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
how to make a vapour absorber?

How are the refrigerants classified? Describe the desirable property of a good refrigerant. Explain in brief the thermodynamics, safe and physical working property of an ideal r

Determine the normal stresses: A short hollow pier 1.6 m × 1.6 m outsides as well as 1.0 m × 1.0 m intersides supports a vertical load of 2000 kN at a point located on a diago

Advantages of Abrasive Jet Cutting Machines 1. Ability to cut brittle or heat sensitive material without damage. 2. Capability to cut intricate holes in material of any hard

Types of Brake The brakes used in motorcycles can be classified as follows: (a) Drum Brake (b) Disc Brake

Question: Characterisation tests have been performed on a four pole, 415 V (line to line), three phase, star connected, 50Hz induction motor, with the following results:

how to calculate flow rate of liquid nitrogen? melting point(-210deg) boiling point(-195.8deg) critical temp(-149.9deg) relative density(0.8) specific gravity(808.5kg/m^3) rel

A "perfect" square (shown in dashed lines in the sketch below) with initial dimensions of 0.5 x 0.5 inch is drawn on the surface of a thin aluminum structure (E = 10x10 6 psi, v =

Flux Cored Arc Welding (FCAW) This is a variation of GMAW where solid wire is replaced by flux cored wire. The equipments and accessories are the same. Generally, flux cored wi