mcgraw hill interactions 2 listening and speaking 4th edition
(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 2 ppt
... 36.87° A )( 0 .20 Ω + j 0.750 Ω ) = 106 .2 2. 6° V The excitation current of this transformer is 106 .2 2. 6° V 106 .2 2. 6° V + = 0.354 2. 6° + 1. 328 ∠ − 87.4° j80 Ω 300 Ω = 1.37∠ − 72. 5° A I EX = IC ... first column being mmf and % the second column being flux load p 22_ mag.dat; mmf_data = p 22( :,1); flux_data = p 22( : ,2) ; % Initialize values S = 1000; Vrms = 120 ; VM = Vrms * sqrt (2) ; NP = 500; % % ... first column being mmf and % the second column being flux load p 22_ mag.dat; mmf_data = p 22( :,1); flux_data = p 22( : ,2) ; % Initialize values S = 1000; Vrms = 24 0; VM = Vrms * sqrt (2) ; NP = 1000; % %...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 1 pptx
... R2 / s ω sync ( RTH + R2 / s ) + ( X TH + X ) 2 ( 26 2 V ) (0.1 72 Ω /1.9 62 ) τ ind = (188.5 rad/s) (0. 321 + 0.1 72 Ω /1.9 62) 2 + (0.418 + 0. 420 )2 ( 26 2 V ) (0.0877 ) τ ind = (188.5 rad/s) (0. 321 ... at slip s = It is τ ind = 3VTH R2 / s ω sync ( RTH + R2 / s ) + ( X TH + X ) 2 ( 26 2 V ) (0.1 72 Ω ) τ ind = (188.5 rad/s) (0. 321 + 0.1 72 Ω )2 + (0.418 + 0. 420 )2 196 = 199 N ⋅ m (d) To determine ... equation for induced voltage is ZP φω 2 a so the required flux is (20 48 cond )(8 poles ) φ 24 00 r/min 2 rad 120 V = ( ) 2 (16 paths) 1r 120 V = 40,960 φ φ = 0.0 029 3 Wb EA = (b) 60 s At rated load,...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 2 ppt
... V A = 24 0 V, then E A = 24 0 V, and n= 9-9 120 V 27 1 V If V A = 180 V, then E A = 180 V, and n= (a) no If V A = 120 V, then E A = 120 V, and n= (a) EA E Ao 24 0 V 27 1 V ( 120 0 r/min ) = 1063 r/min ... know that % % Ea2 K' phi2 n2 Eao2 n2 % - = = -% Ea1 K' phi1 n1 Eao1 n1 % % Ea2 Eao1 % ==> n2 = - n1 % Ea1 Eao2 % % where Ea0 is the internal generated voltage at 120 0 r/min % for ... generated voltage E A = V A = 120 V Therefore, the speed n with a voltage of 120 V would be 22 0 EA n = E Ao no n= 9-10 EA E Ao no = 120 V ( 120 0 r/min ) = 5 02 r/min 28 7 V If the motor is connected...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 3 docx
... 0.75 A = 99 .25 A , and E A = VT − I A R A = 24 0 V − (99 .25 A )( 0.14 Ω ) = 22 6.1 V Therefore, the speed at full load will be n= (c) EA 22 6.1 V no = ( 120 0 r/min ) = 1060 r/min E Ao 25 6 V If Radj ... 0.75 A = 99 .25 A , and E A = VT − I A ( RA + RS ) = 24 0 V − (99 .25 A )(0.14 Ω + 0.05 Ω ) = 22 1.1 V The actual field current will be IF = VT 24 0 V = = 0.75 A RF + Radj 20 0 Ω + 120 Ω and the effective ... when E A rises to E A = 24 0 V − ( 0 .28 4 Ω )(54 A ) = 22 4.7 V If the resistance is cut out when E A reaches 22 8,6 V, the resulting current is IA = 24 0 V − 22 4.7 V = 1 02 A < 135 A , 0.15 Ω so there...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx
... ) = 28 .15 + j 24 .87 Ω 50 + j 2. 40 + j 60 R2 / ( s ) jX ( jX M ) R2 / ( − s ) + jX + jX M 27 0 { I1 { 10-1 Reverse ZB = (a) (1 .28 2 + j 2. 40)( j 60) = 1.185 + j 2. 3 32 Ω 1 .28 2 + j 2. 40 + j 60 The ... j0.5X2 { j2.4 Ω j0.5XM 0.5ZF - ZB = R2 s 0.5 R2 2 s Forward jX2 { 0.5ZB ZF = 0.5 j30 Ω V ZF = j1 .20 Ω j1 .20 Ω j0.5XM j30 Ω ( R2 / s + jX )( jX M ) R2 / s + jX + jX M (50 + j 2. 40)( j 60 ) = 28 .15 ... A 23 0 V − 22 0 .2 V = = 23 1.7 A RA 0.0 423 Ω The power into the dc motor is now (23 0 V) (23 1.7 A) = 53.3 kW, and the power converted from electrical to mechanical form in the dc machine is (22 0.2...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 5 ppsx
... 120 ∠ - 120 ° V = 20 8∠30° V Vbc = Vbn − Vcn = 120 ∠ − 120 ° V - 120 ∠ - 24 0° V = 20 8∠ - 90° V Vca = Vcn − Van = 120 ∠ − 24 0° V - 120 ∠0° V = 20 8∠150° V 28 5 I ab = Vab 20 8∠30° V = = 20 .8∠10° A 10 20 ° Ω ... real and reactive power of each load is P1 = Vφ Q1 = P2 = Z Vφ Z Vφ Q2 = (d) Z Vφ Z cosθ = (25 3 .2 V )2 cos 36.87° = 61.6 kW 2. 5 Ω (25 3 .2 V )2 sin 36.87° = 46 .2 kvar sin θ = 2. 5 Ω cosθ = ( 25 3 .2 V ... cos 27 ° = 25 48 W (d) The air-gap power is PAG,F = I 12 ( 0.5RF ) = (13.0 A ) (13 .29 Ω ) = 22 46 W PAG,B = I 12 ( 0.5 RB ) = (13.0 A ) (0.370 Ω ) = 62. 5 W PAG = PAG,F − PAG,B = 22 46 W − 62. 5 W = 21 84...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 6 ppt
... 3Vφ E A Xd 3V 2 X d − X q sin 2 sin δ + Xd Xq 29 7 3(7 621 )(9890) 3(7 621 ) 0. 62 − 0.40 P= sin δ + (0. 62 )(0.40 ) sin 2 0. 62 P = 364.7 sin δ + 77.3 sin 2 MW A plot ... Physics, 2nd ed., McGraw- Hill, New York, 1991 Sears, Francis W., Mark W Zemansky, and Hugh D Young: University Physics, Addison-Wesley, Reading, Mass., 19 82 cha6 523 9_ch 02. qxd 10/16 /20 03 12: 20 PM ... angle is about 2. 2° different (c) The power supplied by this machine is given by the equation P= 3Vφ E A Xd sin δ + 3V 2 X d − X q sin 2 Xd Xq 3 (27 7 )( 322 ) 3 (27 7) 0 .25 − 0.18 ...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 7 ppt
... three-phase and four-phase stepper motors REFERENCES Fitzgerald, A E., and C Kingsley, Jr Electric Machinery New York: McGraw- Hill, 19 52 National Electrical Manufacturers Association Motors and Generators, ... NEMA, 1993 Veinott, G C Fractional and Subfractional Horsepower Electric Motors New York: McGrawHill, 1970 Werninck, E H (ed.) Electric Motor Handbook London: McGraw- Hill, 1978 ... power system supplying a constant 24 0 V, and the ac machine is connected to a 480-V, 60-Hz infinite bus The dc machine has four poles and is rated at 50 kW and 24 0 V It has a per-unit armature...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 1 pot
... 1 .26 7 1.3 1.333 1.367 1.4 1.433 1.466 1.5 104.4 118.86 1 32. 86 146.46 159.78 1 72. 18 183.98 195.04 20 5.18 21 4. 52 223 .06 23 1 .2 238 24 4.14 24 9.74 25 5.08 25 9 .2 263.74 26 7.6 27 0.8 27 3.6 27 6.14 27 8 27 9.74 ... 25 9 .2 263.74 26 7.6 27 0.8 27 3.6 27 6.14 27 8 27 9.74 28 1.48 28 2.94 28 4 .28 28 5.48 28 6.54 28 7.3 28 7.86 28 8.36 28 8. 82 289 .2 289.375 28 9.567 28 9.689 28 9.811 28 9.950 To use this curve in a MATLAB program, ... rule R4 127 .3 φ TOT = (0.0 023 5 Wb) = 0.00156 Wb 24 .0 + 40.8 + 127 .3 R2 + R3 + R4 φcenter = R2 + R3 24 .0 + 40.8 φTOT = (0.0 023 5 Wb) = 0.00079 Wb 24 .0 + 40.8 + 127 .3 R2 + R3 + R4 φ right = The flux...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 3 ppsx
... 13 .2 kV/ 13 .2 N C + 13 .2 N SE = 13.8 N C 13 .2 N SE = 0.6 N C Therefore, N C / N SE = 22 (b) The power advantage of this autotransformer is S IO N C + N SE N C + 22 N C = = = 23 SW NC NC so 1 /22 ... is Z base,P = (VP )2 = (20 ,000 V )2 = 20 kΩ 20 kVA S The base impedance of this transformer referred to the secondary side is Z base,S = (VS )2 = (480 V )2 S 20 kVA = 11. 52 Ω The open circuit ... are balanced, and the total load is found as: P1 = 120 kW 45 Load Load Q1 = P tan θ = ( 120 kW ) tan cos-1 (0.8) = 90 kvar P2 = 50 kW Q2 = P2 tan θ = (50 kW ) tan cos-1 (0.9 ) = 24 .2 kvar PTOT =...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 4 pot
... = − M cos ω t 2 2 5π π 5π / π /6 VM sin ω t d (ω t ) =− 3VM 3 3 − − = VM = 0. 827 0 VM 2 2 2π The rms voltage is Vrms = Vrms = T v (t ) dt = 2 5π / π /6 3VM 1 ω t − sin 2 t 2 VM sin ω t d ... , Z ) pu on base (Vbase )2 ( Sbase ) (8314 V )2 (1000 kVA) = 0.040 (8314 V )2 (500 kVA) (8314 V )2 (1000 kVA ) = 0.170 X 2, pu = 0.085 (8314 V )2 (500 kVA ) R2,pu = 0. 020 60 The per-unit impedance ... − sin 2 6 3 Vrms = 3VM π 5π π − sin − sin 2 3 Vrms = 3VM π 3 − − − 2 2 = = 3VM π 3 − − − 2 2 3VM π + = 0.8407 VM 2 The resulting ripple factor is r= Vrms VDC − × 100% = 0.8407 VM 0. 827 0 VM...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 5 ppsx
... and the firing angle is in degrees relative to time zero Start Time (ω t) Stop Time (ω t) Positive Phase Negative Phase − T / 12 T / 12 3T / 12 5T / 12 7T / 12 9T / 12 11T / 12 T / 12 3T / 12 ... ω R 2C V B1 = M RC RC Finally, 85 B1 = VM and + ω R 2C B2 = −ω RC VM + ω R 2C Therefore, the forced solution to the equation is vC , f ( t ) = VM ω RC VM sin ωt − cos ωt + ω R 2C + ω R 2C and ... D3, and capacitor C2 will charge to V volts through diode D2 (a) (2) Now, assume that SCR2 and SCR3 are triggered At the instant they are triggered, the voltage across capacitors C1 and C2 will...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps
... equation nm = 120 f e P To make a 50 Hz and a 60 Hz machine have the same mechanical speed so that they can be coupled together, we see that nsync = 120 (50 Hz ) 120 (60 Hz ) = P P2 P2 12 = = P 10 ... Baa = sin(w*t) * (cos(0) + j*sin(0)); Bbb = sin(w*t +2* pi/3) * (cos (2* pi/3) + j*sin (2* pi/3)); Bcc = sin(w*t -2* pi/3) * (cos( -2* pi/3) + j*sin( -2* pi/3)); % Calculate Bnet Bnet = Baa + Bbb + Bcc; % ... number of poles and electrical frequency is nm = 120 f e P The resulting table is Number of Poles f e = 25 Hz 10 12 14 1500 r/min 750 r/min 500 r/min 375 r/min 300 r/min 25 0 r/min 21 4.3 r/min The...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc
... (1800 r/min )( ) = = 60.00 Hz 120 120 f nl2 = The full-load frequency of generator corresponds to a frequency of f fl2 = (a) nm P (1785 r/min )( ) = = 59.50 Hz 120 120 The speed droop of generator ... column being field % current and the second column being line current % (= armature current) load p 52. scc; if_scc = p 52( :,1); ia_scc = p 52( : ,2) ; % Calculate Xs if = 0.001:0. 02: 1; vt = interp1(if_occ,vt_occ,If); ... P (3630 r/min )( ) = = 60.5 Hz 120 120 The full-load frequency of generator corresponds to a frequency of 121 f fl1 = nm P ( 3570 r/min )( ) = = 59.5 Hz 120 120 The no-load frequency of generator...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 8 ppt
... A2 sin δ With a 5% decrease, E A2 = 12, 570 V , and δ = sin −1 E A1 13 ,23 0 V sin δ = sin −1 sin 27 .9° = 29 .5° E A2 12, 570 V Therefore, the new armature current is IA = (e) E A2 − Vφ jX S = 12, 570 29 .5° ... + (34) ( 62. 21 − f ) + (34) ( 62. 21 − f ) + (34) (61.03 − f ) PLOAD = sP1 f nl1 − fsys + sP f nl2 − f sys + sP f nl3 − fsys + sP f nl4 − f sys ( 29 0 MW = ( 34 ) 62. 21 − fsys sys 8. 529 = 24 7.66 − ... is IA = E A2 − Vφ jX S = 11,907∠31.3° − 7044∠0° = 848∠ − 26 .8° A j8.18 With a 15% decrease, E A2 = 11 ,24 6 V , and δ = sin −1 E A1 13 ,23 0 V sin δ = sin −1 sin 27 .9° = 33.4° E A2 11 ,24 6 V Therefore,...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot
... English units is τ load = POUT ωm = ( 21 000 hp)(746 W/hp) ( 120 0 r/min ) 2 rad 60 s τ load = 6-13 = 124 ,700 N ⋅ m 1r 525 2 P 525 2 ( 21 000 hp ) = = 91,910 lb ⋅ ft nm ( 120 0 r/min ) A 440-V three-phase ... 1∠0° − 1.33∠ − 24 .9° = 0.663 20 .2 pu j0.90 In actual amps, this current is I A = (1404 A )( 0.663 20 .2 ) = 931 20 .2 A The reactive power supplied at the conditions of maximum E A and 10 MW power ... 159 I A2 = 50∠36.87° A The internal generated voltage required to produce this current would be E A2 = Vφ − R AI A2 − jX S I A2 E A2 = 120 ∠0° V − j (0.8 Ω )(50∠36.87° A ) E A2 = 147.5∠ − 12. 5°...
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(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 10 pps
... = cos−1 S LR = 14 .2 V = 0 .22 02 Ω 64.5 A 22 00 W = 36. 82 ( 24 .6 V )(64.5 A ) Therefore, RLR = Z LR cos θ LR = ( 0 .22 02 Ω ) cos ( 36. 82 ) = 0.176 Ω ′ ⇒ ⇒ R1 + R2 = 0.176 Ω R2 = 0.071 Ω X LR = ... induction motor at 50 Hz is shown below: IA R1 + Vφ jX1 0 .20 Ω j0.3 42 Ω j 12. 5 Ω jX2 j0.3 42 Ω 0. 120 Ω jXM 1− s R2 s 2. 28 Ω - (a) R2 The easiest way to find the line current (or armature ... (N-m) 500 300 20 0 100 27 00 27 50 28 00 n m 190 28 50 (r/min) 29 00 29 50 3000 This machine is rated at 75 kW It produces an output power of 75 kW at 3.1% slip, or a speed of 29 07 r/min 7-18 A 20 8-V, 60...
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