Transcript Electro Mechanical System

Relationship between P, Q, & S
 Consider a circuit with a source, a load, & appropriate meters.
 P and Q are positive, so the load absorbs both active and
reactive power. The line current I lags behind Eab by an angle .
 Current I has two components Ip & Iq. Value Ip = P/E & Iq = Q/E
 The apparent power S = EI, from which I = S/E
 The phasor diagram shows
2
I 2  I p2  I q2
2
S 
P
Q 
      
E
E
E
2
so S2  P 2  Q 2
S =apparent power [VA|:P =active power [W]:Q =reactive power [var]
 The angle  = arctan Iq / Ip = arctan Q/P
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Power factor
 The power factor of an ac device or circuit is the ratio of the active
power P to the apparent power S. given by the equation:
power factor = P / S = EIP / EI = IP / I = cos 
Where:
P = active power delivered or absorbed by circuit or device [W]
S = apparent power of the circuit or device [VA]
 Power factor is expressed as a simple number, or a percentage.
 The power factor can never be greater than unity (or 100 percent).
 The power factor of a resistor is 100 percent because the apparent
power it draws is equal to the active power.
 The power factor of an ideal coil having no resistance is zero,
because it does not consume any active power.
 The power factor of a circuit or device is simply a way of stating
what fraction of its apparent power is real, or active, power.
 In a single-phase circuit the power factor is also a measure of the
phase angle  between the voltage and current
 If we know the power factor, we can calculate the angle. The
power factor is said to be lagging or leading .
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Power triangle
 Relationship S2 = P2 + Q2 is right-angle
triangle, known as a power triangle.
Following rules apply:
 Active power P absorbed by a circuit or
device is considered to be positive and is
drawn horizontally to the right
 Active power P that is delivered by a
circuit or device is considered to be
negative and is drawn horizontally to left
 Reactive power Q absorbed by a circuit or
device is considered to be positive and is
drawn vertically upwards
 Reactive power Q that is delivered by a circuit or device is
considered to be negative and is drawn vertically downwards
 The concept of the power triangle is useful when solving ac circuits
that comprise several active and reactive power components.
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Sources and loads
 A resistor and capacitor are connected to the source.
 Capacitor acts as a reactive source.
 A wattmeter connected into the circuit will give a positive reading
P = Elp watts, but a varmeter will give a negative reading Q = EIq .
 Source G delivers active power P but receives reactive power Q.
 Two powers flowing in opposite directions over the same line.
 An electrical outlet can act not only as an active or reactive
source, but it may also behave as an active or reactive load.
 It depends upon the type of device connected to the receptacle.
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System with several loads
 Consider, a group of loads connected in
an unusual way to a 380 V source .
 We wish to calculate the apparent power
absorbed & current supplied by source.
 We simply draw a block diagram of the
individual loads, indicating the direction
of active and reactive power flow
 Add all the active powers in a circuit to
obtain the total active power P.
 We can add the reactive powers to obtain
the total reactive power Q.
 Total apparent power S is then found by:
S  P2  Q2
 Adding reactive powers, we assign a +ve
value to those that are absorbed by the
system and a –ve value to those that are
generated (such as by a capacitor)
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System with several loads
 Similarly we assign a +ve value to active
powers that are absorbed and –ve value to
those that are generated (by an alternator)
 we cannot add the apparent powers to
obtain the total apparent power S. Add
them only if their power factors are same
 Let us now solve the circuit :
1. Active power absorbed by the system:
P = (2 + 8 + 14) = +24 kW
2. Reactive power absorbed by the system:
Q1 = (5 + 7 + 8) = +20 kvar
3. Reactive power supplied by the
capacitors:
Q2 = (–9 – 16) = – 25 kvar
4. Net reactive power Q absorbed by the
system:
Q = ( + 20 – 25) = – 5 kvar
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System with several loads
5. Apparent power of the system:
S  P 2  Q 2 
24 2  (5) 2  24.5kVA
6. Because the 380 V source furnishes the
appaent power, the line current is
I = S/E = 24 500/380 = 64.5 A
7. The power factor of the system is
cos L = P/S = 24/24.5 = 0.979 (leading)
 The 380 V source delivers 24 kW of
active power, but it receives 5 kvar of
reactive power.
 This reactive power flows into the
system of the electrical company, where
it is available to create magnetic fields.
 The magnetic fields may be associated
with distribution transformers,
transmission lines, relays etc.
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System with several loads
 Let us make power triangle for the system.
 It is the graphical solution to our problem.
 Starting with the 5 kvar load, we move from
one device to the next around the system.
 We draw the magnitude and direction (up,
down, left, right) of each power vector.
 At the end, we can draw a power vector
from the starting point to the end point.
 Vector having a value of 24.5 kVA.
 The horizontal component of this
vector has a value of 24 kW and,
because it is directed to the right, we
know that it represents power
absorbed by the system.
 The vertical component of 5 kvar is
directed downward, so it represents
reactive power generated by the
system.
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Reactive power without magnetic fields
 Loads sometime absorb reactive power
without creating any magnetic field at all.
 This can happen in electronic power
circuits when the current flow is delayed
by means of a rapid switching device.
 Consider, a circuit in which a 100 V, 60 Hz
source is connected to a 10 Ω resistive
load by means of a mechanical switch.
 The switch opens and closes its contacts
so that current only flows during the latter
part of each half cycle.
 This forced delay causes the current to
lag behind the voltage.
 If we connect a wattmeter and varmeter,
they would read +500 W and + 318 var.
 This corresponds to a lagging power factor (sometimes called
displacement power factor) of 84.4 percent.
 Reactive power is associated with the rapidly operating switch
rather than with the resistor itself.
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