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## 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/s

^{2}

http://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

A

_{1}v

_{1}=A

_{2}v

_{2}

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)**A

_{1}v

_{1}=A

_{2}v

_{2}

v

_{A}=2.0 m/s

r

_{A}=0.025 m

r

_{B}=0.0125 m

v

_{B}=πr

_{A}

^{2}* v

_{A}/ (πr

_{B}

^{2})

=

**8.0 m/s**

**(c)**This is the answer I'm most uncertain about.

From Bernoulli's equation, I derived

(y

_{A}

_{2}-y

_{A}

_{1})=h

_{1}=[(P

_{o}-P

_{1}) - (1/2)*ρv

_{A}

^{2}]/(ρg)

and

(y

_{B}

_{2}-y

_{B}

_{1})=h

_{2}=[(P

_{o}-P

_{2}) - (1/2)*ρv

_{B}

^{2}]/(ρg)

and

P

_{1}=P

_{2}+ρgΔy

where

P

_{0}= atmospheric pressure above water in vertical pipe

P

_{1}= pressure right below vertical pipe #1

P

_{1}= 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 h

_{2}from h

_{1}:

h

_{1}-h

_{2}= [[(P

_{o}-P

_{2}- ρg(0.4625)) - (1/2)*ρv

_{A}

^{2}] - [(P

_{o}-P

_{2}) - (1/2)*ρv

_{B}

^{2}]] / (ρg)

=[[- ρg(0.4625) - (1/2)*ρv

_{A}

^{2}] - [-(1/2)*ρv

_{B}

^{2}]] / (ρg)

=[(1/2)*v

_{B}

^{2}- (1/2)*v

_{A}

^{2}- g(0.4625)] / g

Plugging 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