Lecture notes

1. EE 449T Electrical Power in Ships
Prepared by: Dr Rania Assem
Electrical Power & Control Engineering

2. Per Unit System Calculations
 A key problem in analyzing power systems is the large number of elements such as motors, generators
and transformers, each having a different size and nominal values.
– Especially with transformers, it would be very difficult to continually have to refer impedances to the
different sides of the transformers
 This problem is avoided by a normalization of all variables to be able to compare the performance of
big and small elements.
 This normalization is known as “per unit analysis”.
 Voltages, currents, impedances are divided by “Base values” and are expressed in per unit ot
percentage values
actual quantity
quantity in per unit
base value of quantity

𝑍𝑝𝑢 =
𝑍𝑎𝑐𝑡𝑢𝑎𝑙 (Ω)
𝑍𝑏𝑎𝑠𝑒 (Ω)
× 100%
Example 1

3. Per Unit System Calculations
 For an electric element, we have: Power, Voltage, Current and Impedance
 Usually, the nominal apparent power (S) and the nominal voltage (V) are taken as the base values
for power and voltage
 This means that we get to CHOOSE 𝑆𝑏𝑎𝑠𝑒 and 𝑉𝑏𝑎𝑠𝑒 and any other parameters can be calculated
accordingly
𝑆𝑝𝑢 =
𝑆𝑎𝑐𝑡𝑢𝑎𝑙 (𝑉𝐴)
𝑆𝑏𝑎𝑠𝑒 (𝑉𝐴)
× 100% 𝑉
𝑝𝑢 =
𝑉𝑎𝑐𝑡𝑢𝑎𝑙 (𝑉𝐴)
𝑉𝑏𝑎𝑠𝑒 (𝑉𝐴)
× 100% 𝐼𝑝𝑢 =
𝐼𝑎𝑐𝑡𝑢𝑎𝑙 (𝑉𝐴)
𝐼𝑏𝑎𝑠𝑒 (𝑉𝐴)
× 100% 𝑍𝑝𝑢 =
𝑍𝑎𝑐𝑡𝑢𝑎𝑙 (Ω)
𝑍𝑏𝑎𝑠𝑒 (Ω)
× 100%
Chosen by “YOU”
How to get 𝐼𝑏𝑎𝑠𝑒? 𝑆𝑏𝑎𝑠𝑒 = 3𝑉𝑏𝑎𝑠𝑒𝐼𝑏𝑎𝑠𝑒, 𝐼𝑏𝑎𝑠𝑒 =
𝑆𝑏𝑎𝑠𝑒
3𝑉𝑏𝑎𝑠𝑒
Calculated
How to get 𝑍𝑏𝑎𝑠𝑒? 𝑍𝑏𝑎𝑠𝑒 =
𝑉𝑏𝑎𝑠𝑒
3
𝐼𝑏𝑎𝑠𝑒
=
𝑉𝑏𝑎𝑠𝑒
2
𝑆𝑏𝑎𝑠𝑒
 
base
base
base
MVA
kV
Z
2


4. Per Unit System Calculations
Changing base impedance (Znew)
 What if we have a system that is calculated on a certain base and we need to change it to another
base?? What shall we do then?
 
2
2
base OLD base NEW
NEW OLD
base OLD
base NEW
kV MVA
Z pu Z
MVA
kV
 
 
  
 
 
Before doing all per unit calculations, we need to first draw a single
line diagram to represent our power system

5. Power system Representation and Equations
One-line (single-line) diagrams
 Almost all modern power systems are three-phase systems with the phases of equal amplitude and shifted by 120˚. Since
phases are similar, it is customary to sketch
 Power systems in a simple form with a single line representing all three phases of the real system.
 Combined with a standard set of symbols for electrical components, such one-line diagrams provide a compact way to represent
information
Example: a power system containing two synchronous machines, two loads, two busses, two transformers, and a transmission
line to connect busses together

6. Power system Representation and Equations
Impedance diagrams
In power system fault calculations, it is often that a single-line diagram representing a typical power network in 3- be converted into its per phase
impedance diagram.
(i) A generator can be represented by a voltage source in series with an inductive reactance. The internal resistance of the generator is assumed to be
negligible compared to the reactance.
(ii) The loads are usually inductive represented by resistance and inductance.
(iii) The transformer core is assumed to be ideal, and the transformer may be represented by a reactance only.
(iv) The transmission line is represented by its resistance and inductance, the line-to-ground capacitance is assumed to be negligible.
Generator
G1
G2
Station A Station B
G3
G4
Load
L1
Load
L2
Transformer
T1
Transmission
Line
TL
Transformer
T2
G2
G1
G3 G4
j X1 j X2 j X3 j X4
j
XT1
j
XT2
RL1
j XL1
RL2
j XL2
RTL
j
XTL
Transformer
T1
Transformer
T2
Transmission
Line
TL
Station A Station B
Impedance diagram
Single-line diagram of a power network

7. Per Unit System What to Do?
1. Draw the single line diagram of the power systen
2. Pick a 1 VA base for the entire system, SB
3. Pick a voltage base for each different voltage level, VB. Voltage bases are related by
transformer turns ratios. Voltages are line to neutral.
4. Calculate the impedance base, ZB= (VB)2/SB
5. Calculate the current base, IB = VB/ZB
6. Convert actual values to per unit
Note, per unit conversion affects ONLY magnitudes, not
the angles. Also, per unit quantities have NO
units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)

8. Per Unit Calculations
Determine the per-unit values of the following single-line diagram and draw the impedance diagram.
Example 1
XT1 = 0.1 p.u
5 MVA
Xg = 16%
100 MVA
275 kV/132 kV
50 MVA
132 kV/66 kV
Transmission line
j 3.48 
XT2 = 0.04 p.u Load
40 MW, 0.8 p.f. lagging
Solution:
Chosen base: Always choose the largest rating, therefore Sbase = 100 MVA, V = 66 kV, 132 kV and 275 kV
Per-unit calculations:
Generator G1:
32
.
0
50
100
16
.
0
)
( 


pu
X g p.u.
Transformer T1:
1
.
0
)
(
1 
pu
XT
p.u.
 
2
2
base OLD base NEW
NEW OLD
base OLD
base NEW
kV MVA
Z pu Z
MVA
kV
 
 
  
 
 

9. Per Unit Calculations
Transmission line TL:
0195
.
0
132
100
4
.
3
)
( 2



pu
XTL p.u.
Transformer T2:
08
.
0
50
100
04
.
0
)
(
2 


pu
XT p.u.
Inductive load:
o
actual
Z 87
.
36
12
.
87
8
.
0
10
66
3
10
40
3
10
66
3
6
3








p.u.
)
2
.
1
6
.
1
(
87
.
36
2
66
100
87
.
36
12
.
87
)
( 2
j
or
pu
Z o
o
L 






 
base
base
base
MVA
kV
Z
2
 actual
pu
base
Z
Z
Z


10. Per Unit Calculations
Now, we have all the impedance values in per-unit with a common base and we can now combine all the impedances and determine the overall
impedance.
Load
G
j 0.32 p.u.
j 0.1 p.u. j 0.0195 p.u.
Transformer
T1
Transformer
T2
Transmission Line
TL
j 0.08 p.u.
1.6 p.u..
j 1.2 p.u.
Generator
XT1 = 0.1 p.u
5 MVA
Xg = 16%
100 MVA
275 kV/132 kV
50 MVA
132 kV/66 kV
Transmission line
j 3.48 
XT2 = 0.04 p.u Load
40 MW, 0.8 p.f. lagging

11. THANK YOU