Problems on belt drives
7.B] A belt drive is used to transmit 20kw power from an electric motor to an exhaust fan. The
diameter of motor and fan pulley are 250mm and 1000mm respectively. The speed of
motor shaft is 750 rpm and thickness of belt is 6mm. Determine
i) Speed of the exhaust fan pulley
ii) Velocity of the belt
iii) Torque required to transmit the power
Answer:
(i) Speed of the exhaust fan pulley :
Formula used for velocity ratio is:
N1/N2 = d2+t/d1+t
where,
- d1= diameter of motor
- d2 =diameter of fan pulley
- N1=Speed of the motor in rpm
- N2 = Speed of the fan pulley in rpm
- t=thickness of belt
Therefore, N2 = N1/[d2+t/d1+t ]
N2= 750/[1000+6/250+6 ]
N2=191 rpm
ii) Velocity of the belt:
V = πd1N1 = πd2N2
V= π×250×10-3×750/60
V=9.81 m/s
iii) Torque required to transmit the power:
- Power transmitted by belt is given by:
P=(T2-T1)V,
where (T2-T1)= Tension difference
(T2-T1)=P/V
=20×103/9.8
(T2-T1)= 2040.8N
- Torque Required = T=(T2-T1) × r , where r = radius of motor (r=d/2)
T= 2040.8×250/2×10-3
T=255.1 Nm