Determine the effect on power transmission in the following cases:

i) Initial tension in the belt is increased by 8%

ii) Initial tension in the belt is decreased by 8%

iii) Angle of lap is increased by 8% by the use of an idler pulley, for the same speed and the tension on the tight side, and

iv) Coefficient of friction is increased by 8% by suitable dressing to the friction surface of the belt.

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: High

P= 2.5 kw, $\mu$ =0.3, $\theta$ =165$^{\circ}$, ν = 2.5 m/s

P=($T_{1}$ – $T_{2}$) ν

2500 = ($T_{1}$ – $T_{2}$) ×2.5 or $T_{1}$ – $T_{2}$ =1000N.

$\frac{T_{1}}{T_{2}}=e^{\mu\theta}=e^{0.3\times 165 \frac{\pi}{180}} $=2.37

or $T_{1}$ = 2.37$T_{2}$ - $T_{2}$ =1000 OR $T_{2}$ =729.9N.

$T_{2}$ =729.9 ×2.37 = 1729.9 N.

Initial Tension, $T_{0}$ = ($T_{1}$ + $T_{2}$)/2 =(1729.9+729.9)/2 = 1229.9N.

(i)When initial tension increase by 8%

$T'_{0}$ = 1229.9 × 1.08 = 1328.3N.

$T_{1}$ + $T_{2}$ = 2656.6

As $\mu$ and $\theta$ remain unchanged, $e^{\mu\theta}$ or $\frac{T_{1}}{T_{2}}$ is same,

2.37 $T_{2}$ + $T_{2}$ =2556.6

$T_{2}$ =788.3N, $T_{1}$ =1868.3N

P = ($T_{1}$ –$T_{2}$) ν = (1868.3-788.3)2.5=2700 W or 2.7 kw.

Increase power = (2.7-2.5)/2.5=0.08 or 8%

(ii) When initial tension is decreased by 8%

$T'_{0}$ =1229.9 × (1-0.08) =1131.5

or $T_{1}$ + $T_{2}$ = 2663

3.37$T_{2}$ = 2263

$T_{2}$ = 671.5N, $T_{1}$ = 1591.5N.

P =(1591.5-671.5)2.5=2300W or 2.3kw.

Decrease in power =(2.5-2.3)/2.5 = 0.08 or 8%

(iii) $\frac{T_{1}}{T_{2}}=e^{\mu\theta}$

$T_{1}$ is the same as before, whereas $\theta$ increases by 8%

1729.9/$T_{2}$ =e$^{0.3}$×((165×1.08$\pi$)/180) =2.54

$T_{2}$ =680.5N.

P =(1729.9-680.5)2.5 =2624W or 2.624kw

Increase in power = (2.624-2.5)/2.5 = 0.0496 or 4.96%

(iv) $\frac{T_{1}}{T_{2}}=e^{\mu\theta}=e^{0.3\times 1.08\times \frac{165\pi}{180}}$=2.54

Or $T_{1}$ = 2.54$T_{2}$

$T_{1}$ + $T_{2}$ = 1229.9 × 2 =2459.8

$T_{2}$ = 694.9 N.

$T_{1}$ =694.9 × 2.54 =1764.9 N.

P = (1764.9-694.9)2.5 = 2675W or 2.675kW

Increase in power = (2.675-2.5)/2.5 =0.07 or 7%