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Difference between revisions of "Electrical Formulas"
| Line 1: | Line 1: | ||
| + | ==J=Q x E== | ||
| − | + | Where J = is the energy absorbed in a | |
| − | + | electrical circuit in watt-seconds (joules) | |
| − | Where | ||
| − | electrical circuit in | ||
Q = Coulombs is the quantity of electricity | Q = Coulombs is the quantity of electricity | ||
| Line 14: | Line 13: | ||
Dividing both sides of the equation by time it takes for the quantity Q to flow in a circuits becomes: | 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. | Where P Is the Power in amount of time. | ||
| − | I is the | + | I is the flow of current (coulombs per second, or more commonly amps. 1C/s = 1A) |
| − | E is the | + | E is the EMF measured in Volts |
| Line 51: | Line 50: | ||
To calculate this: | To calculate this: | ||
| − | W = 165V x 175A = 28,875 watts (input | + | W = 165V x 175A = 28,875 watts (input power) |
| − | 28,875 x .83 = 23,966 watts (output | + | 28,875 x .83 = 23,966 watts (output power) |
HP = 23,966 / 746 = 32.1 HP | HP = 23,966 / 746 = 32.1 HP | ||
Latest revision as of 20:03, 22 October 2006
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
