Oil enters a flexible bend in a pipe -- calculate the force needed to hold the bend in place

[Aurelius120]

Homework Statement

Oil enters the bend of a pipe in the horizontal plane with velocity $4 ms–1$ and pressure $280 \times 10^3 {Nm}^{-2}$ as shown in figure. (Take specific gravity of oil as 0.9 and $\sin 37^{\circ}=0.6 )$

Relevant Equations

$$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$$
$$A_1v_1=A_2v_2$$

The pressure is easily calculated from equation of continuity and Bernoullis theorem:
$$A_1v_1=A_2v_2\implies v_2=16ms^{-1}$$
SInce pipe is in horizontal plane, no difference in pressure because of height
$$P_1+\frac{\rho v_1^2}{2}=P_2+\frac{\rho v_2^2}{2}$$
$$P_2=172\times 10^3$$

What I cannot find is the force needed to hold the bend in place?

I tried simply adding $P_1A_1$ and $P_2A_2$. Added just magnitudes and then added them like vectors , neither worked.
Searching a little on the internet, it said that momentum could be used. And I still can't do it

 

[haruspex]

Homework Statement: Oil enters the bend of a pipe in the horizontal plane with velocity $4 ms–1$ and pressure $280 \times 10^3 {Nm}^{-2}$ as shown in figure. (Take specific gravity of oil as 0.9 and $\sin 37^{\circ}=0.6 )$
Relevant Equations: $$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$$
$$A_1v_1=A_2v_2$$

View attachment 345924
The pressure is easily calculated from equation of continuity and Bernoullis theorem:
$$A_1v_1=A_2v_2\implies v_2=16ms^{-1}$$
SInce pipe is in horizontal plane, no difference in pressure because of height
$$P_1+\frac{\rho v_1^2}{2}=P_2+\frac{\rho v_2^2}{2}$$
$$P_2=172\times 10^3$$

What I cannot find is the force needed to hold the bend in place?

I tried simply adding $P_1A_1$ and $P_2A_2$. Added just magnitudes and then added them like vectors , neither worked.
Searching a little on the internet, it said that momentum could be used. And I still can't do it

Consider the two forces, as vectors, that act on the bend from the pressures at each side of the bend.

 

[Aurelius120]

Reaction forces due to oil pressure:
$$\vec F_1=P_1A_1 \hat i=-56\times 10^3\hat i$$
$$\vec F_2=P_2A_2=(8.6\times 10^3) \cos37\hat i-(8.6\times 10^3)\sin 37\hat j$$
Net force on pipe =Sum of Reaction force due to oil pressure= $$|\vec F_1+\vec F_2|=\sqrt{(56000-860\times 8)^2+(860\times 6)^2}\neq\text{ given answer}$$

Consider the two forces, as vectors, that act on the bend from the pressures at each side of the bend.

 

[haruspex]

Reaction forces due to oil pressure:
$$\vec F_1=P_1A_1 \hat i=-56\times 10^3\hat i$$
$$\vec F_2=P_2A_2=(8.6\times 10^3) \cos37\hat i-(8.6\times 10^3)\sin 37\hat j$$
Net force on pipe =Sum of Reaction force due to oil pressure= $$|\vec F_1+\vec F_2|=\sqrt{(56000-860\times 8)^2+(860\times 6)^2}\neq\text{ given answer}$$

How did you arrive at those signs for the $\hat i$ components?

 

[Aurelius120]

How did you arrive at those signs for the $\hat i$ components?

The fluid flows horizontally in $+\hat i$ upto bend so force on it must be in direction of its velocity to make it flow and its reaction force on pipe must be in $-\hat i$
After the bend the fluid flows in $-cos37\hat i+sin37\hat j$ and reaction force will be in opposite direction

 

[haruspex]

The fluid flows horizontally in $+\hat i$ upto bend so force on it must be in direction of its velocity to make it flow and its reaction force on pipe must be in $-\hat i$
After the bend the fluid flows in $-cos37\hat i+sin37\hat j$ and reaction force will be in opposite direction

Pressure does not have a direction. What is the direction of the force that pressure exerts over the area at the aperture of the bend?

 

[Aurelius120]

Pressure does not have a direction. What is the direction of the force that pressure exerts over the area at the aperture of the bend?

$P_2A_2 cos37$? Since force perpendicular to area is taken?

 

[haruspex]

$P_2A_2 cos37$? Since force perpendicular to area is taken?

The force exerted by a pressure p on an area element $\vec dA$ is $p\vec dA$. The direction of those two vectors is normal to the area element. The apertures of the elbow are cross sections of the pipe, so you have two forces at angle of 37°.
Those forces would exist merely from having that as static pressure in the pipe. In addition, the water flow is changing direction. That momentum change requires a force too.

 

[Lnewqban]

Searching a little on the internet, it said that momentum could be used. And I still can't do it

Can you show us your calculations regarding change in momentum?
Generally, the resulting force due to changes of direction of the mass flow velocity is greater in magnitude than the force created by internal pressure.
As correctly explained above, both types of forces need to be added up for calculating the actual reactive force.

 

[Aurelius120]

Sorry for the delay. Had some tests so forgot about the doubts.
Tis the soution given. I dont understand why pressure at bend is not $F_{pipe}\neq F_1+F_2$? Why is there a change of oil momentum?

 

[erobz]

View attachment 346826
Sorry for the delay. Had some tests so forgot about the doubts.
Tis the soution given. I dont understand why pressure at bend is not $F_{pipe}\neq F_1+F_2$? Why is there a change of oil momentum?

Applying Reynolds Transport Theorem with the fluid system momentum as the extensive property we get the"Momentum Equation"

$$ \sum \mathbf{F} = \frac{d}{dt} \int_{cv} \mathbf{v} \rho ~d V\llap{-} + \int_{cs} \mathbf{v} \rho \mathbf{V} \cdot d \mathbf{A} \tag{1} $$

The pressure distribution integrated over each cross section of the flow are forces, and there is a restraint force required so that the control volume is stationary. With assumed uniform pressure/velocity distributions over each area apply (1) for each component direction.

 

[haruspex]

View attachment 346826
Sorry for the delay. Had some tests so forgot about the doubts.
Tis the soution given. I dont understand why pressure at bend is not $F_{pipe}\neq F_1+F_2$? Why is there a change of oil momentum?

Consider the oil passing through the bend in time dt.
It has mass $dm=\rho A_1v_1dt$ and undergoes a change in velocity from $v_1\hat i$ to $v_2(-\cos(\theta)\hat i+\sin(\theta)\hat j)$. Call that change $\vec{\Delta v}$. That implies a momentum change $\vec{\Delta v} dm$. The net force on it is therefore $\vec F_{net}=\vec{\Delta v}\dot m=\rho A_1v_1\vec{\Delta v}$.
Three forces act on it to provide this net force:

• the force due to the pressure at P acting on the area element $A_1\hat i$ : $\vec F_P=P_1A_1\hat i$
• the force due to the pressure at Q acting on the area element $A_2(\cos(\theta)\hat i-\sin(\theta)\hat j)$: $\vec F_Q=P_2A_2(\cos(\theta)\hat i-\sin(\theta)\hat j)$
• the force from the pipe wall: $\vec F_{pipe}$.

$\vec F_{net}=\vec F_P+\vec F_Q+\vec F_{pipe}$.

 

[Aurelius120]

Consider the oil passing through the bend in time dt.
It has mass $dm=\rho A_1v_1dt$ and undergoes a change in velocity from $v_1\hat i$ to $v_2(-\cos(\theta)\hat i+\sin(\theta)\hat j)$. Call that change $\vec{\Delta v}$. That implies a momentum change $\vec{\Delta v} dm$. The net force on it is therefore $\vec F_{net}=\vec{\Delta v}\dot m=\rho A_1v_1\vec{\Delta v}$.
Three forces act on it to provide this net force:

• the force due to the pressure at P acting on the area element $A_1\hat i$ : $\vec F_P=P_1A_1\hat i$
• the force due to the pressure at Q acting on the area element $A_2(\cos(\theta)\hat i-\sin(\theta)\hat j)$: $\vec F_Q=P_2A_2(\cos(\theta)\hat i-\sin(\theta)\hat j)$
• the force from the pipe wall: $\vec F_{pipe}$.

$\vec F_{net}=\vec F_P+\vec F_Q+\vec F_{pipe}$.

Idk why I thought, the pressure at $A_1$ and $A_2$ was going to be exerted on pipe and not liquid. Therefore confused $F_{pipe}=F_1+F_2$