Calculate capacitor for power factor correction | Power factor correction capacitor bank | Power factor correction formula | How to calculate capacitor bank to improve power factor
Content :
1) About power factor correction2) Calculate capacitor for power factor correction
3) Power factor correction capacitor bank
When we talk about ac power, the question of power factor immediately comes into picture. and than Calculate capacitor for power factor correction, Power factor correction capacitor bank, Power factor correction formula, How to calculate capacitor bank to improve power factor ? problems arise in front of us here we discuss all this problems and its solutions.
Most of the loads are inductive in nature and hence have low lagging power factor.
The low power factor is highly undesirable as it causes an increase in current, resulting in additional losses of active power in all the elements of power system from power station generator down to the utilization devices.
In order to ensure most favorable conditions for a supply system from engineering and economical standpoint, it is important to have power factor as close to unity as possible.
In this article we discuss about power factor improvement Technic
There many Technic of improvement of power factor here we discuss static capacitor method
in this method power factor is improved by connecting capacitor parallel to the low power factor load.
A capacitor draws a leading current which neutralized the lagging current component form load current . this rises the power factor of the load.for three phase load capacitor are connected in star or delta connection.
A static capacitor method is mostly use in factories.
So lets start our today's topic How to Calculate capacitor for power factor correction.
Power Factor Improvement
Fig.shows a capacitor connected across the load. The capacitor takes a current Ic which leads the supply voltage V by 90°. The current Ic partly cancels the lagging reactive component of the load current as shown in the phasor I' and its angle of lag is Փ2. It is clear that , is less than previous p.f. angle Փ1.
It is clear that cosՓ2 is improved new power factor
From the phasor diagram, it is clear that after p.f. correction, the lagging reactive component the load is reduced to I' sinՓ2.
I' sinՓ2 = I sinՓ1 - Ic
Ic = I sinՓ1 -I' sinՓ2
Capacitance of capacitor to improve p.f. from cosՓ1 to cosՓ2,
Xc = Ic / 2πf V
Power triangle :
The power factor correction can also be illustrated from power triangle. Thus referring to Fig, the power triangle OAB is for the power factor cosՓ1 , whereas power triangle OAC is for the improved power factor cosՓ2 .
It may be seen that active power (OA) does not change with power factor improvement However, the lagging KVAR of the load is reduced by the pf correction equipment, thus improving the p f. to cosՓ2.
Leading KVAR supplied by p.f correction equipment = BC = AB - AC
= kVAR1 - kVAR2
= OA( tanՓ1 - tanՓ2 )
= kW( tanՓ1 - tanՓ2 )
Here we have value of Փ1,Փ2 and active power kW, so we can find value of KVAR require .
Value of capacitor in Micro Farad
Calculate capacitor for power factor correction :
Example 01
A single phase motor connected to 400 V, 50 Hz supply takes 31.7A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging.
Solution :
Note that the effect of connecting a 94.3 uF capacitor in parallel with the motor. The current taken from the supply is reduced from 31.7 A to 24.65 A without altering the current or power taken by the motor. This enables an economy to be affected in the size of generating plant and in the cross sectional area of the conductors.
Example 02
A 3-phase, 50 Hz, 400 V motor develops 100 H.P. (74.6 kW), the power factor being 0.75 lagging and efficiency 93%. A bank of capacitors is connected in delta across the supply terminals and power factor raised to 0.95 lagging. Each of the capacitance units is built of 4 similar 100 V capacitors. Determine the capacitance of each capacitor.
solution :
Original pf,cosՓ1 = 0 75 lag
Final p.f, cos = 0-95 lag
Motor input, P = output/efficiency = 74 6/0.93 = 80 kW
Փ1 = cosᐨ¹( 0.75 ) = 41.41°
tan = tanᐨ¹ ( 41.41°) = 0 8819
Փ2= cosᐨ¹ ( 0.95 ) = 18.19 °
tanՓ2 = = tanᐨ¹ ( 18.19°) = 0.3288
Leading kVAR taken by the condenser bank
= P (tan P tan )
= 80 (0.8819 - 0.3288)
= 44.25 KVAR
Leading KVAR taken by each of three sets
44.25/3 = 14.75 KVAR
lets find value of capacitor per phase in micro farad
C = 29304 uF
Note : capacitor is always connected in delta connection ,capacitor bank connected in delta is require 3 times less value of capacitance than requires in star connection.
Hope you understand the concept of power factor and Calculate capacitor for power factor correction,Power factor correction capacitor bank,Power factor correction formula for industrial use.
This is my view on after referring book principle of power system by V K Mehta Calculate capacitor for power factor correction,Power factor correction capacitor bank,Power factor correction formula,How to calculate capacitor bank to improve power factor.
This is dummy text. It is not meant to be read. Accordingly, it is difficult to figure out when to end it. But then, this is dummy text. It is not meant to be read. Period.
ConversionConversion EmoticonEmoticon