# Fluid mechanics: water jet impacting an inclined plane

• Guillem_dlc
In summary, the conversation discusses an example of fluid mechanics and a problem statement involving a two-dimensional jet of water impacting an inclined plate. The equations for the continuity equation and generalized Bernoulli equation without a machine are provided. The conversation then focuses on the equation for the quantity movement in the x-axis, where a frictionless condition is assumed and the equation becomes dimensionally inconsistent. The speaker expresses confusion about the equation and its relationship to the size of h, the height of the jet.
Guillem_dlc
Homework Statement
A two-dimensional jet of ##10\, \textrm{cm}## height sprays water at ##10\, \textrm{m}/\textrm{s}## which impacts, without friction, on an inclined plate...

DATA: ##\rho =1000\, \textrm{kg}/\textrm{m}^3##, ##c_1=10\, \textrm{m}/\textrm{s}##, ##h_1=10\, \textrm{cm}=0,10\, \textrm{m}##

Working hypothesis
- Stationary regime
- Uniform properties
- Incompressible flow
Relevant Equations
Continuity equation, Bernoulli equation
I was looking at an example of fluid mechanics and I don't understand this.

Statement figures:

CONTINUITY EQUATION

$$\left. \dfrac{dm}{dt}\right]_{MC}=(\dot{m}_2+\dot{m}_3)-\dot{m}_1=0$$
$$\dot{m}_1=\dot{m}_2+\dot{m}_3$$
$$\rho c_1A_1=\rho c_2A_2+\rho c_3A_3$$
$$\rho c_1 h1=\rho c_2 a1+\rho c_3 b1 \rightarrow \textrm{unit width}$$
$$\boxed{c_1h=c_2a+c_3b}$$

GENERALISED BERNOULLI EQUATION WITHOUT MACHINE
1-2
$$\dfrac{p_2-p_1}{\rho}+\dfrac{c_2^2-c_1^2}{2}+g(z_2-z_1)+w_{\textrm{roz}}=0 \,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{ \begin{array}{l} p_1=p_2=p_{\textrm{atm}}=0 \\ z_1=z_2=0 \\ w_{\textrm{roz}}=0 \end{array}\right.$$
1-3
$$\dfrac{c_2^2-c_1^2}{2}=0\rightarrow c_1=c_2$$
$$\dfrac{p_3-p_1}{\rho}+\dfrac{c_3^2-c_1^2}{2}+g(z_3-z_1)+w_{\textrm{roz}}=0\,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{ \begin{array}{l} p_1=p_3=p_{\textrm{atm}}=0 \\ z_1=z_3=0 \\ w_{\textrm{roz}}=0 \end{array}\right.$$
$$\dfrac{c_3^2-c_1^2}{2}=0\rightarrow c_1=c_3 \quad \boxed{c_1=c_2=c_3=c}$$
Then,
$$c_1h=c_2a+c_3b$$
$$c_1=c_2=c_3=c$$
$$\boxed{h=a+b}$$
Equation Quantity Movement 1-2

$$\overrightarrow{F}_{\textrm{paret}\rightarrow \textrm{fluid}}=(\overrightarrow{c}_2\dot{m}_2+\overrightarrow{c}_3\dot{m}_3)-\overrightarrow{c}_1\dot{m}_1$$
x-axis
$$F_{p\rightarrow f(x)}=c_{2x}\rho c_{2x}A_2+(-)c_{3x}\rho c_{3x}A_3-c_{1x}\rho c_{1x}A_1$$
$$F_{p\rightarrow f(x)}=c^2\rho 1[a-b\cos (\alpha)h]=\tau A=0\rightarrow \textrm{Frictionless}$$
$$\cos (\alpha)h=a-b$$
y-axis
$$F_{p\rightarrow f(y)}=c_{2y}\rho c_{2y}A_2+c_{3y}\rho c_{3y}A_3-(-)c_{1y}\rho c_{1y}A_1$$
$$F_{p\rightarrow f(y)}=c^2\rho h\sin (\alpha)$$
Then,
$$\left\{ \begin{array}{l} h = a+b\\ \cos (\alpha)h=a-b \end{array} \right.$$
$$\alpha :=\dfrac{45\pi}{180}=0.785$$
$$h:=0.1$$
We solve the system for ##a## and ##b##:
$$(0.08536,0.01464)$$
$$\rho := 1000$$
$$c1:=10$$
$$Fpfz:=\rho \cdot c1^2\cdot h\cdot \sin (\alpha)=7071\, \textrm{N}$$

What I don't understand is the following (which has appeared before):
x-axis:
$$\tau A=0\rightarrow \textrm{Frictionless}$$
$$\cos (\alpha)h=a-b$$
$$\alpha :=\dfrac{45\pi}{180}=0.785$$

So the ##h## would be about the size of the 1, wouldn't it?

What is the problem statement? You gave the figures, what is being asked? Just dropping a list of equations without any explanations makes it difficult because anyone trying to help must reverse engineer the question…

vanhees71 and scottdave
erobz said:
What is the problem statement? You gave the figures, what is being asked? Just dropping a list of equations without any explanations makes it difficult because anyone trying to help must reverse engineer the question…
This is the whole statement of the example. I have given you the whole example. It doesn't say more. I guess they are asking you for the fluid wall strength.

Guillem_dlc said:
Homework Statement:: A two-dimensional jet of ##10\, \textrm{cm}## height sprays water at ##10\, \textrm{m}/\textrm{s}## which impacts, without friction, on an inclined plate...

DATA: ##\rho =1000\, \textrm{kg}/\textrm{m}^3##, ##c_1=10\, \textrm{m}/\textrm{s}##, ##h_1=10\, \textrm{cm}=0,10\, \textrm{m}##

Working hypothesis
- Stationary regime
- Uniform properties
- Incompressible flow
Relevant Equations:: Continuity equation, Bernoulli equation

I was looking at an example of fluid mechanics and I don't understand this.

Statement figures:
View attachment 319656
View attachment 319657View attachment 319658
CONTINUITY EQUATION
View attachment 319659
$$\left. \dfrac{dm}{dt}\right]_{MC}=(\dot{m}_2+\dot{m}_3)-\dot{m}_1=0$$
$$\dot{m}_1=\dot{m}_2+\dot{m}_3$$
$$\rho c_1A_1=\rho c_2A_2+\rho c_3A_3$$
$$\rho c_1 h1=\rho c_2 a1+\rho c_3 b1 \rightarrow \textrm{unit width}$$
$$\boxed{c_1h=c_2a+c_3b}$$

GENERALISED BERNOULLI EQUATION WITHOUT MACHINE
1-2
$$\dfrac{p_2-p_1}{\rho}+\dfrac{c_2^2-c_1^2}{2}+g(z_2-z_1)+w_{\textrm{roz}}=0 \,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{ \begin{array}{l} p_1=p_2=p_{\textrm{atm}}=0 \\ z_1=z_2=0 \\ w_{\textrm{roz}}=0 \end{array}\right.$$
1-3
$$\dfrac{c_2^2-c_1^2}{2}=0\rightarrow c_1=c_2$$
$$\dfrac{p_3-p_1}{\rho}+\dfrac{c_3^2-c_1^2}{2}+g(z_3-z_1)+w_{\textrm{roz}}=0\,\,\, \left[ \dfrac{\textrm{J}}{\textrm{kg}}\right] \left\{ \begin{array}{l} p_1=p_3=p_{\textrm{atm}}=0 \\ z_1=z_3=0 \\ w_{\textrm{roz}}=0 \end{array}\right.$$
$$\dfrac{c_3^2-c_1^2}{2}=0\rightarrow c_1=c_3 \quad \boxed{c_1=c_2=c_3=c}$$
Then,
$$c_1h=c_2a+c_3b$$
$$c_1=c_2=c_3=c$$
$$\boxed{h=a+b}$$
Equation Quantity Movement 1-2
View attachment 319660
$$\overrightarrow{F}_{\textrm{paret}\rightarrow \textrm{fluid}}=(\overrightarrow{c}_2\dot{m}_2+\overrightarrow{c}_3\dot{m}_3)-\overrightarrow{c}_1\dot{m}_1$$
x-axis
$$F_{p\rightarrow f(x)}=c_{2x}\rho c_{2x}A_2+(-)c_{3x}\rho c_{3x}A_3-c_{1x}\rho c_{1x}A_1$$
$$F_{p\rightarrow f(x)}=c^2\rho 1[a-b\cos (\alpha)h]=\tau A=0\rightarrow \textrm{Frictionless}$$
$$\cos (\alpha)h=a-b$$
y-axis
$$F_{p\rightarrow f(y)}=c_{2y}\rho c_{2y}A_2+c_{3y}\rho c_{3y}A_3-(-)c_{1y}\rho c_{1y}A_1$$
$$F_{p\rightarrow f(y)}=c^2\rho h\sin (\alpha)$$
Then,
$$\left\{ \begin{array}{l} h = a+b\\ \cos (\alpha)h=a-b \end{array} \right.$$
$$\alpha :=\dfrac{45\pi}{180}=0.785$$
$$h:=0.1$$
We solve the system for ##a## and ##b##:
$$(0.08536,0.01464)$$
$$\rho := 1000$$
$$c1:=10$$
$$Fpfz:=\rho \cdot c1^2\cdot h\cdot \sin (\alpha)=7071\, \textrm{N}$$

What I don't understand is the following (which has appeared before):
x-axis:
$$\tau A=0\rightarrow \textrm{Frictionless}$$
$$\cos (\alpha)h=a-b$$
$$\alpha :=\dfrac{45\pi}{180}=0.785$$

So the ##h## would be about the size of the 1, wouldn't it?

$$F_{p\rightarrow f(x)}=c^2\rho 1[a-b\cos (\alpha)h]=\tau A=0\rightarrow \textrm{Frictionless}$$is dimensionally inconsistent I believe.

erobz
Guillem_dlc said:
This is the whole statement of the example. I have given you the whole example. It doesn't say more. I guess they are asking you for the fluid wall strength.
Fluid wall strength? I'm not sure what that means. Perhaps how thick the jet is in each direction?

I've only encountered questions like: "What is the force to hold the vane given some mounting bracket geometry". It's typically an application of the "Momentum Equation" if fluid mechanics.

vanhees71
hutchphd said:
$$F_{p\rightarrow f(x)}=c^2\rho 1[a-b\cos (\alpha)h]=\tau A=0\rightarrow \textrm{Frictionless}$$is dimensionally inconsistent I believe.
But why is it ##0##?

erobz said:
Fluid wall strength? I'm not sure what that means. Perhaps how thick the jet is in each direction?

I've only encountered questions like: "What is the force to hold the vane given some mounting bracket geometry". It's typically an application of the "Momentum Equation" if fluid mechanics.
So how do you find this ##\alpha## thing?

The fluid wall is the force that the wall exerts on the fluid.

So as there is no friction, I understand that there is not enough tangential force and therefore it is zero.

But then that angle ##\alpha## I don't know what it is and how you find it.

Guillem_dlc said:
But why is it ##0##?
It's zero because there are no forces acting on the fluid jets in that direction. They state the vane is frictionless, correct? They are neglecting pressure change due to elevation change. That is how you get ##c_1 = c_2 = c_3## I believe. So when you are applying the Momentum Equation to the jet the LHS ( sum of the forces) is zero.

erobz said:
It's zero because there are no forces acting on the fluid jets in that direction. They state the vane is frictionless, correct? They are neglecting pressure change due to elevation change. That is how you get ##c_1 = c_2 = c_3## I believe. So when you are applying the Momentum Equation to the jet the LHS ( sum of the forces) is zero.
That's fine.

And what I said before about the ##\alpha##-angle, do you understand that?

Guillem_dlc said:
That's fine.

And what I said before about the ##\alpha##-angle, do you understand that?
##a = b## if ##c_1 = c_2 =c_3##. That is a necessity under the assumption of incompressible flow (continuity).

That is how you get ##0## on the LHS for that direction. The momentum of each stream cancels, and there are no other external forces.
Guillem_dlc said:
The fluid wall is the force that the wall exerts on the fluid.

The Normal Force in the direction of ##y## is acting on the jet ( the normal force acts independent of wall friction). You have to apply the momentum equation in the ##y## direction to determine the normal force.

Last edited:
erobz said:
##a = b## if ##c_1 = c_2 =c_3##. That is a necessity under the assumption of incompressible flow (continuity).

That is how you get ##0## on the LHS for that direction. The momentum of each stream cancels, and there are no other external forces.The Normal Force in the direction of ##y## is acting on the jet. You have to apply the momentum equation in the ##y## direction to determine the normal force.
I don't quite get it.

Guillem_dlc said:
I don't quite get it.
which part?

erobz said:
which part?
This part:

"The Normal Force in the direction of y is acting on the jet ( the normal force acts independent of wall friction). You have to apply the momentum equation in the y direction to determine the normal force."

You have two directions to apply the momentum equation. You have only computed the following:

$$\sum F_x = \cancel{ \frac{d}{dt} \int_{cv} v_x \rho d V\llap{-} }^0 + \sum \dot m_o v_{o_x} - \sum \dot m_i v_{i_x} = 0$$

You need to compute the momentum equation for the ##y## direction.

erobz said:
You have two directions to apply the momentum equation. You have only computed the following:

$$\sum F_x = \cancel{ \frac{d}{dt} \int_{cv} v_x \rho d V\llap{-} }^0 + \sum \dot m_o v_{o_x} - \sum \dot m_i v_{i_x} = 0$$

You need to compute the momentum equation for the ##y## direction.
Ok yes

But how do you know ##a=b##?

Guillem_dlc said:
Ok yes

But how do you know ##a=b##?

Because of continuity if ##c_1 = c_2 = c_3 = c ## implies ##A_1 c_1 = A_2 c_2 = A_3 c_3##.

$$w a c = w b c \implies a = b$$

erobz said:
Because of continuity if ##c_1 = c_2 = c_3 = c ## implies ##A_1 c_1 = A_2 c_2 = A_3 c_3##.

$$w a c = w b c \implies a = b$$
But 2 and 3 add up, don't they?

Guillem_dlc said:
But 2 and 3 add up, don't they?
I'm not following. What do you mean "2 and 3 add up" If you are talking about the velocities, No. If you are talking about the volumetric flowrates, Yes.

erobz said:
I'm not following. What do you mean "2 and 3 add up" If you are talking about the velocities, No. If you are talking about the volumetric flowrates, Yes.
I would say it would be ##A_1c_1=A_2c_2+A_3c_3##

Guillem_dlc said:
I would say it would be ##A_1c_1=A_2c_2+A_3c_3##
Yeah, that's good. What's the problem?

erobz said:
Because of continuity if ##c_1 = c_2 = c_3 = c ## implies ##A_1 c_1 = A_2 c_2 = A_3 c_3##.
But then this is without the areas, isn't it?

Guillem_dlc said:
But then this is without the areas, isn't it?
Oh yeah, I goofed there.

$$Q = A_1c_1 = A_2 c_2 + A_3 c_3$$

I guess we can only say at the moment:

$$b+a = const.$$

Which is the result you get after applying momentum equation in the ##x## direction.

Last edited:
erobz said:
Oh yeah, I goofed there.

$$Q = A_1c_1 = A_2 c_2 + A_3 c_3$$

I guess we can only say at the moment:

$$b+a = const.$$

Which is the result you get after applying momentum equation in the ##x## direction.
Yes, that's why you have to do the finding ##\alpha## thing.

And I don't know where you get ##\dfrac{45\pi}{180}##.

Guillem_dlc said:
Yes, that's why you have to do the finding ##\alpha## thing.

And I don't know where you get ##\dfrac{45\pi}{180}##.
When you say ##\alpha## you mean ##\theta## as labeled in the diagrams?

erobz said:
When you say ##\alpha## you mean ##\theta## as labeled in the diagrams?
That's what I don't know, this is what appears in the notes.

I suppose so.

Yeah, I've got to be honest. I'm coming up short on producing a solution for ##N## without the variable ##\theta## (the angle of the vane).

Your, "notes" imply it's a ##45^{\circ} ## angle. I don't think they solved for that.

erobz said:
Yeah, I've got to be honest. I'm coming up short on producing a solution for ##N## without the variable ##\theta## (the angle of the vane).

Your, "notes" imply it's a ##45^{\circ} ## angle. I don't think they solved for that.
Where does it say it is ##45##?

Where do we use it I mean

Guillem_dlc said:
Where does it say it is ##45##?

Where do we use it I mean
You notes say ##\frac{45 \pi}{180}## that is a ## 45^{\circ}## in radians?

You need ##\theta## it to find the normal force acting on the jet from the vane...unless I'm missing something.

It arises from simultaneously solving the momentum equations in each direction.

erobz said:
You notes say ##\frac{45 \pi}{180}## that is a ## 45^{\circ}## in radians?

You need ##\theta## it to find the normal force acting on the jet from the vane...unless I'm missing something.

It arises from simultaneously solving the momentum equations in each direction.
I don't know, it doesn't explain anything well.

The normal force is:

$$N = \rho A_1 c^2 \sin \theta$$

erobz said:
The normal force is:

$$N = \rho A_1 c^2 \sin \theta$$
Ok perfect thank you very much!

Guillem_dlc said:
Ok perfect thank you very much!
I'm not really sure what's going on. How they determine ##\theta = 45^{\circ}## is a bit of a mystery I feel we haven't solved yet. Sorry, If I didn't turn out to be of much help.

erobz said:
I'm not really sure what's going on. How they determine ##\theta = 45^{\circ}## is a bit of a mystery I feel we haven't solved yet. Sorry, If I didn't turn out to be of much help.
Don't worry, no problem.

I suppose that in an exercise, as soon as it is not an example, everything will be better placed.

erobz
What book is this example from?

Frabjous said:
What book is this example from?
It isn't from a book.

Replies
2
Views
1K
• Classical Physics
Replies
6
Views
309
• Electromagnetism
Replies
2
Views
843
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
30
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
• Set Theory, Logic, Probability, Statistics
Replies
27
Views
3K