3.-Electrotecnia Unit3
Descripción
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS CONTENTS:
Alternating current three-phase circuits
Three-phase systems characteristics Generation of three-phase voltages Three-phase loads -Y and Y- transformation Instantaneous power Three-phase power: S, P and Q Power measurement. Aaron connection Power factor improvement Electrical measurements Exercises 1
2
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
THREE-PHASE SYSTEM CHARACTERISTICS
THREE-PHASE SYSTEM CHARACTERISTICS The electricity grid is made up of four main components: GENERATION: production of electricity from energy sources such as coal, natural gas, hydropower, wind and solar. TRANSMISSION: the transmission system carries the electric power from power plants over long distances to a distribution system. DISTRIBUTION: the distribution system brings the power to the customers. COSTUMERS: these are the consumers of electric power (industry, service sector and residential uses).
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4
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
THREE-PHASE SYSTEM CHARACTERISTICS
GENERATION OF THREE-PHASE VOLTAGES
Instantaneous electric power has a sinusoidal shape with double frequency than voltage or current. SINGLE-PHASE AC CIRCUITS: instantaneous electric power is negative twice a period (power flows from the load to the generator) and positive twice a period, falling to zero. BALANCED THREE-PHASE AC CIRCUITS: instantaneous electric power is constant. Three-phase power never falls to zero. Three-phase electric motors perform better than single-phase AC motors. Three-phase power systems allow two voltage levels (L-L, L-N). When electric power is transmitted, three-phase AC systems require 25% less Cu/Al than single-phase AC systems. 5
Three-phase generators contain three sinusoidal voltage sources with voltages of the same frequency but a 120º-phase shift with respect to each other. This is achieved by positioning three coils separated by 120º angles. There is only one rotor. Amplitudes of the three phases are also equal. The generator is then balanced.
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
INTRODUCTION
INTRODUCTION
6
R
R N
VRS
120º
S T
VRN
VTR VTN T
N
VSN
VST
S
• N: neutral point • R S T (or A B C) direct sequence or sequence RST • VRS, VST, VTR: line voltages or line-to-line voltages • VRN, VSN, VTN: line-to-neutral voltages • Vline = 3 Vline-to-neutral 7
vRN(t) = V0·cos(·t + 90º) V vSN(t) = V0·cos(·t - 30º) V vTN(t) = V0·cos(·t +210º) V 8
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
INTRODUCTION
THREE-PHASE LOAD CLASSIFICATION
R
WYE (two voltages)
IR
R
R
Balanced VRN
VTR
VRS
120º
VTN T
N
Vline 3.Vline to N
VSN
VST
Z
4-wires
O=N Z
Unbalanced
IS
T
T
Z
Z
IT
4-wires DELTA (one voltage)
Usual system
Vphase = Vline-to-neutral
230 volt
Vline
400 volt
Frequency
Z O=N
Z S
IR
IR
50 Hz
IN N
S
3-wires
phase
S
3-wires
IRS
Balanced S
IRS
ITR
Z
IS
R
R
R
R
Z
S
IS
Unbalanced
50 Hz
T
T
IT
ZTR
I ST
I ST S
ITR
ZRS
Z
9
S
T
T
IT
ZST
10
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
BALANCED WYE-CONNECTED LOAD
BALANCED DELTA-CONNECTED LOAD
The wye or star connection is made by connecting one end of each of the three-phase loads together.
This connection received its name from the fact that a schematic diagram of it resembles the Greek letter delta ().
The voltage measured across a single load or phase is known as the phase voltage.
In the delta connection, the line voltage and phase voltage in the load are the same.
The voltage measured between the lines is known as the line-to-line voltage or the line voltage. In a wye-connected system, the line voltage is higher than the load phase voltage by a factor of the square root of 3.
The line current of a delta connection is higher than the phase current by a factor of the square root of 3.
In a wye-connected system, the phase current and line current are the same.
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12
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
-Y TRANSFORMATION
BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD
Z between nodes 1 and 2:
Z nodes 1-2: (1)
Z 1, 2
Z 12 ·(Z 13 Z 23 ) Z1 Z 2 Z 12 ( Z 13 Z 23 )
Z nodes 1-3: (2) Z 1 Z 3
IR
Z 12 ·(Z 13 Z 23 ) Z 12 ( Z 13 Z 23 )
Z 13 ·(Z 12 Z 23 ) Z 12 ( Z 13 Z 23 )
Y:
IN
Z
Z1
Z 13
Z
Stotal 3.Sfase 3.V RN .I *R 3.Vlínia .I línia
Z
S
Z12
IT
Z2
3
2
Z23
T
Z 12 ·Z 13 Z 12 Z 13 Z 23
Z2
(1) + (2) - (3)
Z 12 ·Z 23 Z 12 Z 13 Z 23
Z3
(1) - (2) + (3)
T
P
R
The three currents are balanced.
Balanced loads:
Z 13 ·Z 23 Z 12 Z 13 Z 23
O=N Z
Q
Thus the sum of them is always zero.
Z
S
-(1) + (2) + (3)
Since the neutral current of a balanced, Yconnected three-phase load is always zero, the neutral conductor may be removed with no change in the results.
13
ZY = Z/3
14 T
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD Example A three-phase, RST system (400 V, 50 Hz), has a three-wire Y-connected load for which Z = 1030º · Obtain the line currents and the complex power consumption· R
UNBALANCED FOUR-WIRE WYE-CONNECTED LOAD
90º
Z O=N Z
Z
S
210º
40 3
ZR O=N
S
180º
IT
A 90º
T
ZS
ZT
60º
16000 30º VA 13856.41 (watt) j8000 (VAr) 40 16000 VA 3 40 Ptotal 3·Vl ·I l ·cos 3·400· ·cos 30º 13856.41watt 3 40 Q total 3·Vl ·I l ·sin 3·400· ·sin 30º 8000VAr 3
IN
IS
V RN ZR V SN IS ZS V TN IT ZT I N ( I R I S I T ) 0 IR
N
Stotal 3·Sphase 3·V RN ·I *R 3·(400/ 3 )·(40/ 3 )*
T
IR R
60º
V RN 400/ 3 40 IR A 30º 10 Z 3 30º 60º 40 V SN 400/ 3 A IS 10 30º Z 3 V TN 400/ 3 IT 10 30º Z
º
3.Vlínia .I línia .cos j 3.Vlínia .I línia .sin
S
Z
Z1
V TN V SN V RN IT IS Z Z Z I N (I R IS I T ) 0
O=N IS
Z 23 ·(Z 12 Z 13 ) Z 12 ( Z 13 Z 23 )
From expressions (1), (2) and (3) the result is:
IR
N
1
R
Z3
Z nodes 2-3: (3) Z 2 Z 3
R
Z Y 1, 2 Z 1 Z 2
Stotal V RN ·I *R VSN ·I *S V TN ·I *T
Stotal 3·Vl ·I l 3·400·
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16
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED THREE-WIRE WYE-CONNECTED LOAD
UNBALANCED THREE-WIRE WYE-CONNECTED LOAD
I R IS I T 0
IR R
IR R
1) ZR
V RO V SO V TO 0 ZR ZS ZT
O=N IS
ZS
ZT
S
IT
O=N IS
ZS
ZT
S
V RN V ON V SN V ON V TN V ON 0 ZR ZS ZT
T
V ON
V ON
ZR
2) I R IT
V RO V RN VON ZR ZR
IS
V SO V SN VON ZS ZS
IT
V TO V TN VON ZT ZT
T
V RN V SN V TN ZR ZS ZT V RN ·Y R V SN ·Y S V TN ·Y T 1 1 1 Y R YS Y T ZR ZS ZT 17
V RN V SN V TN ZR ZS ZT 1 1 1 ZR ZS ZT
3) Stotal V RO ·I *R V SO ·I *S V TO ·I *T 18
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED THREE-WIRE WYE-CONNECTED LOAD
BALANCED DELTA-CONNECTED LOAD
Example· A three-phase, RST system (400 V, 50 Hz), has a three-wire unbalanced Y-connected load for which ZR = 100º , ZS = 100º and ZT = 1030º · Obtain the line currents and the total complex power consumption. IR
IR
R
R IRS
ZR
R IS
ZS
S IT
ZT
T
2)
V RN ·Y R V SN ·Y S V TN ·Y T Y R YS Y T ON 90º 0º -30º 230 ·0.1 230 ·0.10º 230 210º ·0.1-30º 40.93114.89º V 30 º 0º 0º 0.1 0.1 0.1 1)
V ON
VRO = VRN –VON = 23090º - 40.93114.89º = 193.6484.90º V
IS = VSO/ZS =
=
S
T
IT
Z
I R I RS I TR IS IST I RS I T I TR IST º
3·Vlínia ·Ilínia ·cos j 3·Vlínia ·Ilínia ·sin
IR = VRO/ZR = 193.6484.90º/100º = 19,3684.90º A 26,45-35.10º
Z
Stotal 3·Sfase 3·V RS ·I *RS 3·Vlínia ·Ilínia
VTO = VTN –VON = 230210º - 40.93114.89º = 237,18-140.10º V 264.54-35.10º/100º
Z
IS
I ST T
VSO = VSN –VON = 230-30º - 40.93114.89º = 264,54-35.10º V 3)
S
V RS Z V ST IST Z V TR I TR Z I RS
ITR
A
P
Q
IT = VTO/ZT = 237.18-140.10º/1030º = 23.72-170.10º A Stot = VRO·IR* + VSO·IS* + VTO·IT* = 15619.56 W + j2812.72 VAr
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Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
UNBALANCED DELTA-CONNECTED LOAD
UNBALANCED THREE-WIRE -CONNECTED LOAD
IR IRS
S
I RS
ITR
ZRS
IS
ZTR
IST I ST S
T
Example· A three-phase, RST system (400 V, 50 Hz), has an unbalanced connected load for which ZRS = 100º , ZST = 1030º i ZTR = 10-30º · Obtain R the line currents and the complex power consumption· 120º V RS 400 40120º A I RS V V V 0º 10 ZRS 120º
R
R
I TR
ZST
I R I RS I TR
V ST ZST
T
IT
V RS Z RS
IS IST I RS
IST
I T I TR IST
V TR ZTR
RN
TR
I TR
VTN
IR
V ST 4000º 30º 4030º A ZST 10
R
R
V TR 400120º 4090º A 30º 10 ZTR
S
I R I RS I TR 77.29105º A
Stotal V RS ·I *RS V ST ·I *ST V TR ·I *TR
S
ITR
ZRS
IS
N VST
T
IRS
RS
VSN
ZTR
I ST
IS IST I RS 77.29 45º A
S
T
T
IT
ZST
I T I TR IST 40.00 150º A Stotal V RS ·I *RS VST ·I *ST V TR ·I *TR 43712.81(W) j0 (VAr) 21
43712.81VA
22
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT. Four-wire load
POWER MEASURMENT. ARON CONNECTION General 3-wire load. Two-wattmeter method (ARON connection)
Balanced wye-connected, four-wire load
Demonstration done for a balanced 3-wire load
R
R
W
W = VRN·IR·cos( VRN - IR)
Z N
LOAD
Ptotal = 3W
O=N Z
Z
S
,
120º
VTN
T
W1 VRT ·IR ·cos( VRT I R ) V·I·cos(30º )
Unbalanced wye-connected, four-wire load R
Z
WS = VSN·IS·cos( VSN - IS)
O=N
WT = VTN·IT·cos( VTN - IT)
R
N
S
T
ZS
WS
VSN S
PTOTAL W1 W2 V·I·[cos( 30º ) cos(30 º )] 3·V·I·cos
ZT
Q TOTAL 3·[W1 W2 ] 3·V·I·[cos( 30º ) cos(30º )] 3·V·I·sin
Ptotal = WR + WS + WT WT
T
N V VST ST
W2 VST ·IS ·cos( VST IS ) V·I·cos(30º )
WR = VRN·IR·cos( VRN - IR)
WR
V VRS SR
VRN
V VTR RT
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Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT. BALANCED LOAD
POWER MEASURMENT: THE TWO-WATTMETER METHOD Aron cyclic permutations
Balanced load, general (Y/D, 3/4 wires). Two-wattmeter method (ARON connection)
W1 V RT SR TS
Ptotal W1 W2 BALANCED
Q total 3·(W1 W2 ) W W2 arctg( 3 1 ) W1 W2
LOAD ,
Unbalanced wye/delta-connected, three-wire load
V ST TR RS
I S T R
Q measurement: cyclic permutations W
UNBALANCED
LOAD
W2 I R S T
V ST TR RS
Ptotal = W1 + W2
,
25
I R S T
W VST ·I R ·cos(VST IR ) Vline ·I line ·cos(90º- ) Vline ·I line ·sin
Q TOT 3W
QTOTAL 3
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
INSTANTANEOUS THREE-PHASE POWER
INSTANTANEOUS POWER: SINGLE-PHASE LOAD
Single-phase load:
26
cosA·cosB = 0·5·[cos(A+B) + cos(A-B)]
p(t) = v(t)·i(t) = V0·cos(w·t + V)·I0·cos(wt + ) p(t) =1/2·V0·I0·cos(V - I) + 1/2·V0·I0·cos(2wt + V + I) watt Constant
v(t)
Oscillates at twice the mains frequency!
Three-phase wye balanced load: p(t) = vRN(t)·iR(t) + vSN(t)·iS(t) + vTN(t)·iT(t) = = 2Vp·cos(wt + V)· 2Ip·cos(wt + )
i(t)
+ 2Vp·cos(wt -120º+ V)· 2Ip·cos(wt -120º+ ) + 2Vp·cos(wt +120º+ V)· 2Ip·cos(wt +120º+ ) = Vp·Ip·cos(V - I) + Vp·Ip·cos(2wt + V + I) + Vp·Ip·cos(V - I) + Vp·Ip·cos(2wt -240º+ V + I)
p(t) Average power = P
+ Vp·Ip·cos(V - I) + Vp·Ip·cos(2wt +240º+ V + I) = 3/2·Vp·Ip·cos(V - I) = 3/2·Vp·Ip·cos = constant!
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Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
INSTANTANEOUS THREE-PHASE POWER
POWER LOSSES: THREE-PHASE/SINGLE PHASE Single-phase line
I
R1 R
v(t)
N
Three-phase line
2
2 P Plosses 2·R1 ·I 2·R1 · 2 load 2 V ·cos
LOAD
V
R1
Pload V cos
Pload 3·V cos
I
2
2
2 Pload P Plosses 3·R2 ·I 3·R2 · R 2 · 2 load 2 2 2 2 ( 3 ) .V ·cos V ·cos
V
i(t)
2R 1 R 2 2
Supposing same losses
l l 1 S3p S1p S1p S3p 2
Single-phase line: 2 conductors of length l and section S1p p(t)
Three-phase line: 3 conductors of length l and section S3p = 1/2S1p pTOTAL = pR(t) + pS(t) + pT(t)
As a result: 29
weight3p-cables = 3/4weight1p-cables
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
Example 1
Example 2
Three-phase balanced RST system for which A1 = 1.633 A, A2 = 5.773 A, W1 = 6928.2 W, W2 = 12000 W, U = 6000 V and Zline=4+j3 · a) Obtain the complex power in the loads, as well as the ammeter A and the voltmeter U1 readings. b) Obtain the value of C to improve the load’s PF to 1, assuming U = 6000 V.Q 3W 12000VAr R
ZL
U1
S
1
A
1
ZL U
Q1 12000VAr 3UI1 sin 1
ZL
1 45º
T
A1
R
W1
A2 W2
C
AS W2
Balanced load 2 inductive
2 53.12º
C
Q 2 3UI 2 sin 2 48000VAr
ITOTAL I1 I 2 1.63390 º 45 º 5.77390 º 53.12 º 5.77490 º 36.85 º A U1, phase I.ZL U phase 5.77490 º 36.85 º (4 j 3)
(6000) 2 Q/3 -36000/3VAr 1/(2 .50.C )
6000 3
and W2 ' W2
AT L
T
R
a) Pt otal W1 ' W2 ' 13855.6 W 2 b) ISR
3492.97 90 º V U1 3U1,phase 6050V 31
400 2 R
V RT SR TS
resulting in :
I R S T
V ST TR RS
I S T R
R
R 23.095 T
S
VSR 400 60 º V 400 60 º 17.32 60 º A I RT RT 17.32 60 º A R 23.095 R 23.095 It results : I R I RT - ISR 3090 º A
c) IS ISR ITS 17.32 60 º
90 º
C 1.06F
W1 ' W1
V
R
S1 2 (P1 P2 ) j(Q1 Q 2 ) 48000 j36000 6000036.87º VA
Q1C ,
T
W1
R
P2 3W2 36000 W 3UI 2 cos 2
S Balanced load 1 capacitive
AR
R S
P1 3UI1 cos 1 12000 W C
T
Three-phase 50 Hz system for which V = 400 V, W1 = -8569.24 W, W2 = -5286.36 W, AS = 21.56 A. Obtain a) the value of R. b) the reading of AR. c) the value of the inductance L W1 W2
S
K
30
VTS 400180º 400 17.32 60 º 8.66 j15 j 90º jX L XL XL
IS 21.56 A 8.66 2 (15
400 2 ) X L 84.308 2 50L XL
The result is : L 0.2684 H
32
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
Example 3
Example 4
Varley phase-sequence indicator. Calculate the voltage in each element and deduce the practical consequences· Three-phase 400 V/50 Hz system· C = 1F XC = 3183 R2bulbs = 2·V2/P = 2·2302/20 = 5290
V RN ·Y R V SN ·Y S V TN ·Y T Y R YS Y T 90º -90º 230 /3183 230-30º /5290 0º 230 210º /5290 0º 1 / 318390 º 1 / 52900 º 1 / 5290 0 º 171.55171.31º V R 90º VR0 VRO = VRN –VON = 230 - 171.55171.31º = 265.3450.28º V V ON
VSO = VSN –VON =
230-30º
-
171.55171.31º
=
394.78-20.91º
V
0
V0N
VT0 T
VTO = VTN –VON = 230210º - 171.55171.31º = 144.00-101.86ª V
A 400 V and 50 Hz three-phase line feeds two balanced loads through a line which has an internal impedance of ZL=0.5 + j·1 · The -connected load has phase impedances whose values are 45+j·30 Ω, whereas the Y-connected load has phase impedances of 15–j·30 Ω· Determine: a) the reading of the ammeter A, b) the reading of the voltmeter V and c) the readings of watt-meters W1 and W2. R
ZL
S
ZL
T
ZL
A
W2 W1
T
90 º
a) I R VS0 S
VRN 400 / 3 13.3082.875 º A ZTOT 17.3657.125 º
c) PLOAD W1 W2 3·13.302 ·16.731 Q LOAD 3·(W1 W2 ) 3·13.30 ·1.154 2
Unit 3. AC THREE-PHASE CIRCUITS
Example 5
Question 1
Three-phase 400 V-50 Hz line. When switch K2 is closed , WA = 4000 W. When K1 and K3 are closed, WA = 28352.6 W and WB = -11647.4 W. Determine: a) R2, b) R1 and c) AT. S
4000 = 400·IS·cos(0º+30º) IS = 11.55 A
WB WA
T N
R2 = VSN/IS = (400/3)/11.55 = 20
AS
AT
K1
R
K2 C1
T
b) K1 and K3 closed:
S
W1
b) V 3·(I·Z// ) 3·(13.30·16.7706) 386.33V
Unit 3. AC THREE-PHASE CIRCUITS
R
S
ZTOT ZL (Z Y / / ZY ) Z L Z// (0.5 j·1) (16.731 j·1.154) 17.365 7.125º
Conductor R is situated where the capacitor is placed, conductor S is situated where the brighter bulb is placed and T is the remaining 33 conductor.
a) K2 closed: WA = VST·IS·cos(VST - IS)
R
V
V RT SR TS
W2 I R S T
V ST TR RS
I S T R
The Aron connection results in: W1 = 4616,1 W, W2 = 4262,5 W
34
An electrical lineman is connecting three single-phase transformers in a Y(primary)Y(secondary) configuration, for a business’s power service. Draw the connecting wires necessary between the transformer windings, and those required between the transformer terminals and the lines. Note: fuses have been omitted from this illustration for simplicity.
K3
R1 R1
R2
PTOT = W1 + W2 = -WB + WA = 40000 W = 2·4002/R1 + 4002/R2 R1 = 10 W1
c) K1 and K3 closed: 180 º
IT1 ITS I RT
VTS VRT 400 R1 R1 10
60 º
400 10
V RT SR TS
60 j·34.64A
W2 I R S T
V ST TR RS
I S T R
180 º
IT2
ITtotal
VTS 400 20 j·0A R2 20 IT1 IT2 (60 j·34.64) (-20 j·0) -80 j·34.64 87.18-156.6º A
This results in AT = 87.18 A
35
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Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
Question 2
Question 3
Identify the primary-secondary connection configuration of these pole-mounted power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).
HV R
S
Identify the primary-secondary connection configuration of these pole-mounted power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).
T
LV
These transformers are connected in a Yy configuration. 37
These transformers are connected in a Yd configuration.
Unit 3. AC THREE-PHASE CIRCUITS
Unit 3. AC THREE-PHASE CIRCUITS
Question 3
Question 4
Identify the primary-secondary connection configuration of these pole-mounted power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).
These transformers are connected in open-delta configuration.
38
One of the conductors connecting the secondary of a three-phase power distribution transformer to a large office building fails when open. Upon inspection, the source of the failure is obvious: the wire overheated at a point of contact with a terminal block, until it physically separated from the terminal. What is strange is that the overheated wire is the neutral conductor, not any one of the ”line” conductors. Based on this observation, what do you think caused the failure? After repairing the wire, what would you do to verify the cause of the failure?
Three single-phase transformers are not normally used because this is more expensive than using one three-phase transformer. However, there is an advantageous method called the open-Delta or Vconnection It functions as follows: a defective single-phase transformer in a Dd three-phase bank can be removed for repair. Partial service can be restored using the openDelta configuration until a replacement transformer is obtained. Three-phase is still obtained with two transformers, but at 57.7% of the original power. 39 This is a very practical transformer application for emergency conditions.
Here’s a hint (“pista”): if you were to repair the neutral wire and take current measurements with a digital instrument (using a clamp-on current probe, for safety), you would find that the predominant frequency of the current is 150 Hz, rather than 50 Hz. This scenario is all too common in modern power systems, as non-linear loads such as switching power supplies and electronic power controls have become more prevalent. Special instruments exist to measure harmonics in power systems, but a simple DMM (digital multimeter) may be used as well to make crude assessments. 40
Unit 3. AC THREE-PHASE CIRCUITS
POWER MEASURMENT. ARON CONNECTION General 3-wire load. Two-wattmeter method (ARON connection)
LOAD ,
p(t) = vRN(t)·iR(t) + vSN(t)·iS(t) + vTN(t)·iT(t) p(t) = vRN(t)·iR(t) + vSN(t)·iS(t) + vTN(t)·[-iR(t) - iS(t)] p(t) = iR(t)·[vRN(t) - vTN(t)] + iS(t)·[vSN(t) - vTN(t)] = vRT(t)·iR(t) + vST(t)·iS(t) Mean value Ptotal = W1 + W2 = VRT·IR·cos(VRT-IR) + VST·IS·cos(VST-IS)
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