor control)
- Peak current: 450 Amps AC for 1 minute (620 Amps DC into the motor control)
- Weight: 35 pounds
- Peak Stall Torque: 75 ft lb.

 

Does it help you to see a torque rating ?

 

Yes, I am under the impression that every time I hear that there is a conversion formula for converting torque into HP . That would mean some 0.12 nM x 200 Ampere = 24 Nm . for a 12 KW input motor. Thus now we can compare a diesel engine really with a stated efficient e-motor. Forgive me if I have the output torque for a max speed motor at 72 Volt DC wrong and is maybe only 19.8 Nm . Which Diesel engine gives only 19.8 to 24 Nm shaft power out, what would that be in HP or Kw. Bert

 

The same sheet says the stall torque is 75 ft. lb, which is slightly more than 100 Nm. That seems like a reasonable value for a light weight electric motor.
What the motor brings at full power cannot be calculated from a torque constant, there should be a graph included.

 

20 Nm and 12 kW would reguire close to 6000 rpm, which way more than typical diesels do. Yanmar 1GM10 has about the same torque (rather constant from 1800 to 3600 rpm), but obviously clearly less power due to lower rpm.

As CDK pointed out the data given is far from being enough. E.g. for the efficiency only one number is given. The efficiency will be clearly less in most operating conditions and 92% is most likely the peak efficiency (not at peak power). Does 92% include the controller?

 

Formulas for Torque, Rotational speed and Power

Power = Rotational speed x Torque
Torque = Power / Rotational speed
Rotational speed = Power / Torque​

Depending on the units used appropriate conversion factors between units may be need.

Another relationship for an electric motor between power in and power out.

Mechanical Power Out is less than Electrical Power In​

This is true when the motor is accelerating, or running at a steady speed. It is possible for mechanical power out to momentarially exceed electrical power in when the motor is deaccelerating.

 

Hi Joakim and CDK and DCockey, It looks like we getting somewhere. The maximum speed is 6500 rpm thus the Yanmar 1GM10 (9 hp only? ) Could this 20 Nm and 12 Kw electric motor reasonable well compare with the Yan Mar 1GM10 diesel?. I think the efficiency is only quoted for the motor. When converting DC to AC, normally one looses quite some efficiency. Except it the motor runs on square waves, like in pulse width switching. Bert

 

No they are not comparable. 1GM10 has only 6.7 kW output and the electric motor probably around 10 kW. 1GM10 is used with e.g. 2.64:1 gear ratio, thus its 18 Nm torque is around 45 Nm (some losses included) at the propeller shaft.

The electric motor could use a 4-4.5:1 gear ratio and thus have clearly more torque at the propeller shaft. 20 Nm at 6500 rpm is 13.6 kW, thus it will not have that at 12 kW electrical input. Maybe 10 kW, which would be 14.7 Nm at 6500 rpm and over 60 Nm at the propeller shaft with 4.5:1 gear ratio.

 

Bert
Attached is a good article that clearly explains, torque, horsepower and rpm.

Power and Torque: Understanding the Relationship Between the Two, by EPI Inc. http://www.epi-eng.com/piston_engine_technology/power_and_torque.htm

As I stated in an earlier thread, torque is a static value, pounds x feet, or Newtons times meters, These are measured static values. Torque being the amount of twist, force times a distance away from the center of a shaft.
In this thread you said "there is a conversion formula to convert torque to horsepower" This is incorrect. There is static torque, twist in the shaft non moving (it could be moving, ie the same twist in the shaft at a PARTICULAR RPM) . So you are unable to convert torque to hp. BUT, if you introduce rpm, speed, force through distance with respect to time, with the formulas that DCockey stated in his response, you can see that in order to involve a value for torque into an equation of horsepower, you need an rpm.
You measure the amount of torque with a dyno at a specific RPM. Then you plug these numbers into the formula to derive the rate of doing work, horsepower, watts.

CDK added, "what a motor brings at full power cannot be calculated from a torque constant, there should be a graph included"
And this graph would have the rpm on one axis. The torque/hp curve.
Note that these curves whether an electric motor or gas or diesel show THAT THESE ARE MAXIMUM NUMBERS THAT THE MOTOR/ENGINE CAN PRODUCE AT THAT RPM
For hydrocarbon burning engines, you would not want to run them at the max hp values at a specific rpm as they use more fuel for a given rpm than if you pull this amount of horsepower off
the engine where it can produce more at max, than what you need it to. (many factors in play, incomplete combustion of the fuel due to lack of available oxygen, high EGT's, and more which affect the thermal efficiency of the burn)
So if a particular engine can produce 50 hp MAX from a HP curve at say 2500 rpm, and you need 50 hp to drive whatever you are driving, you IN MOST CASES gear the system so the engine is running at 3000 rpm (where is could produce 75 hp) but only pull 50 HP off it. (these numbers are only used to illustrate the concept)

So if you are trying to compare a specific electrical motor to a specific diesel engine as an input source into a gear train (which might have to be ratio'd to accommodate the hp required at the prop at your target prop rpm) you need both graphs, pick the horsepower at the shaft that will move your boat, turn a pump, crank a winch, and do the comparison.

You need to compare horsepower to horsepower ( watts to watts)

 

But we have the speed it is 6500 rpm.

 

You have the maximum rpm rotor speed of 6500. Are you wanting to say that at 6500 max rotor rpm, the motor is producing 12 KW output?

If this is so, then a gas/diesel engine needs to produce the same 12 kw output on a continuous basis.
And then you need to supply the gearing (assumed frictionless just to keep it clean) to turn the prop at the same rpm.

 

No I said torque of 24 Nm and 6500 Rpm and INPUT power 12 Kw, I was hoping that you boffins would be able to tell me what the HP/power would be on the axle. I am not a mechanical fellow, although I can use my hands very well. Electronics and software is more my line. I can tell you what my brushless motor could produce at the axle by deducting all the losses in the controller and switching. Bert
P.S. if I knew the flux phi of my magnets.

 

What is meant by "static" in this context? That there is no "time" quantity appearing such as in "revolutions per minute"?

 

Actually there is (almost) enough data to calculate efficiency map. Friction moment (or current) would still be needed and motor driver electronics and cable losses as well. Also it would be necessary to know the temperature of the windings in order to correct for the increase of its resistance. I have an Excel sheet that calculates the output of a BLDC motor based on these values. Using 100% efficiency for the driver and the cables and no friction, I got the following:
With 19.8 Nm output (165 A) the efficiency is 88% at 6500 rpm and lower at lower rpm (e.g. 82% at 4000 and 70% at 2000 rpm). 1770 W heat is produced due to winding resistance. In real life the windings will get hot, resistance increases and more power is wasted. 12 kW input will be reached at 4900 rpm with 85% efficiency (10,2 kW output) or at 15.9 Nm 6500 rpm with 90% efficiency.

If torque is dropped to 10 Nm (83 A) the power loss in the windings drops to 450 W, which would mean 94% efficiency at 6500 rpm. Further reducing torque to 1 Nm (8.3 A) would increase the efficiency to 99% (only 5 W power loss). This is not a real case, since friction will not allow the efficiency to reach so high values.

If we quess that the friction moment is 1% of the maximum output, thus 0.2 Nm, we get the peak efficiency of 92% at 3-10 Nm output. At full power (19.8 Nm) the efficiency drops just a little (87% at 6500 81% at 4000 rpm and 70% at 2000 rpm). At 1 Nm the efficiency is 83 % at 6500 rpm and 81% at 2000 rpm.

If the windings heat up from 20 C to 100 C the resistance increases from 0.065 ohm to 0.085 ohm. With that and the 0.2 Nm friction the efficiency at 12 kW is at 19.8 Nm and 4600 rpm 80% or at 15.3 Nm and 6500 rpm 87%.

The cabling will and the driver will further reduce the efficiency especially at high currents. At low power the driver will again reduce the efficiency.

Note that 72 V can only drive this motor to 5700 rpm at no load. At 19.8 Nm load it is enough for 4900 rpm.

 

Well...boffin or not, the numbers you quote do not end up; somebody is making incorrect statements. Shaft torque (in Nm) times shaft speed (in radians per second) gives the shaft power in W. In your example, 6500 rpm equals 6500 x 2 x pi / 60; i.e. 680.68 rad/sec. Alas, shaft Power is 24 x 680.68 = 16336.3 W. In my world, that is not possible with an input power of 12 kW......