A constant torque 4 kW motor drives a riveting machine. A flywheel of mass 130 kg and radius of gyration 0.5 m is fitted to the riveting machine. Each riveting operation takes 1 second and requires 9000 N-m of energy. If the speed of the flywheel is 420 r.p.m. before riveting, find: 1. the fall in speed of the flywheel after riveting; and 2. the number of rivets fitted per hour
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Given A constant torque 4 kW motor drives a riveting machine. A flywheel of mass 130 kg and radius of gyration 0.5 m is fitted to the riveting machine. Each riveting operation takes 1 second and requires 9000 N-m of energy. If the speed of the flywheel is 420 r.p.m. before riveting, find: 1. the fall in speed of the flywheel after riveting; and 2. the number of rivets fitted per hour
- Given P = 4 kW, m = 130 kg, k = 0.5 m,N1 = 420 r.p.m, ω1 = 2π x 420 / 60 = 44 rad/s
- Let ω2 be the angular speed of the flywheel immediately after riveting.
- So energy supplied by the motor will be
- E2 = 4 kW = 4000 W = 4000 N-m / s (since 1 W = 1 N-m/s)
- Since the riveting operation takes 1 second energy will be
- E1 = 9000 N-m
- For riveting operation energy supplied will be
- So ΔE = E1 – E2
- = 9000 – 4000
- = 5000 N-m
- Now maximum fluctuation energy
- ΔE = ½ x m x k^2 (ω1^2 – ω2^2)
- 5000 = ½ x 130 x (0.5)^2 [ (44)^2 – (ω2)^2]
- 5000 / 16.25 = (1936 – ω2^2)
- So ω2^2 = 1936 – 307.69
- = 1628.31
- Now speed in r.p.m will be
- N2 = 40.35 x 60 / 2π
- = 385.5 r.p.m
- Now number of rivets fitted per hour:
- Since energy of each riveting operation that takes 1 second is 9000 N-m, the number of rivets fitted per hour will be
- E2/ E1 x 3600
- 4000 / 9000 x 3600
- = 1600 rivets.
Reference link will be
https://brainly.in/question/9851745
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Answer:
1600 rivets will be the correct answer
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