ECE 421: Per Unit Examples

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Clare Hodges
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1 ECE 41: Session 14; Page 1/13 Fall 013 ECE 41: Per Unit Examples Define units: MVA 1000kW MW MVA kva kw kvar kw pu 1 Example 1: A three-phase transformer rated 5 MVA, 115/13. kv has per-phase series impedance of ( j0.075) per unit. The transformer is connected to a short distribution line which can be represented by a series impedance per phase of (0.0 + j0.10) per unit on a base of 10 MVA, 13. kv. The line supplies a balanced three-phase load rated 4MVA, 13.kV, with lagging power factor S rated 5MVA V hi 115kV V lo 13.kV V hi Z base_hi Z base_hi 645Ω S rated V lo Z base_lo Z base_lo Ω S rated Z series 0.007pu j0.075pu on transformer rating base. Distribution line: Z line 0.0pu j0.10pu on 10MVA, 13.kV base Load: S load 4MVA V LL_load 13.kV pf 0.85 lagging (a) Draw an equivalent circuit of the system indicating all impedances in per unit. Choose 10MVA, 13. kv as the base of the load. V BLL 10MVA V BLL 13.kV Z B Z B 17.44Ω

2 ECE 41: Session 14; Page /13 Fall 013 Transformer impedance (change of base calculation): V lo Z series_new Z series Z series_new ( i) pu V BLL S rated Line impedance (no change of base needed, already on the correct base) Z line ( i) pu Load: ϕ load acos( pf) ϕ load deg positive since lagging power factor Z load V LL_load S load e j ϕ load Z load ( i)Ω Z load_pu Z load Z B Z load_pu i Per unit equivalent circuit: I load j 0.15 p.u j 0.1 p.u. Vs = 1.0 p.u. Transformer Line Impedance Load.15 + j p.u.

3 ECE 41: Session 14; Page 3/13 Fall 013 (b) With the voltage at the primary side of the transformer held constant at 115kV, the load at the receiving end of the line is disconnected. Find the voltage regulation at the load. V source 115kV LL V source_pu V source V source_pu 1pu V hi Use as reference angle (so 0 deg) I load Z series_new V source_pu I load ( i) pu Z line Z load_pu I load 0.375pu arg I load deg V load_pu I load Z load_pu V load_pu ( i) pu V load_pu 0.937pu arg V load_pu deg V source_pu V load_pu V regulation V load_pu V regulation %

4 ECE 41: Session 14; Page 4/13 Fall 013 Example : A transformer rated 00 MVA, 345 Y/0.5 kv connects a balanced load rated 180 MVA,.5kV, 0.8 power-factor lag to a transmission line. Determine: S r 00MVA V HI 345kV V LO 0.5kV S load 180MVA V load.5kv pf 0.8 lag (a) The rating of each of the three single phase transformers which when properly connected will be equivalent to the above three-phase transformer. V rating_xfmr_hv V rating_xfmr_lv S per_xfmr S r V HI V rating_xfmr_hv kV 3 V rating_xfmr_lv 0.5kV V LO S per_xfmr MVA 3 (b) The complex impedance of the load in per unit in the impedance diagram if the base in the transmission line is 100MVA, 345kV. ϕ load acos( 0.8) ϕ load 36.87deg V load Z load_ohm_lv ASE 100MVA S load e j ϕ load Z load_ohm_lv.81ω 36.87deg arg Z load_ohm_lv

5 ECE 41: Session 14; Page 5/13 Fall 013 V BHI 345kV V LO V BLV V BHI V BLV 0.5kV V HI V BLV Z BLV Z BLV 4.0Ω ASE Z load_pu Z load_ohm_lv Z BLV Z load_pu ( i) pu Z load_pu 0.669pu 36.87deg arg Z load_pu Example 3: Sketch a per unit impedance diagram for the system shown below. Use a 100MVA impedance base, and the generator 1 rated voltage as your reference voltage base. Use pi models for the lines. BUS 1 BUS BUS 3 BUS 4 BUS 5 G1 Line 1 T1 Load G1: 50MVA, 13.8kV G: 0MVA, 14.4kV T1: 40MVA, -Y, 13.:161kV, X = 10% T: 5MVA, Y-, 161kV:13.kV, X = 10% Line T G Load: 45MVA, 0.8pf lagging (Y connected, parallel impedances) Line 1: 100 mile, Z = j0.73 ohm/mi, Y = 5.9*10-6 mho/mi Line : 75 mile, Z = j0.73 ohm/mi, Y = 5.9*10-6 mho/mi

6 ECE 41: Session 14; Page 6/13 Fall 013 Define Base Quantities: Section I (left of T1) 100MVA V B1 13.8kV Line to line voltage V B1 Z B1 Z B Ω I B1 I B A 3V B1 Section II (between T1 and T) Section II (right of T) 161kV 13.kV V B V B1 V B kV V B3 V B V B3 13.8kV 13.kV 161kV V B V B3 Z B Z B 83.31Ω Z B3 Z B Ω I B I B A 3V B I B3 I B A 3V B3

7 ECE 41: Session 14; Page 7/13 Fall 013 Transmission Line Models: Line 1: Length1 100mi Z line1 ( 0.8 j0.73) ohm mi Length1 Z line1 ( 8 73i) ohm Y line1 j mho Length1 Y line1 5.9i 10 4 mho mi Y line1.95i 10 4 mho Line 1 is in section II, so use Zbase Z line1pu Z line1 Z line1pu ( i) pu Z B Note that Ybase is 1/Zbase: Y line1pu Y line1 Z B Y line1pu 0.167ipu We actually need Y/ for the pi model: Y line1pi Y line1pu Y line1pi 0.084ipu Line : length 75mi Z line ( 0.8 j0.73) ohm mi length Z line ( i) ohm Y line j mho length Y line 4.45i 10 4 mho mi Y line.1i 10 4 mho

8 ECE 41: Session 14; Page 8/13 Fall 013 Line is also in section II, so use Zbase Z linepu Y linepu Z line Z linepu ( i) pu Z B Y line Z B Y linepu 0.15ipu We actually need Y/ for the pi model: Y linepi Y linepu Transformer Model Calculations Transformer 1: S T1 Y linepi 0.067ipu 40MVA V T1Low 13.kV V T1hi 161kV X T1 0.10pu Impedance change of base calculation V T1Low X T1new X T1 X T1new 0.9pu Transformer : S T V B1 S T1 5MVA V TLow 13.kV V Thi 161kV X T 0.10pu Impedance change of base calculation V TLow X Tnew X T X Tnew 0.366pu V B3 S T

9 ECE 41: Session 14; Page 9/13 Fall 013 Load Model mags load 45MVA pfload 0.8 lagging V loadrated 161kV ϕ load acos( pfload) ϕ load 36.87deg S load mags load e j ϕ load S load ( 36 7i) MVA Since the load is wye connected with parallel impedances: R load V loadrated Re S R load 70.08Ω V loadrated X load load Im S X load Ω load As a check: 1 1 Z equivload Z equivload 576.0Ω R load jx load 1 Z equivload ( i)Ω 36.87deg arg Z equivload S check V loadrated S check ( 36 7i) MVA Z equivload note complex conjugate in equation Convert to per unit (load in section II): R loadpu R load R loadpu.541pu Z B X loadpu X load X loadpu 3.389pu Z B

10 ECE 41: Session 14; Page 10/13 Fall 013 Per Unit Equivalent Circuit: j0.9pu j 0.58 pu Load j pu j0.366pu Transformer1 Ea1 Line 1 Y/=j pu j 3.39pu.54 pu Line Y/=j pu Transformer Ea Generator1 Generator (b) Suppose G1 is operating at 13.6kV and G is set to operate at the same magnitude. Suppose also, that the two generators are in phase with each other. Determine the phase A line to neutral voltage in Volts and in per unit and the phase A current at the load in Amperes and in per unit. Determine magnitude and phase is each case. Use the generator 1 voltage as your reference angle. Create a Thevenin equivalent circuit looking back to the two sources from the load: Circuit to left of the load: Z left Z left ( i) pu jx T1new Z line1pu V gen1pu 13.6kV V B1 ej0 deg V gen1pu 0.986pu Create a Norton equivalent: I left V gen1pu I left Z left arg I left deg

11 ECE 41: Session 14; Page 11/13 Fall 013 Circuit to right of load: Z right Z right ( i) pu jx Tnew Z linepu V genpu 13.6kV V B3 ej0 deg V genpu 0.986pu Vload Ea Zleft Zleft j 3.39pu.54 pu Zright Ea Zright Vload Zleft Zright Vload Ea Zleft j 3.39pu.54 pu Ea + ( Ea Zright Zleft Zright Zleft + Ea Zright ) Zleft Zright j 3.39pu.54 pu Create a Norton equivalent: I right V genpu I right Z right 8.45 arg I right deg

12 ECE 41: Session 14; Page 1/13 Fall 013 Combine the two parallel impedances from the sources: Z para Z Z para 0.64 left Z right deg arg Z para Combine the two current sources: I para I left I right I para arg I para deg Find the Thevenin equivalent source voltage: V thev I para Z para V thev 0.986pu Z thev Z para 1 1 Z loadequiv Z loadequiv.033pu R loadpu Now we can find the load current: jx loadpu 1 Z loadequiv i 36.87deg arg Z loadequiv I loadpu Z thev V thev I loadpu 0.441pu Z loadequiv arg I loadpu deg And the voltage across the load: V loadpu I loadpu Z loadequiv V loadpu 0.898pu 4.67 arg V loadpu deg

13 ECE 41: Session 14; Page 13/13 Fall 013 Now convert to Ampere and Volts: I loadamps I loadpu I B I loadamps A arg I loadamps deg V ANload V B V loadpu 3 Note that we are using a line to neutral voltage base, since the angles in per unit correspond to the line to neutral voltages. V ANload 87.9kV 4.67 arg V ANload deg V ABload 3V ANload e j30 deg V ABload kV 5.38deg arg V ABload

Mathcad Examples. Define units: MVA 1000kW MW MVA kw kvar kw. pu Entering Phasors in Polar Notation

Mathcad Examples. Define units: MVA 1000kW MW MVA kw kvar kw. pu Entering Phasors in Polar Notation ECE 53: Session ; Page / Fall 07 Mathcad Examples Define units: MVA 000kW MW MVA kva kw kvar kw pu. Entering Phasors in Polar Notation Step : Go to: "Help Menu" -- "Quick Sheets" Step : Choose "Extra Math