1. Technical Presentation on the new Power Saving Unit (PSU) Introduced By View Power

2. Almost 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 reliability Real Power “P” (measured in watts) It accomplishes useful work (e.g., running motors) Reactive Power (measured in var). Electrical Power Types

3. unless you lift the arms! (reactive power) To Simplify Things to all of us... You can’t move the wheelbarrow (active power delivery)

4. Power Factor Power factor (Pf) = The cosine of the angle between Voltage and current signals = cos φ. Current Voltage φ φ Useful current Motor Current Magnetic current φ P =Useful Power (KW) Q Magnetic Power (KVAr) S =Total Power (KVA) Power factor (Pf) = The cosine of the angle between useful power and Total power = cos φ = (P/S)

5. 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 stations Why 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 I 2 R losses in transformers and distribution equipment Reduction of voltage drop in long cables. Extended equipment life – Reduced electrical burden on cables and electrical components

6. Electricity Bill! Penalty Equation For 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) ×0.5 L.E For 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) ×1.0 L.E Numerical Example For 500000 KWHr with PF 0.8, we get Pen=30000 L.E For 500000 KWHr with PF 0.6, we get Pen=160000 L.E

7. Electricity Bill! Bonus For a PF above 0.92, the Bonus per year in L.E is calculated as follows: = KWHr per year × (PF-0.92) ×0.5L.E Numerical Example For 500000 KWHr with PF 0.96, we get Bonus=10000 L.E

8. How 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.

9. Capacitors contained in most power factor correction equipment draw current that leads the voltage, thus producing a leading power factor. If capacitors are connected 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) Q L (VARS) S (VA) System with Power factor correction Q C (VARS) φ S 1 (VA) Q L -Q C (VARS) θ Cosφ = (P/S) Cosθ = (P/S 1 )

10. Now; we can say that Power factor correction succeeded in the following: 1.Decreasing the shift between P and S, thus increasing PF 2.Decreasing the reactive power Q, thus reducing or eliminating the penalty from the electricity bill Power factor correction Failed in 1.Decreasing the active power “P”

11. That was yesterday B-TechB-Tech B-Tech What about Today? Leads a revolution and proudly introduces its new product Power Saving Unit PSU

12. PSU PSU relies on a new technology that uses special capacitors, with unique specifications. PSU Capacitors are chemically treated, such that they gain a negative resistance PSU

13. Non PSU Capacitors PSU Capacitors ≈88 0 S Used for PFC More P consumption S Used for PFC Over-helping source ? Positive Resistance Negative Resistance PFC Active Power Saving PFC P P Q Q

14. Positive versus Negative Resistance R Load + VE R Circuit - VE R Circuit R Circuit = +1 ohm R Load = 9 ohm I=100/(9+1) = 10 Amp. P Load =10 2 × 9 = 900 watt P Circuit = 10 2 × 1= 100 watt P Source = V × I=1000 watt P Source = P Load + P Circuit Conclusion R Circuit acts as over-impeding voltage source R Circuit = -1 ohm R Load = 9 ohm I=100/(9-1)=12.5 Amp. P Load =12.5 2 × 9= 1406.25 watt P Circuit =12.5 2 × 1=156.25 watt P Source = V × I = 1250 watt P Source + P Circuit = P Load Conclusion R Circuit acts as over-helping voltage source

15. PSU Why PSU ? 1- When used with loads supplied from an electrical network, it can save up to 40% of the Active Power 2- When used with loads supplied from Diesel generators, it can effectively save the amount of used fuel 3- When used with inductive loads, it can safely improve the PF, thus eliminating any penalties and gain Bonus 4- It has much less dimensions compared with PFC units 5- It operates automatically to suit different load variations 6- It has much simple design, and needs much less maintenance

16. PSU Why PSU ? 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 five 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 one year 10- The usage of PSU simply reduces the end production marginal cost