Module 6. Power transmission
Lesson 28

28.1 Problem

Design a V belt drive for the following:

Drive: AC motor, operating speed is 1440 rpm and operates for over 10 hours. The equipment driven is a compressor, which runs at 900 rpm and the power transmission is 20kw


Since it is a V belt drive, let us consider belt speed, v = 25 m/sec.

Design power

Pdes = service factor(Csev) * required power (P) = 1.3 * 20kW = 26kW

Fig. 28.1

The value 1.3 is selected from design data book for the given service condition. Now,

Hence, obvious choice for belt section is C


V= π*ds*n / 60 X 100


n=Rotational speed RPM

ds = Diameter of small pulley

dL = Diameter of large pulley

V = Velocity of belt

standard sizes are,

dS=315 mm and dL=530 mm

dS=355 mm and dL = 560 mm.

First combination gives the speed ratio to be 1.68

Second combination gives the speed ratio to be 1.58.

So, it is better to choose the second combination because it is very near to the given speed ratio. Therefore, selected pulley diameters are dS=355 mm and dL= 560 mm.

Center distance, C should be such that, dL < C < 3(dL + dS )

Let us consider, C = 1500 mm, this value satisfies the above condition.

Considering an open belt drive, the belt length,

Inside length of belt = 4444 – 56 = 4388 mm

The nearest value of belt length for C-section is 4394 mm (from design data book)

Therefore, the belt designation is C: 4394/173

Power rating (kW) of one C-section belt

Equivalent small pulley diameter is,

dES =CSR ds = 355 X 1.12 = 398 mm

CSR = 1.12 is obtained from the hand book

For the belt speed of 23 m/sec, the given power rating (KW) =12.1 KW

For the obtained belt length, the length correction factor CV1=1.04

Determination of angle of wrap

For the angle of wrap of 3.00 radian (smaller pulley), the angle of wrap factor, CVW is found to 0.98 for a C section belt.

Therefore, incorporating the correction factors,

2 numbers of C 4394\173 belts are required for the transmission of 20kW.

Last modified: Thursday, 27 September 2012, 10:17 AM