We were
discussing the basic concept of kinetic
energy correction factor and momentum correction factor, power
absorbed in viscous flow and power required to overcome the viscous resistance of journal bearings, in the subject of fluid mechanics, in our
recent posts.

Now we
will go ahead to find out the power required to overcome the viscous resistance
of foot- step bearings, in the subject of fluid mechanics, with the help of
this post.

###
**Power required overcoming the viscous resistance of foot-step
bearings**

When we
need to rotate a vertical shaft inside a fixed bearing, the viscous oil will be
placed between the top of the bearing and bottom surface of the shaft.

Following
figure shows the foot-step bearing, in which a vertical shaft is rotating. In order
to reduce the wear and tear, oil film will be provided between the top of the
bearing and bottom surface of the shaft.

Oil will
offer viscous resistance to the shaft. Foot step bearing is also termed as
pivot bearing.

In case of
foot step bearing or pivot bearing, the radius of the surface of the shaft in
contact with the oil film will not be constant and therefore we will consider an
elementary circular ring of radius r and thickness dr for calculating the
viscous resistance of foot-step bearing.

Let us
consider the following data from above figure.

N = Speed
of the shaft in rpm

t =
Thickness of the oil film

R = Radius
of the shaft

Ï„ = Shear stress in the oil

Now we
will determine the area of elementary circular ring of radius r and thickness
dr and we will have following equation.

Area of
the elementary circular ring = 2Ï€rdr

Shear
stress, Ï„ = Î¼ du/dy

Shear
stress, Ï„ = Î¼ V/t

Angular speed of the shaft, Ï‰ = 2Ï€N/60

V = r x Ï‰ = r x 2Ï€N/60

Shear
stress, Ï„ = Î¼ V/t = Î¼ (r
x 2Ï€N/60) /t

Where,

V = Tangential speed of the shaft
at a radius r

Ï‰ = Angular speed of the shaft

Shear force on the circular ring = shear stress x
area of the elementary circular ring

Shear force on the circular ring = Ï„ x 2Ï€rdr

Shear force on the circular ring, dF = Î¼ (r x 2Ï€N/60) /t x 2Ï€rdr

Shear force on the circular ring, dF = (Î¼/15t) x (Ï€

^{2}Nr^{2}) x dr
Torque
required to overcome the viscous resistance, dT = dF x r

dT = (Î¼/15t) x (Ï€

^{2}Nr^{3}) x dr
Torque
required overcoming the viscous resistance i.e. T will be determined by
integrating the above equation between the limit 0 to R.

###
**Torque required overcoming the
viscous resistance i.e. T **

Now we
will go ahead to start a new topic in the subject of fluid mechanics
i.e. power requirement to overcome the viscous resistance of collar
bearing.

Do you
have any suggestions? Please write in comment box.

###
**Reference: **

Fluid
mechanics, By R. K. Bansal

Image
courtesy: Google

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