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

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Sherman Lambert
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1 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 Symbols" from the list Step 3: Scroll down the page and copy the angle symbol to your Mathcad sheet and use it to define a function as shown below defining the complex phasor in terms of magnitude and angle, where the arguments of the function are the magnitude and angle for a complex number ( magnitude angle) magnitudecos( angle) jmagnitude sin( angle) Step 4: When you want to use this new function to enter a phasor, start typing your expression, before entering the phasor itself, go to the Evaluation Toolbar and select to enter your information as an infix number (select "xfy" from the evaluation toolbar. This will allow you to enter your function in the order you normally use for entering phasors, instead of the normal Mathcad order. You should see three placeholders in your Mathcad sheet - Enter the magnitude of the phasor in the first placeholder - Copy the angle symbol into the middle one - Enter the angle of the phasor in the third placeholder V a ( 45deg) V a i - A lot of people find it easiest to enter this once and then just copy it into other sheets.

2 ECE 53: Session ; Page / Fall 07. Some matrix operations Expanding a matrix: If you have a 3x3 matrix and want to add a row and column while at the stage of entering the matrix initially, make the cell where you want to add rows and columns the active cell Select the add matrix/vector tool from the Matrix tool bar (or type CTRL-M) and enter the number of rows and columns you want to add (note you can add 0 rows and (or more) column or vice versa. The selected number of rows appears below the active cell and the selected number of columns appears to the right of the cell. A Use the submatrix command if you want to pull out part of a matrix Pay attention to the setting for the internal ORIGIN variable (or reassign it if you prefer) ORIGIN B submatrix A 3 B The "a" operator for three phase systems a ( 0deg) a i a i a i a 3

3 ECE 53: Session ; Page 3/ Fall 07 a a 0 a i 3 ( 30deg) i a i a i 3 ( 30deg) i 3 ( 50deg) i a i 3 ( 50deg) i a a.73i 3 ( 90deg).73i a a ( 80deg) Normal balanced three phase set: V AG ( 0deg) V BG ( 0deg) V BG i V CG ( 0deg) V CG i V ABC V AG a V ABC i a i V ABC arg V ABC deg

4 ECE 53: Session ; Page 4/ Fall 07 Example: Sketch a per unit impedance diagram for the system shown below. Use a 00MVA impedance base, and the generator rated voltage as your reference voltage base. Use pi models for the lines. BUS BUS BUS 3 BUS 4 BUS 5 G Line T Load G: 50MVA, 3.8kV G: 0MVA, 4.4kV T: 40MVA, -Y, 3.:6kV, X = 0% T: 5MVA, Y-, 6kV:3.kV, X = 0% Line T G Load: 45MVA, 0.8pf lagging (Y connected, parallel impedances) Line : 00 mile, Z = j0.73 ohm/mi, Y = 5.9*0-6 mho/mi Line : 75 mile, Z = j0.73 ohm/mi, Y = 5.9*0-6 mho/mi Define Base Quantities: Section I (left of T) 00MVA V B 3.8kV Line to line voltage V B Z B Z B.904Ω I B I B A 3V B

5 ECE 53: Session ; Page 5/ Fall 07 Section II (between T and T) Section II (right of T) 6kV 3.kV V B V B V B 68.38kV V B3 V B V B3 3.8kV 3.kV 6kV V B V B3 Z B Z B 83.3Ω Z B3 Z B3.904Ω I B I B 343.0A 3V B I B3 I B A 3V B3 Transmission Line Models: Line : Length 00mi Z line ( 0.8 j0.73) ohm mi Length Z line ( 8 73i) ohm Y line j mho Length Y line 5.9i 0 4 mho mi Y line.95i 0 4 mho Line is in section II, so use Zbase Z linepu Z line Z B Z linepu ( i) pu Note that Ybase is /Zbase: Y linepu Y line Z B Y linepu 0.67ipu

6 ECE 53: Session ; Page 6/ Fall 07 We actually need Y/ for the pi model: Y linepi Y linepu Y linepi 0.084ipu Line : length 75mi Z line ( 0.8 j0.73) ohm mi length Z line ( 54.75i) ohm Y line j mho length Y line 4.45i 0 4 mho mi Y line.i 0 4 mho Line is also in section II, so use Zbase Z linepu Z line Z linepu ( i) pu Z B Y linepu Y line Z B Y linepu 0.5ipu We actually need Y/ for the pi model: Y linepi Y linepu Y linepi 0.067ipu Transformer Model Calculations Transformer : S T 40MVA V TLow 3.kV V Thi 6kV X T 0.0pu

7 ECE 53: Session ; Page 7/ Fall 07 Impedance change of base calculation V TLow X Tnew X T X Tnew 0.9pu Transformer : S T V B S T 5MVA V TLow 3.kV V Thi 6kV X T 0.0pu Impedance change of base calculation V TLow X Tnew X T X Tnew 0.366pu Load Model mags load V B3 S T 45MVA pfload 0.8 lagging V loadrated 6kV 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 As a check: V loadrated Re S R load 70.08Ω V loadrated X load load Im S X load Ω load Z equivload Z equivload 576.0Ω R load jx load 36.87deg arg Z equivload

8 ECE 53: Session ; Page 8/ Fall 07 Z equivload ( i)Ω 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.54pu Z B X loadpu X load X loadpu 3.389pu Z B Per Unit Equivalent Circuit: j0.9pu j 0.58 pu Load j 0.93 pu j0.366pu Transformer Ea Line Y/=j pu j 3.39pu.54 pu Line Y/=j pu Transformer Ea Generator Generator (b) Suppose G is operating at 3.6kV 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 voltage as your reference angle. Create a Thevenin equivalent circuit looking back to the two sources from the load:

9 ECE 53: Session ; Page 9/ Fall 07 Circuit to left of the load: Z left Z left ( i) pu jx Tnew Z linepu V genpu 3.6kV V B ej0 deg V genpu 0.986pu Create a Norton equivalent: I left V genpu I left.986 Z left arg I left deg Circuit to right of load: Z right Z right ( i) pu jx Tnew Z linepu V genpu 3.6kV V B3 ej0 deg V genpu 0.986pu

10 ECE 53: Session ; Page 0/ Fall 07 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.747 Z right 8.45 arg I right deg 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:

11 ECE 53: Session ; Page / Fall 07 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 Z loadequiv Z loadequiv.033pu R loadpu Now we can find the load current: jx loadpu Z loadequiv.67.i 36.87deg arg Z loadequiv I loadpu Z thev V thev I loadpu 0.44pu Z loadequiv 4.54 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 Now convert to Ampere and Volts: I loadamps I loadpu I B I loadamps 5.433A 4.54 arg I loadamps deg

12 ECE 53: Session ; Page / Fall 07 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 5.084kV 5.38deg arg V ABload

ECE 421: Per Unit Examples

ECE 421: Per Unit Examples 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