Transcript 2. Average power

1
Chapter 9.
Steady-state power analysis
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Contents
1. Instantaneous Power: For the special case of steady state sinusoidal signals
2. Average Power : Power absorbed or supplied during one cycle
3. Maximum Average Power Transfer :When the circuit is in sinusoidal steady state
4. Effective or RMS Values :
For the case of sinusoidal signals
5. Power Factor : A measure of the angle between current and voltage phasors
6. Power Factor Correction : How to improve power transfer to a load by “aligning” phasors
7. Complex Power : Measure of power using phasors
8. Single Phase Three-Wire Circuits : Typical distribution method for households and small
loads
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1. Instantaneous power
Instantaneous Power : p(t )   (t )i (t )
In steady State :
v(t )  VM cos( t   v )
i(t )  I M cos( t   i )
p(t )  VM I M cos( t  v ) cos( t  i )
p(t ) 
VM I M
cos( v  i )  cos( 2 t   v  i )
2
  v  i  0
Net power transfer is
positive!
 

2
or 90
Net power transfer is
zero!
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Power vs. voltage
v(t )  VM cos  t
VM I M
cos( v  i )  cos( 2 t   v  i )
p(t ) 
2
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2. Average power
1
P
T
p( t ) 
t0
t0  T
P
t 0 T
 p(t )dt
T
2
t0

2 

  

T 

VM I M
cos( v   i )  cos(2 t   v   i )
2
VM I M
cos( v   i )
2
1
2
If voltage and current are in phase
 v   i  P  VM I M
If voltage and current are in quadrature
v   i  90  P  0
Purely inductive or capacitive
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Example 9.2
Find the average power absorbed by impedance
I
1060 1060

 3.5315( A)
2  j 2 2 245
VM  10, I M  3.53, v  60°, i  15°
P  35.3 cos(45)  12.5W
I  3.5315 [A]
V  IR  2  3.5315  7.0615 [V]
I  3.5315 [A]
V  I  jL  3.5315  290  7.06105 [V]
P0
EMLAB
Determine the average power absorbed by each resistor,
the total average power absorbed and the average power
supplied by the source
Example 9.3
I1 
1245
 345( A)
4
I2 
1245
1245

 5.3671.57( A)
2  j1
5  26.37
7
1
P4   12  3  18W
2
1
P2   2  5.36 2 (W )
2
I  I1  I 2  345  5.3671.57
I  8.1562.10( A)
P
VM I M
cos( v   i )
2
Psupplied
1
 12  8.15  cos(45  62.10)  46.74[W ]
2
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3. Effective or rms Values
root mean square
i (t )
P (t )
R
P
i (t )  I M cos( t   )
1
 V  IR, P  I M2 R
2
If current is DC (i(t )  I dc ) then
If current is periodic with period T
1
P
T
 Pdc  RI dc2
Pav  Pdc
2
 I eff

1 2
I
I M  I eff  M
2
2
I eff
 1 t 0 T 2


t i(t )v(t )dt  R T t i (t )dt 
0
0


2
 RI eff
t 0 T
1

T
t 0 T
2
i
 (t )dt
t0
Definition is valid for ANY periodic
signal with period T
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1
For sinusoidal case : Pav  VM I M cos( v   i )
2
 Pav  Veff I eff cos( v   i )
Example
Compute rms value of the voltage waveform
T 4
Vrms
1

T
t 0 T
2

 (t )dt
t0
v  2t
Vrms
12

(2t ) 2dt

40
2
1  8
  t 3   (V )
3 0 3
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4. Complex power
v(t )  VM cos( t   v )  Re{VM e jv }
i(t )  I M cos( t  i )  Re{ I M e ji }
Vrms  Vrms V , I rms  I rms  i
S  Vrms I*rms
; Complex power
S  Vrms V I rms    i  Vrms I rms (V   i )
S  P  jQ
P  Vrms I rms cos( v   i )
; Average power
Q  Vrms I rms sin( v   i )
P  Re{S}  Re{Vrms I*rms }
; Average power
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Example
Evaluate the average power consumed by the following circuit.
R
L
I rms
VS ,rms
I rms 
C
Vs ,rms

1 

R  j   L 

C


S  VI 
*
Vs ,rmsVs*,rms

1 

R  j   L 

C


 P  Re{S} 


R
2 


1  

R 2    L 

C


Vs ,rmsVs*,rms
R

1 

R 2    L 

C


2
Vs ,rms

1 

j   L 

C


2
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5. Maximum average power transfer
Z TH  RTH  jX TH
VL 
ZL
VOC
Z L  Z TH
IL 
VOC
Z L  Z TH
Z L  RL  jX L
 VOC
1
1
ZL
S L  VL I*L 
VOC 
2
2 Z L  Z TH
 Z L  Z TH
P  Re{S} 
*
2
 1 VOC
 
ZL
2
Z

Z
L
TH

1
Re{VI* }
2
1
| VOC |2 RL
PL 
2 ( RL  RTH ) 2  ( X L  X TH ) 2
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Maximum average power transfer condition
1
| VOC |2 RL
PL 
2 ( RL  RTH ) 2  ( X L  X TH ) 2
PL

 0
X L
  X L   X TH

PL
 0   RL  RTH

RL
*
 Z Lopt  Z TH
max
L
P
1  | VOC |2 

 
2  4 RTH 
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Example
Find ZL for maximum average power transfer. Compute the
maximum average power supplied to the load.
 Z Lopt

*
ZTH
PLmax
1  | VOC |2 

 
2  4 RTH 
8  j 4 (8  j 4)(6  j1) 52  j16



6  j1
37
37
8  j 4 8.9426.57


 1.4716.93
6  j1 6.089.64
ZTH  4 || (2  j1) 
Z L*  1.47  16.93  1.41  j 0.43
VOC  4 
PLmax
2
320
40 
 5.26  9.64
6  j1
6.089.64
1 5.262
 
 2.45(W )
2 4  1.41
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6. Power factor
Instantaneous power in an ac circuit. Positive p represents power
to the load; negative p represents power returned from the load.
Power to a purely resistive load. The peak value of p is VmIm.
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Power to a purely inductive load. Energy stored
during each quarter cycle is returned during the
next quarter-cycle. Average power is zero.
Power to a purely capacitive load.
Average power is zero.
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Power factor
S  S ( v  i )  P  jQ  VI *
Papparent  Vrms I rms : (apparent power)
Z  Z  Z 
V
(V   I )
I
 90   z  0
1
P  VM I M cos( v   i )  Vrms I rms cos( v   i )
2
P
pf 
 cos( v   i )  cos  z
Papparent
current leads
(capacitiv e)
V
P  Vrms  I rms  pf
Power factor
0   z  90
Phase difference
(θv - θi )
Q (Complex power)
leading
-
-
Capacitive
in phase
0
0
Resistive
lagging
+
+
Inductive
current lags
(inductive )
☆ The reference phase angle is that of voltage.
A leading or lagging phase angle is the state that the phase angle of current is leading
or lagging behind that of voltage.
( 전압의 위상을 기준으로 하여 전류의 위상이 lead인지 lag인지 결정됨.)
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Transmission line
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Transmission line loss
House
가정
Electric company
전기 회사
계량기
R1
Power meter
2
PLoss1k I rms
R
+
-
220V
Z
R1
1k
Pav  Vrms I rms  cos 
전선에 의한 손실을 줄이기 위해서 가능한 전류를 적게 흘려야 한다.
To save power losses due to resistance of transmission lines, current should
be kept as small as possible.
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Example
0.R1
1
1k
+
-
PLoss  2 I 2 R  0.2 W
220V
220W
0.R1
1
IL 
1k
220
 1A
220
0.R1
1
1k
+
-
PLoss  2 I 2 R  0.8 W
110V
220W
0.R1
1
1k
IL 
220
 2A
110
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Example 9.10
An industrial load consumes 88 kW at a pf of 0.707 lagging from a 480-V rms line. The transmission line
resistance from the power company’s transformer to the plant is 0.08. Let us determine the power that must be
supplied by the power company (a) under present conditions and (b) if the pf is somehow changed to 0.90 lagging.
(It is economically advantageous to have a power factor as close to one as possible.)
P  Vrms I rms  pf
 cos z  0.707   z  45
(a)
I rms
PL
88  103


 259.3 [ Arms ]
Vrms  pf 480  0.707
2
PS  PL  0.08 I rms
 88,000  0.08  259.32  93.38 kW
(b)
pf  0.9
I rms
PL
88  103


 203.7 [ Arms ]
Vrms  pf 480  0.9
2
PS  PL  0.08 I rms
 88,000  0.08  203.7 2  91.32 kW
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Example
A load operates at 20 kW, 0.8 pf lagging. The load voltage is at 60 Hz. The impedance
of the line is 0.09 + j0.3 . We wish to determine the voltage and power factor at the input
to the line.
S  S ( v   i )  P  jQ  VI *
P  Vrms I rms  pf  S  pf
P  Vrms I rms cos(v  i )
Q  Vrms I rms sin( v   i )
S  VI *  20k  j15k
*
*
20k  j15k
 S   20k  j15k 
I   
 90.9  j 68.2  113.64  36.87 [ Arms ]
 
220
 V   2200 
VS  (0.09  j 0.3)  (90.9  j 68.2)  220  248.64  j 21.1  249.544.86 [Vrms ]
 v   i  41.73
pf  cos( v   i )  0.75 lagging
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7. Power factor correction
S L  PL  jQL
IL
Electrical
source

VL

Industrial
load with
lagging pf
SL
V   I
QL
PL
Simple approach to power factor correction
QL
SL
QC
V   I
PL
Q  VI *  V ( j CV )*   j C V
2
EMLAB
Example
f  60Hz.
Determine the capacitor required to increase the power factor to 0.95 lagging
25
50kW ,VL  2200 rms
pf  0.8 lagging
S  VI *  P  jQ  P1  j tan(V   I )  50k (1  j 0.75)  50,000  j 37,500
 (V   I )  cos 1 (0.8)  36.87
S new  P1  j tan  new   50k (1  j 0.329)  50,000  j16,434
 new  cos 1 (0.95)  18.2
Q  j16,434  j 37,500   j 21,066   j C V
 C  1155F
2
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8. Single-phase three wire circuits
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Disadvantage of single phase two wire system (단상 2선
식의 단점)
I  10 A
R1
rms
R 1k0.05 
100V(rms)
+
-
2
PLoss  I rms
R  10 W
100V(rms) 1kW
R  R1
0.05 
1k
1. If the load increases to 2kW, Ploss becomes 40W.
2. If each wire can sustain heat only to 5W power loss, the cross-section area
should increase 4 times, whereby cost for copper wires increase 4 times.
I rms  20 A
R1
R 1k0.05 
100V(rms)
+
-
2
PLoss  I rms
R  40 W
100V(rms) 2kW
R  0R1
.05 
1k
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Single-phase 3 wire circuits
I rms  10 A
100V(rms)
+
-
100V(rms)
+
-
R1
R 1k0.05 
I rms  0 A
I rms  10 A
100V(rms)
1kW
R1
R 1k0.05 
R1
100V(rms) 1kW
1k
1. Although the power load increases to 2kW, Ploss remains 10W.
2. With 2 times power load, only one copper wire is added.
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Example 9.27
A light-duty commercial single-phase three-wire 60-Hz circuit serves lighting, heating, and
motor loads, as shown below. Lighting and heating loads are essentially pure resistance
and, hence, unity power factor (pf), whereas motor loads have lagging pf.
Unbalanced connection
Im 
10kVA
  36.9  41.7  36.9
240
5k
 41.70
120
I a  I L  I H  I m  41.70  41.70  41.7  36.9  119.4  12.1
I n  I L  I H  41.70  41.70  83.40
IL  IH 
I b  I m  41.7  36.9
S a  V I  (1200)(119.412.1)  14  j 3 [kVA]
Rlines  0.05


Plosses  0.05  | I a |2  | I b |2  | I n |2  1.147kW
*
an a
pf  cos12.1  0.9778 lagging
Sb  Vbn I a*  (1200)( 41.7  36.9)  4  j 3 [kVA]
pf  0.8 lagging
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Balanced connection
I a  I L  I m  41.70  41.7  36.9  79.06  18.4
I b  I a  79.06  18.4
In  0
S a  Van I a*  (1200)(79.0618.4)  9  j 3 [kVA]
pf  cos18.4  0.9487 lagging
Sb  Vbn I b*  (1200)(79.0618.4)  9  j 3 [kVA]
pf  cos18.4  0.9487 lagging
Rlines  0.05


Plosses  0.05  | I a |2  | I b |2  0.625kW
Psaved  0.522kW
$ / year  366(@ 0.08$ / kWh )
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Safety considerations
32
Average effect of 60Hz current from hand to
hand and passing the heart
Required voltage depends on contact, person
and other factors
Typical residential circuit with ground and
neutral
Ground conductor is not needed for
normal operation
이행선
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Grounding for safety considerations
33
Grounding is needed to protect from lightning.
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기기 고장에 의한 short
기기 고장에 의한 short
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Example
Increased safety due to grounding
When switched on the tool case is energized
without the ground connector the
user can be exposed to the full
supply voltage!
Conducting due to wet floor
If case is grounded then the supply is shorted and the fuse acts to open
the circuit
More detailed numbers in a
related case study
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Wet skin
150
150
400
I body 
120
 171mA
701
Ground prong removed
R(dry skin)
R(wet skin)
R(limb)
R(trunk)
15kOhm
150Ohm
100Ohm
200Ohm
Suggested resistances
for human body
1
Can cause ventricular fibrillation
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