Question

A 400 mm diameter shaft is rotating at 200 rpm in a bearing of length 120 mm. If the thickness of oil film is 1.5 mm and the dynamic viscosity of the oil is 0.7 Ns/m2 determine: Power utilization in overcoming viscous resistance

Answer 1

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Answer 2

Answer:

Given: D=400mm

N=200rpm

t=1.5mm=0.0015m

l=120mm=0.12m

μ=0.7Ns/m2

To find : Power=?

Explanation:

Solution:

1 Tangential Velocity of shaft

where,n=speed of shaft in rpm

thus, V=

$( \frac{\pi \times 0.4 \times 200}{60})$

=4.79m/s

2 Torque required T-

we know that,

$τ=(μ \times du/dy)$

for that first find du-change in velocity

therefore, change in velocity is equal to tangential velocity,

and also dy=t=0.0015m

and also dy=t=0.0015m

$τ = (0.7 \times \frac{4.19}{0.0015})$

therefore, τ=1955.3N/m2

3 Shear force(F)-

we know that,

$F = τ \times Area$

$Area = \pi \times d \times l$

so that,

$F = 1955.3 \times \pi \times 0.4 \times 0.12$

we get, F=294.85N

4 viscous Torque,

T=

$T = F \times \frac{d}{2}$

so that,

$T = 294.85 \times \frac{0.4}{2}$

we get, T= 58.97N

5 Power required(P)-

$P = ( \frac{2 \times \pi \times N \times T }{60})$

then,

$P =( \frac{2 \times \pi \times 200 \times 58.97}{60})$

we get,

$P = 1.235 \times {10}^{3} watt$

hence,

$P = 1.235KW$