How do you compare electric motor power to diesel power?

Milehog,
I agree that to achieve the rated HP, diesels need to be run at top speeds where they were rated at.
Real world you can't do that.
That's one of the reasons you have to buy a diesel rated larger than your intended needs.
I'll give another example. 645 series EMD. It's rated to turn close to 1000 RPM, but don't you dare. You'll be looking for another job. EMDs are usually run at 900RPM and the older and longer since last overhaul, the lower RPM you need to run at. Three reasons. It ain't a good idea to be broke down with a blown up engine in middle of the ocean. Two, these are very expensive engines to repair or replace. $250k +/-. Three. Their hourly fuel consumption goes thru the roof the last 100 +/- RPM.

 

I tried and i tried. I can send the message with attachment, but my computer only gets the text. Text says there is attachment, but when I attemp to open it, I get a "STACK ERROR".
Maybe it's because I'm in Mexico and their systems aren't up to it, or my phone is sending the photo weird and confusing my computer.
I'll keep trying.
Wife and i are pleased to learn we can text from phones to computers.
I tried texting to our phones from my email, and we haven't recieved the texts.
Must be a one way street down here.
Thanks loads anyway

 

Another reason these weak horsepower electric trolling motors are pushing my boat where before she had installed a 27 hp diesel, then a 15 hp diesel is: i suspect the 27 HP 3 cyl Albin diesel was more HP than required. In designers spec sheet, Per Brohall claims 8 knots as max, and economical cruise speed about 7 knots. On a 22 ft 2 in waterline length.
sqare root of lwl is 4.708148963941845. Times 1.34 = 6.3 knots. Hull speed. At 8 knots, the factor would be 1.7 times square root lwl. I've never heard of a vessel this resistant to wave making. Even at 7 knots, the factor would be 1.49 exceeding theoretical max of 1.4. So she must have had enough power to exceed hull speed and that takes considerable.
The 5.5 kts with the 15 hp yanmar, previous owners said they achieved could be wrong prop, wrong gears (especially if they used the Albin 27 hp gears) or, reluctance to throttle up.

Anyway, I'm not trying to PROVE magic here. I'm trying to understand why this is working when so many say it shouldn't!

 

daiquieri,
Power (HP) = K x Torque (ft lbs) x RPM.
K is the conversion factor, which might be 63025.
i don't believe it could be 63025, because the indicated power requirement would be huge. Couldn't very well be the inverse either. If so HP would be miniscule. I have know idea what K is. Slip? Efficiency?

 

In this case, K is at unit conversion factor.

1 rpm = 2* Pi Rad / 1 Rev * 1 Minute / 60 Seconds
1 HP = 550 Lb Ft / Sec

Power (HP) = Torque (Ft Lbs) * Rotational Speed (Revolutions / Minute) * 2 * Pi Rad / Revolution * 1 Minute / 60 Secs * 1 HP / 550 Lb Ft

Clearing the units results in:

Power (HP) = Torque (Ft Lbs) * Rotational Speed (RPM) *2 * Pi / (60 * 550)

Power (HP) = Torque (Ft Lbs) * Rotational Speed (RPM) * 0.0001904

 

Thankyou.
so torque would equal power divided by Rotational Speed (RPM) * 0.0001904

in my instance, the motors imput 1 hp +/- and RPM is 750 +/-

750 X 0.0001904 = 0.1428
divide into 1 = 7.002801120448179 or 7 ft lbs torque. This should spin an 11 inch aluminum prop weighing 1 kilo.
Yes or No?

 

power = thrust times speed so
thrust = speed divided by HP. Here HP = 1. So thrust = speed.
Speed's 5 kts. Thrust is cetainly more than 5 lbs.
A kt is per hour. Maybe it needs to be feet per minute.
if so, speed 101.27 ft per minute. I'd say 101 lbs thrust sounds right for these motors the way I've set them up.
Yes or NO?

 

Mr Cockey,
Do you see any glaring errors in my workup?
Granted, actual RPM, thrust, torque are unknowns. All the actual data I have is 5 kts by GPS, 60 amps by clamp on meter divided by two motors = 30 amps each and a nominal 24 volts from two 12v batteries in series. Might be 26 volts.
I have the motor mfg specs of max RPM 1400. Max amps 116. Max thrust 160 lbs for the pair or 80 Lbs each.
Diameter and average pitch of props from mfg specs. Weight and tip pitch I measured.
Is my deduction that each motor is producing 100 lbs thrust at 750 +/- RPM, with 1 HP of input reasonable? And that a total of 200 lbs of thrust at 2 HP would/could/did propel a 30 foot 3800 lb boat at 5 knots? Is it reasonable?
Your assessment will be appreciated

 

Power is power... ALWAYS

Horsepower = torque (in ft lbs) x rpm/5252....

The power to drive your hull at a given speed is the same, it doesn't matter what is driving the prop.

Where the misconception comes in generally is that at lower than design speed some engines (steam, turbine and electric) make a lot more torque than a conventional gasoline engine. A gasoline engine has (generally speaking) a constantly rising power curve. That is the torque curve is pretty flat, and that means at 50% engine speed you can only get 50% power. An electric or turbine can make full power at lower speeds and this results in a decreasing torque curve. In something like a car this means that you can make more power at lower engine speeds. For something like a boat, this could help you get on a plane, since you can get more power at lower engine speed. Since cars don't need full power unless you want to go really fast, you can often exchange a reciprocrating engine for a lower power electric motor and get similar performance.

There is a price to pay for the ability to get full power at lower speed in an electric motor. At lower speed and high torque, the motor current is higher and your efficiency is lower. So the simple answer is if you want to go at hull speed and you size and gear your motor properly it will take the same shaft hp to go the same speed.

 

Ok. What shaft HP do I need to go 5 kts on this 30 ft boat. Obviously 2 HP because I did it. What do the formulas say, cause I can't find one.

 

http://forum.woodenboat.com/archive/index.php/t-2329.html

RonW05-11-2005, 07:47 AM
Here is the chart.
1.1 = 900lbs.
1.2 = 700lbs.
1.3 = 550lbs.
1.34= 500lbs.
1.4 = 450lbs.
1.5 = 350lbs
1.6 = 300lbs.

Here's how to use it. Take your waterline length, 19 and find the square root of it.(calculator)
which is 4.35 and multiply it let's say by the 1.34 which = 5.84 knots, to change to m.p.h. multiply that answer by 1.15 which = 6.7 m.p.h.
The horsepower is one horse(of gas or diesel) to obtain this speed at this waterline length for every 500 lbs of weight.
So in your case your would need about a 5 horse internal combustion engine to run a 2,500 lb boat with a waterline length of 19 ft.
*********************************************************************
To change to electric figure 3 and 1/2 horse gas or diesel engine for every one horse electric, so you would only need 1&1/2 horse electric. Of course this is under ideal conditions with no wind and waves or current.

 

When I send a photo to my email, I don't send any text. Maybe that is confusing your computer.

 

Waterline of my boat is 25 ft couple inches at rest because about 2 ft of new stern is overhang. WL is more than 26 ft underway because water piles up aft. I can't measure it at all underway, and not well at rest because the water lapping.
Using 26 ft, square root is 5.099019513592785. In the table above that would be near the coefficient of 1.1 where 1 hp per 900 lbs displacement is required to achieve speed. At 3800 lbs displ, that would be 4 HP and a bit more to achieve 5 kts. The author says 1 hp electric suffices for 3&1/2 HP combustion.
Thus according to above commentator, I need one and half HP electric to propel my boat 5 kts. And that's what I see happening.

 

Here is what I think I have learned.
Hp indeed =Hp. Hp has to be coverted to use it.
Hp in the wind is coverted to thrust on the mast by the sails, which first converted winds Hp into lift. Hp in wind is also first converted to lift by windmill blades, then thru their wheel arrangement and axle, into torque.
Torque is what spins propellers that convert it to thrust.
Electric motors have more torque at LOW RPM than an equal Hp rated combustion engine, which developes it's torque at high RPM. Since the electric developes more early torque, then a combustion engine developing the SAME torque at LOW RPM as the electric, would logically be larger, and a higher rated Hp.
While a 1 Hp diesel multiplying torque through reduction gears, could provide the same LOW RPM torque as the 1 Hp electric, it has no potential to accelerate, already operating at top RPM.
Conclusion: For relatively low RPM applications requiring high torque, electric motors are equivalent to combustion engines of a higher Hp rating.
Opinions differ as to how much higher. 1:2, 1:3.5, 1:5. Which ratio is more applicable seems to depend on type of electric motor, specific application and auxillary machinery.

 

http://www.propguard.net/test.html

I have these on my hydrofan prop equipped Minn Kota trolling motors.
Various persons attribute improvements in thrust by 25% upto 100%.
I ignored any benefit to thrust in my calculations in previous threads.
I really do believe these help.