On this page , we analyse the effect to supply current when power factor correction capacitors in delta connection is installed in parallel with loads. It is assumed that an initial power factor (pfi) is to be improved to a final power factor (pff). A single phase load is P kW.
Reactive power required to improve power factor from pfi (short for initial power factor) to pff( final power factor) :
Q = P x { ( tan cos -1 pfi ) - (tan cos -1 pff) } kVar
Above diagram shows correct location of power factor meter (PFM). Without capacitor, PFM will display a value of pfi. With the capacitors installed as shown, it will see the improvement in power factor (ie pff). Normally, in practice, Red phase is selected for tapping voltage and current sensors to the PFM which comes in 6-step or 12-step regulator.
Correct tapping point is supply line before capacitor For 11kV and above, voltage sensor to power factor meter is rated V(primary) / root(3) : 110V / root(3) and current sensor is current transformer (CT) of suitable current rating. For pimary voltage of 415V, voltage sensor is directly tapped from the line and neutral, ie 415V/root(3) =240V and current transformer (for current above 100A). CT requirement varies from one district of power company to another. Some district requires that a CT be provided for load current of 60A and above. With current technology, it is possible nowadays that line current of 100A and below can pass through power factor meter without a CT !
Receiving end single phase voltages
Voltage per phase, VRN= VL/1.732 Volts.
Let VRN be the reference voltage => VRN= |VRN| /_0°
VYN lags VRN by 120° or leads by 240° => VYN=VRN /_-120° = VRN /_+ 240°
VBN leads VRN by 120° or lags by 240° => VBN=VRN /_+120° = VRN /_- 240°
Let's first analyze load curents of phases R, Y and B
Let single phase voltage be | Vp | = | VRN | = | VYN | = |VBN|
| Ig | = P/ (Vp x pfi)
Ig = | Ig | /_ - cos -1 pfi, Note: minus sign before cos -1 pfi indicates that Ig lags Vp by an angle of cos -1 pfi (see phasor diagram below)
| Ih |=| Ig | since loads are balanced.
Ih = | Ih | /_ { (- cos -1 pfi ) - 120° }
| Ii |= |Ig| since loads are balanced.
Ii = | Ii | /_ { (- cos -1 pfi ) - 240° }
Secondly, analyzing curents to terminals of delta connected capacitors
Id = Ij - Il = (Sxy /Vxy)* - (Szx / Vzx) * = (-jQ/VL/_30°)* - (-jQ/VL/_(30° - 240°))* = ( Q/VL /_(-90° - 30°) )* - (Q/VL /_(-90° - 30° + 240°) )* = (Q/VL /_-120°)* - (Q/VL /_120°)* = (Q/VL /_+120°) - (Q/VL /_-120°)= 1.732 Q/VL /_90°
where Complex Power S = V x I*. Therefore, I = (S/V) *. Note that I* denotes I conjugate.
Ie = Ik - Ij = (Syz /Vyz)* - (Sxy / Vxy)* = ( -jQ/VL/_(30°-120°) )* - ( -jQ/VL/_30°) * = (Q/VL/_(-90° -30°+120°) )* - (Q/VL /_(-90° - 30°) )* = (Q/VL/_0° )* - (Q/VL /_-120°)*= (Q/VL/_0° ) - (Q/VL /_+120°) = 1.732 Q/VL /_-30°
If = Il - Ik = ( Szx /Vzx)* - (Syz / Vyz) * = (-jQ/VL/_(30°-240°) )* - (-jQ/VL/_(30°-120°)) * = (Q/VL/_(-90° -30°+240°) )* - (Q/VL /_(-90° - 30°+120°) )* = (Q/VL/_120° )* - (Q/VL /_0°)*= (Q/VL /_-120° ) - (Q/VL /_0°) = 1.732 Q/VL /_-120°
Lastly, analyzing utility supply curents of phase R, Y and B.
Ia = Id + Ig = 1.732 Q/VL /_90° + | Ig | /_ - cos -1 pfi
Ib = Ie+ Ih = 1.732 Q/VL /_-30° + | Ih | /_ { (- cos -1 pfi ) - 120° }
Ic = If+ Ii = 1.732 Q/VL /_-120° + | Ii | /_ { (- cos -1 pfi ) - 240° }
Current flow in the neutral line
Line currents in terms of positive, negative and zero sequence components are as follows. Note that subscripts 1 denotes posive sequence, 2 being negative sequence, and 0 zero sequence.
I R = I R1 + I R2 + I R0
I Y = I Y1 + I Y2 + I Y0 = I R1 /_240° + I R2 /_120°+ I R0
I B = I B1 + I B2 + I B0 = I R1 /_120° + I R2 /_240°+ I R0
I R + I Y + I R = I R1 (1 + /_240° + /_120°) + I R2 (1 + /_240° + /_120°) + 3 I R0
IN = - ( I R + I Y + I R ) = - 3 I R0
Let's find value of capacitance,C
Xc = VL / i = 1/(wC)
VL/ (1.732Q/VL) = 1/wC
C= 1.732 Q/(w x VL 2) where w= 2 x pi x f.
Typical Example
A 3-phase 415V supply delivers a balanced 3-phase load of 300kW , power factor lagging 0.75 is to be corrected to 0.87
Load current of line R, Ig = | Ig | /_ - cos -1 pfi = 100kW /(240V x 0.76) /_- cos - 0.75 =555.55 /_-41.4° A
Load current of line Y, Ih = | Ig | /_ (- cos -1 pfi - 120°)= 100kW /(240V x 0.76) /_(-41.4° -120°) =555.55 /_-161.4° A
Load current of line B, Ii = | Ii | /_(- cos -1 pfi - 240°) = 100kW /(240V x 0.76) /_(-41.4° -120°) =555.55 /_-281.4° A
From Table 4D1A of 16th Edition IEE Wiring Regulation (BS 7671), 4 x 1core mm 185 sq PVC Cu carry 296A
Therefore use cable 8 x 1core 185 mm sq PVC Cu (carry 2 x 296A=592A > 555.55A) ok.
Reactive power (single phase), Q=100kW x [ ( tan cos -1 0.75 ) - (tan cos -1 0.87) ] = 31.5kVar
Capacitor line current, Id =1.732 Q/VL /_90° = 131.5 /_90°A
Use 150A 3-pole MCCB and 3 x single core 70 mm sq PVC (171A capacity)
Supply current of line R, Ia = Ig + Id = 555.55 /_-41.4° + 131.5 /_90° = 416.7 - j235.9 =478.8 /_ -29.5°
From Table 4D1A of 16th Edition IEE Wiring Regulation (BS 7671), 4 x 1core 120 mm sq PVC Cu carry 296A
Therefore use cable 8 x 1core 150 mm sq PVC Cu (carry 2 x 262A=524A > 478.8A) ok.
or use cable 4 x 1core 500 mm sq PVC Cu (carry 533A > 478.8A) ok
Current in neutral line = - ( I R + I Y + I R ) = - ( 555.55 /_-41.4° + 555.55 /_-161.4° + 555.55 /_-281.4° ) =0 A (Note: This is true for balanced 3-phase loads. For the case of unbalanced 3-phase loads, current in neutral line is not zero. That's why engineer and contractor are required to check and make sure before works are commissioned that loads of Red, Yellow and Blue lines are balanced)
Note that besides PF is improved from 0.75 to 0.87, smaller sized cables of 2 x 150mm per phase is required to carry supply current when cap bank is installed as compared to bigger cable of 2 x 185 mm sq at load side.
Acknowledgement:
e-circuit-optimizer.com wish to say thanks to Dr Azhar Khairuddin of Electrical Engineering Faculty, Universiti Teknologi Malaysia who helped commenting on Complex Power, current flows in utility supply line, delta connected capacitor line and end load circuit.