We were discussing the various basic concepts such
as Euler’s
Equation of motion, Bernoulli’s
equation from Euler’s equation, derivation
of discharge through venturimeter, derivation of discharge through Orifice
meter and Pitot
tube with the expression of velocity of flow at any point in the pipe
or channel, in the subject of fluid mechanics, in our recent posts.
Today we will see here the resultant force exerted
by flowing fluid on a pipe bend.
Force exerted by a flowing fluid on a pipe – bend
In order to secure the expression of resultant force
exerted by a flowing fluid on a pipe bend, we will use the basic concept of
impulse momentum equation. Before going ahead, it is very important to find out
and read the concept of the momentum equation.
Let us consider that fluid is flowing through a pipe
which is bent as displayed here in following figure. We have considered here
two sections i.e. section 1-1 and section 2-2.
V1 = Velocity of fluid flowing at
section 1
P1 = Pressure of fluid flowing at
section 1
A1 = Area of section 1
V2 = Velocity of fluid flowing at
section 2
P2 = Pressure of fluid flowing at
section 2
A2 = Area of section 2
FX = Force exerted by the flowing
fluid on the pipe bend in X-direction.
FY = Force exerted by the flowing
fluid on the pipe bend in Y-direction.
As we have considered above that Fx and FY are
the forces, exerted by the flowing fluid on the pipe bend in X and Y direction
respectively.
Considering the Newton’s third law of motion, forces
exerted by the pipe bend on the flowing fluid will be - FX and
- FY in
X and Y direction respectively.
There will be some other forces also acting on the
flowing fluid. P1A1 and P2A2 are
the pressure forces acting on the flowing fluid at section 1 and section 2
respectively.
Now we will recall the momentum equation and we will
have following equation for X direction.
Net force acting on fluid in X- direction = Rate of
change of momentum in X- direction
P1A1 – P2A2 Cos
θ – FX = Mass per unit time x change of velocity
P1A1 – P2A2 Cos
θ – FX = ρ Q (Final velocity in X-direction – Initial velocity
in X-direction)
P1A1 – P2A2 Cos
θ – FX = ρ Q (V2 Cos θ – V1)
FX = ρ Q (V1 – V2 Cos
θ) + P1A1 – P2A2 Cos θ
Similarly, we will recall the momentum equation and we will
have following equation for Y direction.
Net force acting on fluid in Y- direction = Rate of
change of momentum in Y direction
– P2A2 Sin θ – FY =
Mass per unit time x change of velocity
– P2A2 Sin θ – FY =
ρ Q (Final velocity in Y-direction – Initial velocity in Y-direction)
– P2A2 Sin θ – FY =
ρ Q (V2 Sin θ – 0)
– P2A2 Sin θ – FY =
ρ Q (V2 Sin θ – 0)
FY = ρ Q (-V2 Sin θ)
– P2A2 Sin θ
Let us determine the resultant force (FR)
acting on pipe bend and angle bend by the resultant force (FR)
with horizontal direction.
We will now find out the "Moment of momentum equation", in
the subject of fluid mechanics, in our next post.
Do you have any suggestions? Please write in comment
box.
Reference:
Fluid mechanics, By R. K. Bansal
Image Courtesy: Google
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