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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Short Transmission Line (33kV)

A short transmission line is considered when line distance is less than 80km and voltage of up to 50kV. Since this is a short transmission line, capacitance effect distributed along the line can be neglected.

This problem example demonstrate how to find receiving end current, sending end current, sending end voltage, sending end power factor,voltage regulation and transmission efficiency.

INPUT DATA
3-Phase supply voltage	33kV
3-phase power to deliver	3.5 MW
Line impedance	4 + j5 Ohm

Let's calculate first single phase voltage Vr, set as reference phasor.

Total impedance per phase = 4 + j5 Ohm = 6.4 /_ 51.3° Volt
Receiving end voltage, Vr=33kV/1.732 =19/_0° kV

Note: Vr is phase voltage Red to neutral

Since loads are balanced, all line currents are the same ,thus I_R =1/3 x 3.5MW/( 19kV x 0.86 ) =72 A or

3-phase current = 3.5MW/(1.732 x 33kV x 0.86) = 72A

I_R lags V_R by an angle of cos ^-1 0.86 = 30.68° (see phasor diagram below)
I_R =72 /_ -30.68° Amps

Note that Sending End Current,Is = Receiving End Current,Ir

Single phase Sending End Voltage, Vs = Ir Z + V_r = (72 /_ -30.68° x 6.4 /_ 51.3°) + 19000/_0° = 455.7/_20.62 + 19000/_0° = 19426.5 + j 160.5 =19427.2 /_0.5° Volts

3-phase Sending End Voltage = 1.732 x 19427.2 V =  33.65 kV. This means that in order to keep 33kV 3-phase at receiving end, a voltage of 33.65 kV must be injected at sending end side, eg at secondary side of 132kV/33kV transformer.

Voltage regulation = (Vs - Vr )/Vr = (19427.2 -19000)/19000 x 100 = 2.2 %

angle of Is + angle of Vs = 30.68° + 0.5° =31.18°

Sending end power factor = cos 31.18°=0.855 lagging.

or alternatively, from phasor diagram above, sending end power factor ={ Vr Cos ø_r + Ir R } / Vs = { (19000 x 0.86 ) + (72 x 4) }/19427.2 = 0.855 lagging

Losses per 3-phase lines = 3 x Ir ² R = 3 x 72 ² x 4 =60.8kW

Efficiency of transmission line = Power _delivered /Power _{sent out} =3500/ (3500 + 60.8) =98.3%

OUTPUT DATA
3-phase sending end voltage	33.65 kV
Voltage regulation	2.2%
Sending end power factor	0.855 lagging
Losses in 3-phase lines	60.8 kW
Efficiency of transmission lines	98.3%

Note: Voltage regulation and efficiency of medium transmission line (end condensor method, PI-method and T-method are covered in e-book

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