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
Track Renewal: Need for Track Renewal Normal maintenance is able to keep up the track standards only to a limited extent. The track gets overaged or the speeds and tr

How would you say environmental studies are multidisciplinary nature?

Find out the mechanical advantage of the machines: A lifting machine may lift 800 N by the application of 100 N. Distance moved by the effort is 100 cm. At the same time load

some suitable examples

Concept of continuum - thermodynamics: Sol: Even the simplification of matter into molecules, atoms, electrons, etc. is too complex a picture for many problems of thermodyna

Basic elements of flashbutt welding A flash butt welding machine is mainly comprised of a Main frame Stationary Platen Moving Platen Clamping mechanism Transf

Q.   Derive the expression of torque developed in closely excited magnetic system. Clearly explain then assumption made.   Sol. Double - Excited System   A doubly - ex

Explain the Cost analysis of radial drilling machine Cost regarding to Manual arrangement of auger extension in paver machine: 1.cost of segmented angle radial drill machin

comparing the different coding systems ?

Component s of Steam Power Plant: There are four components of steam power plant: 1. The boiler: Hot-source reservoir in which the combustion gases raise steam. 2. Engi