Transcript 無效功率

Chap. 10 Sinusoidal Steady-State
Power Calculations
Contents 10.1 Instantaneous Power
10.2
10.3
10.4
10.5
10.6
Average and Reactive Power
The rms Value and Power Calculations
Complex Power
Complex Power Calculations
Maximum Power Transfer
Objectives
1. 了解交流功率觀念、相互關係及如何計算：
◆ 瞬時功率
◆ 平均（實）功率
◆ 無效功率（虛功率） ◆ 複數功率
◆ 功率因數
2. 了解最大實功率傳送至一交流電路負載之情況，並能計算
在此條件下之負載阻抗。
3. 在具有線性變壓器及理想變壓器之交流電路中，能計算所
有形式之交流功率。
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10.1 Instantaneous Power
瞬時功率
p = vi
或
利用三角恆等式：
定值
兩倍頻
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10.2 Average and Reactive Power
P : 平均功率（實功率）
注意:
Average (Real) Power
其中
Q : 無效功率（虛功率）
Reactive Power
單位(Units): 瓦(watt, W) for P and 乏(volt-amp reactive, or VAR) for Q
Power for Purely Resistive Circuits
因純電阻電路之v = i，故可簡化為
且稱為瞬時實功率(instantaneous
real power)。
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Power for Purely Inductive Circuits
因對一純電感性電路而言，其電流落後電
壓之相位角90°，即i = v - 90°，故可簡化
為
且其平均功率為零。
Power for Purely Capacitive Circuits
因對一純電容性電路而言，其電流領先電
壓之相位角90°，即i = v + 90°，故可簡化
為
且其平均功率為零。
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The Power Factor
功率因數角(power factor angle): v - i
功率因數(power factor): pf  cos(θv - θi )
無效功率因數(reactive factor): rf  sin(θv - θi )
由於 cos (v － i ) ＝ cos (i － v )，無法明確描述功率因數角。故用
落後功率因數(lagging power factor) 表示電流落後電壓相位角，
屬電感性。
領先功率因數(leading power factor) 表示電流領先電壓相位角，
屬電容性。
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EX 10.1 Calculating Average and Reactive Power
(a)
(b) By the passive sign convention, the negative value of −100 W
means that the network inside the box is delivering average
power to the terminals.
(c)
Because Q is positive, the network inside the box is absorbing
magnetizing vars at its terminals.
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EX 10.2 Power Calculations Involving Household Appliances
The branch circuit supplying the
outlets in a typical home kitchen is
wired with #12 conductor and is
protected by either a 20 A fuse or
a 20 A circuit breaker.
Assume that the following 120 V
appliances are in operation at the
same time: a coffeemaker, egg
cooker, frying pan, and toaster.
Will the circuit be interrupted by
the protective device?
> 20 A
Yes, the protective device
will interrupt the circuit.
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10.3 The rms Value and Power Calculations
弦波電源之rms值又稱為有效值。
OR, i =
平均功率及無效功率亦
可用有效值表示
The effective value of vs (100 V rms) delivers the
same power to R as the dc voltage Vs (100 V dc).
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The phasor transform of a sinusoidal function may also be expressed in
terms of the rms value. The magnitude of the rms phasor is equal to the
rms value of the sinusoidal function. If a phasor is based on the rms value,
we indicate this by either “rms” or the subscript “eff” adjacent to the phasor
quantity.
EX 10.3 Determining Average Power Delivered to a Resistor
by Sinusoidal Voltage
A sinusoidal voltage having a maximum amplitude of 625 V is
applied to the terminals of a 50  resistor.
(a) Find the average power delivered to the resistor.
(a)
Vrms  Veff  625/ 2  441.94V
2
2


Veff 
441.94
P

R
50
 3906.25 W
(b) Repeat (a) by first finding the current in the resistor.
(b)
I rms  I eff 
625/50
 8.84A
2
P  Ieff  R  8.84 50  3906.25 W
2
2
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10.4 Complex Power
複數功率(complex power):
伏安 (VA)
功率三角形
Power Triangle
視在功率
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EX 10.4 Calculating Complex Power
An electrical load operates at 240 V rms. The load absorbs an
average power of 8 kW at a lagging power factor of 0.8.
a) Calculate the complex power of the load.
b) Calculate the impedance of the load.
(a)
Lagging pf : Q > 0
(b)
Also,
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10.5 Complex Power Calculations
複數功率
=(均方根相量電壓) × (均方根相量電流共軛值)
注意:
Also
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Alternate Forms for Complex Power
(a)
(b)

Z  Veff
Z Z
2
*
 R  jX 
Veff
若Z 為純電阻元件時
P
Veff
R
Z
2
PR
2
Z
2
QX
2
Veff
Z
2
2
若Z 為純電抗元件時
2
Q0
Veff
P0
Q
Veff
2
X
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EX 10.5 Calculating Average and Reactive Power
a) Calculate the load current IL and voltage VL .
b) Calculate the average and reactive power delivered to the load.
c) Calculate the average and reactive power delivered to the line.
d) Calculate the average and reactive power supplied by the source.
或
supplying
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EX 10.6 Calculating Average and Reactive Power
Load 1 absorbs an average power of
8 kW at a leading power factor of 0.8.
Load 2 absorbs 20 kVA at a lagging
power factor of 0.6.
a) Determine the power factor of the two loads in parallel.


10k 


pf  cos tan-1
  cos 26.565  0.8944 lagging
20k 

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EX 10.6 Calculating Average and Reactive Power (Contd.)
b) Determine the apparent power required to supply the loads, the magnitude
of the current, Is , and the average power loss in the transmission line.
c) Given that f = 60 Hz, compute the value of the capacitor that would correct
the power factor to 1 if placed in parallel with the two loads. Recompute
the values in b) for the load with the corrected power factor.
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EX 10.7 Balancing Power Delivered with Power Absorbed
(a)
(b)
(c)
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10.6 Maximum Power Transfer
最大功率轉移之條件
*
Z L  ZTh
Let
Also,
利用微分極值定理之觀念，想得到P 的最大值的條件為
=0
P
P
和
均為零。
RL
X L
X L   X Th
=0
ZL  Z
*
Th
RL  RTh
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The Maximum Average Power Absorbed
*
在 Z L  ZTh
之條件下，負載電流為VTh/2RL，故傳送至負載之最大平均功率
若將載維寧等效電壓改以電壓峯值表示，則
當Z受限制時之最大功率轉移：
(a) 若RL 及XL 被限制在某一範圍時，應先考慮將XL 儘可能調到接近-XTh，則
(b) 若ZL 的大小可改變，但其相位角不可改變時，則在 Z L  Z Th 時，負載可
得到最大功率轉移。
Problem 10.32
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EX 10.8 Maximum Power Transfer without Load Restrictions
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EX 10.9 Maximum Power Transfer with ZL Restrictions
a) No Restrictions on the load impedance.
Z L  3000 j 4000  3000-j4000
*
Pmax
1 102
25


mW  8.33 mW
4 3000 3
b) Restrictions on the load impedance: 0  RL  4000 Ω
-2000Ω  X C  0
Set XC as close to −4000  as possible.
X C  -2000Ω
Set
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EX 10.10 Max Power Transfer with Impedance Angle Restrictions
Restrictions on the load impedance:
Phase angle = −36.87◦
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EX 10.11 A Circuit with an Ideal Transformer
The variable resistor is adjusted
until maximum average power is
delivered to RL .
a) What is the value of RL in ohms?
b) What is the maximum average
power (in watts) delivered to RL?
Open circuit: I2 = 0, hence I1 = 0.
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EX 10.11 A Circuit with an Ideal Transformer (Contd.)
網目方程式:
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