alternating current electricity 15-1 - valparaiso … slides/altcurrslides.pdfprofessional...

Professional Publications, Inc. FERC

15-1Alternating Current ElectricityFrequency and Period

Frequency (linear and angular)

Period = T


15-2aAlternating Current ElectricityAverage Value

Square wave,positive pulse = negative pulse: Xave = 0

Pulse pattern, positive, all the same:

where t is the duration of the pulse andT is the period.

Sawtooth: Xave =

Symmetrical triangular wave: Xave = 0

!

Xave

=tX

max

T

!

1

2X

max


15-2bAlternating Current ElectricityAverage Value

Example (FEIM):A plating tank with an effective resistance of 100 Ω is connected to theoutput of a full-wave rectifier. The applied voltage is sinusoidal with amaximum of 170 V. How long does it take to transfer 0.005 faradays?

!

Vave

=2V

max

"=

(2)(170 V)

"= 108.2 V

!

Iave

=V

ave

R=

108.2 V

100"= 1.082 A

!

t =q

Iave

=

(0.005 F) 96,487A "s

F

#

$ %

&

' (

1.082 A

!

= 446 s


15-3aAlternating Current ElectricityEffective or Root-Mean-Squared (RMS) Value

Effective value of an alternatingwaveform:

For a sinusoidal waveform,

For a half-wave sinusoidal waveform,

Pulse pattern, positive, all the same:

where t is the duration of the pulse andT is the period.

Symmetrical triangular:

Sawtooth:

!

Xrms

=t

TX

max

!

Xrms

=1

3

Xmax

!

Xrms

=1

3

Xmax


15-3bAlternating Current ElectricityEffective or Root-Mean-Squared (RMS) Value

Example 1 (FEIM):A 170 V (maximum value) sinusoidal voltage is connected across a 4 Ωresistor. What is the power dissipated by the resistor?

!

Vrms

=V

max

2

=170 V

2

= 120.2 V

!

P =V

rms

2

R=

120.2 V( )2

4"= 3.61#10

3W


15-3c1Alternating Current ElectricityEffective or Root-Mean-Squared (RMS) Value

Example 2 (FEIM):What is the Irms value for the following waveform?

(A)

(B)

(C)

(D)

!

2

4I

!

3

4I

!

10

4I

!

3

2I


15-3c2Alternating Current ElectricityEffective or Root-Mean-Squared (RMS) Value

!

Irms

=1

T(I(t))2

0

T

" =1

TI

2dt +

1

T

I

2

#

$ % &

' (

2

dtT

2

T

"0

T

2"

!

=I

2

T

T

2

"

# $ %

& ' +

1

T

"

# $ %

& '

I2

4

"

# $

%

& ' T( ) T

2

T

=I

2

2+

I2

4T

"

# $

%

& '

T

2

"

# $ %

& '

!

=I

2

2+

I2

8=

5

8I

2=

10

4I

Therefore, the answer is (C).


15-3dAlternating Current ElectricityEffective or Root-Mean-Squared (RMS) Value

Example 3 (FEIM):A sinusoidal AC voltage with a value of Vrms = 60 V is applied to a purelyresistive circuit. What steady-state voltage would dissipate the samepower as the AC voltage?

(A) 30 V(B) 42 V(C) 60 V(D) 85 V

The answer is right in the problem statement. Since the voltage is givenas an rms value, the equivalent DC voltage is simply 60 V.

Therefore, the answer is (C).


15-4Alternating Current ElectricityPhasor Transforms

Phase angle = θ – φφ as the angle between the reference and voltageθ as the angle between the reference and current

If the phase angle is positive, the signal is called “leading” or “capacitive.”If the phase angle is negative, the signal is called “lagging” or “inductive.”If the phase angle is zero, the signal is called “in phase.”


15-5aAlternating Current ElectricityImpedance

Definitions

Impedance:

Resistor:

Capacitor:

Inductor:


15-5b1Alternating Current ElectricityImpedance

Example (FEIM):For the following circuit:

(a) What is the impedance in polar form?(b) Is the current leading or lagging?(c) What is the voltage across the inductor?



(a)

!

Z = 3"+ j 4001

s

#

$ %

&

' ( (2.5)10*3 H) – j

1

4001

s

#

$ %

&

' ( (625)10*3 F)

+ 9"+40°

!

= 3"+ j1"# j4"+ (9")cos40° + ( j9")sin40°

!

= 3"+ 6.894"+ j(1"# 4"+ 5.785")

!

= 9.894"+ j2.785"

!

Z = (9.894")2+ (2.785")2

= 10.28"

!

Phase angle = tan"1 imaginary

real

#

$ %

&

' (

!

= tan"1 2.785#

9.894#

$

% &

'

( ) = 15.72°

!

Z = 10.28"#15.72°



(b)

!

I =V

Z=

160V"0°

10.28#"15.72°

$

% &

'

( ) = 16.56A"*15.72°

Therefore, the current is lagging (ELI).

(c)

!

VL

= IZL

= (15.56 A "#15.72°)(1$"90°)

!

= 15.56 V "74.28°


15-6Alternating Current ElectricityAdmittance

Multiplying by the complex conjugate,


15-7Alternating Current ElectricityOhm’s Law for AC Circuits

Example (FEIM):What is the current in the capacitor in the following circuits?

!

Z = 2"# j4" = 4.47"$# 63.4°

!

I =V

Z=

180 V"0°

4.47#"$ 63.4°= 40.3 A "63.4°

(a)

!

Zc

= 0.25"#$ 90°

!

I =V

Zc

=180 V"0°

0.25#"$ 90°= 720 A "90°

(b)


15-8aAlternating Current ElectricityComplex Power

P: real power (W)Q: reactive power (VAR)S: apparent power (VA)

• Power Factor:

• Real Power:

• Reactive Power:


15-8b1Alternating Current ElectricityComplex Power

Example (FEIM):For the following circuit, find the (a) real power, and (b) reactive power.(c) Draw the power triangle.

(a) The real power is:

(b) The reactive power is:

!

P =V

R

2

R=

(120 V)2

10 "= 1440 W

!

Q =V

L

2

XL

=(120 V)2

4 "= 3600 VAR


15-8b2Alternating Current ElectricityComplex Power

(c) The power triangle is:

In this example, the impedance of the inductor has a lagging Current sothe current has a negative phase angle. The complex conjugate of thecurrent has a positive phase angle, so the reactive power, Q, is positiveand the power triangle is in the first quadrant.For a leading current (which has a positive phase angle compared tothe voltage) the power triangle has a negative imaginary part and anegative power angle, so it is in the fourth quadrant.


15-9aAlternating Current ElectricityResonance

Parallel ResonanceQuality Factor:

Bandwidth:

Half-power point is when

Series ResonanceQuality Factor:

Bandwidth:

Half-power point is when

!

R = X:"1/2power

= "0

±1

2BW

!

R = X:"1/2power

= "0

±1

2Q


15-9b1Alternating Current ElectricityResonance

Example (FEIM):For the following circuit, find(a) the resonance frequency (rad/s)(b) the half-power points (rad/s)(c) the peak current (at resonance)(d) the peak voltage across each component at resonance

(a)

!

"0

=1

LC

=1

(200#10$6 H)(200#10$12 F)= 5#106 rad

s


15-9b2Alternating Current ElectricityResonance

(b)

!

" 1

2power

= "0

±1

2BW

!

= "0

±R

2L= 5#106 rad

s±

50$

(2)(200#10%6 H)

!

= 5.125"106, 4.875"10

6 rad

s

(c)

!

At resonance, Z = R, I(resonance) =V

R=

20 V "0°

50#= 0.4 A"0°

(d)

!

VR

= I0R = 0.4 A "0°( ) 50#( ) = 20 V "0°

!

VL = I0XL = I0 j"L = (0.4 A #0°) 5$106 rad

s

%

& '

(

) * (200$10+6 H)#90°

!

= 400"90°

!

VC

= "VL

= 400 V #" 90°


15-9cAlternating Current ElectricityResonance

Example (FEIM):A parallel resonance circuit with a 10 Ω resistor has a resonancefrequency of 1 MHz and bandwidth of 10 kHz. What resistor value willincrease the BW to 20 kHz?

C remains the same, so to double the BW.

!

BW ="

0

Q=

"0

"0RC

=1

RC

!

Rnew

=1

2R

old= 5"


15-10aAlternating Current ElectricityTransformers


15-10bAlternating Current ElectricityTransformers

Example 1 (FEIM):What is the voltage V3? Disregard losses.

Therefore, the answer is (D).

!

V2

=N

2

N1

120 V V3

=N

3

NA

V2

!

V3

=N

3

NA

"

# $

%

& '

N2

N1

"

# $

%

& ' 120 V( ) =

300

100

"

# $

%

& '

50

200

"

# $

%

& ' 120 V( ) = 90 V

(A) 45 V(B) 65 V(C) 75 V(D) 90 V


15-10cAlternating Current ElectricityTransformers

Example 2 (FEIM):What is the total impedance in the primary circuit? Assume an idealtransformer.

!

Ztotal

= Zp

+ Z =50

100

"

# $

%

& '

2

(12(+ j4() + 3(+ j(

!

= 4"+ 3"+ j(1"+1") = 7"+ j2"


15-11aAlternating Current ElectricityRotating Machines

Synchronous Machines1. Synchronous motors2. Synchronous generators


15-11bAlternating Current ElectricityRotating Machines

Example (FEIM):A six-pole, three-phase synchronousgenerator supplies current with afrequency of 60 Hz. What is theangular velocity of the rotor in thegenerator?

!

(A) 10"rad

s

!

(B) 40"rad

s

!

(C) 60"rad

s

!

(D) 80"rad

s

!

ns

=120f

p=

60"

2#

!

" =4#f

p

!

=

4"rad

cycle

#

$ %

&

' ( 60

cycle

s

#

$ %

&

' (

6

!

= 40" rad/s

Therefore, the answer is (B).

(Note: Here, Ω does not meanohms. It is the angular velocity.)


15-11cAlternating Current ElectricityRotating Machines

Induction Motors• A constant-speed device that receives power though induction

without using brushes or slip rings

Slip:


15-11dAlternating Current ElectricityRotating Machines

Example (FEIM):A four-pole induction motor operates on a three-phase, 240 Vrms line-to-line supply. The slip is 5%. The operating speed is 1600 rpm. What ismost nearly the operating frequency?(A) 56 Hz(B) 60 Hz(C) 64 Hz(D) 102 Hz

!

ns

= n(1+ s)

!

= 1600 rpm( )(1+ 0.05)

!

= 1680 rpm

Solving the synchronous speed equation for the operating frequency yields

!

f =pn

s

120

!

=4( ) 1680 rpm( )

120

!

= 56 Hz

Therefore, the answer is (A).


15-11eAlternating Current ElectricityRotating Machines

DC Machines• Have a constant magnetic field in the stator; the magnetic field of

rotor responds to the stator field• The magnetic flux produced is described by the equation


15-11fAlternating Current ElectricityRotating Machines

DC Generators• A device that produces DC potential


15-11gAlternating Current ElectricityRotating Machines

DC Generators

For the DC machine equivalent circuit

Terminal Voltage:

For the simplified DC machine equivalent circuitTerminal Voltage:


15-11h1Alternating Current ElectricityRotating Machines

Example (FEIM):A DC generator provides a current of 12 A for a resistive load whenoperating at 1800 rpm. The speed of the armature is reduced, and thenew steady-state current is 10 A. What is most nearly the operatingspeed when the generator reaches steady-state conditions? Ignorearmature resistance.(A) 1500 rpm(B) 1700 rpm(C) 1800 rpm(D) 2200 rpm


15-11h2Alternating Current ElectricityRotating Machines

The current is proportional to the voltage by Ohm’s law. From theequation for the armature voltage, ignoring the armature resistance, thearmature voltage is proportional to the speed.

!

Va

= Kan"

Therefore, the armature current is proportional to the speed.

!

I2

I1

=V

2

V1

=n

2

n1

Rearranging yields the new operating speed.

!

n2

=I

2

I1

n1

Therefore, the answer is (A).

!

=10 A

12 A

"

# $

%

& ' (1800 rpm)

!

= 1500 rpm


15-11iAlternating Current ElectricityRotating Machines

DC Motors• Very similar to a DC generator, only the direction of current

changes

Power (for the motor model in the DC machine equivalent circuit):

• Ignoring the power dissipated as heat for the DC machine equivalent circuit,

Torque:

Mechanical Power:


15-11jAlternating Current ElectricityRotating Machines

Example (FEIM):A DC motor is operating on 100 V with a magnetic flux of 0.02 Wbproduces an output torque of 20 N • m. The magnetic flux is changedto 0.03 Wb. What is most nearly the new steady-state outputtorque? Ignore armature resistance.

!

T1

T2

="

1

"2

!

T2

="

2

"1

T1

!

=0.03 Wb

0.02 Wb

"

# $

%

& ' 20 N (m( )

!

= 30 N "m

The answer is (B).

The motor voltage is not important to the solution because the torqueequation is proportional to the magnetic flux.

!

(A) = 19 N "m

!

(B) = 30 N "m

!

(C) = 32 N "m

!

(D) = 51 N "m


15-12aAlternating Current ElectricityAdditional Examples

Example 1 (FEIM):For the circuit elements shown, draw the conventional impedancediagram.


15-12bAlternating Current ElectricityAdditional Examples

Example 2 (FEIM):What is the steady-state magnitude of the rms voltage across thecapacitor?(A) 15 V(B) 30 V(C) 45 V(D) 60 V

!

Z = 10"+ j(15"#15") = 10"

!

Irms

=V

rms

R=

40 V

10"= 4 A

!

Vc rms

= Irms

Xc

= (4 A)(15") = 60 V



15-12c1Alternating Current ElectricityAdditional Examples

Example 3 (FEIM):For the following circuit, the rms steady-state currents are

!

I1

= 14.4"#36.9° and I2

= 6"90°.

The impedance seen by the voltage source is most nearly

!

(A) 9.5"#18.4°

!

(B) 10.0"#36.9°

!

(C) 10.7"#16.0°

!

(D) 11.0"#7.1°


15-12c2Alternating Current ElectricityAdditional Examples

Note: The preceding calculation is an example of rationalizing thedenominator of a complex number without converting it to polar form.

!

Z = R +1

1

Z1

+1

Z2

= R +Z

1Z

2

Z1+ Z

2

!

= 5"+(4 + j3)(# j12)

4 + j3# j12" = 5"+

36# j48

4# j9

$

% &

'

( )

4 + j9

4 + j9

$

% &

'

( ) "

!

Z = 5"+144"+ 432"+ j(324"#192")

16"+ 81"= 5"+

576"

97"+ j

132"

97"

$

% &

'

( )

!

= 5"+ 5.938"+ j1.361" = 10.938"+ j1.361"

!

Z = (10.938")2+ (1.361")2

= 11.0"

!

tan"1 1.361#

10.938#= 7.1°

!

Z = 11"#7.1°



15-12dAlternating Current ElectricityAdditional Examples

Example 4 (FEIM):The phasor form of I3 is most nearly

!

(A) 10.3 A "#32.5°

(B) 11.8 A "#12.9°

(C) 18.6 A "51.9°

(D) 18.6 A "#51.9°

!

I3

= I1+ I

2

= 14.4 A"# 36.9° + 6 A"90°

= (14.4 A)cos(#36.9°) + j(14.4 A)sin# 36.9° + j6 A

= 11.515 A # j8.646 A + j6 A

= 11.515 A # j2.646 A

!

I3

= (11.515A)2+ (2.646A)2

= 11.8A

tan-1 "2.646A

11.515A= "12.9°

I3

= 11.8A#"12.9°

Therefore, the answer is (B).

alternating current electricity 15-1 - valparaiso … slides/altcurrslides.pdfprofessional...

Documents