c) Energy Method n k=2 n Fk. Vk + Tk. Wk k=2 The force due to pressure is: n = k=2 mk.ak Vk + n k=2 Ik ak Wk (eq. 1) d² 0.038² = (Pressure) x (Surface) = P₁ ×π- = 500 × 103 × π = 566.7 N 4 and F-566.7, V = d = 2.9867, ap = 181.02 m/s² Then, eq.1 =>-566.77 x 2.9867 + T12 × W₂ = mp xap xvp • • Power required: T12 x W₂ = 0.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt Torque required: 1935.74 T12 = = 23.1 N.m 83.776 3 Problem 1 The compressor mechanism shown in the following figure is driven clockwise by a DC electric motor at a constant rate of 800 rpm. In the position shown, the pressure is 500 kPa. a) Find the position of the piston. b) Find the velocity and acceleration of the piston. c) Using Energy Method, determine the torque required from the motor to operate the compressor. The piston has a mass mp = 0.45 kg. Neglect the mass of other links. All dimensions are in cm. 17.8 65 4.4 Solution a) Vector loop of OAB: R₂ R3-R₁₁ = 0 4.4 3.8 500 kPa 17.8 03. 3.8 R3 65 R₁ F=PxS 500 kPa For the slider-crank mechanism: firstly, determine the input angle and link length Angle: • 02 = 180° 65° 115° (input angle) Link length

c) Energy Method n k=2 n Fk. Vk + Tk. Wk k=2 The force due to pressure is: n = k=2 mk.ak Vk + n k=2 Ik ak Wk (eq. 1) d² 0.038² = (Pressure) x (Surface) = P₁ ×π- = 500 × 103 × π = 566.7 N 4 and F-566.7, V = d = 2.9867, ap = 181.02 m/s² Then, eq.1 =>-566.77 x 2.9867 + T12 × W₂ = mp xap xvp • • Power required: T12 x W₂ = 0.45 x 181.02 x 2.98267 + 566.77 x 2.9867 = 1935.74 Watt Torque required: 1935.74 T12 = = 23.1 N.m 83.776 3 Problem 1 The compressor mechanism shown in the following figure is driven clockwise by a DC electric motor at a constant rate of 800 rpm. In the position shown, the pressure is 500 kPa. a) Find the position of the piston. b) Find the velocity and acceleration of the piston. c) Using Energy Method, determine the torque required from the motor to operate the compressor. The piston has a mass mp = 0.45 kg. Neglect the mass of other links. All dimensions are in cm. 17.8 65 4.4 Solution a) Vector loop of OAB: R₂ R3-R₁₁ = 0 4.4 3.8 500 kPa 17.8 03. 3.8 R3 65 R₁ F=PxS 500 kPa For the slider-crank mechanism: firstly, determine the input angle and link length Angle: • 02 = 180° 65° 115° (input angle) Link length

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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