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

[RobL14]

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

$$P + \rho gh + \frac{1}{2}\rho v^2 = C$$

$$A_1 v_1 = A_2 v_2$$

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.

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

$$P_A + \frac{1}{2}\rho v_A^2 = P_0 + \rho gh_A$$

$$P_B + \frac{1}{2}\rho v_B^2 = P_0 + \rho gh_B$$

From 1:

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

$$P_A - P_B = \frac{1}{2}\rho v_B^2 - \frac{1}{2}\rho v_A^2 + \rho gh_A - \rho gh_B$$

After solving:

$$0 = \rho g\left( {d_A - d_B } \right) - \rho g\left( {h_A - h_B } \right) \\$$

$$\left( {h_A - h_B } \right) = \left( {d_A - d_B } \right) = 2.5cm \\$$

So what am I misunderstanding or doing wrong here? I must not "get" this.

 

[Redbelly98]

From 1:

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

Looks good. Just go ahead and calculate P_A-P_B 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?