Per-Unit System of Units

International System of Units

Simscape™ Electrical™ Specialized Power Systems software uses the International System of Units (SI), as described in the following table.

Quantity	Unit	Symbol
Time	second	s
Length	meter	m
Mass	kilogram	kg
Energy	joule	J
Current	ampere	A
Voltage	volt	V
Active power	watt	W
Apparent power	volt-ampere	VA
Reactive power	var	var
Impedance	ohm	Ω
Resistance	ohm	Ω
Inductance	henry	H
Capacitance	farad	F
Flux linkage	volt-second	V. s
Rotation speed	radians per second revolutions per minute	rad/s rpm
Torque	newton-meter	N.m
Inertia	kilogram-meter²	kg.m²
Friction factor	newton-meter-second	N.m.s

What Is the Per-Unit System?

The per-unit system is widely used in the power system industry 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.

$base value in p .u . = \frac{quantity expressed in SI units}{base value}$

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.

$\begin{array}{l} base current = \frac{base power}{base voltage} \\ base impedance = \frac{base voltage}{base current} = \frac{{(base voltage)}^{2}}{base power} \end{array}$

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). However, according to the above definitions, there are as many base values as windings for voltages, currents, and impedances.

The saturation characteristic of saturable transformer is given in the form of an instantaneous current versus instantaneous flux-linkage curve: [i1 phi1; i2 phi2; ... , in phin].

When the Per-Unit system is used to specify the transformer R L parameters, the flux linkage and current in the saturation characteristic must be also specified in pu. The corresponding base values are

$\begin{array}{l} base instantaneous current = (base rms current) \times \sqrt{2} \\ base flux linkage = \frac{(base rms voltage) \times \sqrt{2}}{2 π \times (base frequency)} \end{array}$

where current, voltage, and flux linkage are expressed respectively in volts, amperes, and volt-seconds.

For AC machines, the torque and speed can be also expressed in pu. The following base quantities are chosen:

The base speed = synchronous speed
The base torque = torque corresponding at base power and synchronous speed
$base torque = \frac{base power (3 phases) in VA}{base speed in radians/second}$

Instead of specifying the rotor inertia in kg*m², you would generally give the inertia constant H defined as

$\begin{matrix} H = \frac{kinetic energy stored in the rotor at synchronous speed in joules}{machine nominal power in VA} \\ H = \frac{\frac{1}{2} \times J \cdot w^{2}}{P n o m} \end{matrix}$

The inertia constant is expressed in seconds. For large machines, this constant is around 3 to 5 seconds. An inertia constant of 3 seconds means that the energy stored in the rotating part could supply the nominal load during 3 seconds. For small machines, H is lower. For example, for a 3 HP motor, it can be between 0.5 and 0.7 second.

Example 1: Three-Phase Transformer

Consider, for example, a three-phase two-winding transformer. The following typical parameters could be provided by the manufacturer:

Nominal power = 300 kVA total for three phases
Nominal frequency = 60 Hz
Winding 1: connected in wye, nominal voltage = 25 kV RMS line-to-line
resistance 0.01 pu, leakage reactance = 0.02 pu
Winding 2: connected in delta, nominal voltage = 600 V RMS line-to-line
resistance 0.01 pu, leakage reactance = 0.02 pu
Magnetizing losses at nominal voltage in % of nominal current:
Resistive 1%, Inductive 1%

The base values for each single-phase transformer are first calculated:

For winding 1:

Base power	300 kVA/3 = 100e3 VA/phase
Base voltage	25 kV/sqrt(3) = 14434 V RMS
Base current	100e3/14434 = 6.928 A RMS
Base impedance	14434/6.928 = 2083 Ω
Base resistance	14434/6.928 = 2083 Ω
Base inductance	2083/(2π*60)= 5.525 H

For winding 2:

Base power
300 kVA/3 = 100e3 VA
Base voltage
600 V RMS
Base current
100e3/600 = 166.7 A RMS
Base impedance
600/166.7 = 3.60 Ω
Base resistance
600/166.7 = 3.60 Ω
Base inductance
3.60/(2π*60) = 0.009549 H

The values of the winding resistances and leakage inductances expressed in SI units are therefore

For winding 1: R1= 0.01 * 2083 = 20.83 Ω; L1= 0.02*5.525 = 0.1105 H
For winding 2: R2= 0.01 * 3.60 = 0.0360 Ω; L2= 0.02*0.009549 = 0.191 mH

For the magnetizing branch, magnetizing losses of 1% resistive and 1% inductive mean a magnetizing resistance Rm of 100 pu and a magnetizing inductance Lm of 100 pu. Therefore, the values expressed in SI units referred to winding 1 are

Rm = 100*2083 = 208.3 kΩ
Lm = 100*5.525 = 552.5 H

Example 2: Asynchronous Machine

Now consider the three-phase four-pole Asynchronous Machine block in SI units provided in the Machines library of Fundamental Blocks library. It is rated 3 HP, 220 V RMS line-to-line, 60 Hz.

The stator and rotor resistance and inductance referred to stator are

Rs = 0.435 Ω; Ls = 2 mH
Rr = 0.816 Ω; Lr = 2 mH

The mutual inductance is Lm = 69.31 mH. The rotor inertia is J = 0.089 kg.m².

The base quantities for one phase are calculated as follows:

Base power	3 HP*746VA/3 = 746 VA/phase
Base voltage	220 V/sqrt(3) = 127.0 V RMS
Base current	746/127.0 = 5.874 A RMS
Base impedance	127.0/5.874 = 21.62 Ω
Base resistance	127.0/5.874 = 21.62 Ω
Base inductance	21.62/(2π*60)= 0.05735 H = 57.35 mH
Base speed	1800 rpm = 1800*(2π)/60 = 188.5 radians/second
Base torque (3-phase)	746*3/188.5 = 11.87 newton-meters

Using the above base values, you can compute the values in per units.

Rs= 0.435 / 21.62 = 0.0201 pu Ls= 2 / 57.35 = 0.0349 pu

Rr= 0.816 / 21.62 = 0.0377 pu Lr= 2 / 57.35 = 0.0349 pu

Lm = 69.31/57.35 = 1.208 pu

The inertia is calculated from inertia J, synchronous speed, and nominal power.

$H = \frac{\frac{1}{2} \times J \cdot w^{2}}{P n o m} = \frac{\frac{1}{2} \times 0.089 \times {(188.5)}^{2}}{3 \times 746} = 0.7065 seconds$

If you open the dialog box of the Asynchronous Machine block in pu units provided in the Machines library of Fundamental Blocks (powerlib) library, you find that the parameters in pu are the ones calculated above.

Base Values for Instantaneous Voltage and Current Waveforms

When displaying instantaneous voltage and current waveforms on graphs or oscilloscopes, you normally consider the peak value of the nominal sinusoidal voltage as 1 pu. In other words, the base values used for voltage and currents are the RMS values given above multiplied by $\sqrt{2}$ .

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.
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 X_d 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.

Base power	300 kVA/3 = 100e3 VA
Base voltage	600 V RMS
Base current	100e3/600 = 166.7 A RMS
Base impedance	600/166.7 = 3.60 Ω
Base resistance	600/166.7 = 3.60 Ω
Base inductance	3.60/(2π*60) = 0.009549 H