Determine force from flow on diffuser

[Bluestribute]

Homework Statement

Water exits the 3-in.-diameter pipe at a velocity of 12 ft/s and is split by the wedge diffuser. Assume water is ideal fluid, that is, incompressible and frictionless. Determine the force the flow exerts on the diffuser. Take θ = 30 ∘

Homework Equations

F = ρAV

The Attempt at a Solution

So what I've seen others do, and what tools I've been given so far in the course, here's the numbers I plugged in, and the equation I used:

ρ = 62.4
A = πr²
R = (3/24) <- 3 inch diameter, and this just eliminates one step in the calculator
V = 12
θ = 15 (half of 30)

F = ρAV(Vcosθ). But I'm getting 426, which isn't correct . . .

 

[stockzahn]

F = ρAV(Vcosθ)

The force, which affected the water flow can be calculated as the difference of the momentum flows before and after the diffusor. It seems that you've mixed up the velocities of these two flows (you combined the velocities in horizontal direction of both states in one term).

 

[Bluestribute]

What do you mean? ρAV = (ρAV)1 - (ρAV)2?

 

[Chestermiller]

What direction is the fluid velocity when it is flowing over the diffuser? (a) parallel to the axis (b) parallel to the diffuser surface

Is the mass rate of flow constant in going over the diffuser?

Is the velocity of the fluid on the diffuser constant or changing with axial position along the diffuser?

Is the thickness of the fluid layer on the diffuser constant or changing with axial position along the diffuser?

Chet

 

[Bluestribute]

mass flow rate = ρVA = 62.4 * -12 * π(3/24)²
That's density times velocity dotted with the normal (that's why it's negative) times the area of the circle the water is traveling through.

Would it be constant over the diffuser, but different than the above calculation? Including the velocity? And nothing is changing I don't think with respect to position. The force is different after the wedge, but still constant.

 

[Chestermiller]

mass flow rate = ρVA = 62.4 * -12 * π(3/24)²
That's density times velocity dotted with the normal (that's why it's negative) times the area of the circle the water is traveling through.

Would it be constant over the diffuser, but different than the above calculation? Including the velocity?

What does Bernoulli's equation tell you about the velocity of the fluid as it is passing over the diffuser (i.e., the magnitude of the velocity vector) compared to its value before it hits the diffuser?

 

[Bluestribute]

No idea. We haven't gotten that far . . .

 

[stockzahn]

EDIT: I misread the statement and therefore canceled this post, sorry.

 

[Chestermiller]

The water flow is split up and each part has changed its direction with respect to the one before the diffusor. As mass has changed its state of movement forces must have affected it. To solve the problem I would recommend the following steps:

1) Define a Cartesian coordinate system (horizontal(x)/vertical(y))
2) Find the momentum flows of the water jets before and after the diffusor in x- and y-direction
3) The difference of the momentum flows in each direction will give you the force on the diffusor in the opposite direction

momentum flow:
Your equation calculates the mass flow. To get the momentum flow you have to multiply it (again) with the velocity. To get the momentum flows in the different rectangular directions after the diffusor you can use the angle θ and trigonometry.

hints: Water is more or less incompressible. Look at the thicknesses of the flows after they got split up - they are equal (symmetrical) and their sum is the same as the thickness before the diffusor. What does that mean for

You are giving very good advice here, but you seem to be assuming that the geometry is flat. According to the problem statement, the actual geometry is cylindrical. This is why I was trying to get the OP to use Bernoulli to determine that the magnitude of the fluid velocity along the diffuser is the same as the magnitude of the velocity out of the pipe. As the flow passes over the diffuser, it is the thickness of the water layer that changes (decreases) to conserve mass flow rate. I'm hoping you will continue with your help and advice, modifying it as necessary to address the cylindrical geometry.

Chet

 

[stockzahn]

You are giving very good advice here, but you seem to be assuming that the geometry is flat. According to the problem statement, the actual geometry is cylindrical. This is why I was trying to get the OP to use Bernoulli to determine that the magnitude of the fluid velocity along the diffuser is the same as the magnitude of the velocity out of the pipe. As the flow passes over the diffuser, it is the thickness of the water layer that changes (decreases) to conserve mass flow rate. I'm hoping you will continue with your help and advice, modifying it as necessary to address the cylindrical geometry.

Chet

Dear Chet,

as the statement said "[]...is split by the wedge diffuser." I thought it is a two dimensional problem. Additionally in the drawing the pipe is cut (at the upper side), but it seemed to me that it ends at the lower side without changing its diameter (but where is the diffusor then?). Of course, if it would be 3-dimensional without the diffusor (only the cone/wedge), the thicknesses of the split water flows must decrease along the cone. However there are some additional hints that it is 3-dimensional, as the flows keep in touch with the wedge/cone even after it ends. I probably misunderstood the statement/drawing, I'm sorry for that.

 

[Chestermiller]

Dear Chet,

as the statement said "[]...is split by the wedge diffuser." I thought it is a two dimensional problem. Additionally in the drawing the pipe is cut (at the upper side), but it seemed to me that it ends at the lower side without changing its diameter (but where is the diffusor then?). Of course, if it would be 3-dimensional without the diffusor (only the cone/wedge), the thicknesses of the split water flows must decrease along the cone. However there are some additional hints that it is 3-dimensional, as the flows keep in touch with the wedge/cone even after it ends. I probably misunderstood the statement/drawing, I'm sorry for that.

No problem. Your analysis can still be continued, taking into account, from Bernoulli, that the fluid velocity tangent to the cone is still the same as in the pipe. Please continue.

Chet

 

[Bluestribute]

Wait . . . so are you saying to use ρAV² to find the flow at the inlet and each outlet (but for the outlets, use the cos of half the angle)? And just take the difference of inlet/outlet flows as the force?

 

[Chestermiller]

Wait . . . so are you saying to use ρAV² to find the flow at the inlet and each outlet (but for the outlets, use the cos of half the angle)? And just take the difference of inlet/outlet flows as the force?

Not exactly. That's called the rate of momentum, not the flow. But otherwise, OK.

Chet