Finding centerline speed of a flow through a nozzle

  • #1
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Homework Statement


finding an equation for centerline speed.png



Homework Equations



u=a+ b(x-c)^2

We have two boundary conditions at x=0, u=u(entrance) and x=L, u=u(exit)

Source: Fluid Mechanics by Çengel/Cimbala








The Attempt at a Solution



I cannot understand the parabolic equation in the x-direction.
 

Answers and Replies

  • #2
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relevant equation.png


This is the original solution but it is still hard to understand for me. First I cannot understand general parabolic equation as 1 and why is cet set to 0, a=uentrance and b=(uexit-uentrance)/L^2.

Would you like to explain them?

Thank you.
 
  • #3
SteamKing
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View attachment 99692

This is the original solution but it is still hard to understand for me. First I cannot understand general parabolic equation as 1 and why is cet set to 0, a=uentrance and b=(uexit-uentrance)/L^2.

Would you like to explain them?

Thank you.
Did you set x = 0? What speed did you obtain?

Did you set x = L? What speed did you obtain there?

What about some intermediate location, say x = L/2? What is the speed there?
 
  • #4
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for x=0, u=a+b*(0-c)^2, u=a+b*(c^2)

for x=L, u=a+b*(L-c)^2, u=a+b*(L^2 -2Lc + c^2)

for x=L/2, u=a+b*(L/2 - c)^2, u=a+ b*(L^2 /4 - Lc +c^2)
 
  • #5
SteamKing
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for x=0, u=a+b*(0-c)^2, u=a+b*(c^2)

for x=L, u=a+b*(L-c)^2, u=a+b*(L^2 -2Lc + c^2)

for x=L/2, u=a+b*(L/2 - c)^2, u=a+ b*(L^2 /4 - Lc +c^2)
No, you use Eq. (2) and make the substitutions there. Eq. (1) is just a generic parabola.
 
  • #6
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Meanwhile, would you like to guide me how I can use Latext more effecitively? For example, a link.

Thank you.
 
Last edited:
  • #7
SteamKing
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