past year exam 2.5

Question

3.1. a workpiece with a diameter of 750mm is machined in a lathe. The spindle speed of the lathe is 25r/min and the efficiency of the drive is 95%. Take the cutting-pressure as 800N/mm^2, the cutting depth of the cutting tool as 2,5mm and the feed of the cutting tool 0,5mm per evolution. Calculate the ffg: 3.1.1. the output power at the cutting tool 3.1.2. the input power from the motor 3.2. a grinding wheel having a diameter of 200mm rotates at 3150r/min and exerts a tangential force of 45N on the workpiece. If the motor runs at 1500r/min and the efficiency of the machine is 77%, calculate the ffg: 3.2.1. the output power at the grinding wheel in kW 3.2.2. the input power of the electric motor in kW 3.2.3. the torque on the shaft of the motor in Nm 3.3. a shaft 250mmm in diameter is revolving in a bearing and exerting a force of 75kN on the bearing. If the coefficient of friction between the bearing and the shaft is 0,02 and the shaft rotates at 350r/min, calculate the ffg: 3.3.1. the frictional force 3.3.2. the friction torque 3.3.3. the power lost due to friction

Answer 1

3.1.2 )

Work piece diameter D = 750 mm

D = 750/1000

D = 0.75 m

Spindle speed of the lathe N = 25  rpm

The cutting depth of the cutting tool DOC = 2.5 mm

DOC = 2.5/1000 = 0.0025 m

DOC = D - d = 0.0025 m

D =  Diameter of Work piece before cutting

d =  Diameter of Work piece After cutting

D - d = 0.0025

d = 0.75 - 0.0025

d = 0.7475 m

Area of cutting A = π(D²-d²)

A = π(0.75²-0.7475²)

A = 0.00374 m²

Feed of the cutting tool f = 0.5 mm per revolution

f = 0.5/1000 = 0.0005 m per revolution

The cutting-pressure P = 800 N/mm²

P = 800×10⁶ N/m²

Machine efficiency η = 95%

η = 0.95

The output power at the cutting tool P_o = ?

Cutting force F = PA

Substitute A = 0.00374 and P = 800×10⁶

F = (0.00374)(800×10⁶)

F = 2.992×10⁶ N

D_avg = (D+d)/2

D_avg = (0.75+0.7475)/2

D_avg = 0.74875 m

Torque T = ½FD_avg

T = ½(2.992×10⁶)(0.74875)

T = 1.12×10⁶ N-m

The output power at the cutting tool P_o = 2πNT/60

P_o = 2π(25)(1.12×10⁶ )/60

P_o = 2.93215×10⁶  W

P_o = 2932.15 kW

The input power from the motor P_i = P_o / η

P_i = 2932.15×10³/0.95

P_i = 3086.47 kW

The input power from the motor is 3086.47 kW

Answer 2

3.1.1 )

Work piece diameter D = 750 mm

D = 750/1000

D = 0.75 m

Spindle speed of the lathe N = 25  rpm

The cutting depth of the cutting tool DOC = 2.5 mm

DOC = 2.5/1000 = 0.0025 m

DOC = D - d = 0.0025 m

D =  Diameter of Work piece before cutting

d =  Diameter of Work piece After cutting

D - d = 0.0025

d = 0.75 - 0.0025

d = 0.7475 m

Area of cutting A = π(D²-d²)

A = π(0.75²-0.7475²)

A = 0.00374 m²

Feed of the cutting tool f = 0.5 mm per revolution

f = 0.5/1000 = 0.0005 m per revolution

The cutting-pressure P = 800 N/mm²

P = 800×10⁶ N/m²

Machine efficiency η = 95%

η = 0.95

The output power at the cutting tool P_o = ?

Cutting force F = PA

Substitute A = 0.00374 and P = 800×10⁶

F = (0.00374)(800×10⁶)

F = 2.992×10⁶ N

D_avg = (D+d)/2

D_avg = (0.75+0.7475)/2

D_avg = 0.74875 m

Torque T = ½FD_avg

T = ½(2.992×10⁶)(0.74875)

T = 1.12×10⁶ N-m

The output power at the cutting tool P_o = 2πNT/60

P_o = 2π(25)(1.12×10⁶ )/60

P_o = 2.93215×10⁶  W

P_o = 2932.15 kW

The output power at the cutting tool is 2932.15 kW

 

Answer 3

3.2.1 )

Grinding wheel diameter d_r = 200 mm

d_r = 200/1000 = 0.2 m

r = d/2 = 0.2/2 = 0.1 m

Grinding wheel Speed N_r = 3150 rpm

tangential force F = 45 N

Motor speed of N = 1500 rpm

machine efficiency η = 77%

η = 0.77

The output at the grinding wheel P_g =

Torque of mechine T = rF

Substitute r = 0.1 and F = 45 N

T = (0.1)(45)

T = 4.5 N-m.

Power P = 2πNT/60

Substitute T = 4.5 and N_r = 3150

P = 2π(3150)(4.5) / 60

P = 1484.4 W

P = 1.484 kW

The output power at the grinding wheel P_g is 1.484 kW.

Answer 4

3.2.2 )

Grinding wheel diameter d_g = 200 mm

d_g = 200/1000 = 0.2 m

Grinding wheel Speed N_g = 3150 rpm

Tangential force F = 45 N

Motor speed of N = 1500 rpm

Machine efficiency η = 77%

η = 0.77

The input power of the motor P_i = ?

d_g N_g = d N

d = d_g N_g/N

d = 0.2 * 3150 / 1500

d = 0.42

r = d/2 = 0.42/2 = 0.21

Torque of mechine T = rF

Substitute r = 0.21 and F = 45 N

T = (0.21)(45)

T = 9.45 N-m.

Power output P = 2πNT/60

Substitute T = T = 9.45 and N = 1500

P = 2π(1500)(9.45) / 60

P = 1484.4 W

But P_i = P/η

Substitute η = 0.85 and P = 1484.4

P_i = 1484.4/0.77

P_i =  1927.8 W

P_i =  1.9278 kW

The input power of the electric motor is 1.9278 kW

Answer 5

3.2.3 )

Grinding wheel diameter d_g = 200 mm

d_g = 200/1000 = 0.2 m

r = d_g/2 = 0.2/2 = 0.1 m

Grinding wheel Speed N_g = 3150 rpm

Tangential force F = 45 N

Motor speed of N = 1500 rpm

Machine efficiency η = 77%

η = 0.77

The torque on the shaft of the motor T = ?

Torque of mechine T = rF

Substitute r = 0.1 and F = 45 N

T = (0.1)(45)

T = 4.5 N-m.

The torque on the shaft of the motor is 4.5 N-m

Answer 6

3.3)

The diameter of the shaft is 250 mm .

The radius of the shaft is [ r₂ ]  125 mm .

Amount of force exerted by shaft on bearing is (F) 75 kN .

 The coefficient of friction between the shaft and the bearing is ( μ) 0.02 .

The shaft rotates at the speed of 350 r/min .

The shaft rotates at the speed of 350 r/min = ( 350/60 ) = 5.8333 revolutions per second . 

The axial load W is the amount of force exerted by shaft on bearing  .

So the axial load W is 75 kN .

3.3.1)

The frictional force can be calculated using T = force * radius .

Where T is the friction torque  .

So friction force F = T/r 

Now calculate for friction torque :

The frictional torque T is given by 

T = (2/3)μW [ r₂³ / r₂² ]

T = (2/3)μW [ r₂ ] 

T = (2/3)(0.02) (75 * 1000 ) [ 0.00125 ]

 T = (0.667)(0.02) (75) [ 125 ]

T = (0.0133) (9375)

T = 125 .

So the frictional torque is 125 Nm .

Friction force F = T/r  .

F = 125/0.00125 .

F = 1000 N .

Therefore, the friction force is 1000 N .

 

Answer 7

3.3.2)

The diameter of the shaft is 250 mm .

The radius of the shaft is [ r₂ ]  125 mm .

Amount of force exerted by shaft on bearing is (F) 75 kN .

 The coefficient of friction between the shaft and the bearing is ( μ) 0.02 .

The shaft rotates at the speed of 350 r/min .

The shaft rotates at the speed of 350 r/min = ( 350/60 ) = 5.8333 revolutions per second . 

The axial load W is the amount of force exerted by shaft on bearing  .

So the axial load W is 75 kN .

Now calculate for friction torque :

The frictional torque T is given by 

T = (2/3)μW [ r₂³ / r₂² ]

T = (2/3)μW [ r₂ ] 

T = (2/3)(0.02) (75 * 1000 ) [ 0.00125 ]

 T = (0.667)(0.02) (75) [ 125 ]

T = (0.0133) (9375)

T = 125 .

So the frictional torque is 125 Nm .

 

Answer 8

3.3.3)

The diameter of the shaft is 250 mm .

The radius of the shaft is [ r₂ ]  125 mm .

Amount of force exerted by shaft on bearing is (F) 75 kN .

 The coefficient of friction between the shaft and the bearing is ( μ) 0.02 .

The shaft rotates at the speed of 350 r/min .

The shaft rotates at the speed of 350 r/min = ( 350/60 ) = 5.8333 revolutions per second . 

The axial load W is the amount of force exerted by shaft on bearing  .

So the axial load W is 75 kN .

The frictional torque T is given by 

T = (2/3)μW [ r₂³ / r₂² ]

T = (2/3)μW [ r₂ ] 

T = (2/3)(0.02) (75 * 1000 ) [ 0.00125 ]

 T = (0.667)(0.02) (75) [ 125 ]

T = (0.0133) (9375)

T = 125 .

So the frictional torque is 125 Nm .

Thus the power loss is p = Tω .

p = Tω 

p = T * 2π ( revolutions per second ) 

p = 125 * 2π ( 5.8333 )

p = 125 * (36.6519 )

p = 4581.489 W .

So thus  the power loss is 4581.489 W .