Resultant force at elbow from water flow

In summary: As a summary, the conversation discusses a problem involving water flowing in an elbow that is aligned horizontally at an angle of 135 degrees. The water volume in sections 1 and 2 is 0.2m^3, the elbow weighs 12kg, and the flow rate is 0.4(m^3)/s. The task is to find the resultant force on the elbow. The equations used include continuity, weight = mg, and momentum balance. The solution results in a force of 19246N to the left and 1711N upwards. However, there is some confusion regarding the direction of the resultant force. The expert also provides advice on using TeX for easier readability and encourages the student to solve the problem themselves
  • #36
Take the pipe bend (not the picture in the book, the actual physical pipe bend) and lay it flow on your kitchen table. That's what I'm saying its orientation should be, according to the problem statement.
 
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  • #37
Chestermiller said:
Take the pipe bend (not the picture in the book, the actual physical pipe bend) and lay it flow on your kitchen table. That's what I'm saying its orientation should be, according to the problem statement.
ok , i got the orientation ...for the forces along the y -axis , it should be 0.707(90x10^3)(pi)[(200x10^-3 /2 )^2 ] + Fy = -0.707(1000)(0.4)(12.7) , Fy = 5591N pointing outwards of the book) ?
 
  • #38
Chestermiller said:
Take the pipe bend (not the picture in the book, the actual physical pipe bend) and lay it flow on your kitchen table. That's what I'm saying its orientation should be, according to the problem statement.
is my answer above correct ?
 
  • #39
Chestermiller said:
Here's how I would set up this problem using unit vectors:

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 1 is ##\vec{i}_x##

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 2 is ##\frac{\vec{i}_x}{\sqrt{2}}+\frac{\vec{i}_y}{\sqrt{2}}=\frac{(\vec{i}_x+\vec{i}_y)}{\sqrt{2}}=0.707(\vec{i}_x+\vec{i}_y)##

The pressure force exerted on the fluid in the control volume by the fluid behind it (at cross section 1) is ##P_1A_1\vec{i}_x##

The pressure force exerted on the fluid in the control volume by the fluid ahead of it (at cross section 2) is ##P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)##

The force that the pipe bend exerts on the fluid in the control volume is ##(F_x\vec{i}_x+F_y\vec{i}_y)##, where ##F_x## is the component of the force in the +x direction, and ##F_y## is the component of the force in the +y direction.

The rate of momentum entering the control volume is ##\rho Qv_1\vec{i}_x##

The rate of momentum exiting the control volume is ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)##

The rate of change of momentum of the fluid in the control volume = ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x##

So, the momentum balance on the fluid in the control volume is:
$$P_1A_1\vec{i}_x+P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)+(F_x\vec{i}_x+F_y\vec{i}_y)=-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x$$
So the component of the momentum balance in the x direction is:
$$P_1A_1+(0.707)P_2A_2+F_x=-(0.707)\rho Qv_2-\rho Qv_1$$
And the component of the momentum balance in the y direction is:
$$(0.707)P_2A_2+F_y=-(0.707)\rho Qv_2$$

Chet
Chestermiller said:
Here's how I would set up this problem using unit vectors:

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 1 is ##\vec{i}_x##

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 2 is ##\frac{\vec{i}_x}{\sqrt{2}}+\frac{\vec{i}_y}{\sqrt{2}}=\frac{(\vec{i}_x+\vec{i}_y)}{\sqrt{2}}=0.707(\vec{i}_x+\vec{i}_y)##

The pressure force exerted on the fluid in the control volume by the fluid behind it (at cross section 1) is ##P_1A_1\vec{i}_x##

The pressure force exerted on the fluid in the control volume by the fluid ahead of it (at cross section 2) is ##P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)##

The force that the pipe bend exerts on the fluid in the control volume is ##(F_x\vec{i}_x+F_y\vec{i}_y)##, where ##F_x## is the component of the force in the +x direction, and ##F_y## is the component of the force in the +y direction.

The rate of momentum entering the control volume is ##\rho Qv_1\vec{i}_x##

The rate of momentum exiting the control volume is ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)##

The rate of change of momentum of the fluid in the control volume = ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x##

So, the momentum balance on the fluid in the control volume is:
$$P_1A_1\vec{i}_x+P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)+(F_x\vec{i}_x+F_y\vec{i}_y)=-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x$$
So the component of the momentum balance in the x direction is:
$$P_1A_1+(0.707)P_2A_2+F_x=-(0.707)\rho Qv_2-\rho Qv_1$$
And the component of the momentum balance in the y direction is:
$$(0.707)P_2A_2+F_y=-(0.707)\rho Qv_2$$

Chet
can you explain why there is negative sign for pQv for x-component and y component forces?
 
  • #40
In my analysis, I call ##F_x## the component of the force exerted by the bend on the fluid in the positive x direction, and I call ##F_y## the component of the force exerted by the bend on the fluid in the positive y direction. It says so right in the quote you used.

Chet
 
  • #41
Chestermiller said:
In my analysis, I call ##F_x## the component of the force exerted by the bend on the fluid in the positive x direction, and I call ##F_y## the component of the force exerted by the bend on the fluid in the positive y direction. It says so right in the quote you used.

Chet
so the equation should be
$$P_1A_1+(0.707)P_2A_2+F_x=(0.707)\rho Qv_2-\rho Qv_1$$
And the component of the momentum balance in the y direction is:

$$(0.707)P_2A_2+F_y=(0.707)\rho Qv_2$$
with no negative sign for isn't it for ρQv ??
 
  • #42
I want you to understand that you are using the symbol ##F_x## to represent the component of the force in the negative x direction, and I am using the symbol ##F_x## to represent the component of the force in the positive x direction. We are using the same symbol for two different things. Your ##F_x## is equal to minus my ##F_x##. There is not one way that's right and one way that's wrong.

In your analysis, the force ##\vec{F}## is expressed as ##\vec{F}=-F_x\vec{i}_x-F_y\vec{i}_y##. In my analysis, the force ##\vec{F}## is expressed as ##\vec{F}=+F_x\vec{i}_x+F_y\vec{i}_y##. Do you see the difference?
 
  • #43
Chestermiller said:
I want you to understand that you are using the symbol ##F_x## to represent the component of the force in the negative x direction, and I am using the symbol ##F_x## to represent the component of the force in the positive x direction. We are using the same symbol for two different things. Your ##F_x## is equal to minus my ##F_x##. There is not one way that's right and one way that's wrong.

In your analysis, the force ##\vec{F}## is expressed as ##\vec{F}=-F_x\vec{i}_x-F_y\vec{i}_y##. In my analysis, the force ##\vec{F}## is expressed as ##\vec{F}=+F_x\vec{i}_x+F_y\vec{i}_y##. Do you see the difference?
sorry , i still duno what are you talking about , can you explain it in another way ?
 
  • #44
foo9008 said:
sorry , i still duno what are you talking about , can you explain it in another way ?
Sorry. I'm out of ideas on how to explain it. It's purely math.
 
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