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Chap. 10 Sinusoidal Steady-State Power Calculations

Contents10.1 Instantaneous Power10.2 Average and Reactive Power10.3 The rms Value and Power Calculations10.4 Complex Power10.5 Complex Power Calculations10.6 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

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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

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其中 P : 平均功率（實功率）Average (Real) Power

Q : 無效功率（虛功率）Reactive Power

注意 :

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 (instantaneousreal power) 。

單位 (Units): 瓦 (watt, W) for P and 乏 (volt-amp reactive, or VAR) for Q

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

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功率因數角 (power factor angle): v - i

功率因數 (power factor): ) (cospf iv θ-θ

無效功率因數 (reactive factor): ) sin(rf iv θ-θ

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)

(c)

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.

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?

Yes, the protective device will interrupt the circuit.

> 20 A

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10.3 The rms Value and Power Calculations

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OR, i =

弦波電源之 rms 值又稱為有效值。

The effective value of vs (100 V rms) delivers the same power to R as the dc voltage Vs (100 V dc).

平均功率及無效功率亦可用有效值表示

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EX 10.3 Determining Average Power Delivered to a Resistor by Sinusoidal Voltage

(a)

(b)

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.

V 441.942625/VV effrms

W3906.25

50

441.94V 22eff R

P

(b) Repeat (a) by first finding the current in the resistor.

A 8.842

625/50II effrms

W3906.25508.84I 22eff RP

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.

10.4 Complex Power

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功率三角形Power Triangle

複數功率 (complex power):

伏安 (VA)

視在功率

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EX 10.4 Calculating Complex Power

(a)

(b)

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.

Also,

Lagging pf : Q > 0

10.5 Complex Power Calculations

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注意 :

複數功率=( 均方根相量電壓 ) × ( 均方根相量電流共軛值 )

Also

Alternate Forms for Complex Power

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(a)

(b)

2

2

eff

*

2

eff Z

jXRZZ

Z VV

2

2

eff2

2

eff Z

XQZ

RPVV

若 Z 為純電阻元件時

0 2

eff QR

PV

若 Z 為純電抗元件時

XQP

2

eff 0V

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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.

lagging 0.894426.565cos20k

10ktancospf 1-

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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)

10.6 Maximum Power Transfer

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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 和 均為零。 LR

P

LX

P

= 0

= 0

ThL RR

ThL XX

最大功率轉移之條件*ThL ZZ

*ThL ZZ

The Maximum Average Power Absorbed

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在 之條件下，負載電流為 VTh/2RL ，故傳送至負載之最大平均功率

*ThL ZZ

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 時，負載可 得到最大功率轉移。 Problem 10.32

ThL ZZ

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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.

4000300040003000 -jjZ *L

mW 338 mW 3

25

3000

10

4

1 2

max .P

b) Restrictions on the load impedance:

0Ω 2000

Ω40000

C

L

X-

R

Set XC as close to −4000 as possible. Ω 2000-X C

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 adjusteduntil 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.)

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