presentation

Technical Presentation on the new Power Saving Unit (APSS) Introduced By Smart VisionAlmost all bulk electric power is generated, transported and consumed in an alternating current (AC) network. Elements of AC systems produce and consume two kinds of Power: supports the voltages that must be controlled for system reliabilityReal Power “P” (measured in watts) It accomplishes useful work (e.g., running motors)Reactive Power (measured in var).Electrical Power Typesunless you lift the arms(reactive power)To Simplify Things to all of us...You can’t move the wheelbarrow (active power delivery)Power Factor Power factor (Pf) = The cosine of the angle between Voltage and currentsignals = cos f.CurrentVoltage f fUseful currentMotor CurrentMagneticcurrent fP=Useful Power (KW) Q Magnetic Power (KVAr)S=Total Power (KVA)Power factor (Pf) = The cosine of the angle between useful power andTotal power = cos f = (P/S) Environmental benefit. Reduction of power consumption due to improved energy efficiency. Reduced power consumption means less greenhouse gas emissions and fossil fuel depletion by power stationsWhy improve Power Factor?The benefits that can be achieved by applying the suitable Power Factor correction are: Reduction of Electricity Bills Extra kVA available from the existing supply Reduction of I2R losses in transformers and distribution equipment Reduction of voltage drop in long cables . Extended equipment life – Reduced electrical burden on cables and electrical componentsElectricity BillPenalty EquationFor a PF between 0.72----0.92, the penalty per year in L.E (Pen.) is calculated as follows:Pen.=KWHr per year×(0.92-average PF)×0.5L.EFor a PF below 0.72, the penalty per year in L.E (Pen.) is calculated as follows:Pen.=KWHr per year×(0.92-average PF)×1.0L.ENumerical ExampleFor 500000 KWHr with PF 0.8, we getPen=30000 L.EFor 500000 KWHr with PF 0.6, we getPen=160000 L.EElectricity BillBonusFor a PF above 0.92, the Bonus per year in L.E is calculated as follows:Bonus=KWHr per year×(PF-0.92)×0.5L.ENumerical ExampleFor 500000 KWHr with PF 0.96, we getBonus=10000 L.EHow to improve Power Factor?Power factor correction is achieved by the addition of capacitors in parallel with the connected motor or lighting circuits and can be applied at the equipment, distribution board or at the origin of the installation.Capacitors contained in most power factor correction equipment draw current that leads the voltage, thus producing a leading power factor. If capacitors areconnected to a circuit that operates at a nominally lagging power factor, the extent that the circuit lags is reduced proportionately.So, what is happening?System without Power factor correction P (Watts)QL(VARS)S (VA)System with Power factor correctionQC(VARS)fS1(VA)QL-QC (VARS)?Cosf = (P/S)Cos? = (P/S1)Now; we can say thatPower factor correction succeeded in the following:Decreasing the shift between P and S, thus increasing PFDecreasing the reactive power Q, thus reducing or eliminating the penalty from the electricity bill Power factor correction Failed inDecreasing the active power “P”That was yesterdayB-TechWhat about Today?Leads a revolution and proudly introduces its new productPower Saving UnitAPSSAPSS APSS relies on a new technology that uses special capacitors, with unique specifications. APSS Capacitors are chemically treated, such that they gain a negative resistance APSSNon APSS CapacitorsAPSS Capacitors˜880SUsed for PFCMore P consumptionSUsed for PFCOver-helping source?Positive ResistanceNegative ResistancePFCActive Power SavingPFCPPQQPositive versus Negative ResistanceRLoadRLoad +VERCircuit -VERCircuitRCircuit = +1 ohmRLoad = 9 ohmI=100/(9+1) = 10 Amp.PLoad =102 × 9 = 900 wattPCircuit= 102 × 1= 100 wattPSource = V × I=1000 watt PSource = PLoad + PCircuitConclusionRCircuit acts as over-impeding voltage source RCircuit = -1 ohmRLoad = 9 ohmI=100/(9-1)=12.5 Amp.PLoad =12.52×9= 1406.25 wattPCircuit=12.52×1=156.25 wattPSource = V × I = 1250 watt PSource + PCircuit = PLoadConclusionRCircuit acts as over-helping voltage source APSSWhy APSS ?1- When used with loads supplied from an electrical network, it can save up to 40% of the Active Power2- When used with loads supplied from Diesel generators, it can effectively save the amount of used fuel3- When used with inductive loads, it can safely improve the PF, thus eliminating any penalties and gain Bonus4- It has much less dimensions compared with PFC units 5- It operates automatically to suit different load variations6- It has much simple design, and needs much less maintenance APSSWhy APSS ?7- Our clients no longer need to increase their input power for further loadings, they can increase their loads up to 40 % without any changes in the feeders.8- PSU is being Guaranteed by B-Tech for three complete years, through which professional maintenance and supervision is given 9- Has much competitive Price compared to the percentage saving, where the client mostly regain his investment in about from 12 to 18 monthes 10- The usage of APSS simply reduces the end production marginal cost *