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• Cassio Scheffer

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CHAPTER 5 SOLUTIONS 3/9/10

5-1) a)

Vo, rms =

Vm 2

Vo2,rms

P=

c)

pf =

R

a sin(2a ) a sin(2a ) + = Vrms 1 - + p 2p p 2p

a = 60� � Vo ,rms = 431 V ; I o ,rms =

Vm = 480 2 b)

1-

Vo, rms R

=

431 = 8.61 A 50

4312 = 3708 W 50

=

P P 3708 = = = 0.897 S Vrms I rms ( 480 ) ( 8.61)

d ) I avg , SCR = I rms ,SCR =

Vm 480 2 ( 1 + cosa ) = ( 1 + cos 60�) = 3.24 A 2p R 2p ( 50 ) I o ,rms 2

=

8.61 = 6.1 A 2

�480 � e) I1, rms �0.84 � �= 8.0 A �50 � THDI =

2 I rms - I1,2rms

I1, rms

=

8.612 - 8.0 2 = 0.38 = 38% 8.0

_____________________________________________________________________________________

5-2) Vo , rms =

a)

Vm 2

Vo2, rms

P=

c)

pf =

a sin(2a ) a sin(2a ) + = Vrms 1 - + p 2p p 2p

a = 45� � Vo, rms = 114.4 V ; I o, rms =

Vm = 120 2 b)

1-

R

=

Vo ,rms R

=

114.4 = 5.72 A 20

114.42 = 655 W 20

P P 655 = = = 0.953 S Vrms I rms ( 120 ) ( 5.72 )

d ) I avg , SCR =

Vm 120 2 ( 1 + cos a ) = ( 1 + cos 45�) = 2.30 A 2p R 2p ( 20 )

I rms , SCR =

I o , rms 2

=

5.72 = 4.05 A 2

120 � � e) I1,rms �0.92 � �= 5.53 A �20 � THDI =

2 I rms - I1,2rms

I1,rms

5.72 2 - 5.532 = 0.26 = 26% 5.53

=

_____________________________________________________________________________________ 5-3) P=

Vo2.rms R

�

from Fig . 5.2, I o, rms =

Vo, rms

I SCR , rms = pf =

R

=

I o, rms 2

Vo.rms = PR =

167.3 � 0.7 ޻= 240

a

( 800 ) ( 35)

= 167.3 V

92

167.3 = 4.78 A 35 =

4.78 2

= 3.38 A

P 800 = = 0.70 = 70% S 120 ( 4.78 )

_____________________________________________________________________________________

5-4)

120 = 0.5 ; a �115 from Fig.5.2 240 a sin(2a ) or solving Eq.5-3, 120 - 240 1 - + = 0 � a = 1.99 rad = 114� p 2p With the 240-Vsource, Vo ,rms = 120 V ;

240V source : Vo, peak = 2 ( 240 ) sin ( 114� ) = 310 V 120 V source : Vo , peak = Vm = 2 ( 120 ) = 170V _____________________________________________________________________________________ 5-5)

For P = 200W , Vo ,rms = PR = 200 ( 40 ) = 89.4 V Using Eq. 5-3, 89.4 - 120 1 pf =

a sin(2a ) + = 0 � a = 1.48 rad = 85� p 2p

P P 200 = = = 0.75 = 75%. S Vrms I rms ( 120 ) ( 89.4 / 40 )

For P = 400 W , Vo ,rms = PR = 400 ( 40 ) = 126 V Since 126 V > 120 V of the source, 400 W is not possible. The maximum power available is

1202 = 360 W. The pf is 1.0 for 360 W. 40

_____________________________________________________________________________________

5-6) Using the circuit of Fig. 5-1a, For P = 750W , Vo, rms = PR = 750 ( 32 ) = 154.9 V Using Eq. 5-3, 154.9 - 240 1 -

a sin(2a ) + = 0 � a = 1.703 rad = 97.6� p 2p

For P = 1500 W , Vo , rms = PR = 1500 ( 32 ) = 219 V 219 - 240 1 -

a sin(2a ) + = 0 � 0.986 rad = 56.5� p 2p

Maximum SCR currents are for 1500 W: I SCR ,rms = I SRC ,avg =

Io 2

=

219 / 32 = 4.84 A 2

2 ( 240 ) Vm ( 1 + cos a ) = ( 1 + cos ( 56.5�) ) = 2.62 A 2p R 2p ( 32 )

Vmax = 2(240) = 340 V _____________________________________________________________________________________ 5-7) Using the circuit of Fig. 5-1a, For R = 20W, Vo, rms = PR = 1200 ( 20 ) = 154.9 V Using Eq. 5-3, 154.9 - 240 1 -

a sin(2a ) + = 0 � a = 1.703 rad = 97.6� p 2p

For R = 40W, Vo ,rms = PR = 1200 ( 40 ) = 219 V 219 - 240 1 -

a sin(2a ) + = 0 � 0.986 rad = 56.5� p 2p

Maximum SCR currents are for R = 20 W: I SCR ,rms = I SRC ,avg =

Io 2

=

154.9 / 20 = 5.48 A 2

2 ( 240 ) Vm ( 1 + cos a ) = ( 1 + cos ( 97.6�) ) = 2.34 A 2p R 2p ( 20 )

Vmax = 2(240) = 340 V _____________________________________________________________________________________

5-8) R=

V 2 1202 = = 144 W P 100

a ) P = 75 W : Vrms =

( 144 ) ( 75 )

= 103.9 V

From Fig.5-3, a �1.16 rad = 66.2� b) P = 25 W : Vrms =

( 144 ) ( 25 )

= 60 V

From Fig.5-3, a �1.99 rad = 114�

_____________________________________________________________________________________ 5-9) S1 is on from α to π, and D2 is on from π to 2π. vo ( wt ) = Vm sin wt Vo, rms =

=

1 2p Vm 2

for a �wt �2p

2p

(V � a

1-

sin wt ) d ( wt ) = Vm 2

m

1 a sin ( 2a ) + 2 4p 8p

a sin ( 2a ) + 2p 4p

0 �a‫� ޣ‬ p �

Vm 2

Vo ,rms

Vm 2

_____________________________________________________________________________________ 5-10)

vo (wt ) = Vm sin wt Vo.rms

for a1 < wt < p and for p + a 2 < w t < 2p

p 2p � 1 � 2 � = ( Vm sin wt ) d ( wt ) + ( Vm sin wt ) 2 d ( wt ) � � 2p � a1 p +a 2 � �

�

= Vm

Vo.rms =

Vm 2

�

1 a1 + a 2 sin ( 2a1 ) + sin ( 2a 2 ) + 2 4p 8p

1 -

sin ( 2a1 ) + sin ( 2a 2 ) a1 + a 2 + 2p 4p

_____________________________________________________________________________________

5-11) a) Using Eq. 5-9, Z = 21.3 W; q = 0.561 rad ;

wt = 0.628

i ( wt ) = 7.98sin ( wt - 0.561) - 19.25e -wt /0.628 A

a = 60�= 1.047 rad , b = 3.696 rad = 212� b) I rms = 4.87 A c) I rms , SCR =

4.87 2

= 3.44 A

2 d ) P = I rms R = ( 4.87 ) ( 18 ) = 427 W 2

_____________________________________________________________________________________ 5-12) Using Eq. 5-9,

Z = 26.7 W;

q = 0.601 rad ;

wt = 0.685

i ( wt ) = 6.36sin ( wt - 0.601) - 6.10e -wt /0.685 A

a = 50�= 0.873 rad , b = 3.738 rad = 214� b) I rms = 4.18 A c ) I rms , SCR =

4.18 2

= 2.95 A

2 d ) P = I rms R = ( 4.18 )

2

( 22 ) = 384 W

_____________________________________________________________________________________ 5-13) Using Eq. 5-9,

q = 0.646 rad ;

Z = 15.0 W;

wt = 0.754

i ( wt ) = 11.3sin ( wt - 0.646 ) - 158e -wt /0.754 A

a = 115�= 2.01 rad , b = 3.681 rad = 211� I rms = 2.95 A _____________________________________________________________________________________ 5-14) Using Eq. 5-9,

Z = 14.2 W;

q = 0.561 rad ;

wt = 0.6.28

i ( wt ) = 11.98sin ( w t - 0.561) - 54.1e -wt /0.628 A

a = 70�= 1.222 rad , b = 3.691 rad = 212� I rms = 6.69 A 2 P = I rms R = ( 6.69 ) ( 12 ) = 537 W 2

PSpice: P = AVG(W(R)) in Probe gives 523 W (read at the end of the trace). The difference between PSpice and the theoretical output is because of the nonideal SCR model in PSpice. The PSpice result will be more realistic. The THD is 22.4% from the PSpice output file using Fourier terms through n = 9. _____________________________________________________________________________________

5-15) Use the PSpice circuit of Example 5-3. The .STEP PARAM command is quite useful for determining α. (a) α ≈ 81° for 400 W. (b) α ≈ 46° for 700 W. SINGLE-PHASE VOLTAGE CONTROLLER (voltcont.cir) *** OUTPUT VOLTAGE IS V(3), OUTPUT CURRENT IS I(R) *** **************** INPUT PARAMETERS ********************* .PARAM VS = 120 ; source rms voltage .PARAM ALPHA = 81 ; delay angle in degrees .STEP PARAM ALPHA 10 90 20 ; try several values of alpha. Modify the range for more precision .PARAM R = 15 ; load resistance .PARAM L = 15mH ; load inductance .PARAM F = 60 ; frequency .PARAM TALPHA = {ALPHA/(360*F)} ; converts angle to time delay .PARAM PW = {0.5/F} ; pulse width for switch control ***************** CIRCUIT DESCRIPTION ********************* VS 1 0 SIN(0 {VS*SQRT(2)} {F}) S1 1 2 11 0 SMOD D1 2 3 DMOD ; forward SCR S2 3 5 0 11 SMOD D2 5 1 DMOD ; reverse SCR R 3 4 {R} L 4 0 {L} **************** MODELS AND COMMANDS ******************** .MODEL DMOD D(n=0.01) .MODEL SMOD VSWITCH (RON=.01) VCONTROL 11 0 PULSE(-10 10 {TALPHA} 0 0 {PW} {1/F}) ;control for both switches .TRAN .1MS 50MS 0MS 1u UIC ; one period of output .FOUR 60 I(R) ; Fourier Analysis to get THD .PROBE .END

_____________________________________________________________________________________ 5-16) Modify the PSpice circuit file of Example 5-3. Use the .STEP PARAM command (see Prob. 5-15) for determining α. (a) α ≈ 80° for 600 W. (b) α ≈ 57° for 1000 W. _____________________________________________________________________________________ 5-17) The single-phase voltage controller of Fig. 5-4a is suitable for this application. Equation (5-9) applies for each half-period of the input sine wave. For 250 W delivered to the load, each half period must deliver 125 W. Therefore, the rms value of the current in Eq. (5-9) must be 2.28 A, found by using I2R = 125. A closed-form solution is not possible, but trial-and-error numerical techniques give α ≈ 74°. A similar but perhaps easier method is to use PSpice simulations using the PSpice A/D circuit file in Example 5-3. Modifying the diode model to .MODEL DMOD D(n=.01) to represent an ideal diode, and with trial-and-error values of α, gives α ≈ 74°. The average and rms currents are determined from a numerical integration of the current expression from Eq. (5-9) or from a PSpice simulation. ISCR,avg = 1.3 A, ISCR,rms = 2.3 A. The maximum voltage across the switches is 120√2sin(74°) = 163 V.

5-18) The PSpice circuit file is shown below. The total average load power is three times the power in one of the phase resistors. Enter 3*AVG(W(RA)) in Probe. The results are (a) 6.45 kW for 20°, (b) 2.79 kW for 80°, and (c) 433 W for 115°. Note that the .STEP PARAM command can be used to run the three simulations at once.

THREE-PHASE VOLTAGE CONTROLLER -- R-L LOAD (3phvc.cir) *SOURCE AND LOAD ARE Y-CONNECTED (UNGROUNDED) ********************** INPUT PARAMETERS **************************** .PARAM Vs=480 ; rms line-to-line voltage .PARAM ALPHA=20 ; delay angle in degrees .STEP PARAM ALPHA LIST 20 80 115 .PARAM R=35 ; load resistance (y-connected) .PARAM L = 1p ; load inductance .PARAM F=60 ; source frequency ********************** COMPUTED PARAMETERS ************************** .PARAM Vm={Vs*SQRT(2)/SQRT(3)} ; convert to peak line-neutral volts .PARAM DLAY={1/(6*F)} ; switching interval is 1/6 period .PARAM PW={.5/F} TALPHA={ALPHA/(F*360)} .PARAM TRF=10US ; rise and fall time for pulse switch control *********************** THREE-PHASE SOURCE ************************** VAN 1 0 SIN(0 {VM} 60) VBN 2 0 SIN(0 {VM} 60 0 0 -120) VCN 3 0 SIN(0 {VM} 60 0 0 -240) ***************************** SWITCHES ******************************** S1 1 8 18 0 SMOD ; A-phase D1 8 4 DMOD S4 4 9 19 0 SMOD D4 9 1 DMOD S3 2 10 20 0 SMOD D3 10 5 DMOD S6 5 11 21 0 SMOD D6 11 2 DMOD

; B-phase

S5 3 12 22 0 SMOD ; C-phase D5 12 6 DMOD S2 6 13 23 0 SMOD D2 13 3 DMOD ***************************** LOAD ********************************** RA 4 4A {R} ; van = v(4,7) LA 4A 7 {L} RB 5 5A {R} LB 5A 7 {L}

; vbn = v(5,7)

RC 6 6A {R} ; vcn = v(6,7) LC 6A 7 {L} ************************* SWITCH CONTROL ***************************** V1 18 0 PULSE(-10 10 {TALPHA} {TRF} {TRF} {PW} {1/F}) V4 19 0 PULSE(-10 10 {TALPHA+3*DLAY} {TRF} {TRF} {PW} {1/F}) V3 20 0 PULSE(-10 10 {TALPHA+2*DLAY} {TRF} {TRF} {PW} {1/F}) V6 21 0 PULSE(-10 10 {TALPHA+5*DLAY} {TRF} {TRF} {PW} {1/F}) V5 22 0 PULSE(-10 10 {TALPHA+4*DLAY} {TRF} {TRF} {PW} {1/F}) V2 23 0 PULSE(-10 10 {TALPHA+DLAY} {TRF} {TRF} {PW} {1/F}) ************************ MODELS AND COMMANDS ************************* .MODEL SMOD VSWITCH(RON=0.01) .MODEL DMOD D .TRAN .1MS 50MS 16.67ms 10US UIC .FOUR 60 I(RA) ; Fourier analysis of line current .PROBE .OPTIONS NOPAGE ITL5=0 .END

_____________________________________________________________________________________

5-19) The PSpice input file from Example 5-4 is used for this simulation. In Probe, enter the expression 3*AVG(W(RA)) to get the total three-phase average power in the load, resulting in 368 W. Switch S 1 conducts when the current in phase A is positive, and S4 conducts when the current is negative. _____________________________________________________________________________________ 5-20) The smallest value of α is 120°. The conduction angel must be less than for equal to 60°. The extinction angle is 180°, so α is 120° or greater. _____________________________________________________________________________________ 5-21) THREE-PHASE VOLTAGE CONTROLLER -- R-L LOAD *MODIFIED FOR A DELTA-CONNECTED LOAD *SOURCE IS Y-CONNECTED (UNGROUNDED) ********************** INPUT PARAMETERS **************************** .PARAM Vs=480 ; rms line-to-line voltage .PARAM ALPHA=45 ; delay angle in degrees .PARAM R=25 ; load resistance (y-connected) .PARAM L = 1p ; load inductance .PARAM F=60 ; source frequency ********************** COMPUTED PARAMETERS ************************** .PARAM Vm={Vs*SQRT(2)/SQRT(3)} ; convert to peak line-neutral volts .PARAM DLAY={1/(6*F)} ; switching interval is 1/6 period .PARAM PW={.5/F} TALPHA={ALPHA/(F*360)} .PARAM TRF=10US ; rise and fall time for pulse switch control *********************** THREE-PHASE SOURCE ************************** VAN 1 0 SIN(0 {VM} 60) VBN 2 0 SIN(0 {VM} 60 0 0 -120) VCN 3 0 SIN(0 {VM} 60 0 0 -240) ***************************** SWITCHES ******************************** S1 1 8 18 0 SMOD ; A-phase D1 8 4 DMOD S4 4 9 19 0 SMOD D4 9 1 DMOD S3 2 10 20 0 SMOD D3 10 5 DMOD S6 5 11 21 0 SMOD D6 11 2 DMOD

; B-phase

S5 3 12 22 0 SMOD ; C-phase D5 12 6 DMOD S2 6 13 23 0 SMOD D2 13 3 DMOD ***************************** LOAD ********************************** RA 4 4A {R} ; LA 4A 2 {L} RB 5 5A {R} LB 5A 3 {L}

;

RC 6 6A {R} ; LC 6A 1 {L} ************************* SWITCH CONTROL ***************************** V1 18 0 PULSE(-10 10 {TALPHA} {TRF} {TRF} {PW} {1/F})

V4 19 0 PULSE(-10 10 {TALPHA+3*DLAY} {TRF} {TRF} {PW} {1/F}) V3 20 0 PULSE(-10 10 {TALPHA+2*DLAY} {TRF} {TRF} {PW} {1/F}) V6 21 0 PULSE(-10 10 {TALPHA+5*DLAY} {TRF} {TRF} {PW} {1/F}) V5 22 0 PULSE(-10 10 {TALPHA+4*DLAY} {TRF} {TRF} {PW} {1/F}) V2 23 0 PULSE(-10 10 {TALPHA+DLAY} {TRF} {TRF} {PW} {1/F}) ************************ MODELS AND COMMANDS ************************* .MODEL SMOD VSWITCH(RON=0.01) .MODEL DMOD D .TRAN .1MS 50MS 16.67ms 10US UIC .FOUR 60 I(RA) ; Fourier analysis of line current .PROBE .OPTIONS NOPAGE ITL5=0 .END 40A Ia 0A

SEL>> -40A

I(RA)

50A Source A current 0A

-50A

15ms

20ms

- I(VAN)

25ms

30ms

35ms

40ms

45ms

50ms

Time

_____________________________________________________________________________________ 5-22) The PSpice circuit file modification must include a very large resistor (e.g., one megaohm) connected between the neutral of the load to ground to prevent a “floating node” error because of the series capacitor. The steady-state phase A current has two pulses for each of the switches, assuming that the gate signal to the SCRs is continuously applied during the conduction interval. The rms current is approximately 5.52 A. The total average power for all three phases is approximately 1.28 kW. The THD for the load current is computed as 140% for harmonics through n = 9 in the .FOUR command. However, the current waveform is rich in higher-order harmonics and the THD is approximately 300% for n = 100. It should be noted that this load is not conducive for use with the voltage controller because the load voltage will get extremely large (over 5 kV) because of stored charge on the capacitor.

40A

S1

S1 (1.0000,5.5229)

0A Phase A current SEL>> -40A 2.0KW

S4 I(RA)

S4

RMS(I(RA)) (1.0000,1.2811K)

1.0KW Total average power 0W 0.980s 0.984s AVG(W(RA))*3

0.988s

0.992s

0.996s

1.000s

Time

_____________________________________________________________________________________ 5-23) With the S1-S4 switch path open, the equivalent circuit is as shown. The current in phase A is zero, so the voltage across the phase-A resistor is zero. The voltage at the negative of V 14 is then Vn, and the voltage at the positive of V14 is Va. The voltage across the phase B resistor is half of the voltage from phase B to phase C, resulting in Vn = Vb -

Vb - Vc Vb + Vc = 2 2

Therefore, V14 = Va - Vn = Va -

Vb + Vc 2