How do you compare electric motor power to diesel power?

Thrust = Power / Speed

1) The units have to be consistent. For example power in Thrust in Lb, Power in Lb Ft / Sec (1 HP = 550 Lb Ft / Sec), Speed in Ft / Sec.

2) Power in this equation is the effective power, ie the fraction of the power in the shaft which gets turned into propulsion of the boat. The rest of the shaft power goes into heat in the shaft bearings (small) and churning of the water in the wake of the propeller (not so small). The shaft and propeller efficiency can range from maybe 75% to more typically 60% to 45%. It depends heavily on the propeller size and geometry.

 

Thankyou engineer Cockey
Reworking my calculations:
power = thrust times speed and conversely, thrust = speed divided by HP. Here HP more or less = 2 or 1010 ft lbs. (2 one Hp motors w/2 props)
Speed's 5 kts or 505 ft per min (1 kt = 101 ft min).
505 (fpm speed) divided by 1010 (2 Hp in ft lbs) = 0.5 lbs thrust.
Don't believe it. I'll solve for 1 Hp, as if only one motor did all work.
505 (fpm speed) divided by 550 (1 Hp in flbs)= 0.9181818181818182 thrust? Less than 1 lb thrust? Well, it's twice the thrust I got with 2 motors! I believe in greater thrust with low Hp electric motors but....NAH! Something is screwy here.
I must be getting at absolute minimum, 38 lbs thrust to even move a 3800 lb boat.
Solving for power:
power = thrust times speed. 38 lbs thrust times 505 ft per min =19190 divide by 550 = 34.89090909090909 or 35 Hp. Nah! Wrong! I'll assume I'm screwed up, not the formula. Where am I confused Mr Cockey, or anyone. Daiquiery?

 

I see it
Power = thrust divided by speed...slap my face. WAKE UP!

 

Reworking my calculations:
power = thrust divided by speed and conversely, thrust = speed times HP. Here HP more or less = 2 or 1100 ft lbs. (2 one Hp motors w/2 props)
Speed's 5 kts or 505 ft per min (1 kt = 101 ft min).
1100 (2 Hp in ft lbs) times 505(fpm speed) = 555,500 lbs thrust.
? Wow! Those Kort nozzles are TERRIFIC! Nah! I still don't believe it. I'll solve again for 1 Hp, as if only one motor did all work.
505 (fpm speed) times 550 (1 Hp in flbs)= 277,750 thrust? Well, it's headed right direction but unbelievable! I must be getting at absolute minimum, 38 lbs thrust to even move a 3800 lb boat.
Solving for power:
power = thrust divided by speed. 38 lbs divided by 505 ft per min =0.0752475247524752 Ft lbs .Unbelievable even before converting to Hp. But divided by 550 = 1.368136813681368e-4 Hp. Nah! Wrong! I'll assume I'm still screwed up, not the formula, but the formula is getting VERY suspect!
What's up with this?

 

Power = Thrust * Speed
Double the speed and the power doubles if the thrust remains the same. Double the thrust and the power doubles if the speed remains the same. Double both speed and thrust and the power quadruples.

A consistent set of units is Thrust in Lb, Power in Lb Ft / Sec (1 HP = 550 Lb Ft / Sec), Speed in Ft / Second (not Ft / Minute).

 

Ok thanks...I'll do again

 

Power = Thrust * Speed
Thrust = Power / Speed
Speed = Power / Thrust
Remember that for a boat the shaft and propeller efficiency means that only a fraction of the power out the engine is converted into useful power pushing the boat.

 

1 kts = 1.68780986 feet per second
power = thrust * speed
38 lbs thrust times 8.4390493 fps (5 kts) then divided by 550 conv to Hp=0.583061588 Hp
I'll accept that.
Now 2 Hp = 1100 lbs ft
1100 / 8.4390493 = 130.3464360612279 lbs thrust.
I believe that.
Whew, thanks for straightening me out.

Results:
I'm getting 5 kts with 2 hp electric on my 3800 lb boat.
The 2 motors are capable of producing 3 +/- Hp together.
The 5 blade props I'm using with Kort nozzles produce more thrust than the original 2 blade inefficient weedless prop.
To do so, requires more energy at same RPM or a lower RPM for same energy.
I'm only turning 750 RPM more or less, but far below max RPM of motors. It's unlikely these motors could spin these props at 1400 RPM. I don't need them to.
Thanks Mr Cockey.
Any problems with my workup now? Violating any principles of physics?

 

Because of inefficiency, I'm certain I'm getting less than the theoretical 130 lbs thrust. But apparently I'm getting more than 38 lbs, and enough more to push her to 5 kts. Whats actual thrust developed? I'd sure like to know.

 

or is it I'm getting more than 130 lbs thrust and because of inefficiency I'm only getting 5 kts?
If I wasn't crazy before, I may be headed there!

 

Power (HP) = Torque (Ft Lbs) * Rotational Speed (RPM) * 0.0001904
Torque = Hp / (RPM * 0.0001904)
2 Hp / (750 * 0.0001904) = 2 Hp / 0.1428 = 14.00560224089636 lbs Torque
or is it "like" quantities as coverting Hp to lbs ft:
1100 lbs ft / 0.1428 = 7703.081232492997 lbs torque?
Seems obvious the 14 lbs torque would be more rational. I'm still confused about when I need to convert units of measure.

 

Yobarnacle, I'm on vacation these days and am writing this from my iphone. Since it's a pain in the a$$ to write anything more elaborate than 4-5 lines of text on this tiny screen, I can't help you with your calculations. But I just have to note that it is useless to use all those decimals in your calculations when you're unsure even about your exact engine power, the propeller efficiency or the hull resistance at any given speed. You can use just two decimals in your calcs and be sure that the final result is as true as a one obtained with 10 decimals.
Cheers

 

thanks. Usually I use decimals to 2 significant places. Since I was using an on screen calculator, it was simple to copy result complete.
I was trying to deduce the missing information by calculations. Obviously resulting data is in-exact. But good ball park figures are obtained, and principles can be learned, determined, and compared.
I appreciate your input. Without double checking my math, do you see any where idiotically I'm flying in face of expert opinion?

 

Power units are lb ft per sec or lb ft / sec. 1 HP = 550 lb ft per sec. The per sec is required.

1 HP or power is needed to lift 1 lb 550 feet every second or 550 lbs 1 foot in every second or 55 lbs 10 feet every second (assuming the lifting mechanism is 100% efficient).

Torque is ft lbs. Thrust is lbs.

You can't simply convert torque to thrust. The mechanism which converts rotational (shaft) power to linear thrust/force has to be considered.

 

Assuming your math is correct and the motors output equals your estimate and the boat speed is 5 knts then IF the propellers were 100% efficient the thrust would be 130 lbs. But for several reasons the propeller efficiency is not 100% and probably in the range of 40% to 60%. If propeller efficiency is 50% and the other assumptions are correct then thrust would 65 lbs.