Open belt drive:
An open belt drive connects the two pulleys 120cm and 50cm diameter on parallel shafts which are apart 4m. The maximum tension in belt is 1855.3N. Coefficient of the friction is 0.3. The driver pulley having diameter 120cm runs at 200rpm.
Calculate
(i) The power transmitted
(ii) Torque on each shafts.
Sol: Given data:
D_{1} = Diameter of the driver = 120cm = 1.2m
R_{1} = Radius of the driver = 0.6m
N_{1} = Speed of the driver in R.P.M. = 200RPM
D_{2 } = Diameter of the driven or Follower = 50cm = 0.5m
R_{2 } = Radius of the driven or Follower = 0.25m
X = Distance between the centers of two pulleys = 4m
µ = Coefficient of friction = 0.3
T_{1 } = Tension in the tight side of the belt = 1855.3N Calculation for power transmitting:
Let
P = The maximum power transmitted by belt drive
= (T_{1}-T_{2}).V/1000 KW ...(i)
Here,
T_{2} = Tension in slack side of belt
V = Velocity of the belt in m/sec.
= pDN/60 m/sec, D is in meter and N is in Rotation per minute ...(ii)
For T_{2},
We use relation Ratio of belt tension = T_{1}/T_{2} = e^{µθ} ...(iii)
But angle of contact is not given,
let
θ = Angle of contact and, θ = Angle of lap
for the open belt, Angle of contact (θ) = P - 2a ...(iv)
Sina = (r_{1} - r_{2})/X = (0.6 - 0.25)/4
a = 5.02° ...(v)
By using the equation (iii),
θ = P - 2a = 180 - 2 X 5.02 = 169.96°
= 169.96° X P/180 = 2.97 rad ...(iv)
Now by using the relation (iii)
1855.3/T_{2} = e(0.3)(2.967)
T_{2} = 761.8N ...(vii)
For finding velocity, using equation (ii)
V = (3.14 X 1.2 X 200)/60 = 12.56 m/sec ...(viii)
For finding Power, using the equation (i)
P = (1855.3 - 761.8) X 12.56
P = 13.73 KW .......ANS
We know that,
1. Torque exerted on driving pulley = (T_{1} - T_{2}).R_{1}
= (1855.3 - 761.8) X 0.6
= 656.1Nm .......ANS
2. Torque exerted on driven pulley = (T_{1} - T_{2}).R_{2}
= (1855.3 - 761.8) X 0.25
= 273.4.1Nm .......ANS