Bài giảng Power Circuits and Electromechanics - Lecture 1 - Nguyễn Quang Nam

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ECE430

Power Circuits and Electromechanics

Dr. Nam Nguyen-Quang

Fall 2009

http://www4.hcmut.edu.vn/~nqnam/lecture.php

Lecture 1

Review of power

 Assuming sinusoidal voltage and current, i.e.

 





 





v t V_mcos t _v

i t  I_mcos t _i

 Instantaneous power (i = I_mwhen t = 0)

 

   





 

p t  v t i t V_mI_mcos t _v_icos t

 Average (also real or active) power over one period T = 2/

V_mI_m









P 

cos _v_iV_rmsI_rmscos _v_i

where V_rmsand I_rmsare root mean square (rms) voltage and current,

respectively.  = _v _iis called power factor angle, and cos() is called

power factor (PF).

Lecture 1

Review of phasor

 Sinusoidal quantities can be represented as phasors, for example

V  V_rms_v

I  I_rms_i

Phase angle

Magnitude

Lagging power factor

Leading power factor

_i

_v

_i

_v

An inductive load exhibits a lagging PF, and a capacitive load has a

leading PF.

Lecture 1

Class examples

 Ex. 2.1: Write phasor forms of v(t) & i(t), and find the average power P





 210cos

 25cos

t  20⁰

 20

cos  32.14 W

50⁰



t  30⁰ V  1030⁰







 I  5  20⁰

 50⁰

(lagging PF)

  _v_i 30 





P  10











 Ex. 2.2: Recalculate the average power P with new i(t)





 25cos



t 90⁰ I  5  90⁰



120⁰



 25 W

(generating power!)

P 





cos





Lecture 1

Review of complex power

 Define reactive power as

V_mI_m

Q 

sin

_v_i



 V_rmsI_rmssin

_v_i



 The instantaneous power can be expressed





 P  Pcos



2t



Qsin



2t



 P



1 cos



2t





Qsin



2t



j_i

 As V  V_rmse ^jand I  I_rmse , it can be seen that

P  Re



V  I ^*



 V_rmsI_rmscos



_v_i



Q  Im

 V_rmsI_rmssin



_v_i

 Complex power is defined as

S 



V  I^*

 P  jQ



Lecture 1

Review of complex power (cont.)

 In power calculations, rms values are always used. Hence, the

subscript rms will be omitted hereafter

P  VI cos



_v_i



Q  VI sin

S VI



_v_i



 And magnitude of complex power is

 To distinguish S, P, and Q, their units are voltamperes (VA), watts

(W), and voltampere reactive (VAR), respectively

 Alternative forms of complex power

S  ZII ^* I ²Z  I ²



R  jX  P  jQ



Z  R  jX

V  ZI

P  I ²R

Hence

Q  I ²X

Lecture 1

Class examples

 Ex. 2.4: Find complex power for given v(t) and i(t)





 210cos



t 10⁰



V 1010⁰

 I  20  20⁰



VI ^*

 220sin



t  70⁰



S 









1010⁰





2020⁰



 20030⁰173.2 j100 VA

P 173.2 W

Q 100 VAR

 Ex. 2.5 and 2.6: see text book

Lecture 1

Conservation of complex power

 Series circuit





S V  I  V₁V₂...V_nI  S₁ S₂... S_n

 Parallel circuit





S V  I V I₁ I₂... I_n S₁ S₂... S_n

 Total complex power is the sum of complex powers in individual

elements. Most loads are connected in parallel. Conservation of complex

power

P  P  P  ... P

Q  Q₁ Q₂ ... Q_n

 Power triangle: see example 2.7

Lecture 1

Class examples

 Ex. 2.7: Find complex power in the form of a power triangle

S VI^*



10010⁰





10  26.8⁰



100036.8⁰ 800 j600 VA

Hence

Q  600 VAR

P  800 W

VI 1000 VA

Since  > 0, current lags voltage,

Q = 600

VAR

and the load is an inductive one.

36.8⁰

P = 800 W

 Ex. 2.8, 2.9 and 2.10: see text book

Lecture 1

Specifying power in a load

 Power absorbed by the load can be specified by a combination of

three of the following six quantities: V, I, PF (lag or lead), S, P, Q.

 If and are specified it is equivalent to specifying V, I, and PF

 Another way is to specify V, PF, and P

I 

Q  VI sin

S  P  jQ

V cos

 The third way is to specify V, PF, and S: I is calculated from V and S,

then Q can be calculated from S and PF

 The final way is to specify V, P, and Q: S is calculated from P and Q,

then PF is calculated from P and S

Lecture 1

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