Fluid Dynamics Water Pipe

Gear300

Through a pipe 15.0cm in diameter, water is pumped from the Colorado River up to Grand Canyon Village, located on the rim of the canyon. The river is at an elevation of 564m, and the village is at an elevation of 2096m (a) What is the minimum pressure at which the water must be pumped if it is to arrive at the village? (b) If 4500m^3 are pumped per day, what is the speed of the water in the pipe? (c) What additional pressure is necessary to deliver this flow?

(Assume that the freefall acceleration and the density of air are constant over this range of elevations).

I can get (a) and (b) easily. For (a) the minimum pressure would be 1atm + the pressure necessary to balance the weight of the water in the pipe. For (b), one can use the equation of continuity to find the speed. (c) is what gets me. I'm supposing its asking the question in respect to (b)...but isn't the rate of volume flow and speed explicitly independent of the pressure: (PV)/t = p[power] = Fv, P*(V/t) = Fv, P*(V/t)/v = F, P*A = F, P = F/A....so then the speed and volume flow can be a varying value under the same pressure. How would I interpret what (c) is asking?
The answer to (a) is ~1atm + 15.0MPa, (b) is ~2.95m/s, and (c) is 4.34kPa

Related Introductory Physics Homework Help News on Phys.org

alphysicist

Homework Helper
Hi Gear300,

From the equations of fluid flow, we can know two things: the speed is related to the volume flow rate (that's how the equation of continuity is set up), and the speed is related to the pressures (from Bernoulli's equation). So from part b, you used the term from the equation of continuity to go from volume flow rate to speed; for c you can use Bernoulli's equation to go from speed to pressure.

For your argument, let me accept the equations as they are and just talk about your chain of reasoning. I don't think your equations lead to the conclusion you gave. I believe your final equation is indicating that the ratio of pressure to force is independent of any particular value of speed you might have. In other words, if the area is constant, then the ratio of P/F is a constant (equal to 1/A), no matter what else is going on in the problem. That does not mean, however, that the particular value of P is independent of whatever else is in the problem.

So if v=20m/s, that gives one value of P, and P=F/A; if v=30m/s, that gives another value of P, but P still equals F/A (it's just a different F).

Gear300

Hello alphysicist. I see, I see...I forgot to take into account the values for force...and in Bernoulli's equation there is a particular relation between pressure and speed (and it even explicitly shows it...funny how I overlooked that). Thanks for the help.

Gear300

Wait...I just took a whack at the problem. I'm still stuck. Since the pipe is taken to have a constant diameter, then the speed of the water should be constant throughout the pipe.
Po + K/V + Uo/V = Pf + K/V + Uf/V, in which K is kinetic energy, U is gravitational potential, Po is pressure applied at bottom and Pf is pressure at top (atmospheric pressure). At the bottom, there is a pressure Po and U can be taken to be 0...so
Po + K/V = Pf + K/V + U/V, and since the velocity is constant, Po = Pf + U/V. The value for Po here apparently comes out to the same value as the one for part (a)...which would imply 0Pa of additional pressure (which contradicts the given answer). Any help?

alphysicist

Homework Helper
Wait...I just took a whack at the problem. I'm still stuck. Since the pipe is taken to have a constant diameter, then the speed of the water should be constant throughout the pipe.
Po + K/V + Uo/V = Pf + K/V + Uf/V, in which K is kinetic energy, U is gravitational potential, Po is pressure applied at bottom and Pf is pressure at top (atmospheric pressure). At the bottom, there is a pressure Po and U can be taken to be 0...so
Po + K/V = Pf + K/V + U/V, and since the velocity is constant, Po = Pf + U/V. The value for Po here apparently comes out to the same value as the one for part (a)...which would imply 0Pa of additional pressure (which contradicts the given answer). Any help?
Right, that's a bit tricky to work through. But the way I'm reading the problem, we are not supposed to apply Bernoulli's equation to two ends of the pipe.

One side of Bernoulli's equation is applied to the river water, and the other side of the equation is applied to the top of the pipe. Then the initial velocity would be zero.

So the idea is, for part b, we were asking what pressure (work per volume) is required to give a certain potential energy per volume to the water. And for part c, we are asking what additional pressure is required to give a certain additional kinetic energy/volume to the water.

Does that make sense?

Gear300

Right, that's a bit tricky to work through. But the way I'm reading the problem, we are not supposed to apply Bernoulli's equation to two ends of the pipe.

One side of Bernoulli's equation is applied to the river water, and the other side of the equation is applied to the top of the pipe. Then the initial velocity would be zero.

So the idea is, for part b, we were asking what pressure (work per volume) is required to give a certain potential energy per volume to the water. And for part c, we are asking what additional pressure is required to give a certain additional kinetic energy/volume to the water.

Does that make sense?
Oh...I understand now...the pressure needed to give the water the boost in speed. Its been solved at last...thanks for all the help.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving