Angle of contact(θ) =165° = 165° X Π/180 = 2.88rad
Velocity of belt(V) = 1700m/min = 28.33m/sec
Width of belt(b) = 20cm
Thickness of belt(t) = 8mm 0.8cm
density of belt = 1gm/cm3
Maximum permissible stress(f) = 250 N/cm2
Coefficient of friction(µ) = 0.25 Maximum Power = ?
We know that, Tmax = σ.b.t
= 250 X 20 X 0.8 = 4000N ......(i)
Since Unit mass (m) = ρ.b.t.L
= 1/1000 X 20 X 0.8 X 100 = 1.6Kg .... (ii)
Since velocity(V) is given,
So we don't find the velocity using formula TC =1/3 Tmax = mV2, then Using the formula, TC = mV2, For finding TC.
= 1.6(28.33)2 = 1284 N ...(iii)
Using the formula, Tmax
= T1 + Tc , for finding T1
4000 = T1 + 1284
T1 = 2716N ...(iv)
Now, For T2, Using relation Ratio of belt tension = T1/T2 = eµθ
2716/T2 = e(0.25)(2.88) T2 = 1321 N ...(v)
Now Maximum Power Transmitted = (T1–T2).V/1000 KW
P = (2716 – 1321) X 28.33/1000 KW
P = 39.52KW.