Bernoulli/ Continuity Eq'n problem

[jamesbiomed]

First of all, thanks to all the PF mentors out here, especially TSny and pgardn, who have made physics doable and are helping me accomplish my dreams! Even when putting in the work its not easy to get all this stuff!

Homework Statement

A large water tank has an inlet pipe and an outlet pipe. The inlet pipe has a diameter of 3 cm and is 1 m above the bottom of the tank. The outlet pipe has a diameter of 9 cm and is 7 m above the bottom of the tank. A volume of 0.6 m3 of water enters the tank every three minutes at a gauge pressure of 1 atm.

(a) What is the velocity of the water in the outlet pipe?

Homework Equations

A1V1=A2V2

DV/Dt=AV (for both)

The Attempt at a Solution

I've done three practice problems, and gotten them all right. The difference between those and this one, is that the volume of water enters over three minutes, instead of one. So I think that's part of where I'm messing up.

My method:

Knowing DV/Dt = Av, I plugged and chugged:

.6/180=(.045^2)pi*v1=> v1=.524 m/s. This exact method worked before, but for this problem, the answer is .589 m/s. Does anyone see where I messed up?

 

[jamesbiomed]

Apparently this is a tough one :)

 

[TSny]

Hello Jamesbiomed. If the problem stated that the volume flow rate in the outlet pipe is the same as the volume flow rate in the inlet pipe then I think your answer would be correct. But it could be that the amount of water in the tank is changing. Then the outflow rate would not match the inflow rate. It doesn't seem to me that there is enough information given. I guess there are additional questions being asked in the problem besides part (a). Perhaps those additional parts give some hints as to any assumptions that you are supposed to make.

 

[jamesbiomed]

Hey TSny! The only other part is a follow up:

(b) What is the gauge pressure in the outlet pipe?

So the gauge pressure will be different (which would be true whether the volume flow rate is the same or not I think)

Like I said, in three practices, setting volume flow rate equal to AV gave me the correct answer. So I would guess it they mean the rate should be equal. Since I don't know initial "v" I don't see another way to go about it if not.

 

[Nickie]

Hi there! It's great to hear that you have been successful with similar problems before. It's important to recognize that each problem is unique and may require a different approach. In this case, it looks like you may have missed a key factor in the problem: the difference in height between the inlet and outlet pipes. This difference in height will affect the velocity of the water in the outlet pipe, as it will experience gravitational acceleration as it flows downwards.

To solve this problem, you can use the Bernoulli equation, which relates the pressure, density, and velocity of a fluid at two different points in a system. In this case, we can set the inlet point at the top of the inlet pipe and the outlet point at the top of the outlet pipe. The equation is as follows:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

Where P is the pressure, ρ is the density, v is the velocity, and h is the height.

Plugging in the given values, we get:

1 atm + (1/2)(1000 kg/m^3)(v1)^2 + (1000 kg/m^3)(9.8 m/s^2)(1 m) = 1 atm + (1/2)(1000 kg/m^3)(v2)^2 + (1000 kg/m^3)(9.8 m/s^2)(7 m)

Simplifying and solving for v2, we get:

v2 = √[(v1)^2 + 2gh]

Plugging in the value of v1 that you calculated (.524 m/s), we get:

v2 = √[(.524 m/s)^2 + 2(9.8 m/s^2)(7 m)] = .589 m/s

This matches the answer you were given, so it looks like you were on the right track! Just be sure to pay attention to all the given information in the problem and use the appropriate equations. Keep up the good work!