EXAMPLE
45 kW are transmitted by 2 m long wedge belts which operate at full capacity. The diameter of the motor pulley is 180 mm, and it rotates at 1440 rpm; the output pulley is 400 mm diameter.
Determine the effect of belt number on life if the belts are (a) SPA; (b) SPB.
Given : P = 45 kW ; n1 = 24 Hz ; so, from ( 1) v = πD1 n1 = π (0.18)(24) = 13.6 m/s
Geometry : D1 = 180 mm ; D2 = 400 mm ; L = 2000 mm so, from ( 2a) ( 2b) and ( 2c) :-
C = 533 mm ; γ = 0.208 rad ; θ1 = θmin = 2.726 rad
Since the belts operate at full capacity, slip is imminent and ( 5a) applies. Assume f = 0.512, so
kθ = 1 - e - fθmin = 1 - e- 0.512 ∗ 2.726 = 0.752
and, in ( 5a) P/kθ v = 45 ∗ 103 / 0.752 ∗13.6 = 4400 N
Properties of SPA from Table 1 ; F = 6235 N ; M = 87.48 Nm ; ρ = 0.1287 kg/m ; m = 13.0
and, in ( 5a) ρv2 = 0.1287 ∗ 13.62 = 23.8 N
( P/kθvz + M/D1 + ρv2 )/F = ( 4400/z +87.48/0.180 +23.8 )/ 6235 = 0.7057/z + 0.0818
( P/kθvz + M/D2 + ρv2 )/F = ( 4400/z +87.48/0.400+23.8 )/ 6235 = 0.7057/z + 0.0389
So, from ( 5a) with z = 6 belts for example :-
L/vT = ( 0.7057/6 + 0.0818 )13.0 + ( 0.7057/6 + 0.0389 )13.0 = 8.23 E-10
∴ T = L / 8.23 E-10 v = 2.0 / 8.23 E-10 ∗ 13.6 = 1.79 E8 s = 50 k.hours (kh)
Repeating these latter steps with different numbers of belts, and then repeating all calculations in a similar manner for SPB belts, illustrates the marked effect of number of belts upon the resulting belt life, thus :
Number of belts | 1 | 2 | 3 | 4 | 5 | 6 | 8 | 10 | 12 | ||
SPA life ( kh ) | 6.1E-7 | 1.6E-3 | 0.11 | 1.6 | 11 | 50 | 400 | 1700 | 4800 | [ 26kh = | |
SPB life ( kh ) | 3.0E-5 | 0.035 | 1.1 | 9.5 | 40 | 120 | 500 | 1300 | 2600 | 10 years ] |
The trends exemplified here - in particular the marked effect of belt number on life - should be borne in mind when selecting a belt drive without the aid of software.