Calculations:Electrical Design Guideline/Basic Electrical Engineering Formulas

Basic Electrical Engineering Formulas

List of Symbols

V - Voltage (volts)

I - Current (amps)

R - Resistance (ohms)

X - Reactance (ohms)

Z - Impedance (ohms)

W Real Power (watts)

θ - Phase angle whose cosine is the power factor

eff - Efficiency

Direct Current (DC) Formulas

Basic Formulas

Resistance

$R={V \over I}$

$R={V^{2} \over P}$

$R={P \over I^{2}}$

Volts

$V=I\times R$

$V={P \over I}$

$V={\sqrt {P\times R}}$

Power

$P=V\times I$

$P=I^{2}\times R$

$P={V^{2} \over R}$

Current

$I={P \over V}$

$I={V \over R}$

$I={\sqrt {P \over R}}$

Alternating Current (AC) - Single Phase

Note: V denotes line to neutral voltage.

Basic Formulas

Impedance Z (ohms)

$Z={V \over I}={\sqrt {R^{2}+X^{2}}}$

Volts V (volts)

$V={I\times Z}$

Real Power P (watts)

$P=V\times I\times cos\phi$

Power Factor

$pf=cos\phi$

$pf={P \over S}$

Apparent Power S (volt-ampere)

$S=V\times I$

Reactive Power Q (volt-ampere-reactive)

$Q=V\times I\times sin\phi$

Real Power P (watts)

$P=V\times I\times cos\phi$

Voltage Drop

$V_{d}=2\times (I\times R\times cos\phi +I\times X\times sin\phi )$

where:

$V_{d}$ = voltage drop

$cos\phi$ = load power factor

$sin\phi$ = load reactive factor

$X$ = reactance

$R$ = resistance

Alternating Current (AC) - Three Phase

Note: V denotes line to neutral voltage.

Basic Formulas

Apparent Power S

$S={\sqrt {3}}\times V\times I$

$S={\sqrt {P^{2}+Q^{2}}}$

Real Power P

$P={\sqrt {3}}\times V\times I\times cos\phi$

Reactive Power Q

$Q={\sqrt {3}}\times V\times I\times sin\phi$

Power Factor pf

$pf={P \over S}=cos\phi$

Voltage Drop

$V_{d}={\sqrt {3}}\times (I\times R\times cos\phi +I\times X\times sin\phi )$

where:

$V_{d}$ = voltage drop

$cos\phi$ = load power factor

$sin\phi$ = load reactive factor

$X$ = reactance

$R$ = resistance

Motors

1 horsepower (hp) = 746 watts.

Note: Motor hp rating relates to motor mechanical output. To determine motor input kVA requirements, the motor efficiency and power factor must be accounted for. In general, for preliminary or rough load calculations, assume:

1 kVA of electrical input power for 1 hp of motor.

Example

Condition: A motor control center with a total connected horsepower of 337 hp can be assumed to require 337 kVA of input power. This is a conservative value, particularly for larger motors.

$Torque={{hp\times 5250} \over {RPM}}$

Fans and Blower Horsepower Equation

The following equation determines the required horsepower to drive the fan or blower element. This equation does not compensate for temperature, density or airflow characteristics of any particular fan or blower.

$P={{(Q\times p)} \over {(229\times \mu )}}$

or

$P={{(Q\times Pf)} \over {(33000\times \mu )}}$

or

$P={{(Q\times Pw)} \over {(6356\times \mu )}}$

Where:

P = Power, hp

Q = Flow Rate, cfm

p = Pressure, lb/in²

Pf = Pressure, lb/ft²

Pw = Water Gauge, Inches

μ= Efficiency coefficient

Pump Horsepower Equation

$hp={{GPM\times head\times specific-gravity} \over {3960\times eff}}$

Motors (Single Phase):

$hp={{(V\times I\times eff\times pf)} \over 746}$

Motors (3 phase):

Synchronous Speed:

$N_{s}={{120\times frequency} \over {No.-of-Poles}}$

$hp={{{\sqrt {3}}\times (V\times I\times eff\times pf)} \over {746}}$

Power Factor Correction

The size of the capacitor needed to increase the power factor from pf1 to pf2 with the initial kVA given is: $kVAR=kVA\times {{{\sqrt {1-{pf_{1}}^{2}}}-1} \over {pf_{2}{\sqrt {1-{pf_{2}}^{2}}}}}$