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If I have an electric vehicle on an incline (m=17,300 kg) I want to figure out how much energy it would generate by regenerative braking.

I've used $a= g \cdot \sin({\theta}) $ and $f=m\cdot a$.

at 05 deg incline, acceleration is 0.85 m/s/s; this creates a force of 11114.972 N
at 10 deg incline, acceleration is 1.70 m/s/s; this creates a force of 22145.352 N

So bearing these values in mind if I don't want to accelerate and maintain a set speed I will need to create a braking force of 11114.972 N at 5 deg incline - this in turn will keep the machine at the same speed (ignoring friction, bumpy/uneven surface, etc.).

How can I figure out the Watts generated for the regenerative braking force of 11114.972 N?

The DC motor (I think it's 3 phase) is a 105 kW motor rated at 350 V and 300 A, max torque is 190 Nm @ 285 V drawing 285 A. (That's all the information I have for the motor.) Efficiency of the motor would be 0.85 (Assumed).

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  • $\begingroup$ What IS it? Tram? Cable car, ...? $\endgroup$ – Russell McMahon Jun 25 '15 at 11:51
  • $\begingroup$ You should post answers separately rather than editing them into the question. See more discussion here on Meta SE. $\endgroup$ – Air Jul 9 '15 at 19:18
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If I have an electric vehicle on an incline (m=17,3oo kg) I want to figure out how much energy it would generate by regenerative braking ... and maintain a set speed.

Change in potential energy per time is as good an approach as using kinetic energy.
The answers should be the same :-).

E= mgh

  h = change in vertical axis  
  m = mass
  g = gravitational acceleration.

In one second power = energy numerically.
Say g = 10 (more than good enough for 'playing')

1 kph = 1000 metres in 3600 seconds.
1 kph = 0.278 m/s (or 1 m/s = 3.6 kph)

V_vertical / V_vehicle = sin(slope angle) V_vertical = V_vehicle x sin(slope_angle)

Mass = 17300 kg
First calculate for 1 metre/second vertical rate of change to get a general feel.
Power = energy change in one second at constant V
P = mgh = 17300 x 10 x 1 = 173,000 Joule/1s
= 173 kW at 1 m/s !!!!

At your example 45 degrees and 15 kph

V_vertical = 15 x sin(45) = 10.6 kph
10.6 kph = 10.6/3.6 m/s = 2.95 m/s vertically.
So power = 2.95 x 173 kW = 510 kW. Change g to 9.8 instead of 10 and I get 500 kW
= your result.

This tells you that under the stated conditions your motor would be trying to regenerate at about 5 x it's rated power. It is almost certain that it would not achieve this, and quite likely that it would be damaged if you ran it in that manner. 500 kW (or even 100 kW) of magic smoke would be very impressive indeed!.

It also tells you that the motor could not drive the vehicle up a 45 degrees slope at that rate. Maximum speed would be about 105 kW/500 kW x 15 kph = 3 kph

This is not a vast surprise when you consider what you are asking for.
Braking at constant V at 15 kph is equivalent to lifting a 17 ton vehicle ~= 10+ cars vertically at 3 m/S

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Useful rule of thumb

Energy  ~= kg x metres lifted vertically x 10 x efficiency 

This is good for energy required to lift mass OR power delivered by falling masses.

Power is energy per unit time so power can be used instead of energy is rates are in meters per second rather than total metres moves.

So, for water wheel

Power = litres/second x head in metres x 10 x efficiency 

For pump

Power = litres/second x head in metres x 10 / efficiency 

Note that efficiency Z = power out / power in in both cases, but

Efficiency for water wheel = Power delivered / Power from water flow
Efficiency for pump = Power in outlet water flow / Power used to drive pump

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  • $\begingroup$ One question: with the line: Power = 2.95 m/s x 173 kW = 510 kW, the only inconsistency is the units P=m/s x kW = kW ... but in fact multiplying m/s with kW will equal to kW m/s? can you confirm? The numbers are right - but the units in the calculation is a bit confusing... $\endgroup$ – 3kstc Jun 26 '15 at 4:10
  • $\begingroup$ never mind - it's 173kw per 1m/s thus the m/s cancel out leaving only kW... All good.. $\endgroup$ – 3kstc Jun 26 '15 at 4:14
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The simple answer is based upon your speed. $P = F*v$, so if you're going at 2 m/s, Wolfram gives you 22.4 kW. Quite a bit of energy to start with! Of course, your final design will eat up some of this power due to energy conversion.

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  • $\begingroup$ This is how I did it: acceleration=gravity·sin(incline) then I used the acceleration to find the force; f=m·a;since I know the radius of the tyre (0.525m), i can work out the torque: Torque=wheelradius·force; converting a preset speed of 10kph to rad/s of the wheel 5.29rad/s I know torque and rad/s I can work out power generated = Power = τ·ω = Nm·rad/s . This is how I did it... Is this approach right? $\endgroup$ – 3kstc Jun 25 '15 at 2:29
  • $\begingroup$ This approach is just fine - but, since $r\omega = v$, my method saves a step (and allows you to answer a post that doesn't tell you the wheel radius!) $\endgroup$ – Mark Jun 25 '15 at 2:54
  • $\begingroup$ the values I got are unbelievably huge :/ for example at a 45 degree incline maintaining a speed of 15 kph will result to 500kW of power being generated. Here is my working out - if you could double check I would be most appreciative: a=gsin(theta)=9.81*sin(45)= 6.94m/s/s then f=ma=17300*6.94= 120005N calc torque, t=dist.x force = 1200005*0.525= 63003Nm 15 kph ->4.16m/s ->1.26 rev/s -> 7.94 rad/s ; 63003Nm * 7.94= 500.02 kW I've got a excel sheet if you could kindly have a look-! Many thanks Mark! $\endgroup$ – 3kstc Jun 25 '15 at 3:46
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    $\begingroup$ That works out - but the site isn't meant for double checking work. But, I will verify you are doing the math right. Bear in mind, you started with a lot of weight, and 500 kW is only 20% more than the power of a standard diesel engine. Since you're dealing with something of about that weight, you'd need devices of that size to control it. $\endgroup$ – Mark Jun 25 '15 at 4:11
  • $\begingroup$ I've never dealt with such a problem before! I just thought 500kW is alot of power generated, such high power will definately damage the electric motor right? but I also need to factor in efficieny ratio - etc. Many thanks for your help! $\endgroup$ – 3kstc Jun 25 '15 at 4:57
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The calculation perfectly right, bu the normal electric vehicle weights just 1/10th on the weight that you have considered. Hence the results are scaled in that ratio (10x more than what the real world EV could generate)... Hope that helps!

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