Per Unit (PU) System

Usually in the analysis of power systems, actual values of quantities are expressed as fractions of
reference quantities such as rated or full-load values. These fractions are called per unit (p.u.).
Actual value in any unit
The p.u. value of any quantity = (1)
Base or reference value in the same unit
Advantages:
1. 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.
2. The use of √3′s in three-phase calculations is reduced.
3. By the choice of appropriate voltage bases the solution of networks containing several
transformers is facilitated.
4. Per-unit values lend themselves more readily to digital computation.
• Resistance and Impedance
𝑉𝑏
𝑅𝑏 = (2)
𝐼𝑏

where 𝑅𝑏 is the base value of resistance, 𝑉𝑏 is the base value of voltage and 𝐼𝑏 the base value of
current.
From (1) and (2)
𝑅(𝛺) 𝑅(𝛺) 𝑅(𝛺).𝐼𝑏 Voltage drop across 𝑅 at base current
𝑅𝑝.𝑢. = 𝑅 =𝑉 = = (3)
𝑏 (𝛺) 𝑏 /𝐼𝑏 𝑉𝑏 Base voltage

and

𝑅(𝛺). 𝐼𝑏2 Power loss at base current

𝑅𝑝.𝑢. = = (4)
𝑉𝑏 𝐼𝑏 Base power or voltage − 𝑎𝑚𝑝𝑒𝑟𝑒𝑠

2
Power loss (p.u.) at 𝐼𝑝.𝑢. current = 𝑅𝑝.𝑢. 𝐼𝑝.𝑢. (5)

𝑍(𝛺) 𝑍(𝛺) 𝑍(𝛺)𝐼𝑏

𝑝. 𝑢. 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 = = = (6)
𝑍𝑏 (𝛺) 𝐵𝑎𝑠𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑉𝑏
𝐵𝑎𝑠𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
• Three-Phase Circuits
In three-phase circuits, a p.u. phase voltage has the same numerical value as the corresponding
p.u. line voltage.
For example, with a measured line voltage of 100 kV and a rated line voltage of 132 kV,
100
p. u. value = = 0.76
132
100 132
The equivalent phase voltages are kV and kV and hence the p.u. value is again 0.76.
√3 √3

The actual values of 𝑅, 𝑋𝐿 and 𝑋𝐶 for lines, cables, and other apparatus are phase values.
It is convenient in a.c. circuit calculations to work in terms of base volt-amperes, (𝑆𝑏𝑎𝑠𝑒 ). Thus
𝑆𝑏𝑎𝑠𝑒 = 𝑉𝑏𝑎𝑠𝑒 × 𝐼𝑏𝑎𝑠𝑒 × √3
where 𝑉𝑏𝑎𝑠𝑒 is the line voltage and 𝐼𝑏𝑎𝑠𝑒 is the line current in a three-phase system.
Hence
𝑆𝑏𝑎𝑠𝑒
𝐼𝑏𝑎𝑠𝑒 = (7)
√3𝑉𝑏𝑎𝑠𝑒

Equation (7) shows that if 𝑆𝑏𝑎𝑠𝑒 and 𝑉𝑏𝑎𝑠𝑒 are specified, then 𝐼𝑏𝑎𝑠𝑒 is determined. Only two base
quantities can be chosen from which all other quantities in a three-phase system are calculated. Thus,
2
𝑉𝑏𝑎𝑠𝑒 /√3 𝑉𝑏𝑎𝑠𝑒 /√3 𝑉𝑏𝑎𝑠𝑒
𝑍𝑏𝑎𝑠𝑒 = = = (8)
𝐼𝑏𝑎𝑠𝑒 𝑆𝑏𝑎𝑠𝑒 /√3𝑉𝑏𝑎𝑠𝑒 𝑆𝑏𝑎𝑠𝑒

Hence
𝑍(Ω) × 𝑆𝑏𝑎𝑠𝑒
𝑍𝑝.𝑢. = 2 (9)
𝑉𝑏𝑎𝑠𝑒

Equation (9) shows that 𝑍𝑝.𝑢. is directly proportional to the base VA and inversely proportional to
the base voltage squared.
𝑍𝑝.𝑢. to a new base VA can be calculated as

𝑆𝑛𝑒𝑤 𝑏𝑎𝑠𝑒
𝑍𝑝.𝑢. (𝑛𝑒𝑤 𝑏𝑎𝑠𝑒) = 𝑍𝑝.𝑢. (𝑜𝑙𝑑 𝑏𝑎𝑠𝑒) × (10)
𝑆𝑜𝑙𝑑 𝑏𝑎𝑠𝑒

𝑍𝑝.𝑢. to a new voltage base can be calculated as

2
𝑉𝑜𝑙𝑑 𝑏𝑎𝑠𝑒
𝑍𝑝.𝑢. (𝑛𝑒𝑤 𝑏𝑎𝑠𝑒) = 𝑍𝑝.𝑢. (𝑜𝑙𝑑 𝑏𝑎𝑠𝑒) × 2 (11)
𝑉𝑛𝑒𝑤 𝑏𝑎𝑠𝑒
• Transformers
Consider a single-phase transformer in which the total series impedance of the two windings
𝑍
referred to the primary is 𝑍1 . Then the p.u. impedance 𝑍𝑝.𝑢. = 𝑉 /𝐼1 , where 𝐼1 , and 𝑉1, are the base
1 1

values of the primary circuit.

The ohmic impedance referred to the secondary is

𝑍2 = 𝑍1 𝑁 2 (𝛺)
and this in p.u. notation is
𝑍1 𝑁 2
𝑍𝑝.𝑢. =
𝑉2 /𝐼2
where 𝑉2 and 𝐼2 are base voltage and current of the secondary circuit.
If they are related to the base voltage and current of the primary by the turns ratio of the transformer
then.
𝐼1 1 𝑍1 𝐼1
𝑍𝑝.𝑢. = 𝑍1 𝑁 2 ∗ =
𝑁 𝑉1 𝑁 𝑉1
where
𝑁2 𝑉2 𝐼1
𝑁= = =
𝑁1 𝑉1 𝐼2
Hence provided the base voltages on each side of a transformer are related by the turns ratio, the
p.u. impedance of a transformer is the same whether considered from the primary or the secondary
side.
• Per Unit Impedance Diagram of a Power System
From a one-line diagram of a power system the impedance diagram can be directly drawn by the
following steps:
1. Choose an appropriate common MVA (or kVA) base for the system.
2. Consider the system to be divided into a number of sections by the transformers.
Choose an appropriate kV base in one of the sections. Calculate kV bases of other sections in the
ratio of transformation.
3. Calculate per unit values of voltages and impedances in each section and connect them up as per
the topology of the one-line diagram. The result is the single-phase per unit impedance diagram.
Example:
In the network of figure below, two single-phase transformers supply a 10 kVA resistance load at
200 V. Show that the p.u. load is the same for each part of the circuit and calculate the voltage at point
D.

Solution:
2002
𝑇ℎ𝑒 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = =4𝑉
10 × 103
In each of the circuits A, B, and C a different voltage exists, so that each circuit will have its own
base voltage:
100 V in A, 400 V in B and 200 V in C.
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 simple p.u. framework
breaks down. The same volt-ampere base is used for all the circuits as 𝑉1 𝐼1 and 𝑉2 𝐼2 on each side of
a transformer and is taken in this case as 10 kVA. The per unit impedances of the transformers are
already on their individual equipment bases of 10 kVA and so remain unchanged.
The base impedance in C
2
𝑉𝑏𝑎𝑠𝑒 2002
𝑍𝑏𝑎𝑠𝑒 = = =4𝛺
𝑏𝑎𝑠𝑒 𝑉𝐴 10000
4
𝑇ℎ𝑒 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑝. 𝑢. )𝑖𝑛 𝐶 = = 1 𝑝. 𝑢.
4
 In B the base impedance is
2
𝑉𝑏𝑎𝑠𝑒 4002
𝑍𝑏𝑎𝑠𝑒 = = = 16 𝛺
𝑏𝑎𝑠𝑒 𝑉𝐴 10000
and
𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 (𝑖𝑛 𝑜ℎ𝑚𝑠)𝑟𝑒𝑓𝑒𝑟𝑟𝑒𝑑 𝑡𝑜 𝐵 = 4 × 𝑁 2 = 4 × 22 = 16 𝛺
Hence
16
𝑡ℎ𝑒 𝑝. 𝑢. 𝑙𝑜𝑎𝑑 𝑟𝑒𝑓𝑒𝑟𝑟𝑒𝑑 𝑡𝑜 𝐵 = = 1 𝑝. 𝑢.
16
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 the following figure. 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). The base current
at voltage level C of this single-phase circuit
𝑏𝑎𝑠𝑒 𝑉𝐴 10000
𝐼𝑏𝑎𝑠𝑒 = = = 50 𝐴
𝑉𝑏𝑎𝑠𝑒 200

The corresponding base currents in the other circuits are 25 A in B, and 100 A in A.
The actual load current is 200V/4 Ω = 50 𝐴 = 1 𝑝. 𝑢.
Hence the supply voltage 𝑉𝑆
𝑉𝑆 = 1(𝑗0.1 + 𝑗0.15) + 1 p. u.

𝑉𝑆 = √12 + 0.252 = 1.03 p. u. = 1.03 × 100 = 103 V

The voltage at point D
𝑉𝐷 = 1(𝑗0.15) + 1 p. u.

𝑉𝐷 = √12 + 0.152 = 1.012 p. u. = 1.012 × 400 = 404.8 V

It is a useful exercise to repeat this example using ohms, volts and amperes. A summary table of the
transformation of the circuit into per unit is shown below.
Example:
For the following three-phase system, the ratings and reactances of the various components are shown.
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. Calculate the terminal voltage of the synchronous machine.

Solution:
The voltage bases of the various circuits are decided by the nominal transformer voltages, that is,
11, 132, and 33 kV. A base of 100 MVA will be used for all circuits. The reactances are expressed on
the appropriate voltage and MVA bases.
2
V𝑏𝑎𝑠𝑒 (132 × 103 )2
𝐵𝑎𝑠𝑒 𝑖𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 = = = 174 Ω
𝑆𝑏𝑎𝑠𝑒 100 × 106
Hence
𝑗100
𝑡ℎ𝑒 𝑝. 𝑢. 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 = = 𝑗0.575 p. u.
174

100
𝑃𝑒𝑟 𝑢𝑛𝑖𝑡 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑛𝑑𝑖𝑛𝑔 − 𝑒𝑛𝑑 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 = 𝑗0.1 × = 𝑗0.2 𝑝. 𝑢.
50
100
𝑃𝑒𝑟 𝑢𝑛𝑖𝑡 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑐𝑒𝑖𝑣𝑖𝑛𝑔 − 𝑒𝑛𝑑 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 = 𝑗0.12 × = 𝑗0.24 𝑝. 𝑢.
50
50 × 106
𝐿𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = = 1203 𝐴
√3 × 30 × 103 × 0.8
100 × 106
𝐵𝑎𝑠𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑓𝑜𝑟 33 𝑘𝑉, 100 𝑀𝑉𝐴 = = 1750 𝐴
√3 × 33 × 103
Hence
 1203
𝑡ℎ𝑒 𝑝. 𝑢. 𝑙𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = = 0.687 𝑝. 𝑢.
1750

30
𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 𝑏𝑢𝑠𝑏𝑎𝑟 = = 0.91 𝑝. 𝑢.
33

𝑉𝑆 = 0687 × (0.8 − 𝑗0.6)(𝑗0.2 + 𝑗0.575 + 𝑗0.24) + (𝑗0.91 + 𝑗0) = 1.328 + 𝑗0.558 𝑝. 𝑢.

𝑉𝑆 = √1.3282 + 0.5582 = 1.44 𝑝. 𝑢. = 1.44 × 11 𝑘𝑉 = 15.84 𝑘𝑉

Example:
Obtain the per unit impedance (reactance) diagram of the following power system.

Solution:
The per phase impedance diagram of the power system has been drawn in the following figure.
Simplifying assumptions:
1. Line capacitance and resistance are neglected. 2. Assume that the impedance diagram is meant
for short circuit studies. Current drawn by static loads under short circuit conditions can be
neglected. Loads A and B are therefore ignored.
A common three-phase MVA base of 30 and a voltage base of 33 kV line-to-line are selected on
the transmission line. Then the voltage base in the circuit of generator 1 is 11 kV line-to-line and
that in the circuits of generators 2 and 3 is 6.2 kV. The per unit reactances of various components
are calculated as
 20.5×30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑙𝑖𝑛𝑒 = = 0.564 p. u
332
15.2×30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑇1 = = 0.418 p. u
332
16×30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑇2 = = 0.44 p. u
332
1.6×30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐺1 = = 0.396 p. u
112
1.2×30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐺2 = = 0.939 p. u
6.22
0.56×30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐺3 = = 0.437 p. u
6.22

Example:
The reactance data of generators and transformers is usually specified in pu (or percent) values,
based on equipment ratings rather than in actual ohmic values as given in the previous example; while
the transmission line impedances may be given in actual values. Let us resolve the previous example
assuming the following pu values of reactances.
Transformer T1: 0.209; Transformer T2: 0.220; Generator G1: 0.435; Generator G2: 0.413;
Generator G3: 0.3214
Solution:
With a base MVA of 30, base voltage of 11 kV in the circuit of generator 1 and base voltage of 6.2
kV in the circuit of genera tors 2 and 3 as used in the previous example, the pu values of the reactances
of transformers and generators are calculated as
30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑇1 = 0.209 × 15 = 0.418 p. u
30
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑇2 = 0.22 × 15 = 0.44 p. u
10.52
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐺1 = 0.435 × = 0.396 p. u
112
30 6.62
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐺2 = 0.413 × 15 × 6.22 = 0.939 p. u
30 6.62
𝑇ℎ𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐺3 = 0.3124 × 25 × 6.22 = 0.437 p. u