文章快速检索 高级检索
 北京化工大学学报(自然科学版)  2017, Vol. 44 Issue (6): 124-128  DOI: 10.13543/j.bhxbzr.2017.06.020 0

### 引用本文

LIANG Zheng, WANG Yun, ZHANG Ying, ZHOU YuHong, JIA Hui, BU JianDong. Research on some queries of reactive power in sinusoidal ac circuits[J]. Journal of Beijing University of Chemical Technology (Natural Science), 2017, 44(6): 124-128. DOI: 10.13543/j.bhxbzr.2017.06.020.

### 文章历史

1. 公安海警学院 基础部, 浙江 宁波 315801;
2. 北京化工大学 理学院, 北京 100029;
3. 海纳国际集团, 费城 PA19004, 美国;
4. 中材地质工程勘查研究院有限公司, 北京 100102

Research on some queries of reactive power in sinusoidal AC circuits
LIANG Zheng 1, WANG Yun 2, ZHANG Ying 3, ZHOU YuHong 1, JIA Hui 1, BU JianDong 4
1. Department of Basic Courses, China Maritime Police Academy, Ningbo, Zhejiang 315801, China;
2. Faculty of Science, Beijing University of Chemical Technology, Beijing 100029, China;
3. Susquehanna International Group, Philadelphia PA 19004, USA;
4. CNBM Geological Engineering Exploration Academy Co. Ltd., Beijing 100102, China
Abstract: Reactive power is one of important physical quantities in physics and engineering. In this paper, the concept of reactive power is classified in linear, time invariant, or lumped parameter circuits with ideal sinusoidal steady-state excitation. Two equivalent methods for the definition of reactive power are presented first. The instantaneous power of impedance is decomposed into active power and reactive power. Accurate expressions of active and reactive components of the instantaneous power are given for both series models and parallel models of impedance. Results based on our calculations and analysis of the reactive component of typical circuits give the conclusions. The reactive energy of inductive and capacitive load throughput can be reversed in one-port circuits of ideal linear capacitors and inductors in series or parallel, RLC in series or parallel, and impedance series. In contrast, reactive energy cannot be reversed in circuits of inductive impedance and capacitive impedance in parallel, or in circuits with three-phase loads. In addition, reactive energy can flow between the power supply and each load, and also flow between each load. Finally, the throughput amplitude of the whole load-side network is not necessarily equal to the algebraic sum of the reactive power of each of the loads in the network.
Key words: sinusoidal AC circuit    reactive power    concept discrimination

1 原始概念辨析 1.1 无功功率的理想定义

1.2 有功分量与无功分量的正确表述

 $p = {u_{\rm{S}}}i = 2UI\sin \left( {\omega t} \right)\sin \left( {\omega t - \varphi } \right) = UI\left[ {\cos \varphi - \cos \left( {2\omega t - \varphi } \right)} \right]$ (1)

 $\begin{array}{l} \;\;\;\;\;\;p = {u_{\rm{S}}}i = 2UI\sin \left( {\omega t} \right)\sin \left( {\omega t - \varphi } \right) = 2UI\sin \left( {\omega t} \right)\\ \left( {\sin \omega t\cos \varphi - \cos \omega t\sin \varphi } \right) = 2UI\cos \varphi {\sin ^2}\left( {\omega t} \right) - \\ UI\sin \varphi \sin \left( {2\omega t} \right) \end{array}$ (2)

 $\begin{array}{l} \;\;\;\;\;\;p = {u_{\rm{S}}}i = 2UI\sin \left( {\omega t} \right)\sin \left( {\omega t - \varphi } \right) = 2UI\sin \left[ {\left( {\omega t - } \right.} \right.\\ \left. {\left. \varphi \right) + \varphi } \right]\sin \left( {\omega t - \varphi } \right) = 2UI\cos \varphi {\sin ^2}\left( {\omega t - \varphi } \right) - \\ UI\sin \varphi \sin \left( {2\omega t - 2\varphi } \right) \end{array}$ (3)

2 正弦稳态电路中无功功率计算分析 2.1 感性负载与容性负载无功分量的相位关系

 图 1 RL与C并联的电路模型 Fig.1 Single port circuit of RL parallel with C

 ${i_{\rm{C}}} = \sqrt 2 \frac{U}{{{X_{\rm{C}}}}}\sin \left( {\omega t + {{90}^ \circ }} \right)$

 ${p_{\rm{C}}} = {u_{\rm{C}}}{i_{\rm{C}}} = {u_{\rm{S}}}{i_{\rm{C}}} = \frac{{{U^2}}}{{{X_{\rm{C}}}}}\sin \left( {2\omega t} \right)$

 ${i_{\rm{L}}} = \sqrt 2 \frac{U}{{\sqrt {{R^2} + X_{\rm{L}}^2} }}\sin \left( {\omega t - \varphi } \right)$

 ${u_{\rm{L}}} = \sqrt 2 \frac{{U{X_{\rm{L}}}}}{{\sqrt {{R^2} + X_{\rm{L}}^2} }}\sin \left( {\omega t - \varphi + {{90}^ \circ }} \right)$

 ${p_{\rm{L}}} = {u_{\rm{L}}}{i_{\rm{L}}} = \frac{{{U^2}{X_{\rm{L}}}}}{{{R^2} + X_{\rm{L}}^2}}\sin \left( {2\omega t - 2\varphi } \right)$

 $\begin{array}{l} \;\;\;\;\;\;{p_{{\rm{X}}1}} = {u_{{\rm{X}}1}}{i_{{\rm{X}}1}} = \frac{{2{U^2}{X_1}}}{{R_1^2 + X_1^2}}\sin \left( {\omega t - {\varphi _1} + {{90}^ \circ }} \right)\sin \left( {\omega t - } \right.\\ \left. {{\varphi _1}} \right) = \frac{{{U^2}{X_1}}}{{R_1^2 + X_1^2}}\sin \left( {2\omega t - 2{\varphi _1}} \right) \end{array}$ (4)

Z2的非耗散吞吐功率为

 $\begin{array}{l} \;\;\;\;\;\;{p_{{\rm{X}}2}} = {u_{{\rm{X}}2}}{i_{{\rm{X}}2}} = \frac{{2{U^2}{X_2}}}{{R_2^2 + X_2^2}}\sin \left( {\omega t - {\varphi _2} - {{90}^ \circ }} \right)\sin \left( {\omega t - } \right.\\ \left. {{\varphi _2}} \right) = \frac{{{U^2}{X_2}}}{{R_2^2 + X_2^2}}\sin \left( {2\omega t - 2{\varphi _2} - {\rm{ \mathsf{ π} }}} \right) \end{array}$ (5)

2.2 无源网络总吞吐幅度与各元件无功功率代数和

 $\begin{array}{l} \;\;\;\;\;\;{p_{{\rm{total}}}} = {p_{\rm{C}}} + {p_{\rm{L}}} = \frac{{{U^2}}}{{{X_{\rm{C}}}}}\sin \left( {2\omega t} \right) + \frac{{{U^2}{X_{\rm{L}}}}}{{{R^2} + X_{\rm{L}}^2}}\sin \left( {2\omega t - } \right.\\ \left. {2\varphi } \right) = M\sin \left( {2\omega t} \right) + N\sin \left( {2\omega t - 2\varphi } \right) \end{array}$ (6)

 $\begin{array}{l} \;\;\;\;\;\;{p_{{\rm{total}}}} = M\sin \left( {2\omega t} \right) + N\cos \left( {2\varphi } \right)\sin \left( {2\omega t} \right) - N\sin \\ \left( {2\varphi } \right)\cos \left( {2\omega t} \right) = \left[ {M + N\cos \left( {2\varphi } \right)} \right]\sin \left( {2\omega t} \right) - \left[ {N\sin } \right.\\ \left. {\left( {2\varphi } \right)} \right]\cos \left( {2\omega t} \right) \end{array}$ (7)

 ${A_1} = \sqrt {{M^2} + {N^2} + 2MN\cos \left( {2\varphi } \right)}$ (8)

 $\begin{array}{l} \;\;\;\;\;\;{p_{{\rm{X}}1}} + {p_{{\rm{X}}2}} = \frac{{{U^2}{X_1}}}{{R_1^2 + X_1^2}}\sin \left( {2\omega t - 2{\varphi _1}} \right) + \frac{{{U^2}{X_2}}}{{R_2^2 + X_2^2}}\sin \\ \left( {2\omega t - 2{\varphi _2} - {\rm{ \mathsf{ π} }}} \right) \end{array}$ (9)

$J=\frac{{{U}^{2}}{{X}_{1}}}{R_{1}^{2}+X_{1}^{2}}, K=\frac{{{U}^{2}}{{X}_{2}}}{R_{2}^{2}+X_{2}^{2}}$，则式(9)可以简化为

 ${p_{{\rm{X}}1}} + {p_{{\rm{X}}2}} = J\sin \left( {2\omega t - 2{\varphi _1}} \right) + K\sin \left( {2\omega t - 2{\varphi _2} - {\rm{ \mathsf{ π} }}} \right)$ (10)

 ${A_2} = \sqrt {{J^2} + {K^2} - 2JK\cos \left( {2{\varphi _1} - 2{\varphi _2}} \right)}$ (11)

2.3 三相电路中的无功功率分析

 图 2 三相负载Y型联接电路模型 Fig.2 Three-phase load circuit of type Y connection

 ${i_{\rm{A}}} = \sqrt 2 \frac{U}{{\left| {{Z_{\rm{A}}}} \right|}}\sin \left( {\omega t - {\varphi _{\rm{A}}}} \right) = \sqrt 2 {I_{\rm{A}}}\sin \left( {\omega t - {\varphi _{\rm{A}}}} \right)$

 ${i_{\rm{B}}} = \sqrt 2 {I_{\rm{B}}}\sin \left( {\omega t - {{120}^ \circ } - {\varphi _{\rm{B}}}} \right)$
 ${i_{\rm{C}}} = \sqrt 2 {I_{\rm{C}}}\sin \left( {\omega t + {{120}^ \circ } - {\varphi _{\rm{C}}}} \right)$

 $\begin{array}{l} \;\;\;\;\;\;{p_{\rm{A}}} = {u_{\rm{A}}}{i_{\rm{A}}} = 2U{I_{\rm{A}}}\sin \left( {\omega t} \right)\sin \left( {\omega t - {\varphi _{\rm{A}}}} \right) = 2U{I_{\rm{A}}}\sin \\ \left[ {\left( {\omega t - {\varphi _{\rm{A}}}} \right) + {\varphi _{\rm{A}}}} \right]\sin \left( {\omega t - {\varphi _{\rm{A}}}} \right) = 2{P_{\rm{A}}}{\sin ^2}\left( {\omega t - {\varphi _{\rm{A}}}} \right) + \\ {Q_{\rm{A}}}\sin \left( {2\omega t - 2{\varphi _{\rm{A}}}} \right) \end{array}$ (12)
 $\begin{array}{l} \;\;\;\;\;\;{p_{\rm{B}}} = {u_{\rm{B}}}{i_{\rm{B}}} = 2U{I_{\rm{B}}}\sin \left( {\omega t - {{120}^ \circ }} \right)\sin \left( {\omega t - {{120}^ \circ } - } \right.\\ \left. {{\varphi _{\rm{B}}}} \right) = 2U{I_{\rm{B}}}\sin \left[ {\left( {\omega t - {{120}^ \circ } - {\varphi _{\rm{B}}}} \right) + {\varphi _{\rm{B}}}} \right]\sin \left( {\omega t - {{120}^ \circ } - } \right.\\ \left. {{\varphi _{\rm{B}}}} \right) = 2{P_{\rm{B}}}{\sin ^2}\left( {\omega t - {{120}^ \circ } - {\varphi _{\rm{B}}}} \right) + {Q_{\rm{B}}}\sin \left( {2\omega t - 2{\varphi _{\rm{B}}} + {{120}^ \circ }} \right) \end{array}$ (13)
 $\begin{array}{l} \;\;\;\;\;\;{p_{\rm{C}}} = {u_{\rm{C}}}{i_{\rm{C}}} = 2U{I_{\rm{C}}}\sin \left( {\omega t + {{120}^ \circ }} \right)\sin \left( {\omega t + {{120}^ \circ } - } \right.\\ \left. {{\varphi _{\rm{C}}}} \right) = 2{P_{\rm{C}}}{\sin ^2}\left( {\omega t + {{120}^ \circ } - {\varphi _{\rm{C}}}} \right) + {Q_{\rm{C}}}\sin \left( {2\omega t - 2{\varphi _{\rm{C}}} - {{120}^ \circ }} \right) \end{array}$ (14)

 $\begin{array}{l} {q_{总}} = {q_{\rm{A}}} + {q_{\rm{B}}} + {q_{\rm{C}}} = {Q_{\rm{A}}}\sin \left( {2\omega t - 2{\varphi _{\rm{A}}}} \right) + {Q_{\rm{B}}}\sin \\ \left( {2\omega t - 2{\varphi _{\rm{B}}} + {{120}^ \circ }} \right) + {Q_{\rm{C}}}\sin \left( {2\omega t - 2{\varphi _{\rm{C}}} - {{120}^ \circ }} \right) = \\ \left[ {{Q_{\rm{A}}}\cos \left( {2{\varphi _{\rm{A}}}} \right) + {Q_{\rm{B}}}\cos \left( {{{120}^ \circ } - 2{\varphi _{\rm{B}}}} \right) + {Q_{\rm{C}}}\cos \left( {{{120}^ \circ } + } \right.} \right.\\ \left. {\left. {2{\varphi _{\rm{C}}}} \right)} \right]\sin \left( {2\omega t} \right) + \left[ {{Q_{\rm{B}}}\sin \left( {{{120}^ \circ } - 2{\varphi _{\rm{B}}}} \right) - {Q_{\rm{C}}}\sin \left( {{{120}^ \circ } + } \right.} \right.\\ \left. {\left. {2{\varphi _{\rm{C}}}} \right) - {Q_{\rm{A}}}\sin \left( {2{\varphi _{\rm{A}}}} \right)} \right]\cos \left( {2\omega t} \right) \end{array}$

q整理成标准正弦时域函数形式，可得总非耗散功率的幅值：

 ${A_3} = \sqrt {Q_{\rm{A}}^2 + Q_{\rm{B}}^2 + Q_{\rm{C}}^2 + 2{Q_{\rm{A}}}{Q_{\rm{B}}}\cos \left( {2{\varphi _{\rm{A}}} - 2{\varphi _{\rm{B}}} + {{120}^ \circ }} \right) + 2{Q_{\rm{B}}}{Q_{\rm{C}}}\cos \left( {2{\varphi _{\rm{B}}} - 2{\varphi _{\rm{C}}} + {{120}^ \circ }} \right) + 2{Q_{\rm{C}}}{Q_{\rm{A}}}\cos \left( {2{\varphi _{\rm{C}}} - 2{\varphi _{\rm{A}}} + {{120}^ \circ }} \right)}$ (15)

 ${\varphi _{\rm{A}}} = {\varphi _{\rm{B}}} = {\varphi _{\rm{C}}}$
 ${Q_{\rm{A}}} = {Q_{\rm{B}}} = {Q_{\rm{C}}}$

 ${\varphi _{\rm{A}}} = {\varphi _{\rm{B}}} = {\varphi _{\rm{C}}} = \varphi$
 ${Q_{\rm{A}}} = {Q_{\rm{B}}} = {Q_{\rm{C}}} = Q$

 $\begin{array}{l} {A_3} = \\ \sqrt {{Q^2} + {Q^2} + {Q^2} + 2{Q^2}\cos {{120}^ \circ } + 2{Q^2}\cos {{120}^ \circ } + 2{Q^2}\cos {{120}^ \circ }} = 0 \end{array}$

A3=0说明此种工况下电源与负载间没有无功能量的流动，无功能量只在负载间流动，故任一瞬时三相电源输出的总功率总是等于各相有功功率之和；同理，通过计算式(12) ~ (14)中第一项之和可知三相总有功分量在任一瞬时为常数(即三相电源的总功率为常数)。文献[11]佐证了这一结论。

3 结论

(1) 仅就单频正弦稳态电路而言，元件的电压、电流之间存在的正交分量导致了无功功率，对应不同元件模型的瞬时功率可分解为不同形式的有功分量和无功分量。

(2) 对典型电路的无功分量计算分析表明：感性负载和容性负载的能量吞吐不一定反相互补，无功能量既可能在电源与负载间流动、也可能只在负载间流动，或二者兼而有之，即电源与负载间存在功率交换不是无功功率产生的必要条件；无源网络的总吞吐幅度不一定等于网络中各负载无功功率的代数和；经典无功功率定义是有局限性的。

 [1] 秦文萍, 王鹏, 韩肖清, 等. 电力系统有功和无功功率充裕度评估[J]. 电力系统自动化, 2014, 38(1): 28-33. Qin W P, Wang P, Han X Q, et al. Adequacy evaluation of active and reactive power in power system[J]. Automation of Electric Power Systems, 2014, 38(1): 28-33. DOI:10.7500/AEPS20130529009 (in Chinese) [2] 董大伟. 浅谈无功功率的产生和补偿技术[J]. 中国电力教育, 2011(27): 104-105. Dong D W, Discussion on reactive power generation and compensation technology[J]. China Electric Power Education, 2011(27): 104-105. (in Chinese) [3] 秦曾煌. 电工学(上册)[M]. 7版. 北京: 高等教育出版社, 2009, 108-181. Qin Z H. Electro technician (the first volume)[M]. 7th ed. Beijing: Higher Education Press, 2009: 108-181. (in Chinese) [4] 李翰荪. 简明电路分析基础[M]. 北京: 高等教育出版社, 2014, 457-462. Li H S. Fundamentals of concise circuit analysis[M]. Beijing: Higher Education Press, 2014: 457-462. (in Chinese) [5] 燕庆明. 电路分析基础[M]. 3版. 北京: 高等教育出版社, 2012, 173-220. Yan Q M. Electric circuit analysis[M]. 3rd ed. Beijing: Higher Education Press, 2012: 173-220. (in Chinese) [6] Alexander C K, Sadiku M N O. Fundamentals of electric circuits[M]. Beijing: China Machine Press, 2013: 457-488. [7] 包国荣. 论正确理解无功功率概念[J]. 咸宁师专学报, 1993(2): 43-47. Bao G R, On correctly understanding the concept of reactive power[J]. Journal of Xianning Teachers College, 1993(2): 43-47. (in Chinese) [8] 田社平, 陈洪亮. 关于无功功率的讨论[J]. 电气电子教学学报, 2012, 34(1): 23-35. Tian S P, Chen H L, Discussion on reactive power[J]. Journal of Electrical & Electronic Engineering Education, 2012, 34(1): 23-35. (in Chinese) [9] 汪小娜, 单潮龙, 何坊, 等. 对无功功率物理意义的讨论[J]. 物理与工程, 2014, 24(2): 34-36. Wang X N, Shan C L, He F, et al. Disscussion on physical significance of reactive power[J]. Physics and Engineering, 2014, 24(2): 34-36. (in Chinese) [10] 王茂海, 孙元章. 三相电路中功率现象的解释及无功功率的分类[J]. 中国电机工程学报, 2003, 23(10): 63-66. Wang M H, Sun Y Z, Analysis of power phenomenon and classification of reactive power in three-phase circuit[J]. Proceedings of the CSEE, 2003, 23(10): 63-66. DOI:10.3321/j.issn:0258-8013.2003.10.013 (in Chinese) [11] Alok Kumar, Reactivepower control in electrical power transmission system[J]. International Journal of Engineering Trends and Technology, 2013, 4(5): 1707-1709.