The Real! MUSTANG MACH E is here

Anything to do with the electric or hybrid world

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pdq67
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Re: The Real! MUSTANG MACH E is here

Post by pdq67 »

I haven't looked but do know about it!

So just how much is this electric pony gonna cost me??

As well as, when can I pick it up at my Dealer?

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Re: The Real! MUSTANG MACH E is here

Post by Belgian1979 »

Like with all cars hp says nothing. It's a mathematical equasion that is based on torque. The more torque at low rpm the higher the calculated hp will be. Since all electric motors have a lot of hp at very low rpm, evidently the amount of HP is going to be high.

I do not debating that an electric motor is better in terms of torque produced at low rpm (I'm an electrical engineer by training) but they consume huge amounts of energy to do so. And as always the problems remains where to get the energy from. As most have to use huge batteries that weigh a lot the comparison is not clear cut.
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Re: The Real! MUSTANG MACH E is here

Post by gruntguru »

Belgian1979 wrote: Mon Oct 19, 2020 5:04 amI do not debating that an electric motor is better in terms of torque produced at low rpm (I'm an electrical engineer by training) but they consume huge amounts of energy to do so.
You mean an electrician right?

An "engineer" (with a "degree") would recognise this is nonsense.
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Re: The Real! MUSTANG hwMACH E is here

Post by Belgian1979 »

You can assume anything you want. I don't care.

Are you going to debate that an electric motor (any type) when being in a high torque, low speed situation is going to draw a lot of amps ? Are you going to debate that drawing that much amps means a lot of power is consumed, power that needs to come from a lot of batteries?
Are you saying that the vehicle, which is heavier due to the battery weight is going to consume less power to accelerate than an equal vehicle equiped with a combustion engine that weighs less?

How do you assume they derived the hp produced by the car on that race track? Do you think the weight is factored in or not?
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Re: The Real! MUSTANG hwMACH E is here

Post by gruntguru »

Belgian1979 wrote: Thu Oct 22, 2020 1:56 pm You can assume anything you want. I don't care.

Are you going to debate that an electric motor (any type) when being in a high torque, low speed situation is going to draw a lot of amps ? Are you going to debate that drawing that much amps means a lot of power is consumed, power that needs to come from a lot of batteries?
Engineers (with a degree) learn thermodynamics including the first law. Actually high school physics teaches the law of conservation of energy. Extending that law to electric motors tells us that the electrical power drawn from the battery is equal to the mechanical power output of the motor plus the waste heat generated.

Furthermore, if you look at the motor efficiency map below, the efficiency tells us the relationship between the electric power in and the mechanical power out. So if you look at the left edge of the map and take the data point at 500 Nm of torque and 300 rpm, the mechanical power is 15.7 kW (peak power for this motor is about 120 kW so 15.7kW is not much).

The efficiency is 82% so the electrical power consumed by the motor is 15.7/0.82 = 19.1 kW. From say a 500v battery, that would be 15.7 x 1000/500 = 31.4 Amps.

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Re: The Real! MUSTANG MACH E is here

Post by Belgian1979 »

So we're ignoring the questions, ok, whatever.

BTW: the electric motors in the EV's (most use AC motors due to their better power density) are driven by 3 phase power, so the formula you used is wrong :D (P = 1,73 * U * I) More so, most cars run their motors at 350V or 375v.

So try to recalculate and then come back before you start discussing the first thermodynamical law.
Also, if you know anything about electric motors, you would know that they have a current draw in a low rpm high torque situation which is up to +-9X the nominal draw. Your graph is completely out of context as it doesn't show what the current draw is when trying to accelerate from a dead stop. Even less when constantly trying to accelerate. The more loaded an electric motor is, the slip increases and the more current is drawn.

Oh and maybe answer me how they got to the 1400 HP they claim for the car. Maybe give me the weight and the continuous torque it produced to get to the number.
I'm also going to add another one, what is the constant acceleration that the car sees and how much torque is needed to get it to accelerate that fast? (Don't forget the 1400 hp that they gave as a power value).

Oh, and tell me how they regulate power output on that electric motor so they achieve the highest amout of torque at every rpm and how does that affect torque and current draw :lol:
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Re: The Real! MUSTANG MACH E is here

Post by Belgian1979 »

some considerations:

Assuming a well built asynchronous motor with about 1% slip, it would turn around 2970 rpm (with 2 poles) at optimal working rpm.
I read that the car in the subject has 2300 ft lbs torque or 3.162,5 Nm. This means that it would produce 983,10 kW at that rpm and assuming 50 Hz AC.

So the correct calculation for nominal current that flows is (983,10 X 1000)/(375 x 1,73) = 1.515,38 A in total or 216,48 A per motor (as there are apparently 7 motors in the car)

Assuming that the current draw during start would be only a modest 6x the nominal current draw this would equate to around 1298 A per motor....even assuming they use a VFD to modulate the motors and allow 33% of current to flow it still is huge...The batteries don't last long is my 'uneducated guess'.
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Re: The Real! MUSTANG MACH E is here

Post by Belgian1979 »

And then this:
They claim 1419 HP for the car. This can only be the case when they work with 60 Hz. The synchronous RPM is 3600 in that case, or 3564 rpm at 1% slip.
This also means that they are using the breakway torque for acceleration, which is roughly with 10-12% slip and when assuming 11% slip, one gets close to the 1419 HP they claim.
However, the current draw at breakaway torque is still X4 to X5 the nominal current draw of the motors.
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Re: The Real! MUSTANG MACH E is here

Post by gruntguru »

Belgian1979 wrote: Sat Oct 24, 2020 2:57 pmBTW: the electric motors in the EV's (most use AC motors due to their better power density) are driven by 3 phase power, so the formula you used is wrong :D (P = 1,73 * U * I)
So you think I should use a 3 phase AC formula to calculate the current draw from a battery?
From say a 500v battery, that would be 15.7 x 1000/500 = 31.4 Amps.
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Re: The Real! MUSTANG MACH E is here

Post by gruntguru »

Belgian1979 wrote: Sat Oct 24, 2020 2:57 pmAlso, if you know anything about electric motors, you would know that they have a current draw in a low rpm high torque situation which is up to +-9X the nominal draw. Your graph is completely out of context as it doesn't show what the current draw is when trying to accelerate from a dead stop.
I know nothing about electric motors.

I gave you an example at 300rpm because that is the limit of the graph. If installed in a car with a single reduction gear, 300 rpm would be about 5 mph.

- You say 9X current draw? OK, lets say 180 kW instead of 19.1 (horribly pessimistic because we know from the graph that at 5 mph the battery draw is 19.1 kW so the average power draw between zero and five mph will be much less than 180 kW).

- Time to accelerate for 0 - 5 mph at full throttle? Lets say one second.

- Energy consumed? 180kW x 1sec / 3600sec/hr = 0.05 KW.hr. (A 50 kW.hr battery pack will supply 1000 of these launches or 9,000 if the battery draw is only 19.1 kW.)

How do you think this compares to the energy required to launch a vehicle with a combustion engine?
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Re: The Real! MUSTANG MACH E is here

Post by Belgian1979 »

Image
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edit: tried to post some diagrams but since photobucket doesn't provide the service anymore it doesn't work.

https://www.researchgate.net/figure/Typ ... g2_3171041

I don't know the specifics of how they control these motors, but typically when you want to control the current draw at start or acceleration as it might be too high. Therefore you would modulate the frequency and/or voltage. That would reduce the current draw depending on the torque you require. In order to get maximum acceleration you would want the motor to produce its maximum torque and you let it accelerate freely up to its maximum rpm. Best or highest torque is not at the nominal rpm (at synchronous speeds nominal would be 3000 rpm at 50 Hz and 3600 rpm at 60 Hz). By closely controlling the voltage and or the frequency they can control both power output and speed)

What you have to understand with an electric motor is that it gives its (close to) maximum torque right from 0 rpm contrary to what you see with an IC motor. So you do not need a transmission as you can connect the motors directly to the wheel. So if your electric motors maximum speed would be 2970 rpm you just multiply that by the circumference of the wheel and you have the maximum speed that the car can reach (pretty high).
The negative side is that the current draw is very high until you reach your nominal rpm (which is where you also have your maximum efficiency). So assuming you would be cruising, they would modulate both the frequence and voltage to get the motor to stay in its optimal working range (intersection between the variable load curve and the torque curve of the motor)
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Re: The Real! MUSTANG MACH E is here

Post by Belgian1979 »

When you use VFD the curves look something like the ones on this page:

https://www.researchgate.net/figure/Tor ... _251869545
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Re: The Real! MUSTANG MACH E is here

Post by Belgian1979 »

Bear in mind I'm not designing electric cars for a living, so I can only speculate on what they do.

When reading up on the subject I note that the Tesla's seem to have a reduction gear to increase torque to the wheels. This obviously would reduce the top end speed substantially (I read somewhere that they use 9.34:1 reduction gear - not sure if correct though).

Anyway the only way that they can get enough top speed beyond the nominal rpm of the motor is to increase the frequency of the AC power. Beyond the nominal rpm point, they will not be able to still increase the voltage as it will have reached its maximum. What will happen is, they will drive the motor at say 200% or 300 % of the nominal frequency (50*200% = 100 Hz or 50 * 300 % = 150 Hz). Going over 100% is ok, but it has a drawback as you're magnitizing iron and the more frequency you use the higher the losses. (inductors impedance is frequency dependant). Since they cannot increase the voltage anymore, the torque is reduced even though speed can still increase (up to a point). At speeds higher than nominal obviously efficiency will decrease as a result.

As far as that Mustang is concerned I'm not sure if they have a reductor gear or not since they have 7 electric motors in place. The fact remains though that the amount of torque needed for a run like that requires a lot of current and the battery will be depleted very fast.
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Re: The Real! MUSTANG MACH E is here

Post by gruntguru »

The question of how quickly the battery depletes is actually very simple. Far easier to analyse using energy than current. All you need to know is the mechanical power required, the motor rpm and the conversion (motor, controller, inverter etc) efficiency at each stage. The energy draw at launch is actually fairly insignificant for two reasons:

1. Motor power output (= torque x speed) is zero at zero rpm.
2. The time spent at zero rpm once current starts flowing is also zero and the time spent getting the motor up to an efficient speed (with tyres spinning in the Mustang case) is very short.

Here is an example - only 416 hp and 12.3 sec 1/4 mile but the energy required for the quarter mile was only 1.1 kW.hr. Interestingly, more than half of that energy was recoverable if there was enough runoff for the Tesla to use regenerative braking. Total energy cost for the 1/4 mile? 6 cents.

https://www.dragtimes.com/blog/tesla-mo ... e-and-cost

The Mustang has 3.5 times the power but won't use 3.5 times the energy for two reasons:

1. The car is traction limited for the early part of the run and can't use full power for at least half the time (not distance).
2. The ET is less than for the Tesla, so energy (= power x time) will reduce.

On the other hand the Mustang wastes more energy in wheelspin than the Tesla.

I would estimate the Mustang energy at closer to 2 x the Tesla so about 2.5 kW.hr
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Re: The Real! MUSTANG MACH E is here

Post by gruntguru »

Belgian1979 wrote: Mon Oct 26, 2020 12:18 pm Bear in mind I'm not designing electric cars for a living, so I can only speculate on what they do.

When reading up on the subject I note that the Tesla's seem to have a reduction gear to increase torque to the wheels. This obviously would reduce the top end speed substantially (I read somewhere that they use 9.34:1 reduction gear - not sure if correct though).
Almost all (if not all) EV's use a reduction gear. To get high power from a compact lightweight motor, it must turn high rpm - much higher than the wheel rpm at Vmax.
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