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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) ?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 ?