Joós, G. IGEE 402 – Power System Analysis MID-TERM EXAM – Sample Instructions: 1. - Duration: 75 minutes. - Material allowed: a crib sheet (double sided 8.5 x 11), calculator. - Attempt all questions. Make any reasonable assumption. - Return the completed answer sheet attached. QUESTION 1 (25 points) A single phase 600 V, 60 Hz feeder supplies 2 loads in parallel: (a) a heating load with a 10 Ω resistance; (b) a 70 kVA motor load, with a power factor of 0.70 (lagging). Compute the total current, power, reactive power and power factor. Draw the V-I vector diagram. A capacitor is connected in parallel with the loads to increase the power factor to 0.95. Compute the reactive power supplied by the capacitor and the value of the capacitor bank in µF. 2. QUESTION 2 (25 points) A short circuit test on a single phase 2400/240 V, 50 kVA, 60 Hz transformer yields the following results seen from the high side: 66 V, 20.8 A, 750 W. The open circuit test done from the low voltage side gives: 240 V, 5.97 A, 213 W. Give the approximate equivalent circuit referred to the high side, ignoring the magnetizing branch. For a 50 kVA inductive load (0.75 power factor) fed at 240 V, compute the current and voltage on the high side. Using as a base the transformer rated values, find: (a) the transformer parameters in pu for the approximate equivalent circuit; (b) the current and voltage on the high side; (c) the voltage regulation; (d) the efficiency (include core losses). 3. QUESTION 3 (25 points) Three identical single-phase 69 kV/6.9 kV, 60 Hz, 33 MVA distribution transformers, with a leakage reactance of 0.10 on the transformer base, are fed from the 69 kV bus. They supply two three phase 12 kV loads connected in parallel: a 60 MW, 0.8 power factor (lagging) load (assume a Y connection) and a capacitor bank of three elements (-j15 Ω) connected in ∆ (hint: convert into Y). Compute: (a) total line current; (b) the total power and total reactive power drawn from the feeder; (c) the resulting power factor. Draw the single line pu circuit, using a 100 MVA, 12 kV base. 4. QUESTION 4 (25 points) A 500 km, 500 kV, 60 Hz transmission line has a series impedance of z = 0.033 + j0.35 Ω/km and a shunt reactance of y = j4.42 x 10-6 S/km. Assume the line is lossless. Calculate: (a) the ABCD parameters; (b) the nominal π equivalent circuit parameters; (c) the surge impedance loading. The line delivers 1200 MW at a unity power factor and at a 0.95 pu voltage. Using the ABCD parameters, compute: (a) the sending end voltage and δ angle; (b) the no load receiving end voltage for the rated sending end voltage. Compute the sending end voltage and δ angle using the nominal π equivalent and ignoring the shunt elements. QUESTION 1 Vl := 600 600 R := 10 Ir := S := 70000 pf := 0.70 φ⋅ 180 Pr := Vl⋅ Ir 10 4 = 45.573 Pm = 4.9 × 10 π Ir = 60 Pr = 3.6 × 10 Pm := S⋅ pf φ := acos( pf ) Im := Im = 116.667 Qm := S⋅ sin( φ) Irc := Ir Imc := Im⋅ cos( φ) + Im sin( φ) ⋅ i Imc = 81.667 + 83.317i Itc := Irc + Imc Itc = 141.667 + 83.317i Itc = 164.351 180 ⋅ arg( Itc) = 30.461 π a := 4 S Vl 4 Qm = 4.999 × 10 180 π cos( arg( Itc) ) = 0.862 4 Pt := Pr + Pm Pt = 8.5 × 10 Qt := Qm Qt = 4.999 × 10 Stc := Pt + Qt⋅ i Stc = 8.5 × 10 + 4.999i × 10 pft := 4 Pt Pt Xc := Vl⋅ Vl Qcap Vl Xc⋅ i Itcc := Itc + Icc C := 1 (60⋅ 2⋅ π ⋅ Xc) 4 Stc = 9.861 × 10 a⋅ acos( pfc) = 18.195 Qtcor := Stcor ⋅ sin( acos( pfc ) ) pfc Qtcor = 2.794 × 10 4 a⋅ arg( Stc) = 30.461 pft = 0.862 pfc := 0.95 Stc Stcor := Icc := 4 4 Qcap := Qt − Qtcor Qcap = 2.205 × 10 4 Xc = 16.325 Icc = −36.753i Itcc = 141.667 + 46.564i −4 C = 1.625 × 10 a⋅ arg( Itcc) = 18.195 QUESTION 2 - Transformers Calculations in A and V Vpr := 2400 Vsr := 240 Sr := 50000 Vpsc := 66 Ipsc := 20.8 Psc := 750 Vpsc Zpsc := Zpsc = 3.173 Ipsc Rpsc := Psc Xpsc := ( Zpsc⋅ Zpsc − Rpsc⋅ Rpsc) Sl := 50000 Vlp := Ilp := Rpsc = 1.734 Ipsc⋅ Ipsc Xpsc = 2.658 pf := 0.75 Vpr ⋅ Vl Vsr Vsr ⋅ Ils Vpr Ils := a⋅ acos( pf ) = 41.41 Sl φ := acos( pf ) Vl := 240 Ils = 208.333 Vl Ilpc := Ilp⋅ ( cos( φ) − i⋅ sin( φ) ) Ilpc = 15.625 − 13.78i Ilpc = 20.833 3 Vp := Vlp + Ilpc⋅ ( Rpsc + i⋅ Xpsc) Vp = 2.464 × 10 + 17.638i Vp = 2.464 × 10 Vp 3 Vpr = 1.027 Calculation on a pu basis Vpbase := Vpr Ipbase := r := 3 Sbase := Sr Sbase Zpbase := Vpbase Rpsc x := Zpbase Vpbase Ipbase Xpsc Zpbase 4 Vpbase = 2.4 × 10 Sbase = 5 × 10 Ipbase = 20.833 Zpbase = 115.2 r = 0.015 x = 0.023 ip := 1 ⋅ ( cos( φ) − i⋅ sin( φ) ) ip = 0.75 − 0.661i vp := 1 + ip⋅ ( r + i⋅ x ) vp = 1.027 + 7.349i × 10 Vreg := efficiency := Vreg = 0.027 1 pin := Re vp⋅ ip + pcore ( pout pin −3 vp = 1.027 ( vp − 1 ) pout := 1 ⋅ pf ip = 1 ) pin = 0.769 efficiency = 0.975 pcore := 213 50000 QUESTION 3 - Three-phase circuits Vline := 12000 Plphase := 60⋅ 10 6 pf := 0.8 3 φ := −acos( pf ) Illine := Vlphase := a⋅ φ = −36.87 Slphase := Slphase ⋅ ( cos( φ) + i⋅ sin( φ) ) Vlphase Vline 3 Vlphase = 6.928 × 10 3 Plphase 7 Slphase = 2.5 × 10 pf 3 Illine = 2.887 × 10 − 2.165i × 10 3 3 Illine = 3.608 × 10 Icphase := Vlphase 3 Icphase = 1.386i × 10 15 −i⋅ 3 Icline := Icphase 3 Iline := Illine + Icline Iline = 2.887 × 10 − 779.423i 3 Iline = 2.99 × 10 a⋅ arg( Iline) = −15.11 cos( arg( Iline) ) = 0.965 Pt := 3 ⋅ Slphase ⋅ cos( φ) Pt = 6 × 10 Qt := 3 ⋅ Slphase ⋅ sin( −φ) − 3 ⋅ Icline ⋅ Vlphase Qt = 1.62 × 10 Sbase := 100 ⋅ Load Capacitor 10 6 Vbase := 3 12000 p := 20⋅ 10 p = 0.6 Sbase xc := 15 3 ⋅ Zbase Ibase := 3 6 xc = 3.472 7 pf = 0.8 7 a⋅ atan Qt Pt Sbase Vbase = 15.11 Zbase := Vbase Ibase Zbase = 1.44 QUESTION 4 - Transmission lines −6 rl := 0.033 xl := 0.35 X := ixl⋅ l X = 175i Y := iyl⋅ l Y = 2.21i × 10 xl yl Zc := β := yl := 4.42⋅ 10 l := 500 Vl := 500 Vl = 288.675 3 −3 Zc = 281.399 SIL := xl⋅ yl ( Vl⋅ Vl) SIL = 888.418 Zc β = 1.244 × 10 −3 Lossless ABCD A := cos( β ⋅ l) A = 0.813 B := i⋅ Zc⋅ sin( β ⋅ l) B = 163.936i C := i⋅ sin( β ⋅ l) C = 2.07i × 10 Zc A := 1 + Y⋅ Medium line −3 X A = 0.807 2 B := X B = 175i C := Y⋅ 1 + Y⋅ 3 Vrated := 500 ⋅ 10 X 4 6 Vrl := 0.95⋅ 500 ⋅ Vsl := A⋅ Vrl + B⋅ Ir 10 3 3 5 Ir = 1.459 × 10 5 5 Vsl = 2.212 × 10 + 2.552i × 10 5 a arg( Vsl) = 49.086 Vrated 3 Vrnl := A⋅ Vsr Vregl := Vrnl = 2.329 × 10 ( Vsl − Vrl ) Vrl 5 Vrl = 2.742 × 10 3 Vrl = 2.742 × 10 Vsl = 3.378 × 10 Vsr := −3 Srated := 1200⋅ 10 pf := 1 Srated 1 Ir := 3 Vrl C = 1.996i × 10 Vregl = 0.232 5

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