## Thursday, 14 June 2018

We were discussing the various basic concepts such as Euler’s Equation of motionBernoulli’s equation from Euler’s equationderivation 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 – P2ACos θ – FX = Mass per unit time x change of velocity
P1A1 – P2ACos θ – FX = ρ Q (Final velocity in X-direction – Initial velocity in X-direction)
P1A1 – P2ACos θ – FX = ρ Q (VCos θ – V1)
FX = ρ Q (V– V2 Cos θ) + P1A1 – P2ACos 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
– P2ASin θ – FY = Mass per unit time x change of velocity
– P2ASin θ – FY = ρ Q (Final velocity in Y-direction – Initial velocity in Y-direction)
– P2ASin θ – FY = ρ Q (VSin θ – 0)
– P2ASin θ – FY = ρ Q (VSin θ – 0)
FY = ρ Q (-VSin θ) – P2ASin 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