Electric Circuits II - Philadelphia University...Electric Circuits II AC Circuit Power Analysis 1...

Electric Circuits II AC Circuit Power Analysis

1

Dr. Firas Obeidat

Dr. Firas Obeidat – Philadelphia University

2

1

• Conservation of AC power

1 • Power Triangle

1 • Power Factor Correction

Table of Contents


3

Conservation of AC Power

The principle of conservation of power applies to ac circuits can be explained

by considering AC parallel circuit and AC series circuit.

AC parallel circuit

two load impedances Z1 and Z2 are connected in

parallel across an ac source V. KCL gives

The complex power supplied by the source is

𝐒 = 𝐕𝐫𝐦𝐬𝐈𝐫𝐦𝐬∗ = 𝐕𝐫𝐦𝐬(𝐈𝟏𝐫𝐦𝐬

∗+𝐈𝟐𝐫𝐦𝐬∗) = 𝐕𝐫𝐦𝐬𝐈𝟏𝐫𝐦𝐬

∗+𝐕𝐫𝐦𝐬𝐈𝟐𝐫𝐦𝐬∗= 𝐒𝟏 + 𝐒𝟐

Where S1 and S2 denote the complex powers delivered to loads Z1 and Z2.

The complex, real, and reactive powers of the sources equal the respective sums

of the complex, real, and reactive powers of the individual loads.

𝐈𝐫𝐦𝐬 = 𝐈𝟏𝐫𝐦𝐬 +𝐈𝟐𝐫𝐦𝐬


4

Conservation of AC Power

AC series circuit

If the loads are connected in series with the voltage

source, KVL yields

The complex power supplied by the source is

𝐒 = 𝐕𝐫𝐦𝐬𝐈𝐫𝐦𝐬∗ = (𝐕𝟏𝐫𝐦𝐬+𝐕𝟐𝐫𝐦𝐬)𝐈𝐫𝐦𝐬

∗= 𝐕𝟏𝐫𝐦𝐬𝐈𝐫𝐦𝐬∗+𝐕𝟐𝐫𝐦𝐬𝐈𝐫𝐦𝐬

∗= 𝐒𝟏 + 𝐒𝟐

Where S1 and S2 denote the complex powers delivered to loads Z1 and Z2.

𝐕𝐫𝐦𝐬 = 𝐕𝟏𝐫𝐦𝐬 +𝐕𝟐𝐫𝐦𝐬

From the above discussion, it can be concluded that whether the loads are

connected in series or in parallel (or in general), the total power supplied by the

source equals the total power delivered to the load. Thus, in general, for a source

connected to N loads,


5

Power Triangle

The three quantities average power, apparent power, and reactive power can be

related in the vector domain by

𝐒 = 𝐕𝐫𝐦𝐬𝐈𝐫𝐦𝐬∗=P+jQ

𝐰𝐡𝐞𝐫𝐞 𝐏 = 𝑃∠0, 𝐐𝐋 = 𝑗𝑄𝐿= 𝑄𝐿∠90, 𝐐𝐂 = −𝑗𝑄𝐶= 𝑄𝐶∠ − 90

When solving problems involving power, P

values can be added to get PT, and Q values

to get QT (where Q is positive for inductive

elements and negative for capacitive).

For an inductive load, the phasor power S is

𝐒=P+j𝑸𝑳

For an capacitive load, the phasor power S is

𝐒=P-j𝑸𝑪 (a)

(b)

As in figure (a)

As in figure (b)


6

Power Triangle

Example: The P and Q values for a circuit are shown

in the figure.

a. Determine the power triangle.

b. Determine the rms current supplied by the source.

PT = 700 + 800 + 80 + 120 = 1700 W

QT = 1300 − 600 − 100 − 1200

QT = −600 VAR

ST = PT+jQT=1700-j600=1803∠-19.4o

(a)

(b) 𝐼𝑟𝑚𝑠 =𝑆

𝑉𝑟𝑚𝑠=

1803 VA

120 V= 15 𝐴𝑟𝑚𝑠

The net load is capacitive


7

Power Triangle

Example: The circuit shows a load being fed by a voltage source through a

transmission line. The impedance of the line is represented by (4+j2)Ω the impedance

and a return path. Find the real power and reactive power absorbed by: (a) the source,

(b) the line, and (c) the load.

𝐒𝐬 = 𝐕𝐬𝐫𝐦𝐬𝐈𝐫𝐦𝐬∗

The real power is 2163.5 W and the reactive power is 910.8 VAR (leading).

(a) For the source

(b) For the line


8

Power Triangle

The real power is 455.4 W and the reactive power is

227.76 VAR (lagging).

(c) For the load

𝐒𝐥𝐢𝐧𝐞 = 𝐕𝐥𝐢𝐧𝐞𝐫𝐦𝐬𝐈𝐫𝐦𝐬∗

𝐒𝐋 = 𝐕𝐋𝐫𝐦𝐬𝐈𝐫𝐦𝐬∗

The real power is 1708 W and the

reactive power is 1139 VAR (leading).

Note that SS=Sline+SL


9

Power Triangle

Example: A generator (its voltage in rms) supplies power to an electric heater, an

inductive element, and a capacitor as in the figure.

a. Find P and Q for each load.

b. Find total active and reactive power supplied by the generator.

c. Draw the power triangle for the combined loads and determine total apparent

power.

d. Find the current supplied by the generator.

(a) The components of power are as follows:

(b)


10

Power Triangle

(c)

The apparent power is 3081VA

(c) Generator current is

𝐼 =𝑆𝑇

𝑉=

3081VA

120 V= 25.7 𝐴𝑟𝑚𝑠


11

Power Triangle

Example: In the circuit, Z1=60∠-30oΩ and Z2=40∠45oΩ. Calculate the total: (a)

apparent power, (b) real power, (c) reactive power, and (d) pf, supplied by the source

and seen by the source.


12

Power Triangle


13

Power Triangle

Example: For the circuit, calculate the complex power for each load then draw the

power triangle.


14

Power Triangle

Example: for the circuit find (the values are given in rms)

a. Find the average power, apparent power, reactive power, and pf for each branch.

b. Find the total number of watts, volt-amperes reactive, volt-amperes, and the power

factor of the system. Sketch the power triangle.

𝑃1 = 12 × 60W = 720W

𝑝𝑓1 = 1

𝑄1 = 0 VAR

(a) ► Bulbs

𝑆1 = 𝑃1 = 720 VA

𝑃2 = 6.4 kW

𝑄2 = 0 VAR

► Heating elements

𝑆2 = 𝑃2 = 6.4 kVA

𝑝𝑓2 = 1


15

Power Triangle

𝑝𝑓3 = 0.72 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

► The motor

η =𝑃𝑜

𝑃𝑖→ 𝑃𝑖 =

𝑃𝑜

η=

5 × 746 W

0.82= 4548.78 W

𝑃3 = 𝑃𝑖 = 4548.78W

𝑃3 = 𝑆3𝑐𝑜𝑠θ → 𝑆3 =𝑃3

𝑐𝑜𝑠θ=

4548.78 W

0.72= 6317.75 VA

𝜃 = 𝑐𝑜𝑠−10.72 = 43.95𝑜

𝑄3 = 𝑆3𝑠𝑖𝑛θ = 6317.75 × 𝑠𝑖𝑛43.95𝑜 = 6317.75 × 0.694 = 4384.71 VAR (L)

► The capacitive load

Irms =Vrms

Z=

208∠0

9 − j12=

208∠0

15∠ − 53.13= 13.87∠53.13

𝑃4 = 𝐼𝑟𝑚𝑠2𝑅 = 13.872 × 9 = 1731.39 W

𝑄4 = 𝐼𝑟𝑚𝑠2𝑋𝐶 = 13.872 × 12 = 2308.52 VAR (C)


16

Power Triangle

𝑆4 = 𝑃42 + 𝑄4

2 = 1731.392 + 23.8.522 = 2885.65 VA

𝑝𝑓4 =𝑃4

𝑆4=

1731.39 W

2885.65 VA= 0.6 𝑙𝑒𝑎𝑑𝑖𝑛𝑔

(b) 𝑃𝑇 = 𝑃1 +𝑃2 +𝑃3 +𝑃4

𝑃𝑇 = 720 + 6400 + 4548.78 + 1731.39

𝑃𝑇 = 13400.17 W

𝑄𝑇 = 𝑄1 +𝑄2 +𝑄3 +𝑄4

𝑄𝑇 = 0 + 0 + 4384.71 − 2308.52

𝑄𝑇 = 2076.19 VAR (𝐿)

𝑆𝑇 = 𝑃𝑇2 + 𝑄𝑇

2 = 13400.172 + 2076.192

𝑆𝑇 = 13560.06 VA

𝑝𝑓𝑇 =𝑃𝑇

𝑆𝑇=

13400.17 W

13560.06 VA= 0.988 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

𝜃 = 𝑐𝑜𝑠−10.988 = 8.89𝑜


17

Power Triangle for Transformers

A transformer is a magnetically coupled circuit,

that is, a circuit in which the magnetic field

produced by time-varying current in one circuit

induces voltage in another.

Where

n : is the turns ratio or transformer ratio.

V1: is the voltage across the primary winding.

V2: is the voltage across the secondary winding.

N1: is the number of turns in the primary.

N2: is the number of turns in the secondary.

When n=1, the transformer is an isolation transformer. If

n>1, the transformer is step-up transformer, as the voltage is

increased from primary to secondary (V2>V1). If n<1, the

transformer is step-down transformer, since the voltage is

decreased from primary to secondary (V2<V1).


18


Example: A 25 kVA transformer feeds a load with 12 kW. If the pf=0.6 lagging.

(1) Find the loading percentage for the maximum load that can be connected to the

transformer.

(2.a) If a unity pf load is added to the same transformer, find the power which can be

added before the transformer becomes fully loaded.

(2.b) If the added load has pf=0.866 leading (instead of unity pf load), find the

apparent power which can be added so the transformer becomes fully loaded.

𝑆𝑡𝑟𝑎𝑛𝑠. = 25𝑘VA

𝑆𝑙𝑜𝑎𝑑1 =𝑃𝑙𝑜𝑎𝑑1

𝑝𝑓1=

12 𝑘W

0.6= 20 𝑘VA

%𝑙𝑜𝑎𝑑𝑖𝑛𝑔 =𝑆𝑙𝑜𝑎𝑑

𝑆𝑡𝑟𝑎𝑛𝑠.× 100% =

20 𝑘VA

25 𝑘VA× 100% = 80%

𝜃1 = 𝑐𝑜𝑠−10.6 = 53.1𝑜

𝑄𝑙𝑜𝑎𝑑1 = 𝑆𝑙𝑜𝑎𝑑1𝑠𝑖𝑛θ = 20000 × 𝑠𝑖𝑛53.1𝑜 = 16 kVAR

(1)

(2.a) If a unity pf load is added to the transformer means that the load is resistive load, so.

𝑄𝑙𝑜𝑎𝑑2 = 0 Pf2=cos𝜃2=1

Transformer apparent power is

Load1 apparent power is


19


𝐒𝑙𝑜𝑎𝑑 = 𝐒𝑙𝑜𝑎𝑑1 + 𝐒𝑙𝑜𝑎𝑑2 = 𝑃𝑙𝑜𝑎𝑑1 + 𝑗𝑄𝑙𝑜𝑎𝑑1 + 𝑃𝑙𝑜𝑎𝑑2 + 𝑗𝑄𝑙𝑜𝑎𝑑2

𝐒𝑙𝑜𝑎𝑑 = (𝑃𝑙𝑜𝑎𝑑1 + 𝑃𝑙𝑜𝑎𝑑2) + 𝑗(𝑄𝑙𝑜𝑎𝑑1 + 𝑄𝑙𝑜𝑎𝑑2)

Before the transformer becomes fully loaded means that total load complex power equal

to transformer complex power

𝐒𝑡𝑟𝑎𝑛𝑠. = 𝐒𝑙𝑜𝑎𝑑 = (𝑃𝑙𝑜𝑎𝑑1 + 𝑃𝑙𝑜𝑎𝑑2) + 𝑗(𝑄𝑙𝑜𝑎𝑑1 + 𝑄𝑙𝑜𝑎𝑑2

0

)

𝐒𝑡𝑟𝑎𝑛𝑠. = 𝐒𝑙𝑜𝑎𝑑 = 𝑃𝑙𝑜𝑎𝑑 + 𝑗𝑄𝑙𝑜𝑎𝑑

Total load complex power is

𝑆𝑡𝑟𝑎𝑛𝑠.2 = 𝑃𝑙𝑜𝑎𝑑

2 + 𝑄𝑙𝑜𝑎𝑑2 = (𝑃𝑙𝑜𝑎𝑑1 + 𝑃𝑙𝑜𝑎𝑑2)2+(𝑄𝑙𝑜𝑎𝑑1)2

252 = (12 + 𝑃𝑙𝑜𝑎𝑑2)2+(16)2

625 − 256 = 144 + 24𝑃𝑙𝑜𝑎𝑑2 +𝑃𝑙𝑜𝑎𝑑22

𝑃𝑙𝑜𝑎𝑑22 + 24𝑃𝑙𝑜𝑎𝑑2 − 225 = 0

𝑃𝑙𝑜𝑎𝑑2 = 7.2 kW 𝑜𝑟 − 31.21 kW

Taking the positve value ∴ 𝑃𝑙𝑜𝑎𝑑2 = 7.2 kW


20


(2.b) Pf2=cos𝜃2=0.866

𝑃𝑙𝑜𝑎𝑑2 = 𝑆𝑙𝑜𝑎𝑑2𝑐𝑜𝑠30 = 0.866𝑆𝑙𝑜𝑎𝑑2

𝑄𝑙𝑜𝑎𝑑2 = 𝑆𝑙𝑜𝑎𝑑2𝑠𝑖𝑛30 = 0.5𝑆𝑙𝑜𝑎𝑑2

𝑃𝑙𝑜𝑎𝑑 = 𝑃𝑙𝑜𝑎𝑑1 + 𝑃𝑙𝑜𝑎𝑑2 = 12 + 0.866𝑆𝑙𝑜𝑎𝑑2

𝑄𝑙𝑜𝑎𝑑 = 𝑄𝑙𝑜𝑎𝑑1 − 𝑄𝑙𝑜𝑎𝑑2 = 16 − 0.5𝑆𝑙𝑜𝑎𝑑2

S𝑙𝑜𝑎𝑑2 = 𝑃𝑙𝑜𝑎𝑑

2 + 𝑄𝑙𝑜𝑎𝑑2

𝐒𝑙𝑜𝑎𝑑 = (𝑃𝑙𝑜𝑎𝑑1 + 𝑃𝑙𝑜𝑎𝑑2) + 𝑗(𝑄𝑙𝑜𝑎𝑑1 + 𝑄𝑙𝑜𝑎𝑑2)

𝐒𝑙𝑜𝑎𝑑 = (12 + 0.866𝑆𝑙𝑜𝑎𝑑2) + 𝑗(16 − 0.5𝑆𝑙𝑜𝑎𝑑2)

252 = (12 + 0.866𝑆𝑙𝑜𝑎𝑑2)2+(16 − 0.5𝑆𝑙𝑜𝑎𝑑2)2

𝑆𝑙𝑜𝑎𝑑2 = 12.8 kVA

𝑄𝑙𝑜𝑎𝑑2 = 0.5𝑆𝑙𝑜𝑎𝑑2 = 0.5 × 12.8 = 6.4 kVA

𝑃𝑙𝑜𝑎𝑑2 = 0.866𝑆𝑙𝑜𝑎𝑑20.866 × 12.8 = 11 𝑘𝑊

𝜃2 = 𝑐𝑜𝑠−10.866 = 30𝑜


21

Power Factor Correction

Power factor correction is the process of increasing the power factor without

altering the voltage or current to the original load.

Most domestic loads (such as washing machines, air conditioners, and

refrigerators) and industrial loads (such as induction motors) are inductive and

operate at a low lagging power factor. Although the inductive nature of the load

cannot be changed, we can increase its power factor.

Most loads are inductive loads as shown in fig(a). A load’s Power factor is

improved or corrected by installing a capacitor in parallel with the load, as

shown in fig(b).


22


The circuit in fig(a) has a power factor of cosθ1,

while the circuit in fig(b) has a power factor of cosθ2.

It is evident that adding the capacitor has caused the

phase angle between the supplied voltage and

current to reduce from θ1 to θ2, thereby increasing

the power factor.

From the magnitudes of the vectors, with the

same supplied voltage, the circuit in fig(a) draws

larger current IL than the current I drawn by the

circuit in fig(b).

Perspective (A)

Perspective (B)

If the original inductive load has apparent power

We want to increase the power factor from θ1 to θ2

without altering the real power


23


The new reactive power is

The reduction in the reactive power is caused by the

shunt capacitor

But

The value of the required shunt capacitance C is determined as

If the load is capacitive, an inductor should be connected across the load for

power factor correction. The required shunt inductance L can be calculated

from


24


Example: When connected to a 120 Vrms, 60-Hz power line, a load absorbs 4kW at a

lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to

0.95.


25


Example: Suppose an industrial client is charged a penalty if the plant power factor

drops below 0.85. The equivalent plant loads are as in the figure. The frequency is 60

Hz.

a. Determine PT and QT.

b. Determine what value of capacitance is required to bring the power factor up to

0.85.

c. Determine generator current before and after correction.

a.

𝑄1 = 𝑄𝑇 = 132 𝐾𝑉𝐴𝑅


26


b. 𝜃2 =cos-1(0.85)=31.8o

𝑄2 =PT tan𝜃2 =146tan31.8o=90.5 KVAR

𝑄𝐶 =Q1-Q2=132-90.5=41.5 KVAR

C =Q

C

𝛚Vrms

2=41.5×103

2π×60×6002 = 306 μF

c.

ST =196.8 KVA

I =S

T

𝑽=

196.8 𝐾𝑉𝐴

600 V= 328 A

S2 =171.8 KVA

I =S

T

𝑽=

171.8 𝐾𝑉𝐴

600 V= 286 A

Electric Circuits II - Philadelphia University...Electric Circuits II AC Circuit Power Analysis 1...

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