How do you compare electric motor power to diesel power?

Engineer Cockey kindly gave me several formula one of which was torque =power times RPM. 60 amps @ 24 volts=1440 watts. This is for 2 motors and props, so each would be recieving 720 watts. If 750 watts is a horsepower, then each gets 0.96 hp juice.
The two most popular trolling motors are motorguide and Minn Kota. MotorGuide claims 95% efficiency for their motors and Minn Kota claims they're more efficient than the competitors. Hype! What efficiency do these motors operate at? I'd love to believe 95% but lots of 3 phase motors claim 85%. So I'll assume it's 85% or better. Gives us .816 shaft hp each motor. Torque at .816 hp times 750 RPM (small increase over 700 rpm cause dont believe only 15% slip), computes 571.2 lbs torque producing 200 lbs of thrust at 5 kts each motor.

Is it reasonable to believe these motors produce 571.2 lbs torque? Is my math messed up somewhere?
Have to research more

 

yep! messed up. was to divide not multiply.

 

.816 hp divided by 750 RPM = 0.001088 lbs torque. That can't be right! I can sneeze harder than that!
HELP!

 

http://bbs.trailersailor.com/forums/articles/index.cgi/noframes/read/23

the above link suggests one lb of thrust for every 100 lbs of displacement for propelling a boat with a trolling motor. At 3800 lbs displacement, that would be 38 lb thrust trolling motor just to get it to move. Seems the 160 lb thrust of the un-modified Minn Kotas would do the job. And then I modified them. I'm totally ignoring the 25% increase in thrust theorectically claimed as a benefit of Kort nozzles.

 

Stumble claimed got a couple knots on a 3600 lb 30 ft Olson with a 40 lb thrust trolling motor on this forum.
http://www.boatdesign.net/forums/outboards/using-trolling-motor-outboard-28179.html

 

Mr Cockey,
Is that formula
Torque=power divided by RPM in error?
I have a 3hp router turns 10,000 RPM. If this is the correct formula, then torque =.0003 ft lbs and I guarantee it's a LOT more than that. If it were'nt soft start, the centrifugal force at start up would tear it from your hands.
Please HELP!

 

I found correct formula on net. It's:
Torque (lb.in) = 63,025 x Power (HP) / Speed (RPM)

So 63025 times .816 hp = 51428.4 divided by 750 RPM = 68.57 in lbs or 5.714 ft lbs.
The NET GAIN MOTOR spec sheet posted by Boston gives 10 ft lbs torque for a motor drawing 72 volts @ 96.8 amps and 3919 RPM.
So I reckon 5.714 ft lbs of torque from these little 24 volt motors is darn good and certainly can spin the 11 inch aluminum props.

Have I messed up or overlooked anything? Please comment

 

http://continuouswave.com/ubb/Forum4/HTML/005903.html

this link says power= torque times RPM. same as Mr Cockey

it also saysL
power = thrust times speed.

 

so dividing both sides of equation by thrust gives speed= power divided by thrust.
That's high thrust at given power=lower speed. Ok. High thrust, low pitch props provide lower speed at a given RPM.

 

Good plan as those nozzles create added wetted surface and drag that is often overlooked.

 

(80 lbs thrust times 750 RPM) divided by 63025 to convert to HP = 0.9520031733439111 HP
Real close to the 24 volts times 30 amps ( for 1 motor) divided br 750 = .96 HP

 

Thanks Milehog. Do you see any glaring errors in my math or reasoning? other than those I caught and corrected?

 

I'm in no position to correct the math or theory. The link Chuck provided in his post above is by no means the only example of electric HP being greater than gas-diesel HP claims I've seen debunked.

Earlier in another thread you refrenced smaller electric pumps outworking bigger Detroit Diesel powered pumps. You stated the diesels were rated at 2800 rpm but only allowed or able to turn 2200 rpm. Those old Jimmies have to be wound tight to produce usable power, they are designed to bounce off the governor (and sound like they are about to explode) all day long. If they were geared wrong, mated to the wrong pump or throttled back you prolly weren't seeing seeing their true potential.
Is it possible this experience has biased your assessment of power systems?

 

No, power is not torque times RPM. The correct formula is:
Power (W) = Torque (Nm) x Angular speed (rad/s).
Since rad/s is proportional to RPM, Nm is proportional to ft lbs and Watts are proportional to HP, then the formula can be written as:
Power (HP) = K x Torque (ft lbs) x RPM.
K is the conversion factor, which might be 63025, but I am not sure. I always work in SI units - less chance of making errors. If I need to get the result in Imperial units, then I do the conversion of the final result, from SI to Imperial.
Cheers

 

thanks daiquieri,
I seem to find formulae, including ones engineer Cockey gave me, written in their most basic form with no explanation of the variable and appropriate units of measure.
of all the research I did in recent 24 hours, I still couldn't find a formula or method of determining neither how much HP nor torque was needed to swing a prop of given diameter, pitch, and blade area, number of.
Do you have a method or know of one?
I downloaded javaprop but it's not working for me.