Force require to move water through a pipe?

[inveni0]

I'm writing the third in a series of Young Adult Sci-Fi/Fantasy novels, and I'm trying to calculate how much force would be required in the following:

Imagine a pipe that is 1.25 centimeters in diameter and 3000 meters long. The pipe lies perfectly flat, but a meter of the pipe at each end is turned upward at a 90 degree angle. Now, the pipe is filled completely with water so that the water is level with the tops of the upturned ends.

My question is, how much force would it take to drive a plunger down into one end and force half a meter of water out of the other? Is the mass of the water in the 3000 meter pipe taken into consideration? Or is only the mass of the water in the upturned ends considered? Could this be done by hand?

Thank you for any help you can give! This problem just contains more physics than I'm familiar with.

 

[Evanish]

I think it would depend on how fast you are driving the water out. If you are doing it slowly I think it would take a lot less force than if you are doing it quickly. If you are doing it very slowly, ignoring things like friction, I think the force would be almost equal to the weight of the displaced water. So at first not much force would be required, but as you displace more water the force needed would increase.

 

[jbriggs444]

Evanish is correct. If you are willing to plunge slowly, the 3000 meters of pipe in between is irrelevant. The needed force is the same as supporting lifting a column of water 1 meter in height. At an inside diameter of 1.25 centimeters, that's a bit over 1 kilogram mass -- a force of about 12 Newtons.

(area = pi r^2 which comes to about 1.22 cm^2. Multiply by 1000 for 1.2 kg per meter).

The pipe contains in the neighborhood of 3600 kilograms of water. Getting that water moving will take time. As a rough order of magnitude estimate, 12 Newtons divided by 3600 kilograms means that you'll start out with an acceleration of about three millimeters per second per second -- push down for one second and you've managed 1.5 millimeters and have achieved a velocity of 3 millimeters/second. After 10 seconds, you'd have managed 15 centimeters. After about 20 seconds you'd have the water where you wanted it, but it would be continuing to move, sloshing up out of the other end, even if you stopped pushing.

Trying to stop that moving water suddenly could be difficult. http://en.wikipedia.org/wiki/Water_hammer

 

[inveni0]

I understand why moving faster would require more force, but there are some other things I don't understand. For instance, wouldn't we use the formula for volume of a cylinder to figure out the mass of the water? (V = pi*r²h) If that's the case, wouldn't it be V = 3.142 * 0.0125² * 3000 = 1.42m³? If water weighs roughly 1000kg per m³, that would mean the water weighs about 1400kg. Is my math off?

Secondly, how fast would a human be able to push the water? Is there a formula for that? How is the acceleration of the water related to the force required to accelerate it (directly, squared, cubed, etc)?

 

[anorlunda]

How is the acceleration of the water related to the force required to accelerate it (directly, squared, cubed, etc)?

This is exactly the right place to ask that question, Newton's second law, f=ma, force equals mass times acceleration.

 

[inveni0]

This is exactly the right place to ask that question, Newton's second law, f=ma, force equals mass times acceleration.

Would friction in this scenario be minimal enough to discount? Over 3km, it seems like friction might play a large role through the horizontal length of the pipe.

 

[anorlunda]

Again, it depends on how fast you are talking about. Friction could be negligiable or it could be huge. Tell us, are you picturing this man plunging the plunger in 1/2 second or 5 seconds, or what?

 

[jbriggs444]

I understand why moving faster would require more force, but there are some other things I don't understand. For instance, wouldn't we use the formula for volume of a cylinder to figure out the mass of the water? (V = pi*r²h) If that's the case, wouldn't it be V = 3.142 * 0.0125² * 3000 = 1.42m³?

Fair question. Let's go through it slowly.

${\pi}r^2$ is the correct formula for area where r is the pipe radius. We've been given the pipe diameter, 1.25 cm. The radius is half that. 0.625 cm or 0.00625 meters. Square and multiply by pi and you have 1.22 square centimeters or 0.000122 square meters.

Multiply by 3000 meters and you have about 0.37 cubic meters.

Multiply by 1000 liters per cubic meter and that's 370 liters of volume.

Water is 1 kg per liter, so that's 370 kilograms.

Apparently my initial estimate of 3600 was out by a factor of 10. You were right to question it

 

[inveni0]

Again, it depends on how fast you are talking about. Friction could be negligiable or it could be huge. Tell us, are you picturing this man plunging the plunger in 1/2 second or 5 seconds, or what?

Let's say he needs to displace the water by moving the plunger 1/2 meter in 10 seconds.

Fair question. Let's go through it slowly.

${\pi}r^2$ is the correct formula for area where r is the pipe radius. We've been given the pipe diameter, 1.25 cm. The radius is half that. 0.625 cm or 0.00625 meters. Square and multiply by pi and you have 1.22 square centimeters or 0.000122 square meters.

Multiply by 3000 meters and you have about 0.37 cubic meters.

Multiply by 1000 liters per cubic meter and that's 370 liters of volume.

Water is 1 kg per liter, so that's 370 kilograms.

Apparently my initial estimate of 3600 was out by a factor of 10. You were right to question it

I see one of the discrepancies is that I said 1.25cm diameter yet I was dumbly plugging that in as radius.

 

[anorlunda]

OK, 0.5 meters in 10 seconds. With .00625 square meter cross section, that is a flow rate of 0.0003125 cubic meters per second. If you assume constant acceleration, then the final flow rate will be twice the average flow rate. You can use ths to calculate the acceleration.

Now, unless the man's feet are strapped to the ground, the maximum force he can put on a plunger is his own weight. That force divided by the cross sectional area is the pressure at the plunger.

Now do a Google search on pipe loss calculator. Use it to calculate the pressure loss at the average flow rate. Compare that with the pressure at the plunger and you can see how significant the friction is. (Friction loss is not linear with flow, but you can probably ignore that.)

We are not going to do the work for you. But you seem to be catching on quickly, so I think you will succeed. Post again if you get stuck.

 

[inveni0]

OK, 0.5 meters in 10 seconds. With .00625 square meter cross section, that is a flow rate of 0.0003125 cubic meters per second. If you assume constant acceleration, then the final flow rate will be twice the average flow rate. You can use ths to calculate the acceleration.

Now, unless the man's feet are strapped to the ground, the maximum force he can put on a plunger is his own weight. That force divided by the cross sectional area is the pressure at the plunger.

Now do a Google search on pipe loss calculator. Use it to calculate the pressure loss at the average flow rate. Compare that with the pressure at the plunger and you can see how significant the friction is. (Friction loss is not linear with flow, but you can probably ignore that.)

We are not going to do the work for you. But you seem to be catching on quickly, so I think you will succeed. Post again if you get stuck.

I guess what's sticking me is that I'm not sure what to consider and what to ignore. I imagine I need to consider the friction, yet ignore the mass of the water in the 2.998km of pipe and only consider the mass of the water in the upturned ends. However, because I'm also forcing water down on one end, would the weight of that water cancel out the weight of the water at the opposite end (the amount of cancellation being reduced as my end fills with air and the other end remains filled)? Or am I literally trying to push 370kg of water with a 1.25cm plunger?

If I go with my gut, I'd say that I would need a final acceleration of 2.55m/s², which would require less than 22lbf against the plunger. (370kg x 2.55m/s² = 22lbf) But this seems like a really low requirement if I'm considering all of that mass, unless the flow rate that was calculated has canceled out some of that mass automatically.

Am I being dense?

 

[anorlunda]

You are perhaps overcomplicating it a bit, but not too much. A pipe loss calculator such as http://www.freecalc.com/fricfram.htm[/PLAIN] allows you to put in all sorts of complications such as elbows, and vertical differences, and pipe surface. Have a look at it.