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MINISTRY OF EDUCATION OF THE REPUBLIC OF MOLDOVA
Technical University of Moldova
Theoretical bases of the electrical engineering department
THEORETICAL BASES OF THE ELECTRICAL ENGINEERING
Computer-assisted laboratory guide
Chisinau 2015
The given manual is intended for the help of the students of High Educational Institutions for non electrical specialty, which studies the electrical engineering in English, to perform the laboratory works, including the application of the computer. It can also be used by the post-graduate students and beginning teachers to study the terminology in English both training the skills of their realizations and making up the reports . The manual also contains some items with information about using the MULTISIM 2001 program, which is necessary to carry out the laboratory works.
Author: Mihail Kiorsak, univ. prof., Dr. hab. Sc., Mariana Ababii, univ. lector The responsible redactor: Referent:
© , T.U.M., 2015
The order of the admission to execute the laboratory works
In order to be admitted to execute the next laboratory work, the students should beforehand present and sustain the report on the previous laboratory work, which was done. To be familiarized with its contents, to know basic formulas and theoretical material for execution of the given laboratory work. To have prepared the necessary tables for the initial data and for record the results of measurements.
The experimental part of laboratory work the students carry out independently under teacher's supervision with the respect of the safety rules.
The safety precautions for execution of the laboratory works
At assembling the given electrical circuit for the given laboratory work, first of all, the consecutive circuits of current should be assembled, which includes the current circuits of wattmeters (varmeters, voltampermeters), phasemeters and ampermeters , at the second, it should be assambled the parallel voltage circuits of the wattmeters (varmeters, voltampermeters), phasemeters and voltmeters.
After the assembling the electrical circuit, before its switching on to a source of the electric power, it should be checked up by the teacher and only after this, it can be switched to the power source.
In case of any malfunctions, the circuit should be immediately switched off from the power source, with the message to teacher about malfunction.
Processing of results
The received experimental data are writing in the given tables and are shown to the teacher for confirmation, before dismantle the circuit.
If the experimental data are unsatisfactory, it is necessary to repeat the experiments, untill getting the satisfactory data.
The obtained experimental data are partially processed in the laboratory and finally at home. The students prepare a report on the carried out laboratory work by respecting: the requirements to the laboratory report, technical design standards and the standards of presentation of electrical elements. The paper should contain: the theme and purpose of the work, electric diagrams, tables with experimental data, calculation results tables, calculating formulas used with a calculating variant for measurement (experiment), graphs and necessary vector diagrams on millimeter paper, respecting the chosen scale, the experimentally obtained graphs at the oscilloscope or computer, work conclusions. The report is presented and sustained by the student before making the next laboratory work.
THE USE OF COMPUTING TECHNOLOGY
Laboratory works involve the providing modern computing technology through mathematical modeling of physical processes to the computer, by using Multisim-2001 program, taking place in reality in all its diversity and complexity, which makes it possible to study the respective material more deeply.
MULTISIM 2001 PROGRAM OVERVIEW
Input file of the program represents the circuit scheme with elements parameters, their name, nodes and if it is necessary, explanatory text (Fig. 1).
Fig.1 The aspect of an initial scheme of Multisim 2001 program
The introduction process of the circuit elements in Multisim 2001 program is simple and fast due to the way of graphic representation of components. It is like the drawing of a principle electric scheme.
By pressing a button from a group selection buttons (Figure 2) it opens a submenu where you can find the necessary components: sources, passive components, diodes, analog circuits, indicators, etc., electro-mechanical elements bar.
Fig. 2. Bar of elements
Once all components are selected, they are iterconnected for making the needed circuit. For this, the cursor is positioned on the terminal of the element, it presses the left mouse button and then the cursor moves to the second terminal pressing the left button again to make the connection. If you need to connect two wires which intersect, instead of the intersection they put the connection node.
After having been drawn the scheme, the circuit is saved. To do this, it presses the button to activate the pictogram of a floppy disk or from the File menu, by choosing the Save command and showing the saved file name and the place where it will be saved.
- Pictogram of the save button
Multisim 2001 program provides to the user 11 virtual instruments of measurement
which are represented in (Fig. 3)
Fig. 3. Virtual instruments of measurement
Due to the outside vision of the measure devices "as real" is very simple to work with them.
Instruments menu contains: multimeter, operating generator, wattmeter, oscilloscope etc.
After the mounting of the scheme for simulating it switch on to the source, it is necessary to press the run button, or press F5.
Laboratory work nr. 1
THE STUDY OF ELECTRIC ENERGY TRANSPORTING THROUGH THE DIRECT CURRENT LINE
The purpose of the work. The study of DC line modes of operation and building the main characteristics of the circuit for the variation of load resistance from 0 to ∞.
General concepts
For DC electric energy transporting from the generator to the consumer they are used two-wire power lines. Since the power line has always a resistance R , there are voltage loss across it ΔU, and losses of power ΔP and energy ΔW inside it.
For the scheme shown in Figure 1.1 it can be written the following relations:
1. Voltage losses in the electric line:
ΔU = U1 - U2 = R I, (1.1)
Where: U1- input voltage at the beginning of the line, V;
U2- output voltage at the end of the line V.
- wires resistance, Ω
ρ – specific resistance of the material, 2Ohm mm
m
l – line length, m;
S – cross-section of the conductor, mm2.
2. From (1.1) we obtain the relation between voltages U1 and U2:
U 1= U2 + ∆U = U2 + R I (1.2)
3. The electrical power given by the electric energy source, W:
P1 = U1I = P2 + ∆P, (1.3)
Where: 22 sP = R I - power consumed by the load, W;
22ΔP = RI - power losses in the line, W.
4. The line ratio is calculated using the following formula:
2 1
1 1 1
P P - ΔP ΔPη= ×100%= ×100%= 1- ×100%
P P P ,
or 2 2
1 1
U ×I Uη= ×100%= ×100%
U ×I U (1.4)
In the power lines they are possible the following operating modes:
а) normal operating mode or working mode (0 < Rs < ∞)
According to Ohm's law, the current : 1
S
UI =
R+R , (1.5)
where:
Rs – resistance of the load , Ω;
R – electrical line resistance, Ω.
U1 and R – are constant quantities, so the current in the circuit depends on the value of the load resistance;
b) open mode of operation (Rs = ∞), the consumer (load) is switch off.
I=0; ∆U=0; ∆P=0; P1=P2=0; U 2= U1 ;
с) short circuit mode of operation (Rs = 0), the consumer (load) is shorted.
U2=0; 1UI =R
; 1ΔU =U =R I ;
21ΔP =I R =P ; 2
2 SP =R I =0 .
From the relation 22 1 1P =P - ΔP =U ×I - R I , we can see, that if the Rs do not
changes, power P2 – is a function of the square of current.
When the Rs = R, the load is into the accordance with a wire resistance, the current 1 sc
s
U II = =
2R 2 and the power 2sc2
2 s 2max
IP =R ×I = =P
4 has the max value.
Order to perform the work:
1) mount the electric circuit shown on (Fig. 1.1) or at the computer in Multisim 2001 program (Fig. 1.2)
Fig. 1.1. The mounted scheme
R1
50ohm
50%
100OhmKey = a
R2 30.000 V
+
-
15.000 V+ -
15.000 V+
-
0.300 A+-
J1
Key = SpaceV1
30V
Fig. 1.2. Arrangement of the scheme in Multisim 2001 program.
2) varying load resistance Rs from maximum value to zero, to study the line at different operating modes, from the open mode (Rs = ∞) to short circuit mode (Rs = 0), maintaing input voltage U1 = const., (recommended by the teacher).
3) according the experimental data to calculate:
P1 – power supplied by the circuit source; P2 – power received by the consumer (load);
Rs – load resistance; R – electrical line resistance; ∆U – voltage losses on the line;
∆P – power loss in the line; η – ratio of the line.
Experimental data and calculation results to introduce in Table 1.1
Table of measurements
Table 1.1
Nr. Measurements Calculations
I1,A U1,V U2,V ∆U,V P1,W P2,W ∆P,W η,%
1
2
....
....
....
....
10
4) To build dependency graphics ∆U=f(I); U2=f(I); P1=f(I); ∆P=f(I); P2=f(I); η =f(I).
5) To make work conclusions :
Questions:
1. What is the power loss in lines and how does this value depend on the geometric parameters of the power line?2. How do the voltage losses and power losses depend in transmission lines on the load?3. Under what conditions the power transmitted to the load is maximum?4. How does the ratio depend on the load resitance?
5. How can electrical line resistance be determined?
References: References: [5 p.41-43], [10 p.103-105], [9 task 1-20, p.26]
Laboratory work nr. 2
THE STUDY OF ELECTRIC CIRCUIT THROUGH THE METHOD OF VOLTAGE EQUIVALENT GENERATOR
The purpose of the work: Experimental verification of the method of voltage equivalent generator.
General concepts
The method of voltage generator is applied to determine the current or voltage only in one branch of electric circuit. The essence of the method is that, the total circuit (Fig. 2.1) is replaced by an active dipole (Fig. 2.2) of which terminals ab is connected the chosen branch with the resistance R3.
Fig. 2.1. Circuit scheme
Fig. 2.2. Equivalent scheme of the VEG
In relation to the chosen branch ab with R3 resistance, the active dipole can be replaced by a voltage equivalent generator (VEG) with two parameters: equivalent electromotive force (EMF) Ee and internal resistance Re.
VEG parameters can be determined experimentally and analytically. For experimental determination of VEG parameters are made the following modes:
а) open circuit mode (branch with R3 resistance is disconnected). With the voltmeter is measured the voltage between ab terminals. This voltage is equal to the EMF of the equivalent generator:
(2.1)b) short circuit mode (ab branch resistance is zero - R3 =0). In this case with the ampermeter it measures the short circuit current Iab.s.c in this branch.
Using the obtained results in this experience is calculated the internal Re resistance of the equivalent generator. According to the equivalent scheme (fig.2.2), the current in the branch with R3 resistance is determined according to Ohm's law:
. (2.2)
From formula (2.2) into the short circuit mode R3=0, we obtain:
, when: . (2.3)
The determination of equivalent generator parameters analytically.
If the electrical circuit (Fig. 2.1) parameters are known, the calculation of equivalent generator
parameters can be done in the following sequence:а) the resistance R3 is disconnected, the branch with the resistance R3 is switch off
(open circuit mode, R3 = ∞ (fig.2.3), it determines the potentials of the terminals (nodes) a and b.
According Ohm's Law it determines the currents I ' and I”:
; . (2.4)
Fig. 2.3 The scheme of the open circuit mode of operation
If we assume conventionally, that the potential of the node c is equal to zero (the node is connected to the earth φc = 0), we obtain:
; .
Therefore, EMF of the VEG is:
. (2.5)
b) The input resistance from the terminals ab of the VEG as passive dipole Rab in= Re (Fig. 2.4):
Fig. 2.4. Equivalent resistance scheme for calculation of internal resistance of the VEG.
Having calculated VEG parameters , the current through the resistence R 3 in the branch ab (Fig.2.2), is determined according to Ohm's law as:
(2.7)
The power transmitted from the VEG to the load (R 3 ) can be expressed as follows:
(2.8)
By changing the load resistance R3 we can obtain the mode which corresponds to maximum power transmitted to the load when R3 = Re.
From the mathematical point of view this confirmation, that the power P3 has
maximum value when R3 = Re , can be determined to take the derivate from the
expression (2.8)
Maximum power transmitted to the load will be equal to:
(2.9)
The order of work execution:
1) Mount the electrical circuit shown in the (Fig. 2.5) or at the computer using Multisim 2001 program (Fig. 2.6)
Fig. 2.5. The mounted scheme
2) Connect the circuit and set EMF values of the electrical source in the range indicated by the teacher : E1 = 16 ... 22V, E2 = 24 ... 30V, keeping them constant during the experiments.
V140V
V220V
R1
50ohm
R2
80ohm
Key = SpaceS2
Key = SpaceS3
R420ohm
R5100ohm
R3
20ohm
22.538 V+ -
0.020n A+ -
40.000 V+
- -20.000 V
+
-
Fig. 2.6 Arrangement of the scheme in Multisim 2001program
3) Set the switch K1 in average position (open circuit mode of operation). Insert in the Table 2.1 indication of the V3 voltmeter (EMF of the equivalent generator).4) Transfer switch K1 in position 1-1 (short circuit mode) and incert in Table 2.1 A4 ampermeter indication (Iab s.c. – short circuit current).5) Transfer switch K1 in position 2-2 (operating mode) and incert in Table 2.1 of the V3 voltmeter and of the A4 ampermeter indications (Uab and Iab – voltage and current through R3 resistance).
Determine the values of Re, R3 and insert the results in Table 2.1
Table of measurements
Table 2.1
Modes of operations
Measured values
Calculated values
Uab Iab R
V A Ω
Open circuit mode
Short circuit
mode
Operating mode
6) With the help of the obtained results verify the equivalent generator method according to the formula:
. .3
3 int . 3
abm ge
e r ab
UEI
R R R R
Compare the obtained value of the current I3 with that measured experimentally in
the operation mode.
7) Switch off and remove the circuit, install electrical circuit with an EMF source and by the "voltmeter - ampermeter" determine the values of all resistors that were used in the previous circuit.
Write the results in table 2.2.
8) Using EMF values E1 and E2 and values of the resistances R1 ... R5 from the table 2.2, calculate and Re=Rab intr.
Table for calculating resistances
Table 2.2
Resistances R1 R2 R3 R4 R5
U, V
I, A
, Ω
9) Calculate the power P3 which is given from the source to the load (given by VEG to the R3) when R3 is variable and find the condition of maximum power transfer, when R3=Re
Questions:
1. Describe the VEG method.
2. How can you determine internal resistance and EMF of the VEG?
3. In which conditions the power transmitted from VEG to the load has maximum value? Derive the formula for determining this condition.
... References: References: [5 p.38-42], [10p.96-103], [9 p.11-12, task 1-60, p.55]
Laboratory work nr. 3
Rheostat, Inductance Coil and Capacitor in Electric Circuits of Direct and Alternating Currents
The purpose of the work: To study the phenomena, which takes place in the electrical circuit, when the rheostat, inductance coil and capacitor are connected in turn at the direct and alternating current sources.
General concepts
When the rheostat (active resistance R), inductance coil (inductance L) and capacitor (capacitance C) are connected to the source of direct current by measuring the voltage applied across them with the voltmeter of direct current and the currents through them by ampermeter of direct current, we can observe that the current through the condenser C is equal to zero, because the direct current doesn’t flow through the condenser.
In accordance of the Ohm’s law, the resistances of the rheostat (active resistance R), and inductance coil (inductance L) are:
R=Ur/Ir – for the rheostat and RL=UL/IL =ρl/F for the inductance coil, because
the coil into the direct current circuits has only an active resistance. In those formulas Ur, UL and Ir, IL - the voltages across and the current through the rheostat and coil; ρ –the specific resistance of the material of the wires; l, F- the length and the cross section of the coil wire.
When the rheostat (active resistance R), inductance coil (inductance L) and capacitor (capacitance C) are connected to the source of alternating sinusoidal current in accordance of the Ohm’s law, the resistances R of the rheostat, impedances of the inductance coil ZL and capacitor ZC are:
R=Ur/Ir – for the rheostat; ZL=UL/IL for the inductance coil and ZC =Uc/Ic for the capacitor , because the coil and capacitor (the sinusoidal current flows through the capacitor) into the sinusoidal current circuits have also the reactive
resistance respectively XL = ωL=2πf, XC= 1/ ωC=1/2πfC and can be find as: 2 2
L L LX = Z - R , c c c2 2X = Z - R , where 2
L2
L LZ = R +X and 2 2C c cZ = R +X . Whence
LL = X ω , c
1C = ωX .
The resistance R of the rheostat can be also find as: R=Pr/Ir2 , where Pr – the
power disseminated into the resistance R, computed or measured by the wattmeter . Power factor: Pcosφ = UI and Parccosφ = UI .
The order of work execution:
1) Mount the scheme shown n fig. 3.1:
Fig.3.1. Experimental scheme
2) By connecting on turn to demountable jacks of the circuit rheostat R, rheostat R both inductance L and condenser C to the direct current and alternating sinusoidal current, write down in the table 3.1 for each case the indications of the ampermeter, wattmeter and voltmeter .
3) Calculate the required values indicated in the table , using the given up formulas .
Measured and calculated parameters Table 3.1
Elements of circuit
Type of current
Measured CalculatedU I P cos R Z X L CV A W gr. mH
nF
RheostatInductanc
e coil
Capacitorall in turn
4) Draw the vectors diagrams of the currents, voltages, triangles of resistance and power for each case.
Questions:
1. Why can’t the direct current flow through the condenser and the coil in the direct current circuits hasn’t the reactive resistance?
2. Explain the character and direction of the currents reference of applied voltages across the resister, inductance and capacitance ?
3. In what kind of energy is the electrical energy transformed into resistance ?
References: : [1 p.107-109], [6p.104-107], [4 p.11-12, task 1-60]
Laboratory work nr. 4
Voltage resonance into sinusoidal single phase electrical circuits
The purpose of the work: to study the phenomena of voltage resonance, which can occur in the sinusoidal single phase electrical circuits.
General conceptsThe voltage resonance phenomena can occur into the sinusoidal single phase
electrical circuits, when the elements: resistence R, inductance L and capacitance C are connected in series and inductive reactance XL is equal to capacitive reactance XC (XL= Xc).
In this case L CZ = R + j(X - X ) = R ;
R L C L CU = U + U + U = R I+ j(X - X ) I = R I and
L CL CU = jX I = -U = - jX I
and can be 6-7 times bigger that voltage .
RU = U from the
source The resonance frequency :
01 L
ω =L C
.
The order of work execution:
1) Mount the electrical circuit shown in the (Fig. 4.1) ), or at the computer using Multisim 2001 program (Fig. 4.2)
PW*
*
PV1PV3
PV
PA
127
V
RC3C3 RL1 L1
Fig.4.1. Experimental scheme
V2
113.14V 50Hz 0Deg
0.858 A+ -
80.002 V+
-
v I
XWM1
C1
30uFKey = a
41%
215.069 V+ -
217.281 V+ -
L2
800mH
R2
30ohm
R1
1kohm
Crez=12uF
Crez=41% din 30uF
Uc Ub
Itot
Uint
Ptot
E
condensator bobina
Fig. 4.2 Arrangement of the scheme in Multisim 2001program
1) 1) Obtain the voltage resonance condition XL =XC by maintaining constant U1 =30…35V, inductance L = Const (inductance switch in position 12-16) and changing the capacitance C (take 11-12 values) till voltages across inductance (UL) and capacitance (UC) are equal between them. Near the resonance point take five values changing capacitance with step 1 F, outside the resonance point change capacity over 4-5 F (4-5 F variance step).
Table of measurements and results Tabel 4.1
Nr.
Measured Calculated
Obs
erva
ţii
com
men
ts
I P Ub Uc Zb Rb Xb Xc L C cosφ φ
A W V V Ω Ω Ω Ω mH μF °
1
...
...
12
At the resonance the current (I) has maximum value (I=U1/R). Further rising of capacitance (C) is followed by decreasing of the current (I1) and appearance discrepancy between voltages (UL) and (UC).
3) During the experiment, record the devices indications shown in fig.4.1 and write there in table 4.1.
4) Calculate the required values indicated in the table 4.1, using the formulas:
cos =P/(U1 I); = arc cos (P/(U1 I)); R=P/I2 ; Z= U1/I; XC=UC/I; XL=UL/I; L=XL
Measured and Calculated parameters Table 4.1
Measured CalculatedI U1 UL UC P C cos Z R XC XL LA V V V W F grad mHn
1234567
5) Draw graphs of dependences: I=f(C); UL =f(C); UC= f(C); = f(C) and the vector diagrams for: XL=XC; XL>XC and XL<XC.
Questions:
1. The conditions of voltage resonance?
2. How we can obtain the voltage resonance?
3. Which are the main characteristics of voltage resonance?
4. Can the voltage resonance taken pleases in to the branched circuit? Which is the resonance condition?
References: [1 p.109-111], [5 p.38-42], [10p.96-103], [9 p.11-12, task 1-60, p.55]
Laboratory work nr. 5
Current resonance into sinusoidal single phase electrical circuits
The purpose of the work: to study the phenomena of current resonance, which can occur in the sinusoidal single phase electrical circuits.
General conceptsThe current resonance phenomena can occur into the sinusoidal single phase
electrical circuits, when the elements: resistence R, inductance L and capacitance C are connected in parallel and inductive susceptance bL is equal to capacitive susceptance bC
(bL= bc).In this case, the full conductance
L CY = g - j(b - b ) = g , when 1g = R - admitance of the circuit,
LL
1 1b = =X ωL and cC
1b = = ωCX - inductive and capacitive susceptances of the
circuit.
R L C RL CI = I + I + I = gU - j(b - b )U = I , where
; ;
R L CL L C CI = U / R = gU I = U / X = b U I = U / X = b U;
L CI = - I and can be 6-
7 times bigger that current
RI = I from the source. The resonance frequency can be
finding using the same formula:
01 L
ω =L C
.
The order of work execution:
1) Mount the electrical circuit shown in the (Fig. 5.1), or at the computer using Multisim 2001 program (Fig. 5.2)
Fig. 5.1. Experimental scheme
V2
141.42V 50Hz 0Deg
0.147 A+ -
99.999 V+
-
v I
XWM1
R11kohm
Uint
Ptot
Itot
~E Crez=41% de la 30uFCrez=12uF
L1800mH
R330ohm
Bobina
C2
30uFKey = a 41%
0.399 A+
- 0.395 A
+
-
Ic Ib
Condensator
Fig. 5.2 Arrangement of the scheme in Multisim 2001 program
2) Keeping with laboratory autotransformer voltage 1U = 40..50 V, measure and register indicated values in tabel.5.1:
a) for inductance (inductance switch in position 12-16) and capacity (take 11-12valori C). Near the resonance point change the C by step 1 F, besides the resonance point – change by step of 4-5 F.
b) for capacity and (change the inductance L from position 10 till position 20 by inductance switcher).
Tabel 5.1
Nr.
Measured Calculated
Obs
erva
ţii
com
men
tsU I Ib Ic P Yb gb bb bc L C
cosφ
φ
V A A A W Sm Sm Sm Sm mH μF °
1
3) Construct vector diagrams of the circuit, using points 2 for three values of capacity:
; ; .
4) Build on graph paper graphs of the columns 'measured' of tables 5.1 depending on the capacity .
Questions:
5. The conditions of current resonance?
6. How we can obtain the current resonance? The resonance frequency ?
7. Which are the main characteristics of current resonance?
8. In which circuits are used the resonance phenomena?
References: [1 p.123-127], [5 p.38-42], [10 p.96-103], [9 p.11-12, task 1-60, p.55], [8]
Laboratory work nr. 6
Three Phases Sinusoidal Alternating Current Circuits(Y and connections)
The purpose of the work: to study the quantities which characterize three-phases alternating current circuits in different modes of operation.
General concepts
Three phases of sinusoidal current circuits can be connected in <<Y>> -wye and << >>- delta.
<<Y -Y>> wye- wye connection with neutral wire (fig.6.1).
0
EA
EC
EB BC
AIA
IB
IC
Za
ZbZc
01
bc
a
Ua
Uab
ZA
ZC
ZN
ZB
IN
Fig.6.1 <<Y>> -wye connection of the three phases electrical circuits
The relations between linear UL and phases UPh voltages and currents IL and IPh , when three phases of electrical circuits are balanced ( a b cZ = Z = Z ) and connected in <<Y>> -wye connection, are :
L PhU = 3U and L PhI = I .
The powers : L LS = 3U I - apparent power; L LP = 3U I cos φ - active;
L LQ = 3U I sin φ - reactive power. There are the followers relation between S, P, Q :
L L L LS = P + jQ = 3U I cos φ+ j 3U I cos φ ; 2 2 QPS = P + Q = =cosφ sinφ .
The power factor : SPcosφ = , where Qφ = arctg P .
When three phase circuit is connected in <<Y-Y>> (bows, source and load are connected in “Y” connection) with the neutral wire we have basically two cases: - balanced, symmetrical three phase circuit, in the phases A, B, C the load is
symmetrical : a b cZ = Z = Z , the currents of the phases are symmetrically with equal
modules and can be find as:
AA
EI =
Z;
o- j120B AI = I e
C
oj120AI = I e ;
A B C L Ph= =I I I = I = I
Here A a AZ Z Z , where AZ and aZ are the impendances of the wire and of the load
of phase A.
The current through neutral wire
'N A B C N 0 0I = I + I + I = Y U 0 .
The phases and linear voltages at the load:
; ;
A B ca A A b B B c C cU = E - Z I U = E - Z I U = E - Z I ,
; ;
ab ab abU = Ua-Ub U = Ua-Ub U = Ua-Ub and modules are equal
between them and equal consequently to the phase Ph.LU and linear LLU voltages of the load:
a b Ph.LU = U = Uc = U ;
Lab ab ab LU = U = U = U .
Of cores LLU = 3 Ph.LU .
- unbalanced, unsymmetrical three phase circuit, in the phases A, B, C the load is unsymmetrical : a b c Z Z Z . The phases currents are un symmetrically:
A B CI I I and can be find as:
; ; ,
'' 'A CB 0000 00
A B C
A a B b C c
E -U E -UE -UI = I = I =
Z + Z Z + Z Z + Z
where '
A A B B C C
0 0
A B C N
E Y + E Y + E YU =
Y + Y + Y + Y; ; ;
A B AA a B b C c
1 1 1Y = Y = Y =
Z + Z Z + Z Z + Z;
NN
1Y =
Z - summary conductances of the phases A,B,C and neutral wire.
The current through neutral wire
'N A B C N 0 0I = I + I + I Y U 0 .
The phases and linear voltages at the load:
; ;
A B ca A A b B B c C cU = E - Z I U = E - Z I U = E - Z I ,
; ;
ab ab abU = Ua-Ub U = Ua-Ub U = Ua-Ub and there modules are not
equal between them .
When the neutral wire is cut, in both cases, symmetrical or unsymmetrical load
N A B CI = I + I + I = 0 , but '
0 0U =0 if the load is symmetrical and
'
A A B B C C
0 0
A B C N
E Y + E Y + E YU =
Y + Y + Y + Y if the load is unsymmetrical.
<<Y - >> wye- delta connection (fig.5.2).
Fig.6.2 <<Y - >> wye- delta connection of the three phases electrical circuits
The relations between linear UL and phases UPh voltages and currents IL and IPh at the load , when the load is balanced ( ab bc caZ = Z = Z ) and connected in << >> -delta connection, are :
L PhU = U and L PhI = 3I .
The powers S,P, Q can be find as in the case <<Y>> -wye connection. For calculation the currents we shod transform the load connected in << >> in to the equivalent <<Y>> -wye connection :
; ; ,ab ca ab bc ca bca b c
ab bc ca ab bc ca ab bc ca
Z Z Z Z Z ZZ = Z = Z =
Z + Z + Z Z + Z + Z Z + Z + Z
where aZ , bZ , cZ - the equivalent impendances of the phases for the load, connected in equivalent <<Y>> -wye connection. After those transformation , the currents and voltages can be find as in the precedent case when three phase circuit is connected in <<Y-Y>> .
The phases currents of the load can be find as:
; ; ,
ab bc caab bc ca
ab bc ca
U U UI = I = I =
Z Z Z where
abU ,
bcU ,
caU - the linear voltages of the load.
The order of work execution:
1. <<Y>> -wye connection
1. 1. Mount the electrical circuit shown in the (Fig. 6.3) ), or at the computer using Multisim 2001 program (Fig. 6.4)For this circuit we should studied the quantities which characterize three-phases
alternating current “wye” connected circuit (load) in different modes of operations:- balanced load mode of operation with neutral wire;
- unbalanced load mode of operation with neutral wire;-unbalanced load mode of operation without neutral wire.
Fig.6.3 <<Y>> -wye connected study circuit
Fig. 6.4 Arrangement of the scheme in Multisim 2001 program
1.2. Introduce the measured and calculated values of the lines and phases voltages and currents including neutral current for balanced (symmetrical) and unbalanced (unsymmetrical) load with and without neutral wire in the table 6.1.
Table 1Modes of operation
Variants UAB UBC UCA UAX UBY UCZ IA IB IC IN
V V V V V V A A A ASymmetricalload with
neutral wire
12
Unsymmetrical load with
neutral wire
12
Unsymmetrical load without neutral wire
12
1.3 For wall cases draw the vectors diagrams of voltages and currents.
2. << >> -delta connection
1. 1. Mount the electrical circuit shown in the (Fig. 6.5) ), or at the computer using Multisim 2001 program (Fig. 6.6)For this circuit we should studied the quantities which characterize three-phases
alternating current “ ” connected circuit (load) in two modes of operations:- balanced load mode of operation;
- unbalanced load mode of operation.
Fig.6.5 << >> - delta connected study circuit
Fig. 6.6 Arrangement of the scheme in Multisim 2001 program
2.1. Introduce the measured and calculated values of the lines and phases voltages and currents in the table 6.2.
Table 1Modes of operation
Variants UAB UBC UCA UAX UBY UCZ IA IB IC IN
V V V V V V A A A ASymmetrical
load 12
Unsymmetrical load
12
1.3 For those two cases draw the vectors diagrams of voltages and currents.
Questions:
1.What are the relationships between the lines and phases voltages and currents in a three phase balanced (simmetrical) and unbalanced (unsimmetrical) circuits ? 2.In which cases appeare the current in neutral wire and voltage betwin neutrals of the source and the load is bigeer? Haw realy to evoid in three phase power circuits the dangerous values of this voltage? 3. The particularities of the operation of the three phase circuit when the phase conductor cuted, phasor diagram in this case? 4. The particularities of the operation of the three phase circuit when the phase conductor is shorted to the eath, phasor diagram in this case?
References: [1p.137-149], [5 p.38-42], [10 p.256-277], [9 p.11-12, task 1-60, p.55], [7]
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8. Louis Denton. Theoretical and Practical Electrical Engineering: Comprising a Course of Lectures Given at the Bliss Electrical School Upon the Principles and . Both Direct and Alternating Current Apparatus: 2011.
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