Answer:
The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp
Explanation:
Polar moment of Inertia
[tex](I_p)s = \frac{\pi(0.55)4}2[/tex]
   = 0.14374 in 4
Maximum sustainable torque on the solid circular shaft
[tex]T_{max} = T_{allow} \frac{I_p}{r}[/tex]
     =[tex](14 \times 10^3) \times (\frac{0.14374}{0.55})[/tex]
     = 3658.836 lb.in
     = [tex]\frac{3658.836}{12}[/tex] lb.ft
    = 304.9 lb.ft
Maximum sustainable torque on the tubular shaft
[tex]T_{max} = T_{allow}( \frac{Ip}{r})[/tex]
     = [tex](14 \times10^3) \times ( \frac{0.13101}{0.55})[/tex]
     = 3334.8 lb.in
     = [tex](\frac{3334.8}{12} )[/tex] lb.ft
     = 277.9 lb.ft
Maximum sustainable power in the solid circular shaft
[tex]P_{max} = 2 \pi f_T[/tex]
     = [tex]2\pi(2.1) \times 304.9[/tex]
     = 4023.061 lb. ft/s
     = [tex](\frac{4023.061}{550})[/tex] hp
     = 7.315 hp
Maximum sustainable power in the tubular shaft
[tex]P _{max,t} = 2\pi f_T[/tex]
      = [tex]2\pi(2.1) \times 277.9[/tex]
      = 3666.804 lb.ft /s
      = [tex](\frac{3666.804}{550})[/tex]hp
      = 6.667 hp
