Question

A 20kW electric motor, which is running at 43 rev/s, is used to drive a crusher that operates 24 hours per day. An engineer has specified 212 mm and 315 mm pitch diameters for the small pulley and large pulley respectively; and C 3454 V-belt type.

a) What should be the conservative service factor for this belt and pulley system?

b) Calculate the center distance between the pulleys?

c) Calculate allowable power, design power, and required number of belts. (Take design factor nd as 1.)

d) Calculate centrifugal tension, forces at tension side and loose side, initial tension, and factor of safety. (Take friction coefficient including wedge effect as 0.5123.)

Answer 1

To solve this problem i am using MITSUBOSHI DESIGN MANUAL -

a)

We are having a crusher and electric motor with power transmission = 20 KW

From table service correctiion factor table -

@Working hour 24 hours/day

Service correction factor Ko = 1.5

conservative service factor K = Ko + Ki + Ke

Where Ki be the idler correction factor = 0 (Because idler not present)

Ke be the envireonment correction factor = 0 (Environment type not mentioned)

Therefore,

K = Ko + Ki + Ke = 1.5

b)

As motor directy mounted to the small pulley so having speed of 43 rps

C 3454 V-belt type- Wrapped type V-Belt.

Effective length = 3454 mm

Center diatance between the pulley -

Where

Lp = Belt pitch length

Dp & dp be the large and small pulley pitch diameter.

Dp = 315 mm

dp = 212 mm

Therefore,

Center diatance between the pulley -

c)

Design power -

Pd = Pt x K

Where Pt = Power transmission

K = Service factor =1.5

Pt = 20 KW

Therefore

Pd = 20 x 1.5 = 30 KW

Actual power rating = P

P = (Ps + Pa)Kc

Kc = Kt x Kl

Where

Kc = Power rating correction factor

Kt = Arc of contact correction factor

(Dp - dp)/C = 315 - 212/1312.085 =

Contact angle on small pulley = 180

Therefore at above value Kt = 1 (DESIGN MANUAL - MITSOBOSHI)

Kl = Belt length corrction factor

Ps = Basic power rating

Pa = Additional power rating

For Belt pitch length 3456 - Kl = 0.98 (For C 3454 V-belt type)

Therefore,

Kc = Kt x Kl = 1 x 0.98

Kc = 0.98

Now

Actual power rating = P

P = (Ps + Pa)Kc

Now from power rating - (USE POWER RATING TABLE DESIGN MANUAL)

Small pulley speed = 43 rev/s = 2580 rpm

And small pulley pitch dia = 212 mm

Basic power rating = Ps = 17.67 KW

Additional power rating = Pa =

At Small pulley speed = 43 rev/s = 2580 rpm

And speed ratio = 315/212 = 1.48

Pa = 0.95 KW

Thereforre,

Actual power rating = P

P = (Ps + Pa)Kc = (17.67 + 0.95) x 0.98 = 18.24 KW

Next,

Number of belt = n

Actual power rating = 18.24 KW

Design Power = Pd = 30 KW

No. of belt required is 2.

d)

Calculate centrifugal tension, and factor of safety. (Take friction coefficient including wedge effect as 0.5123.) -

Belt speed = V

Small pulley speed = 43 rev/s = 2580 rpm

Tight side tension =

As belt weight is not given so ignore or neglect it.

Pd = design power

n = Number of belt

v = speed

Kt = Arc of contact correction factor

(Dp - dp)/C = 315 - 212/1312.085 =

Contact angle on small pulley = 180

Therefore at above value Kt = 1 (DESIGN MANUAL - MITSOBOSHI)

Slack side tension =

As belt weight is not given so ignore or neglect it.

Initial tension =

Pt = Power transmission = 20 KW

Mass per unit length of belt is not given so rest parameters can't be evaluated.

Centrifugal tension = Tc

I have mentioned the values as well as table from where i have taken the values.

So if you found any values that is differ from your data book please ask in comment section.

and If you have any doubt in any parameter or in any values or in concept too. Please ask in comment section.

Thank you.