Question #202465

The power transmitted between two shafts 3.5 metres apart by a cross belt drive round the two pulleys 600 mm and 300 mm in diameters, is 6 kW. The speed of the larger pulley (driver) is 220 r.p.m. The permissible load on the belt is 25 N/mm width of the belt which is 5 mm thick. The coefficient of friction between the smaller pulley surface and the belt is 0.35. Determine : 1. necessary length of the belt ; 2. width of the belt, and 3. necessary initial tension in the belt.


1
Expert's answer
2021-06-04T04:45:02-0400

L=π2(d1d2)+2𝑥+(d1d2)24𝑥=π2(0.60.3)+(2×3.5)+(0.60.3)24×3.5=8.472mL = \fracπ2 (d_1 − d_2) + 2𝑥 + \frac{(d1−d2)^2}{4𝑥}=\fracπ 2(0.6 − 0.3) +(2 × 3.5) + \frac{(0.6−0.3)^2}{4×3.5}= 8.472m

v=πd1N160=π×0.6×22060=6.911m/sv =\frac{ πd_1N_1}{ 60 }= \frac{π × 0.6 × 220}{ 60 }= 6.911m/s

α=sin1(02.6×325.3)=7.387°α = sin^{−1} (\frac{02.6×−3}{2*5.3}) = 7.387°

θ=(1802α)×π180=(180(2×7.387))×π180=3.399radθ = (180 − 2α) × \frac{π}{ 180} = (180 − (2 × 7.387)) × \frac{π}{ 180 }= 3.399 rad

T1T2=eμθ=e0.3×3.399=3.286.a\frac{T1}{ T2} = e^{μθ} = e^{0.3×3.399} = 3.286 … … . a

P=(T1T2)×v6000=(T1T2)×6.911P = (T_1 − T_2) × v → 6000 = (T_1 − T_2) × 6.911

T1T2=60006.911=868.181NT1=868.181+T2.b→ T_1 − T_2 =\frac{ 6000}{ 6.911 }= 868.181N → T_1 = 868.181 + T_2 … … . b

sub a in b

868.181+T2T2=3.286T2=379.781N\frac{868.181 + T_2}{ T_2} = 3.286 → T_2 = 379.781N

T1=868.181+379.781=1247.926NT_1 = 868.181 + 379.781 = 1247.926N

b=T1T=1247.92625=49.9mmb = \frac{ T1}{ T′ }= \frac{ 1247.926}{ 25 }= 49.9mm

To=T1+T22=1247.926+379.782=889.08Tₒ = \frac{ T1 + T2}{ 2} =\frac{1247.926 + 379.78}{ 2 }= 889.08


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