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56 PROBLEM SOLUTIONS: Chapter 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 T ∝ Vt sin δRF f P = ωf T ∝ Vt sin δRF part part part part (a): (b): (c): (d): Reduced to 31.1◦ Unchanged Unchanged 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 Lab = − (Laa − Lal ) = −2.25 mH Ls = (Laa − Lal ) + Lal = 7.08 mH Problem 5.4 part (a): Laf = √ Vl−l,rms √ = 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 Ia = 40 × 103 √ = 59.1 A 0.85 × 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 √ − j4.15 × 59.1e−j31.8 = 136 E − 56.8◦ V The field current can be calculated from the magnitude of the generator voltage √ 2Eaf If = = 11.3 A ωLaf part (b): ˆaf = 266 E − 38.1◦ V; If = 15.3 A ˆaf = 395 E − 27.8◦ V; If = 20.2 A part (c): 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 E − 43.7◦ V; If = 16.3 A ˆaf = 416 E − 31.2◦ V; If = 22.0 A part (c): Problem 5.7 part (a): Laf = √ Vl−l,rms √ = 49.8 mH 3ωIf part (b): 600 × 103 Iˆa = √ = 151 A 2300 ˆaf = Va − jXs Iˆa = 1.77 E − 41.3◦ V √ 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): Zbase = Vbase (26 × 103 )2 = = 0.901 Ω Pbase 750 × 106 Ls = Xs,pu Zbase = 4.88 mH ω Lal = Xal,pu Zbase = 0.43 mH ω part (b): part (c): Laa = (Ls − Lal ) + Lal = 3.40 mH Problem 5.9 part (a): SCR = AFNL = 0.520 AFSC part (b): Zbase = (26 × 103 )2 /(800 × 106 ) = 0.845 Ω Xs = = 2.19 pu = 1.85 Ω SCR part (c): Xs,u = AFSC = 1.92 pu = 1.62 Ω AFNL, ag 59 Problem 5.10 part (a): SCR = AFNL = 1.14 AFSC part (b): Zbase = 41602 /(5000 × 103 ) = 3.46 Ω Xs = = 1.11 pu = 3.86 Ω SCR part (c): Xs,u = AFSC = 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 4160 Defining Zs = Ra + jXs = 0.038 + j4.81 Ω 4160 |Eaf | = |Va − Zs Ia | = | √ − Zs Ia | = 4349 V, line − to − neutral Thus If = AFNL 4349 √ 4160/ = 306 A 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 At rated √ power, unity power factor, the armature current will be Ia = 5000 kW/( 4160 V) = 694 A The power dissipated in the armature winding will then equal Parm = × 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 and thus If = AFNL 3194 √ 4160/ = 319 A At 125◦ C, the field-winding resistance will be Rf = 0.279 234.5 + 125 234.5 + 75 = 0.324 Ω 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 Qmax = Eaf,max − Va = 0.875 per unit Xs Problem 5.17 part (a): Zbase = Xs = Vbase = 5.29 Ω Pbase = 0.595 per-unit = 3.15 Ω SCR 62 part (b): Using generator convention for current part (c): Eaf = 150 = 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): Eaf = 700 = 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): Xs = 226 = 0.268 per-unit 842 part (b): P = 0.875 and S = P/0.9 = 0.972, both in per unit The powerfactor 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 − P = 4.24 MVA part (c): Now |Eaf | = 1.0 per unit δ = sin−1 P Xs |Eaf | Va = 13.6◦ 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 δt = sin−1 P X∞ Vt V∞ = 12.6◦ 65 and Vt ejδt − V∞ Iˆa = = 0.578 3.93◦ per-unit jX∞ √ Ibase = Pbase /( 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 Pmax = V∞ Eaf = 1.04 per-unit Xs noindent Hence, full load can be achieved This will occur at δ = sin−1 Xs Eaf Vinfty = 74.6◦ part (b): The generator base impedance is 1.31 Ω Thus, X∞ = 0.14/1.31 = 0.107 per unit Now Pmax = V∞ Eaf = 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): P = V2 V∞ Eaf sin δ + ∞ Xd 1 − Xq Xd sin 2δ = 1.037 sin δ + 0.173 sin 2δ 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∞ P = V2 V∞ Eaf sin δ + ∞ X 1 − XQ XD sin 2δ = 0.934 sin δ + 0.136 sin 2δ 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, Pmax = Vt2 1 − Xq Xq = 0.21 = 21% This maximum power occurs for δ = 45◦ Id = Iq = Vt cos δ = 0.786 per-unit Xd Vt sin δ = 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 − P = 1.32 per-unit Problem 5.30 P = V∞ Eaf V2 sin δ + ∞ Xd 1 − Xq Xd sin 2δ 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 P = V∞ Vt 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 f= 3000 × n × poles = = 150 Hz 120 120 Problem 5.35 part (a): Because the load is resistive, we know that P 4500 =√ = 13.5 A 3Va 3192 √ part (b): We know that Eaf = 208/ = 120 V Solving Ia = Eaf = Va2 + (Xs Ia )2 for Xs gives Xs = −V2 Eaf a = 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 Iˆa = Ea ωKa = Ra + Rb + jωLa Ra + Rb + jωLa Thus |Iˆa | = ωKa = (Ra + Rb )2 + (ωLa)2 Ka La 1+ Ra +Rb ωLa Clearly, Ia will remain constant with speed as long as the speed is sufficient to insure that ω >> (Ra + Rb )/La ... occurs at a power angle of 75 Problem 5. 27 68 Problem 5. 28 Problem 5. 29 Problem 5. 30 For Eaf = 0, Pmax = Vt2 1 − Xq Xq = 0.21 = 21% This maximum power occurs for δ = 45 Id = Iq = Vt cos δ =... = 0.8 75 per unit Xs Problem 5. 17 part (a): Zbase = Xs = Vbase = 5. 29 Ω Pbase = 0 .59 5 per-unit = 3. 15 Ω SCR 62 part (b): Using generator convention for current part (c): Eaf = 150 = 0. 357 per-unit... V∞ Eaf = 1.04 per-unit = 0.934 per-unit = 1 35 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