lowering the motor rpm with Voltage ?
- Thread starter jmygann
- Start date
tostino
10 kW
You lose power (and a lot of it) by doing that, but yes you can.
Jeremy Harris
100 MW
To a certain extent you can trade volts for amps, and vice versa, but the resistive losses mean that you don't get a linear exchange rate for the power produced.
Here's a worked example that illustrates this, with some made up numbers for a motor:
Let's say the motor has a winding resistance of 40 mohms, the controller, wiring and battery pack has an effective resistance of 20 mohms, so we have a total circuit resistance of 60 mohms.
If the motor has a Kv of 150, then at 48V it will have a no load rpm of 7200. If you pull 2000 watts from this motor at 48V, then it will draw around 42 amps and the losses from total circuit resistance will be around 104 watts.
If the voltage is now dropped to 24V, then the no load rpm will drop to 3600. If you pull 2000 watts from this motor at 24V then it will draw twice the current, around 84 amps. The losses from total circuit resistance will rise to around 416 watts though, four times the losses at 48V.
If you were to run the same motor at 12V, with the same output power of 200 watts, then the losses would quadruple again to a massive 832 watts.
In general, if you want a lot of power then it's much more efficient to increase the voltage and then gear the motor down.
Jeremy
Here's a worked example that illustrates this, with some made up numbers for a motor:
Let's say the motor has a winding resistance of 40 mohms, the controller, wiring and battery pack has an effective resistance of 20 mohms, so we have a total circuit resistance of 60 mohms.
If the motor has a Kv of 150, then at 48V it will have a no load rpm of 7200. If you pull 2000 watts from this motor at 48V, then it will draw around 42 amps and the losses from total circuit resistance will be around 104 watts.
If the voltage is now dropped to 24V, then the no load rpm will drop to 3600. If you pull 2000 watts from this motor at 24V then it will draw twice the current, around 84 amps. The losses from total circuit resistance will rise to around 416 watts though, four times the losses at 48V.
If you were to run the same motor at 12V, with the same output power of 200 watts, then the losses would quadruple again to a massive 832 watts.
In general, if you want a lot of power then it's much more efficient to increase the voltage and then gear the motor down.
Jeremy
Lagoethe
10 W
- Joined
- Nov 10, 2010
- Messages
- 89
But, if he drives slower, Speed /2 then Needed Power/2. So 1000W, that makes the same circuit and the same losses, if the use is the same (I mean low torque and acceleration)
About notation:
How do you write simplified Power: P ? (W)
Speed: S, V or C ?(ms-1)
Acceleration: A ? (ms-²)
Tension: U ? (V)
Intensity: I ? (A)
rhitee05
10 kW
Lagoethe said:
Actually, the power required varies roughly as the square of speed, so at 1/2 the speed you would need about 1/4 the power. This assumes that air resistance is the primary factor, so it won't be accurate for slower speeds, but should be fairly true if you're comparing say 30 MPH to 15 MPH.
The comparison Jeremy makes is still true at lower power draws. At 1000W the 24V system will draw 42A and have 104W of loss. A 48V system would draw only 21A and have 27W of loss. It's still a factor of 4 difference because the losses vary as current squared, so 2x the current means 4x the loss.
Jeremy Harris
100 MW
rhitee05 said:
Actually, the power required varies as the cube of speed, not the square of it. Aerodynamic drag (the major component at moderate to high speeds) increases by the square of speed, but the power needed to overcome it has another squared term in it, hence the cube law relationship.
Here's a worked example to illustrate this:
Say that a bike creates 100N of drag at 10mph. 10mph is about 4.47 m/S. To overcome 100N of total drag the bike propulsion system needs to deliver 100 x 4.47 = 447 watts.
If it's speed is increased to 20 mph (around 8.94 m/S) then the drag increases by a factor of 4 to 400N. The power needed to overcome 400N of total drag is 400 x 8.94 = 3576 watts.
Jeremy
rhitee05
10 kW
Thanks for the correction, Jeremy. I forgot that power if force times velocity, thus related to velocity cubed. So 1/2 the velocity will only require 1/8 the power.
Jeremy Harris
100 MW
Motor specs and characteristics aren't that easy to work through from the data most manufacturers provide, so terms like "most efficient" are only relevant under a quoted set of circumstances.
Let's take an Etek-like motor as an example (I don't have the specs for the actual Etek, as it's been out of production for a few years now) and work through the "most efficient" operating points for two voltages. The Mars ME0708 is very similar to the old Etek in some ways, so let's use those curves. This motor is specced from 24V to 48V, like the Etek and has about the same Kv and power rating.
From the 24V CW performance chart this motor will deliver about 80lbs in of torque at its best efficiency operating point, spinning at around 1650 rpm, with an efficiency of about 85%. This means it will be delivering around 1.56kW and drawing around 75 amps (about 1.8kW) from the supply.
We can also run exactly the same motor at its "most efficient" point on 48V. Looking at the 48V CW performance chart this same motor will now deliver the same 80lbs in of torque at its best efficiency operating point, spinning at around 3,300 rpm with an efficiency of about 86%. This means it will be delivering around 3.12kW and drawing about 77 amps (about 3.7kW) from the supply.
These two worked examples show how the same motor, run at it's "best efficiency" point can deliver twice the power when run on twice the voltage, with no additional heat load on the motor and no loss of efficiency (in fact the motor is slightly more efficient at the higher voltage).
Generally, motors don't have a voltage rating, even though the manufacturer may put a number on a sheet somewhere. There are really two limiting factors to a motors ultimate performance; how much current can it take before overheating and how fast can it spin before breaking. The maximum current determines the torque the motor will deliver and the rpm is determined by the applied voltage.
Jeremy
Let's take an Etek-like motor as an example (I don't have the specs for the actual Etek, as it's been out of production for a few years now) and work through the "most efficient" operating points for two voltages. The Mars ME0708 is very similar to the old Etek in some ways, so let's use those curves. This motor is specced from 24V to 48V, like the Etek and has about the same Kv and power rating.
From the 24V CW performance chart this motor will deliver about 80lbs in of torque at its best efficiency operating point, spinning at around 1650 rpm, with an efficiency of about 85%. This means it will be delivering around 1.56kW and drawing around 75 amps (about 1.8kW) from the supply.
We can also run exactly the same motor at its "most efficient" point on 48V. Looking at the 48V CW performance chart this same motor will now deliver the same 80lbs in of torque at its best efficiency operating point, spinning at around 3,300 rpm with an efficiency of about 86%. This means it will be delivering around 3.12kW and drawing about 77 amps (about 3.7kW) from the supply.
These two worked examples show how the same motor, run at it's "best efficiency" point can deliver twice the power when run on twice the voltage, with no additional heat load on the motor and no loss of efficiency (in fact the motor is slightly more efficient at the higher voltage).
Generally, motors don't have a voltage rating, even though the manufacturer may put a number on a sheet somewhere. There are really two limiting factors to a motors ultimate performance; how much current can it take before overheating and how fast can it spin before breaking. The maximum current determines the torque the motor will deliver and the rpm is determined by the applied voltage.
Jeremy
rhitee05
10 kW
Jeremy, I'm trying to follow your numbers and I'm a little confused by your conclusion:
Without checking to make sure any of the figures you posted make sense for a Mars/Etek motor, I agree with the numbers you posted for power at the given torque/RPM/voltage/current figures. That is, I agree that 80 lb-in of torque at 1650 RPM is 1.56 kW, 75A at 24V is 1.8 kW, etc.
I can't match your numbers for efficiency. For 3.12kW out and 3.7kW in I get about 84% efficiency, not 86% in the 48V case. So the motor is slightly less efficient at the higher voltage (not by much). I also get closer to 87% at 24V. That's a minor quibble, one or two percent is not that important.
But, I really don't follow why you say that there is no additional heat load on the motor at the higher voltage. Putting out twice the power, at the same efficiency, means twice the heat. 3.7-3.12kW = 580W heat vs. 1.8-1.56kW = 240W heat. Perhaps you meant to compare the case of equal output power at the two voltages? In that case similar efficiency means similar losses in the motor, but lower current should mean lower losses elsewhere in the system.
Jeremy Harris said:
Without checking to make sure any of the figures you posted make sense for a Mars/Etek motor, I agree with the numbers you posted for power at the given torque/RPM/voltage/current figures. That is, I agree that 80 lb-in of torque at 1650 RPM is 1.56 kW, 75A at 24V is 1.8 kW, etc.
I can't match your numbers for efficiency. For 3.12kW out and 3.7kW in I get about 84% efficiency, not 86% in the 48V case. So the motor is slightly less efficient at the higher voltage (not by much). I also get closer to 87% at 24V. That's a minor quibble, one or two percent is not that important.
But, I really don't follow why you say that there is no additional heat load on the motor at the higher voltage. Putting out twice the power, at the same efficiency, means twice the heat. 3.7-3.12kW = 580W heat vs. 1.8-1.56kW = 240W heat. Perhaps you meant to compare the case of equal output power at the two voltages? In that case similar efficiency means similar losses in the motor, but lower current should mean lower losses elsewhere in the system.
Jeremy Harris
100 MW
Eric,
I just read/estimated the numbers off the performance chart, and then rounded them, so there's an inevitable error of around 2% or so. Here's a link to the pdf plots: http://lib.store.yahoo.net/lib/yhst-57437235823410/ME0708data.pdf.
I guess I should have been more specific when referring to 'heat load', once more the danger of trying to simplify for a non-technical reader has been spotted by another who's technically savvy! What I was really referring to was just the I²R losses in the motor windings, but I didn't want to confuse the OP with extra detail. As this is a PM motor, it has a near-constant torque per unit current, so the current drawn for 80lbs ins at 24V is pretty close to being exactly the same current as 80lbs ins at 48V. The I²R loss will therefore be near-enough the same for both voltages.
The real point I was trying to make to the OP was that there's really no such figure as a "best efficiency voltage" for a motor, by showing that the efficiency for the same torque, but markedly different power levels, at two widely spaced voltages was pretty much the same. Maybe I wasn't clear enough.
Jeremy
I just read/estimated the numbers off the performance chart, and then rounded them, so there's an inevitable error of around 2% or so. Here's a link to the pdf plots: http://lib.store.yahoo.net/lib/yhst-57437235823410/ME0708data.pdf.
I guess I should have been more specific when referring to 'heat load', once more the danger of trying to simplify for a non-technical reader has been spotted by another who's technically savvy! What I was really referring to was just the I²R losses in the motor windings, but I didn't want to confuse the OP with extra detail. As this is a PM motor, it has a near-constant torque per unit current, so the current drawn for 80lbs ins at 24V is pretty close to being exactly the same current as 80lbs ins at 48V. The I²R loss will therefore be near-enough the same for both voltages.
The real point I was trying to make to the OP was that there's really no such figure as a "best efficiency voltage" for a motor, by showing that the efficiency for the same torque, but markedly different power levels, at two widely spaced voltages was pretty much the same. Maybe I wasn't clear enough.
Jeremy
rhitee05
10 kW
No worries. I agree that the important point is that the efficiency is roughly the same across different voltages.
rhitee05
10 kW
The system efficiency will likely still be higher in the higher voltage system for a given power output as losses in the controller and wiring are proportional to current squared, so those losses will be lower at a higher voltage (for a given power output). You'd have to balance this advantage with any additional mechanical complexity such as gearing to see if it's worth it. Even if it didn't make a significant difference in the system level efficiency, reducing the controller heating might be valuable. The higher-voltage system would also have a higher peak power capability.
Jeremy Harris
100 MW
jmygann said:
As Eric has rightly said, for any given power a higher voltage, lower current, system is likely to be more efficient and have lower losses.
Here's another worked example to try and help get this message across. Let's say that you need 2kW to get your bike to go at the maximum speed you want. Let's also assume that you're using an Etek type motor, as we have the specs for it (see post above). Let's also assume that you have a battery pack, controller, connectors and wiring with a total internal resistance of 30mohms (all stuff we use has some resistance, these figures are probably typical for a ebike type set-up)
Example 1 - 48V system, 2kW power output
At 48V and 2kw output the motor will draw about 50 amps, and will be using around 2.4kW to deliver your needed power. The motor itself will be wasting around 400 watts. The controller, batteries, connectors and cables will waste another 75 watts from I²R losses (power wasted = current² x resistance). The total wasted power for this 48V system, delivering 2kW, is therefore around 475 watts.
Example 2 - 24V system, 2kW power output
At 24V and 2kW output the motor will draw about 100 amps, and will be using around 2.4kW, as before, to deliver your needed power. The motor itself will be wasting around 400 watts. The controller, batteries, connectors and cables will waste another 300 watts from I²R losses (power wasted = current² x resistance). The total wasted power for this 24V system, delivering 2kW, is therefore around 700 watts.
Hopefully these worked examples may help show why running at higher voltage, and hence lower current, for any given power level is a good thing. The losses for the 24V system above are significantly greater (700 watts) than those for the 48V system delivering the same power (475 watts).
In terms of total system electrical efficiency, at around 2kW power output the 48V system is around 80.1% efficient, the 24V system is only around 74.1% efficient.
If you increased the power output, you would see these differences get progressively worse. For a 3kW power output, for example, the 48V system losses rise to about 769 watts whilst the 24V system losses increase to about 1,441 watts. Total electrical system efficiencies for 3kW power output are around 79.6% for 48V and only 67.5% for 24V.
Jeremy
Samba
100 W
jmygann said:
Nice fairing. Did you make it or are they for sale?
