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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Litar kawalan

Ditunjukkan di atas ialah litar kawalan asas yang terdiri dari off push button, on push button, normally off relay contact, indication lamp dan relay coil.

Bekalan voltan auxillary bagi sesebuah litar kawalan boleh direka berdasarkan kesesuaian sistem elektrik. Ada 12Vdc, 24Vdc, 30Vdc, 48Vdc, 24Vac, 240Vac. Dalam contoh di atas bekalan auxilliary ialah 240 Vac.

Apabila on push button ditekan(tutup), walau sesaat, relay coil R1 akan ter"energize" dan serentak dengan itu normally open relay R1 akan tutup (self-hold).

Apabila off push button ditekan (buka), relay R1 yang dalam keadaan ter"energize" akan putus bekalan dan melepaskan "self-holding" contact. Sistem akan off sehinggalah "on push button" ditekan semula.

Jika kita memahami konsep operasi litar ringkas ini, kita boleh reka beratus, bahkan beribu litar kawalan untuk pelbagai aplikasi seperti kawalan sesebuah sistem penghawa dingin, sistem kawalan lampu isyarat jalan, sistem kawalan motor dalam sesebuah kilang dan bermacam-macam lagi.

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Contoh 1 : Kawalan 3 nos split unit Air-conditioner

Litar kawalan di atas direka untuk kawalan operasi tiga unit A/con.

Tiga buah thermostat digunakan, TM1 untuk unit no 1, TM2 untuk unit no 2 dan TM3 untuk unit no 3. Tiga nilai suhu yang diprogram iaitu 19 ° C, 22° C dan 25 ° C.

Suis pemula litar kawalan ditutup dan jika didapati suhu melebihi 19°C thermostat TM1 akan activate dan relay R1 akan energize. Split unit no 1 akan ON. Seterusnya apabila suhu naik dan mencapai bacaan 22°C thermostat TM2 akan activate dan split unit A/con no 2 akan ON. Sekarang kedua-dua A/con no 1 dan A/con no 2 sedang ON. Akhir sekali apabila suhu mencecah 25 ° C thermostat TM3 activate dan A/con no 3 akan ON. Sekarang ketiga-tiga A/con sedang ON.

Apabila suhu jatuh bawah 25°C relay R3 akan melepaskan self-holding contact R3 dan A/con no 3 akan OFF. Seterusnya apabila suhu jatuh lagi ke bacaan bawah 22 °C A/con no 2 akan OFF dan tinggallah satu sahaja iaitu unit A/con no 1 akan terus ON.

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Penurunan voltan (voltage drop)

Semakin jauh beban dari punca bekalan semakin banyaklah voltan yg turun.

Penurunan voltan(VD)= beban arus(A) x volt-drop figure (mV/A/M) x jarak (m)

% VD= VD x 100 /415, based on 415V

% VD= VD x 100 /240, based on 240V

Penurunan voltan tidak dapat dielakkan kerana pengalir(conductor) kabel yg berbeza saiz mempunyai angkatap(constant) VD dalam unit mV per Ampere per meter yang berbeza.
Peraturan pendawaian elektrik (IEE Wiring Regulations UK, Edisi 17) menyatakan had maksima penurunan voltan yang dibenarkan ialah 4%

Contoh kiraan VD
Katakan beban 3-fasa ialah 70kW, faktor kuasa 0.8 lagging, bersambung ke DB-A sejauh 20 meter
Arus beban = 70000/(1.73 x 415 x 0.8) = 121.9AA

Pilih kabel 1 no 4-core PVC/SWA/PVC armoured untuk direntang di dalam tanah. Muatan arus kabel ialah 135A (Table 4D4A, 17th Edition IEE Wiring Regulations, UK) dan volt-drop figure 1.1mV/A/m (Table 4D4B, 17th Edition IEE Wiring Regulations, UK)

VD sebahagian (sectional) dari DB-X ke DB-A = 121.9 A x 1.1 mV/A/M x 20m = 2681mV =2.7V = (2.7/415)x 100% = 0.65%

Anggap % VD sebahagian(sectional) dari transformer ke DB-X telahpun dikira dan didapati nilainya = 1.5%,

Oleh itu VD kumulatif di DB-A = 1.5 + 0.65 = 2.15 % < 4% (ok!)

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Overview of Voltage & current - magnitude & phase angle

Always set Voltage (red phase to neutral) as reference phasor.ie V_rn= |V_rn|/_0°

The next Voltage drawn in clockwise rotation (yellow phase to neutral, Vyn) is displaced by 120° with respect to reference phasor. So V_yn= |V_yn|/_120°

The next Voltage again drawn in clockwise rotation after Vyn is (blue phase to neutral, Vbn) displaced by 240° with respect to reference phasor. So V_bn= |V_bn|/_240°

Let's start calculating load currents.

Assume a balanced 3-phase load of 900kW is delivered from 415V_ac 3-phase source at power factor 0.8 lagging.

Load per phase = 900kW/3 =300kW

3-phase current magnitude = 900kW / (√3 x 0.415 kV x 0.8) = 1565 A

Load current always lags it's phase Voltage by an angle of cos ^-1pf where pf is power factor.

Angle = cos ^-1pf = cos ^-1 0.8 = 36.9 °

Phase current I_R= 1565 /_-36.9° A. The -ve sign shows that the current I_R lags phase Voltage |V_rn|/_0° by 36.9 °

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