Chap. 10 Sinusoidal Steady-State Power Calculations 2018/11/15 Chap. 10 Sinusoidal Steady-State Power Calculations Contents 10.1 Instantaneous Power 10.2 Average and Reactive Power 10.3 The rms Value and Power Calculations 10.4 Complex Power 10.5 Complex Power Calculations 10.6 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。

10.1 Instantaneous Power 或 瞬時功率 p = vi 利用三角恆等式： 定值 兩倍頻 2 2

10.2 Average and Reactive Power Average (Real) Power 其中 注意: 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)。 3 3

Power for Purely Inductive Circuits 2018/11/15 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。 4 4

The Power Factor 功率因數角(power factor angle): v - i 2018/11/15 The Power Factor 功率因數角(power factor angle): v - i 功率因數(power factor): 無效功率因數(reactive factor): 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。 5 5

EX 10.1 Calculating Average and Reactive Power 2018/11/15 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. 6 6

EX 10.2 Power Calculations Involving Household Appliances 2018/11/15 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. 7 7

10.3 The rms Value and Power Calculations 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). 8 8

2018/11/15 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) (b) Repeat (a) by first finding the current in the resistor. (b) 9 9

10.4 Complex Power 複數功率(complex power): 伏安 (VA) 功率三角形 Power Triangle 視在功率 10

EX 10.4 Calculating Complex Power 2018/11/15 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, 11 11

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 12

Alternate Forms for Complex Power 2018/11/15 Alternate Forms for Complex Power (a) (b) 若Z 為純電阻元件時 若Z 為純電抗元件時 13 13

EX 10.5 Calculating Average and Reactive Power 2018/11/15 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 14 14

EX 10.6 Calculating Average and Reactive Power 2018/11/15 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. 15 15

EX 10.6 Calculating Average and Reactive Power (Contd.) 2018/11/15 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. 16 16

EX 10.7 Balancing Power Delivered with Power Absorbed 2018/11/15 EX 10.7 Balancing Power Delivered with Power Absorbed (a) (b) (c) 17 17

10.6 Maximum Power Transfer 最大功率轉移之條件 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。 = 0 = 0 18

The Maximum Average Power Absorbed 2018/11/15 The Maximum Average Power Absorbed 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 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 19 19

EX 10.8 Maximum Power Transfer without Load Restrictions 2018/11/15 EX 10.8 Maximum Power Transfer without Load Restrictions 20 20

EX 10.9 Maximum Power Transfer with ZL Restrictions 2018/11/15 EX 10.9 Maximum Power Transfer with ZL Restrictions a) No Restrictions on the load impedance. b) Restrictions on the load impedance: Set XC as close to −4000  as possible. Set 21 21

EX 10.10 Max Power Transfer with Impedance Angle Restrictions 2018/11/15 EX 10.10 Max Power Transfer with Impedance Angle Restrictions Restrictions on the load impedance: Phase angle = −36.87◦ 22 22

EX 10.11 A Circuit with an Ideal Transformer 2018/11/15 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. 23 23

EX 10.11 A Circuit with an Ideal Transformer (Contd.) 2018/11/15 EX 10.11 A Circuit with an Ideal Transformer (Contd.) 網目方程式: 24 24