A belt drive consists of two V-belts in parallel, on grooved pulleys of the same size. The angle of the groove is 30^{\circ}. The cross-sectional area of each belt is 750 mm ^{2} and\mu .=0.12. The density of the belt material is 1.2 Mg / m ^{3} and the maximum safe stress in the material is 7 MPa. Calculate the power that can be transmitted between pulleys 300 mm diameter rotating at 1500 r.p.m. Find also the shaft speed in r.p.m. at which the power transmitted would be maximum.
Question : A belt drive consists of two V-belts in parallel, on grooved...
Given : 2 \beta=30^{\circ} \text { or } \beta=15^{\circ} ; \alpha=750 mm ^{2}=750 \times 10^{-6} m ^{2} ; \mu=0.12 ; \rho=1.2 Mg / m ^{3} =1200 kg / m ^{3} ; \sigma=7 MPa =7 \times 10^{6} N / m ^{2} ; d=300 mm =0.3 m ; N=1500 r.p.m.
Power transmitted
We know that velocity of the belt,
v=\frac{\pi d . N}{60}=\frac{\pi \times 0.3 \times 1500}{60}=23.56 m / sand mass of the belt per metre length,
m=\text { Area } \times \text { length } \times \text { density }=750 \times 10^{-6} \times 1 \times 1200=0.9 kg / m\therefore Centrifugal tension,
T_{ C }=m \cdot v^{2}=0.9(23.56)^{2}=500 NWe know that maximum tension in the belt,
T=\text { Maximum stress } \times \text { cross-sectional area of belt }=\sigma \times a=7 \times 10^{6} \times 750 \times 10^{-6}=5250 N
\therefore Tension in the tight side of the belt,
T_{1}=T-T_{ C }=5250-500=4750 NLet T_{2} = Tension in the slack side of the belt.
Since the pulleys are of the same size, therefore angle of contact, \theta=180^{\circ}=\pi rad.
We know that
2.3 \log \left(\frac{T_{1}}{T_{2}}\right)=\mu . \theta \operatorname{cosec} \beta=0.12 \times \pi \times \operatorname{cosec} 15^{\circ}=1.457
\log \left(\frac{T_{1}}{T_{2}}\right)=\frac{1.457}{2.3}=0.6334 \text { or } \frac{T_{1}}{T_{2}}=4.3 …(Taking antilog of 0.6334)
and T_{2}=\frac{T_{1}}{4.3}=\frac{4750}{4.3}=1105 N
We know that power transmitted,
P=\left(T_{1}-T_{2}\right) v \times 2 \ldots(\because \text { No. of belts }=2)
=(4750-1105) 23.56 \times 2=171752 W =171.752 kW
Shaft speed
Let N_{1} = Shaft speed in r.p.m., and
v_{1}= Belt speed in m/s.
We know that for maximum power, centrifugal tension,
T_{C}=T / 3 \text { or } m\left(v_{1}\right)^{2}=T / 3 \text { or } 0.9\left(v_{1}\right)^{2}=5250 / 3=1750
\therefore \left(v_{1}\right)^{2}=1750 / 0.9=1944.4 \text { or } v_{1}=44.1 m / s
We know that belt speed \left(v_{1}\right),
44.1=\frac{\pi d . N_{1}}{60}=\frac{\pi \times 0.3 \times N_{1}}{60}=0.0157 N _{1}
\therefore N_{1}=44.1 / 0.0157=2809 \text { r.p.m. }