AC Drives II Lab Repport 1

8/13/2019 AC Drives II Lab Repport 1

1/7


2/7

1 STATEMENT OF OBJECTIVE To observe real power flow when sender and receiver voltages are out of phase.

To observe reactive power flow when sender and receiver voltages are not equal in

magnitude.

2 INTRODUCTIONTransmission lines are designed and built to deliver electric power. Power flows from the

generator to the load but, the sender and the receiver ends may become reversed. Power in such a

line may flow in either direction depending upon the system load conditions which vary throughout

the day. Then, how can we attempt to understand and solve the flow of electric power under such

variable conditions?

We can obtain meaningful answers by turning to the voltage E1 and E2at each end of a line

having a reactance of Xohms. If we allow these voltages to have any values and any relative phase

angle, we can cover all possible loading conditions which may occur. The voltage drop along the line

is E1-E2; consequently, for a line having a reactance X, the current I can be found by the equation

=1 2

If we know the value of E1and E2, and the phase angle between them, it is a simple matter to

find the current I, knowing the reactanceXof the line. From this knowledge we can calculate the real

and reactive power which is delivered by the source and received by the load.

3 REAL POWER FLOW3.1 DESCRIPTION OF EXPERIMENTAL SETUPTwo ideal voltage sources designated as machine 1 and machine 2 were connected through an

impedance Z as shown in figure 3-1.

Fig 3-1

Where E1 = 120 V at -5

E2 = 100 V at 0

Z = 1+j7

I Z

E1 E2


3/7

3.2 PROCEDURETo carry out the experiment, we wrote a MATLAB program for the system of Fig 3-1 such that

the phase angle of voltage source 1 was changed from its initial value by 30 in step of 5. The

voltage magnitudes of the two sources and the phase angle of source 2 were kept constant.

At each value of phase angle , the program computed the complex power for each source

and the line loss. And plotted P1, P2 and P lossvs.

Here is the script executed in MATLAB to perform the simulation.

a=( - 35: 5: 25) ; %a i s t he phase angl e of E1E1=120*exp( 1i *a*pi / 180) ;

E2=100*exp( 1i *0*pi / 180) ; Z=1+1i *7; I =( E1- E2) . / Z; %assumed di r ect i on f r om Machi ne1 to Machi ne2S1=E1. *conj ( - I ) ; %si nce t he i nput cur r ent t o machi ne1 i s - I S2=E2. *conj ( I ) ; %si nce t he i nput curr ent t o machi ne2 i s +I Sl oss=Z. *( abs( I ) . 2) ; P1=r eal ( S1) ; P2=r eal ( S2) ; Q1=i mag( S1) ; Q2=i mag( S2) ; Pl oss=r eal ( Sl oss); Ql oss=i mag( Sl oss) ; subpl ot ( 2, 1, 1) ; pl ot ( a, P1, ' - or ' , a, P2, ' - xg' , a, Pl oss, ' - xb' )

xl abel ( ' E1 phase angl e ( i n Deg) ' ) yl abel ( ' Act i ve power ( i n W) ' ) hl eg1=l egend( ' Machi ne 1' , ' Machi ne2' , ' Li neLoss' , ' Locat i on' , ' Nor t hEast Out si de' ) ; gr i d onsubpl ot ( 2, 1, 2) ; pl ot ( a, Q1, ' - or ' , a, Q2, ' - xg' , a, Ql oss, ' - xb' ) xl abel ( ' E1 phase angl e ( i n Deg) ' ) yl abel ( ' React i ve power ( i n VAR) ' ) hl eg2=l egend( ' Machi ne 1' , ' Machi ne2' , ' Li neLoss' , ' Locat i on' , ' Nor t hEast Out si de' ) ; gr i d on


4/7

3.3 RESULTSAfter executing the script we got the following plots:

Fig 3-2

3.4 DISCUSSION OF RESULTSBy definition we know that the power absorbed by a circuit element is

= = + = |||| cos + |||| sin

Where is the phase angle by which Ilags V, and the current Iis entering the circuit element.

As we can see in the script, the complex power formula of machine 1 has a minus sign for I,because we assumed that the current actually leaves machine 1. From the active power plot, we see

that machine 1 has positive real power for less than 0, means that machine 1 absorbs real power.

While machine 2 has negative real power for less than 0, means that machine 2 delivers real

power. Thus, we can say that the real power flows from machine 2 to machine 1. Now, when is

greater than 0, machine 1 delivers real power and machine 2 absorbs real power. In this case, the

real power flows from machine 1 to machine 2. Notice that the power loss in the line is always

positive, which means that the line absorbs real power.

From the reactive power plot, we see that machine 1 has negative reactive power for all

values, means that machine 1 delivers reactive power. While machine 2 has positive reactive powerfor all values, means that machine 2 absorbs reactive power. Thus, we can say that the reactive


5/7

power flows from machine 1 to machine 2. We can see that the variation of has no effect on the

direction of flow of reactive power. Notice that the reactive power in the line is always positive,

which means that the line absorbs reactive power.

4 REACTIVE POWER FLOW4.1 DESCRIPTION OF EXPERIMENTAL SETUPWe will use the same system as for real power simulation.

4.2 PROCEDURETo carry out the experiment, we wrote another MATLAB program for the system of Fig 3-1

such that the voltage magnitude E1of voltage source 1 was changed from 70V to 130V in step of 5V.

The voltage magnitude of source 2 and the phase angles of the two sources were kept constant.

At each value of voltage magnitude E1, the program computed the complex power for eachsource and the line loss. And plotted Q1, Q2 and Qlossvs. E1

Here is the script executed in MATLAB to perform the simulation.

a=( 70: 5: 130) ; %a i s t he RMS of E1E1=a*exp( - 5*1i *pi / 180) ; E2=100*exp( 1i *0*pi / 180) ; Z=1+1i *7; I =( E1- E2) . / Z; %assumed di r ect i on f r om Machi ne1 to Machi ne2S1=E1. *conj ( - I ) ; %si nce t he i nput cur r ent t o machi ne1 i s - I S2=E2. *conj ( I ) ; %si nce t he i nput curr ent t o machi ne2 i s +I Sl oss=Z. *( abs( I ) . 2) ; P1=r eal ( S1) ; P2=r eal ( S2) ; Q1=i mag( S1) ; Q2=i mag( S2) ; Pl oss=r eal ( Sl oss); Ql oss=i mag( Sl oss) ; subpl ot ( 2, 1, 1) ; pl ot ( a, P1, ' - or ' , a, P2, ' - xg' , a, Pl oss, ' - xb' ) xl abel ( ' E1 magni t ude ( i n Vol t s) ' )

yl abel ( ' Act i ve power ( i n W) ' ) hl eg1=l egend( ' Machi ne 1' , ' Machi ne2' , ' Li neLoss' , ' Locat i on' , ' Nor t hEast Out si de' ) ; gr i d onsubpl ot ( 2, 1, 2) ; pl ot ( a, Q1, ' - or ' , a, Q2, ' - xg' , a, Ql oss, ' - xb' ) xl abel ( ' E1 magni t ude ( i n Vol t s) ' ) yl abel ( ' React i ve power ( i n VAR) ' ) hl eg2=l egend( ' Machi ne 1' , ' Machi ne2' , ' Li neLoss' , ' Locat i on' , ' Nor t hEast Out si de' ) ; gr i d on


6/7

4.3 RESULTSAfter executing the script we got the following plots:

Fig 4-1

4.4 DISCUSSION OF RESULTSFrom the real power plot, we see that machine 1 has positive real power for all E1values;

means that machine 1 absorbs real power. While machine 2 has negative real power for all E1values,

means that machine 2 delivers real power. Thus, we can say that the real power flows from machine

2to machine 1. We can see that the variation of E1 has no effect on the direction of flow of real

power. Notice that the power loss in the line is always positive, which means that the line absorbs

real power.

From the reactive power plot, we see that machine 1 has positive real power for E1less than

100V; means that machine 1 absorbs reactive power. While machine 2 has negative reactive power

for E1 less than 100V, means that machine 2 delivers reactive power. Thus, we can say that the

reactive power flows from machine 2 to machine 1. Now, when E1 is greater than 100V, machine 1

delivers reactive power and machine 2 absorbs reactive power. In this case, the reactive power flows

from machine 1 to machine 2. Notice that the reactive power in the line is always positive, which

means that the line absorbs reactive power.


7/7

5 CONCLUSIONSIn general, we can determine the P and Q absorbed or delivered by any ac circuit simply by

regarding the circuit as enclosed in a box with entering current from the positive polarity of the

voltage, and applying the definition of S. When positive values are found means absorbed power.

Real power can only flow over a line if the sender and receiver voltages are out of phase. The

direction of power flow is from the leading to the lagging voltage end. While reactive power flows

from the high-voltage to the low-voltage side.

Documents

AC Drives II Lab Repport 1