UPS basics
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What is a UPS?
UPS is a power supply system with energy storage which ensures a permanent supply of the load in case of a supply voltage breakdown (according to EN 50091-1).
What is a UPS needed for?
For protection against data loss and system damage due to power failures, voltage dips, voltage spikes, undervoltage, overvoltage, switching spikes, interference voltages, frequency changes and harmonic distortion.
This list is also referred to as the - 9 voltage problems -
What sort of different UPS systems exist?
a) VFI - systems
Former designation: Online systems or also continuous converter systems
VFI = Output Voltage and Frequency Independent from mains supply.
The UPS output is independent of mains voltage and frequency fluctuations.
Protects against all 9 voltage problems and is therefore the professional protection level for critical applications.
The load is continuously supplied by an inverter - irrespective of the status of the alternating current input.
Quality of output voltage in accordance with IEC 62040-3:
Classification 1:
Within a time period from 100 µs to 5 µs, the output voltage is not allowed to deviate from the tolerance range of +/- 30% under any conditions!
This can only be achieved with true online systems.
b) VI - systems
Former designation: Line-interactive systems (also delta technology)
Output voltage independent from mains supply
The UPS output is dependent on mains frequency fluctuations, but the mains voltage is prepared by electronic / passive voltage control units.
Only protects against 5 of the 9 voltage problems (power failure, voltage dips, voltage spikes, undervoltage and overvoltage), therefore it represents medium-grade protection at a favourable price.
Applications:
Network systems in the office environment.
The mains frequency is also the frequency of the load (beware when supply is from a diesel generator). A low-impedance mains short circuit can result in a brief supply outage due to the switching procedure within the UPS.
Quality of output voltage in accordance with IEC 62040-3: Classification 2
c) VFD - systems
Former designation: Offline systems
Output Voltage and Frequency Dependent on mains supply.
The UPS output is dependent on mains voltage and frequency fluctuations.
Only protects against 3 of the 9 voltage problems (power failure, voltage dips, voltage spikes) and is the least expensive solution.
Applications: Individual work stations in the office environment.
An electronically stabilised voltage only supplies the load when the input AC voltage is outside the tolerance range. A switching operating within the UPS is required in each case - resulting in a supply outage!
Quality of output voltage in accordance with IEC 62040-3: Classifcation 3What is the crest factor?
The crest factor is the quotient of the peak value / r.m.s. value of an electrical parameter and for sinusoidal sequences is c = 1.414
Computer power supply units (SMPSs) accept a highly distorted current - the current peak is significantly higher than in a sinusoidal sequence. The required peak current for supplying computers, for example, is described by the crest factor and can have values of c = 3. The required peak current for supplying computers, for example, is described by the crest factor and can have values of c = 3.How is it possible to find the required crest factor of a UPS?
This question is closely connected to the required nominal power of a UPS (see next question).
How is it possible to find the required UPS power?
The load values are not usually uniform, but are made up of several terminal units which have to be protected. Here is an example:
UPS loads Power Current crest factor cos phi* Making current PC, server; monitors, printers, (SMPSs) 4500 VA 3 0,95 kap 1,5 x Inenn Air-conditioning units (motors) 3000 VA 1,41 0,8 ind 5 x Inenn Lighting 2000 VA 1,41 0,9 ind 6 x Inenn Miscellaneous 1500 VA 2 1 1 x Inenn Sum 11000 VA 2,14 0,95 ind 1 x Inenn (1) (2) (3) (4)
Calculation for this example:
(1) 4500 VA + 3000 VA + 2000 VA + 1500 VA = 11000 VA
(2) [4500 VA x 3 + 3000 VA x 1,41 + 2000 VA x 1,41 + 1500 VA x 2] / 11000 VA = 2,14
(3) { [4500 VA x 0,05 - 3000 VA x 0,2 - 2000 VA x 0,1] / 11000 VA } + 1 = 0,95ind. (4) [4500 VA x 1,5 + 3000 VA x 6 + 2000 VA + 1500 VA] / 11000 VA = 2,57 -> Ipeak = 2,57 x = 3,64
*The power of a UPS system is specified in VA. This power includes what is referred to as the reactive power. The power factor, also referred to as cos phi, is decisive in terms of establishing the correct dimensions. In practice, values used for the calculation are:
• between 0.6 and 0.7 (offline and line interactive UPS)
• between 0.7 and 0.8 (online continuous converter UPS)
Rule of thumb
UPS power (VA) x Power factor (cos phi) = UPS power (W)How do I select the UPS?
A) Depending on the possible level of damage in case of a data loss / production stoppage, critical applications require exclusively online UPSs, classification 1, in accordance with IEC 62040-3 (double conversion). You have to calculate the possible damage by conducting a risk analysis with your customer.
B) All other aspects such as low purchase and operating costs (efficiency) are - strictly speaking - secondary and must take second place to damage avoidance.
Continuation of the example above:Type PROTECT 1. 33 PROTECT 1. 33 bei 11 kVA required Nominal output 15 kVA 15 KVA > 11 kVA Crest factor 3 4,09 > 2,14 Power factor
cos phikap. ind.
0 - 1 - 0kap. ind.
0 - 1 - 0> 0,95 ind. Overload behaviour for starting / making current 4,09 >3,64 The selected UPS is suitable!
How do you find out the power factor?
There is only one solution: Collect the connection data of the loads that are to be used. UPSs from AEG PSS do not have a limitation on the cos phi in the inductive range, so this value is not critical. If the cos phi of the total loads is highly capacitive, it is recommended that this value should be calculated exactly because the nominal power of the UPS will be limited in this case.
The power of a UPS system is specified in VA. This power includes what is referred to as the reactive power. The power factor, also referred to as cos phi, is decisive in terms of establishing the correct dimensions. In practice, values used for the calculation are:
• between 0.6 and 0.7 (offline and line interactive UPS)
• between 0.7 and 0.8 (online continuous converter UPS)
Rule of thumb
UPS power (VA) x Power factor (cos phi) = UPS power (W)Output power in kVA or kW?
The nominal power of a UPS is defined by two values:
1. Apparent output power S in kVA
2. By the power factor e.g. cos phi = 0.8
The output power P in kW is defined by
cos phi = P / S → P = S x cos phi
Example:
Apparent output power S = 100 kVA at cos phi = 0.8
Output power P = 100 kVA x 0.8 = 80 kW
See also: The difference between watts and VAHow can the power losses of a UPS be calculated?
The decisive factors here are the max. required output power Pa that has to be supplied by the UPS (in kW) and the efficiency of the entire UPS, i.e. AC - AC at this operating point (e.g. 70% of nominal power).
Example:
Pv = Pa (kW !) x ( 1- ) / η
Pa = 11000 VA bei cos phi = 0,95
Pa = 11000 VA x 0,95 = 10.450 W
η = 92,5% in the operating point at 11000 VA
Pv = 10.450 W x (1 - 0,925) / 0,925 = 848 WHow high are the operating costs of a UPS?
A) Energy costs (example)
UPS nominal power: 330 kVA; average capacity utilisation:
75%
Pw load = 330 kVA x cos phi x 75%
330 kVA x 0.8 x 0.75 = 198 kW
Efficiency η in this operating point: 93.8%
Pv UPS = Pw load x ( 1 - η) / η = 198 kW x ( 1 - 0.938) / 0.938 = 13.1 kW
Energy costs / year assuming an energy price of 5 cents / kWh and 8760 h / year:
E-costs / year: 13.1 kW x 8760 h x 0.05 EUR = 5738 EUR / year
B) Other costs
Maintenance costs, e.g. minimised and possible to calculate with our service contracts. Replacement of wearing parts:
Batteries after 3, 5, 8, 10 years depending on the EuroBat class
• fan after 4.5 - 8 years, depending on type
• aluminium electrolyte capacitors after > 10 years, depending on operating temperatureWhy should the UPS (inverter) deliver a high short-circuit current?
A high short-circuit current means that the inverter can trip load breakers without switching over to the mains. This permits selective switch-off of the faulty load.
In addition, loads with high making currents (e.g. motors and transformers) can be switched on directly from the UPS without needing to use the mains for assistance.What advantage is provided by a UPS with a short-circuit-proof output?
If there is a fault (terminal short circuit) at the UPS output, the current is limited to a maximum value that will not damage the device, meaning that the UPS will be immediately ready for use again after the external cause of the malfunction has been rectified.
The required copper cross sections are defined in VDE 0100 part 540:
Is an air-conditioning unit required for the UPS system?
There is no need for the installation room to be air-conditioned. However, the installation room must be sufficiently ventilated to allow the heat losses from the UPS system to be dissipated. However, the installation room must be sufficiently ventilated to allow the heat losses from the UPS system to be dissipated.
What must be considered when earthing the UPS?
Earthing of a device is a protective measure and serves to avoid impermissibly high touch voltages at freely accessible metal parts. Earthing of a device is a protective measure and serves to avoid impermissibly high touch voltages at freely accessible metal parts. UPS devices and systems with medium and high power levels must be connected to the existing PE systems using cables with sufficient cross section and identified with green/yellow insulation. Identified earthing screws (PE) are located on the devices.
Nominal current of the devices (A) < 24 32 54 98-158 198 292 391 528... Copper cross section (mm2) of the PE conductor 0,75 1,5 4 16-16 25 50 75 120 What are system disturbances?
System disturbances are caused by "harmonic currents" emitted by an electric load. System disturbances are caused by "harmonic currents" emitted by an electric load. System disturbances are caused by "harmonic currents" emitted by an electric load. System disturbances are caused by "harmonic currents" emitted by an electric load.
What harmonic currents are permitted with which mains impedances?
If the mains voltage is impermissibly distorted, this can disrupt other connected loads!
Draft standard IEC 6100-3-4 (for currents > 16A ; DRAFT status) therefore defines a ratio between the "power capacity" of a power system and the power of the causal element, as follows:
Short circuit ratio Rsce = Ssc / Sequ
Ssc : Short circuit power of the mains at the
mains connection point = Unom² / Z
Z = Impedance at the mains connection point
Sequ: Rated apparent power of the load = U x I equ
The mains impedance Z must be requested from the local power utility.
The higher the short circuit power of the mains (Ssc) compared to the rated power of the load (S equipment), the more harmonic currents are permitted (currents > 16 A):Ssc >= S equ
RatioTHD* (%) 33 26 - for devices with single-phase connections 66 16 for devices with three-phase / symmetrical connections 120 18 175 25 250 35 350 48 450 58 600 70 * THD: Total harmonic distortion