Fluid Mechanics Force Body Diagrams

[Valour549]

Homework Statement

Water flows steadily through a 3.25-in-diameter pipe and discharges through a 1.25-in-diameter nozzle to atmospheric pressure. The flow rate is 24.5 gpm. Calculate the minimum static pressure required in the pipe to produce this flow rate. Evaluate the axial force of the nozzle assembly on the pipe flange.

Relevant Equations

ρQ*V1+P1*A1 = ρQ*V2+P2*A2

Let me start off by saying that I have found (or is given) all of these: ρ, Q, V1, V2, P1, P2, A1, A2 (V being the velocity here). So no problem with Bernoulli or the Continuity equation calculations.

I am just struggling with drawing the FBD in order to evaluate the axial force, Fx

I know we need to write an x-momentum equation to the control volume: ρQ*V1+P1*A1 = ρQ*V2+P2*A2

But I don't know which side the Fx should be added on in the above equation. It says "axial force of the nozzle assembly on the pipe flange", so if I had to guess, Fx would point towards the left. But the bigger problem here is that I'm not even sure whether the terms I have on the LHS (and RHS) causes a force that acts to the left or right of the control volume.

 

[Lnewqban]

You have the same mass of fluid per unit of time everywhere.
What mass has a greater impulse?

 

[erobz]

Homework Statement: Water flows steadily through a 3.25-in-diameter pipe and discharges through a 1.25-in-diameter nozzle to atmospheric pressure. The flow rate is 24.5 gpm. Calculate the minimum static pressure required in the pipe to produce this flow rate. Evaluate the axial force of the nozzle assembly on the pipe flange.
Relevant Equations: ρQ*V1+P1*A1 = ρQ*V2+P2*A2

Let me start off by saying that I have found (or is given) all of these: ρ, Q, V1, V2, P1, P2, A1, A2 (V being the velocity here). So no problem with Bernoulli or the Continuity equation calculations.

I am just struggling with drawing the FBD in order to evaluate the axial force, Fx
View attachment 327329
I know we need to write an x-momentum equation to the control volume: ρQ*V1+P1*A1 = ρQ*V2+P2*A2

But I don't know which side the Fx should be added on in the above equation. It says "axial force of the nozzle assembly on the pipe flange", so if I had to guess, Fx would point towards the left. But the bigger problem here is that I'm not even sure whether the terms I have on the LHS (and RHS) causes a force that acts to the left or right of the control volume.

In words:

Sum of the external forces acting on the control volume = Rate of Momentum Accumulation within the control volume + Net rate of momentum efflux through the control surface.
$$ \sum \boldsymbol F = \frac{d}{dt} \int_{cv} \rho \boldsymbol v ~dV\llap{-} + \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A $$

$$ \text{Momentum Equation}$$

 

[Chestermiller]

Homework Statement: Water flows steadily through a 3.25-in-diameter pipe and discharges through a 1.25-in-diameter nozzle to atmospheric pressure. The flow rate is 24.5 gpm. Calculate the minimum static pressure required in the pipe to produce this flow rate. Evaluate the axial force of the nozzle assembly on the pipe flange.
Relevant Equations: ρQ*V1+P1*A1 = ρQ*V2+P2*A2

Let me start off by saying that I have found (or is given) all of these: ρ, Q, V1, V2, P1, P2, A1, A2 (V being the velocity here). So no problem with Bernoulli or the Continuity equation calculations.

I am just struggling with drawing the FBD in order to evaluate the axial force, Fx
View attachment 327329
I know we need to write an x-momentum equation to the control volume: ρQ*V1+P1*A1 = ρQ*V2+P2*A2

But I don't know which side the Fx should be added on in the above equation. It says "axial force of the nozzle assembly on the pipe flange", so if I had to guess, Fx would point towards the left. But the bigger problem here is that I'm not even sure whether the terms I have on the LHS (and RHS) causes a force that acts to the left or right of the control volume.

If +F is included on the left side of the equation , it represents the force that the nozzle exerts on the fluid in the positive x-direction.

 

[Valour549]

You have the same mass of fluid per unit of time everywhere.
What mass has a greater impulse?

The mass at the nozzle exit because it has a higher velocity?

If +F is included on the left side of the equation , it represents the force that the nozzle exerts on the fluid in the positive x-direction.

When you say the force the nozzle exerts on the fluid, is that at point 1 or point 2? The question says the axial force of the nozzle assembly on the pipe flange, isn't the pipe flange that rectangular piece between the pipe and the nozzle? I guess I'm just not understanding the whole picture of the forces involved. Would really appreciate a diagram.

 

[erobz]

Yes, you need to draw a diagram.

Notice that the jet discharges to $P_2 = 0 \rm{gage}$. The pressure $P_1$ acts on the left side of the control volume. Given $P_2 = 0 \rm{gage}$, can $P_1$ alone satisfy the requirement that requirement that the control volume is not accelerating?

 

[Valour549]

Yes, you need to draw a diagram.

Notice that the jet discharges to $P_2 = 0 \rm{gage}$. The pressure $P_1$ acts on the left side of the control volume. Given $P_2 = 0 \rm{gage}$, can $P_1$ alone satisfy the requirement that requirement that the control volume is not accelerating?

I know P2=0. When you say P1 acts on the left side of the CV, that could be one of two possibilities below and I really don't know which.

I know that you guys are trying to help users derive the answer without telling them, but the issue here is I already have the answer and the equation from the manual. But I just cannot for the life of me understand what the hell it's drawing. This isn't for a homework I need to hand in or anything, I'm just reading through the book's examples. Please just explain to me how the forces are acting on the CV with arrows.

 

[Chestermiller]

The mass at the nozzle exit because it has a higher velocity?

When you say the force the nozzle exerts on the fluid, is that at point 1 or point 2? The question says the axial force of the nozzle assembly on the pipe flange, isn't the pipe flange that rectangular piece between the pipe and the nozzle? I guess I'm just not understanding the whole picture of the forces involved. Would really appreciate a diagram.

It is not at point 1 or point 2. It is the axial force the nozzle exerts on the flowing fluid between points 1 and 2.

 

[erobz]

The nozzle is held stationary by the pipe. The pipe is applying a force to the nozzle (through the flange). They are asking for the force the nozzle applies to the pipe (through the flange).

 

[Valour549]

So combining what the two of you helpers have said:

The pipe applies a force to the nozzle through the flange, and then the nozzle then applies it to the fluid flowing through it? And this is necessary for the fluid to have a higher velocity exiting the nozzle (as compared to the slower velocity when entering)?

 

[erobz]

So combining what the two of you helpers have said:

The pipe applies a force to the nozzle through the flange, and then the nozzle then applies it to the fluid flowing through it? And this is necessary for the fluid to have a higher velocity exiting the nozzle (as compared to the slower velocity when entering)?

My understanding is they are freeing the control volume from the pipe, replacing the effect of the pipe with an external force on the control volume $R_x$. They are asking for the newtons 3rd law compliment of that force $-R_x$ which is the force the nozzle (the control volume) applies to the pipe.

 

[Valour549]

My understanding is they are freeing the control volume from the pipe, replacing the effect of the pipe with an external force on the control volume $R_x$. They are asking for the newtons 3rd law compliment of that force $-R_x$.

OK so... since ρQ*V1+P1*A1 turns out to be greater than ρQ*V2 (P2=0), the force acting on the CV (or nozzle) by the pipe must be to the left (negative X direction).

And therefore the force on the pipe by the nozzle is to the right (positive X direction)?

 

[erobz]

OK so... since ρQ*V1+P1*A1 turns out to be greater than ρQ*V2 (P2=0), the force acting on the CV (or nozzle) by the pipe must be to the left (negative X direction).

And therefore the force on the pipe by the nozzle is to the right (positive X direction)?

The pressures are absolute in the diagram. They are asking for the force the nozzle applies to the pipe, which is $-R_x$

Sum of the forces = rate of momentum exiting control volume - rate of momentum entering control volume.

 

[Valour549]

Except Rx turns out to be negative and -Rx turns out to be positive right?

How do you represent the terms ρQ*V1 and ρQ*V2 in the FBD?

 

[erobz]

Except Rx turns out to be negative and -Rx turns out to be positive right?

That’s what your solution says. It seems wrong to me, I’ll check it if I get a chance later.

How do you represent the terms ρQ*V1 and ρQ*V2 in the FBD?

They are usually presented in a separate diagram, called a “momentum diagram”

 

[Chestermiller]

OK so... since ρQ*V1+P1*A1 turns out to be greater than ρQ*V2 (P2=0), the force acting on the CV (or nozzle) by the pipe must be to the left (negative X direction).

And therefore the force on the pipe by the nozzle is to the right (positive X direction)?

Correct

 

[erobz]

Yeah, $R_x$ comes out negative.