Question

A belt embraces the shorter pulley by an angle of 165º and runs at a speed of 1700 m/min, Dimensions of the belt are Width = 20cm and thickness = 8mm. Its density is 1gm/cm³. Determine the maximum power that can be transmitted at the above speed, if the maximum permissible stress in the belt is not to exceed 250N/cm² and µ = 0.25.

Answer 1

Given date: 

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 T_C =1/3 Tmax = mV², then Using the formula, TC = mV², For finding T_C.

= 1.6(28.33)2 = 1284 N ...(iii) 

Using the formula, T_max 

= T₁ + T_c , for finding T₁

4000 = T₁ + 1284

T1 = 2716N ...(iv) 

Now, For T₂, Using relation Ratio of belt tension = T₁/T₂ = e^µθ 

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.

Comments

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