Formula for maximum power transmitted by belt:
Derive formula for maximum power transmitted by belt when centrifugal tension is considered.
Sol: Let T_{1 } = Tension on tight side T_{2} = Tension on slack side v = Linear velocity of belt
Then the power transmitted can be given by the equation
Power transmitted will be maximum if d(P)/dv = 0
Thus differentiating equation with respect to V and equating to zero for maximum power, we get
Equation (iv) gives velocity of belt at which maximum power is transmitted.
From equation (iv) T_{max} = 3T_{c } ...(v)
Hence when power transmitted is maximum, the centrifugal tension would be around 1/3^{rd }of the maximum tension.
We know that T_{max} = T_{1 } + T_{c}
Hence condition for transmission of maximum power is:
T_{c }= 1/3 T_{max}, and
T_{1} = 2/3T_{max} ...(viii)
NOTE: The net driving tension in the belt = (T_{1} - T_{2})