The discussion focuses on calculating the resultant force at an elbow in a horizontal pipe with water flow at a 135-degree angle. The flow rate is 0.4 m³/s, and the elbow weighs 12 kg. The x-component of the force was calculated as 19170 N to the left, while the y-component was determined to be 1711 N upwards, resulting in a total resultant force of 19246 N. Participants emphasized the importance of correctly applying trigonometric functions and the Bernoulli equation in the calculations.
PREREQUISITESEngineers, fluid mechanics students, and anyone involved in the design or analysis of piping systems will benefit from this discussion, particularly those focusing on fluid flow dynamics and force calculations in elbows.
I interpreted the problem statement as saying that the entire section of pipe was horizontal, so that the weight of the water would be in the vertical z direction.foo9008 said:yes , for (cos-135), i would get -0.707 , since there is also negative sign in front of 90x10^3 , -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) , i got positive 1999N
how about the weight of water , should it be included in the calculation ? -1000(9.81)(0.2) ?
you mean the pipe is placed in x( left to right ) and y direction(inside or outside the book) only ?Chestermiller said:I interpreted the problem statement as saying that the entire section of pipe was horizontal, so that the weight of the water would be in the vertical z direction.
How do you feel about the approach I used in terms of unit vectors? Do you feel that it would make things easier, or no particular advantage?
I don't know what you mean. What I'm saying is that, from the problem statement, I thought the centerline of the entire bend is horizontal.foo9008 said:you mean the pipe is placed in x( left to right ) and y direction(inside or outside the book) only ?
foo9008 said:you mean the pipe is placed in x( left to right ) and y direction(inside or outside the book) only ?
Chestermiller said:I don't know what you mean. What I'm saying is that, from the problem statement, I thought the centerline of the entire bend is horizontal.
so , 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) ?Chestermiller said:I don't know what you mean. What I'm saying is that, from the problem statement, I thought the centerline of the entire bend is horizontal.
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) ?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 ?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.
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?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
so the equation should beChestermiller 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
sorry , i still duno what are you talking about , can you explain it in another way ?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'm out of ideas on how to explain it. It's purely math.foo9008 said:sorry , i still duno what are you talking about , can you explain it in another way ?