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UNITUNITUNITUNIT----IIIIIIII:::: THREE PHASE CIRCUIT ANDTHREE PHASE CIRCUIT ANDTHREE PHASE CIRCUIT ANDTHREE PHASE CIRCUIT AND
POWER MEASUREMENTPOWER MEASUREMENTPOWER MEASUREMENTPOWER MEASUREMENT
Contents:
Three phase voltage generation and its waveform, Star and delta balanced systems, Relationship between phase and line quantities, phasor diagram, power in a three phase circuits, three phase 4 wire system, Difference between neutral and ground conductors, Safety measures in electrical system, types of wiring, Active and Reactive
Power measurement in single and three phase balanced system. (7 Hrs)
2.1 Introduction
The AC power (alternating current) is a kind of electricity in which the flow of current is frequently changing directions. At the beginning of the 1900 year, AC power supply is used for businesses as well as homes. The system of the power supply is categorized into two type’s namely single phase power supply, as well as three phase power supply. For most industrial and businesses settings, three-phase supply is used to run the high loads, whereas homes are generally supplied by a single phase, because home appliances require less power. Let us discuss upon the difference between single phase and three phase power supply.
2.1.1 Difference between Single Phase & Three Phase Supply
Both the power systems like single phase as well as three-phase uses AC power is always in the directions of alternating. Single Phase Supply
In the field of electrical, single phase supply is the delivery of AC power by a system in which all the supply voltages change in simultaneously. This type of power supply sharing is used when the loads (home appliances) ate generally heat and lighting with some moderate electric motors. When a single phase supply is connected to an AC motor doesn’t generate a rotating magnetic field , single phase motors require extra circuits (starter explained in Unit-6) for working, but such electric motors are rare over in rating of 10 kW. In every cycle, a single phase system voltage achieves a peak-value two times; the direct power is not stable. The benefits of choosing a single phase supply include the following. The design is less complex, Design cost is less, Most efficient AC power supply for up to 1 kW to 5 kW, Single Phase AC Power Supply is most competent for up to 2000 watts.
Three Phase Supply
The three-phase power supply may includes four wires such as one neutral as well as three conductors. The three conductors are away from phase & space 120º distant from each other. Three phase power supplies are utilized as a single-phase AC power supply. For the small load, 1-phase AC as Phase and neutral can be chosen from the 3-phase AC power supply system. This supply is constant and not at all totally falls to zero. The power of this system can be illustrated in two configurations namely star connection (or) delta connection. The connection of star configuration is used for Power distribution at low voltage along with a neutral wire. Choosing a single phase (or) three-phase system mainly depends on the power requirements of a particular application.
2.1.2 Advantages of Three Phase System
- The rating of three-phase motor and three-phase transformer are about 1.5 times greater than single-phase motor or transformer with a similar frame size.
- To transmit certain amount of power at a given voltage over a given distance, three-phase transmission line requires less amount of copper than single-phase line. This reduces the cost of material required, hence, becomes economical.
- The power delivered by a single-phase system pulsates. The power falls to zero, three times during each cycle. The power delivered by a three-phase circuit pulsates also, but it never falls to zero. So in three-phase system, power delivered to the load is same at any instant. thus three-phase system gives steady output.
- Three-phase motors are self starting, as the magnetic field produced by three-phase supply is rotating. But the magnetic field produced by single-phase system is pulsating, so most of the single-phase motors are not self starting.
- Power factor of three-phase motor is greater than single-phase motor for same rating.
- For converting machines like rectifiers, the d.c. output voltage becomes smoother if number of phases are increased.
2.2 Generation of Three Phase voltage/ Current & Waveforms
If a single coil placed in magnetic field is set in motion then we have seen that alternating emf is induced in a coil as explained in Unit-I. If three coils Namely R, Y and B displaced by an angle of 120ο^ from each other, placed in a changing magnetic field and rotated, then alternating emf is induced in all the three coils. If this three coils are equal and placed in same Magnetic field, therefore Max. value of EMF induced the them will be Same. These Emfs are given by
X 3 R 1
R 3 X 1
X 2 R 2
(c) Balanced Load : The load is said to be balanced when loads in each phase are equal in magnitude and identical in nature.
2.3 Star Connected balanced systems
Derive the relations between line values and phase values of voltage and current for a 3- phase, balanced, star-connected, lagging power factor load. Sketch the phasor diagram of these quantities. Derive also powers consumed by load. Consider the balanced star-connected load. Line voltages, VL = VRY = VYB = VBR {means supply is balanced} Line currents, IL = IR = IY = IB Phase voltages, Vph = Vr = Vy = Vb Phase currents, Iph = Ir = Iy = Ib
1. Voltage relation: To derive relation between VL and Vph, let us consider line voltage VL = VRY. Applying Kirchhoff’s Voltage Law Between lines R and Y i.e. loop R-N-Y-R, we have, - V (^) r+ V (^) y+ V (^) RY= 0
Hence, V RY = V r- V y= V r + ( - V y) ……. (i)
Phasor sum of Phase voltage V r & ( - V y) is Line voltage VRY. See in Phasor dia.
Impedance of all three phases same in magnitude and nature.
Z 3 =10Ω Z 1 =10Ω
Z 2 =10Ω
VBR Vb - Vy VRY
Ib=IB φο 60 ο -Vr φο^ Vr = Ph. Voltage (Ref) Ir=IR φο Iy=IY - Vy VRY B Vy -Vb C O 30 ο A VYB Vr Phasor Diagram for STAR connection To find Voltage relation
Similarly, V^ YB = V^ y- Vb^ (ii)
and V BR = Vb - V r (iii)
The phasor diagram will give relation between line voltage and phase voltage. Consider the balanced Star-connected load of inductive in nature then current per phase will lag to voltage per phase. Consider Phase current lags phase voltage by an angle φ.
The perpendicular is drawn from point A on phasor OB representing VL. OC = CB =
VL
Angle between Vr and –Vy is 60ο. So ∠AOC = 30ο^ (i.e. OB bisects 60ο^ angle between Vr and –Vy)
From ∆ AOC, cos 30ο^ =
V r
V
OA
OC = RY
V r
V RY
V
V^3
ph^ =^ L
VL =^3 Vph (1)
2. Current Relation: Since each supply line is connected to one phase of the load, obviously IL= Iph (2)
Thus, line voltage is 3 times the phase voltage and line current and phase currents are same.
3. Power Relation: If load is balanced then Total power consumed by Three Phase ckt., P3-ph = 3 × (Power consumed by single phase) = 3 × (Vph Iph cosφ)
Example: 02 Three similar coils takes a power of 3 kW at 0.6 power factor when connected in star to a 3 phase, 415 volt, 50 Hz supply. Calculate: (1) Line currents and (2) Resistance and inductance of each coil.
Solution: Given data, Total power in 3-Ph, P= 3 kW, Given power factor, cosφ= 0.6 Line voltage, VL = 415V and f= 50Hz
Total active power consumed, P = 3 VL IL cosφ
= 3 × 415 × IL ×0.
∴ IL =
3 V cos
P
L × ×
= 6.956 Amp
In star connection, IL= Iph ∴∴∴∴ Iph= 6.956 Amp
For star connection, we know that VL = 3 Vph
∴ Vph = 3
VL = =239.6V
Phase current, Iph = ph
ph Z
V (^) ∴∴∴∴ Zph =
- 956
I
V
ph
ph (^) = = 34.44 Ω
We know that Power factor, cosφ = ph
ph Z
R (^) (here pf given)
∴ Resistance per phase of ckt., Rph = Zph cosφ= 34.44 ×0.6= 20.67 Ω
Impedance per phase, Zph = R ph^2 + X Lph^2
∴ XLph = Z^ ph^2^ −^ Rph^2 =^34.^442 −^20.^672 = 27.55 Ω
XL = 2πf L ∴ L = 2 π f = 10027.^55 π
X Lph
= 0.0877 H or 87.7 mH
Example: 03 A balanced star connected load is supplied by 3-phase, 415 V, 50 Hz supply. Current in phase is 20Amp and lags 30 ο^ behind its phase voltage. Find: (a) Power consumed by load (b) Circuit parameters (c) Load p.f.
Rph=? VL= 415V Zph=? Xph L=?=?
VL= 415V
Solution: Given data, Phase current, Iph= 20A , power factor angle, φ= 30° (Iph lags) Line voltage, VL = 415V and f= 50Hz In star connection, IL= Iph ∴∴∴∴ IL= 20 Amp
For star connection, we know that VL = 3 Vph
∴ Vph = 3
VL = =239.6V
Impedance per phase, Zph= VI^23920.^6 ph
ph (^) = = 11.98 Ω
- Load power factor, cosφ= cos( 30 °)= 0.866 lag
2) Total active power consumed, P = 3 VL IL cosφ
= 3 × 415 × 20 ×0.866 =12449.63 W ≈ 12500 Watts
- We know that power factor, cosφ = ph
ph Z
R =0.
∴ Resistance per phase of ckt., Rph = Zph cosφ= 11.98×0.866= 10.375 Ω
Impedance per phase, Zph = Rph^^2 + X Lph^2
∴ XLph = Z ph^2^ − Rph^2 = 11. 982 − 10. 3752 = 5.98997 Ω= 6 Ω
XL = 2πf L ∴ L = X 2 π^ Lph f^ = 1006 π= 0.0191 H or 19.1 mH
2.4 Delta Connected balanced systems
Derive the relations between line values and phase values of voltage and current for a 3- phase, balanced, Delta-connected, lagging power factor load. Sketch the phasor diagram of these quantities. Derive also power consumed by load.
Iph=20A IL= Iph Rph=? VL= 415V Zph=? Xph=? L=? IL= 20A VL= 415V
IL= 20A
The perpendicular is drawn from point A on phasor OB representing IL. OC = CB = I 2^ L Angle between Ir and –Ib is 60ο. So ∠AOC = 30ο^ (i.e. OB bisects 60ο^ angle between Ir and –Ib)
From ∆ AOC, cos 30ο^ =
I ph
I
OA
OC = L
I ph
I
3 = L
I
I^3
ph^ =^ L
IL = 3 Iph ------------ (2)
3. Power Relation: If load is balanced then Total power consumed by Three Phase ckt., P = 3 × (Power consumed by single phase) = 3×(Vph Iph cosφ)
= 3×( VLIL 3 cosφ) {... VL=Vph & IL= 3 Iph}
P = 3 VL IL cosφ Watts
Similarly total reactive power , Q = 3 VL IL sinφ VAr
Similarly total apparent power , S = P^2 + Q^2 = 3 VL IL VA
Example: 04 [here load of Ex.1 taken in delta] Three identical coils each having resistance of 15 ΩΩΩΩ and inductance of 0.03H are connected in Delta across a three phase 400V, 50Hz supply. Calculate: (a) Line and Phase Voltage (b) Line and Phase Current
IB
Vb =VBR
Ib
- Iy
- φο Ir O Vr φο^ Vr (Ref)= VRY Ir Ir 30 ο^ A IY φο^60 ο Iy - Ib C IR - Ib Vy = VYB B IR
Phasor Diagram for DELTA connection To find Current relation
(c) Power Factor and Power Factor Angle (d) Active and Reactive Power Consumed Solution: Given data, Resistance per phase, Rph= 15Ω , Inductance per phase, L= 0.03 H (a) Given Line voltage, VL = 400V and f= 50Hz
∴ Inductive reactance per phase, Xph = 2πfL= =2π× 50 ×0.03= 9.425Ω ∴∴∴∴ Impedance per phase, Zph = R^2 + X L^2 ==== 15 2 + 9. 4252 = 17.72Ω For Delta connection, we know that VL = Vph ∴ Vph = 400 V
(b) Phase current, Iph = (^) ZV 17400. 72 ph
ph (^) = = 22.57 A
Also in Delta connection, IL= 3 Iph = 3 ×22.57 = 39.1 Α= 39.1 Α= 39.1 Α= 39.1 Α ∴∴∴∴ IL=39.1 A
(c) Power factor of each phase of load = (^) ZR 1715. 72 ph
ph (^) = = 0.846 lag {since load is R-L}
Power factor angle, φ= cos-1(0.866) = 32.22 οοοο^ {Current Iph lags Vph by 32.22ο}
(d) Active Power consumed, P = 3 VL IL cosφ
= 3 × 400 ×39.1×0.846 = 22,917.53 Watts ( appox. 3 times as in Ex.1 )
We have, P∆ = 3 P hence proved.
and Reactive power consumed, Q = 3 VL IL sinφ
= 3 × 400 ×39.1×sin(32.22ο) = 14,443.23 VAr
Example: 05 (here Impedance in Rectangular form) A three phase 400 V, 50 Hz supply is connected to a delta connected load which has impedance of (15 + j12) Ω in each phase. Find : (1) line current (2) phase current (3) active power (4) apparent power Solution: Given data, Resistance per phase, Rph= 15 Ω , Inductive reactance per phase, Xph = 12 Ω Given Line voltage, VL = 400V and f= 50Hz
VL= 400V Xph = 9.425Ω Rph Zph =17. Rph= 15Ω Xph VL= 400V
VL= 400V Xph = 12Ω Rph Zph= 19.21∠38.65ο Rph= 15Ω Xph VL= 400V
Iph2 = ph
ph Z
V =
ph
L Z
V
IL2= 3 Iph2= 3
ph
L Z
V
Active Power Consumed in Delta, P∆ = 3 VL1 IL2 cosφ
P∆ = 3 VL1 3
ph
L Z
V (^) cosφ
P∆ = 3
ph
L1^2 Z
V (^) cosφ -----------------------------(2)
Put ph
L1^2 Z
V (^) cosφ = P ,^ Active Power Consumed in Star connection in Eq” (2) We get, P∆ = 3 P hence proved.
Example: 07 (here Impedance in Polar form) The three equal impedances of each of 10 ∠∠∠∠ 60 ο^ Ohms are connected in star across 3- phase, 400 volts 50Hz supply. Calculate : (1) Line Voltage and Phase Voltage (2) Power Factor and Active Power consumed (3) If the same three impedances are connected in delta to the same source of supply what is the active power consumed? Solution: Given data, Zph = 10 ∠∠∠∠ + 60 ο^ =Zph∠φ = 5 + j8.66 = R+ j XLph {if Z in polar form, then V/Z becomes easy} Given Line voltage, VL = 400V and f= 50Hz
For star connection, we know that VL = 3 Vph
∴ Vph = 3
VL = =230.94 V
If voltage per phase is taken as reference, then V ph= Vph∠ 0 ο^ = 230.94 ∠∠∠∠ 0 οοοο
(2) Phase current, I ph= °
° ∠
Z
V
ph
ph (^) = 23.094 ∠∠∠∠ -60 οοοο (^) A {i.e. if (^) V ph is taken as ref. then Iph lag}
In star connection, IL= Iph ∴∴∴∴ IL=23.094∠-60ο^ Amp Power factor of each phase of load= cosφ = cos( 60 ο)= 0.5 lag {since load is R-L}
Active Power Consumed, P = 3 VL IL cosφ
= 3 × 400 ×23.094×0.5 = 7999.996 ≅≅≅≅ 8000 Watts or 8 kW
(3) Power drawn by load in Delta, P∆= Three times power drawn by same load in Star
P∆ = 3 P= 3× 8kW = 24 kW Other method Same can verified as, if load impedance Zph = 10 ∠∠∠∠ + 60 ο^ is connected in delta then,
VL = Vph = 400V, if Vph is taken as ref. then V ph= Vph∠ 0 ο= 400 ∠∠∠∠ 0 οοοο
Phase current, I ph= °
° ∠
Z
V
ph
ph (^) = 40 ∠∠∠∠ -60 οοοο (^) A {here also Iph lag}
In delta connection, IL=^3 Iph ∴∴∴∴ IL= 3 ×40 = 69.28 Amp
Power factor of each phase of load= cosφ = cos( 60 ο)= 0.5 lag {since load is R-L}
Active Power Consumed, P = 3 VL IL cosφ
= 3 × 400 ×69.28×0.5 = 24000 Watts or 24 kW
Example: 08 Three 100 ΩΩΩΩ non inductive resistance are connected in (1) Star (2) Delta across a 400 V , 50 Hz, 3 phase supply. Calculate the power taken from supply system in each case. In the event of one of three resistances getting open- circuited, what would be the value of total power taken from the mains in each of the two cases? Solution: Given data, Zph = 100 Ω Given Line voltage, VL = 400V and f= 50Hz
For star connection, we know that VL = 3 Vph
∴ Vph = 3
VL = =230.94 V
Phase current, I ph= VZ 230.94 100
ph
ph (^) = = 2.3094 A
Here Vph & Iph are in phase, since impedance per phase is non-inductive resistor. In star connection, IL= Iph ∴∴∴∴ IL=2.3094 Amp Power factor of each phase of load= cosφ = cos( 0 ο)= 1 {since load is pure R}
Active Power Consumed, P = 3 VL IL cosφ
= 3 × 400 ×2.3094×1= 1599.999 ≅≅≅≅ 1600 Watts or 1.6 kW
(2) Power drawn by load in Delta, P∆= Three times power drawn by same load in star P∆ = 3 P= 3× 1.6kW = 4.8 kW Other method Same can verified as, if Zph = 100 Ω is connected in delta then, VL = Vph = 400V,
IL
Rph VL
∴ Impedance per phase, Zph = VI 12440. 7 ph
ph (^) = = 34.64 Ω
We know that power factor, cosφ = ph
ph Z
R
∴ Resistance per phase of ckt., Rph = Zph cosφ= 34.64×0.537= 18.6 Ω
Impedance per phase, Zph = Rph^2 + X Cph^2
∴ XCph = Z^ ph^2 −^ Rph^2 =^34.^642 −^18.^62 = 29.22 Ω
XC = 2 π^1 fC = 1001 π C ∴ C = 100 π ×^129. 22 = 1.089 ×××× 10 -4^ F
2.5 Three-phase, 4 Wire System
This system employs four conductors and is widely used in all areas where it is considered economical to supply large amounts of energy for industrial and domestic purposes.
The system is shown in Figure below – R, Y and B are the active conductors and N is the neutral which is connected to the “star point” of the transformer. It is usual for the “star point” to be earthed as shown.
Ref:-https://electrical-engineering-portal.com/current-systems-voltage-levels
Fig.- Three-phase system with earthed neutral
Three-phase system with earthed neutral
A four - wire system allows you to have phase voltages or line to ground voltages. ... Three phase four wire system can be used as single phase by utilizing one phase and neutral. But in three phase three wire there is no neutral.
Generally on Power distribution side three phase four wire system is more preferable, yet if the load is industrial loading(balanced mainly by induction motor/furnace etc.) three phase three wire is used. In three phase four wire system, with domestic unbalanced load, as unbalanced current get path through neutral, the phase/line voltage applied to appliances remains as per requirement, and hence the better performance is obtained. As neutral does not shift and remains at ground potential better voltage regulation is obtained even in fault condition, on any phase, and complete black out can be avoided.
This system can feed both balanced three phase loads as well as unbalanced three loads and single phase loads. The neutral wire provides a path for out-of-balance current. Therefore, even though currents are unbalanced, phase voltages are balanced. Three phase four wire system is the common distribution system for domestic supplies because it gives a choice of single phase voltage as well as three phase voltages; and allows unbalances too.
2.6 Three Difference between neutral and ground conductors
Interestingly, neutral and earth are both grounded wires with one subtle difference,
that is neutral is grounded where the transformer is located and earth is a local
ground beside a house with an access to electricity.
Live wire is the one which emits electrons from it, capable of flowing electron to conducting material. Neutral wire is the one where no electricity is produced this is wire from where voltage is measured with live wire. In power transformer there are three windings for three phase all three windings one end are tied together and this tied together point is called neutral or central point of three windings. Our earth is a good conductor of electricity. Many time while live wire connected to instrument may get touched to instrument body causing electrical shock to person to prevent this shock earthing is connected to the body of instrument and other end of earthing wire is inserted in the earth 10 feet deep with copper plate and coals to get perfect earth. So if
Figure shows the line and phase voltages in a three-
phase system. The neutral point is usually earthed at the
supply end (for protection and safety reasons) and each
live conductor is then at a definite potential to earth.
The standard voltage between actives is 415V, while the
voltage between any one of the actives, (A, B and C
respectively) and the neutral is 240V.
