Upload
ahmad-wahi
View
225
Download
0
Tags:
Embed Size (px)
Citation preview
1
CHAPTER 2 (Part 2)
Single-Phase Circuits
BEKG 2433 1
BEKG 2433 2
7. Effect of frequency variations
8. Power in AC circuits P, Q & S
9. Power to resistive, inductive, capacitive loads
12. Apparent power (S)
13. Relation between P,Q & S
14. Power to resistance and reactance loads
15. Power Factor
16. Measurement of Power
This part will cover the subtopic as below:
2
BEKG 2433 3
AC Circuit with Series Loads
VS
Z3 Z2 Zn Z1
Total Impedance : ZT = Z1 + Z2 + Z3 +…….+ Zn
Is
Source Current : IS = VS / ZT
Voltage Across each Impedance : Vn = IS Zn
1
321
1.....
111
n
TZZZZ
Z
BEKG 2433 4
AC Circuit with Parallel Loads
VS
Total Impedance :
Is
Source Current : IS = I1 + I2 + I3 +……+In
Current Across each Impedance : In = VS / Zn
Z1 Z3 Z2 Zn
I1 I3 I2 In
3
BEKG 2433 5
7. Effect of frequency
Variation of Inductive Reactance with frequency
XL= ω L = 2 π f L
Inductive reactance is directly proportional to frequency & inductance
“If frequency is doubled, reactance doubles”
“If frequency is halved, reactance halves and so on”
“If inductance is doubled, reactance doubles”
“If inductance is halved, reactance halves and so on”
BEKG 2433 6
Example:
A circuit has 50 ohms inductive reactance. If
both the inductance and the frequency are
doubled, what is the new XL?
Answer: 200Ω
4
BEKG 2433 7
Variation of Capacitive Reactance with frequency
Capacitive reactance is inversely proportional with frequency & Capacitance
• The higher the frequency, the lower the reactance, and vice versa
• At 0Hz, capacitive reactance is infinite
• If the capacitance is doubled, reactance will be halves
CfCXC
2
11
BEKG 2433 8
8. Power in AC Circuits
• What is power?
• Electrical appliances rated in Watt, HP (horsepower), VA (volt-
ampere) or VAR.
• In DC circuits, the equation for power, P = VI Watts.
• In AC circuits, we have real power or active power (P), reactive power in
VAR (Q) and apparent power in VA (S).
• Active power is the power that does useful work- light a lamp, heat, turning
of an electric motor, etc
• Reactive power is the power that does not do useful work or “phantom
power”. Normally exist when the load contains inductor or capacitor
• Power is defined as the rate of flow of energy past a
given point
5
Complex Power
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 9
Another way to easily calculate power
Real Power, P (a scalar with unit: Watt)
RMS based – thermally equivalent to DC power
Reactive Power, Q (a vector with unit: Var)
Oscillating power into & out of the load because of its reactive elements
(L &C)
Positive value for inductive (lagging) & negative fro capacitive(leading)
Complex Power, S (a vector with unit: VA) - overall power delivered
into/out of the system
Apparent power – is the magnitude of the complex power.
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 10
Voltage phasor:
Current phasor:
Complex Power Formulas
vV rmsV
iI rmsI
S Q
P
sinIV
cosIV
sinIVcosIV
IVIV
rmsrms
rmsrms
rmsrmsrmsrms
rmsrmsrmsrms
**
Q
P
jQPS
jS
IVVIS iviv
iInote rmsI*:
. ivfplet
Thus, the complex power is the product of the voltage & the conjugate
of the current:
22powerapparent of magnitude QPSS
6
Power Factor & Power Triangle
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 11
• A ratio between real power & apparent power :
• Power factor can only have value from 0 to 1.
• Low power factor : value close to 0 (e.g 0.2, 0.3)
• indicates poor utilization of electric
• High power factor : value close to 1 (0.9, 0.8, 1, )
• electrical capacity is being utilized effectively
• Unity power factor : value = 1
• electrical capacity is being utilized at the MOST effective
•A measurement of how efficient a facility uses electrical energy.
)cos( ..factor power ivfp
)( anglefactor power ivpf
Sfp
P
powerapparent
power real.
The Power Factor & Power Triangle..cont’d
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 12
VIcosP
tanP
sin VIQ
VIS
P
Qtan
P.
cospowerapparent
power real.
1
22
iv
QP
P
Sfp
fp
This triangle is to summarize & give relations of the real, reactive, apparent and power factor concept.
7
The Power Factor & Power Triangle..cont’d
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 13
( )=δ2
( )=δ1
Where δ = power factor phase Angle For δ1 : θv > θi For δ2 : θv < θi
Power Consumption in Devices
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 14
2Resistor Resistor
2Inductor Inductor L
2
Capacitor Capacitor C
Capaci
Resistors only consume real power:
,
Inductors only "consume" reactive power:
,
Capacitors only "generate" reactive power:
1.C
P I R
Q I X
Q I X XC
Q
2
Capacitortor
C
(Note-some define negative.). CXV
X
8
Applying Ohms Law in Complex power formulas
*
2
*
2
*
**
22**
S
VZ
Z
V
Z
VVVIS
or
IjXIRZIIVIS
ZIV
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 15
22 and
: note
XRZ
ZjXRZ
R
X
Z
Impedance triangle
Conservation of Power (balance power)
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 16
At every node (bus) in the system:
Sum of real power into node must equal zero,
Sum of reactive power into node must equal zero.
This is a direct consequence of Kirchhoff’s current
law, which states that the total current into each node
must equal zero.
Conservation of real power and conservation of reactive
power also follows since S = VI*.
9
The Complex Power Balance
The total complex/apparent power delivered to the loads in parallel
Is the sum of the complex powers delivered to each branch:
V Z3 Z2 Z1
I3 I2 I1 I
*
3
*
2
*
1
*
321
* ][
VIVIVI
IIIVVIS
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 17
Example 1
V Z3 Z2 Z1
I3 I2 I1 I V = 120000 V
Z1 = 60 + j0
Z2 = 6 + j12
Z3 = 30 – j30
Find the power absorbed by each load and the total
complex power
Answers:
varW
varW
varW
varW
000,72000,96
000,24000,24
000,96000,48
0000,24
3
2
1
jS
jS
jS
jS
17/09/2015 UNIVERSITI TEKNIKAL MALAYSIA MELAKA 18
10
BEKG 2433 19
9. Power to Pure Resistive Loads
RIR
VIV
VVIVttIV
dtt
IVdttIVP
tItVtitvtp
rmsrms
rmsrmsmmmm
mm
mmmm
mm
22
2
0
2
0
2
0
2
2224
2sin
22
)2
2cos1(
2sin
2
)sin)(sin()()()(
General equation for the power
BEKG 2433 20
10. Power to Pure Inductive Loads
Energy stored - power flows from source to load
Energy released - power goes back to the source
04
2cos
2
)2
2sin(
2cossin
2
)cos)(sin()()()(
2
0
2
0
2
0
tIV
dtt
IVdtttIVP
tItVtitvtp
mm
mmmm
mm
Power merely absorbed and returned in load due to its reactive properties
11
BEKG 2433 21
where V = IXL or I = V/XL
L
LLX
VXIVIQ
22
Unit for reactive power is VAR
(By convention, the reactive power for inductance is positive or leading)
We can see that inductors dissipates zero power, yet the fact that it drops voltage and draws current gives the deceptive impression that it actually does dissipate power.
This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is (unfortunately) the capital letter Q
BEKG 2433 22
11. Power to Capacitive Loads
Energy stored - power flows from source to load
Energy released - power goes back to the source
04
2cos
2
)2
2sin(
2cossin
2
)cos)(sin()()()(
2
0
2
0
2
0
tIV
dtt
IVdtttIVP
tItVtitvtp
mm
mmmm
mm
Again, power is merely absorbed and returned in load due to its reactive properties
12
BEKG 2433 23
C
CCX
VXIVIQ
22 Unit for reactive power is VAR
(By convention, the reactive power for capacitance is negative or lagging)
Again we can see that reactive load such as capacitor dissipate zero power
and the fact that it drops voltage and draws current gives the same
deceptive impression that it actually do dissipate power.
BEKG 2433 24
12. Apparent Power (S)
• We have seen that in pure resistive circuit – Active power, real power, average power (P) appear
• We also have seen that in pure inductive or pure capacitive circuit – Reactive power, (QL) or (QC) appear
• How do both active & reactive power interact when we have R & L , R & C, or R, L & C in a circuit?
• The combination of reactive power and active power is called Apparent power, and it is the product of a circuit's voltage and current, power being supplied to the load without reference to phase angle
S = V I Unit for apparent power is VA
13
13. Relation between P,Q & S
BEKG 2433 25
22CQPS
22LQPS
Purely Resistive (Unity)
Resistive + Inductive (Lagging)
Resistive + Capacitive (Leading)
jXL
“Power Triangle”
Z = R
X = 0
V = IR S = P
Q = 0
jIXL
θ P = V2/R
jQL =V2/XL θ
R
Z = R+jXL
jIXC
θ
IR
V = IZ IZ
θ
P
θ R
jXC
Z = R-jXL
θ IR
IZ
V = IZ
jQC =V2/XC
BEKG 2433 26
14. Power in a circuit with Resistance and Reactance loads
)2cos(2
1cos
2
1
)2cos(cos2
1
)sin()(sin(
)()()(
tIVIV
tIV
tItV
titvtp
mmmm
mm
mm
)sin(
sin
tII
tVV
m
m
CosIVIV
IVP rmsrmsmm
mm cos22
cos2
1
0
cosVIP
R
X1tan
14
Summary
BEKG 2433 27
22 QPIVS
P – Active power, real power or average power. (Watt)
P = V I cos
Q – Reactive power. QL or QC. (Unit: VAR)
Q = V I sin = P tan
S – Apparent power. (Unit: VA)
Where, V & I is the rms value & θ is the phase angle between V and I
**Apparent power – is the magnitude of the complex power. Complex Power:
sinIV
cosIV
sinIVcosIV
IVIV
rmsrms
rmsrms
rmsrmsrmsrms
rmsrmsrmsrms
**
Q
P
jQPS
jS
IVVIS iviv
BEKG 2433 28
Example 1:
A coil having a resistance of 6Ω & an inductance of 0.03H is connected across a 50V, 60Hz supply. Calculate:
a) The current
b) The phase angle between the current & the applied voltage
c) The apparent power
d) The active power
Answer :
a) 3.91 (A) b) 62.05°
b) 195.5 (VA) c) 91.61 (W)
15
BEKG 2433 29
Example 2: A coil of inductance 159.2mH and resistance 40 Ω is connected in
parallel with a 30 µF capacitor across a 240V, 50Hz supply.
Calculate:
(a) the current in the coil and its phase angle
(b) the current in the capacitor and its phase angle,
(c) the supply current and its phase angle,
(d) the circuit impedance,
(e) the power consumed,
(f) the apparent power, and
(g) the reactive power.
(h) draw the phasor diagram
Answer: a) 3.75/-51.35° b) 2.26 / 90° c) 2.43/-15.98°
d) 98.76Ω e)560.66 (W) f)582.3 (VA) g)160.56 (VAR)
BEKG 2433 30
• The power factor of an AC electric power system is defined as the ratio of the real power (P) flowing to the load to the apparent power (S) .
Power Factor, PF = P/S = cos θ
where θ is the angle between V & I
• In an electric power system, a load with low power factor draws more current than a load with a high power factor, for the same amount of useful power transferred. The higher currents increase the energy lost (I2R) in the distribution system, and require larger wires and other equipment
• Linear loads with low power factor (such as induction motors) can be corrected with a passive network of capacitors or inductors
15. Power factor
16
BEKG 2433 31
PF can be categorized into: Unity, Lagging or Leading
If I lags V, the pf is lagging (Inductive)
If I leads V, the pf is leading (Capacitive)
Example pf = 0.85 lagging
Power factor
PF can be improve or corrected by adding the Capacitive or Inductive
loads.
BEKG 2433 32
Basically the load being supplied consists of resistance and inductance.
This will lead to the appearance of an active (P) and reactive (Q) power (S = P +jQ)
Our aim is to have a value of apparent power (S) , closes to the active power (P) so that the
excessive current drawn from the supply can be reduced.
This can be done by placing reactance of opposite type in parallel to the load so that the
positive Q can be cancelled by the negative Q and vice versa.
The reason to do this correction is to improve system loading, to reduce “copper loss” and
eliminate power factor penalty
Power factor correction
17
BEKG 2433 33
,
+ QC
How????
old
P
Qold Sold
new
P
Qnew
Snew
Example
BEKG 2433 34
A load (L1) absorbs an average power of 8kW at a lagging power factor of 0.6 from 230V, 50Hz supply. Calculate:
a) the value of capacitance (at L2) to bring up the power factor to 0.85
b) the old and new supply current
V(t)
L1 L2
I
18
BEKG 2433 35
,
PL1 = 8kW PFold = 0.6
old = cos-10.6 = 53.1
Qold = P tan old
= 8000 x tan(53.1)
= 10.667kVAR
P = V I cos , Q = V I sin = P tan 22 QPIVS
PL1 = 8kW PFnew = 0.85
new = cos-10.85 = 31.8
Qnew = P tan new
= 8000 x tan(31.8)
= 4.956kVAR
old
P
Qold Sold
new
P
Qnew
Snew
BEKG 2433 36
We need to add capacitance to reduce the Qold to Qnew
QC = Qold – Qnew
= (10.667 – 4.956)kVAR
= 5.711kVAR
But since c
cCX
VXIQ
22
Therefore 269107115
2303
22
.VAR.
)V(
Q
VX
c
c
F..HzX
Cc
75343269502
11
CX c
1Also since
Hence
19
BEKG 2433 37
For the current, we know
cosVIP
Therefore
A.
.V
kW
cosV
PI
old
old
9757
60230
8
A.
.V
kW
cosV
PI
new
new
9240
850230
8
BEKG 2433 38
16. Measurement of Power
• What do we need to measure voltage?
• What do we need to measure Current?
Voltmeter / Multimeter
Ammeter / Multimeter
• What do we need to measure power? Wattmeter
• Like voltmeter & Ammeter, the wattmeter can be analog or digital
Digital:
Numerical
Readout
Analog:
Pointer
on a
scale
20
BEKG 2433 39
• To help understand the concept of power measurement, consider the following figures:
• Instantaneous load power is the product of load voltage times load current, an average power is the average of this.
• Therefore, wattmeter should consist of:
Current sensing circuit Voltage sensing circuit
Multiplier circuit Averaging circuit
BEKG 2433 40
• The following figure shows a simplified representation of a wattmeter:
• Connection of a wattmeter:
• Basically there are 4 points, 2 for current measurement (in series), another 2 for voltage measurement (in parallel)
In series
In parallel
21
BEKG 2433 41
Example:
For the circuit of the following figure, what does the wattmeter indicate if:
a. Vload = 100 0 V and Iload = 1560 A
Answer:
a. 750 W
b. 1409.5 W
WHY???
b. Vload = 100 10 V and Iload = 1530 A
BEKG 2433 42
Example:
Determine the wattmeter reading:
Answer:
600 W
22
BEKG 2433 43
It should be noted that the wattmeter reads only for circuit elements on the load side of the meter. In addition, if the load consists of several elements, it reads the total Real/Active power.
Example:
Answer:
750 W