## Per-Unit System of Units

### 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 typically 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.

$$\text{basevalueinp}\text{.u}\text{.=}\frac{\text{quantityexpressedinSIunits}}{\text{basevalue}}$$

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.

$$\begin{array}{l}\text{basecurrent=}\frac{\text{basepower}}{\text{basevoltage}}\\ \text{baseimpedance=}\frac{\text{basevoltage}}{\text{basecurrent}}\text{=}\frac{{\text{(basevoltage)}}^{2}}{\text{basepower}}\end{array}$$

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 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

$$\begin{array}{l}\text{baseinstantaneouscurrent=(basermscurrent)}\times \text{}\sqrt{2}\\ \text{basefluxlinkage=}\frac{(\text{basermsvoltage)}\times \text{}\sqrt{2}}{2\pi \times (\text{basefrequency)}}\text{}\end{array}$$

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

$$\text{basetorque=}\frac{\text{basepower(3phases)inVA}}{\text{basespeedinradians/second}}$$

Instead of specifying the rotor inertia in kg*m^{2}, you
would generally give the inertia constant *H* defined as

$$\begin{array}{c}H=\frac{\text{kineticenergystoredintherotoratsynchronousspeedinjoules}}{\text{machinenominalpowerinVA}}\\ H=\frac{\frac{1}{2}\times J\cdot {w}^{2}}{Pnom}\end{array}$$

The inertia constant is expressed in seconds. For large machines, this constant is
around 3–5 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 0.5–0.7
seconds.

### Example 1: Three-Phase Transformer

Consider, for example, a three-phase two-winding transformer with these manufacturer-provided, typical parameters:

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 a three-phase, four-pole Asynchronous Machine block in SI units. 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.m^{2}.

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 (three-phase) | 746*3/188.5 = 11.87 newton-meters |

Using the 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.

$$H=\frac{\frac{1}{2}\times J\cdot {w}^{2}}{Pnom}=\frac{\frac{1}{2}\times 0.089\times {(188.5)}^{2}}{3\times 746}=0.7065\text{seconds}$$

If you open the dialog box of the Asynchronous Machine block in pu units provided in the Machines library of the Simscape™ Electrical™ Specialized Power Systems Fundamental Blocks library, you find that the parameters in pu are the ones calculated.

### 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 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 3–300 kVA power range, the leakage reactance varies approximately 0.01–0.03 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 0.03–0.12 pu, whereas the winding resistances vary between 0.005–0.002 pu.

Similarly, for salient pole synchronous machines, the synchronous reactance

*X*_{d}is generally 0.60–1.50 pu, whereas the subtransient reactance*X'*_{d}is generally 0.20–0.50 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 considering the transformer ratios.