Thursday, October 16, 2014

Single phase and Three phase formula's used in electrical engineering

Electrical Formulas

Electrical engineering required certain formula which must be known for an electrical engineer and even a non engineer person so to know certain parameters of appliances around everyone.
Lets discuss 1st about basic parameters used in electrical systems:-
I = Amperes
E = Volts
kW = Kilowatts
kVA = Kilo volt-Amperes
HP = Horsepower
% eff. = Percent Efficiency
pf = Power Factor

For Single-Phase load here are the formula’s as below:-
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)

For findingThree-Phase parameters formula’s are as below:-
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