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How to avoid low power factor?


  1. As far as possible, over excited synchronous motors have to be used in place of induction motors.
  2. High speed induction motors have better power factor than low-speed induction motors. They are smaller in frame size and economical too. 
  3. The induction motors have maximum power factor when fully loaded. So we should try use induction motors at full load condition.

What are the causes of low power factor?

Various Causes of Low Power factor:
  1. Most of the AC motors are induction motor type. They have low lagging power factor. During light loads the induction motors will work at a power factor or 0.2 to 0.3 and during full load condition they will work at 0.8 to 0.9.
  2. Arc lamps, electric discharge lamps and industrial heating furnaces operate at low lagging power factor.
  3. Very low lagging power factorof agriculture motor pump set.
  4. The load on the power system is varying. It will be high during morning and evening and will be low at other times. The supply voltage is increased during low load period, which increases the magnetization current. It results in low power factor.

What are the disadvantages of Low Power Factor?

Low Power Factor Disadvantages:

In AC circuits, power consumed depends on the power factor. Thus the power factor plays an important role in AC circuits. For instant, we know that;

Power in a Three Phase AC Circuit = P = √3 V x I CosФ
And Current in a Three Phase AC Circuits = I = P / (3 V x CosФ)
I ∝1 /CosФ….… (1)

Also,
Power in a Single Phase AC Circuits = P = V x I CosФ
And Current in a Three phase AC Circuits = I = P / (V x CosФ)
I ∝ 1/CosФ……… (2)

It is clear from both equations (1) an (2) that Current “I” is inversely proportional to CosФ i.e. Power Factor. In other words, When Power Factor increases, Current Decreases, and when Power Factor decreases, Current Increases.

Now, In case of Low Power Factor, Current will be increased, and this high current will cause cause the following disadvantages:

1.) Large Line Losses (Copper Losses):
We know that Line Losses is directly proportional to the squire of Current “I2
Power Loss = I2xR i.e., the larger the current, the greater the line losses i.e. I>>Line Losses
In other words,  
Power Loss = I2xR = 1/CosФ….. Refer to Equation “I ∝ 1/CosФ”….… (1)
Thus, if Power factor = 0.8, then losses on this power factor =1/CosФ2 = 1/ 0.8= 1.56 times will be greater than losses on Unity power factor.

2.) Large kVA rating and Size of Electrical Equipments:
As we know that almost all Electrical Machinery (Transformer, Alternator, Switchgears etc) rated in kVA. But, it is clear from the following formula that Power factor is inversely proportional to the kVA i.e.
CosФ = kW / kVA

Therefore, The Lower the Power factor, the larger the kVA rating of Machines also, the larger the kVA rating of Machines, The larger the Size of Machines and The Larger the size of Machines, The Larger the Cost of machines.

3.) Greater Conductor Size and Cost:
In case of low power factor, current will be increased, thus, to transmit this high current, we need the larger size of conductor. Also, the cost of large size of conductor will be increased.

4.Poor Voltage Regulation and Large Voltage Drop:
Voltage Drop = V = IZ.
Now in case of Low Power factor, Current will be increased. So the Larger the current, the Larger the Voltage Drop.

Also Voltage Regulation = V.R = (VNo Load – VFull Load)/ VFull Load
In case of Low Power Factor (lagging Power factor) there would be large voltage drop which cause low voltage regulation. Therefore, keeping Voltage drop in the particular limit, we need to install Extra regulation equipments i.e. Voltage regulators.

5.Low Efficiency:
In case of low Power Factor, there would be large voltage drop and large line losses and this will cause the system or equipments efficiency too low. For instant, due to low power factor, there would be large line losses; therefore, alternator needs high excitation, thus, generation efficiency would be low. 

6.) Penalty from Electric Power Supply Company on Low Power factor
Electrical Power supply Company imposes a penalty of power factor below 0.95 lagging in Electric power bill. So you must improve Pf above 0.95. 

Basic Motor Formulas and Calculations

Here are some  motor formulas that may be useful.

Calculating Motor Speed:
A squirrel cage induction motor is a constant speed device. It cannot operate for any length of time at speeds below those shown on the nameplate without danger of burning out.

To Calculate the speed of a induction motor, apply this formula:

Srpm = 120 x F
               P
Srpm = synchronous revolutions per minute.
120
   = constant
F
       = supply frequency (in cycles/sec)
P
       = number of motor winding poles

Example: What is the synchronous of a motor having 4 poles connected to a 60 hz power supply?

Srpm = 120 x F
                P
Srpm
 = 120 x 60
                4
Srpm
 = 7200
              4
Srpm
 = 1800 rpm

Calculating Braking Torque:
Full-load motor torque is calculated to determine the required braking torque of a motor.

To Determine braking torque of a motor, apply this formula:

T = 5252 x HP
           rpm

T      = full-load motor torque (in lb-ft)
5252
 = constant (33,000 divided by 3.14 x 2 = 5252)
HP
    = motor horsepower
rpm
 = speed of motor shaft

Example: What is the braking torque of a 60 HP, 240V motor rotating at 1725 rpm?

T = 5252 x HP
           rpm
T
 = 5252 x 60
          1725
T
 = 315,120
         1725
T
 = 182.7 lb-ft

Calculating Work:
Work is applying a force over a distance. Force is any cause that changes the position, motion, direction, or shape of an object. Work is done when a force overcomes a resistance. Resistance is any force that tends to hinder the movement of an object. If an applied force does not cause motion the no work is produced.
To calculate the amount of work produced, apply this formula:

W = F x D

W = work (in lb-ft)
F
  = force (in lb)
D
  = distance (in ft)

Example: How much work is required to carry a 25 lb bag of groceries vertically from street level to the 4th floor of a building 30' above street level?

W = F x D
W
 = 25 x 30
W
 = 750 -lb

Calculating Torque:
Torque is the force that produces rotation. It causes an object to rotate. Torque consist of a force acting on distance. Torque, like work, is measured is pound-feet (lb-ft). However, torque, unlike work, may exist even though no movement occurs.

To calculate torque, apply this formula:

T = F x D

T = torque (in lb-ft)
F
 = force (in lb)
D
 = distance (in ft)

Example: What is the torque produced by a 60 lb force pushing on a 3' lever arm?

T = F x D
T
 = 60 x 3
T
 = 180 lb ft


Calculating Full-load Torque:
Full-load torque is the torque to produce the rated power at full speed of the motor. The amount of torque a motor produces at rated power and full speed can be found by using a horsepower-to-torque conversion chart. When using the conversion chart, place a straight edge along the two known quantities and read the unknown quantity on the third line.

To calculate motor full-load torque, apply this formula:

T = HP x 5252
           rpm

T = torque (in lb-ft)
HP
 = horsepower
5252
 = constant
rpm
 = revolutions per minute

Example: What is the FLT (Full-load torque) of a 30HP motor operating at 1725 rpm?

T = HP x 5252
           rpm
T
 = 30 x 5252
          1725
T
 = 157,560
          1725
T
 = 91.34 lb-ft


Calculating Horsepower:
Electrical power is rated in horsepower or watts. A horsepower is a unit of power equal to 746 watts or 33,0000 lb-ft per minute (550 lb-ft per second). A watt is a unit of measure equal to the power produced by a current of 1 amp across the potential difference of 1 volt. It is 1/746 of 1 horsepower. The watt is the base unit of electrical power. Motor power is rated in horsepower and watts.
Horsepower is used to measure the energy produced by an electric motor while doing work.

To calculate the horsepower of a motor when current and efficiency, and voltage are known, apply this formula:

HP = V x I x Eff
               746

HP = horsepower
V
    = voltage
I
     = curent (amps)
Eff.
 = efficiency

Example: What is the horsepower of a 230v motor pulling 4 amps and having 82% efficiency?

HP = V x I x Eff
               746
HP
 = 230 x 4 x .82
                746
HP
 = 754.4
           746
HP
 = 1 Hp

Eff = efficiency / HP = horsepower / V = volts / A = amps / PF = power factor
Horsepower Formulas
To Find
Use Formula
Example
Given
Find
Solution
HP
HP = I X E X Eff.
               746
240V, 20A, 85% Eff.
HP
HP = 240V x 20A x 85%
                    746
HP=5.5
I
I = HP x 746
     E X Eff x PF
10HP, 240V,
90% Eff., 88% PF
I
I = 10HP x 746
      240V x 90% x 88%
I = 39 A

To calculate the horsepower of a motor when the speed and torque are known, apply this formula:

HP = rpm x T(torque)
           5252(constant)

Example: What is the horsepower of a 1725 rpm motor with a FLT 3.1 lb-ft?

HP = rpm x T
             5252
HP
 = 1725 x 3.1
              5252
HP
 = 5347.5
           5252
HP
 = 1 hp
  

Calculating Synchronous Speed:
AC motors are considered constant speed motors. This is because the synchronous speed of an induction motor is based on the supply frequency and the number of poles in the motor winding. Motor are designed for 60 hz use have synchronous speeds of 3600, 1800, 1200, 900, 720, 600, 514, and 450 rpm.

To calculate synchronous speed of an induction motor, apply this formula:

rpmsyn = 120 x f
                    Np

rpmsyn = synchronous speed (in rpm)
f
           = supply frequency in (cycles/sec)
Np
       =  number of motor poles

Example: What is the synchronous speed of a four pole motor operating at 50 hz.?

rpmsyn = 120 x f
                  Np
rpmsyn = 120 x 50
                   4
rpmsyn = 6000
                   4
rpmsyn = 1500 rpm