Problem solutions: Chapter 5

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Nội dung Text: Problem solutions: Chapter 5

56 PROBLEM SOLUTIONS: Chapter 5 Problem 5.1 Basic equations are T ∝ ΦR Ff sin δRF . Since the ﬁeld current is constant, Ff is constant, Note also that the resultant ﬂux is proptoortional to the terminal voltage and inversely to the frequency ΦR ∝ Vt /f . Thus we can write Vt sin δRF T ∝ f P = ωf T ∝ Vt sin δRF part (a): Reduced to 31.1◦ part (b): Unchanged part (c): Unchanged part (d): Increased to 39.6◦ Problem 5.2 part (a): The windings are orthogonal and hence the mutual inductance is zero. part (b): Since the two windings are orthogonal, the phases are uncoupled and hence the ﬂux linkage under balanced two-phase operation is unchanged by currents in the other phase. Thus, the equivalent inductance is simply equal to the phase self-inductance. Problem 5.3 1 Lab = − (Laa − Lal ) = −2.25 mH 2 3 Ls = (Laa − Lal ) + Lal = 7.08 mH 2 Problem 5.4 part (a): √ 2 Vl−l,rms Laf = √ = 79.4 mH 3ωIf part (b): Voltage = (50/60) 15.4 kV = 12.8 kV. Problem 5.5 part (a): The magnitude of the phase current is equal to 40 × 103 Ia = √ = 59.1 A 0.85 × 3 460 and its phase angle is − cos−1 0.85 = −31.8◦ . Thus
57 Iˆa = 59.1e−j31.8 ◦ Then ˆaf = Va − jXs Iˆa = 460 E √ − j4.15 × 59.1e−j31.8 = 136 − 56.8◦ V ◦ 3 The ﬁeld current can be calculated from the magnitude of the generator voltage √ 2Eaf If = = 11.3 A ωLaf part (b): ˆaf = 266 − 38.1◦ V; If = 15.3 A E part (c): ˆaf = 395 − 27.8◦ V; If = 20.2 A E Problem 5.6 The solution is similar to that of Problem 5.5 with the exception that the sychronous impedance jXs is replaced by the impedance Zf + jXs . part (a): ˆaf = 106 − 66.6◦ V; If = 12.2 A E part (b): ˆaf = 261 − 43.7◦ V; If = 16.3 A E part (c): ˆaf = 416 − 31.2◦ V; If = 22.0 A E Problem 5.7 part (a): √ 2 Vl−l,rms Laf = √ = 49.8 mH 3ωIf part (b): 600 × 103 Iˆa = √ = 151 A 3 2300 ˆaf = Va − jXs Iˆa = 1.77 − 41.3◦ V E √ 2Eaf If = = 160 A ωLaf
58 part (c): See plot below. Minimum current will when the motor is operating at unity power factor. From the plot, this occurs at a ﬁeld current of 160 A. Problem 5.8 part (a): 2 Vbase (26 × 103 )2 Zbase = = = 0.901 Ω Pbase 750 × 106 Xs,pu Zbase Ls = = 4.88 mH ω part (b): Xal,pu Zbase Lal = = 0.43 mH ω part (c): 2 Laa = (Ls − Lal ) + Lal = 3.40 mH 3 Problem 5.9 part (a): AFNL SCR = = 0.520 AFSC part (b): Zbase = (26 × 103 )2 /(800 × 106 ) = 0.845 Ω 1 Xs = = 2.19 pu = 1.85 Ω SCR part (c): AFSC Xs,u = = 1.92 pu = 1.62 Ω AFNL, ag
59 Problem 5.10 part (a): AFNL SCR = = 1.14 AFSC part (b): Zbase = 41602 /(5000 × 103 ) = 3.46 Ω 1 Xs = = 1.11 pu = 3.86 Ω SCR part (c): AFSC Xs,u = = 0.88 pu = 3.05 Ω AFNL, ag Problem 5.11 No numerical solution required. Problem 5.12 part (a): The total power is equal to S = P /pf = 4200 kW/0.87 = 4828 kVA. The armature current is thus 4828 × 103 Iˆa = √ (cos−1 0.87) = 670 29.5◦ A 3 4160 Deﬁning Zs = Ra + jXs = 0.038 + j4.81 Ω 4160 |Eaf | = |Va − Zs Ia | = | √ − Zs Ia | = 4349 V, line − to − neutral 3 Thus 4349 If = AFNL √ = 306 A 4160/ 3 part (b): If the machine speed remains constant and the ﬁeld current is not reduced, the terminal voltage will increase to the value corresponding to 306 A of ﬁeld current on the open-circuit saturation characteristic. Interpolating the given data shows that this corresponds to a value of around 4850 V line-to-line.
60 Problem 5.13 Problem 5.14 √ power, unity power factor, the armature current will be Ia = At rated 5000 kW/( 3 4160 V) = 694 A. The power dissipated in the armature winding will then equal Parm = 3 × 6942 × 0.011 = 15.9 kW. The ﬁeld current can be found from 4160 |Eaf | = |Va − Zs Ia | = | √ − Zs Ia | = 3194 V, line-to-neutral 3 and thus 3194 If = AFNL √ = 319 A 4160/ 3 At 125◦ C, the ﬁeld-winding resistance will be 234.5 + 125 Rf = 0.279 = 0.324 Ω 234.5 + 75 and hence the ﬁeld-winding power dissipation will be Pfield = If2 Rf = 21.1 kW. The total loss will then be Ptot = Pcore + Parm + Pfriction/windage + Pfield = 120 kW Hence the output power will equal 4880 kW and the eﬃciency will equal 4880/5000 = 0.976 = 97.6%.
61 Problem 5.15 part (a): part (b): AFNL = 736 A. AFSC = 710 A. part (c): (i) SCR = 10.4, (ii) Xs = 0.964 per unit and (iii) Xs,u = 1.17 per unit. Problem 5.16 For Va = 1.0 per unit, Eaf,max = 2.4 per unit and Xs = 1.6 per unit Eaf,max − Va Qmax = = 0.875 per unit Xs Problem 5.17 part (a): 2 Vbase Zbase = = 5.29 Ω Pbase 1 Xs = = 0.595 per-unit = 3.15 Ω SCR
62 part (b): Using generator convention for current part (c): 150 Eaf = = 0.357 per-unit 420 For Va = 1.0 per-unit, Eaf − Va Iˆa = = 1.08 90◦ per-unit = 1.36 90◦ kA jXs using Ibase = 1255 A. part (d): It looks like an inductor. part (e): 700 Eaf = = 1.67 per-unit 420 For Va = 1.0 per-unit, Eaf − Va Iˆa = = 1.12 − 90◦ per-unit = 1.41 − 90◦ kA jXs In this case, it looks like a capacitor.
63 Problem 5.18 Problem 5.19 part (a): It was underexcited, absorbing reactive power. part (b): It increased. part (c): The answers are the same. Problem 5.20 part (a): 226 Xs = = 0.268 per-unit 842 part (b): P = 0.875 and S = P/0.9 = 0.972, both in per unit. The power- factor angle is − cos−1 0.9 = −25.8◦ and thus Iˆa = 0.875 − 25.8◦ . ˆaf = Va + jXs Iˆa = 1.15 11.6◦ per-unit E ˆaf | = 958 A. The rotor angle is 11.6◦ and the The ﬁeld current is If = AFNL|E reactive power is Q = S 2 − P 2 = 4.24 MVA part (c): Now |Eaf | = 1.0 per unit. −1 P Xs δ = sin |Eaf | = 13.6◦ Va ˆaf = 1.0 13.6◦ . and thus E ˆaf − Va E Iˆa = = 0.881 6.79◦ jXs Q = Imag[Va Iˆa∗ ] = −0.104 per-unit = −1.04 MVAR
64 Problem 5.21 ˆaf − Va E Va Eaf Iˆa = =j + (sin δ − j cos δ) jXs Xs Xs The ﬁrst term is a constant and is the center of the circle. The second term is a circle of radius Eaf/Xs . Problem 5.22 part (a): (i) (ii) Vt = V∞ = 1.0 per unit. P = 375/650 = 0.577 per unit. Thus P X∞ δt = sin−1 = 12.6◦ Vt V∞
65 and Vt ejδt − V∞ Iˆa = = 0.578 3.93◦ per-unit jX∞ √ Ibase = Pbase /( 3 Vbase ) = 15.64 kA and thus Ia = 9.04 kA. (iii) The generator terminal current lags the terminal voltage by δt /2 and thus the power factor is pf = cos−1 δt /2 = 0.998 lagging (iv) ˆaf | = |V∞ + j(X∞ + Xs )Iˆa | = 1.50 per-unit = 36.0 kV,line-to-line |E part (b): (i) Same phasor diagram (ii) Iˆa = 0.928 6.32◦ per-unit. Ia = 14.5 kA. (iii) pf = 0.994 lagging (iv) Eaf = 2.06 per unit = 49.4 kV, line-to-line. Problem 5.23 part (a):
66 part (b): part (c): Problem 5.24 part (a): From the solution to Problem 5.15, Xs = 0.964 per unit. Thus, with V∞ = Eaf = 1.0 per unit
67 V∞ Eaf Pmax = = 1.04 per-unit Xs noindent Hence, full load can be achieved. This will occur at −1 Xs δ = sin = 74.6◦ Eaf Vinfty part (b): The generator base impedance is 1.31 Ω. Thus, X∞ = 0.14/1.31 = 0.107 per unit. Now V∞ Eaf Pmax = = 1.04 per-unit = 0.934 per-unit = 135 MW (X∞ + Xs Problem 5.25 Follwing the calculation steps of Example 5.9, Eaf = 1.35 per unit. Problem 5.26 Now Xd = .964 per unit and Xq = 0.723 per unit. Thus part (a): 2 V∞ Eaf V∞ 1 1 P = sin δ + − sin 2δ = 1.037 sin δ + 0.173 sin 2δ Xd 2 Xq Xd An iterative solution with MATLAB shows that maximum power can be achieved at δ = 53.6◦ . part (b): Letting XD = Xd + X∞ and XQ = Xq + X∞ V∞ Eaf V2 1 1 P = sin δ + ∞ − sin 2δ = 0.934 sin δ + 0.136 sin 2δ X 2 XQ XD An iterative solution with MATLAB shows that maximum power that can be achieved is 141 Mw, which occurs at a power angle of 75◦ . Problem 5.27
68 Problem 5.28 Problem 5.29 Problem 5.30 For Eaf = 0, Vt2 1 1 Pmax = − = 0.21 = 21% 2 Xq Xq This maximum power occurs for δ = 45◦ . Vt cos δ Id = = 0.786 per-unit Xd Vt sin δ Iq = = 1.09 per-unit Xq
69 and thus Ia = Id2 + Iq2 = 1.34 per unit. S = Vt Ia = 1.34 per-unit Hence Q= S 2 − P 2 = 1.32 per-unit Problem 5.30 V∞ Eaf V2 1 1 P = sin δ + ∞ − sin 2δ Xd 2 Xq Xd The generator will remain synchronized as long as Pmax > P . An iterative search with MATLAB can easily be used to ﬁnd the minimum excitation that satisﬁes this condition for any particular loading. part (a): For P = 0.5, must have Eaf ≥ 0.327 per unit. part (b): For P = 1.0, must have Eaf ≥ 0.827 per unit. Problem 5.32 part (a): part (b): We know that P = 0.95 per unit and that V∞ Vt P = sin δt Xbus and that Vˆt − V∞ Iˆa = jXt It is necessary to solve these two equations simultaneously for Vˆt = Vt δt so that both the required power is achieved as well as the speciﬁed power factor
70 angle with respect to the generator terminal voltage. This is most easily done iteratively with MATLAB. Once this is done, it is straightforward to calculate Vt = 1.02 per-unit; Eaf = 2.05 per-unit; δ = 46.6◦ Problem 5.33 part (a): Deﬁne XD = Xd + Xbus and XQ = Xq + Xbus . (i) Eaf,min = Vbus − XD = 0.04 per-unit Eaf,max = Vbus + XD = 1.96 per-unit (ii) part (b):
71 part (c): Problem 5.34 n × poles 3000 × 6 f= = = 150 Hz 120 120 Problem 5.35 part (a): Because the load is resistive, we know that P 4500 Ia ==√ = 13.5 A 3Va 3192 √ part (b): We know that Eaf = 208/ 3 = 120 V. Solving Eaf = Va2 + (Xs Ia )2 for Xs gives 2 −V2 Eaf a Xs = = 3.41 Ω Ia part (c): The easiest way to solve this is to use MATLAB to iterate to ﬁnd the required load resistance. If this is done, the solution is Va = 108 V (line-to-neutral) = 187 V (line-to-line). Problem 5.36 Ea ωKa Iˆa = = Ra + Rb + jωLa Ra + Rb + jωLa Thus ωKa Ka |Iˆa | = = (Ra + Rb )2 + (ωLa)2 La 1 + RωL a +Rb a Clearly, Ia will remain constant with speed as long as the speed is suﬃcient to insure that ω >> (Ra + Rb )/La