Question

The impeller shaft of a fluid agitator transmits 28 kW at 440 rpm. If the allowable shear stress in the impeller shaft must be limited to 80 MPa, determine(a) the minimum diameter required for a solid impeller shaft.(b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 40 mm.(c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)

Answer 1

Answer:

a) 34 mm

b) 39 mm

c) 93.16%

Explanation:

power transmitted P = 28 kW 28000 W

angular speed N = 440 rpm

angular speed in rad/s Ω = 2$\pi$N/60

Ω = (2 x 3.142 x 440)/60 = 46.08 rad/s

allowable shear stress τ = 80 MPa = 80 x $10^{6}$ Pa

torque T = P/Ω = 28000/46.08 = 607.64 N-m

a)  for the minimum diameter of a solid shaft, we use the equation

τ$d^{3}$= $\frac{16T}{ \pi}$

80 x $10^{6}$ x $d^{3}$ =  $\frac{16*607.64}{3.142}$ = 3094.28

$d^{3}$ = 3094.28/(80 x $10^{6}$) = 0.0000386785

d = $\sqrt[3]{0.0000386785}$ ≅ 0.034 m = 34 mm

b) For a hollow shaft with outside diameter D = 40 mm = 0.04 m

we use the equation,

T = $\frac{16}{\pi }$ x τ x $\frac{D^{4} - d^{4}}{D^{4} }$

where d is the internal diameter of the pipe

607.64 =  $\frac{16}{3.142}$ x 80 x $10^{6}$ x  $\frac{0.04^{4} - d^{4}}{0.04^{4} }$

3.82 x $10^{-12}$ = $0.04^{4} - d^{4}$

$d^{4}$ = $\sqrt[4]{2.56*10^{-6} }$

d = 0.039 m = 39 mm

c) we assume weight is proportional to cross-sectional area

for solid shaft,

area = $\pi r^{2}$

r = diameter/2 = 34/2 = 17 mm

area = 3.142 x $17^{2}$ = 907.92 mm^2

for hollow shaft, radius is also gotten as before

external area =  $\pi r^{2}$ = 3.142 x $20^{2}$ = 1256.64 mm^2

internal diameter =  $\pi r^{2}$  = 3.142 x $19.5^{2}$ = 1194.59 mm^2

true area of hollow shaft = external area minus internal area

area = 1256.64 - 1194.59 = 62.05 mm^2

material weight saved is proportional to 907.92 - 62.05 = 845.87 mm^2

percentage weight saved is proportional to  845.87/907.92 x 100%

= 93.15%