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.

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**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.

V

_{1}= Velocity of fluid flowing at section 1
P

_{1}= Pressure of fluid flowing at section 1
A

_{1}= Area of section 1
V

_{2}= Velocity of fluid flowing at section 2
P

_{2}= Pressure of fluid flowing at section 2
A

_{2}= Area of section 2
F

_{X}= Force exerted by the flowing fluid on the pipe bend in X-direction.
F

_{Y}= Force exerted by the flowing fluid on the pipe bend in Y-direction.
As we have
considered above that Fx and F

_{Y}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 - F

_{X}and - F_{Y}in X and Y direction respectively.
There will
be some other forces also acting on the flowing fluid. P

_{1}A_{1}and P_{2}A_{2}are the pressure forces acting on the flowing fluid at section 1 and section 2 respectively.
Net force
acting on fluid in X- direction = Rate of change of momentum in X- direction

P

_{1}A_{1}– P_{2}A_{2 }Cos θ – F_{X}= Mass per unit time x change of velocity
P

_{1}A_{1}– P_{2}A_{2 }Cos θ – F_{X}= ρ Q (Final velocity in X-direction – Initial velocity in X-direction)
P

_{1}A_{1}– P_{2}A_{2 }Cos θ – F_{X}= ρ Q (V_{2 }Cos θ – V_{1})
F

_{X}= ρ Q (V_{1 }– V_{2}Cos θ) + P_{1}A_{1}– P_{2}A_{2 }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

– P

_{2}A_{2 }Sin θ – F_{Y}= Mass per unit time x change of velocity
– P

_{2}A_{2 }Sin θ – F_{Y}= ρ Q (Final velocity in Y-direction – Initial velocity in Y-direction)
– P

_{2}A_{2 }Sin θ – F_{Y}= ρ Q (V_{2 }Sin θ – 0)
– P

_{2}A_{2 }Sin θ – F_{Y}= ρ Q (V_{2 }Sin θ – 0)
F

_{Y}= ρ Q (-V_{2 }Sin θ) – P_{2}A_{2 }Sin θ
Let us
determine the resultant force (F

_{R}) acting on pipe bend and angle bend by the resultant force (F_{R}) 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.

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**
Reference:**

Fluid
mechanics, By R. K. Bansal

Image
Courtesy: Google