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learn:courses:real-analog-chapter-12:start [2017/01/19 19:25] – created Marthalearn:courses:real-analog-chapter-12:start [2022/10/04 11:38] (current) – external edit 127.0.0.1
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 ====== Real Analog: Chapter 12 ====== ====== Real Analog: Chapter 12 ======
 +[[{}/learn/courses/real-analog-chapter-11/start|Back to Chapter 11]]
 +-->Chapter 12 Materials#  
 +  * Lecture Material:
 +    * {{ :learn:courses:real-analog-chapter-12:lecture31.ppt |Lecture 31 PowerPoint Slides}}: Sinusoidal steady-state power, instantaneous & average power, reactive power, complex power, power factor
 +    * {{ :learn:courses:real-analog-chapter-12:lecture32.ppt |Lecture 32 PowerPoint Slides}}: Review: AC power analysis (average & complex power, power triangles, RMS values, power factor), power factor correction
 +    * [[http://www.youtube.com/watch?v=LQAhsOeR73c&list=PLDEC730F6A8CDE318&index=34&feature=plpp_video| Lecture 31 Video]]
 +    * [[http://www.youtube.com/watch?v=t36bRysjZCA&list=PLDEC730F6A8CDE318&index=35&feature=plpp_video| Lecture 32 Video]]
 +  * Chapter 12 Video:
 +    * [[http://www.youtube.com/watch?v=k0mdwQvn2iw&list=PL170A01159D42313D&index=25&feature=plpp_video| Lab 12 Video 1]]: AC Power & Power Factor: An example of the role of power factor in the transmission of AC power. Power factor correction is used to improve the efficiency of power transmission.
 +  * {{ :learn:courses:real-analog-chapter-12:real-analog-chapter-12.pdf | Chapter 12 Complete PDF}}
 +    * {{ :learn:courses:real-analog-chapter-12:lab12p4p1.pdf |Lab 12.4.1}}
 +  * {{ :learn:courses:real-analog-chapter-12:realanalog-exercisesolutions-chapter12.pdf |Exercise Solutions}}: Chapter 12 exercise solutions
 +  * {{ :learn:courses:real-analog-chapter-12:homework12.docx |Homework}}: Chapter 12 homework problems
 +
 +
 +<--
 +
 +
 ====== 12. Introduction and Chapter Objectives ====== ====== 12. Introduction and Chapter Objectives ======
 In this chapter we will address the issue of power transmission via sinusoidal (or AC) signals. This topic is extremely important, since the vast majority of power transmission in the world is performed using AC voltages and currents. In this chapter we will address the issue of power transmission via sinusoidal (or AC) signals. This topic is extremely important, since the vast majority of power transmission in the world is performed using AC voltages and currents.
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 Solving this for $I_{eff}$ results in: Solving this for $I_{eff}$ results in:
  
-$$I_{eff} = I_{RMS} = \sqrt{\frac{1}{T} \int\limits_{t_0}^{t_0 + T} i^2 (t)dt} (Eq. 12.12)+$$I_{eff} = I_{RMS} = \sqrt{\frac{1}{T} \int\limits_{t_0}^{t_0 + T} i^2 (t)dt} (Eq. 12.12)$$
  
 And the effective current is the square root of the mean of the square of the time-varying current. This is also called the RMS (or root-mean-square) value, for rather obvious reasons. And the effective current is the square root of the mean of the square of the time-varying current. This is also called the RMS (or root-mean-square) value, for rather obvious reasons.
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   * The RMS value of a sinusoidal signal $f(t) = F_m \cos(\omega t + \theta)$ is given by:   * The RMS value of a sinusoidal signal $f(t) = F_m \cos(\omega t + \theta)$ is given by:
  
-$$f_{RMS} = \frac{F_m}{\sqrt{2}$$+$$f_{RMS} = \frac{F_m}{\sqrt{2}}$$
  
   * The above formula __cannot__ be used for any signal other than a pure sinusoid with no offset.   * The above formula __cannot__ be used for any signal other than a pure sinusoid with no offset.
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 $$\underline{V} = Z_L \cdot \underline{I}  (Eq. 12.18)$$ $$\underline{V} = Z_L \cdot \underline{I}  (Eq. 12.18)$$
 +
 +Where $\underline{V}$ is the voltage phasor across the load, $\underline{I}$ is the current phasor through the load, and $Z_L$ is the load impedance. Thus, the difference between the voltage and current phase angles is simply the phase angle of the load impedance: $\theta_v - \theta_i = \angle \theta_{ZL}$. Therefore, the load impedance sets the power factor. If the load is purely resistive, $\theta_v = \theta_i$, the power factor is one, and the average power is the same as the apparent power. If the load is purely imaginary (as with purely inductive or purely capacitive loads) the power factor is zero and there is no average or real power absorbed by the load.
 +
 +==== Section Summary ====
 +  * //Apparent power// and //power factor// provide an alternate method for characterizing the average power delivered to the load. If the average power is:
 +
 +$$P= \frac{V_m I_m}{2} \cos(\theta_v - \theta_i)$$
 +
 +The apparent power is $\frac{V_m I_m}{2}$ and the power factor is $\cos(\theta_v - \theta_i)$.
 +
 +==== Exercises ====
 +  - The current and voltage delivered to a load are $i(t) = 2 \cos(100t)$ and $v(t) = 120 \cos(100t+65^{\circ})$, respectively. What are the apparent power and power factor of the power delivered to the load?
 +  - A load consumes 100kW with a power factor $pf = 0.85$ (lagging). If the load current is 256A (RMS), find the load voltage.
 +  - An industrial plant has a load which consumes 20kW of power from a $220V_{RMS}$ line. If the power factor is 0.9 (lagging), what is the difference in angle between the load voltage and the load current?
 +
 +
 +----
 +
 +===== 12.5: Complex Power =====
 +Apparent power, average power, reactive power, and power factor can all be represented simultaneously in a single parameter called complex power. If we define complex power as:
 +
 +$$S= \frac{\underline{VI}^{*}}{2}  (Eq. 12.19)$$
 +
 +Where $\underline{V}$ is the phasor representing the voltage, $\underline{I}$ is the phasor representing the current, and the superscript * denotes complex conjugation (simply changing the sign on the imaginary part of the phasor). 
 +
 +If we substitute the magnitude and phase angle representations for the phasors in equation (12.19), we obtain (since complex conjugation simple changes the sign of the phase angle of a complex number):
 +
 +$$S = \frac{1}{2} (V_m \angle \theta_v) (I_m \angle - \theta_i) = \frac{V_m I_m}{2} \angle (\theta_v - \theta_i)  (Eq. 12.20)$$
 +
 +So that the complex power $S$ is a complex number with magnitude $\frac{V_m I_m}{2}$ (or, equivalently, $V_{RMS} I_{RMS}$) and phase angle $\theta_v - \theta_i$. It is easy to see that the magnitude of the comples power is simply the apparent power.
 +
 +If we represent $S$ in rectangular coordinates, we obtain:
 +
 +$$S=P+jQ  (Eq. 12.21)$$
 +
 +Where $P$ is the average power:
 +
 +$$P=\frac{V_m I_M}{2} \cos(\theta_v - \theta_i) = V_{RMS} I_{RMS} \cos(\theta_v - \theta_i)$$
 +
 +As before, and Q is the reactive power:
 +
 +$$Q= \frac{V_m I_m}{2} \sin(\theta_v - \theta_i) = V_{RMS} I_{RMS} \sin(\theta_v - \theta_i)$$
 +
 +Also as before. 
 +
 +The complex power, real power, reactive power, and apparent power can be represented graphically in the complex plane as a //power triangle//, as shown in Figure 12.4 below.
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12d.png |Figure 12.4.}}
 +
 +One important thing to note about Figure 12.4 is that this figure differs from a phasor diagram – the components shown on the power triangle are __not__ phasors, since they do not provide magnitude and phase information about sinusoidal signals. The vectors shown in Figure 12.4 are simply complex numbers.
 +
 +We conclude this chapter with an example.
 +
 +
 +----
 +
 +==== Example 12.1 ====
 +For the circuit below,
 +  - Find the average power delivered by the source
 +  - Find the power absorbed by the resistor
 +  - Find the apparent, real, and reactive powers delivered by the source
 +  - Sketch a power triangle for the source
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12e.png |Example image 1.}}
 +
 +The frequency domain circuit is shown below.
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12f.png |Example image 2.}}
 +
 +The current delivered by the source is therefore: 
 +
 +$$\underline{I} = \frac{100 \angle 0^{\circ}}{(10-j8)\Omega} = 7.8 \angle 38.66^{\circ} A$$
 +
 +  * The average power delivered by the source is therefore:
 +
 +$$P=\frac{(100V)(7.8A)}{2} \cos(0^{\circ} - 38.66^{\circ}) = 305W$$
 +
 +  * The power absorbed by the resistor is the same as the average power delivered by the source, so $P_{10\Omega}=305W$, which is the same as $R \cdot I^2_{RMS}$.
 +  * The apparent power is $\frac{(100V)(7.8A)}{2}=391VA$, the reactive power is $Q = \frac{(100V)(7.8A)}{2} \sin(0^{\circ}-38.66^{\circ}) = -244VAR$, and the real power is simply the average power, $P=305W$.
 +  * The power triangle is shown below:
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12g.png |Example image 3.}}
 +
 +----
 +
 +=== Section Summary ====
 +  * It is often convenient to express power as a complex number, S. This quantity expresses, simultaneously, the average power, the reactive power, the apparent power, and the power factor. Complex power can be determined from the voltage and current phasors as:
 +
 +$$S=\frac{\underline{VI}^{*}}{2}$$
 +
 +Where * denotes complex conjugation.
 +
 +  * As with any other complex number, complex power can be expressed in either rectangular or polar coordinates.
 +    * In rectangular coordinates, the complex power readily provides the real power ($P$) and reactive power ($Q$):
 +
 +$$S=P+jQ$$
 +
 +    * In polar coordinates, the complex power readily provides the apparent power and power factor:
 +
 +$$S=\frac{V_m I_m}{2} \angle (\theta_v - \theta_i)$$
 +
 +Where the apparent power is $\frac{V_m I_m}{2}$ and the power factor is given by $\cos(\theta_v - \theta_i)$.
 +
 +==== Exercises ====
 +**1.** The current and voltage delivered to a load are $i(t) = 2 \cos(100t-30${\circ})$ and $v(t) = 120 \cos(100t+65^{\circ})$, respectively. What is the complex power delivered to the load?
 +
 +**2.** For the circuit shown, $i(t) = 2 \cos(100t)$ and $v(t) = 120 \cos(100t+65^{\circ})$. Find:
 +  - The complex power delivered by the source
 +  - The average power delivered by the source
 +  - The power dissipated by the resistor
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12h.png |Example image 1.}}
 +
 +**3.** For the circuit shown, find
 +  - $i(t)$
 +  - The complex power delivered by the source
 +  - The average power delivered by the source
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12i.png |Example image 2.}}
 +
 +
 +----
 +
 +===== 12.6: Power Factor Correction =====
 +In previous sections, we saw that power can be considered to be real (average power) or imaginary (reactive power). Reactive power is power which is provided to energy storage elements such as inductors or capacitors; this power is returned to the power source, since inductors and capacitors have no energy dissipation mechanism. The power factor provides a metric for assessing the amount of real power relative to the reactive power delivered to a load. The higher the power factor, the larger the amount of real power relative to reactive power – ideally, a power factor of one means that no reactive power is provided to the load.
 +
 +Power companies in general cannot charge customers for power which is returned to the power company, so delivering reactive power to a customer is not productive from a power company’s standpoint. In fact, since transmission losses will typically result in real power losses due to transmission of reactive power, the power company actually __loses__ power when transmitting reactive power, for which no one pays! Unsurprisingly, this is not popular with power companies.
 +
 +Large power users, such as factories, may have requirements placed upon them by the power company to provide a minimum power factor for their loads. If the factory cannot meet the required power factor, the power company can refuse to supply power or charge an increased rate for the power they do provide. This chapter illustrates how to re-design an inductive load to increase its power factor.
 +
 +Most large power users’ loads are inductive in nature. Therefore, in this section, we will only consider power factor correction for inductive loads. We illustrate the overall process in the context of an example.
 +
 +
 +----
 +
 +==== Example 12.2 ====
 +Determine the power factor for the circuit below if $vs(t)=100 \cos(377t)$. Re-design the load so that the power factor is one.
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12j.png |Example image 1.}}
 +
 +The load in this case consists of a $100\Omega$ resistor in parallel with a 1H inductor. To determine the power factor, we need to determine the current delivered by the source; we do this by determining the equivalent impedance of the load, $Z_L$:
 +
 +$$Z_L = \frac{(100 \Omega)(j377 \Omega)}{100 \Omega + j377 \Omega} = 96.67 \angle 14.86^{\circ}$$
 +
 +Therefore, the current phasor delivered by the source is:
 +
 +$$\underline{I}_s = \frac{\underline{V}_s}{Z_L} = \frac{100 \angle 0^{\circ}}{96.67 \angle 14.86^{\circ}} = 1.03 \angle - 14.86^{\circ}$$
 +
 +The power factor is:
 +
 +$$pf = \cos(\theta_v - \theta_i) = \cos(0^{\circ} - (-14,86^{\circ})) = 0.967 \text{(lagging)}$$
 +
 +Note that the power factor is lagging, since current lags voltage, this is because the load is inductive in nature.
 +
 +We can change the power factor of the load by adding a capacitor in parallel with the inductive load. The impedance of the capacitor can be used, in essence, to cancel out the inductive impedance of the load. The re- designed circuit is as shown below. The capacitance, C, must be chosen to provide the desired power factor.
 +
 +{{ :learn:courses:real-analog-chapter-12:chapter12k.png |Example image 2.}}
 +
 +In order to determine the necessary size of the capacitor, we first determine the reactive power of the uncompensated system. Without the capacitor, the reactive power is:
 +
 +$$Q= \frac{V_m I_m}{2} \sin(\theta_v - \theta_i) = \frac{(100V)(1.03A)}{2} \sin(14.86^{\circ}) = 13.26 VAR$$
 +
 +Our goal is to use the capacitor to change this reactive power so that the resulting power factor is as desired, without __changing the real power delivered to the load__. The capacitor will induce a negative reactive power; in order to achieve a power factor of exactly one, the reactive power introduced by the capacitor must exactly cancel the reactive power of the original load. Thus, the capacitor’s reactive power must be:
 +
 +$$Q_C = -13.26 VAR$$
 +
 +For the parallel combination of the resistor, capacitor, and inductor, we can claim:
 +
 +$$|Q_C| = \frac{V^2_{RMS}}{|Z_C|} = \frac{V^2_{RMS}}{|\frac{1}{\omega C}|} = \omega CV^2_{RMS}$$
 +
 +Where $|Q_C|$ is the magnitude of the desired reactive power, and $|Z_C|$ is the magnitude of the capactior's impedance. Solving the above expression for the desired capacitance provides:
 +
 +$$C = \frac{|Q_C|}{\omega V^2_{RMS}} = \frac{13.26VAR}{(377 \text{rad/sec}) (\frac{100}{\sqrt{2}^V})} = 4 \mu F$$
 +
 +And a $7 \mu F$ capacitor placed in parallel with the load will give the desired $pf = 1$.
 +
 +
 +----
 +
 +==== Section Summary ====
 +  * A small power factor means that a large portion of the power delivered by the power company to a user is in the form of reactive power. The user does not pay for reactive power, since it is essentially returned to the power company. The power company, however, incurs costs in exchanging reactive power with the user, since transmission of the reactive power results in power dissipation. Power companies, therefore, may require that a user meet a minimum power factor requirement. 
 +  * The power factor of an inductive load can be increased by placing a capacitor in parallel with the load. This process is called power factor correction. The goal is to increase the power factor without changing the average power delivered to the load.
  
  
 +[[{}/learn/courses/real-analog-chapter-11/start|Back to Chapter 11]]
 +[[{}/learn/courses/real-analog/start|Back to Landing Page]]