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FAQs Uninterruptible Power Supplies This paper contains general inform ation on questions relating to find the right UPS. Don't hesitate to send us an em ail [email protected] or call us +49 (0) 2902 763 154. We support you at finding the right UPS for your application. What is a UPS? A UPS is a pow er supply system w ith an energy store that ensures the load continues to be supplied even if the supply voltage fails (in accordance w ith EN 50091-1). What is a UPS needed for? For protection against data loss and system damage due to pow er failures, voltage dips, voltage spikes, undervoltage, overvoltage, sw itching spikes, interference voltages, frequency changes and harmonic distortion. This list is also referred to as the - 9 voltage problems - What sort of different UPS system s exist? a) VFI - system s 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 w ith IEC 62040-3: Classification 1: Within a time period from 100 µs to 5 µs, the output voltage is not allow ed to deviate from the tolerance range of +/- 30% under any conditions! This can only be achieved w ith true online systems. b) VI - system s 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 (pow er failure, voltage dips, voltage spikes, undervoltage and overvoltage), therefore it represents medium-grade protection at a favourable price. Applications: Netw ork systems in the office environment. The mains frequency is also the frequency of the load (bew are w hen supply is from a diesel generator). A low -impedance mains short circuit can result in a brief supply outage due to the sw itching procedure w ithin the UPS. Quality of output voltage in accordance w ith IEC 62040-3: Classification 2 c) VFD - system s 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 (pow er failure, voltage dips, voltage spikes) and is the least expensive solution. Applications: Individual w ork stations in the office environment. An electronically stabilised voltage only supplies the load w hen the input AC voltage is outside the tolerance range. A sw itching operating w ithin the UPS is required in each case - resulting in a supply outage! Quality of output voltage in accordance w ith IEC 62040-3: Classifcation 3 What 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 pow er 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 pow er of a UPS (see next question) How is it possible to find the required UPS pow er? The load values are not usually uniform, but are made up of several terminal units w hich have to be protected. Here is an example: UPS loads Pow er Current crest factor Cos phi* Making current PC, server, mmonitors, printers (SMPSs) 4500 VA 3 0,95 kap 1,5 x Inenn Air-conditioning untis (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) (2) (3) (4) 4500 VA + 3000 VA + 2000 VA + 1500 VA = 11000 VA [4500 VA x 3 + 3000 VA x 1,41 + 2000 VA x 1,41 + 1500 VA x 2] / 11000 VA = 2,14 { [4500 VA x 0,05 - 3000 VA x 0,2 - 2000 VA x 0,1] / 11000 VA } + 1 = 0,95ind. [4500 VA x 1,5 + 3000 VA x 6 + 2000 VA + 1500 VA] / 11000 VA = 2,57 -> Ipeak = 2,57 x = 3,64 * The pow er of a UPS system is specified in VA. This pow er includes w hat is referred to as the reactive pow er. The pow er factor, also referred to as cos phi, is decisive in terms of establishing the correct dimensions. In practice, values used for the calculation are: • betw een 0.6 and 0.7 (offline and line interactive UPS) • betw een 0.7 and 0.8 (online continuous converter UPS) Rule of thumb UPS pow er (VA) x Pow er factor (cos phi) = UPS pow er (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 w ith IEC 62040-3 (double conversion). You have to calculate the possible damage by conducting a risk analysis w ith 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. Nominal output Crest factor 15 kVA 3 PROTECT 1. bei 11 kVA 15 KVA 4,09 Power factor cos phi kap. ind. 0-1- 0 kap. ind. 0-1 - 0 > 0,95 ind. 4,09 >3,64 Overload behaviour for starting / making current required > 11 kVA > 2,14 The selected UPS is suitable! How do you find out the pow er 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 pow er of the UPS w ill be limited in this case. The pow er of a UPS system is specified in VA. This pow er includes w hat is referred to as the reactive pow er. The pow er factor, also referred to as cos phi, is decisive in terms of establishing the correct dimensions. In practice, values used for the calculation are: • betw een 0.6 and 0.7 (offline and line interactive UPS) • betw een 0.7 and 0.8 (online continuous converter UPS) Rule of thumb: UPS pow er (VA) x Pow er factor (cos phi) = UPS pow er (W) Output pow er in kVA or kW? The nominal pow er of a UPS is defined by tw o values: 1. Apparent output pow er S in kVA 2. By the pow er factor e.g. cos phi = 0.8 The output pow er P in kW is defined by cos phi = P / S → P = S x cos phi Example: Apparent output pow er S = 100 kVA at cos phi = 0.8 Output pow er P = 100 kVA x 0.8 = 80 kW See also: The difference betw een w atts and VA How can the pow er losses of a UPS be calculated? The decisive factors here are the max. required output pow er 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 pow er). 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 W How high are the operating costs of a UPS? A) Energy costs (example) UPS nominal pow er: 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 w ith our service contracts. Replacement of w earing 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 temperature Why should the UPS (inverter) deliver a high short -circuit current? A high short-circuit current means that the inverter can trip load breakers w ithout sw itching over to the mains. This permits selective sw itch-off of the faulty load. In addition, loads w ith high making currents (e.g. motors and transformers) can be sw itched on directly from the UPS w ithout needing to use the mains for assistance. What advantage is provided by a UPS w ith 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 w ill not damage the device, meaning that the UPS w ill 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. How ever, the installation room must be sufficiently ventilated to allow the heat losses from the UPS system to be dissipated. How ever, the installation room must be sufficiently ventilated to allow the heat losses from the UPS system to be dissipated. 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 w ith medium and high pow er levels must be connected to the existing PE systems using cables w ith sufficient cross section and identified w ith green/yellow insulation. Identified earthing screw s (PE) are located on the devic es. Nominal current devices (A) of the < 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. What harm onic currents are perm itted w ith w hich m ains im pedances? 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 betw een the "pow er capacity" of a pow er system and the pow er of the causal element, as follow s: Short circuit ratio Rsce = Ssc / Sequ Ssc : Short circuit pow er of the mains at the mains connection point = Unom² / Z Z = Impedance at the mains connection point Sequ: Rated apparent pow er of the load = U x I equ The mains impedance Z must be requested from the local pow er utility. The higher the short circuit pow er of the mains (Ssc) compared to the rated pow er of the load (S equipment), the more harmonic currents are permitted (currents > 16 A): Ssc >= S equ Ratio 33 THD* (%) 26 66 16 120 175 250 350 450 600 18 25 35 48 58 70 * THD: Total harmonic distortion - for devices with single-phase connections for devices with three-phase / symmetrical connections