# Power factor | Expression for economic electrical power factor

0
26 ## What is power factor?

Power factor is the value of cosine angle (cosθ) between AC voltage (V) and current (I). In other words, it can be said (cosθ) is the ratio of actual electrical energy (KW) and apparent energy (KVA), it is said that, cosθ = KW/KVA. If the value of this power factor is less, current flows more as a result copper loss $$(I^2R)$$ increases, the output becomes less, drop (IR) increases. Electric machine, in other words, welding of motor transformer gets warm and within few days, electric machinery and appliance become inactive. Low valued power factor is harmful to both customers and supply authorities.
The power factor of low qualities can be developed in three ways –

#### Static capacitor or capacitor bank

In a system, three capacitors are connected in star or delta, then connect parallel with the three-phase line and can promote the value of power factor. In this system, the loss is less. There are no rolling foundlings, so less look after is needed. It can set easily and no strong foundation is needed. Overall, it works well in bad weather. There are also some disadvantages of this system, they are- life span is 8 to 10 years. If the voltage is more from bated voltage for any reason, then it ruins easily. Besides, it is not easy to repair.

#### Synchronous condenser

A synchronous motor becomes over excited when in current leading and works like a capacitor. And this overexcited synchronous motor is operated by connecting in line for promoting without load, then it is called synchronous condenser. This system has some advantages, they are- by changing its feed excitation the value of current can bring in actual value, the thermal stability of motor wilding is much and can remove the fault easily. But this system also has some disadvantages, they are- as it is a motor, so power loss happens in proportional rate, the repair cost is high, create sound. In case of rating, if go over 500 KVA the cost is more than a static capacitor. Besides, there needs a helper instrument for starting. Three-phase advancer is used very less to improve the value of the power factor.

Usually, to develop the value of power factor of large induction motor, this phase advancer is seen to use. The cause of being the low value of induction motor is the exciting current motor is lagging 90° angular distance. Therefore, when the motor accepts from this current line, the angular distance increases between the supply voltage and line current. As a result, the deterioration of it happens. But if phase advancer is used with a large induction motor, the supply of rotor circuit exciting current of motor can get from phase advancer. In it, only the effective current should accept from the motor line. At that time, being the same direction of line current and supply voltage, the value of the power factor stands in the highest position.

### The expression for most economic factor of electrical power

Most economical power factor is the value, which makes the highest annual savings by developing its value. When a customer promotes his/her power factor, then his/her demand for maximum KVA decreases and charge savings of maximum KVA demand saves. Assuredly, when it becomes developed then capitalizing also invested for refinement of the equipment of it. Therefore, the customer considers the cost of every year in the annual interest and depreciation of mentioned investment. So, the maximum demand for KVA charge of customer’s annual net savings (-) is equal to the annual cost of it refinement equipment.
Suppose, a customer accepts P KW load in cosθ and the annual charge is X per KVA for maximum demand. The customer promotes the power factor in cosθ2 by setting refinement equipment of it. To guess, the annual cost for refinement equipment of it is = Dollar (\$): (Y) in per KVAR. The power triangle OAB is the main power factor of cosθ, and triangle OAC is the promoted factor of cosθ2.

When the power factor is cosθ1, then maximum KVA demand
KVA1= P/cosθ1
= Psecθ1
When is the cosθ2, then maximum KVA demand
KVA2= P/cosθ2
= Psecθ2
The highest demand is the charge in annual savings = x(KVA1-KVA2)
= Dollar x(Pcosθ1-Psecθ2)
= Dollar xP(secθ1-secθ2) …….. (1)
When power factor cosθ1, then reactive power, KVAR1=Ptanθ1
When cosθ2, then reactive power, KVAR2=Ptanθ2
Leading using power factor equipment KVAR
supply = P(tanθ1-tanθ2)………. (2)
Yearly cost for refinement equipment of power factor
= Dollar Py(tanθ1-tanθ2)
Net annual saving (S)= from equation (1) and (2)
S= xP(secθ1-secθ2) – yP(tanθ1-tanθ2)
In this equation only θ2 variable, when others quantity is fixed. So, net annual savings will be maximum if the above-mentioned equation will be zero in contingently differentiate of θ2. $$d/dθ_2$$ (S)=0

$$d/dθ_2$$ [xP(secθ1-secθ2) – yP(tanθ1-tanθ2)]=0
$$d/dθ_2$$ (xPsecθ1)- $$d/dθ_2$$(xPsecθ2)-$$d/dθ_2$$ (yPtanθ1)+$$d/dθ_2$$ (yPtanθ2)=0
0 – xPsecθ2tanθ2 – 0 + $$yPsec^2θ_2$$=0
Psecθ2 (-xtanθ2 + ysecθ2)=0
Therefor, -xtanθ2+ysecθ2=0
tanθ2= Y/x secθ2
sinθ2=y/x
Most economical power factor
$$cosθ_2=√(1-sin^2 θ_2 )$$
= $$√[1-(y/x)^2]$$