- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 1 (17.74 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 2 (14.31 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 3 (16.51 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 4 (13.82 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 5 (16.48 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 6 (23.06 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 7 (27.95 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 8 (27.95 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 9 (36.2 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 10 (28.18 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 11 (28.18 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 12 (45.34 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 13 (20.1 KiB) Viewed 37 times
- The Customers Fin M200 Fen Foot Ke Spad 5 Then W Q2 Force And Power For Steady Speed Up The Acceleration Is Zero 14 (12.55 KiB) Viewed 37 times
Q2. Force and power for steady speed up: The acceleration is zero, so from (1) we find T = 2000 N and Fn = 4000 N What poner is required from the motor [1 mark] The power required from the motoris Number w Hint Penalty Activity 4 The customer CR 00 View Hent
Q3. Force and power required for decceleration: You need to rewitations (1) and (2) with change of sign for the section The tension in the cabin Nur N [1 mark] The force required from the motor is Number N [1 mark] The power required from the motor amb w [1 mark) Conclusions The greatest power is required from the motor when conting per te mest power is regured when slowing down Event steady spend the motor has to do work to counter the contraty we re ignoring triction in all of these caniculators in reality we would also need to as a safety factor, so the motor is not it hull capacity of the time Hint Ponaty Act 4 the customers DO
Q4. Imagine we now go down a floor with the same acceleration and speed charatonistics What is the net power suched through the motor The netpower supplied through the motor for the Whole DEGOS LUD and down) Number w [1 mark] We nood a system for delivery of power to the motor which allows power to flow both ways to the motor and from the motor This one of the topictwo cover in the second your course Electrical Energy Systems and Contro Ponaty Activity the customers 4 0 Ve Hint
Q5. Motor Seleobon The cable is wound onto a pulley wheel 05 min radius Calculate the magnitude of the angular speed of the puley wheet w for the stat steady speed between floors [1 mark] The angular speed of the pulley wheel is Number rad s Hint Penalty Activity. The customer lifts 05 00 View Hint
06. Gear and Motor solection For a 50 Hz supply frequency, the motor speed is 50 revolutions per second What gearbox is required to stop this down to 2 rad.-1 at the pulley wheel? What is the maximum torque to be developed by the motor Round off your answers to one decimal place The gear ratio required is Number [1 mark] The maximum torque to be developed by the motor is Number Nm [1 mark] Hint Penalty Activity 4 The customer is 06 0.0 View Hint
The customer lifts Say each it comes up to 600 kg (that's the car and a few people). The iftis operated by a motor and a counterweight, as shown in Figure 1. We want to calculate the force the DC motor must provide to power the litt socording to our chent's specifications. We want to specify the power rating of the motor FEN m = 200 kg FEN 2009 m = 600 kg 600g Figure 1 Force diagram for the lift system, adapted from 1 T is the tension in the cables gia the acceleration due to gravity which wew take as 10 ms? Fis the force exerted by the motor - Here's the specification Total lift mass is 600 kg, counterweight is 200 kg • Acceleration of 2 ms -2 over 0.5 s to a speed of 1 ms -1 Steady speed between the floors of 1 ms -1 for 45 • Decolorato at 2 ms over 0.5 8 to a stop
Speed 15 1 1 2 3 4 5 5 Time (s) Figure 2: Speed-time characteristic of the motor to raise the lit by one floor We need to work out the requirements for the motor over these 5 seconds, to move up one floor. Q1. Force and power required for acceleration upwards: We can write two simultaneous equations, one for the lift going up, one for the counterweight going down. When the lift is going up, its acceleration a is in the positive direction (upwards). The counterweight is accelerating downwards at the same rate, it's acceleration is negative because of the direction downwards. For the counterweight I - 2009 = -200 a 1) For the lift T + F - 600 g = 600 a (2) Solving, we find T = 1600 N and hence F = 5600 N. How much power do we need? Enter your answer as an integer. [1 mark] w The power required from the motor is Number
Say each lift cames up to 600 kg (that's the car and a few people) The itt is operated by a motor and a courterweight as shown in Figure 1 We want to calculate the force F the DC motor must provide to power the it according to our chant's specifications We want to specify the power rating of the motor TA F&N m = 200 kg FRN 200g m = 600 K 600g Figure 1. Force diagram for the litt system, adapted from u T is the tension in the cables, gis the acceleration due to gravity which we'll take as 10 msFa the force exerted by the motor Here's the specification • Total lift mass is 600 kg, counterweight is 200 kg • Accoleration of 2 me -2 over 0.5 8 to a speed of 1 ms Steady speed between the doors of 1 ms -1 for 48 Decelerate at 2 ms over 0.5 to a stop
Speed (ms) 1 1 2 3 4 4 5 Time (s) Se 2 Speed-two characteristic of the motor to bylo We need to work out thereurements for the motor over these 5 seconds to move on Boot Q1. Forco and power required for acceleration word We can write to smutaneous out on one for the litt going up, one for the counter weight going when the Congo, its acceleration is in the positive direction (upwards). The counterweight is cool down words the same rate its acceleration is nogave cause of the direction dowards For the counterweight T- 2009 200 For the all T + P 600 - 600 3) Song, we find T-1600 N and hence R5600 M How much power do we needy Enter your worn intogor [1 mark]
Q3. Force and power required for decceleration: You need to rewrite equations (1) and (2) with change of sign for the acceleration The tonsion in the cable is Nimber N [1 mark] The force required from the motor is Number N [1 mark) The power required from the motor is Number w [1 mark] Conclusions: The greatest power is required from the motor when accelerating upwards, the smallest power is required when slowing down Even at steady speed the motor has to do work, to counter the acceleration due to gravity We're ignoring friction in all of these calculations in reality we would also need to add a safety factor, so the motor is not at capacity al the time
04. Imagine we now go down a floor with the same acceleration and speed charaleristics What is the net power supplied through the motor? The net power supplied through the motor for the whole process (up and down) Is Number w [1 mark]