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Per Unit Calculations A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer impedances to the different sides of the transformers This problem is avoided by a normalization of all variables. This normalization is known as per unit analysis. actualquantity quantity in perunit base value ofquantity 1

Per unit system

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Page 1: Per unit system

Per Unit CalculationsA key problem in analyzing power systems is the large

number of transformers. – It would be very difficult to continually have to refer

impedances to the different sides of the transformersThis problem is avoided by a normalization of all

variables.This normalization is known as per unit analysis.

actual quantityquantity in per unit base value of quantity

1

Page 2: Per unit system

Per Unit System

2

(1) The per-unit values for transformer impedance, voltage and current are identical when referred to the primary and secondary (no need to reflect impedances from one side of the transformer to the other, the transformer is a single impedance).(2) The per-unit values for various components lie within a narrow range regardless of the equipment rating.(3) The per-unit values clearly represent the relative values of the circuit quantities. Many of the ubiquitous scaling constants are eliminated.(4) Ideal for computer simulations

Page 3: Per unit system

Per Unit Conversion Procedure, 11. Pick a 1 VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3. Calculate the impedance base, ZB= (VB)2/SB

4. Calculate the current base, IB = VB/ZB

5. Convert actual values to per unit

Note, per unit conversion affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts) 3

Page 4: Per unit system

Per Unit Solution Procedure

1. Convert to per unit (p.u.) (many problems are already in per unit)

2. Solve3. Convert back to actual as necessary

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Page 5: Per unit system

Per Unit ExampleSolve for the current, load voltage and load power in the circuit shown below using per unit analysis with an SB of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV, respectively.

Original Circuit5

Page 6: Per unit system

Per Unit Example, cont’d2 2

2 2

2 2

8 (kV) 0.64100MVA

80 (kV) 64100MVA

16 (kV) 2.56100MVA

LeftB

MiddleB

RightB

Z

Z

Z

Same circuit, withvalues expressedin per unit.

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Page 7: Per unit system

Per Unit Example, cont’d

2*

1.0 0 0.22 30.8 p.u. (not amps)3.91 2.3271.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

L

LL L L

G

Ij

V

VS V I

ZS

7

Page 8: Per unit system

Per Unit Example, cont’dTo convert back to actual values just multiply the per unit values by their per unit base

Actual

Actual

Actual

Middle

ActualMiddle

0.859 30.8 16 kV 13.7 30.8 kV

0.189 0 100 MVA 18.9 0 MVA

0.22 30.8 100 MVA 22.0 30.8 MVA100 MVA 1250 Amps

80 kV0.22 30.8 275 30.8

L

L

G

B

V

S

S

I

I

8

Page 9: Per unit system

Three Phase Per Unit

1. Pick a 3 VA base for the entire system, 2. Pick a voltage base for each different voltage

level, VB,LL. Voltages are line to line. 3. Calculate the impedance base

Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases

3BS

2 2 2, , ,3 1 1

( 3 )3

B LL B LN B LNB

B B B

V V VZ

S S S

Exactly the same impedance bases as with single phase using the corresponding single phase VA base and voltage base!9

Page 10: Per unit system

Three Phase Per Unit, cont'd

4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

33 3 3B B B

B LL B LN B LN

S S SI IV V V

Exactly the same current bases as with single phase!

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Page 11: Per unit system

Three Phase Per Unit Example•Solve for the current, load voltage and load power in the previous circuit, assuming:

–a 3 power base of 300 MVA, –and line to line voltage bases of 13.8 kV, 138 kV and 27.6 kV (square root of 3 larger than the 1 example voltages)–the generator is Y-connected so its line to line voltage is 13.8 kV.

Convert to per unitas before. Note the system is exactly the same!

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Page 12: Per unit system

3 Per Unit Example, cont'd

2*

1.0 0 0.22 30.8 p.u. (not amps)3.91 2.3271.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

L

LL L L

G

Ij

V

VS V I

ZS

Again, analysis is exactly the same!12

Page 13: Per unit system

3 Per Unit Example, cont'd

LActual

ActualLActualG

MiddleB

ActualMiddle

0.859 30.8 27.6 kV 23.8 30.8 kV

0.189 0 300 MVA 56.7 0 MVA

0.22 30.8 300 MVA 66.0 30.8 MVA300 MVA 125 (same cu0 Amps

3138 kV

0.22 30

rrent

8

!)

.

V

S

S

I

I

Amps 275 30.8

Differences appear when we convert back to actual values

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