Open belt drive, Mechanical Engineering

Open belt drive:

An open belt drive connects the two pulleys 120cm and 50cm diameter on parallel shafts which are apart 4m. The maximum tension in belt is 1855.3N. Coefficient of the friction is 0.3. The driver pulley having diameter 120cm runs at 200rpm.
(i) The power transmitted
(ii) Torque on each shafts.

Sol: Given data:

D1  = Diameter of the driver = 120cm = 1.2m

R1  = Radius of the driver = 0.6m

N1  = Speed of the driver in R.P.M. = 200RPM

D2  = Diameter of the driven or Follower = 50cm = 0.5m

R2  = Radius of the driven or Follower = 0.25m

X = Distance between the centers of two pulleys = 4m

µ = Coefficient of friction = 0.3

T1  = Tension in the tight side of the belt = 1855.3N Calculation for power transmitting:


P = The maximum power transmitted by belt drive

= (T1-T2).V/1000 KW                                                                                                       ...(i)


T2  = Tension in slack side of belt

V = Velocity of the belt in m/sec.

= pDN/60 m/sec, D is in meter and N is in Rotation per minute                                                ...(ii)

For T2,

We use relation Ratio of belt tension = T1/T2  = eµθ                                                                                              ...(iii)

But angle of contact is not given,


θ = Angle of contact and, θ = Angle of lap

for the open belt, Angle of contact (θ) = P - 2a                                                                               ...(iv)

Sina = (r1  - r2)/X = (0.6 - 0.25)/4

a = 5.02°                                                                                                                             ...(v)

By using the equation (iii),

θ = P - 2a = 180 - 2 X 5.02 = 169.96°

= 169.96° X P/180 = 2.97 rad                                                                                ...(iv)

Now by using the relation (iii)

1855.3/T2 = e(0.3)(2.967)

T2 = 761.8N                                                                                                                        ...(vii)

For finding velocity, using equation (ii)

V = (3.14 X 1.2 X 200)/60 = 12.56 m/sec                                                              ...(viii)

For finding Power, using the equation (i)

P = (1855.3 - 761.8) X 12.56

P = 13.73 KW                                                            .......ANS

We know that,

1. Torque exerted on driving pulley = (T1  - T2).R1

= (1855.3 - 761.8) X 0.6

= 656.1Nm                                                               .......ANS

2. Torque exerted on driven pulley   = (T1  - T2).R2

= (1855.3 - 761.8) X 0.25

= 273.4.1Nm                                                            .......ANS

Posted Date: 10/18/2012 8:22:37 AM | Location : United States

Related Discussions:- Open belt drive, Assignment Help, Ask Question on Open belt drive, Get Answer, Expert's Help, Open belt drive Discussions

Write discussion on Open belt drive
Your posts are moderated
Related Questions
calculate the vertical reaction at A in Newtons

Q. Safety and Loss Prevention? Are there special venting requirements for flammables and toxic materials? Air conditioning, heating and ventilation: • are non-sparking fa

Evaluate elongation of bar of varying cross section: How you evaluate elongation of bar of varying cross section:? Sol.: Take bar made up of different lengths and which



Find out change in length of circular bar of uniform taper Sol.: The stress at any cross section can be found by dividing load by area of cross section and extension can be


Illustarte the following (a) Acceptance sampling (b) OC-Curve

Q. Process of Corrosion in BFW Tank? The feed to the BFW Tank is hot and will be flashing when it enters the tank. The vapours will be corrosive when the tank is open to atmosp