written 7.1 years ago by
teamques10
★ 69k
|
•
modified 6.9 years ago
|
P= 7.5 kw =7500 w; d =1.2 m; N =250rpm. ;θ =165∘ =165 ×(π/180) =2.88 rad; μ =0.3 ; σ =1.5 MPa = 1.5×106 N/m2 ; ρ = 1 Mg//m3 =1 ×10/6 g//m3 =1000 kg//m3 ;t = 10 mm =0.01 m.
Let b =width of the belt in meters,
T1 =Tension in the tight side of the belt in N, and
T2 =Tension in the slack side of the belt in N.
We know that velocity of the belt,
ν=πdN60=π×1.2×25060 = 15.71 m/s.
and power transmitted (P),
7500 =(T_{1} –T_{2})ν = (T_{1} –T_{2}) 15.71
T_{1} –T_{2} = 7500/15.71 = 477.4 N. …(i)
We know that
2.3log(\frac {T_{1}}{T_{2}}) =\mu\theta =0.3×2.88 =0.864
log(\frac {T_{1}}{T_{2}})=0.864/2.3
or \frac {T_{1}}{T_{2}} =2.375 ….(ii) …(Taking antilog of 0.3756)
From equation (i) and (ii) T_{1} =824.6 N T_{2} =347.2 N.
We know that mass of the belt per meter length,
m = area × length × density =b.t.l.p
= b×0.01×1×1000 =10b kg
Centrifugal tension,
T_{C} = mν^{2} =10b(15.71)^{2} =2468b N
And maximum tension in the belt,
T =\sigmab.t =1.5×10^{6} × b ×0.01 =15000b N
We know that
T =T_{1} +T_{C} or 15000b =824.6 +2468b
15000b -2468b =824.6 or 12532b =824.6
b =824.6/12532 =0.0658 m = 65.8 mm.