Electric Power – Part 4 – AC Power Equation

 

Illustration of power flow direction

Let us consider an induction motor operating at 1 kVA @ 0.8 p.f. lagging.  The motor supply voltage is given to be 240V.  (For simplicity, let us assume that it is a single-phase system).

The current direction is assumed to be flowing into the motor.  This is the normal convention since the active power flows into the motor.

We have,

 S  =  1 ∟+cos^-1(0.8)  kVA

      =  1 ∟+36.87º  kVA

      =  (0.8 + j 0.6)  kVA

We have used a positive angle for the complex power (S), since the ‘inductive’ reactive power flowing into the load is positive as per the assumed current direction.

(If the power factor is given to be leading, then we need to use a negative angle, if the current direction is assumed to be flowing into the motor.)

We now have,

P = +0.8 kW   and    Q = +0.6 kvar

Since P and Q are positive values, both active and reactive power is flowing in the direction of the assumed current direction.

Let us now calculate the current flowing into the motor, assuming the terminal voltage as our reference vector.

We have,

V = 240 ∟0º  V

Using Equation 1 (S = V I*) , we have,

 I  =  (S / V)*

    =  ( 1000 ∟+36.87º  / 240 ∟0º )*

    =  4.167 ∟-36.87º  A

Note that the current is lagging the voltage.  Hence, the motor is absorbing inductive (lagging) reactive power.

Overview of AC power flow terminologies

For a novice electrical engineer, the AC power flow concepts and terminologies could be quite daunting.  The following overview provides a useful summary.

Complex Power (S)

The term power (with no qualifications) in AC systems is normally presumed to mean the complex power (S) and is expressed in the following forms:

S  =  (P + j Q)  =  S ∟φ  =  V I*

Active power (P)

The term active power (P) is the average power flow over one cycle.  This is the power generated by energy conversion (for example, a hydro generator) and this ‘electric’ power is available for conversion to other forms of energy.  This is the ‘useful’ electric power.  The value of ‘P’ is calculated as below.

P = Pav = Real part of ( S )  = V I cos(φ)  W (watts)

Reactive power (Q)

The term reactive power refers to the ‘oscillatory part’ of the AC power flow.  It quantifies the power flow due to energy storage elements, namely inductances and capacitances.  The value ‘Q’ is defined as below.

Q = Imaginary part of ( S )  = V I sin(φ)   var (volt-ampere-reactive)

Apparent power (S)

The magnitude (S) of the AC power is called the apparent power.  It is calculated as given below.

S  = Modulus of (V I*)

    = √(P*P + Q*Q)

    = V I    VA (volt-amperes)

Even though the magnitude of complex power is a mathematical quantity, it is popularly used by electrical engineers to specify the power rating of the electrical equipment.  In practice, the apparent power specifies the current rating of the electrical equipment, since electrical equipment is normally operated at a specified voltage based on the supply.

The term ‘apparent power’ is rarely used in practice.  Instead, the terms ‘rated power’ or ‘power rating’ are commonly used.  For example, we say the induction motor is rated at 1 kVA at the rated voltage of 415 V.

Power factor (p.f.)

The specification of ‘apparent power’ (S) essentially specifies the ‘current rating’ of the electrical equipment.  It does not specify the ‘active power’ (P) rating.  To facilitate specification of active power,  the ‘power factor‘ is specified.

The power factor (p.f.) is defined as the ratio of ‘active power’ to ‘apparent power’.

p.f.  = P / S

It is common to specify the power rating of electrical equipment as apparent power rating (normally in kVA) along with rated power factor (p.f.).  Alternatively, active power rating (normally in kW) and the rated power factor (p.f.) is also specified.

The unit for power factor is not relevant as it is a ratio of powers!