Help with a more difficult Fluid Dynamics Problem involving Bernoulli's equation

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SUMMARY

This discussion centers on a fluid dynamics problem involving Bernoulli's equation and the continuity equation. The problem involves a water pipe tapering from a diameter of 5.0 cm at point A to 2.5 cm at point B, with water entering at a velocity of 2.0 m/s. The user calculated the velocity at point B to be 8 m/s and attempted to find the height difference between two vertical pipes connected to the ends of the tapered pipe. The user concluded that the height difference (h_A - h_B) equals the diameter difference (d_A - d_B), resulting in a height difference of 2.5 cm, which raised questions about the validity of their assumptions.

PREREQUISITES
  • Understanding of Bernoulli's equation
  • Knowledge of the continuity equation in fluid dynamics
  • Familiarity with pressure differences in fluid systems
  • Basic concepts of hydrostatics and height differences in fluid columns
NEXT STEPS
  • Review the derivation and application of Bernoulli's equation in varying cross-sectional areas
  • Study the implications of pressure differences in fluid dynamics
  • Examine how to accurately apply the continuity equation in practical scenarios
  • Learn about the effects of height differences on fluid pressure and flow rates
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Students studying fluid dynamics, engineers working with hydraulic systems, and anyone needing to solve complex problems involving Bernoulli's equation and fluid flow in pipes.

RobL14
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Homework Statement



Consider a water pipe that tapers down from a diameter d_A = 5.0 cm at end A to a diameter d_B = 2.5 cm at end B. At each end a vertical pipe that is open to the air at the top is attached to the pipe. Assume that water flows through the pipe at high enough pressure that the vertical pipes 1 and 2 are partially filled with water.

If the water enters the pipe at point A with a velocity v_A = 2.0 m/s, what is the height difference between the levels in the two vertical pipes? If the difference is not zero, please indicate which pipe has the higher level.

Homework Equations



<br /> P + \rho gh + \frac{1}{2}\rho v^2 = C<br />

<br /> A_1 v_1 = A_2 v_2 <br />

The Attempt at a Solution



I already found using the continuity equation that v_b = 8 m/s.

In the first equation, I am applying Bernoulli's equation; the water begins in the wider pipe A and then flows into the thinner pipe B, which is raised a height (d_A - d_B) relative to pipe A.

In the second and third equations, I've applied Bernoulli's equation with the assumption that I know P_A and P_B, which I don't, but having three equations will let me cancel them. Using the first three equations, I can eliminate (P_A - P_B) and solve for (h_A - h_B). When I do that, though, I get that (h_A - h_B) = (d_A - d_B) = 2.5cm, which can't be right.

<br /> P_A + \frac{1}{2}\rho v_A^2 = P_B + \rho g\left( {d_A - d_B } \right) + \frac{1}{2}\rho v_B^2 \\ <br />

<br /> P_A + \frac{1}{2}\rho v_A^2 = P_0 + \rho gh_A<br />

<br /> P_B + \frac{1}{2}\rho v_B^2 = P_0 + \rho gh_B<br />

From 1:

<br /> P_A - P_B = \rho g\left( {d_A - d_B } \right) + \frac{1}{2}\rho v_B^2 - \frac{1}{2}\rho v_A^2 <br />From 2+3:

<br /> P_A - P_B = \frac{1}{2}\rho v_B^2 - \frac{1}{2}\rho v_A^2 + \rho gh_A - \rho gh_B <br />

After solving:

<br /> 0 = \rho g\left( {d_A - d_B } \right) - \rho g\left( {h_A - h_B } \right) \\ <br />

<br /> \left( {h_A - h_B } \right) = \left( {d_A - d_B } \right) = 2.5cm \\ <br />

So what am I misunderstanding or doing wrong here? I must not "get" this.
 
Last edited:
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RobL14 said:
From 1:

<br /> P_A - P_B = \rho g\left( {d_A - d_B } \right) + \frac{1}{2}\rho v_B^2 - \frac{1}{2}\rho v_A^2 <br />
Looks good. Just go ahead and calculate PA-PB from this.

Also, are you sure the two ends are at different heights? Does a figure you have clearly show that to be the case?
 

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