Let’s understand the concept of per unit system by solving an example. In the one-line diagram below, the impedance of various components in a power system, typically derived from their nameplates, are presented. The task now is to normalize these values using a common base.

Now that you have carefully examined the system and its parameters, the equivalent impedance diagram for the above system would look something like the following.

Resistive impedance for most components have been ignored. Rotating machines have been replaced with a voltage source behind their internal reactance. Capacitive effects between lines and to ground are ignored as well.

To obtain the new normalized per unit impedances, first we need to figure out the base values (Sbase, Vbase, Zbase) in the power system. Following steps will lead you through the process.

### Step 1: Assume a system base

Assume a system wide S_{base} of 100MVA. This is a random assumption and chosen to make calculations easy when calculating the per unit impedances.

So, S_{base} = 100MVA

### Step 2: Identify the voltage base

Voltage base in the system is determined by the transformer. For example, with a 22/220kV voltage rating of T1 transformer, the V_{base} on the primary side of T1 is 22kV while the secondary side is 220kV. It does not matter what the voltage rating of the other components are that are encompassed by the V_{base} zone.

See figure below for the voltage bases in the system.

### Step 3: Calculate the base impedance

The base impedance is calculated using the following formula:

Z_{base}=\frac{{kV_{base}}^2}{S_{base MVA}} Ohms…..(1)

For T-Line 1: Z_{base}=\frac{(220)^2}{100}= 484 Ohms

For T-Line 2: Z_{base}=\frac{(110)^2}{100}= 121 Ohms

For 3-phase load: Z_{base}=\frac{(11)^2}{100}= 1.21 Ohms

### Step 4: Calculate the per unit impedance

The per unit impedance is calculated using the following formulas:

Z_{p.u.}=\frac{Z_{actual}}{Z_{base}} …..(2)

Z_{p.u._{new}}=Z_{p.u._{old}}(\frac{S_{base_{new}}}{S_{base_{old}}})(\frac{V_{base_{old}}}{V_{base_{new}}})^2 …..(3)

The voltage ratio in equation (3) is not equivalent to the transformers voltage ratio. It is the ratio of the transformer’s voltage rating on the primary or secondary side to the system nominal voltage on the same side.

For T-line 1 using equation (2): X_{l1_{p.u.}}=\frac{48.4}{484}= 0.1 pu

For T-line 2 using equation (2): X_{l2_{p.u.}}=\frac{65.43}{121}= 0.5 pu

For 3-Phase load:

Power Factor: \cos^{-1}(0.6)=\angle{53.13}

Thus, S_{3\phi}(load)=57\angle{53.13}

Z_{act}=\frac{(V_{rated})^2}{\overline{S}^*}= \frac{10.45^2}{57\angle{-53.13}}= 1.1495+j1.53267 Ohms

Per unit impedance of 3-phase load using equation (2)= \frac{1.1495+j1.5326}{1.21} = 0.95+j1.2667 pu

For generator, the new per unit reactance using equation (3)

X_{sg}= 0.18(\frac{100}{90})(\frac{22}{22})^2= 0.2 pu

For transformer T1: X_{t1}= 0.1(\frac{100}{50})(\frac{22}{22})^2 = 0.2 pu

For transformer T2: X_{t2}= 0.06(\frac{100}{40})(\frac{220}{220})^2 = 0.15 pu

For transformer T3: X_{t3}= 0.064(\frac{100}{40})(\frac{22}{22})^2 = 0.16 pu

For transformer T4: X_{t4}= 0.08(\frac{100}{40})(\frac{110}{110})^2 = 0.2 pu

For Motor, X_{sm}= 0.185(\frac{100}{66.5})(\frac{10.45}{11})^2 = 0.25 pu

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To view full load amps due to motor load and inductive load at Bus 2, see this post.

### Summary

- Assume a Sbase for the entire system.
- The Vbase is defined by the transformer and any off-nominal tap setting it may have.
- Zbase is derived from the Sbase and Vbase.
- The new per unit impedance is obtained by converting the old per unit impedance on old base values to new ones. See equations (2) and (3).

samanthaplease help: Three transformers each rated 25 MVA, 38. 1 /3.81 kV are connected star-delta with a balanced load of three 0.6?, Y-connected resistors. Choose a base of 75 MVA, 66 kV for the high-voltage side of the transformer and specify the base for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then, determine the load resistance R_L i n ohms referred to the high-voltage side and the per-unit value of this resistance on the chosen base.

daranplease add more

RustyShouldn’t the 3-ph load be multiplied by (100/57)?

Mian FaizanLove it.. A job well done. 🙂

PAUL PEREZHELP! How do determine the equivalent three phase impedance of three single phase transformer? Example. Three each of 10 kva transformer, 7.52kv-240V, wiith an individual impedance of 4%. What is the equivalent impedance when they are connected in three phase at 30 kva.

Vanjinathan Pgood thing, to see a dedicated power system website.

sayonsomHow does the calculations change if the transformers are Y-Y or Y-Delta connected in different parts of the network?

Aleen Mohammed@sayonsom

What we developed in this exercise is a positive sequence network. A transformer’s winding configuration does not affect it. The positive sequence network is good for calculating balanced load current and 3 phase faults (not involving ground).

The zero sequence network however does get affected by transformer winding configuration. Delta and ungrounded Y have a big impact in the design of zero sequence impedance network.

You will need a zero sequence network for unsymmetrical fault current analysis like L-G or L-L-G etc.

I am getting into symmetrical components with this discussion.

ShaliniAwesome !!! Best explanation ever.. thank u

Nazgood job. its a comprehensive article. Appreciate your hard work

JathinThanks a lot.. Nice elaboration..