Question

An oil of vicosity 7 poise is used for lubrication between shaft and sleeve. The diameter of shaft is 0.6 m and it rotates is 360 rpm. Calculate the power lost in oil for a sleeve length of 160mm. The thickness of oil film is 1.0mm *

1 point

a) 25.31 kW

b) 50.62 kW

c) 37.97 kW

d) 12.65 kW

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Answer 1

The correct answer is option : (a) 25.31 kW

Given : viscocity (μ) = 7 poise = 0.7 N.s/m²

diameter of the shaft = 0.6 m

speed (N) = 360 rpm

sleeve length (L) = 160 mm or 0.16 m

oil film thickness (l) = 1 mm or 0.001 m

Power absorbed in over coming the shear force is given by,

P = torque (T) x angular velocity (ω)

⇒ torque (T) = force x radius

⇒ T = shear stress (τ) x contact area x radius

⇒ T = μ x (v/) x 2πr x L x r

( also we know, v = rω ; ω = 2πN / 60 )

    T = (π²D³NLμ) / (120 $l$)

∴ P = (π³D³N²Lμ) / (3600 $l$) Watt

⇒ power required = ( π³ x 0.6³ x 360² x 0.16 x 0.7) / (3600 x 0.001)

   ∴ P = 27003 W or 27 kW.

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