Given that
D1 = Diameter of the driver = 45cm = 0.45m
R1 = Radius of the driver = 0.225m
D2 = Diameter of the driven = 20cm = 0.2m
R2 = Radius of the driven = 0.1m
X = Distance between the centers of two pulleys = 1.95m
T1 = Maximum permissible tension = 1000N
µ = Coefficient of friction = 0.20
N1 = Speed of the driver(Larger pulley) in R.P.M. = 100RPM
Since we know that,
Power Transmitted = (T1–T2).V/1000 Kw ...(i)
Tension is in KN and V is in m/sec
First ve find the velocity of the belt,
V = Velocity of the belt in m/sec.
Here we take diameter and RPM of larger pulley
= πDN/60 m/sec, D is in meter and N is in RPM
= (3.14 X0.45 X 100) /60
= 2.36m/sec.......(ii)
Now Ratio of belt tension, T1/T2 = eµ.θ ...(iii)
Here we don't know the value of θ, For θ, first find the value of α, by the formula,
Angle of Lap for cross belt α = sin–1(r1 + r2)/X
= sin–1(0.225 + 0.1)/1.95
= 9.59° ...(iv)
Now Angle of contact (θ) = Π + 2α ----- for cross belt
θ = Π + 2 X 9.59°
= 199.19°
= 199.19°(Π/180°) = 3.47rad ...(v)
Now putting all the value in equation
(iii) We get 1000/T2 =e(0.2)(3.47) T2 = 498.9 N ...(vi)
Using equation (i), we get
P = [(1000 – 498.9) X 2.36 ]/1000
P = 1.18KW.