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Water Velocity and Pressure From 5 degree Gradient At 200 m

  • Thread starter morrobay
  • Start date
  • #1
morrobay
Gold Member
703
105
1, Problem Statement.
A resevoir releases water into a 2 meter diameter drain pipe that is 200 m long with a 5ο gradient drop. Assume pipe is full without friction. If v0 is zero what is velocity at 200 meter discharge ?
And what is the pressure in atm ? The height at pipe end is - 17 meters. A 17 m drop from resevoir to end of 200 meter pipe

Homework Equations


For vertical time , y = 1/2 at2 = 1,86 sec
For vertical velocity vy = v = √2ay = 18m/sec
Pressure = Force/Area = mass*acc/area = 628000 kg * 9,8m/sec2/3.14 m2
Pressure = p0 + ρ h g
Assume water density = 1

The Attempt at a Solution

[/B]
Sine 50 = .087. = 18m/s/v = 204m/sec ? or is it sine 50 (18m/s) = 1.5m/sec ?
Not sure about pressure formula to atm

All Im looking for are the equations for velocity and pressure at pipe exit from above data. http://www.calctool.org/CALC/eng/civil/hazen-williams_g
 
Last edited:

Answers and Replies

  • #2
morrobay
Gold Member
703
105
1, Problem Statement.
A resevoir releases water into a 2 meter diameter drain pipe that is 200 m long with a 5ο gradient drop. Assume pipe is full without friction. If v0 is zero what is velocity at 200 meter discharge ?
And what is the pressure in atm ? The height at pipe end is - 17 meters. A 17 m drop from resevoir to end of 200 meter pipe

Homework Equations


For vertical time , y = 1/2 at2 = 1,86 sec
For vertical velocity vy = v = √2ay = 18m/sec
Pressure = Force/Area = mass*acc/area = 628000 kg * 9,8m/sec2/3.14 m2
Pressure = p0 + ρ h g
Assume water density = 1

The Attempt at a Solution

[/B]
Sine 50 = .087. = 18m/s/v = 204m/sec ? or is it sine 50 (18m/s) = 1.5m/sec ?
Not sure about pressure formula to atm

All Im looking for are the equations for velocity and pressure at pipe exit from above data. http://www.calctool.org/CALC/eng/civil/hazen-williams_g

Attempt # 2.
This is a velocity vector problem t0 v0 = 0 with theta 50 with x axis
v0x = v0 cos θ
v0y = v0 sin θ
vy = v0 sin θ +gt
The only component I can solve for is vy =√2gh == 18,25 m/sec
So I am missing a = ax + ay
 
  • #3
morrobay
Gold Member
703
105
The acceleration for this problem is g sin 50 and final velocity = v2 = 2(g *sin50 *200m) = 18.5 m/sec
 
Last edited:
  • #4
billy_joule
Science Advisor
1,200
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Was a diagram provided? What pressure is it asking for?
For a problem like this it's generally safe to assume no acceleration occurs in the pipe, and pressures at the reservoir surface and pipe outlet are both atmospheric.

Are you familiar with Bernoulli's equation?
 
  • #5
morrobay
Gold Member
703
105
this should be solution
Was a diagram provided? What pressure is it asking for?
For a problem like this it's generally safe to assume no acceleration occurs in the pipe, and pressures at the reservoir surface and pipe outlet are both atmospheric.

Are you familiar with Bernoulli's equation?
p = 1/2 ρv2 +ρ g h = constant But not familiar enough with crossover units in order to convert the above velocity , 18.5 m/sec, full 2 m diameter pipe to pressure in atmospheres at pipe outfall. I agree with assumptions above. The question is based on a submarine outfall about 17m below sea level from the 200m pipe that is gravity fed from reservoir on shore at sea level . At outfall the pressure in the ocean surroundings in atmospheres is p = p0 = ρgh about 2.7 atmospheres. So I want to know if outfall pressure from pipe is greater than 2.7 atmospheres. Otherwise a pump would be required for storm water drainage from reservoir.
 
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  • #6
billy_joule
Science Advisor
1,200
330
Can you answer these questions?

Was a diagram provided? What pressure is it asking for?

this should be solution

p = 1/2 ρv2 +ρ g h = constant But not familiar enough with crossover units in order to convert the above velocity , 18.5 m/sec, full 2 m diameter pipe to atmospheres at pipe outfall. I agree with assumptions above. The question is based on a submarine outfall about 17m below sea level from the 200m pipe that is gravity fed from reservoir on shore at sea level.
If the reservoir surface is also at sea level then clearly no water would flow.
But, if my interpretation is correct, from the given information it can be found that the reservoir surface is slightly above sea level so there is a pressure difference so water will flow. Though, the question is a little ambiguous/contradictory on this point. Have you provided the question word for word?

I don't know what 'crossover units' are, you can use any units you like in the Bernoulli's, as long as you are consistent.

At outfall the pressure in the ocean surroundings in atmospheres is p = p0 = ρgh about 2.7 atmospheres. So I want to know if outfall pressure from pipe is greater than 2.7 atmospheres. Otherwise a pump would be required for storm water drainage from reservoir.
Why are you bringing up a pump? Is there more to the question statement?
If you want to know if the reservoir is free draining to the ocean all you have to look at is if sea level is lower than the reservoir level. That is all the information you need.
 
  • #7
morrobay
Gold Member
703
105
Can you answer these questions?






If the reservoir surface is also at sea level then clearly no water would flow.
But, if my interpretation is correct, from the given information it can be found that the reservoir surface is slightly above sea level so there is a pressure difference so water will flow. Though, the question is a little ambiguous/contradictory on this point. Have you provided the question word for word?

I don't know what 'crossover units' are, you can use any units you like in the Bernoulli's, as long as you are consistent.



Why are you bringing up a pump? Is there more to the question statement?
If you want to know if the reservoir is free draining to the ocean all you have to look at is if sea level is lower than the reservoir level. That is all the information you need.
https://www.physicsforums.com/threads/submarine-outfall-for-storm-water-drainage-pipe.839527/

The question is related to this situation. No diagram is given. The reservoir ( one atm ) on shore is slightly above sea level and gravity released storm water enters 200 m long pipe to submarine out fall at 17 meters below sea level. where the surrounding pressure is about 2.7 atm. So since pressure at reservoir is less than pressure at submarine outfall I am asking if the pressure from velocity will be greater than 2.7 atm
 

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