ENGR_400_Module_6_Assignment

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Embry-Riddle Aeronautical University *

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Electrical Engineering

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Jan 9, 2024

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Jason L. Wilson 11/25/2023 6.3 - Chapter Assignment ENGR 400 11.2 - A single-phase transformer is rated at 2kVA, 240/120V. The transformer is fully loaded by an inductive load of 0.8 power factor lagging. Compute the following: a. Real Power delivered to the load P = Scos ( θ ) P = 2 ( 0.8 ) P = 1.6 kW b. Load Current I L = S V I L = 2000 120 I L = 16.67 A 11.4 - Three single-phase transformers are connected in wye/delta configuration. Each single- phase transformer has an identical number of turns in the primary and secondary windings. If a line-to-line voltage of 480V is applied on the wye windings, compute the line-to-line voltage on the delta windings. V L − Y = 480 V V ph − Y = V L √ 3 = 480 √ 3 V ph − Y = V ph − Δ V L − Δ = V ph − Δ V L − Δ = 480 √ 3 ¿ 277.13 V 11.8 - A single-phase 10kVA, 2300/2230 V two-winding transformer is connected as an autotransformer to step up the voltage from 2300 to 2530 V. a. Draw the schematic diagram of the autotransformer showing the winding connections and all voltages and currents at full load. I 1 = 10 2300 I 1 = 4.378 A

Jason L. Wilson 11/25/2023 I 2 = 10 230 I 2 = 43.4783 A I all = I 1 + I 2 4.378 + 43.4783 I all = 47.8261 A b. Find the kVA rating of the autotransformer. Do not allow the currents of windings to exceed their rated values. S = 47.8261 ( 2300 ) S = 110 kVA 11.9 - A single-phase, 240/120 V transformer has the following parameters: R 1 = 1 Ω ; R 2 = 0.5 Ω ; X 1 = 6 Ω; X 2 = 2 Ω R 0 = 500 Ω; X 0 = 1.5 kΩ A load of 10 Ω at 0.8 power factor lagging is connected across the low-voltage terminals of the transformer. The voltage measured across the load side is 110V. Compute the following: a. Load voltage referred to the primary side N 1 N 2 = 240 120 Turnsratioequals , 2 V 2 ' = V 2 ( N 1 N 2 ) V 2 ' = 220 V b. Currents of the primary and secondary winding cosθ = 0.8

Jason L. Wilson 11/25/2023 θ = cos − 1 ( 0.8 ) θ = 36.87 ° Z L = 10 ∠ θ Z L = 10 ∠ 36.87 ° I 2 = V 2 R L I 2 = 110 10 ∠ 36.87 ° I 2 = 11 ∠ − 36.87 ° A Primary winding I 2 ' = I 2 N 1 N 2 I 2 ' = 11 ∠ − 36.87 ° 2 I 2 ' = 5.5 ∠ − 36.87 ° A SecondaryWinding c. Source voltage E 2 = V 2 + I 2 ( R 2 + j X 2 ) E 2 = 110 +( 11 ∠ − 36.87 )( 0.5 + j 2 ) E 2 = 110 +( 8.8 − j 6.6 )( o .5 + j 2 ) E 2 = 110 +( 17.6 + j 14.3 ) Secondary winding, E 2 = 128.4 ∠ 6.39 °V E 1 = E 2 ( N 1 N 2 ) Primary winding, E 1 = 256.8 ∠ 6.39 ° V I o = 256.8 ∠ 6.39 °V 500 ∨ ¿ ( j 1.5 k ) I o = 256.8 ∠ 6.39 °V 450 ∨ ¿ ( j 150 k )

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Jason L. Wilson 11/25/2023 I o = 256.8 ∠ 6.39 ° V 474.34 ∠ 18.43 ° I o = 0.54 ∠ − 12.04 ° I 1 = ( 0.54 ∠ − 12.04 ° ) +( 5.5 ∠ − 36.87 ° ) I 1 = 4.93 − j 3.413 I 1 = 6 ∠ − 34.7 ° A V 1 = ( 6 ∠ − 34.7 ° ) ( 1 + j 6 ) +( 256.8 ∠ 6.39 ° ) V 1 = 280.631 + j 54.263 V 1 = 285.92 ∠ 11.04 ° d. Voltage regulation VR = 285.92 − 220 220 ∗ 100 VR = 29.97% e. Load power S 2 = 110 ( 11 ∠ − 36.87 ° ) S 2 = 1.21 ∠ 36.87 ° f. Efficiency of the transformer P ¿ = 285.92 ( 6 ) cos ( 11.04 − ( − 34.7 ) ) P ¿ = 285.92 ( 6 )( 0.70 ) P ¿ = 1197.28 W P cu = 6 2 ( 1 ) +( 11 2 )( 0.5 ) P cu = 96.5 W P fe = 256.8 2 500 P fe = 131.9 W η = 1197.28 − 96.5 − 131.9 1197.28 ∗ 100

Jason L. Wilson 11/25/2023 η = 80.92% 11.11: The transformer in figure 11.6 consists of one primary winding and two secondary windings. The numbers of turns of the windings are: N 1 = 10,000 ; N 2 = 5,000 , ? N 3 = 1,000. A voltage source of 120 V is applied to the primary winding. The load of winding N 2 consumes 600 W and 300 VAr inductive power. The load of winding N 3 consumes 24 W and 36 VAr capacitive power. Compute the following: a. Voltages of the secondary windings E 2 = ( 120 ) ( 5000 10000 ) E 2 = 60 V E 3 = 120 ( 1000 10000 ) E 3 = 12 V b. Currents of all windings I 2 = 600 + j 300 60 I 2 = 11.18 ∠ 26.75 ° A I 3 = 24 − j 36 12 I 3 = 3.61 ∠ − 56.31 ° A I 1 = ( 11.18 ∠ 26.75 ° ) 5000 + ( 3.61 ∠ − 56.31 ° ) 1000 10000 I 1 = 5 + 2.5 j + 0.2 − 0.3 j 5.2 + j 2.2 I 1 = ¿