Diameter of pulley=400 mm, Diameter of smaller pulley=250 mm, Distance between two pulleys=2 m, Coefficient of friction between smaller pulley surface and belt = 0.4, Maximum tension when the belt is on the point slipping=1200 N.

Find the power transmitted at speed of 10 m/s. it is desired to increase the power. Which of the following two methods you will select?

1. Increasing the initial tension in the belt by 10 percent.
2. Increasing the coefficient of friction between the smaller pulley surface and belt by 10 percent by the application of suitable dressing on the belt.

Find also, the percentage increasing in power possible in each case.

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: Medium

Given :- $d_{1}$ =400 mm =0.4m; $d_{2}$ =250 mm= 0.25 m; x= 2 m; $\mu$= 0.4;

T=1200N; ν =10m/s.

Power transmitted

We know that for an open belt drive,

$\sin\alpha=\frac{(r_{1}-r_{2})}{x}=\frac{(d_{1}-d_{2})}{2x}$ =(0.4+0.25)/(2×2) = 0.0375 or $\alpha$ = 2.150

Angle of contact, $\theta$ = 180$^{\circ}$ -2$\alpha$= 1800-(2×2.150) = 175.70

=175.7×($\pi$/180) =3.067 rad.

Let $T_{1}$ = Tension in the tight side of the belt, and

$T_{2}$ = Tension in the slack side of the belt

Neglecting centrifugal tension,

$T_{1}$ = T=1200N ….(Given)

We know that

$2.3log\frac{(T_{1}}{T_{2})}=\mu\theta$= 0.4×3.067 =1.2268

$log\frac{(T_{1}}{T_{2})}$ = 1.2268/2.3 = 0.5334 or T1/T2 = 3.41 …(taking antilog of 0.5334)

and $T_{2}$ = $T_{1}$/3.41 =1200/3.41 = 352N.

We know that power transmitted

P= ($T_{1}$ – $T_{2}$)ν = (1200 - 352)10 = 8480W =8.48kw.

Power transmitted when initial tension is increased by 10%

We know that initial tension,

$T_{0}$ = ($T_{1} +T_{2})$ = (1200+352)/2 = 776N.

Increased initial tension,

$T'_{0}$ =776 + (776×10)/100 =853.6N

Let $T_{1}$ and $T_{2}$ be the corresponding tensions in the tight side and slack side of the belt respectively.

$T'_{0}$ = ($T_{1}$ + $T_{2}$)/2

= $T_{1}$ +$T_{2}$ = 2$T'_{0}$ =2× 853.6 =1707.2N ….(i)

OR

Since the ratio of tension is constant, therefore

($T_{1}$/$T_{2}$) =3.41 ….(ii)

$T_{1}$=1320.2N ; and $T_{2}$ = 387N

Power transmitted,

P = ($T_{1}$ –$T_{2}$)ν = (1320.2-387)10 =9332 W =9.332 kW.

Power transmitted when coefficient of friction is increased by 10%

When know that coefficient of friction,

$\mu$=0.4

Increased coefficient of friction.

$\mu '$= 0.4+0.4×(10/100) =0.44

Let $T_{1}$ and $T_{2}$ be the corresponding tension in the tight side and slack side respectively.

We know that

$2.3log\frac{(T_{1}}{T_{2})}=\mu '\theta$=0.44×3.067 =1.3495

$log\frac{(T_{1}}{T_{2})}$ = (1.3495/2.3) =0.5867 or $\frac{(T_{1}}{T_{2})}$ =3.86 …(iii) …..(Taking antilog of 0.5867)

Here the initial tension is constant, i.e.

$T_{0}$ = ($T_{1}$ +$T_{2}$)/2 or $T_{1}$ + $T_{2}$ =2$T_{0}$ =2×776=1552N …(iv)

From equation (iii) and (iv),

T =1232.7N and $T_{2}$ = 319.3N

Power transmitted,

P =( $T_{2}$ -$T_{2}$)ν =(1232.7-319.3)10 =9134W =9.134kW.

Since the power transmitted by increasing the initial tension is more, therefore in order to increase the power transmitted we shall adopt the method of increasing the initial tension.

Percentage increase in power:

We know that percentage increase in power when the initial tension is increased

= (9.332-8.48)/8.48 × 100 =10.05%

And percentage increase in power when coefficient of friction is increased

= (9.134-8.48)/8.48 × 100 =7.7%