Analyze Fundamental components of power systems using objects oriented programming

1. 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****

2. 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.

3. 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

4. MATLAB CODES
Question 1

6. 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)

7. 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

8. TESTED RESULTS
Question 1
Question 1:
a) 𝑅𝑒𝑓𝑒𝑟 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑖𝑚𝑝𝑒𝑑𝑒𝑛𝑐𝑒 𝑡𝑜 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 ∶
𝑎 =
𝑁1
𝑁2
=
2400
240
= 10
𝑅𝑠 𝐼
= 𝑎2
𝑅𝑠 = (10)2(0.002) = 0.2Ω
𝑋𝑠 𝐼
= 𝑎2
𝑋𝑠 = (10)2(0.045) = 0.45Ω
𝑉𝑠 𝐼
= 𝑎2
𝑉𝑠 = (10)2(240) = 2400
𝑅 𝐸𝑄 = 𝑅𝑠 𝐼
+ 𝑅𝑝 = 0.4
j𝑋 𝐸𝑄 = 𝑗𝑋𝑠 𝐼
+ 𝑗𝑋𝑝 = 𝑗0.9
𝑍 𝐸𝑄 = 𝑅 𝐸𝑄 + 𝐽𝑋 𝐸𝑄 = 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°
𝐼𝑠 𝐼
=
𝐼𝑠
𝑎
=
625<−36.87
10
= 62.5 < −36.87𝐴
𝑉𝑝 = 𝑉𝑠 𝐼
+ (𝑅 𝐸𝑄 + 𝑄 𝐸𝑄)𝐼𝑠 𝐼
= (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°
𝐼𝑠 𝐼
=
𝐼𝑠
𝑎
=
312.5<−36.87
10
= 31.25 < −36.87𝐴
𝑉𝑝 = 𝑉𝑠 𝐼
+ (𝑅 𝐸𝑄 + 𝑄 𝐸𝑄)𝐼𝑠 𝐼
= (2400 < 0) + [0.4 + 𝑗0.9](31.25 < −36.87)
= 2426.77 < 0.35 V

9. 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑅𝑒𝑔𝑙𝑎𝑡𝑖𝑜𝑛: 𝑉𝑅 =
2426.77−2400
2400
× 100 = 1.11%
d) 𝐼𝑠 =
150𝑘𝑉𝐴
240
< 36.87 = 625 < 36.87°
𝐼𝑠 𝐼
=
𝐼𝑠
𝑎
=
625<36.87
10
= 62.5 < 36.87𝐴
𝑉𝑝 = 𝑉𝑠 𝐼
+ (𝑅 𝐸𝑄 + 𝑄 𝐸𝑄)𝐼𝑠 𝐼
= (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

10. f) Linear Transformer with a source and load Simulation and Output
a) Voltage Output

11. b) Current Input

12. 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

13.  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
𝑋 𝑛𝑒𝑤 = 𝑋 𝑜𝑙𝑑 (
𝑆 𝑛𝑒𝑤
𝑆 𝑜𝑙𝑑
) (
𝑉𝑜𝑙𝑑
𝑉𝑛 𝑒𝑤
)
2
o Generator 1:
𝑋 𝑏 𝐺1 = (0.09) (
100𝑀
90𝑀
) (
20𝑘
20𝑘
)
2
= 0.1𝑝𝑢
o Transformer 1:
𝑋 𝑏 𝑇1 = (0.16) (
100𝑀
80𝑀
) (
20𝑘
20𝑘
)
2
= 0.2𝑝𝑢
o Transformer 2:
𝑋 𝑏 𝐺2 = (0.2) (
100𝑀
90𝑀
) (
20𝑘
20𝑘
)
2
= 0.22𝑝𝑢
o Generator 2:
𝑋 𝑏 𝐺2 = (0.09) (
100𝑀
90𝑀
) (
20𝑘
20𝑘
)
2
= 0.2.1
o Base impedance of transmission lines
𝑍 𝑏 𝐿𝑖𝑛𝑒 =
𝑉𝑏
2
𝑆
=
(200𝑘)2
100
= 400Ω
o Obtaining the pu value from ohms:

14. 𝑋 𝐿𝑖𝑛𝑒 =
𝑍 𝑏 𝑙𝑖𝑛𝑒
𝑍 𝑏 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
𝑍 𝐿𝑜𝑎𝑑 (𝑝𝑢) = (
𝑍𝑙𝑜𝑎𝑑
𝑍 𝑏 𝑙𝑖𝑛𝑒
) =
(0.5∠ − 53.13∘)
400
= 0.75 + 𝑗1.0𝑝𝑢
Equivalent Circuit

15. 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.

16. 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.