University of the South Pacific

Faculty of Science & Technology

School of Engineering & Physics

Electrical & Electronics Engineering Discipline

EE321 POWER SYSTEM ANALYSIS

LAB 2: Analyze fundamental components of power systems using object

oriented programming

Group Members: Bernadette Pesamino s11102091

Seci Durivou s11098325

Tevita Daivalu s 11090790

Eddie Arukelana s1109****

AIM

Analyze important power system components by making the necessary manual

calculations and comparing the results with the results simulated obtained in MATLAB

and Simscape Power Systems.

Finds how Transformer and efficiency behaves and implementing it in Matlab.

ALGORITHM

Question 1

Finding the primary Voltage:

Primary Current

Equivalent Circuit Referred to the

high Voltage Side

Power Factor

Leading or

Lagging

Leading

<+36.87

Lagging

<-36.87

Primary Voltage

Voltage Regulations

MATLAB CODES

Question 1

Question 2

X_G1 = 0.09;

%Transformer 1 Values

Sb_T1 = 80*10^6;

Vb_T1 = 20*10^3;

X_T1 = 0.16;

%Transformer 2 Values

Sb_T2 = 80*10^6;

Vb_T2 = 20*10^3;

X_T2 = 0.2;

%Generator 2 Values

Sb_G2 = 90*10^6;

Vb_G2 = 18*10^3;

X_G2 = 0.09;

%Line

Vb_line = 200*10^3;

Xold_line = 120;

%Load

Vb_load = 200*10^3;

Sb_load = (48*10^6)+j*(64*10^6);

%System Reactance for equivalent circuit

Xnew_G1 = X_G1 * (Sbnew/Sb_G1)*((Vbnew/Vb_G1) ^2)

Xnew_T1 = X_T1 * (Sbnew/Sb_T1)*((Vbnew/Vb_T1)^2)

Xnew_T2 = X_T2 * (Sbnew/Sb_T2)*((Vbnew/Vb_T2)^2)

Xnew_G2 = X_G2 * (Sbnew/Sb_G2)*((Vb_G2/Vbnew)^2)

Zb_line = (Vb_line)^2/Sbnew;

X_line = Xold_line/Zb_line

Zb_load = (Vb_load)^2/Sb_load;

Z_loadpu = Zb_load/400

TESTED RESULTS

Question 1

Question 1:

a) 𝑅𝑒𝑓𝑒𝑟 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑖𝑚𝑝𝑒𝑑𝑒𝑛𝑐𝑒 𝑡𝑜 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 ∶

𝑎 =𝑁1

𝑁2=

2400

240= 10

𝑅𝑠𝐼 = 𝑎2𝑅𝑠 = (10)2(0.002) = 0.2Ω

𝑋𝑠𝐼 = 𝑎2𝑋𝑠 = (10)2(0.045) = 0.45Ω

𝑉𝑠𝐼 = 𝑎2𝑉𝑠 = (10)2(240) = 2400

𝑅𝐸𝑄 = 𝑅𝑠𝐼 + 𝑅𝑝 = 0.4

j𝑋𝐸𝑄 = 𝑗𝑋𝑠𝐼 + 𝑗𝑋𝑝 = 𝑗0.9

𝑍𝐸𝑄 = 𝑅𝐸𝑄 + 𝐽𝑋𝐸𝑄 = 0.4 + 𝑗0.9

b)𝐴𝑡 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 0.8 𝑝𝑓 𝑙𝑎𝑔𝑔𝑖𝑛𝑔: 𝑆 = 150 < −36.87° 𝑘𝑉𝐴

𝐼𝑠 =150𝑘𝑉𝐴

240< −36.87 = 625 < −36.87°

𝐼𝑠𝐼 =𝐼𝑠

𝑎=

625<−36.87

10= 62.5 < −36.87𝐴

𝑉𝑝 = 𝑉𝑠𝐼 + (𝑅𝐸𝑄 + 𝑄𝐸𝑄)𝐼𝑠𝐼 = (2400 < 0) + [0.4 + 𝑗0.9](62.5 < −36.87)

= 2453.934 < 0.7004V

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑅𝑒𝑔𝑙𝑎𝑡𝑖𝑜𝑛: 𝑉𝑅 =2453.934−2400

2400× 100 = 2.24%

c)50% 𝑜𝑓 150𝑘𝑉𝐴: 150𝑘𝑉𝐴 × 50% = 75𝑘𝑉𝐴

𝐼𝑠 =75𝑘𝑉𝐴

240< −36.87 = 312.5 < −36.87°

𝐼𝑠𝐼 =𝐼𝑠

𝑎=

312.5<−36.87

10= 31.25 < −36.87𝐴

𝑉𝑝 = 𝑉𝑠𝐼 + (𝑅𝐸𝑄 + 𝑄𝐸𝑄)𝐼𝑠𝐼 = (2400 < 0) + [0.4 + 𝑗0.9](31.25 < −36.87)

= 2426.77 < 0.35 V

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑅𝑒𝑔𝑙𝑎𝑡𝑖𝑜𝑛: 𝑉𝑅 =2426.77−2400

2400× 100 = 1.11%

d) 𝐼𝑠 =150𝑘𝑉𝐴

240< 36.87 = 625 < 36.87°

𝐼𝑠𝐼 =𝐼𝑠

𝑎=

625<36.87

10= 62.5 < 36.87𝐴

𝑉𝑝 = 𝑉𝑠𝐼 + (𝑅𝐸𝑄 + 𝑄𝐸𝑄)𝐼𝑠𝐼 = (2400 < 0) + [0.4 + 𝑗0.9](62.5 < 36.87)

= 2387.07 < 1.43V

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑅𝑒𝑔𝑙𝑎𝑡𝑖𝑜𝑛: 𝑉𝑅 =2387.07−2400

2400× 100 = −0.53%

Regulation and Efficiency

f) Linear Transformer with a source and load Simulation and Output

a) Voltage Output

b) Current Input

Q2

Flowchart:

Algorithm

Draw an impedance diagram for the electric power system shown in Figure

26 showing all impedances in per unit on a 100-MVA base. Choose 20 kV as the

voltage base for generator. The three-phase power and line-line ratings are

given below.

G1: 90 MVA 20 kV X = 9%

T1: 80 MVA 20/200 kV X = 16%

T2: 80 MVA 200/20 kV X = 20%

G2: 90 MVA 18 kV X = 9%

Line: 200 kV X = 120Ω

Load: 200 kV S = 48 MW +j64 Mvar

Calculate the

base voltages

each side of the

transformers

Calculate the

generator &

transformer

reactance

Calculate the

base impedance

for the

transmission line

Calculate the per

unit reactance of

the line

Calculate the

load impedance

in ohms

Calculate the

load impedance

in per unit

Draw equivalent

impedance diagram

using calculated

reactance in per unit

values

Base voltages

𝑉𝑏 𝑅1 = 20𝑘 (200𝑘

20𝑘) = 200𝑘𝑉 (Secondary side of T1)

𝑉𝑏 𝑅2 = 200𝑘𝑉 (secondary side of T1 & primary side of T2)

𝑉𝑏 𝑅3 = 200𝑘 (200𝑘

20𝑘) = 20𝑘𝑉 (secondary side of T2)

Reactances (per unit) of the generators & transformers using a new base

of 100MVA

𝑋𝑛𝑒𝑤 = 𝑋𝑜𝑙𝑑 (𝑆𝑛𝑒𝑤

𝑆𝑜𝑙𝑑) (

𝑉𝑜𝑙𝑑

𝑉𝑛𝑒𝑤)

2

o Generator 1:

𝑋𝑏 𝐺1 = (0.09) (100𝑀

90𝑀) (

20𝑘

20𝑘)

2

= 0.1𝑝𝑢

o Transformer 1:

𝑋𝑏 𝑇1 = (0.16) (100𝑀

80𝑀) (

20𝑘

20𝑘)

2

= 0.2𝑝𝑢

o Transformer 2:

𝑋𝑏 𝐺2 = (0.2) (100𝑀

90𝑀) (

20𝑘

20𝑘)

2

= 0.22𝑝𝑢

o Generator 2:

𝑋𝑏 𝐺2 = (0.09) (100𝑀

90𝑀) (

20𝑘

20𝑘)

2

= 0.2.1

o Base impedance of transmission lines

𝑍𝑏 𝐿𝑖𝑛𝑒 = 𝑉𝑏

2

𝑆=

(200𝑘)2

100= 400Ω

o Obtaining the pu value from ohms:

𝑋𝐿𝑖𝑛𝑒 =𝑍𝑏 𝑙𝑖𝑛𝑒

𝑍𝑏 𝑙𝑖𝑛𝑒=

120Ω

400Ω= 0.38𝑝𝑢

Load impedance of the line:

𝑍𝑙𝑜𝑎𝑑 =(𝑉𝑏)2

𝑆=

(200𝑘)2

48𝑀𝑊 + 𝑗64𝑀𝑣𝑎𝑟=

40𝑀

80𝑀∠53.13∘= 0.5∠ − 53.13∘ ≃ 0.3 − 𝑗0.399Ω

𝑍𝐿𝑜𝑎𝑑 (𝑝𝑢) = (𝑍𝑙𝑜𝑎𝑑

𝑍𝑏 𝑙𝑖𝑛𝑒) =

(0.5∠ − 53.13∘)

400= 0.75 + 𝑗1.0𝑝𝑢

Equivalent Circuit

DISCUSSION

- The efficiency of a transformer is low as appeared in the figure, however it

increment to a steady incentive as it gets steady.

- Then again the chart of the regulation is diminishing, also, it gets to the base

an incentive after short little time.

- For a decent transformer productivity ought to be high and on the other

regulation control to be little.

- The drawing of the equivalent impedance diagram was done using

calculated reactance in per unit.

CONCLUSION

To conclude, the power systems component were analysis using the Per Unit

systems and the manual calculations were compared with results simulated in

Matlab and Simscape Power Systems as displayed in the result section. In the

underlying state, the efficiency of a transformer is low as appeared in the figure,

however it increment to a steady incentive as it gets steady. Then again the chart

of the regulation is diminishing, also, it gets to the base an incentive after short

little time. So we can conclude that for a decent transformer productivity ought to

be high and on the other regulation control ought to be little.