IGEE 402 Power System Analysis. MID-TERM EXAM Sample

Transcription

1 Joós, G. IGEE 402 Power System Analysis MID-TERM EXAM Sample Instructions: - Duration: 75 minutes. - Material allowed: a crib sheet (double sided 8.5 x, calculator. - Attempt all questions. Make any reasonable assumption. - Return the completed answer sheet attached.. QUESTION (25 points A single phase 600 V, 60 Hz feeder supplies 2 loads in parallel: (a a heating load with a 0 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 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, 2 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.. QUESTION (25 points Three identical single-phase 69 kv/6.9 kv, 60 Hz, MVA distribution transformers, with a leakage reactance of 0.0 on the transformer base, are fed from the 69 kv bus. They supply two three phase 2 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 (-j5 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 00 MVA, 2 kv base. 4. QUESTION 4 (25 points A 500 km, 500 kv, 60 Hz transmission line has a series impedance of z = j0.5 Ω/km and a shunt reactance of y = j4.42 x 0-6 S/km. Assume the line is lossless. Calculate: (a the ABCD parameters; (b the nominal equivalent circuit parameters; (c the surge impedance loading. The line delivers 200 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 equivalent and ignoring the shunt elements.

2 QUESTION := 600 R := 0 Ir 600 := Pr := Ir Ir = 60 Pr = S := pf := 0.70 Pm := S pf φ := acos( pf Im := S φ 80 = Pm = Im = Qm := S sin( φ Qm = Irc := Ir i Imc := Im cos φ + Imsin φ Imc = i Itc := Irc + Imc Itc = i Itc = cos( arg( Itc = := Pr + Pm = Qt := Qm Qt = Stc := + Qt i Stc = i 0 4 Stc = a arg( Stc = 0.46 pft := pft = Stc pfc := 0.95 a acos( pfc = 8.95 Stcor := Qtcor := Stcor sin( acos( pfc pfc Qtcor = Qcap := Qt Qtcor Qcap = Xc := Xc = 6.25 Qcap Icc := Icc = 6.75i Xc i arg ( Itc = a := Itcc := Itc + Icc Itcc = i a arg( Itcc = 8.95 C := C = ( 60 2 Xc 4

3 QUESTION 2 - Transformers Calculations in A and V Vpr := 2400 Vsr := 240 Sr := Vpsc := 66 Ipsc := 20.8 Psc := 750 Vpsc Zpsc := Zpsc =.7 Ipsc Psc Rpsc := Rpsc =.74 Ipsc Ipsc Xpsc := ( Zpsc Zpsc Rpsc Rpsc Xpsc = Sl := pf := 0.75 a acos( pf = 4.4 φ := acos( pf := 240 Vpr p := Vsr Ils Sl := Ils = 208. Vsr Ilp := Vpr Ils Ilpc := Ilp ( cos( φ i sin( φ Ilpc = i Ilpc = 20.8 Vp := p + Ilpc ( Rpsc + i Xpsc Vp = i Vp = Vp Vpr =.027 Calculation on a pu basis Vpbase := Vpr := Sr Vpbase = = Vpbase Ipbase := Zpbase := Ipbase = 20.8 Zpbase = 5.2 Vpbase Ipbase Rpsc Xpsc r := x := r = 0.05 x = 0.02 Zpbase Zpbase ( i sin( φ ip := cos φ ip = i ip = vp := + ip ( r + i x vp = i 0 vp =.027 ( vp Vreg := Vreg = pout := pf pin := Re( vp ip + pcore pin = pout efficiency := efficiency = pin pcore :=

4 QUESTION - Three-phase circuits ine := 2000 Plphase ine := pf := 0.8 phase := phase = Plphase φ := acos( pf a φ = 6.87 Slphase := Slphase = pf Slphase Illine := ( phase cos ( φ + i sin( φ Illine = i 0 Illine = phase Icphase := i 5 Icphase =.86i 0 Icline := Icphase Iline := Illine + Icline Iline = i Iline = cos( arg( Iline = := Slphase cos( φ = a arg( Iline = 5. Qt := Slphase sin( φ Icline phase Qt = a atan Qt = Vbase := Vbase := Ibase := Zbase := Vbase Ibase Zbase =.44 Load p := p = 0.6 pf = 0.8 Capacitor 5 xc := xc =.472 Zbase

5 QUESTION 4 - Transmission lines rl := 0.0 xl := 0.5 yl := l := 500 := 500 X := ixl l X = 75i Y := iyl l Y = 2.2i 0 xl ( Zc := Zc = SIL := SIL = yl Zc β := xl yl β = = Lossless ABCD A := cos β l A = 0.8 B := i Zc sin β l B = 6.96i C := i sin β l C = 2.07i 0 Zc Medium line A := + Y X 2 A = B := X B = 75i C := Y + Y X 4 Vrated := Srated := C =.996i 0 pf := Ir := Srated Vrl 0 Vrl := Vrl = Vrl = Ir = Vsl := A Vrl + BIr Vsl = i 0 5 Vsl = a arg( Vsl = Vrated Vsr := Vrnl := A Vsr Vrnl = Vregl := ( Vsl Vrl Vrl Vregl = 0.22