The monitoring functions with a generator
Due to the specific characteristics of the generator and its regulation, the proper operating parameters of the generator set must be monitored when special loads are implemented.
The behaviour of the generator is different from that of the transformer:
- The active power it supplies is optimised for a power factor = 0.8
- At less than power factor 0.8, the generator may, by increased excitation, supply part of the reactive power
Capacitor bank
An off-load generator connected to a capacitor bank may self-excite, consequently increasing its overvoltage.
The capacitor banks used for power factor regulation must therefore be disconnected. This operation can be performed by sending the stopping setpoint to the regulator (if it is connected to the system managing the source switchings) or by opening the circuit-breaker supplying the capacitors.
If capacitors continue to be necessary, do not use regulation of the power factor relay in this case (incorrect and over-slow setting).
Motor restart and re-acceleration
A generator can supply at most in transient period a current of between 3 and 5 times its nominal current.
A motor absorbs roughly 6 In for 2 to 20 s during start-up.
If the sum of the motor power is high, simultaneous start-up of loads generates a high pick-up current that can be damaging. A large voltage drop, due to the high value of the generator transient and subtransient reactances will occur (20% to 30%), with a risk of:
- Non-starting of motors
- Temperature rise linked to the prolonged starting time due to the voltage drop
- Tripping of the thermal protection devices
Moreover, all the network and actuators are disturbed by the voltage drop.
Application
(see Fig. N7)
A generator supplies a set of motors.
Generator characteristics: Sn = 130 kVA at a power factor of 0.8, Un=500V
In = 150 A
x’d = 20% (for example) hence Isc = 750 A.
- If the Σ Pmotors is 45 kW (45% of generator power)
Calculating voltage drop at start-up:
Σ PMotors = 45 kW, Im = 81 A, hence a starting current Id = 480 A for 2 to 20 s.
Voltage drop on the busbar for simultaneous motor starting:
ΔU = 55%
which is not tolerable for motors (failure to start).
- If the Σ Pmotors is 20 kW (20% of generator power)
Calculating voltage drop at start-up:
Σ PMotors = 20 kW, Im = 35 A, hence a starting current Id = 210 A for 2 to 20 s.
Voltage drop on the busbar:
ΔU = 10%
which is high but tolerable (depending on the type of loads).
Restarting tips
- If the Pmax of the largest motor > , a soft starter must be installed on this motor
- If Σ Pmotors > , motor cascade restarting must be managed by a PLC
- If Σ Pmotors > , there are no restarting problems
Non-linear loads – Example of a UPS
Non-linear loads
These are mainly:
- Saturated magnetic circuits
- Discharge lamps, fluorescent lights
- Electronic converters
- Information Technology Equipment: PC, computers, etc.
These loads generate harmonic currents: supplied by a Generator Set, this can create high voltage distortion due to the low short-circuit power of the generator.
Uninterruptible Power Supply (UPS)
(see Fig. N8)
The combination of a UPS and generator set is the best solution for ensuring quality power supply with long autonomy for the supply of sensitive loads.
It is also a non-linear load due to the input rectifier. On source switching, the autonomy of the UPS on battery must allow starting and connection of the Generator Set.
UPS power
UPS inrush power must allow for:
- Nominal power of the downstream loads. This is the sum of the apparent powers Pa absorbed by each application. Furthermore, so as not to oversize the installation, the overload capacities at UPS level must be considered (for example: 1.5 In for 1 minute and 1.25 In for 10 minutes)
- The power required to recharge the battery: This current is proportional to the autonomy required for a given power. The sizing Sr of a UPS is given by: Sr = 1.17 x Pn
Figure N9 defines the pick-up currents and protection devices for supplying the rectifier (Mains 1) and the standby mains (Mains 2).
Nominal power | Current value (A)
Mains 1 with 3Ph battery |
Mains 2 or 3Ph application |
---|---|---|
Pn (kVA) | 400 V - I1 | 400 V - Iu |
40 | 86 | 60.5 |
60 | 123 | 91 |
80 | 158 | 121 |
100 | 198 | 151 |
120 | 240 | 182 |
160 | 317 | 243 |
200 | 395 | 304 |
250 | 493 | 360 |
300 | 590 | 456 |
400 | 793 | 608 |
500 | 990 | 760 |
600 | 1,180 | 912 |
800 | 1,648 | 1,215 |
Generator Set/UPS combination
- Restarting the Rectifier on a Generator Set
The UPS rectifier can be equipped with a progressive starting of the charger to prevent harmful pick-up currents when installation supply switches to the Generator Set (see Fig. N10).
- Harmonics and voltage distortion
Total voltage distortion τ is defined by:
where Uh is the harmonic voltage of order h.
This value depends on:
- The harmonic currents generated by the rectifier (proportional to the power Sr of the rectifier)
- The longitudinal subtransient reactance X”d of the generator
- The power Sg of the generator
We define the generator relative short-circuit voltage, brought to rectifier power, i.e. t = f(U’Rcc).
Note 1: As subtransient reactance is great, harmonic distortion is normally too high compared with the tolerated value (7 to 8%) for reasonable economic sizing of the generator: use of a suitable filter is an appropriate and cost-effective solution.
Note 2: Harmonic distortion is not harmful for the rectifier but may be harmful for the other loads supplied in parallel with the rectifier.
Application
A chart is used to find the distortion τ as a function of U’Rcc (see Fig. N11).
The chart gives:
- Either τ as a function of U’Rcc
- Or U’Rcc as a function of τ
From which generator set sizing, Sg, is determined.
Example: Generator sizing
- 300 kVA UPS without filter, subtransient reactance of 15%
The power Sr of the rectifier is Sr = 1.17 x 300 kVA = 351 kVA
For a τ < 7%, the chart gives U’Rcc = 4%, power Sg is:
- 300 kVA UPS with filter, subtransient reactance of 15%
For τ = 5%, the calculation gives U’Rcc = 12%, power Sg is:
Note: With an upstream transformer of 630 kVA on the 300 kVA UPS without filter, the 5% ratio would be obtained.
The result is that operation on generator set must be continually monitored for harmonic currents.
If voltage harmonic distortion is too great, use of a filter on the network is the most effective solution to bring it back to values that can be tolerated by sensitive loads.