Electrical Formulas

Here i discuss some of important electrical formulas.it very useful for daily jobs in electrical engineering.

Glossary :-

I = Amperes

E = Volts

kW = Kilowatts

kVA = Kilo volt-Amperes

HP = Horsepower % eff. = Percent Efficiency

pf = Power Factor


Single-Phase

TO FIND :-

  • Amperes when kVA is known –>   I = kVA x 1000 / E
  • Amperes when horsepower is known –>  ( HP x 746) / ( E  x  % eff.  x pf )
  • Amperes when kilowatts are known –>  ( kW x 1000 ) / ( E x pf )
  • Kilowatts  –>  ( I x E x pf ) /1000
  • Kilovolt-Amperes  –>  ( I x E ) / 1000
  • Horsepower  –>  ( I x E x % eff. x pf  ) / 746
  • Watts   –>  E x I x pf
  • Energy Efficiency  –>  Load Horsepower x 746 / Load Input kVA x 1000
  • Power Factor  @ cos θ –>  Power Consumed /Apparent Power ( W / VA ) @ ( kW / kVA )

Two-Phase

TO FIND :-

  • Amperes when kVA is known –>   I = ( kVA x 1000 )  / ( E x 2 )
  • Amperes when horsepower is known –>  ( HP x 746) / ( E  x  2  x % eff.  x pf )
  • Amperes when kilowatts are known –>  ( kW x 1000 ) / ( E x 2 x pf )
  • Kilowatts  –>  ( I x E x 2 x pf ) /1000
  • Kilovolt-Amperes  –>  ( I x E x 2 ) / 1000
  • Horsepower  –>  ( I x E x 2 x % eff. x pf  ) / 746
  • Watts   –>  E x I x 2 x pf
  • Energy Efficiency  –>  Load Horsepower x 746 / Load Input kVA x 1000
  • Power Factor  @ cos θ –>  Power Consumed /Apparent Power ( W / VA ) @ ( kW / kVA )

Three-Phase

TO FIND :-

  • Amperes when kVA is known –>   I = ( kVA x 1000 )  / ( E x 1.73 )
  • Amperes when horsepower is known –>  ( HP x 746) / ( E  x  1.73  x % eff.  x pf )
  • Amperes when kilowatts are known –>  ( kW x 1000 ) / ( E x 1.73 x pf )
  • Kilowatts  –>  ( I x E x 1.73 x pf ) /1000
  • Kilovolt-Amperes  –>  ( I x E x 1.73 ) / 1000
  • Horsepower  –>  ( I x E x 1.73 x % eff. x pf  ) / 746
  • Watts   –>  E x I x 1.73 x pf
  • Energy Efficiency  –>  Load Horsepower x 746 / Load Input kVA x 1000
  • Power Factor  @ cos θ –>  Power Consumed /Apparent Power ( W / VA ) @ ( kW / kVA )


Others Formula

 

  • kW = hp x .746
  • Torque in lb-ft = hp x 5250 / rpm
  • Motor synchronous speed in rpm = 120 x Hz / number of poles
  • Three-phase full-load amp= hp x .746 / 1.73 x kV x effi ciency x power factor
  • Rated motor kVA = hp (.746) / efficiency x power factor
  • kW loss = hp (.746) (1.0 – effi ciency) / efficiency
  • kVA in-rush = percent in-rush x rated kVA
  • Approximate voltage drop (%) = motor kVA in-rush x transformer impedance / transformer kVA
  • Stored kinetic energy in kW-sec = 2.31 x (total Wk2) x rpm2 x 107
  • Inertia constant (H) in seconds = stored kinetic energy in kW-seconds / hp (.746)
  • Conversion factors: CV = (metric hp) = 735.5 watts = 75 kg-m/sec Wk2 (lb-ft) = 5.93 x GD2 (kg-m2)
  • Ventilating-air requirements: 100-125 cfm of 400C air at 1/2-in. water pressure for each kW of loss
  • Degrees C = (Degrees F-32) x 5/9
  • Degrees F = [(Degrees C) x 9/5 ] + 32