past year exam 2.2

Question

3.1. The tangential cutting force exerted on the bar held in the lathe chuck is 2400N. The bar is 50mm in diameter and a torque of 12Nm is required to overcome friction of the lathe spindle in the bearings. Calculate the following: 3.1.1. Torque required at the driving pulley on the machine 3.1.2. Driving force required on the machine pulley if it has an effective diameter of 240mm 3.1.3. the minimum power required at the machine pulley if it has to revolve at 200r/min 3.2. a lathe is being driven by a motor providing a maximum input of 2,5kW at 1500r/min. At maximum power the machine efficiency is 85%. The minimum velocities of the lathe spindle are 3500r/min and 30r/min respectively. Determine the torque at maximum power: 3.2.1. at the driving shaft of the motor 3.2.2. at the driving spindle of the lathe at maximum speed 3.2.3. at the driving spindle of the lathe minimum speed

Answer 1

3.1.1 )

Bar diameter d_b = 50 mm

d_r = 50/1000 = 0.05 m

r = d/2 = 0.05/2 = 0.025 m

Torque required to overcome friction of the lathe spindle T_f = 12 N-m

Tangential force F = 2400 N

Torque T = rF

T = (0.025)(2400)

T = 60 N-m

Torque required at the driving pulley on the machine is 60 N-m

Answer 2

3.1.2 )

Bar diameter d_b = 50 mm

d_b = 50/1000 = 0.05 m

r = d/2 = 0.05/2 = 0.025 m

Torque required to overcome friction of the lathe spindle T_f = 12 N-m

Tangential force F = 2400 N

Effective diameter d_e = 240 mm

d_e = 240/1000 = 0.24 m

r_e= 0.24/2 = 0.12 m

Driving force required on the machine pulley if it has an effective diameter of 240mm is F_e = ?

By maintaining torque constant

Torque T = rF = r_eF_e

Substitute r_e = 0.12 , F = 2400 , r = 0.025

(0.025)(2400) = (0.12)F_e

F_e = 60/0.12

F_e = 500 N

Driving force required on the machine pulley if it has an effective diameter of 240mm is " 500 N "

Answer 3

3.1.3)

Bar diameter d_b = 50 mm

d_b = 50/1000 = 0.05 m

r = d/2 = 0.05/2 = 0.025 m

Speed N = 200

Torque of is required to overcome friction T_friction = 12 N-m

The minimum power required at the machine pulley for 200 rpm P_min = ?

Tangential force F = 2400 N

Torque T = rF

T = (0.025)(2400)

T = 60 N-m

P = 2πNT/60

Substitute N = 200 and T = 60 N

P = 2π(200)(60)/60

P = 1256.637 W

1 HP  = 746.27 W

So P_unit = 746.27 W

Power due to friction P_friction = 2πNT_friction/60

Substitute N = 200 and T_friction=12

P_friction = 2π(200)(12)/60

P_friction = 251.33 W

Gross power P_gross = P + P_friction

P_gross = 1256.637 + 251.33

P_gross = 1507.96 W

Minimum Power P_min = P_unit + P_gross

P_min = 746.27 + 1507.96

P_min = 2254.234 W

The minimum power required at the machine pulley for 200 rpm is 2254.234 W

Answer 4

3.2)

Maximum input P = 2.5 kW

Speed N = 1500 rpm

At maximum power the machine efficiency η = 85 %

Maximum Speed N_max = 3500 rpm

Minimum Speed N_min= 30 rpm

The torque at maximum power T = ?

P = 2πNT/60

From above relation Power is directly proportional to speed

At maximum power , Speed also maximum.

So P_max = 2πN_maxT/60

Substitute N_max = 3500 , P = 2.5×10³

2.5×10³ = 2π×3500T/60

T = (2500×60)/(2π×3500)

T = 6.82 N-m

The torque at maximum power is 6.82 N-m

 

Answer 5

3.2.1)

Maximum input P = 2.5 kW

Speed N = 1500 rpm

At maximum power the machine efficiency η = 85 %

Maximum Speed N_max = 3500 rpm

Minimum Speed N_min= 30 rpm

The torque at the driving shaft of the motor T = ?

P = 2πNT/60

Substitute N = 1500 , P = 2.5×10³

2.5×10³ = 2π×1500T/60

T = (2500×60)/(2π×1500)

T = 15.9 N-m

The torque at the driving shaft of the motor is 15.9 N-m

Answer 6

3.2.2)

Maximum input P_max = 2.5 kW

Speed N = 1500 rpm

At maximum power the machine efficiency η = 85 %

Maximum Speed N_max = 3500 rpm

Minimum Speed N_min= 30 rpm

The torque at the driving spindle of the lathe at maximum speed T = ?

P = 2πNT/60

From above relation Power is directly proportional to speed

At maximum power , Speed also maximum.

We need to take maximum power into consideration.

At spindle power losses included.So 85 % of total power is remained.

Power at spindle P_s = ηP_max

P_s = 0.85×2.5×10³

P_s = 2125 W

So P_s = 2πN_maxT/60

Substitute N_max = 3500 , P_s = 2125

2125 = 2π×3500T/60

T = (2500×60)/(2π×1500)

T = 5.8 N-m

The torque at the driving spindle of the lathe at maximum speed is 5.8 N-m

Answer 7

3.2.3)

Maximum input P_max = 2.5 kW

Speed N = 1500 rpm

At maximum power the machine efficiency η = 85 %

Maximum Speed N_max = 3500 rpm

Minimum Speed N_min= 30 rpm

The torque at the driving spindle of the lathe at minimum speed T = ?

P = 2πNT/60

At spindle power losses included.So 85 % of total power is remained.

Power at spindle P_s = ηP_max

P_s = 0.85×2.5×10³

P_s = 2125 W

But at minimum speed , torque is maximum.So T = T_max

P_s = 2πN_minT_max/60

T_max = P_s×60/(2πN_min)

Substitute N_min = 30 , P_s = 2125

T_max = 2125×60/(2π×30)

T_max = 795.775 N-m

The torque at the driving spindle of the lathe at minimum speed is 795.775 N-m