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A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 9 ft. The angular speed of the small pulley is 1750 rev/min as it delivers 2 hp. The service is such that a service factor Ks of 1.25 is appropriate. (a) Find Fc, Fi, (F1)a, and F2, assuming operation at the maximum tension limit. (b) Find Ha, nfs, and belt length.

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Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

[tex]H_{nom}=2hp=1491.4W[/tex]

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

[tex]w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m[/tex]

[tex]V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s[/tex]

[tex]F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N[/tex]

[tex](F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N[/tex]

[tex]T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm[/tex]

[tex]F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N[/tex]

[tex]F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N[/tex]

b)

[tex]H_a=1491*1.25=1863.75W[/tex]

[tex]n_f_s=\frac{H_a}{H_{nom}K_S }=1[/tex]

dip = [tex]\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm[/tex]

batolisis

A) The values of  Fc, Fi, ( F1)a  and F2

  • Fc = 4.15 N
  • Fi = 435.65 N
  • ( F1 )a = 640 N
  • F2 = 239.6 N

B ) The values of  Ha, nfs  and belt length

  • Ha = 1863.75 N
  • nfs = 1
  • belt length = 11.8 mm

Given data :

Angular velocity ratio of large pulley = 0.5

width of polyamide ( b ) = 6 inches = 0.1524 m

pulley width ( d ) = 2 inches = 0.0508 m

center to center distance ( L ) = 9 ft =  2.7432 m

angular speed of small pulley ( n ) = 1750 rev/min

power of small pulley ( Hnom ) = 2 hp = 1491.4 Watts

Ks = 1.25

γ = 9500 N/m³

g = 9.81 m/s²

t = 0.0013 mm

A) Determine the values of Fc, Fi, F1a and F2

w = γ * b * t

   = 9500 * 0.1524 * 0.0013 = 1.88 N/m

V = [tex]\frac{\pi dn}{60}[/tex] = ( π * 0.0508 * 1750 ) / 60 = 4.65 m/s  

T = 10.17 Nm ( calculated value )

i) Fc = [tex]\frac{wV^2}{g}[/tex]  ---- ( 1 )

Insert values into equation ( 1 )

Fc = ( 1.88 * 4.65² ) / 9.81

    = 4.15 N

ii) Fi = [tex]\frac{(F1)a+F2 }{2}[/tex]  - Fc ----- ( 2 )

where ( F1 )a = 0.1524 * 6000 * 0.7 * 1 = 640 N

             F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex] =  640 - [tex]\frac{2 *10.17}{0.0508}[/tex]  = 239.6 N

back to equation ( 2 )

Fi = [ ( 640 + 239.6 ) / 2 ] - Fc

   =  435.65 N

iii) ( F1 )a = b * Fa * Cp * Cv

              = 0.1524 * 6000 * 0.7 * 1  = 640 N

iv) F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex]

         = 640 - [tex]\frac{2 *10.17}{0.0508}[/tex]  = 239.6 N

B) Find the values of Ha, nfs and belt length

i) Ha = 1491 * 1.25

        = 1863.75 W

ii) nfs = [tex]\frac{Ha}{Hnom*Ks } = 1[/tex]

iii) belt length ( dip ) = [tex]\frac{L^2w}{8fi}[/tex]  

                                 = ( 2.7432 * 1.88 ) / 435.65  = 11.8 mm

Hence we can conclude that the values of the variables in the question is as listed above

Learn more about polyamide  : https://brainly.com/question/6130438

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