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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. 4.4 65° 17.8 Solution a) Vector loop of OAB: R2 R3-R₁ = 0 4.4 y 3.8 500 kPa 17.8 ძვ 3.8 R 65 R₁ F₁ =PxS 500 kPa x For the slider-crank mechanism: firstly, determine the input angle and link length Angle: • 02 180°-65° = 115° (input angle) Link length ⚫ a = 4.4cm, Then, calculate the output angle b = 17.8 cm, c = 0. 031 =sin-1 (asin 2 12.946° (rejected) 032 =sin¹ a sin ₂ += 167.05° (Accepted, open mechanism) The position of the piston w.r.t. O da cose₂-b cos 03 15.488 cm b) Velocity analysis Input velocity, W₂ = -800×2 60 -83.776 rad/s W3 = a cose₂ b cose3 W2 -8.98 rad/s • d = -aw₂sine₂+ bw3sin03 = 298.267 cm/s Acceleration analysis Input acceleration, a2 = 0 (constant speed) • α3 a a₂ cos ₂-a w² sin ₂+bw sin631594.8 rad/s² b cos 03 • ä= -aα₂ sin 02-aw cos 02 + bα3 sin 03 + bw cos 03 = 18102 cm/s² c) Energy Method FK VK+ Tk. Wk = Σ m Ik ak Wk (eq. 1) k=2 The force due to pressure is: d² 0.038² F = (Pressure) x (Surface) = P₁ ×π 500 x 103 xπ = 566.7 N 4 and F-566.7, = 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: T12= 1935.74 83.776 = 23.1 N.m 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. 4.4 65° 17.8 Solution a) Vector loop of OAB: R2 R3-R₁ = 0 4.4 y 3.8 500 kPa 17.8 ძვ 3.8 R 65 R₁ F₁ =PxS 500 kPa x For the slider-crank mechanism: firstly, determine the input angle and link length Angle: • 02 180°-65° = 115° (input angle) Link length ⚫ a = 4.4cm, Then, calculate the output angle b = 17.8 cm, c = 0. 031 =sin-1 (asin 2 12.946° (rejected) 032 =sin¹ a sin ₂ += 167.05° (Accepted, open mechanism) The position of the piston w.r.t. O da cose₂-b cos 03 15.488 cm b) Velocity analysis Input velocity, W₂ = -800×2 60 -83.776 rad/s W3 = a cose₂ b cose3 W2 -8.98 rad/s • d = -aw₂sine₂+ bw3sin03 = 298.267 cm/s Acceleration analysis Input acceleration, a2 = 0 (constant speed) • α3 a a₂ cos ₂-a w² sin ₂+bw sin631594.8 rad/s² b cos 03 • ä= -aα₂ sin 02-aw cos 02 + bα3 sin 03 + bw cos 03 = 18102 cm/s² c) Energy Method FK VK+ Tk. Wk = Σ m Ik ak Wk (eq. 1) k=2 The force due to pressure is: d² 0.038² F = (Pressure) x (Surface) = P₁ ×π 500 x 103 xπ = 566.7 N 4 and F-566.7, = 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: T12= 1935.74 83.776 = 23.1 N.m 3

Elements Of Electromagnetics
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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Note: The question is solved, but I want a comprehensive explanation of how to use the energy method and where we got these numbers from... I want a complete explanation of the part C energy method, please quickly.
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. 4.4 65° 17.8 Solution a) Vector loop of OAB: R2 R3-R₁ = 0 4.4 y 3.8 500 kPa 17.8 ძვ 3.8 R 65 R₁ F₁ =PxS 500 kPa x For the slider-crank mechanism: firstly, determine the input angle and link length Angle: • 02 180°-65° = 115° (input angle) Link length
Transcribed Image Text: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. 4.4 65° 17.8 Solution a) Vector loop of OAB: R2 R3-R₁ = 0 4.4 y 3.8 500 kPa 17.8 ძვ 3.8 R 65 R₁ F₁ =PxS 500 kPa x For the slider-crank mechanism: firstly, determine the input angle and link length Angle: • 02 180°-65° = 115° (input angle) Link length
⚫ a = 4.4cm, Then, calculate the output angle b = 17.8 cm, c = 0. 031 =sin-1 (asin 2 12.946° (rejected) 032 =sin¹ a sin ₂ += 167.05° (Accepted, open mechanism) The position of the piston w.r.t. O da cose₂-b cos 03 15.488 cm b) Velocity analysis Input velocity, W₂ = -800×2 60 -83.776 rad/s W3 = a cose₂ b cose3 W2 -8.98 rad/s • d = -aw₂sine₂+ bw3sin03 = 298.267 cm/s Acceleration analysis Input acceleration, a2 = 0 (constant speed) • α3 a a₂ cos ₂-a w² sin ₂+bw sin631594.8 rad/s² b cos 03 • ä= -aα₂ sin 02-aw cos 02 + bα3 sin 03 + bw cos 03 = 18102 cm/s² c) Energy Method FK VK+ Tk. Wk = Σ m Ik ak Wk (eq. 1) k=2 The force due to pressure is: d² 0.038² F = (Pressure) x (Surface) = P₁ ×π 500 x 103 xπ = 566.7 N 4 and F-566.7, = 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: T12= 1935.74 83.776 = 23.1 N.m 3
Transcribed Image Text:⚫ a = 4.4cm, Then, calculate the output angle b = 17.8 cm, c = 0. 031 =sin-1 (asin 2 12.946° (rejected) 032 =sin¹ a sin ₂ += 167.05° (Accepted, open mechanism) The position of the piston w.r.t. O da cose₂-b cos 03 15.488 cm b) Velocity analysis Input velocity, W₂ = -800×2 60 -83.776 rad/s W3 = a cose₂ b cose3 W2 -8.98 rad/s • d = -aw₂sine₂+ bw3sin03 = 298.267 cm/s Acceleration analysis Input acceleration, a2 = 0 (constant speed) • α3 a a₂ cos ₂-a w² sin ₂+bw sin631594.8 rad/s² b cos 03 • ä= -aα₂ sin 02-aw cos 02 + bα3 sin 03 + bw cos 03 = 18102 cm/s² c) Energy Method FK VK+ Tk. Wk = Σ m Ik ak Wk (eq. 1) k=2 The force due to pressure is: d² 0.038² F = (Pressure) x (Surface) = P₁ ×π 500 x 103 xπ = 566.7 N 4 and F-566.7, = 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: T12= 1935.74 83.776 = 23.1 N.m 3
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