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Electrical Formulas

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Revision as of 20:03, 22 October 2006 by 71.142.72.99 (talk)
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J=Q x E

Where J =  is the energy absorbed in a
         electrical circuit in watt-seconds (joules)
     Q = Coulombs is the quantity of electricity
         which is transferred between two points
     E = Voltage or Electro Motive Force


Dividing both sides of the equation by time it takes for the quantity Q to flow in a circuits becomes:

     J/T = (Q x E)/T   or  P = I x E

Where P Is the Power in amount of time.

     I is the flow of current (coulombs per second, or more commonly amps. 1C/s = 1A)
     E is the EMF measured in Volts


Simplify to:

     W = I x E   (watts or energy - no time)
     Wh = I x E  (watt hour - amount for 1 hour)


Example:

     If you are charging a battery pack at 240 volts AC at 50 amps AC, then 240V x 50A = 12 KW  or 12,000 watts (input into the charger)

If you charge for 1 hour, than this will become 12 KWH or if you charge for 1 second, than the cost of Power will be 12 Kw/3600 secs (3600 seconds in a hour).

If you drive the EV for one hour at 50 battery amps and 100 volts, than you will use about 50 ADC x 100 VDC = 5KW Per Hour or 5KWH.

The HP (horsepower) use, where 1 HP = 746 watts:

     HP = Watts/ 746  = 5000 / 746 = 6.7 HP


If you look on some motor labels on a motor, or in the specifications of that motor, You may see data as:

165V @ 175A   32HP   6000 RPM    83% Effiency.

To calculate this:

W = 165V x 175A = 28,875 watts (input power)
   28,875 x .83 = 23,966 watts (output power)
HP = 23,966 / 746 = 32.1 HP


Roland