Transcript power - Deeteekay Community

ELECTRICAL TECHNOLOGY
ET 201
 Explain and calculate average power,
apparent power, reactive power
 Calculate the total P, Q and S and
sketch the power triangle.
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Power in AC Circuit
• Power is distributed into the resistance and
reactance in AC circuit.
• The power delivered to a load at any instant is
defined by the product of the applied voltage and
the resulting current:
p  vi
• Since v and i are sinusoidal quantities
v  Vm sin t   
i  I m sin t
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Power in AC Circuit
v  Vm sin t   
i  I m sin t
The chosen v and i include all possibilities because
• If the load is purely resistive:
 = 0°
• If the load is purely inductive:
 = 90°
• If the load is purely capacitive:
 = - 90°
• For a network that is primarily inductive,  is positive
(v leads i or i lags v)
• For a network that is primarily capacitive,  is negative
(i leads v)
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Power in AC Circuit
• Positive power means that power has been
distributed from supply into circuit.
• Negative power means that the power has been
distributed from circuit into supply.
• There are three type of power in AC circuit:
i) Average power, P
ii) Apparent Power, S
iii) Reactive Power, Q
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Power in AC Circuit
Average Power, P
• The average power (real power)
is the power delivered to the load
and dissipated by the load.
P  Vrms I rms cos 
Where,
OR
[watt, W]
θ : phase angle between Vrms and Irms
P  I rms R
2
;
Vrms
P
R
2
[watt, W]
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14.5 Power Factor (Review)
P  Vrms I rms cos 
Average Power, P
• For a purely resistive load;
T  0
Hence;
FP  cos  1
P  Vrms I rms
• For purely inductive or purely capacitive load;
T  90
FP  cos  0
Hence;
P0
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Power in AC Circuit
Apparent Power, S
• From analysis of DC networks
(and resistive elements above),
it would seem apparent that the
power delivered to the load is
simply determined by P = VI, with no concern for the
components of the load.
• However, in Chapter 14 (Lecture 10) the power factor
(cos θ) of the load has a pronounced effect on the power
dissipated, less pronounced for more reactive loads.
• Therefore P = VI is called apparent power, S.
S  Vrms I rms
[volt-amperes, VA]
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Power in AC Circuit
Apparent Power, S
OR
S  I rms Z
2
; S
Vrms
Z
2
• The average power to the load is;
However;
Therefore
[volt-amperes, VA]
P  VI cos
S  VI
P  S cos
• The power factor of a system Fp is
P
Fp  cos  
S
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Power in AC Circuit
Reactive Power, Q
• In general, the reactive power associated with any circuit is
defined to be
Q  Vrms I rms sin 
[volt-ampere reactive, VAR]
Where, θ : phase angle between Vrms and Irms
OR
Q  I rms X
• For the resistor,
2
[volt-ampere reactive, VAR]
Q  0 VAR
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Power in AC Circuit
Q  Vrms I rms sin 
Reactive Power, Q
• For a purely inductive circuit,
QL  Vrms I rms
OR
QL  I rms X L
2
vL leads iL by 90°
;
Vrms
QL 
XL
2
Since the apparent power, S = VI , and
the average power for inductor, P = 0
P 0
Fp  cos   
0
S VI
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Power in AC Circuit
Q  Vrms I rms sin 
Reactive Power, Q
• For a purely capacitive circuit,
QC  Vrms I rms
OR
QC  I rms X C
2
iC leads vC by 90°
;
Vrms
QC 
XC
2
Since the apparent power, S = VI , and
the average power for capacitor, P = 0
P 0
Fp  cos   
0
S VI
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19.7 Power Triangle
• The three quantities average power P, apparent
power S and reactive power Q can be related in the
vector domain by
with
P  P 0
Q L  QL  90
QC  QC   90
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19.7 Power Triangle S  P 2  Q 2
• For an inductive load,
the phasor power S, as it
is often called, is defined
by
S = P + jQL
• For a capacitive load,
the phasor power S is
defined by
S = P - jQC
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19.7 Power Triangle
• If a network has both capacitive and inductive
elements, the reactive component of the power
triangle will be determined by the difference
between the reactive power delivered to each.
• If QL  QC, the resultant power triangle will be similar
to the inductive load power diagram.
• If QC  QL, the resultant power triangle will be similar
to the capacitive load power diagram.
• That the total reactive power is the difference
between the reactive powers of the inductive and
capacitive elements.
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19.7 The Total P, Q, and S
The total number of watts PT, volt-amperes reactive QT,
and volt-amperes ST, and the power factor Fp of any
system can be found using the following procedure:
1.
Find the real (average) power and reactive power for each
branch of the circuit.
2.
The total real power of the system (PT) is the sum of the
average power delivered to each branch
3.
The total reactive power (QT ) is the difference between the
reactive power of the inductive loads and that of the capacitive
loads.
4.
The total apparent power is ST2
5.
The total power factor is PT / ST.
= PT2 + QT2.
2
S  P2  Q
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19.7 The Total P, Q, and S
There are two important points in the
previous slide:
• First, the total apparent power, ST must be
determined from the total average PT and total
reactive powers QT and cannot be determined
from the apparent powers of each branch.
• Second, and more important, it is not necessary
to consider the series-parallel arrangement of
branches. In other words, the total real PT , total
reactive QT , or total apparent power ST is
independent of whether the loads are in series,
parallel, or series-parallel.
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19.7 The Total P, Q, and S
Example 19.3
1. Find the total number of watts PT, volt-amperes reactive QT,
and volt-amperes ST and draw the power triangle.
2. Find the power factor Fp
3. Find the current in phasor form.
17
19.7 The Total P, Q, and S
Example 19.3 – Solution
1.
Load 1
S  P2  Q2
P1  100 W;
S1  100  j 0
Q1  0 VAR
 100 VA
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19.7 The Total P, Q, and S
Example 19.3 – solution (cont’d)
Load 2
S  P2  Q2
P2  200 W;
S 2  200  j 700
Q2  700 VAR
 728 VA
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19.7 The Total P, Q, and S
Example 19.3 – solution (cont’d)
Load 3
P3  300 W;
Q3  1500 VAR
S  P2  Q2
S3  300  j1500
 1529.71 VA
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19.7 The Total P, Q, and S
Example 19.3 – solution (cont’d)
Total
PT  P1  P2  P3
 100  200  300
PT  600 W
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19.7 The Total P, Q, and S
Example 19.3 – solution (cont’d)
Total
QT  Q1  Q2  Q3
 0  700  1500
 800 VAR
QT  800 VAR(C)
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19.7 The Total P, Q, and S
Example 19.3 – solution (cont’d)
The power triangle;
Total
PT  600 W
QT  800 VAR(C)
ST  PT  jQT
 600  j800
ST  1000  53.13 VA
Note: ST is NOT equal to sum of each branch!!
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19.7 The Total P, Q, and S
Example 19.3 – solution (cont’d)
2. The power factor FP
PT
Fp 
ST
600 W

1000 VA
Fp  0.6 leading (C )
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19.7 The Total P, Q, and S
Example 19.3 – solution (cont’d)
Fp  0.6 leading (C )
3. The current
ST  1000 VA
VI  1000 VA
1000 VA
I
100 V
I  10 A
Since Fp is leading  I leads E, predominantly capacitive circuit.
I  10 A   53.13
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