Course: SCEP516 / DE5401
New Zealand Diploma of Engineering Level 6
Fault Study – Per Unit System
The Per Unit System
In the analysis of power networks instead of using actual values of quantities it is usual to express
them as fractions of reference quantities, such as rated or full-load values. These fractions are called
per unit (denoted by p.u.) and the p.u. value of any quantity is defined as:
per unit value 
the actual value (any unit )
a base reference value ( same unit )
Some books and jurisdictions express the p.u. value as a percentage. This approach is equally valid
for example 160% = 1.6 p.u.
Per-Unit System – An Introduction
The per unit system provides a handy means of expressing the magnitude of a quantity in
terms of a convenient unit, called the per-unit base of the system. For example, suppose the
average weight of rugby playing adults in Hamilton is 100kg. If this 100kg is chosen as the
base unit then a rugby player from Auckland weighing 60kg would weigh:
Per Unit value 
60kg
kg base
The answer is 0.6 p.u.
Using this arbitrary weight as a base, the weight of any individual may be compared in terms of this
base weight. To generalize, a per-unit system of measurement consists of selecting one or more
convenient measuring bases and comparing similar things against the selected base.
The following link provides lecture that may be useful.
http://www.google.co.nz/url?sa=t&rct=j&q=per+unit+system&source=web&cd=4&ved=0CD0QtwIw
Aw&url=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3Dn7mySlSlcfY&ei=Um43T7X5DYajiQf
vs-WCAg&usg=AFQjCNEN4lVVUH0e5Fbln3c1eofaVzL7VA&cad=rja
In the power system industry the per unit system is widely used to express values of voltages,
currents, powers, and impedances of various power equipment. It is mainly used for transformers
and AC machines. For a given quantity (voltage, current, power, impedance, torque, etc.) the per
unit value is the value related to a base quantity.
Generally the following two base values are chosen:


The base power = nominal power of the equipment
The base voltage = nominal voltage of the equipment
All other base quantities are derived from these two base quantities. Once the base power and the
base voltage are chosen, the base current and the base impedance are determined by the natural
laws of electrical circuits.
2
The Per-Unit System with One Base
If the size of only one quantity is used as the common basis, the per-unit system is said to have a
single base. The base may be a power, a voltage, a current or a machine rotation speed/RPM. For
example, suppose that three motors have power ratings of 25kW, 40kW, and 150kW. Using an
arbitrary base power P of 50kW the corresponding per-unit ratings are then 25 kW/50 kW = 0.5,
40kW/50kW = 0.8 and 150 kW /50 kW = 3. Thus, in this per-unit system where the base is 50kW, the
three motors have power ratings of 0.5, 0.8, and 3 pu, respectively.
Using a base power of 15kW the respective per-unit ratings would be 25 kW /15kW, 40kW /15kW =
2.67 and 150kW /15kW =10. It is therefore important to know the magnitude of the base of the perunit system. If we do not know its value, the actual values of the quantities we are dealing with
cannot be calculated.
The per-unit method can also be applied to impedances. Consider, for example, the circuit below is
composed of several resistors, capacitors, and inductors having the impedances shown, If we decide
to use an impedance of 1500 ohms as the base, the per-unit impedances are as follows:
R1 ( pu ) 
3500
 2.33
1500
R2 ( pu ) 
450
 0.30
1500
X L ( pu ) 
4800
 3.2
1500
X C ( pu ) 
3000
 2.0
1500
The per-unit circuit following contains the same elements as the real circuit, but the impedances are
now expressed in per-unit values. We can solve this circuit as we would any other circuit.
3
For example, if vector notation is used, the per-unit circuit is that shown as follows:
Per-Unit System with Two Bases
The per-unit system becomes particularly useful when two bases are used. The bases are usually a
base voltage EB and a base power PB. If the selected base voltage is 4kV and the selected base power
500kW it is possible to progress those base values forward into convenient calculations. Note that
the two base values can be selected quite independently of each other. By using the voltage/power
per- unit system is that it automatically establishes a corresponding base current mid base
impedance. Thus the base current is:
IB 
P
base power
 B
base voltage E B
and the base impedance is
ZB 
base voltage E B

base current I B
For example, if the base voltage is 4kV and the base power is 500 kW, the base current is:
IB 
PB 500000

 125 A
EB
4000
4
And the base impedance is:
ZB 
E B 4000

 32
IB
125
In effect, by selecting the voltage/power per— unit system we also get a base current and base
impedance. Consequently, the 2-base per-unit system really provides a 4-base per-unit system.
Example
A 400Ω resistor carries a current of 60 A. Using the above base values, calculate:
1. The per-unit resistance
2. The per-unit current
3. The per-unit voltage across the resistor
4. The per-unit power dissipated in the resistor
5. The actual K and F of the resistor
Solution
The per-unit resistance is:
R(pu) = 400Ω/32Ω= 12.5
The per-unit current is:
I(pu) = 60 A/125 A= 0.48
The per-unit voltage across the resistor is:
E(pu)
=
=
=
I(pu) )x R(pu)
0.48 x 12.5
6
The per-unit power is:
P(pu)
=
=
=
E(pu) x I(pu)
6 x 0.48
2.88
The actual voltage across the resistor is:
E
=
=
=
EB x E(pu)
4 kV x 6
24kV
5
The actual power dissipated hi the resistor is:
P
=
=
=
PB x P(pu)
500kW x 2.88
1440kW
Example
A 7.2 kV source delivers power to a 24 resistor and a 400kW electric boiler.
Draw the equivalent per unit circuit diagram.
Use the same base values as in the Example above calculate:
1. The per-unit E(pu), R(pu), P(pu)
2. The per-unit current I2(pu)
3. The per-unit line current IL(pu)
4. The per-unit power absorbed by the resistor
5. The actual power absorbed by the resistor
6. The actual line current
IL
I2
I1
E
R=
24Ω
Boiler
7200V
P
400kW
Solution
The per-unit line voltage is:
E1 ( pu ) 
7.2kV
 1.8
4kV
The per-unit resistance is:
R( pu ) 
24
 0.75
32
The per-unit power of the boiler is:
6
P( pu ) 
400kW
 0.8
500kW
The per-unit current I2:
I 2 ( pu ) 
E ( pu ) 1.8

 2.4
R( pu ) 0.75
The per-unit current I1:
I1( pu ) 
P( pu ) 0.8

 0.444
E ( pu ) 1.8
The per-unit current IL:
I L ( pu)  I1 ( pu)  I 2 ( pu)  0.444  2.4  2.844
The per-unit power in the resistor:
P( pu)  E( pu)  I 2 ( pu)  1.8  2.4  4.32
The actual power in the resistor:
P2  PB  P( pu)  500kw  4.32  2160kW
The actual line current:
I 2  I B  I L ( pu)  125  2.884  355.5 A
I(pu) =
2.844
E(pu)=
I1(pu)
I2(pu)
=0.44
4
= 2.4
R(pu) =
Boiler
0.75
1.8
P(pu)=
0.8
7
Why Use the Per Unit System Instead of the Standard SI Units?
Here are the main reasons for using the per unit system:

When values are expressed in pu, the comparison of electrical quantities with their "normal"
values is straightforward.
For example, a transient voltage reaching a maximum of 1.42 pu indicates immediately that
this voltage exceeds the nominal value by 42%.

The values of impedances expressed in pu stay fairly constant whatever the power and
voltage ratings.

The apparatus considered may vary widely in size; losses and volt drops will also vary
considerably. For apparatus of the same general type the p.u. volt drops and losses are in
the same order regardless of size.

The use of √3 in three-phase calculations is reduced.

By the choice of appropriate voltage bases the solution of networks containing several
transformers is facilitated.

Per unit values lend themselves more readily to automatic computation.

For a transformer with multiple windings, each having a different nominal voltage, the same
base power is used for all windings (nominal power of the transformer).
For example, for all transformers in the 3 kVA to 300 kVA power range, the leakage reactance varies
approximately between 0.01 pu and 0.03 pu, whereas the winding resistances vary between 0.01 pu
and 0.005 pu, whatever the nominal voltage. For transformers in the 300 kVA to 300 MVA range, the
leakage reactance varies approximately between 0.03 pu and 0.12 pu, whereas the winding
resistances vary between 0.005 pu and 0.002 pu.
Similarly, for salient pole synchronous machines, the synchronous reactance Xd is generally between
0.60 and 1.50 pu, whereas the subtransient reactance X'd is generally between 0.20 and 0.50 pu.
It means that if you do not know the parameters for a 10 kVA transformer, you are not making a
major error by assuming an average value of 0.02 pu for leakage reactances and 0.0075 pu for
winding resistances.
The calculations using the per unit system are simplified. When all impedances in a multivoltage
power system are expressed on a common power base and on the nominal voltages of the different
subnetworks, the total impedance in pu seen at one bus is obtained by simply adding all impedances
in pu, without taking into consideration the transformer ratios.
Generally the following two base values are chosen:

The base power = nominal power of the equipment

The base voltage = nominal voltage of the equipment
8
R p.u. 

R()
baseR()
R()
basevoltage(Vb ) / BaseCurrent ( I b )

Voltage drop across R at base or rated current
(base or rated voltage)
By rearranging, the numerator becomes Ib x RΩ which equals the voltage drop across R at base
current, and the denominator is simply Vb.
R p.u. 
R() I b2
Vb I b
R p.u. 
Power Loss at base current
base power or VA
Therefore the power loss (p.u.) at base or rated current  R p.u.
Power loss (p.u.) at Ip.u. current  R p.u.  I p2.u.
Similarly:
P.U . impedance  impedance in ohms /
base voltage
base current
Example
A d.c. series machine rated at 200 V, 100A has an armature resistance of 0.1Ω and field resistance of
0.15Ω. The friction and windage loss is 1500W. Calculate the efficiency when operating as a
generator.
Solution
(Note that Vbase selected as 200 and Ibase selected as 100, thus Zbase = 200/100)
So series resistance = 0.1Ω + 0.15Ω and expressing as a per unit quantity following:
Total series R p.u. 
0.25
(200 / 100)
= 0.125 p.u.
9
Friction and windage loss
(Note that Vbase selected as 200 and Ibase selected as 100, thus Pbase = 200 x 100)

1500
200  100
= 0.075 p.u.
At the rated load, the series-resistance loss
= I2 x0.125 p.u.
and the total loss
= 0.125+0.175 = 0.2 p.u.
As the output = 1 p.u., the efficiency
1

1  .02
= 0.83 p.u.
Three-Phase Circuits
A p.u. phase voltage has the same numerical value as the corresponding p.u. line voltage. With a line
voltage of 100 kV and a rated line voltage of 132 kV, the p.u. value is 0.76. The equivalent phase
voltages are 100//3 kV and 132/13 kV and hence the p.u. value is again 0.76. The actual values of R,
XL and X for lines, cables and other apparatus are phase values. When working with ohmic values it
is less confusing to use the phase values of all quantities. In the p.u. system three-phase values of
voltage, current and power can be used without undue anxiety about the result being a factor of J3
incorrect.
I Base 
baseVA
3 Vbase
Vbase
Base Z 
Vb


3
I base
(assuming a Star system)
Vb
3
3
I b  Vb
3

Vb2
3 VbIb
10
(base line voltage) 2

baseVA
It should be noted that the same value for Zbase is obtained using purely phase values.
Z p.u.  Z () 
baseVA
(base voltage) 2
i.e. is directly proportional to the volt-ampere base and inversely to (base voltage)2.
 baseVold
Z p.u. (newbase)  Zp.u.(original base)  
 baseVnew

2
 baseVAnew
 
 baseVAold

Fault Studies
The fault analysis of a power system is required in order to provide information for the selection of
switchgear, setting of relays and stability of system operation. A power system is not static but
changes during operation (switching on or off of generators and transmission lines) and during
planning (addition of generators and transmission lines).
Faults usually occur in a power system due to insulation failure, flashover, physical damage or
human error. These faults may either be three phase in nature involving all three phases in a
symmetrical manner, or may be asymmetrical where usually only one or two phases may be
involved, or to system ground (earth). Faults may also be caused by either short-circuits to earth or
between live conductors, or may be caused by broken conductors in one or more phases.
Balanced three phase faults may be analysed using an equivalent single phase circuit. With
asymmetrical three phase faults, the use of symmetrical components help to reduce the complexity
of the calculations as transmission lines and components are by and large symmetrical, although the
fault may be asymmetrical.
Fault analysis is usually carried out in per-unit quantities or percentage quantities as they give
solutions which are somewhat consistent over different voltage and power ratings, and operate on
values of the order of unity.
In an electric power system, a fault is any abnormal electric current. For example, a short circuit is a
fault in which greatly current exceeds the normal load. In power systems, protective devices detect
fault conditions and operate circuit breakers and other devices to limit the loss of service due to a
failure.
The prospective short circuit current of a fault can be calculated for power systems using the per
unit system and this is the main focus of this introductory course. Ultimately the design of systems
to detect and interrupt power system faults is the main objective of power system protection.
Example
11
A substation has three step-down 33/11kV transformers in parallel, with the 11kV transformer
incomers feeding an 11kV busbar. Transformer 1 is rated at 15MVA and has 6% impedance,
transformer 2 is rated 10MVA and has 8% impedance and transformer 3 is 12MVA with impedance
8% .
a) Draw an appropriate schematic diagram for the substation, annotated with ratings and values.
b) Calculate the prospective short circuit current at the 11kV bus using the Per Unit system. Ignore
incomer cable resistance and reactance.
33kV
T1
15MVA
6%
T2
10MVA
8%
T3
15MVA
8%
Fault bus
11kV
The single line diagram is shown above.
Using the Per Unit system the first activity (Step 1) is choose a base MVA , 10MVA is chosen
for this example.
Having chosen a base MVA now convert impedances to the base (Step 2).
Zpu(newbase)  Zpu(oldbase) 
Zpu(T1) 
6 10
  0.04 pu
100 15
Zpu(T 2) 
8 10
  0.08 pu
100 10
Zpu(T 3) 
8 10
  0.0667 pu
100 12
MVA(newbase)
MVA(old )
Now Step 3 combine and simplify the network. For this simple system using the formula for
three parallel impedances
1
1
1
1



Zequiv Z1 Z 2 Z 3
12
1
1
1
1



 52.43 Ohms 1
Zequiv 0.04 0.08 0.0667
Thus Zequiv = 0.0191
Step 4 is to calculate the MVA fault level at the fault bus.
FaultLevel ( MVA) 
MVAbase
10

 524MVA
Zequiv
0.0191
Step 5 is to calculate the Fault Level in kA (engineers virtually always use kA and MVA in
fault studies)
Ifl 
MVA
Vll  3

524 x10 6
11x10 3  3
 27.52kA
Transformers
Consider a single-phase transformer in which the total series impedance of the two windings
referred to the primary is Z1 following.
Equivalent circuit of single-phase transformer.
Then the p.u. impedance
= I1Z1/V1
where I1 and V1 are the rated or base values. The total impedance referred to the secondary
= Z1xN2
and in p.u. notation
I 
 Z1 N 2  2 
 V2 
 Z1 N 2
I1
1

N V1 N
13

Z1 I1
V1
Hence the p.u. impedance of a transformer is independent of the winding considered.
In a circuit with several transformers care has to be taken regarding the different voltage levels.
Consider the network in Figure 2; in this two single-phase transformers supply a 10 kVA resistance
load, the load voltage being maintained at 200 V. Hence the load resistance is (2002/10,000), i.e. 4 Ω
In each of the circuits A, B, and C a different voltage exists, so that each circuit will have its own base
voltage, i.e. 100 V in A, 400 V in B, 200 V in C.
Network with two transformers—p.u. approach.
Although it is not essential for rated voltages to be used as bases, it is essential that the voltage
bases used be related by the turns ratios of the transformers. If this is not so the whole p.u.
framework breaks down. The same volt-ampere base is used for all the circuits as V111 = V212 on
each side of a transformer and is taken in this case as 10 kVA.
The base impedance in C:

200 2
10,000
=4Ω
The load resistance (p.u.) in C

4
4
=1 p.u.
In B the base impedance

400 2
10000
= 16Ω
and the load resistance referred to B
=4x22=16Ω
14
Hence the p.u. load referred to B
= 1 p.u.
Similarly, the p.u. load resistance referred to A is also 1 p.u. Hence, if the voltage bases are related
by the turns ratios the load p.u. value is the same for all circuits. An equivalent circuit may be used
as shown in Figure 3.
Figure 3 Equivalent circuit with p.u. values, of network in Figure 2.
Let the volt-ampere base be 10 kVA; the voltage across the load (VR) is 1 p.u. (as the base voltage in
C is 200 V, if the load voltage had been maintained at 100 V, VR would be 0.5 p.u.). The base current

(VA) b
Vb

1000
200
= 50A inC
The corresponding currents in the other circuits are 25A in B, and 100A in A. The actual load current
= 50/50 = 1 p.u. (in phase with VR the reference phasor).
Hence the supply voltage Vs
= 1(jO.1+jO.15)+1 p.u.
V=1.03p.u.
= 1.03 x 100 = 103 V.
The voltage at point D in Figure 3
= 1 +jO.15 xl = 1 +jO.15
= 1.0 12 p.u. modulus
= 1.012x400=404.8V.
It is a useful exercise to repeat this example using ohms, volts, and amperes.
15
Example System
Line
diagram of
Example
System
In the
the
Figure above
schematic
diagram of a
radial
transmission
system
is
shown.
The
ratings and reactances of the various components are shown, along with the nominal transformer
line voltages. A load of 50 MW at 0.8 p.f. lagging is taken from the 33 kV substation which is to be
maintained
at
30 kV. It is required to calculate the terminal voltage of the synchronous machine. The line and
transformers may be represented by series reactances. The system is three-phase.
It will be noted that the line reactance is given in ohms; this is usual practice. The voltage bases of
the various circuits are decided by the nominal transformer voltages, i.e. 11, 132, and 33 kV. A base
of 100 MVA will be used for all circuits. The reactances (resistance is neglected) are expressed on the
appropriate voltage and MVA bases. Base impedance for the line

132 2 x10 6
100 x10 6
= 174Ω
Hence the p.u. reactance
j100

174
= JO.575
Per unit reactance of the sending-end transformer
16

100
 j 0.1
50
= jO.2
Per unit reactance of the receiving-end transformer

100
 j 0.12
50
= j0.24
Load current

50  10 6
3  30  10 3  0.8
= 1200 A
Base current for 33 kV, 100 MVA

100  10 6
3  33  10 3
= 1750 A
Hence, p.u. load current

1200
1750
= 0.685
Voltage of load busbar

30
33
= 0.91 p.u.
The equivalent circuit is shown in Figure below
Equivalent Circuit
17
Vs=0.685(0.8-j0.6)(j0.2+j0.575+j0.24)+(0.91+j0)
= 1.33 +j0.555 p.u.
Vs= 1.44p.u.
or this equals 1.44x 11kV = 15.84kV
18
19
Revision
Single Phase Systems
20
21
Some Notes and Worked Examples - Handwritten Material (Note 3.3)

Per Unit values

Selected P.U. fault studies
22