Real and delta-connected capacitive loads

Calculate load end currents.

Load per phase= 1320kW/3 = 440kW

Current magnitude = 440/ { (0.415/sqrt(3) ) x 0.74} = 2481.5A

θ= cos ^-1 0.74 = 42.3°

I_R= 2481.5 /_-42.3° A

Since loads are balanced,

I_Y=2481.5 /_(-42.3-120) ° A = 2481.5 /_-162.3° A

IB = 2481.5 /_(-42.3-240) ° A = 2481.5 /_-282.3° A.
Since load current is 2481.5 A, use 3000A 43kA TPN busbar.

Given line voltage is 415 V.

Let V_Rn be the reference phasor ,V_Rn= (415 /1.732) /_0°= 240/_ 0° V
V_Yn= (415 /1.732) /_0°= 240/_ -120° V
V_Bn= (415 /1.732) /_0°= 240/_ -240 ° V

V_RY= 415/_ 30° V .........Note: Why 30° ? Textbook says so !
V_YB= 415/_ (30-120)° = 415 /_ -90° V
V_BR= 415 /_ (30-240)° = 415/_ -210° V

Calculate power factor correction capacitor:
Take, for example, max load demand (MD) of Main Switchboard (normal supply) of 1320kW
Assume power factor is to be corrected from expected low value, pfi =0.74 to final power factor pff=0.85

Using basic trigonometry as shown in Power Diagram above, required reactive power to compensate low power factor of 0.74 to 0.85 can be easily obtained as:

y =1320 x tan 31.8^o=818kVar
(x + y) = 1320 x tan 42.3^o = 1200kVar

Therefore x = 1200 - 818= 382 kVar capacitors have to be installed.

A 6-step or 12-step PF regulator is used to control amount of reactive load to be connected in parallel between incomer supply line and load end.

Reactive load per phase = 1/3 x 382 kVar = 127.3 kVar

3-phase currents to delta-connected capacitors = 382 kVar /(1.732 x 415V) =531A.
Let's prove it

Calculate phase current in capacitor circuits.

I_RY= (S_RY/V_RY)^* = [127.3 /_-90° kVar/415/_30° V ] ^*= 306.7/_ 120° A

I_YB= (S_YB/V_YB)^* = [ 127.3/_-90° kVar/ 415/_ -90° V ] ^*= 306.7/_0° A

I_BR= (S_BR/V_BR)^* = [ 127.3/_-90°kVar/ 415/_ -210° V] ^*= 306.7/_-120°

Note: ^* denotes conjugate.

Calculate line currents to cap banks.

I_Rd = I_RY - I_BR = 306.7/_ 120 - 306.7/_-120 = 531/_90°

I_Yd= I_{Y B}- I_RY =306.7/_0 - 306.7/_120°= 531/_ -30°

I_Bd=I_BR - I_YB= 306.7/_-120 - 306.7/_0 =531/_ -150°

It's proven that line currents to delta-connected capacitors is 531A.

Suggest to use 600A 4-pole MCCB

From Table 4D1A of Wiring Regulations (BS7671-17th Edition),

Cable size , therefore, must have capacity of > 600 A

Try use 2 nos 240 mm sq PVC cables per phase = 346 A x 2= 692A per phase > ACB size of 600A. (OK!)

Therefore for the 3-phase incoming supply, use 8 x 1core 240 mm sq PVC cables in trunking.

Calculate supply currents.

I_R,sum = I_R + I_Rd = 2481.5 /_-42.3° + 531/_90° = 1835.4 -j1139.1 = 2160 /_31.8°
I_Y,sum = I_Y + I_Yd = 2481.5 /_-162.3° + 531/_ -30° = 1904.2 -j1020.0 = 2160 /_28.2°A

I_B,sum = I_B + I_Bd = 2481.5 /_-282.3° + 531/_ -150° = 68.7 + j2159 = 2160/_88.2°A

When adding capacitors close to loads end, we see a saving in current consumption. Instead of supplying 2481.5A to loads, only 2160A supply current is required !

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