Looking through the archives, I found another old post where I was backing into the propellor efficiency (not just prop slip) of my electric conversion:
Hi Redu.
But we already know that the answer has to be 2450W, assuming 100% efficiency between the batteries and the propeller. Personally I know that some energy is lost in the cabling, controller, motor and gearbox. Obviously not much, so for now let's keep using 100% efficient driveline, if we don't, the predictions are even farther off.
There are only two variables in your lower equation that we don't actually know, the propeller efficiency and the actual drag at 5kts. We also know that your applet is calculating for low drag hulls, I would believe that my full keel hull had more drag than the wirefram model that the applet generated, but for now let's use your calulated figures.
My simplified version of the formula for my boat is:
Pe * 0.0023 * Drag * 5kts = 2.45kW
I messed around with your web page and none of the prismatic coefficient changes generated drag forces of less than 170lbs so I'm good with that figure.
Now we've got: Pe * 0.0023 * 170lbs * 5kts = 2.45kW
So if the drag force on my hull is 170lbs at 5kts, then my propeller efficiency must be 1.253 or 79.8% at 5kts. Does anyone here believe that my propeller is 80% efficient at 5kts?
We already said that the drag was probably an underestimate, but apparently not. To make the formula work with a 60% prop efficiency, the drag has to be only 128lbs at 5kts. And that drag is not only hull drag, My measurements of battery load included all of the drag above the water line of the cabin and the rig.
And it get worse again. Let's say that my driveline is 83% efficient in turning electricity into shaft power. My drive line has a calculated efficiency of 83%, multiplying the stated efficiencies of the wiring, controller, motor, gearbox, thrust bearing and stuffing box. That means that the 2450W coming out of batteries should only put 2035W of power into the propeller shaft. Now we're at: Pe * 0.0023 * Drag * 5kts = 2.035kW. Using 60% Pe, we get 106lbs of total drag at 5 kts.
Here's the weird part. A few years ago, James at Propulsion Marine did some towing tests of his Catalina 30 with a scale in the tow line. A Catalina 30 is a little wider than my hull, but it's a fin keel instead of my full keel so I'm going to guess that our total drag is pretty close. In his experiments, the Catalina 30 put 100 lbs of tension on the tow line at 5kts (he did have his propeller off, so no extra prop drag). So let's do the excercise again using info that has been observed in the real world.
I measured 2450W of battery power used to sustain 5kts. Applying the 83% efficiency to the driveline, power at the shaft is 2035W. Using the 100lbs of drag, we get the following:
Pe * 0.0023 * 100 lbs * 5kts = 2.035kW or Pe = 1.77 or 56%
Prop efficiency of 56% at 5 knots is much more believable. So I also believe that we have confirmed that the drag of my hull at 5kts must be around 100 lbs, certainly nowhere near 170 lbs which requires a prop efficiency of more than 96%.
Since we already know the results, we can use my boat to validate any calculated power or drag predictions. What I need to see to believe a power prediction calculator is one that predicts something close to 2035W to drive my 30' 10,200 lb ketch at 5 kts. A drag calcuator that predicts close to 100 lbs of drag would also lead to the same answer. I invite everyone to try to poke holes in this analysis, in academic circles, that is known as peer review and I appreciate any help in making my work more accurate.
Fair winds,
Eric
Marina del Rey, CA
But we already know that the answer has to be 2450W, assuming 100% efficiency between the batteries and the propeller. Personally I know that some energy is lost in the cabling, controller, motor and gearbox. Obviously not much, so for now let's keep using 100% efficient driveline, if we don't, the predictions are even farther off.
There are only two variables in your lower equation that we don't actually know, the propeller efficiency and the actual drag at 5kts. We also know that your applet is calculating for low drag hulls, I would believe that my full keel hull had more drag than the wirefram model that the applet generated, but for now let's use your calulated figures.
My simplified version of the formula for my boat is:
Pe * 0.0023 * Drag * 5kts = 2.45kW
I messed around with your web page and none of the prismatic coefficient changes generated drag forces of less than 170lbs so I'm good with that figure.
Now we've got: Pe * 0.0023 * 170lbs * 5kts = 2.45kW
So if the drag force on my hull is 170lbs at 5kts, then my propeller efficiency must be 1.253 or 79.8% at 5kts. Does anyone here believe that my propeller is 80% efficient at 5kts?
We already said that the drag was probably an underestimate, but apparently not. To make the formula work with a 60% prop efficiency, the drag has to be only 128lbs at 5kts. And that drag is not only hull drag, My measurements of battery load included all of the drag above the water line of the cabin and the rig.
And it get worse again. Let's say that my driveline is 83% efficient in turning electricity into shaft power. My drive line has a calculated efficiency of 83%, multiplying the stated efficiencies of the wiring, controller, motor, gearbox, thrust bearing and stuffing box. That means that the 2450W coming out of batteries should only put 2035W of power into the propeller shaft. Now we're at: Pe * 0.0023 * Drag * 5kts = 2.035kW. Using 60% Pe, we get 106lbs of total drag at 5 kts.
Here's the weird part. A few years ago, James at Propulsion Marine did some towing tests of his Catalina 30 with a scale in the tow line. A Catalina 30 is a little wider than my hull, but it's a fin keel instead of my full keel so I'm going to guess that our total drag is pretty close. In his experiments, the Catalina 30 put 100 lbs of tension on the tow line at 5kts (he did have his propeller off, so no extra prop drag). So let's do the excercise again using info that has been observed in the real world.
I measured 2450W of battery power used to sustain 5kts. Applying the 83% efficiency to the driveline, power at the shaft is 2035W. Using the 100lbs of drag, we get the following:
Pe * 0.0023 * 100 lbs * 5kts = 2.035kW or Pe = 1.77 or 56%
Prop efficiency of 56% at 5 knots is much more believable. So I also believe that we have confirmed that the drag of my hull at 5kts must be around 100 lbs, certainly nowhere near 170 lbs which requires a prop efficiency of more than 96%.
Since we already know the results, we can use my boat to validate any calculated power or drag predictions. What I need to see to believe a power prediction calculator is one that predicts something close to 2035W to drive my 30' 10,200 lb ketch at 5 kts. A drag calcuator that predicts close to 100 lbs of drag would also lead to the same answer. I invite everyone to try to poke holes in this analysis, in academic circles, that is known as peer review and I appreciate any help in making my work more accurate.
Fair winds,
Eric
Marina del Rey, CA
--- In electricboats@yahoogroups.com, redu <reino.urala@...> wrote:
>
> >
> Hi John,
> The problem is that these "power required" calculators don't
actually work for many of our boats. We had a long discussion last
year about this, I even discussed the problem with Dave Gerr
himself. Dave admitted that his formulas won't work at the lower
speeds and powers that we typically operate at.
> Here's my proof. The caculator that you suggested says that my 30'
ketch with a 24' LWL and 10,200 lb displacement would need more than
6.81kW or 9.13hp. But I know that the boat only draws 2500W from the
batteries to motor at 5kts, less than 37% of the predicted value.
> >
>
>
> One should use eg. Michlet for better hull drag estimation. Michlet was
> actually made for narrow hulls estimation, but it works pretty well for
> wider hulls too.
>
> An easier way is to use this one on my web page (JavaApplet capable
> browser required):
> http://boats.yyrtti.com/performance/hulldrag/hulldrag.html
> Give length, beam, weight, max speed, and you get predicted drag curve
> (not wetted trasom stern or pointer stern hull). You get 170lb at 5kn
> for your example ketch.
> Additionally you must calculate required power from speed and drag
> figures, and you must know your prop efficiency:
> Power = Drag * Speed / prop efficiency
> Prop efficiency may vary 30% to 60% depending ...
> Drag to Power equation (european units way):
> 2 * 0.001 * 0.5148 * 0.4536 * 9.81 * 170 lb * 5 kn = 3.89 kW
> clarifications:
> 2 is: 50% prop efficiency assumed
> 0.001 is: Watts to Kilowatts
> 0.5148 is: Knots to m/s
> 0.4536 is: lb to kg
> 8.81 is: kg to Newton
> Simplified version in case of 50% efficiency and lb and knots:
> 0.0046 * 170 lb * 5 kn = 3.89 kW
>
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Posted by: ewdysar@yahoo.com
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