ENGR_400_Module_7_Assignment

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Embry-Riddle Aeronautical University *

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400

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Mechanical Engineering

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Jan 9, 2024

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docx

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Jason L. Wilson 12/1/2023 7.3 - Chapter Assignment ENGR 400 Problem 12.10 The speed of a 12-pole, 480 V, 60 Hz three-phase induction motor at full load is 560 rpm. The motor has the following parameters: R₁ = 0.1Ω; R = 0.5 Ω; X + X = 5 Ω a. Compute the developed torque. η s = 120 ( 60 ) 12 η s = 600 rpm S = 600 560 600 S = 0.067 T = 3 ( 60 ) 2 π ( 600 ) ( ( 480 3 ) 2 ( 0.1 + 0.5 0.067 ) 2 + 5 2 ) ( 0.5 0.067 ) T = 0.048 ( 934.42 )( 7.46 ) T = 334.60 N M b. While the torque is unchanged, the voltage is changed to reduce the speed of the motor to 520 rpm. Compute the new voltage. S = 600 520 600 S = 0.133 334.60 = 3 ( 60 ) 2 π ( 600 ) ( ( V L 3 ) 2 ( 0.1 + 0.5 0.133 ) 2 + 5 2 ) ( 0.5 0.133 ) 334.60 = ( 0.179 ) ( ( V L 3 ) 2 39.89 )
Jason L. Wilson 12/1/2023 ( V L 3 ) 2 = 334.60 ( 39.89 ) 0.179 ( V L 3 ) 2 = 74565.32 V L = 74565.32 V L = 273.07 3 V L = 472.97 V Problem 12.16 A wind turbine has 50 m long blades. The far stream wind speed is 15 m/s and the coefficient of performance Cp is 30% at the operational pitch angle. The induction generator is six-pole, 60 Hz machine and is rotating at 1260 rpm. The mechanical efficiency of the turbine is 85% and the electrical efficiency of the generator is 90%. Compute the following: a. Power captured by the blade P = 1 2 ( 1.23 )( π 50 2 )( 15 3 )( 0.3 ) P = 4890576.189 W P = 4.89 MW b. Slip of the generator S = ( 120 ( 60 ) 6 ) 1260 120 ( 60 ) 6 S =− 0.05 c. Power delivered to the stator P s = 0.85 ( 4.89 ) P s = 4.15 MW d. Output electric power of the wind turbine P out = 4.15 ( 0.9 ) P out = 3.74 MW e. Rotor Copper Loss P g = 3.74 1 0.05
Jason L. Wilson 12/1/2023 P g = 3.94 P cu ( R ) = 0.05 ( 3.94 ) P cu ( R ) = 0.197 MW f. Stator Losses P cu ( S ) = 4.15 3.94 P cu ( S ) = 0.21 MW g. Total efficiency of the system η = 3.74 4.89 100 η = 76.5% Problem 12.19 A synchronous generator is connected directly to an infinite bus. The voltage of the infinite bus is 15 kV. The excitation of the generator is adjusted until the equivalent field voltage Ef is 14 kV. The synchronous reactance of the machine is 5 Ω. Compute the following: a. Pullover power (maximum power) P max = ( 14 15 ) sin ( 90 ) 5 P max = 42 MW b. Equivalent excitation voltage that increases the pullover power by 20% P = 1.2 ( 42 ) P = 50.4 MW E f = 50.4 ( 5 ) 15sin ( 90 ) E f = 16.8 kV Problem 12.25 An industrial load is connected across a three-phase, Y-connected source of 4.5 KV (line-to-line). The real power of the load is 160 kW, and its power factor is 0.8 lagging: a. Compute the reactive power of the load. S = 160 ( 0.8 ) S = 200 k θ = cos 1 ( 0.8 ) θ = 36.86 ° sin ( θ ) = 0.60
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