L#8 Per unit calculations-2

# L#8 Per unit calculations-2 - Per Unit Calculations Masoud...

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Per Unit Calculations Masoud Fathizadeh, PhD, PE Department of Engineering Technology Purdue Calumet Hammond, Indiana 46323

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2 Per-Unit System In the per-unit system, the voltages, currents, powers, impedances, and other electrical quantities are expressed on a per-unit basis by the equation: Quantity per unit = Actual value Base value of quantity It is customary to select two base quantities to define a given per-unit system. The ones usually selected are voltage and power.
3 Per-Unit System Assume: Then compute base values for currents and impedances: rated b V V rated b S S b b b V S I b b b b b S V I V Z 2

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4 Per-Unit System And the per-unit system is: b actual u p V V V . . b actual u p I I I . . b actual u p S S S . . b actual u p Z Z Z . . % 100 % . . u p Z Z Percent of base Z
5 Example 1 An electrical lamp is rated 120 volts, 500 watts. Compute the per-unit and percent impedance of the lamp. Give the p.u. equivalent circuit. Solution: (1) Compute lamp resistance power factor = 1.0 8 . 28 500 ) 120 ( 2 2 2 P V R R V P 0 8 . 28 Z

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6 Example 1 (2) Select base quantities (3) Compute base impedance (4) The per-unit impedance is: VA S b 500 V V b 120 8 . 28 500 ) 120 ( 2 2 b b b S V Z . . 0 1 8 . 28 0 8 . 28 . . u p Z Z Z b u p
7 Example 1 (5) Percent impedance: (6) Per-unit equivalent circuit: % 100 % Z . . 0 1 u p Z . . 0 1 u p V S

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8 Example 2 An electrical lamp is rated 120 volts, 500 watts. If the voltage applied across the lamp is twice the rated value, compute the current that flows through the lamp. Use the per- unit method. Solution: V V b 120 . . 0 2 120 240 . . u p V V V b u p . . 0 1 . . u p Z u p
9 Example 2 The per-unit equivalent circuit is as follows: . . 0 1 u p Z . . 0 2 u p V S . . 0 2 0 1 0 2 . . . . . . u p Z V I u p u p u p A V S I b b b 167 . 4 120 500 A I I I b u p actual 0 334 . 8 167 . 4 0 2 . .

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L#8 Per unit calculations-2 - Per Unit Calculations Masoud...

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