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

a) 34 mm

b) 39 mm

c) 93.16%

Step-by-step 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)`=
`(16T)/( \pi)`

80 x
`10^(6)` x
`d^(3)` =
`(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 =
`(16)/(\pi )` x τ x
`(D^(4) - d^(4))/(D^(4) )`

where d is the internal diameter of the pipe

607.64 =
`(16)/(3.142)` x 80 x
`10^(6)` x
`(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%