Download presentation

Presentation is loading. Please wait.

1
Since Therefore Since

2
**You can see that that the frequency of the Instantaneous**

power is twice the frequency of the voltage or current

3
**The instantaneous power**

real (or average) power (Watts) Which is the actual power absorb by the element Examples Electric Heater , Electric Stove , oven Toasters, Iron …etc reactive power Which is the reactive power absorb or deliver by the element Reactive power represents energy stored in reactive elements (inductors and capacitors). Its unit is Volt Ampere Reactive (VAR)

4
**The instantaneous power**

real (or average) power (Watts) reactive power (Volt Ampere Reactive (VAR) )

5
Complex Power Previously, we found it convenient to introduce sinusoidal voltage and current in terms of the complex number the phasor Definition Let the complex power be the complex sum of real power and reactive power

6
**Advantages of using complex power**

- We can compute the average and reactive power from the complex power S - complex power provide a geometric interpretation were The geometric relations for a right triangle mean the four power triangle dimensions ( , P, Q, ) can be determined if any two of the four are known

7
Power Calculations were Is the conjugate of the current phasor Circuit Since Similarly

8
**Power Calculations Summery**

Circuit

9
**This implies that in any circuit, conservation of average power and **

In any circuit, conservation of complex power is achieved This implies that in any circuit, conservation of average power and Conservation of reactive power are achieved However, the apparent power (the magnitude of the complex power) is not conserved

10
**Ex:6.13 Determine the average and reactive power delivered by the source.**

The phasor current leaving the source is The average power delivered by the source is:

11
**The reactive power delivered by the source is:**

And the complex power delivered by the source is

12
**The voltage across the elements are:**

Determine the average power and reactive power delivered to each element The voltage across the elements are: Thus the complex power delivered to each element is

13
**show that conservation of complex power, average power, and reactive power is achieved.**

14
**6.6.1 Power Relations for the Resistor**

The voltage and current are in phase so Average power is: Reactive power is zero for resistor

15
**6.6.1 Power Relations for the Inductor**

The voltage leads the current by 90 so that

16
**6.6.1 Power Relations for the Capacitor**

The current leads the voltage by 90 so that

17
**Recall the Instantaneous power p(t)**

The power factor Recall the Instantaneous power p(t) The angle qv - qi plays a role in the computation of both average and reactive power The angle qv - qi is referred to as the power factor angle We now define the following : The power factor

18
The power factor Knowing the power factor pf does not tell you the power factor angle , because To completely describe this angle, we use the descriptive phrases lagging power factor and leading power factor Lagging power factor implies that current lags voltage hence an inductive load Leading power factor implies that current leads voltage hence a capacitive load

19
6.6.2 Power Factor Since

20
Circuit

21
Circuit

22
EX:6.15 Determine the average and reactive powers delivered to the load impedance and the power factor of the load Average power

23
Average power

24
reactive powers

25
reactive powers OR

26
EX:6.15 Determine the average and reactive powers delivered to the load impedance and the power factor of the load OR

27
**This could also be calculated from the complex power **

delivered to the load The power factor of the load is: The load is lagging because the current lags the voltage

28
**A typical power distribution circuit**

The consumer is charged for the average power consumed by the load The load requires a certain total apparent power

29
**The powers consumed in the line losses**

Ex Suppose that the load voltage figure is 170V The line resistance is 0.1 ohm The load requires 10KW of average power. Examine the line losses for a load power factor of unity and for a power factor of 0.7 lagging. The load current is obtained from For unity power factor this is For power factor of 0.7 The powers consumed in the line losses 720 W extra power to be generated if pf is 0.7 to supply the load

30
**Power Factor Correction**

Ex: 6.17: in Ex 6.16 determine the value of capacitor across the load to correct the power factor from 0.7 to unity if power frequency is 60Hz. From Ex 6.16 For power factor of 0.7 power factor 0.7 lagging The current through the added capacitor is: Hence the total current Unity power factor Imaginary component of the line current is zero

32
**The average power delivered to the load is:**

33
**6.6.3 Maximum Power Transfer**

Source-load Configuration Determine the load impedance so that maximum average power is delivered to that load. Represent the source and the load impedances with real and imaginary parts: The load current is:

34
**The average power delivered to the load is:**

Since the reactance can be negative and to max value, we choose leaving Differentiate with respect to RL and set to zero to determine required RL which is RL= RS Hence: In this case the load is matched to the source. The max power delivered to the load becomes:

35
**6.6.4 Superposition of Average Power**

Average power computation when circuit contains more than one source

37
**The instantaneous power delivered to the element is**

Substituting Using the identity

38
Using the identity

39
**The instantaneous power becomes**

Average powers delivered individually by the sources Suppose that the two frequencies are integer multiples of some frequency as The instantaneous power becomes

40
**Averaging the instantaneous over the common period**

THUS: we may superimpose the average powers delivered by sources of different frequencies, but we may not, in general, apply superposition to average power if the sources are of the same frequency. where

41
**Ex 6.18: Determine the average power delivered by the two sources of the circuit**

Hence the average power delivered by the voltage source is This can be confirmed from average powers delivered to the two resistors

42
**of average powers delivered individually by each source**

By current division: The voltage across the current source is Hence the average power delivered by the current source is This may be again confirmed by computing the average power delivered to the Two resistors: Since frequencies are not the same, total average power delivered is the sum of average powers delivered individually by each source

43
**EX 6.19: Determine the average power delivered by the two sources**

Since both sources have the same frequency, we can’t use superposition. So we include both sources in one phasor circuit. The total average power delivered by the sources is equal to the average power delivered to the resistor We use superposition on the phasor circuit to find the current across the resistor

44
The phasor current is: Hence the average power delivered to the resistor is Note that we may not superimpose average powers delivered to the resistors by the individual sources We can compute this total average power by directly computing the average power delivered by the sources from the phasor circuit The voltage across the current source is The average power delivered by voltage source is The average power delivered by the current source is The total average power delivered by the sources is

45
**6.6.5 Effective (RMS) Values of Periodic Waveforms**

Sinusoidal waveform is one of more general periodic waveforms Apply a periodic current source with period T on resistor R The instantaneous power delivered to the resistor is The average power delivered to the resistor is Hence the average power delivered to the resistor by this periodic waveform can be viewed as equivalent to that produced by a DC waveform whose value is This is called the effective value of the waveform or the root-mean-square RMS value of the waveform

46
Ex 6.20 Determine the RMS value of the current waveform and the average power this would deliver to resistor The RMS value of the waveform is Hence the average power delivered to the resistor is

47
**RMS voltages and currents in phasor circuits**

The sinusoid has a RMS value of Hence the average power delivered to a resistor by a sinusoidal voltage or current waveform is In general, the average power delivered to an element is Therefore, if sinusoidal voltages and currents are specified in their RMS values rather than their peak values, the factor ½ is removed from all average-power expressions. However, the time-domain expressions require a magnitude multiplied by square root of 2 Since X is the peak value of the waveform. Common household voltage are specified as 120V. This is the RMS value of the peak of 170V.

48
**Ex 6.21 Determine the average power delivered by the source and the time-domain current i(t)**

Phasor circuit with rms rather than peak The phasor current is Hence the average power delivered by the source is The time-domain current is

49
**Commercial Power Distribution**

50
**Time domain representation**

54
Using Vectors

58
**We are going to investigate the transmission of power from **

6.9.1 Wye-Connected Loads We are going to investigate the transmission of power from three phase generator to two types of load: I – WYE connected load Hence the current returning through the neutral wire is zero, and the neutral may be removed

59
Power calculation For a balanced wye-connected load, the voltages across the individual loads are the respective phase voltages whether the neutral is connected or not. The power delivered to the individual loads is three times the power delivered to an individual load, because the individual loads are identical. No ½ factor in power expression because values are rms

60
**Since the line voltage is more accessible than the phase voltage**

Power in terms of line-to-line voltages and load current gives

61
**Hence, the average power delivered to each load is**

Example 6.24 Consider a balanced, wye-connected load where each load impedance is the phase voltages are 120 V. Determine the total average power delivered to the load. The line currents are Hence, the average power delivered to each load is The total average power delivered to the load is

62
Example 6.25 If the line voltage of a balanced, wye-connected load is 208 V and the total average power delivered to the load is 900 W, determine each load if their power factors are 0.8 leading. Thus The phase voltage is Thus the magnitude of the individual load impedance is Since the power factor is 0.8 leading (current leads voltage; voltage lags current) Thus the individual loads are

63
Example 6.25 If the line voltage of a balanced, wye-connected load is 208 V and the total average power delivered to the load is 900 W, determine each load if their power factors are 0.8 leading. Thus The phase voltage is Thus the magnitude of the individual load impedance is Since the power factor is 0.8 leading (current leads voltage; voltage lags current) Thus the individual loads are

Similar presentations

Presentation is loading. Please wait....

OK

Sinusoidal Steady-State Power Calculations

Sinusoidal Steady-State Power Calculations

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google