How much current or volts or watts are needed to power electric cars?
Wednesday, April 22nd, 2009 at
12:05 am
I am working on the X prize for 100 mpg car. I have a unheard of way to generate electric current. I need to know how much current or volts or thier multiplied result watts is needed to have a 1-3 ton car move from 0-60 in 12 secs and all the way up to extreme 300 mph in a normal range of acceleration. Basically, how much does the other cars tend to use?
My way doesnt use plugged power or batteries….and yes it does exist!
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Tagged with: acceleration • batteries • car move • cars • secs • volts • watts • X-Prize
Filed under: Electric Cars
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Are you sure you want to go through with this? If you can't do these calculations yourself or have ready access to someone who can, your chance of success is dim indeed.
To move 2 metric tonnes of vehicle and payload from zero to 100 kilometers per hour (about 62 miles per hour or 27.8 meters per sec) in 12 seconds, assuming a constant acceleration and neglecting air resistance and other frictional losses – a highly questionable assumption at 100 km/hr – an energy of about 400 kilowatt-seconds (joules) at a peak power of about 65 kilowatts when the car reaches maximum speed. The average power will be half the peak power.
The power necessary to maintain that speed will depend a lot on the body design and the aerodynamic drag associated with it. At the speeds encountered in automobilies, the drag increases with the square of the car's speed. The drag will be noticeable at 100 km/hr and will be about 25 times worse at your proposed top speed of about 500 km/hr. For a top speed like that, you will have some serious aerodynamic drag and stability issues to contend with.
One way to estimate the frictional air drag on your own vehicle would be to take your car out on the road with a passenger with a stopwatch, take the car up to 65 mph, throw the car into neutral (or disengage the clutch) and time how long it takes the car to slow to 55 mph. You must also find a way to estimate or actually weigh the car with passengers.
Once you have done this, the power needed to maintain that speed can be calculated from the formula
P ? m*v*?v / t
If P is the necessary power in watts,
then m is the mass of the car, passengers, and other payload in kilograms (pounds * 0.4536 kg/lb)
v is the (average) speed in meters per second (60 MPH -> 26.8 m/sec)
?v is the speed change (10 MPH -> 4.47 m/sec)
and t is the time it takes the car to slow down, in seconds. You might be able to find the mass of the car itself on the plate riveted to the center column on the edge of the driver's side door or the center column just to the left and behind the driver.
Nuclear reactors have proven to be unrelaible for personal automobiles!!
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