Question |
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Momentus |
| Subject: |
Where does the Energy go? |
| Question: |
Where does the energy go?
Imagine a heavy train, running along at 100 kph. A small electric car accelerates from rest to catch up with the train, taking 10 units of energy from the battery. The car then uses regenerative braking to bring the car back to rest, recovering the stored kinetic energy, recharging the battery.
Accelerate the car again, this time taking the car on board the moving train when their speeds match.
Now whilst the car is on the train, it is accelerated to 100kph along the train corridor, using a further 10 units of energy. 20 units of energy have been taken from the battery.
Now for the catch.
a) If the car accelerates from the back of the train and exits from the front, it will be travelling at 200 kph when it touches the track. Then regenerative braking can bring it to rest, recharging the battery, recovering the 20 units of energy.
b) If however the car accelerates from the front of the train and exits from the rear, it will be at rest when it touches the track. No regenerative braking, no energy recovery.
As a gedankenexperiment, scenario b) can be reiterated until the amount of energy transferred becomes very large. How will that manifest itself? Will the mass be heated up?
Where does the energy go?
All thoughts are welcome.
Momentus
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| Date: |
21 January 2015
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Answers (Ordered by Date)
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| Answer: |
Nitro - 21/01/2015 17:02:39
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| | Into accelerating the train
kind regards
NM
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| Answer: |
Harry K. - 21/01/2015 18:19:10
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| | I´m conform with Nitro but will add:
- decelerating the train in Scenario a)
- accelerating the train in Scenario b)
"Gedankenexperiment" - German Expression. Did not know that there is no English translation. :-)
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Blaze - 21/01/2015 19:08:20
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| | I agree with Nitro and Harry. Basically what you have described is how a rocket works. This is easier to see is scenario b. In scenario b you have ejected an accelerated mass (the car) to accelerate the rocket (the train). The car has to push against something to accelerate. It pushes against the train through the friction between its wheels and the platform of the train which accelerates the train.
regards,
Blaze
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| Answer: |
Momentus - 22/01/2015 14:27:08
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| | Where does the energy go? 2
“Into accelerating the train”
If only it was that simple. Energy is expressed as the square of velocity.
The energy passed to the train is proportional to the square of the change in velocity of the train. As change in velocity is directly proportional to mass, the change in the train’s velocity is small, and the square of small is miniscule and the mass of the train times miniscule is insignificant. Harry may wish to enlighten us as to real values in energy exchange of an Australian ore train being accelerated by a toy car (he is good at that sort of thing) or you can take my word for it, too many zeros after the decimal point to be written as a number without using minus powers!
If you wish to include this insignificant amount of energy in the accounting, it must be done at all stages. When the car accelerates to the rear of the train, there is an additional drain from the battery to accelerate the train, over and above the 10 units required to change the MV of the car. My intention in setting up the scenario, knowing that squares were involved, was to reduce the energy transfer to and from the train to an insignificant amount compared to the energy transfer to and from the car. And then ignore it, which I shall continue to do.
There is energy taken from the battery to accelerate the car. Not taken from or passed to the train, energy taken from the battery, used to accelerate the car. Real actual energy transfer, out of battery, energy used for the sole purpose of changing the momentum of the car’s mass, regardless of the direction of acceleration.
That energy can be recovered in a), but not in b). It remains as an inherent property of the mass in some form or another in b). The significance of that fact in the design and operation of an inertial drive is crucial. Take the wheels off the car and use your inertial drive to accelerate the car in this scenario and the question is the same, where does the energy go.
The question, down here on the surface of planet earth is a scientific oddity, only of philosophical interest. In space as you use your inertial drive it becomes reality.
Momentus
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Momentus - 22/01/2015 14:41:36
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| | Hi Harry
Gedankenexperiment
The English language excels at borrowing from other languages to express nuances of meaning. Gedankenexperiment does translate literally as thought experiment, but it means more than that and derives from Einstein’s extensive use of the concept to stimulate scientific thought.
Wikipedia has a definition, giving Schrodinger’s cat as an example.
Regards Momentus
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Harry K. - 22/01/2015 15:26:14
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| | Momentus
No problem at all to calculate this. Give me the mass and velocity of the train and mass with end velocity of the car. If you like you can give me friction losses as well if you want to take friction into consideration.
The calculation is very simple and will show that output energy is equal to input energy.
Thanks for the clarification of "gedankenexperiment".
Regards,
Harald
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| Answer: |
Momentus - 22/01/2015 18:49:22
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| | Hi Harry
“Conveying 82000 wet tonnes of iron ore, the 7•3 km train was formed of 682 wagons hauled by all eight of BHPIO's General Electric AC6000CW diesel locomotives. Gross weight was 99734 tonnes. “
For the purposes of this calculation assume 100,000 tonnes for the train and 500gm for a model car including batteries.
The calculation is to initially establish how much the velocity of the train changes when the velocity of the car changes from 0 to 100 kph
From that figure calculate how much energy is used to bring about the change in velocity of the train.
Please ignore friction, or if you must factor it in afterwards.
Thank you in advance
Momentus
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| Answer: |
Harry K. - 22/01/2015 19:45:59
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| | Momentus,
Fine that the train is not too heavy! :-)
Forgot one thing although this factor is not very important for this issue: in which time span should the car accelerate from 0 to 100 kmph? This is only necessary to know for calculating the power.
Thanks
Harald
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Harry K. - 22/01/2015 20:47:48
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| | Momentus:
I assumed an accelertion time of 10s. However, as mentioned before, that is not very important. Here my calculations:
RAW DATA
Mass of train (kg) 1.00E+08
Length of train (m) 7 300.00
Train velocity (km/h - m/s) 100.00 27.7778
Mass of car (kg) 0.50
Car velocity (km/h - m/s) 100.00 27.7778
Acceleration time (s) 10.00
CALCULATIONS
Kinetic energy train (Nm) 3.86E+10
Kinetic energy car (Nm) 192.90
Acceleration car (m/s^2) 2.7778E+00
Kinetic power (W) 19.290123
Scenario a) car accelerates from the back of the train and exits from the front
Train velocity - NEW (km/h - m/s) 99.99999975 27.77777771
Velocity difference (km/h - m/s) -2.50E-07 -6.94E-08
Acceleration train (m/s^2) 2.7778E+00
Scenario b) car accelerates from the front of the train and exits from the rear
Train velocity - NEW (km/h - m/s) 100.00000025 27.77777785
Velocity difference (km/h - m/s) 2.50E-07 6.94E-08
Acceleration train (m/s^2) 2.7778E+00
The calculations are a copy and paste from Excel and thus I hope it will be well formated here.
Regards,
Harald
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| Answer: |
Harry K. - 22/01/2015 21:30:00
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| | Sorry, there was mistake in my calculations regarding acceleration calculation for the train. Here the corrected calculations:
RAW DATA
Mass of train (kg) 1.00E+08
Length of train (m) 7 300.00
Train velocity (km/h - m/s) 100.00 27.7778
Mass of car (kg) 0.50
Car velocity (km/h - m/s) 100.00 27.7778
Acceleration time (s) 10.00
CALCULATIONS
Kinetic energy train (Nm) 3.86E+10
Kinetic energy car (Nm) 192.90
Acceleration car (m/s^2) 2.7778E+00
Kinetic power (W) 19.290123
Scenario a) car accelerates from the back of the train and exits from the front
Train velocity - NEW (km/h - m/s) 99.99999975 27.77777771
Velocity difference (km/h - m/s) -2.50E-07 -6.94E-08
Acceleration train (m/s^2) -6.9444E-09
Scenario b) car accelerates from the front of the train and exits from the rear
Train velocity - NEW (km/h - m/s) 100.00000025 27.77777785
Velocity difference (km/h - m/s) 2.50E-07 6.94E-08
Acceleration train (m/s^2) 6.9444E-09
Regards,
Harald
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Momentus - 24/01/2015 11:49:44
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| | Hi Harry,
Thank you for the calculations I am pleased that they confirm my supposition.
The car gains energy and maybe, as E=MC^2, there is an increase in the mass of the car and that is where the energy is conserved.
Momentus.
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