Help me out the prop size for a permanent magnet motor

[BertKu]

Can we together try to work out what the best type of propeller is for a brushless permanent magnet motor powertrain.
The difference: A permanent magnet electric motor normally has a pulse width controller, controlling the speed. To look at it in detail, we have for example: 100 Ampere, 48 Volt and 4,8 KW motor at 1400 revs switching at 1 millisecond on and 9 millisecond off. It means that during the 1 millisecond the full current and voltage gives the full power and torque. With other words, 4,8 KW torque for 1 milli second. The prop will thus every 1, 11,21,31,41,51 milli Second get a boost and turns a little bit. The prop will thus turn 140 revs. If I make a prop with the maximum diameter for 140 revs. The blade will turn for 1 milli second and pushes the hull 1 millisecond forward. But if I have a ratio of 6 milliseconds on and 4 milliseconds off, I still have the full torque pushing the same blade, but just for 5 millisecond longer.
Here is my dilemma. Do I make a prop for maximum power, i.e. 4.8 KW ( 6.4 HP) and put this into a “propcalc” software program which is designed for IC (Internal combustion) engines. Or should I just calculate it as follow. 1 HP equals 75Kg lifting in 1 second 1 meter. Thus if my boat displacement is 500 kg, losses ignored, I need at least a prop which pushes and claws through the water with a blade area which can comply with that.
Thus, if I like to go at full speed 5 knots, I need to push 500 kg x 5 knots x 1.852 Km = 4630 in 3600 seconds. Thus 4630 : 3600 = 1,111 Hp
i.e. I need 1.111 HP. But I will have losses and extra resistance, not many people can predict what I will encounter. Thus I take it 200% extra, thus I need at least 3,3 HP I am thus able to calculate roughly what the blade area and angle and diameter has to be at 3,3 HP. Luckily I have 6.4 HP available. Thus theoretically I can have the maximum diameter and run at half available average power.
Half average power means 700 Revs . If my angle of attack clawing through the water is 45 degree, and need to do 9.26 km in one hour (5 knots) , I need 154,3 meter per minute or 0,22 meter per ONE prop revolution. 0.22 meter = 220 cm forward in one turn. That will be a hell of a big blade.
What is your comment. The torque will handle any size prop as well as the available average power and maximum power. I have proven and explained that at the top. Should I go for a 20 inch prop ??
Bert

 

[johneck]

Bert, You are on the right track, but maybe not quite on the right train. What you are trying to calculating is what naval architects call effective horsepower (EHP). This is the useful work done by the propulsion system. I believe that your calculation would be correct for lifting the boat, but not necessarily for pushing the boat. The formula for EHP is Rt*V, where Rt is the total resistance of the vessel.

The losses you describe can be rooled up into a single propulsive coefficient to calculate brake horsepower required. A propulsive coefficient of 0.50 ( which is your 200% factor) is certainly conservative for an easily driven (ie long, thin, clean) vessel, but may not be conservative for a barge or something boxy and slow.

It doesn't matter how much power you have available, only how much you will use. So for your problem of 3.3 HP for 5 kts we can calculate an optimal propeller for any given RPM. For 700 RPM, optimal diameter is about 13-14". The pitch required will vary depending on the mean speed of the flow where the propeller is located. For a propeller mounted behind a vessel this could be much slower than the vessel speed due to boundary layer. If I assume a 20% factor, I calculate a pitch of 14". The loading on the blades is very low for this case and not much blade area will be required (DAR 0.50)

Estimating vessel resistance is of course the difficult part of this problem, but there are many ways to get a rough estimate based on wetted surface, basic vessel attributes, etc.

 

[BertKu]

Bert, You are on the right track, but maybe not quite on the right train. What you are trying to calculating is what naval architects call effective horsepower (EHP). This is the useful work done by the propulsion system. I believe that your calculation would be correct for lifting the boat, but not necessarily for pushing the boat. The formula for EHP is Rt*V, where Rt is the total resistance of the vessel.

The losses you describe can be rooled up into a single propulsive coefficient to calculate brake horsepower required. A propulsive coefficient of 0.50 ( which is your 200% factor) is certainly conservative for an easily driven (ie long, thin, clean) vessel, but may not be conservative for a barge or something boxy and slow.

It doesn't matter how much power you have available, only how much you will use. So for your problem of 3.3 HP for 5 kts we can calculate an optimal propeller for any given RPM. For 700 RPM, optimal diameter is about 13-14". The pitch required will vary depending on the mean speed of the flow where the propeller is located. For a propeller mounted behind a vessel this could be much slower than the vessel speed due to boundary layer. If I assume a 20% factor, I calculate a pitch of 14". The loading on the blades is very low for this case and not much blade area will be required (DAR 0.50)

Estimating vessel resistance is of course the difficult part of this problem, but there are many ways to get a rough estimate based on wetted surface, basic vessel attributes, etc.

Thanks John, I will take your advise. Just for my own piece of mind, is there a difference using brushless motor data and normal IC data in a propeller calucation software program?. Bert

 

[johneck]

No, not directly. fixed pitch propeller is designed to absorb a given power at a given RPM and speed of advance. The way that a designer would use the engine data (torque vs RPM) is by deciding the best operating point to select. This is especially true for a vessel with several different operating points, ie loading conditions, cruise vs loiter vs top speed or a towing conditon, etc. Each of these might have an optimal propeller design, so it is necessary to look at each and make sure that the design will allow all the conditions to be met without overloading the engine or having a hydrodynamic issue with the prop. For many (most) applications, the design point is the rated power at rated RPM and max speed. At any other operating point, you get whatever you get.
With the PM motor it is actually easier to make the design work, since there is almost full torque available at any RPM. this means that there is great flexibility in selecting a design point without having an overload condition.

 

[BertKu]

No, not directly. fixed pitch propeller is designed to absorb a given power at a given RPM and speed of advance. The way that a designer would use the engine data (torque vs RPM) is by deciding the best operating point to select. This is especially true for a vessel with several different operating points, ie loading conditions, cruise vs loiter vs top speed or a towing conditon, etc. Each of these might have an optimal propeller design, so it is necessary to look at each and make sure that the design will allow all the conditions to be met without overloading the engine or having a hydrodynamic issue with the prop. For many (most) applications, the design point is the rated power at rated RPM and max speed. At any other operating point, you get whatever you get.
With the PM motor it is actually easier to make the design work, since there is almost full torque available at any RPM. this means that there is great flexibility in selecting a design point without having an overload condition.

Thank you for the explanation.
Bert

 

[BertKu]

With the PM motor it is actually easier to make the design work, since there is almost full torque available at any RPM. this means that there is great flexibility in selecting a design point without having an overload condition.

John, a quick question. What is the minimum space between the top of the 14inch prop and the surface of the water? Also when I build a tunnel around the prop does that affect the minimum distance. Many thanks
Bert

 

[johneck]

It is a quick question, but I don't have a quick answer. The problem of ventilation of a propeller is very complex. Adding something over (or around) the propeller does change the situation, it should reduce the distance required to avoid ventilation. Sorry I can't be much help.

 

[BertKu]

It is a quick question, but I don't have a quick answer. The problem of ventilation of a propeller is very complex. Adding something over (or around) the propeller does change the situation, it should reduce the distance required to avoid ventilation. Sorry I can't be much help.

Not a problem John, I will then take the same as prescribed between hull and screw i.e. 20% of screw diameter. If I am correct, somewhere I have read that a prop in a tunnel is more efficient. If I am wrong, it is not the end of it. What I will do now, I will make the construction so, that I can lower or lift the motor and prop. The advantage is, should I test the boat out in shallow water, I can then lift the prop higher. Bert

 

[sailor0000]

Electric motor controller

Bert, your description of how a speed controller works is perfectly accurate. However, you do not have to consider all this. The frequency of the pulses is so high that the inductance of the coils and the mass (inertia) of the rotor averages everything out. In your example, 10% "on" and 90% "off", this is exactly the same as connecting the motor to a 4,8 Volt power supply. You can actually connect a voltmeter and it will read 4,8 Volt.

However, watch out: This does NOT mean the rpm will be 140! Depending on the load (torque), current and rpm will vary greatly at 4,8 Volt. 140 rpm may well be your no-load speed at 4,8 Volt, but no more.

Hope this helps.

 

[BertKu]

Bert, your description of how a speed controller works is perfectly accurate. However, you do not have to consider all this. The frequency of the pulses is so high that the inductance of the coils and the mass (inertia) of the rotor averages everything out. In your example, 10% "on" and 90% "off", this is exactly the same as connecting the motor to a 4,8 Volt power supply. You can actually connect a voltmeter and it will read 4,8 Volt.

Thanks sailor. My biggest mistake I made, was the way the manufacturers quote the AH rating of a battery. Because the battery get pulsed at very high peak currents, a 200Ah battery is no longer 200Ah. In my case I have 24 Ah batteries placed in series and parallel. Thus 288Ah in total. Wrong.
The 20 hour rate is 1.2 Ampere for 20 hours = 24 AH , but because I will load the battery with maybe 10 Ampere per batter pulsed (average 1,2 Amp), the AH rating is maybe 18 - 20 AH. This is another reason why I should increase my Lithium battery packs. Lithium you can draw up to 5C to 40 C (nano technology) and the capacity is nearly the same. This is different to Lead Acid batteries.

Hope this helps.

Thanks

 

[sailor0000]

PWM controller #

Bert, the details of how a PWM controller works are tough stuff, and pretty much impossible to understand without an education in electrical engineering. Here's what happens: The VOLTAGE is pulsed, if you were to connect an oscilloscope to the motor you'd indeed see the pulses on the screen. However, because of the "inductance" of the coils in the motor, this pulsed voltage DOES NOT induce a pulsed CURRENT (AMPS). If you were able to visualize the current, you'd see a slightly wobbly straight line. It's totally counter-intuitive, I know, but that's what happens.

So, the batteries DO NOT get pulsed. The current (amps) will be almost constant, and the same as if your motor was connected to a battery with a lower voltage (20% pulses = battery with 20% voltage). You don't have to worry about that.

And you're absolutely right about the AH rating (capacity) of your batteries. The faster you discharge them, the less "juice" you can get out of them. My batteries have 2 ratings: 245 AH @ 20 hours, and 200 AH @ 5 hours. 5 hours, that's a 40 amp current, so I'm guessing if I were to draw 100 amps, I'd get 150 AH or less out of them. So yes, this is your "mistake". You simply didn't have the information, how could you know?

A word of caution, again (sorry): Lithium batteries can (and will) burn fiercely if not handled properly. Overcharging, deeply discharging, mechanically damaging them comes with a high risk of a really bad fire. I'm aware of one documented case of a hybrid boat (a Greenline) burning on a Swiss lake, and of a couple of big cargo planes crashing because of lithium batteries igniting in the cargo hold. So please be careful if you decide to use them.

 

[BertKu]

Bert, the details of how a PWM controller works are tough stuff, and pretty much impossible to understand without an education in electrical engineering. Here's what happens: The VOLTAGE is pulsed, if you were to connect an oscilloscope to the motor you'd indeed see the pulses on the screen. However, because of the "inductance" of the coils in the motor, this pulsed voltage DOES NOT induce a pulsed CURRENT (AMPS). If you were able to visualize the current, you'd see a slightly wobbly straight line. It's totally counter-intuitive, I know, but that's what happens.

that is the reason why it is so much fun to write the software in assembler and not in C or C++ . To make printed circuit boards and use the Microchip PIC128F with 6 x PWM on the chip. It saves me to use PWM drivers or to make a PWM circuit with some analogue components.

So, the batteries DO NOT get pulsed. The current (amps) will be almost constant, and the same as if your motor was connected to a battery with a lower voltage (20% pulses = battery with 20% voltage). You don't have to worry about that.

Unfortunately here I do not agree with you. If you open a brushless motor, you will find some hundred parallel soldered short pieces of winding wire with shellac, in how the 3 coils are made up. That is the reason why a brushless motor has such high torque. By conducting the stranded wires for a short period, you get the full current for that short period. That you will get an EMF because of switching the current off is a different story. I will not go into that complex calculations. But as your battery get "shortened" for a very short time. You cannot use the 20 hour AH rate but, merely what it would have been at that short period of time. Example , my motor draws 50 Ampere for 1/10 of a second, I have to express myself, ; a current pulse of 50 Ampere, however only an average current of 5 Ampere. Thus my AH ratings is no longer 288AH but +/- 180 AH. What I do agree with you, you may give me English lessons, as I sometimes express myself clumsy. Pulsed should have been indicated as burst. The battery gives bursts of current to the circuit etc.etc.

And you're absolutely right about the AH rating (capacity) of your batteries. The faster you discharge them, the less "juice" you can get out of them. My batteries have 2 ratings: 245 AH @ 20 hours, and 200 AH @ 5 hours. 5 hours, that's a 40 amp current, so I'm guessing if I were to draw 100 amps, I'd get 150 AH or less out of them. So yes, this is your "mistake". You simply didn't have the information, how could you know?

I should have, I work long enough with batteries and electronics.

A word of caution, again (sorry): Lithium batteries can (and will) burn fiercely if not handled properly. Overcharging, deeply discharging, mechanically damaging them comes with a high risk of a really bad fire. I'm aware of one documented case of a hybrid boat (a Greenline) burning on a Swiss lake, and of a couple of big cargo planes crashing because of lithium batteries igniting in the cargo hold. So please be careful if you decide to use them.

I have 40 x LiFePo4 here already for 3 years. According to the manufacturer, I can hit a nail 1 inch long, 3 mm thick into the can, and I shall not have any problems other than to watch for the electrolyte, which is dangerous. I can throw the battery from 1 meter on concrete and it shall not give any problem. etc. etc. Not all Lithium batteries are explosive. LiFePo4 are quite save.
Bert

 

[sailor0000]

About the controller: I was replying to your first paragraph, which I believe describes in fact a DC PWM controller with a brushed DC motor. A controller for a brushless motor actually is a 3-phase variable frequency inverter, which works differently, for sure. As fas as I know, they are designed to minimize the AC component in the current they draw from the battery. Doesn't your controller have big capacitors in the DC circuit?

About lithium batteries: Good for you, you seem to know what you're doing. I found that sometimes people are not aware of the risk, this is why I spoke up.

 

[BertKu]

About the controller: I was replying to your first paragraph, which I believe describes in fact a DC PWM controller with a brushed DC motor. A controller for a brushless motor actually is a 3-phase variable frequency inverter, which works differently, for sure. As fas as I know, they are designed to minimize the AC component in the current they draw from the battery. Doesn't your controller have big capacitors in the DC circuit?

No, the currents are too high, up to 100 Ampere. However I was planning to use super capacitors over the battery and asked a local supplier to stock them for me. I haven't heard from him, but it is summer holidays here. I was looking at the 2000 Farad Maxwell Capacitor and have 10 or maybe 18 in serial. That will solve my problem with lower AH, as the capacitor will act as a buffer. It should increase my AH. I think it should be worthwhile to spent that extra money.

About lithium batteries: Good for you, you seem to know what you're doing. I found that sometimes people are not aware of the risk, this is why I spoke up.

Yes, you are correct. Nothing wrong with a reminder. Sometimes it is good to point it out, to make sure the correct Lithium battery is bought and used. Bert

 

[sailor0000]

Hey Bert,

I can't help it, I'm still thinking about this. My experience is in 2 fields: A long time ago, I used to do research on train engines with 3-phase AC motors. We had this setup: the mains, a rectifier, a DC intermediate circuit ("Gleichstrom-Zwischenkreis" in German) with big capacitors and inductances, an inverter with variable frequency and voltage, the motor. The job of the DC circuit was to "decouple" the inverter from the rectifier. I also built and flew RC model airplanes with brushed and brushless DC motors. ALL the controllers had big capacitors right across the wires to the battery, to avoid pulsing the battery and to reduce electromagnetic interference wich might cause RC problems.

I'm not familiar with the controllers currently available for motors in the 1 kW to 10 kW power range. That's why I'm so surprised your controller doesn't have big capacitors in the DC circuit. Would they have a modern trick, maybe a software thing, to avoid pulsing the battery? Or do they use a very high pulse frequency? Do you happen to know the pulse frequency of your controller?

Anyway, I have a feeling something's not adding up, but I can't put my finger on it.

Any thoughts?

Gerold