209--Power system lab Manual by Prof.Mrs. V. K. Thombare

Sheet Report 1

Object

Power System Components

Objectives

To study-

1.      Single line Diagram of power system

2.      D.C .Radial Distribution System

3.      A.C. Distribution Radial System

4.      Ring main system

5.      Interconnected System

Theory

The conveyance of electric power from a power station to consumes’ premises is known as Electric supply system. A electrical supply system consist of three principal components viz. the power station, the transmission lines, and the distribution system.

            Electrical power is produced at the power stations which are located at favorable places, generally quite away from the consumers. It is transmitted over large distance to load centers with the help of conductors known as transmission lines. Finally, it is distributed to a large number of small and big consumers through a distribution network.

            The electric supply system can be classified into i) d.c. or a.c. system (ii) overhead or underground system. Figure shows the layout of a typical a.c. power supply scheme by a single line diagram.

(i)                  Generating Station:

In the figure G.S. represents the generating station where electric power is produced by 3-phase alternators operating in parallel. The usual generating voltage is 11kv.

(ii)                Transmission System:

The transmission of electric power at high voltage has several advantages including the saving of conductor material and high transmission efficiency.

(a)    Primary Transmission:

The electric power at 132kv is transmitted by 3-phase, 3-wire overhead system to the outskirts of the city. This forms the primary transmission.

(b)   Secondary Transmission:

The primary transmission line terminates at the receiving station (RS) which usually lies at the outskirts of the city. At the receiving station, the voltage is reduced to 33kv by step down transformer.

(iii)                Distribution System:

(a)    Primary Distribution:

The secondary transmission line terminates at the sub-station (SS) where voltage is reduced from 33kv to11kv, 3-phase, 3-wire. The 11kv lines run along the important road sides of the city.

(b) Secondary Distribution:

The electric power from primary distribution line(11kv) is delivered to distribution substation(DS). These sub-stations are located near the consumers locality and step down voltage to 400V, 3-phase, 4-wire for secondary distribution.  

   

.

Conclusion

Questions

1.       What is electrical power system? Draw a single line diagram of a typical a.c power supply scheme

2.       What are the advantages and disadvantage of d.c transmission over a.c transmission?

3.       Explain the following systems of distribution:

(i)                  Radial Systems

(ii)                Ring main systems

(iii)               Interconnected systems

Sheet Report 2

Object

Insulator

Objectives

To understand Construction and ratings of Insulators.

Theory

1. Pin Type Insulator:

As the name suggest, the pin type insulator is secured to the cross-arm on the pole. There is groove on the upper end of the insulator for housing the conductor. The conductor passes through this groove and is bound by the annealed wire of the same material as the conductor.

Pin type insulator are used for transmission and distribution of electric power at voltage up to 33kv. 

2. Suspension Type Insulator:

The Suspension type of insulator consist of a number of porcelain discs connected in series by metal links in the form of strings. The conductor is suspended at the bottom end of the string while the other end of the string is secured to the cross arm of the tower. Each unit or disc is designed for low voltage, say 11kv. The number of the disc in series would obviously depend upon the working voltage.

3.      Stain Insulator:

  When there is dead end of the line or there is corner or sharp curve, the line is subjected to greater television. In order to relieve the line of excessive tension, strain insulators are used. The discs of strain insulator are used in the vertical plane. When the tension in lines is exceedingly high, as at long river spans, two or more strings are used in parallel.

4.      Shackle Insulator:

In early days, the shackle insulator were used as strain insulators. But now a days, they are frequently used for low voltage distribution lines. Such insulator can be used either in a horizontal position or in a vertical position. They can be directly fixed to the pole with a bolt or to the cross arm.

Conclusion

Questions

1.       Discuss the advantages & Disadvantages of

i)                    Pin Type Insulator

ii)                   Suspension Type Insulator

2.       Why are insulators used with overhead lines? Discuss the desirable properties of Insulators.

3.       Explain the terms:

i)                    Stain Insulator

ii)                   Shackle Insulator

Sheet Report 3

Object

Underground cables

Objectives

To understand the Construction and Classification of Cables,

Theory

Underground cables:

.

Underground Cables:

An underground cable essentially consists of one or more conductors covered with suitable insula- tion and surrounded by a protecting cover.

Constuction of Cable:

(i)     Cores or Conductors. A cable may have one or more than one core (conductor) depending upon the type of service for which it is intended. For instance, the 3-conductor cable shown in Fig. 11.1 is used for 3-phase service. The conductors are made of tinned copper or alu- minium and are usually stranded in order to provide flexibility to the cable.

(ii)     Insulatian. Each core or conductor is provided with a suitable thickness of insulation, the thickness of layer depending upon the voltage to be withstood by the cable. The commonly used materials for insulation are impregnated paper, varnished cambric or rubber mineral compound.

(iii)   Metallic sheath. In order to pro- tect the cable from moisture, gases or other damaging liquids (acids or alkalies) in the soil and atmosphere, a metallic sheath of lead or aluminium is provided over the insulation as shown in Figure.

(iv)      Bedding.   Over the metallic sheath is applied a layer of bedding which consists of a fibrous material like jute or hessian tape. The purpose of bedding is to protect the metallic sheath against corrosion and from mechanical injury due to armouring.

(v)     Armouring. Over the bedding, armouring is provided which consists of one or two layers of galvanised steel wire or steel tape. Its purpose is to protect the cable from mechanical injury while laying it and during the course of handling. Armouring may not be done in the case of some cables.

(vi)   Serving.  In order to protect armouring from atmospheric conditions, a layer of fibrous material (like jute) similar to bedding is provided over the armouring. This is known as  Serving

Belted Cables—Upto 11Kv

Screened Cables—Upto 22Kv to 66Kv

Pressure Cables—Beyond 66Kv.

Conclusion

Questions

1.       Compare the merits and demerits of underground systems versus overhead system.

2.      Explain the terms:

i)                    Belted Cables

ii)                   Screened Cables

iii)                 Pressure Cables

Experiment no. 1

Object

GMD & GMR, Inductance and Capacitance of 3 Phase Transposed Line

Objectives

To Find Out GMD & GMR, Inductance and Capacitance of 3 Phases Transposed Line

Theory

Consider the three-phase line shown in Figure, Each of the conductors has a radius of r and their centers form an equilateral triangle with a distance D between them. Assuming that the currents are balanced, we have

I_a+I_b+I_c=0

Consider a point P external to the conductors. The distance of the point from the phases a, b and c are denoted by D_pa, D_pband D_pcrespectively.

Fig: Three-phase symmetrically spaced conductors and an external point P

Let us assume that the flux linked by the conductor of phase-a due to a current I_a includes the internal flux linkages but excludes the flux linkages beyond the point P . Then from (1.18)we get

The flux linkage with the conductor of phase-a due to the current I_b , excluding all flux beyond the point P , is given by as

Similarly the flux due to the current I_c is

Therefore the total flux in the phase-a conductor is

The above expression can be expanded as

we get,

Substituting the above expression in eqn.

Now if we move the point P far away, then we can approximate D_pa  D_pb  D_pc.. Therefore their logarithmic ratios will vanish and we can write as

Hence the inductance of phase-a is given as

C_A=  F/m

.

Apparatus

MATLAB software, Power system analysis By Hadi sadad Tool Box

Procedure

1.Open MATLAB Window

2.Open matlab program from Hadi sadat ref example no.4.4

3.Execuite the program

4.Check the output

5.Verify the output with manual calculation.

Example

A 735KV three phase transposed line is composed of 4 ACSR 954,000 cmil 45/7 rail conductor per phase with horizontal conductor configuration as shown in fig. Bundle spacing is 46 cm. Use ACSR in MATLAB to obtain the conductor size and the electrical characteristics for rail conductor. Find out the inductance and capacitance per phase km of the line.

[Ref: Example 4.4, Power System Analysis, Hadi Sadat]

Manually calculation

MATLAB Input:

[GMD, GMRL, GMRC] = gmd;

L=0.2*log(GMD/GMRL)        

C = 0.0556/log(GMD/GMRC)   

Output:

Number of three-phase circuits            Enter     

 ------------------------------            -----     

     Single-circuit line                         1       

     Double-circuit vertical configuration       2       

     Double-circuit horizontal configuration     3       

     To quit                                     0 

           b                                                

              O                            For spacing unit use

             / \                           m within quotes or  

            /   \                          ft within quotes.   

           /     \                                             

         D12     D23                                           

         /         \             a           b           c     

        /           \            O----D12----O----D23----O     

      a/             \c                                        

      O------D13------O    OR     ----------D13----------      

Enter spacing unit within quotes 'm' or 'ft'-> 'ft'

Enter row vector [D12, D23, D13] = [45.5 45.5 89]

Cond. size, bundle spacing unit: Enter 'cm' or 'in'-> 'cm'

Conductor diameter in cm = 2.959

Geometric Mean Radius in cm = 1.173

No. of bundled cond. (enter 1 for single cond.) = 4

Bundle spacing in cm = 46

 

Result

GMD  = 56.90332 ft

 GMRL =  0.65767 ft  

 GMRC =  0.69696 ft

L = 0.8921

C = 0.0126

Conclusion

Questions

1.       What is Transposition

2.       Expression for the inductance per phase for a 3-phase symmetrical transposed overhead transmission line

3.       Expression for the inductance per phase for a 3-phase unsymmetrical transposed overhead transmission line.

4.       Derive an expression for flux linkage

i)                    Due to single current carrying conductor

ii)                   In parallel current carrying conductor

Experiment no. 2

Object

GMD & GMR, inductance and capacitance of double circuit 3 phase Transposed line with vertical conductor configuration.

Objectives

To find out GMD & GMR, inductance and capacitance of double circuit 3 phase Transposed line with vertical conductor configuration

Theory

Consider the three-phase transposed line shown in Figure In this the charges on conductors of phases a, b and c are q_a, q_band q_c respectively. Since the system is assumed to be balanced we have

Fig:Charge on a three-phase transposed line

Using superposition, the voltage V_ab for the first, second and third sections of the transposition are given respectively as

Then the average value of the voltage is

This implies

The GMD of the conductors is given in (1.42). We can therefore write

Similarly the voltage V_ac is given as

For a set of balanced three-phase voltages

Therefore we can write

For bundles conductor

Apparatus

MATLAB software, Power system analysis By Haadi sadad Tool Box

Procedure

1.Open MATLAB Window

2.Open matlab program from Haadi sadat ref example no.4.4

3.Execuite the program

4.Check the output

5.Verify the output with manual calculation.

Example:

A 345 kv double circuit 3 phase Transposed line is of two ACSR ,1,431,000-cmil,45/7 bobolink conductors per phase with vertical conductor configuration as shown in fig. The conductor have a diameter of 1.427 in and a GMR of 0.564 in. The bundle spacing is 18 in. find out the  inductance and capacitance per phase of the line.

[Ref: Example 4.5, Power System Analysis, Haadi Sadat]

Manually calculation

MATLAB Code                  

[GMD, GMRL, GMRC] = gmd;

L=0.2*log(GMD/GMRL)        

C = 0.0556/log(GMD/GMRC)

MATLAB Output                                               

Number of three-phase circuits            Enter     

     ------------------------------            -----     

     Single-circuit line                         1       

     Double-circuit vertical configuration       2       

     Double-circuit horizontal configuration     3       

     To quit                                     0          

                           a             c`(a`)                        

 Circuit Arrangements      O-----S11-----O         For spacing   unit use

 --------------------      |                       m within quotes or  

 (1)  abc-c`b`a`          H12                      ft within quotes.   

 (2)  abc-a`b`c`        b  |                b`(b`)                     

                        O--------S22--------O                          

                            |                                          

                           H23                                         

                           c|           a`(c`)                         

                            O----S33----O                              

 Enter (1 or 2)  -> 2

 

Enter spacing unit within quotes 'm' or 'ft'-> 'm'

Enter row vector [S11, S22, S33] = [11 16.5 12.5]

Enter row vector [H12, H23] = [7 6.5]

Cond. size, bundle spacing unit: Enter 'cm' or 'in'-> 'in'

Conductor diameter in inch = 1.427

Geometric Mean Radius in inch = 0.564

No. of bundled cond. (enter 1 for single cond.) = 2

Bundle spacing in inch = 18

 

Result

GMD  = 12.03227 m

GMRL =  1.03122 m  

GMRC =  1.09366 m

L = 0.4914

C =0.0232

Conclusion

Questions

1.       Different types of conductor

2.       What is GMD & GMR

3.       Define

i)                    Skin Effect

ii)                   Proximity Effect

4.       Expression for the inductance per phase for a 3-phase symmetrical transposed overhead transmission line.

5.       Expression for the inductance per phase for a 3-phase unsymmetrical transposed overhead transmission line

Experiment no. 3

Object

Sending end voltage ,voltage Regulation and power of  Short Line

Objectives

Student should able to calculate ABCD Parameters, Sending end current,Voltage,Active & Reactive

Power & voltage Regulation using Nominal π Method

Theory

 Short transmission lines. When the length of an overhead transmission line is upto about 50 km and the line voltage is comparatively low (< 20 kV), it is usually considered as a short transmission line. Due to smaller length and lower voltage, the capacitance effects are small and hence can be neglected. Therefore, while studying the performance of a short transmisison line, only resistance and inductance of the line are taken into account.

Consider a per phase short transmission line model shown in fig. where,

V_S &  I_S sending end voltage and current

V_R &  I_R receving end voltage and current

If a three-Phase load with apparent power S_R(3Φ) is connected at the end of the transmission line, the receving end current is obtained by,

The phase voltage at the sending end is

                                1

Since shunt capacitance is neglected in short transmission line,  

                                              2

The transmission line is represented by a two port network (ABCD parameter) given below

                       3

By comparing equation 1,2 and 3

         4

Apparatus

MATLAB software, Power system analysis By Haadi sadad Tool Box

Procedure

1.Open MATLA Window

2.Open matlab program from Haadi sadat ref example no.5.1

3.Execuite the program

4.Check the output

5.Verify the output with manual calculation.

Example

 [Ref: Example 5.2, Power System Analysis, Haadi Sadat]

Manually calculation

MATLAB Code

VRLL=220; VR = VRLL/sqrt(3);

Z = (0.15+j*2*pi*60*1.3263e-3)*40;

disp('(a)')

SR=304.8+j*228.6;

IR = conj(SR)/(3*conj(VR)); IS = IR;

VS = VR + Z*IR;

VSLL = sqrt(3)*abs(VS)

SS = 3*VS*conj(IS)

REG = (VSLL - VRLL)/VRLL*100

Eff = real(SR)/real(SS)*100

disp('(b)')

SR=304.8-j*228.6;

IR = conj(SR)/(3*conj(VR)); IS = IR;

VS = VR + Z*IR;

VSLL = sqrt(3)*abs(VS)

SS = 3*VS*conj(IS)

REG = (VSLL - VRLL)/VRLL*100

Eff = real(SR)/real(SS)*100

                                                        

Output

a)

VSLL =250.0186

SS =3.2280e+002 +2.8858e+002i

REG =13.6448

Eff =  94.4252

(b)

VSLL =210.2884

SS =3.2280e+002 -1.6862e+002i

REG = -4.4144

Eff = 94.4252                      

Conclusion

Questions

1. Analysis of short transmission Line and also evaluate constant for it.

Experiment no. 4

Object

Receiving end voltage, power, voltage regulation of medium line

Objectives

Student should able to calculate ABCD Parameters, Sending end current,Voltage,Active & Reactive

Power & voltage Regulation using Nominal T Method

Theory

Medium transmission lines are modeled with lumped shunt admittance. There are two different representations - nominal-p and nominal-T depending on the nature of the network.

Nominal-p Representation:

In this representation the lumped series impedance is placed in the middle while the shunt admittance is divided into two equal parts and placed at the two ends. The nominal-p representation is shown in Figure. This representation is used for load flow studies, as we shall see later. Also a long transmission line can be modeled as an equivalent p-network for load flow studies.

Figure: Nominal-p  representation

Let us define three currents I₁, I₂ and I₃ as indicated in Fig. 2.3. Applying KCL at nodes M and N we get

---------------------------(1)

Again

---------------------------(2)

Substituting (2) in (1) we get

------------------------------(3)

Therefore from (2) and (2) we get the following ABCD parameters of the nominal-p representation

Apparatus

MATLAB software, Power system analysis By Haadi sadad Tool Box

Procedure

1.Open MATLA Window

2.Open matlab program from Haadi sadat ref example no.5.2

3.Execuite the program

4.Check the output

5.Verify the output with manual calculation.

Example

A 345 kv three phase transmission line is 130 km long. The resistance per phase is 0.036Ω per km and inductance per phase is 0.8mH per km. The shunt capacitance is 0.0112μF per Km. The receiving end load is 270 MVA with 0.8 P.F. lagging at 325 kv. Use the medium line model to find out the voltage, voltage regulation  and power at sending end.

[Ref: Example 5.2, Power System Analysis, Haadi Sadat]

Manually calculation

MATLAB Code

r = .036; g = 0; f = 60;

L = 0.8;       % milli-Henry

C = 0.0112;    % micro-Farad

Length = 130;  VR3ph = 325;

VR = VR3ph/sqrt(3) + j*0;   % kV (receiving end phase voltage)

[Z, Y, ABCD] = rlc2abcd(r, L, C, g, f, Length);

AR = acos(0.8);

SR = 270*(cos(AR) + j*sin(AR));     %   MVA (receiving end power)

IR = conj(SR)/(3*conj(VR));         %  kA (receiving end current)

VsIs = ABCD* [VR; IR];              %      column vector [Vs; Is]

Vs = VsIs(1);

Vs3ph = sqrt(3)*abs(Vs);            % kV(sending end L-L voltage)

Is = VsIs(2); Ism = 1000*abs(Is);   %     A (sending end current)

pfs= cos(angle(Vs)- angle(Is));     %  (sending end power factor)

Ss = 3*Vs*conj(Is);                 %     MVA (sending end power)

REG = (Vs3ph/abs(ABCD(1,1)) - VR3ph)/VR3ph *100;

fprintf(' Is = %g A', Ism), fprintf('  pf = %g\n', pfs)

fprintf(' Vs = %g L-L kV\n', Vs3ph)

fprintf(' Ps = %g MW', real(Ss)),

fprintf('  Qs = %g Mvar\n', imag(Ss))

fprintf(' Percent voltage Reg. = %g\n', REG)

Output

Enter 1 for Medium line or 2 for long line --> 1

 Nominal pi model

 ----------------

 Z = 4.68 + j 39.2071 ohms

 Y = 0 + j 0.000548899 Siemens

0.98924     + j 0.0012844     4.68        + j 39.207      

 ABCD =                                                             

         -3.5251e-007 + j 0.00054595    0.98924     + j 0.0012844   

 Is = 421.132 A  pf = 0.869657

 Vs = 345.002 L-L kV

 Ps = 218.851 MW  Qs = 124.23 Mvar

 Percent voltage Reg. = 7.30913

Conclusion

Questions

1. Evaluate the generalized circuit constants for Medium π Network.

2. Voltage Regulator

3.Method of Voltage control

Experiment no. 5

Object

Receiving end voltage, power, voltage regulation of medium line

Objectives

Use the medium line model to find out the voltage, current, voltage regulation and power at receiving end..

Theory

In this representation the shunt admittance is placed in the middle and the series impedance is divided into two equal parts and these parts are placed on either side of the shunt admittance. The nominal-T representation is shown in Figure Let us denote the midpoint voltage as V_M. Then the application of KCL at the midpoint results in

Figure: Nominal-T representation

Rearranging the above equation can be written as

-----------------------------(1)

Now the receiving end current is given by

                   -----------------------------------(2)

Substituting the value of V_M from (1) in (2) and rearranging we get

Furthermore the sending end current is

-----------------------------------(3)

Then substituting the value of V_M from (1) in (3) and solving

Then the ABCD parameters of the T-network are

Apparatus

MATLAB software, Power system analysis By Haadi sadad Tool Box

Procedure

1.Open MATLA Window

2.Open matlab program from Haadi sadat ref example no.5.3

3.Execuite the program

4.Check the output

5.Verify the output with manual calculation.

Example

A 345 kv three phase transmission line is 130 km long.The series impedance z = .036 + j* 0.3Ω per phase per km, the shunt admittance is y=j4.22*10^-6. The sending end voltage is 345 kv and sending end current is 400 A at 0.95 p.f. lag. Use the medium line model to find out the voltage, current, voltage regulation  and power at receiving end.  

[Ref: Example 5.3, Power System Analysis, Haadi Sadat]

Manually calculation

MATLAB Code

z = .036 + j* 0.3;  y = j*4.22/1000000;  Length = 130;

Vs3ph = 345;  Ism = 0.4;  %KA;

As = -acos(0.95);

Vs = Vs3ph/sqrt(3) + j*0;       % kV (sending end phase voltage)

Is = Ism*(cos(As) + j*sin(As));

[Z,Y, ABCD] = zy2abcd(z, y, Length);

VrIr = inv(ABCD)* [Vs; Is];      %        column vector [Vr; Ir]

Vr = VrIr(1);

Vr3ph = sqrt(3)*abs(Vr);         % kV(receiving end L-L voltage)

Ir = VrIr(2); Irm = 1000*abs(Ir);%     A (receiving end current)

pfr= cos(angle(Vr)- angle(Ir));  %  (receiving end power factor)

Sr = 3*Vr*conj(Ir);              %     MVA (receiving end power)

REG = (Vs3ph/abs(ABCD(1,1)) - Vr3ph)/Vr3ph *100;

fprintf(' Ir = %g A', Irm), fprintf('  pf = %g\n', pfr)

fprintf(' Vr = %g L-L kV\n', Vr3ph)

fprintf(' Pr = %g MW', real(Sr))

fprintf('  Qr = %g Mvar\n', imag(Sr))

fprintf(' Percent voltage Reg. = %g\n', REG)

Output

Output:

Enter 1 for Medium line or 2 for long line --> 1

 Nominal pi model

 ----------------

 Z = 4.68 + j 39 ohms

 Y = 0 + j 0.0005486 Siemens

         0.9893      + j 0.0012837     4.68        + j 39      

 ABCD =                                                            

         -3.5213e-007 + j 0.00054567    0.9893      + j 0.0012837    

Result

 Ir = 441.832 A  pf = 0.887501

 Vr = 330.68 L-L kV

 Pr = 224.592 MW  Qr = 116.612 Mvar

 Percent voltage Reg. = 5.45863

Conclusion

Questions

1. Evaluate the generalized circuit constants for Medium T Network.

2. Analysis of short transmission Line and also evaluate constant for it.

Experiment no. 6

Object

Surge Impedance Loading

Objectives

Use the medium line model to find out the voltage, current, voltage regulation and power at sending  end..

Theory

Surge impedance load is  the load (of unity power factor) that can be delivered by the line of negligible resistance.

 When the line is loaded by being terminated with an impedance equal to its characteristic impedance, the receiving end current is

                                                             1

For a lossless line Z_C is purely resistive. The load corresponding to the surge impedance at rated voltage is kown as the surge impedance loading (SIL) given by

                                       2

Where, V_R is the receiving end voltage in kV and Z_C is the surge impedance in ohms, and SIL is the surge impedance loading or natural loading of the line

                                              3

The above  expression gives a limit of the maximum power that can be delivered by a line and is useful in designing the transmission line. This can be used for the comparison of loads that can be carried on the transmission lines at different voltages

From the above expression power transmitted through a long transmission lines can be either increased by increasing the value of the receiving end line voltage (V_R ) or by reducing the surge impedance (Z_C). Voltage transmission capability is increased day by day; this is the most commonly adopted method for increasing the power limit of the heavily loaded transmission line. But there is a limit beyond which is neither economical nor practical to increase the receiving end line voltage

By applying some methods such as introducing series capacitors (capacitors in series with the transmission line) or shunt capacitors (capacitors in parallel with transmission lines) can be used to reduce the value of surge impedance (Z_C).

            At natural as impedance is purely resistive,

                                                       4            

                                             5
                                               6

Example                              

   

[Ref: Example 5.5, Power System Analysis, Haadi Sadat]

Manually calculation

MATLAB Code

L=0.97; C=0.0115; lngth=300; SR=800+j*600; VRLL = 500;

w=2*pi*60;

beta=w*sqrt(L*C*1e-9)

Zc = sqrt(L/C*1e3)

vel=1/sqrt(L*C*1e-9)

lambda=vel/60

betal=beta*lngth;

VR=VRLL/sqrt(3);

IR=conj(SR)/(3*conj(VR));

VS=cos(betal)*VR+j*Zc*sin(betal)*IR;

VSLL=sqrt(3)*abs(VS)

IS = j/Zc*sin(betal)*VR+cos(betal)*IR;

ISM=abs(IS)*1000

SS=3*VS*conj(IS)

A=abs(cos(betal));

REG=(VSLL/A - VRLL)/VRLL*100

Output

beta = 0.0013

Zc = 290.4270

vel = 2.9941e+005

lambda = 4.9902e+003

VSLL = 617.5458

ISM = 902.3314

SS = 8.0000e+002 +5.3992e+002i

REG = 32.8766

Conclusion

Questions

Explain SIL.

Experiment no. 7

Object

Complex power flow

Objectives

Student should able to calculate complex power flow for each source and losses using MATLAB

Theory

Apparatus

MATLAB software, Power system analysis By Haadi sadad Tool Box

Procedure

1.Open MATLA Window

2.Open matlab program from Haadi sadat ref example no.1.1

3.Execuite the program

4.Check the output

5.Verify the output with manual calculation.

Example

1)      Two voltage sources V1=120∟-5 V and V2 =100∟0 V are connected by short line of impedance Z=1+j7Ω as shown in fig. above. Determine the real and reactive power supplied or received by each source and the power loss in the line.

2)      Write a MATLAB program for a system of above Example such that the phase angle of source 1 is changed from its initial value by ±30⁰ in steps of 5⁰ . Voltage magnitude of the two sources and voltage phase angle of sources 2 is to be kept constant. Compute the complex power for each source and the line losses. Tabulate the real power and plot p1 , p2 and p_L versus voltage phase angle δ.

[Ref: Example 2.5 and 2.6,  Power System Analysis, Haadi Sadat]

Manually calculation

MATLAB Code

1)

R = 1; X = 7; Z = R +j*X;

V1 = 120*(cos(-5*pi/180) + j*sin(-5*pi/180));

V2 = 100+j*0;

I12 = (V1 - V2)/Z, I21 = -I12;

S12 = V1*conj(I12), S21 = V2*conj(I21)

SL = S12 + S21

PL = R*abs(I12)^2, QL = X*abs(I12)^2

2)

E1=input('Source # 1 Voltage Mag. = ');

a1=input('Source # 1 Phase Angle  = ');

E2=input('Source # 2 Voltage Mag. = ');

a2=input('Source # 2 Phase Angle  = ');

R=input('Line Resistance = ');

X= input('Line Reactance = ');

Z= R + j*X;                                      % line impedance

a1 = (-30+a1:5:30+a1)';    % change a1 by +/- 30 deg., col. array

a1r = a1*pi/180;                       % convert degree to radian

k=length(a1);

a2=ones(k,1)*a2;        % create col. array of same length for a2

a2r = a2*pi/180;                       % convert degree to radian

V1=E1.*cos(a1r) + j*E1.*sin(a1r);

V2=E2.*cos(a2r) + j*E2.*sin(a2r);

I12 = (V1 - V2)./Z;  I21=-I12;

S1= V1.*conj(I12);  P1 = real(S1);  Q1 = imag(S1);

S2= V2.*conj(I21);  P2 = real(S2);  Q2 = imag(S2);

SL= S1+S2;        PL = real(SL);  QL = imag(SL);

Result1=[a1, P1,  P2, PL];

disp('   Delta 1    P-1       P-2       P-L ')

disp(Result1)

plot(a1, P1,  a1, P2,  a1, PL), grid

text(-26, -550, 'P1'), text(-26, 600,'P2'), text(-26, 100, 'PL')

xlabel('Source #1 Voltage Phase Angle'), ylabel('P, Watts')

Output

I12 =  -1.0733 - 2.9452i

S12 =-9.7508e+001 +3.6331e+002i

S21 =1.0733e+002 -2.9452e+002i

SL =   9.8265 +68.7858i

PL =9.8265

QL =   68.7858

2)

Source # 1 Voltage Mag. = 120

Source # 1 Phase Angle  = -5

Source # 2 Voltage Mag. = 100

Source # 2 Phase Angle  = 0

Line Resistance = 1

Line Reactance = 7

Conclusion

Experiment no. 8

Object

Load curve

Objectives

Student should able to calculate load factor using MATLAB

Theory

Load Curves

The curve showing the variation of load on the power station with respect to (w.r.t) time is known as a load curve.

The load on a power station is never constant; it varies from time to time. These load variations during the whole day (i.e., 24 hours) are recorded half-hourly or hourly and are plotted against time on the graph. The curve thus obtained is known as daily load curve as it shows the variations of load w.r.t. time during the day.

Importance. The daily load curves have attained a great importance in generation as they sup-ply the following information readily :

(i) The daily load curve shows the variations of load on the power station during different hours of the day.

(ii) The area under the daily load curve gives the number of units generated in the day. Units generated/day = Area (in kWh) under daily load curve.

(iii) The highest point on the daily load curve represents the maximum demand on the station on that day.

(iv) The area under the daily load curve divided by the total number of hours gives the average load on the station in the day.

(v) The ratio of the area under the load curve to the total area of rectangle in which it is con-tained gives the load factor.

(vi)  The load curve helps in selecting* the size and number of generating units.

(vii)  The load curve helps in preparing the operation schedule of the station.

Important Terms and Factors

(i) Connected load. It is the sum of continuous ratings of all the equipments connected to supply system.

(ii) Maximum demand : It is the greatest demand of load on the power station during a given period.

(iii) Demand factor. It is the ratio of maximum demand on the power station to its connected load i.e.

(iv) Average load. The average of loads occurring on the power station in a given period (day or month or year) is known as average load or average demand.  

(v) Load factor. The ratio of average load to the maximum demand during a given period is known as load factor i.e.

(vi) Diversity factor. The ratio of the sum of individual maximum demands to the maximum demand on power station is known as diversity factor i.e.,

(vii) Plant capacity factor. It is the ratio of actual energy produced to the maximum possible energy that could have been produced during a given period i.e.,

(viii) Plant use factor. It is ratio of kWh generated to the product of plant capacity and the number of hours for which the plant was in operation i.e.

Apparatus

MATLAB software, Power system analysis By Haadi sadad Tool Box

Procedure

1.Open MATLA Window

2.Open matlab program from Haadi sadat ref example no.1.1

3.Execuite the program

4.Check the output

5.Verify the output with manual calculation.

Example

The daily load on power system is given below. Use barcyle function to obtain a plot of the daily load curve. Using the given data compute the average load and daily load factor.

[Ref: Example 1.1, Power System Analysis, Haadi Sadat]

Manually calculation

MATLAB Code

data = [ 0   2    6

         2   6    5

         6   9   10

         9  12   15

        12  14   12

        14  16   14

        16  18   16

        18  20   18

        20  22   16

        22  23   12

        23  24    6];

P =data(:,3);                               % Column array of load

Dt = data(:, 2) - data(:,1);    %  Column array of demand interval

W = P'*Dt;                    % Total energy, area under the curve

Pavg = W/sum(Dt)                                    % Average load

Peak = max(P)                                          % Peak load

LF = Pavg/Peak*100                           % Percent load factor

barcycle(data)                              % Plots the load cycle

xlabel('Time, hr'), ylabel('P, MW'), grid

Output

Pavg =   11.5417

Peak =18

LF =64.1204

Load curve

Conclusion

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