Transcript Power factor (Pf)

Technical Presentation on the new
Power Saving Unit (PSU)
Introduced By View Power
Electrical Power Types
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:
Real Power “P” (measured in watts)
It accomplishes useful work (e.g., running motors)
Reactive Power (measured in var).
supports the voltages that must be controlled for system reliability
To Simplify Things to all of us...
You can’t move the wheelbarrow
(active power delivery)
unless you lift the arms!
(reactive power)
Power Factor
Power factor (Pf) = The cosine of
the angle between Voltage and current
signals = cos φ.
Power factor (Pf) = The cosine of
the angle between useful power and
Total power = cos φ = (P/S)
Current
Motor Current
Magnetic
current
S=Total Power (KVA)
Q
Magnetic
Power
(KVAr)
Useful current
P=Useful Power (KW)
Why improve Power Factor?
The benefits that can be achieved by applying the suitable Power
Factor correction are:
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
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 components
.
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
For a PF below
as follows:
0.72,
L.E
the penalty per year in L.E (Pen.) is calculated
Pen. = KWHr per year × (0.92-average PF) × 1.0
L.E
Numerical Example
For 500000 KWHr with PF 0.8, we get
For 500000 KWHr with PF 0.6, we get
Pen=30000 L.E
Pen=160000 L.E
Electricity Bill!
Bonus
For a PF above 0.92, the Bonus per year in L.E is calculated as follows:
Bonus
= KWHr per year × (PF-0.92) ×
0.5
L.E
Numerical Example
For 500000 KWHr with PF 0.96, we get
Bonus=10000 L.E
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.
So, what is happening?
System with
without
Power
Power
factor
factor
correction
correction
Cosφ = (P/S)
P (Watts)
Cosθ = (P/S1)QC(VARS)
φ θ
QL-QC (VARS)
S1(VA)
QL(VARS)
S (VA)
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.
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”
That
yesterday
What was
about
Today?
B-Tech
Leads a revolution and proudly introduces its new product
Power Saving Unit
PSU
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
Non PSU Capacitors
Used for PFC
Q
Positive Resistance
S
≈880
More P consumption
PFC
P
PSU Capacitors
Negative Resistance
Used for PFC
Active Power Saving
S
Over-helping Q
source
PFC
?
P
Positive versus Negative Resistance
RLoad
RLoad
+VE
-VE
RCircuit
RCircuit
RCircuit = +1 ohm
RLoad
= 9 ohm
I=100/(9+1) = 10 Amp.
PLoad =102 × 9 = 900 watt
PCircuit= 102 × 1= 100 watt
PSource = V × I=1000 watt
RCircuit = -1 ohm
RLoad
= 9 ohm
I=100/(9-1)=12.5 Amp.
PLoad =12.52×9= 1406.25 watt
PCircuit=12.52×1=156.25 watt
PSource = V × I = 1250
watt
PSource = PLoad + PCircuit
PSource + PCircuit = PLoad
Conclusion
RCircuit acts as over-impeding
voltage source
Conclusion
RCircuit acts as over-helping
voltage source
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
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