Why Is My Fluid Mechanics Solution Incorrect?

[iSimon]

Homework Statement

Task number 2.
The problem is that my solution is wrong but the lecturer didnt want to explain what is wrong, she insists on doing it by myself. But to be honest I'm not really good at it and I just need someone either explain it or solve or at least check it and tell me what is wrong there.

Homework Equations

The Attempt at a Solution

 

[Chestermiller]

What is the pressure at depth z below the surface? What is the direction of the pressure acting on the flat ends of the cylinder, and what is the direction of the pressure acting on the curved surface of the cylinder?

Chet

 

[iSimon]

I would really like to respond to that, but to be honest I don't know how. I didnt know I have to include anything about pressure there :/

 

[Chestermiller]

I would really like to respond to that, but to be honest I don't know how. I didnt know I have to include anything about pressure there :/

Well the forces on these surfaces are the result of the water pressure acting on them. Are you familiar with the equation p = p_o+ρgz, where p_o is the pressure at the air interface, ρ is the water density, and z is the depth?

 

[iSimon]

I am not, sooo...
P1=ρgz=1000*10*0.9=9000 kg/m3
P2=ρgz=1000*10*2.1=21000 kg/m3

I just don't know from where will I get p_o.

 

[iSimon]

Ummm I have problem with units now. p_o is F/s where F = P*s (s is the area of base)
F1= p1*s=qgh1*s ;
F2= p2*s=qgh2*s

F1 = 10179 kg/(m^2 * s^2)
F2 = 23751 kg/(m^2 * s^2)

V = (h2-h1)*s=(2.1-0.9)*1.131=1.3572

F = qgV= 1000kg/m^3 * 10 kg/s^2 * 1.3572m^2 = 13572

I made somewhere a mistake again. Units are totally wrong.

 

[Chestermiller]

I am not, sooo...
P1=ρgz=1000*10*0.9=9000 kg/m3
P2=ρgz=1000*10*2.1=21000 kg/m3

I just don't know from where will I get p_o.

These are the pressures at the axis of the cylinder at the two ends faces (over and above atmospheric pressure). But the pressure is not uniform on the two end faces. It is lower on parts of the faces that are shallower in depth, and higher on the parts of the faces that are deeper. Fortunately for you, the variations with depth cancel out, and you can use the pressure at the axis as the average pressure on the end faces. Knowing the diameter of the cylinder, what is the area of each end face? What is the direction of the pressure force on each end face?
(a) parallel to the axis of the cylinder
(b)perpendicular to the axis of the cylinder
(c) in the vertical direction or
(d) in some other direction?
What is the magnitude and direction of the force on each end face?
In terms of the length of the cylinder, what is the vector sum (resultant) of the forces on the end faces?

 

[iSimon]

Area is equal to 1.131 m^2.
Direction of pressure force on each face will be perpendicular to the axis of cylinder. And the force can be split into two forces each like I did in the 2nd picture.
Oh bloody hell.
Average pressure is 15000 kg/m^3.
I can't really imagine how the forces vectors should look like. Force on the bottom of cylinder will be directed upwards and the one from top face will be directed opposite. Vector sum will be F= (p+ ρgH - p) = ρHAg

Is that finally it?

 

[Chestermiller]

Area is equal to 1.131 m^2.
Direction of pressure force on each face will be perpendicular to the axis of cylinder.

No. The pressure force on any surface is perpendicular to that surface. So the pressure force on each end face is parallel to the axis of the cylinder.

You got the average pressure on each of the two faces correct: 21000 Pa (check your previous units) and 9000 Pa. You got the areas of the two faces correct, so what are the magnitudes of each of the forces on the two faces? What is the net of these two forces, and which way is it pointing:
(a) up and to the left
(b) down and to the right
(c) some other direction
[/QUOTE]

 

[iSimon]

Magnitudes at each face is:
F1 = 10179 kg/(m^2 * s^2)
F2 = 23751 kg/(m^2 * s^2)

And the net of these two forces is... I think that the bottom Force is pointed up and left. And the top force is down and right.

 

[Chestermiller]

Magnitudes at each face is:
F1 = 10179 kg/(m^2 * s^2)
F2 = 23751 kg/(m^2 * s^2)

And the net of these two forces is... I think that the bottom Force is pointed up and left. And the top force is down and right.

Those units should be Newtons. You are correct about the directions of the two forces. Which one wins out, and, numerically, by how much?

 

[iSimon]

The lower one has greater value, and the resultant will be 13 572 N.

 

[iSimon]

And how to calculate the force on the curved surface?

 

[Chestermiller]

The lower one has greater value, and the resultant will be 13 572 N.

Excellent. So the net resultant force from the two end faces is 13572 N, and it is up and to the left, parallel to the axis of the cylinder.

Now let's turn attention to determination of the resultant pressure force on the curved face of the cylinder. There are two methods by which we can determine this force.
1. Integrate the pressure vectorially over the area of the surface.
2. Use Archimedes principle to determine the total force vector exerted by the water on the cylinder, and then vectorially subtract the forces on the end faces.

Which method do you prefer? (One of these methods is easy, and the other is hard)

Chet

 

[iSimon]

I think the 2nd one may seem more complicated but calculating integral is beyond my reach. We didnt have integrals yet.

 

[Chestermiller]

I think the 2nd one may seem more complicated but calculating integral is beyond my reach. We didnt have integrals yet.

Method 2 is the simpler method. From Archimedes principle, what is the overall force exerted by the water on the cylinder, and what is its direction?

Chet

 

[iSimon]

The force is directed upwards and its value is equal to the weight of our cylinder. Which is ummm, 13572 * sin30 degree?

 

[Chestermiller]

The force is directed upwards and its value is equal to the wights of our cylinder. Which is ummm, 13572 * sin30 degree?

You're right about the direction, but not the magnitude. The magnitude of the force is equal to the weight of the displaced volume of water (which is equal to the volume of the cylinder). What is the volume of the cylinder? What is the density of water? What is the mass of the displaced water? What is the weight of the displaced water?

Chet

 

[iSimon]

Volume of cylinder is V=3.14*(0.6^2) * 2.4 = 2.714m^3
Density of water is 1000kg/m^3
mass of water will be the the same as volume of cylinder so 2.714kg also.
And to be honest my english isn't that good, mass and weight? Isnt that the same? :/So it should be: N= g*m*sin30= 2.714 kg * sin30 * 10 m/s^2 = 13.57 N?

 

[Chestermiller]

Volume of cylinder is V=3.14*(0.6^2) * 2.4 = 2.714m^3
Density of water is 1000kg/m^3
mass of water will be the the same as volume of cylinder so 2.714kg also.

No. The mass of the water will be the volume times the density = 2714 kg

And to be honest my english isn't that good, mass and weight? Isnt that the same? :/

No. The weight is a force (in Newtons). It is equal to the mass times g (i.e., mg). (I think you may have learned that in your course). So, again, what is the weight of the displaced water? This will then be equal to the total upward force of the water on the cylinder.

Chet

 

[iSimon]

If weight is a force where mass is multiplied with g, then weight is equal to 27140 kg*m/s^2. 1kg*m/s^2 = 1N

 

[Chestermiller]

If weight is a force where mass is multiplied with g, then weight is equal to 27140 kg*m/s^2. 1kg*m/s^2 = 1N

OK. Good. So you have a total force of 27140 N acting upward on the cylinder, and you have a net force of 13572 N acting on the ends of the cylinder, and pointing up and to the left, parallel to the axis of the cylinder. So the total force has a unit vector in the vertical direction, and the force on the ends has a unit vector pointing up and to the left, parallel to the axis of the cylinder. Do you know how to resolve these forces into components in the vertical and horizontal directions? You will need to do this if you want to determine the difference between the overall force and the net force on the ends, since this difference must be the force on the curved surface.

Chet

 

[iSimon]

To determine vertical force I will have to multiply 13572 * sin30 degree.
For horizontal its 13573 * cos30 degree.

 

[iSimon]

So the force acting on the curved surface is 27140 N + 13572 *sin30 = 33926 N?

 

[Chestermiller]

So the force acting on the curved surface is 27140 N + 13572 *sin30 = 33926 N?

No, you need to take the difference between the total force and the net force on the ends to get the force on the curved surface. In other words, the net force on the ends plus the force on the curved surface equals the overall force.

So, vertical component of curved surface force = 27104 -13572 * sin 30 = 20318 upward
Horizontal component of curved surface force = 13572 * cos 30 =11753 to the right.

Relative to the axis of the cylinder, what direction do you think this force is pointing?
(a) parallel to the axis
(b) perpendicular to the axis
(c) some other direction.

Chet

 

[iSimon]

The force that is pointing at curved surface is upward and right. The force is placed somewhere on the curved surface, up and right. Its perpendicular to the axis of cylinder.

 

[Chestermiller]

The force that is pointing at curved surface is upward and right. The force is placed somewhere on the curved surface, up and right. Its perpendicular to the axis of cylinder.

Excellent! Excellent! Excellent!

With regard to the placement of the force: It is not possible to identify any specific location at which the resultant force is applied on the curved surface. The force is distributed over the entire curved surface, and at each location on the curved surface, the force per unit area (pressure) is pointing radially inward toward the axis of the cylinder. But the pressure is not uniform around the circumference of the cylinder. This gives rise to the resultant force up and to the right.

Chet

 

[iSimon]

I am really sorry, I just consulted this task. My lecturer told me it's wrong again. Gave me some explanation which to be honest is quite helpful but I will need to draw it. I will manage to do it later when I get home :)

 

[Chestermiller]

I am really sorry, I just consulted this task. My lecturer told me it's wrong again. Gave me some explanation which to be honest is quite helpful but I will need to draw it. I will manage to do it later when I get home :)

I must have mis-interpreted the question (it was a little ambiguous with regard to the figure). For the interpretation that we used, I am confident that we got the correct answer.

Chet

 

[iSimon]

I struggled with replying here because its hard to explain what my teacher wants me to do but again, I am not really able to do it myself. And I don't quite understand her demands. Maybe you guys are going to help me. This is what she said:
I should try to project lines from top and bottom of radius line and project them to the right, do the same with top and bottom of whole cylinder to receive figure I tried to make on computer. Its a composition of rectangular and 2 elipses. Then from trygonometry I should receive heights of this new figure. But I don't really understand how can I get values of forces exerted by water.
Any ideas? Please?