past year exam 1

Question

An open flat-belt drive is used to transmit 25kW. The driving pulley has a diameter of 200mm and rotates at 200 r/min, while the driven pulley rotates at 266,667 r/min. the contact angle on the driven pulley is 176 degrees and the centre distance between the pulley shafts is 1,5m assume the coefficient of friction of 0,37. Determine the following: 2.1. belt speed 2.2. Ratio of belt tension 2.3. tension in the slack side of the belt 2.4. tension in the tight side of the belt 2.5. length of belt for the drive 2.6. length of the belt if the drive is changed to a crossed-belt drive 3. A conveyor belt drive placed at an inclination angle of 15 degrees has a length of 120m between the loading point and discharge point. The belt speed is 2,5m/s and is subjected to a friction force of 8,5kN. Assume the contact angle on the driving pulley to be 220 degrees. The coefficient of friction is 0,25 and the efficiency of the drive is 85%. The maximum tension in the belt is 27,49 kN. Calculate the quantity of rock transported in per tonne. 4. A conveyor belt drive, placed at an inclination angle of 12 degrees, has a capacity of 300 tonnes of rock per hour and length of 120mm between the resistance is 8kN. Assume that the contact angle on the driving pulley is 240 degrees and the coefficient of friction is 0.3. Determine the following: 4.1. the power equired from the driving motor 4.2. the tension in the slack side of the belt 4.3. the tension in the tight side of the belt. 5. Calculate the quantity of rock, in tones per hour, which can be transported by a belt conveyor with the following particulars: Maximum tension in the belt=35,5kN Contact angle of the belt on the driving pulley=185 degrees Frictional force=2,5kN Belt speed= 95m/min Delivery height=35m Coefficient of friction=0,3 6. The ffg are the particulars of an open flat-belt drive between a motor and a machine: Belt speed=900m/min Centre distance between shafts=4,5m Speed of motor=1400r/min Contact angle on machine pulley=178degrees Thickness of belt=5mm Maximum allowable tension=160N per cm width of belt Coefficient of friction=0,3 Calculate the following: 6.1. the diameter of pulleys 6.2. the width of belt in mm 6.3. the length of belt in meter

Answer 1

(2.1)

The Power transmitted by the drive = 25 KW.

The Diameter of the driving pulley d1= 200 mm

Belt Speed

Therefore the Belt Speed = 2.09 m/sec.

Answer 2

(2.2)

Contact angle a = 176°

a = 176 * 0.0174532925 = 3.0712 radians

Coefficient of Friction μ = 0.37.

Ratio of the tension on tight and slack side

Therefore the ratio of tension is 3.1153.

Answer 3

(2.3)

Contact angle a = 176°

a = 176 * 0.0174532925 = 3.0712 radians

Coefficient of Friction μ = 0.37.

The Power transmitted by the drive = 25 KW = 25000 N-m/sec.

The Diameter of the driving pulley d1= 200 mm

Belt Speed .

The force = power / velocity

Force = (25000) / 2.09

Force = 11,961.7 N

Tension on slack side 

Therefore tension on the slack side = 5654.29 N. 

Answer 4

(2.4)

Contact angle a = 176°

a = 176 * 0.0174532925 = 3.0712 radians

Coefficient of Friction μ = 0.37.

The Power transmitted by the drive = 25 KW = 25000 N-m/sec.

The Diameter of the driving pulley d1= 200 mm

Belt Speed .

The force = power / velocity

Force = (25000) / 2.09

Force = 11,961.7 N

Tension on tight side 

Therefore tension on the tight side = 17,616.39 N. 

Answer 5

(2.5)

The Diameter of the driving pulley d1= 200 mm = 200*10^-3 m.

The speed of the driving pulley = 200 r/min

The speed of the driven pulley = 266.667 r/min

Centre distance between pulley shafts C= 1.5 m.

When there is no slip between the belt and pulley then

Therefore the diameter of the driven pulley d2= 150 mm. 

d2 = 150 * 10^-3 m

Length of the open flat-belt drive 

Therefore the length of the open flat-belt drive is 3.55m.

Answer 6

(2.6)

The Diameter of the driving pulley d1= 200 mm = 200*10^-3 m.

The speed of the driving pulley = 200 r/min

The speed of the driven pulley = 266.667 r/min

Centre distance between pulley shafts C= 1.5 m.

When there is no slip between the belt and pulley then

Therefore the diameter of the driven pulley d2= 150 mm. 

d2 = 150 * 10^-3 m

Length of the cross-belt drive 

Therefore the length of the cross-belt drive is 3.57m.

Answer 7

3)

Contact Angle 'a' is 220° = 220 * (π/180) = 3.8397 radians

The relation between tension and contact angle is 

Where T1 is Maximum tension in the belt = 27.49 kN .

         a = contact angle =3.8397 radians

           µ is Coefficient of Friction = 0.25

Therefore the Tension in slight side is 10.526 kN

 

Effective Driving Force Fe = T1 - T2

Fe = 27.49 - 10.526

Fe = 16.964 kN

 

Therefore Total Force Acting in lifting the load is

F  = Effective Driving force - Frictional Force

F = 16.964 - 8.5 = 8.464 kN

 

For inclined planes the resultant force is 

F = mg sin ϴ + μ mg cos ϴ

Where ϴ is angle of inclination is = 15°

8.464 = m(g sin ϴ + μ mg cos ϴ)

8.464 = m(9.8 sin 15 + (0.25) (9.8) cos 15)

8.464 = m(2.536 + 2.366)

8464 = m(4.9025)

m = (8464)/(4.9025)

m = 1726 

The quantity of rock transported per tonne is 1726/1000 = 1.726 kg per ton.