Fluid Flow: Water Pipe with Vertical Pipes

[JohanM]

Homework Statement

Consider a water pipe that tapers down from a diameter dA= 5.0 cm at
end A to a diameter dB = 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 as shown in the Figure. (Not to scale, the pipes are much taller than
shown.) Assume that water flows through the pipe at high enough pressure that the vertical pipes
1 and 2 are partially filled with water. You may take g=10 m/s²http://uploadnow.org/image/328813-physics.jpeg

(a) In which pipe will the water level be higher or will it be at the same height in both pipes?
Explain. Does this depend on the direction of the water flow?

(b) If the water enters the pipe at point A with a velocity vA=2.0 m/s, what is its velocity when it
exits at point B?

(c) If the water enters the pipe at point A with a velocity vA=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

A₁v₁=A₂v₂
Bernoulli's Equation
P=ρgh

The Attempt at a Solution

(a) My answer is that because end A will have higher pressure than end B due to Bernoulli's principle and the equation above, the water level in column 1 will be higher than column 2.
I'm just not entirely sure that it's correct.

(b) A₁v₁=A₂v₂
v_A=2.0 m/s
r_A=0.025 m
r_B=0.0125 m

v_B=πr_A² * v_A / (πr_B²)
= 8.0 m/s

(c) This is the answer I'm most uncertain about.
From Bernoulli's equation, I derived
(y_A2-y_A1)=h₁=[(P_o-P₁) - (1/2)*ρv_A²]/(ρg)
and
(y_B2-y_B1)=h₂=[(P_o-P₂) - (1/2)*ρv_B²]/(ρg)
and
P₁=P₂+ρgΔy

where
P₀ = atmospheric pressure above water in vertical pipe
P₁ = pressure right below vertical pipe #1
P₁ = pressure right below vertical pipe #2
Δy = difference in height between A and B

I then found the difference in height between the two openings, A and B, to be 0.4625,
and, still assuming that column #1 has a higher water level, subtracted h₂ from h₁:

h₁-h₂ = [[(P_o-P₂ - ρg(0.4625)) - (1/2)*ρv_A²] - [(P_o-P₂) - (1/2)*ρv_B²]] / (ρg)

=[[- ρg(0.4625) - (1/2)*ρv_A²] - [-(1/2)*ρv_B²]] / (ρg)

=[(1/2)*v_B² - (1/2)*v_A² - g(0.4625)] / gPlugging in
v_A=2.0
v_B=8.0
g=10

I get Δh=2.5375 m

As I said, I'm very uncertain about the way I approached this last part, so help would be greatly appreciated!

--Johan

 

[edgepflow]

I think part (c) is a little off. You should get a nice clean expression for the column height difference.

Try writing Bernoulli's equation: Pa + (1/2)ρva^2 = Pb + (1/2)ρvb^2

Solve for Pa - Pb and note this is equal to ρg(h1 - h2).

 

[JohanM]

edgepflow: Could you explain how Pb-Pa=ρg(h1 - h2)?
Solving for Pb-Pa in Bernoulli's equation, I get
Pb-Pa=(1/2)ρ(va^2-vb^2)

I'm not sure how this ties to the height of the water in the columns. To me it seems to only relate the velocities, pressures, and heights of the two ends, A and B.

 

[edgepflow]

edgepflow: Could you explain how Pb-Pa=ρg(h1 - h2)?
Solving for Pb-Pa in Bernoulli's equation, I get
Pb-Pa=(1/2)ρ(va^2-vb^2)

I'm not sure how this ties to the height of the water in the columns. To me it seems to only relate the velocities, pressures, and heights of the two ends, A and B.

The expression Pa-Pb=ρg(h1 - h2) is basically the same as P=ρgh you listed in Relevant Equations. It relates static pressure difference in a fluid column (Pa - Pb) to the height difference of the fluid column (h1 - h2). So the value of Pb - Pa is the pressure differerence due to the velocity difference. And now you can find h1 - h2.

 

[JohanM]

edgepflow: Thanks! It's so much clearer now, I appreciate your quick response.

P_A-P_B=(1/2)ρ(v_B²-v_A²)
P_A-P_B=ρg(h₁-h₂)
(1/2)ρ(v_B²-v_A²) = ρg(h₁-h₂)
(h₁-h₂)=Δh=(v_B²-v_A²)/(2g)
Δh=3.0 m

EDIT: I'm assuming my answers to (a) and (b) are correct since you didn't comment on them. Is that right?