Need help with finding velocity of a liquid exiting from a pipe

[yang09]

Homework Statement

Part 1:
A liquid of density 1392 kg/m3 flows with speed 1.94m/s into a pipe of diameter 0.28m.
The diameter of the pipe decreases to 0.05m at its exit end. The exit end of the pipe is
4.2 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.4 atm. What is the velocity v2 of the liquid flowing out of the exit end of the pipe? Assume the
viscosity of the fluid is negligible and the fluid is incompressible. The acceleration of gravity is 9.8 m/s2 and Patm = 1.013 × 105 Pa.
Answer in units of m/s.

Part 2:
Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe?
Answer in units of Pa.

Homework Equations

Part 1:
A1V1 = A2V2

Part 2:
P1 + .5(ρ)(v1^2) + (ρ)(g)(y1) = P2 + .5(ρ)(v2^2) + (ρ)(g)(y2)

The Attempt at a Solution

Part 1:
I used the equation A1V1 = A2V2. For the area, I used A = ∏ r^2. My A1V1 was (∏)(.14^2)(1.94m/s). My A2 was (∏)(.025^2). I got an answer, 60.8384 m/s, but that seems wrong. It can't be going that fast. What am I doing wrong? Am I supposed to use another equation or something?

Part 2:
I know how to use Bernoulli's Equation, but just a quick clarification. The y1 would be 0 right since pipe 1 is at a height of 0 since its where the liquid starts? And the y2 would be 4.2 m since pipe 2 is where the water exits from.

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[vela]

Homework Statement

Part 1:
A liquid of density 1392 kg/m3 flows with speed 1.94m/s into a pipe of diameter 0.28m.
The diameter of the pipe decreases to 0.05m at its exit end. The exit end of the pipe is
4.2 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.4 atm. What is the velocity v2 of the liquid flowing out of the exit end of the pipe? Assume the
viscosity of the fluid is negligible and the fluid is incompressible. The acceleration of gravity is 9.8 m/s2 and Patm = 1.013 × 105 Pa.
Answer in units of m/s.

Part 2:
Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe?
Answer in units of Pa.

Homework Equations

Part 1:
A1V1 = A2V2

Part 2:
P1 + .5(ρ)(v1^2) + (ρ)(g)(y1) = P2 + .5(ρ)(v2^2) + (ρ)(g)(y2)

The Attempt at a Solution

Part 1:
I used the equation A1V1 = A2V2. For the area, I used A = ∏ r^2. My A1V1 was (∏)(.14^2)(1.94m/s). My A2 was (∏)(.025^2). I got an answer, 60.8384 m/s, but that seems wrong. It can't be going that fast. What am I doing wrong? Am I supposed to use another equation or something?

Other than the number of sig figs, your answer looks fine.

Part 2:
I know how to use Bernoulli's Equation, but just a quick clarification. The y1 would be 0 right since pipe 1 is at a height of 0 since it's where the liquid starts? And the y2 would be 4.2 m since pipe 2 is where the water exits from.

You're free to choose where you say y=0. One possibility is to say y=0 at the entrance to the pipe, as you have done. The exit is 4.2 m lower than the entrance, so y₂ should be negative, right?

 

[yang09]

But 60.8384 is wrong. I entered it and it says it is incorrect.

For part 2, I don't think y2 would be negative because the height differs by an amount of 4.2m. But I'm not sure, what do you think?

 

[vela]

What exactly did you enter? If you entered too many or too few digits, the system will probably mark your answer as wrong. How many significant figures should your answer have?

To answer your other question, how would you change your calculation if the pipe's exit was 4.2 m above the entrance? Are the speeds the same because the height difference is the same in both cases, as you seem to imply?

 

[yang09]

I'm using Quest and I enter all numbers from my calculator. I first entered 60.8384 and then entered 60.83840001. They both wrong answers.

If the pipe was 4.2 m above, then the y2 would still be 4.2 meters. The value of y is just the height difference between the entrance and the exit pipes. But the speed of the fluid would be different at each point. The speed at the exit is going to be faster than the initial

 

[vela]

I'm using Quest and I enter all numbers from my calculator. I first entered 60.8384 and then entered 60.83840001. They both wrong answers.

I suggest you review the concept of significant figures.

If the pipe was 4.2 m above, then the y2 would still be 4.2 meters. The value of y is just the height difference between the entrance and the exit pipes. But the speed of the fluid would be different at each point. The speed at the exit is going to be faster than the initial

So you're saying if you enter exactly the same numbers into the same equation, you'll get different answers for the two cases?

 

[yang09]

I'm not exactly understanding what you're saying. Can you explain yourself again.

 

[vela]

Sorry, I was thinking of a slightly different problem, so my questions didn't make sense in the context of this problem.

Suppose you had a large tank filed with water to a depth of 1 m, and there's a hose connected to bottom of the tank. The other end of the hose is open to the atmosphere. If the hose is left lying on the ground so that the end of the hose is 1 meter below the surface of the water in the tank, the water will drain out at a certain rate which you can find by applying Bernoulli's equation between the surface of the water in the tank and the end of the hose. Now, suppose you were to hold the end of hose at a height of 2 meters so that it's 1 meter above the surface of the water in the tank. Nothing changes at the surface of the water between the two situations, and the pressure at the end of the hose is still 1 atm. So according to what you said, it will drain at exactly the same rate when you hold the hose up because there's the same 1 meter difference in height between the surface of the water and the end of the hose. Is that what really happens?

 

[yang09]

I'm still a bit hazy, but check if this is right.
If you were to hold the hose up so that it is 1 m above the surface of the water, then no water would drain out because the hose is facing upwards. Since its facing upwards, then there is no momentum for the water to drain through the hose resulting in no draining of the water.
Is this correct?

 

[vela]

Water won't drain because the end of the hose is above the surface of the water. I'm not sure if that's what you meant by "facing upward." The point is, it makes a big difference if the end of the hose is 1 meter above or 1 meter below the surface of the water in the tank.

So in your problem, you can't just blindly say that y₂=4.2 m because the difference in height is 4.2 m. You have to include a positive or negative sign to differentiate between the cases where the end of the pipe is 4.2 m above the entrance and 4.2 m below.

 

[yang09]

Thanks for all the help Vela. It turns out I was doing the problem right, but I just entered the answer into the wrong question. That's pretty stupid of me, but again, thanks for all the help.

 

[vela]

Don't you hate it when that happens?

 

[yang09]

I know. This has happened to me before many times so I'll have to do better next time. Because of this mistake, I already lost 2 points. But hey Vela, can you help me out on my other post. It's called "Need help with Pistons." I don't know where to start so can you give me some tips on where to start. Thanks again