# Homework Help: Static Pressure and Dynamic Pressure to Atmospheres

1. Nov 7, 2015

### morrobay

1. The problem statement, all variables and given/known data
Calculate the dynamic water pressure at submarine outfall from a pipe 2m diameter , 500 m length at depth of 52 meters with velocity 41m/sec. Pipe is at 100 relative to x axis. Water is released from reservoir on shore at sea level.
Then calculate the static water pressure at depth of 52 meters in ocean surroundings.
density,ρ = 1000kg/m3
acceleration = gsinθ
v = √2(sinθ)(g)( length)
1 atmosphere = 1.01 * 105 Newtons/m2
2. Relevant equations
Static pressure = p + p0 + ρgh
Dynamic pressure = p = 1/2 ρv2

3. The attempt at a solution
The dynamic pressure = 846528/101000 = 8.38 atm
The static pressure = 511560/101000 = 5.06 atm

2. Nov 8, 2015

### insightful

Somewhat confusing, e.g., a 500m pipe starting at sea level and going down at a 10o angle does not end up at a depth of 52m. Could you post a diagram?

3. Nov 8, 2015

### billy_joule

Other threads on the same topic for more background:

Where did this velocity come from?

I don't know anything about submarine outfall pipes but I'd bet the flow velocities are not ~150 km/hr!

4. Nov 8, 2015

### morrobay

Sorry that should have been 300 m : sine 100 = .173 = y/300 m = 52m.
And I also made two other careless/typos:
Static pressure, p = p0 +ρgh. I forgot to add po of 1 atm so correction to 6 atm
Dynamic pressure p = 1/2 ρv2 correction to 5.06 atm.
Because I made error on velocity: v = √2*sin 100 * g * length pipe = 32m/sec

From another reference/table : For every 10m of depth add one atm to pressure. So 52/10 + 1 is a good number above arrived at through calculation

5. Nov 8, 2015

### billy_joule

That's not right either.
You can assume no acceleration occurs in the pipe and find velocity via Bernoulli's eqn.
As I said in one of your other threads on this topic, the flow velocity is dependant only on the difference in height between the surface of the reservoir and the sea.
If the sea level is equal to the reservoir then no flow occurs.
Your problem statement implies sea level = reservoir level so no flow occurs in which case dynamic pressure = 0 at all points.

6. Nov 8, 2015

### morrobay

Actually my calculation is about right for velocity, 114km/hr .@ 31m/s. It is correct but seems too high. Engineers ?
http://www.calctool.org/CALC/eng/civil/hazen-williams_g
diameter 2 m
drop 52 m
pipe length 300 m

This velocity formula is based on a cart sliding down a ramp. Are you saying that gravity has no effect
on the water flow in pipe at 100 drop/incline ?

7. Nov 8, 2015

### billy_joule

That calculator assumes open ended/free draining pipe. That's not the situation you have.

Yes I am. Water will flow up the pipe if the sea level rises above the reservoir level, as explained in your first thread. The incline of the pipe is not relevant to the direction or velocity of flow.

What matters is the pressure difference, and after three threads on this topic we still don't know if the reservoir level and sea level are the same!

I think it's fairly safe to assume sea level is lower but you've never clearly stated the height difference.
Water will only flow from one to the other if there is a height difference.
In you second thread you said it was a frictionless pipe, in which case nothing at all about the pipe itself (angle, diameter, length, colour, owners dog's name etc ) has any effect on the flow velocity. The flow velocity depends only on the height difference.

To get anywhere on your problem you'll need to make a serious effort to learn something about the topic at hand, posting new threads with the same problem won't get you anywhere.

http://physics.info/flow/

8. Nov 8, 2015

### morrobay

This is an evolving study and I will be looking at above reference.Its best to only consider the last posts above. The reservoir water level on the beach would be about 3m above sea level at high tide and about 6 m at low tide. You say above the flow of water only depends on pressure differences. Dynamic water pressure is related to velocity. So do you want to do a calculation on the pressure differences here from above data ? The question originated here : http://www.thaivisa.com/forum/topic/827246-beach-road-drainage-pipe-installation-to-begin-in-june/
From my calculations the static pressure of 6 atm at 52m depth is greater that dynamic pressure 5.06.
So water would not flow. So again I would like to see some calculations on these pressure differences.

9. Nov 8, 2015

### SteamKing

Staff Emeritus
Since it's an evolving study, you should also realize that the ocean isn't composed of fresh water. Sea water has an average density of 1025 kg / m3.

10. Nov 9, 2015

### morrobay

With the static pressure of 6 atm at 52 m depth
Suppose the water level in reservoir on beach is 5m above sea level.
Then would velocity = √ 2*sin 100 * g* 5m = 4m/sec
Then dynamic pressure = 1/2 ρv2 . is only .08 atm
If not then what is the calculation for dynamic pressure at outfall ?
Note Im asking this question before referring to reference above for future calculation check.

11. Nov 9, 2015

### billy_joule

That doesn't make sense, dynamic pressure and flow go hand in hand, you can't have one without the other.

I explained in post #5 (and #7) why that is not the right way to find velocity - water is incompressible so can't accelerate in the pipe.

Did you follow the link in post #7? Can you apply Bernoulli's equation?
To get an accurate answer you really need the all parameters - there will be check valves, grates and other fittings required, tidal, ocean current and storm surge data, allowances for corrosion etc etc (these are guesses, I'm not a civil engineer)

This sort of multi million dollar project would involve teams of professional engineers and the design work alone, before a single spade is lifted, would cost tens or maybe hundreds of thousands of dollars in man hours.
From your link it looks like you want to show the project is doomed to failure, that's not going to happen without a detailed design and considerable (and expensive..) help from experts in the field to assess that design..

12. Nov 9, 2015

### morrobay

From top : If the static pressure at 52 m depth is greater than dynamic pressure at outfall. Then I would expect that an opposing pressure would "block" outflow ?
.p1 + 1/2ρv2 + ρgy1 = p2 + 1/2ρv2 + ρgy2
All Im looking for is a textbook answer , ie just the pure physics. I have been going over the chapters on fluid statics and dynamics in my text
( Halliday⋅ Resnick) and see no examples of this particular situation .
If you can show dynamic pressure at outfall pipe from variables above with Bernoulls's equation I would appreciate it. Again I am only interested in the physics here.Disregard engineering variables. If you had ever spent much time in Thailand you would understand why I am expecting failure.
Basically an" engineer." 4 yrs Uni might be be equivalent to a two year technician in any Western country + India and China.
Also what is the concept here ? p + 1/2 ρv2 + ρgy = constant.
Im going to try and solve this but not sure on some of the values here : velocity

13. Nov 9, 2015

### billy_joule

That's not how static & dynamic pressures work.

Bernoullis equation:

P1 + ρgh1 + ½ρv12= P2 + ρgh2 + ½ρv22

Let's choose our two points along a stream line to be the free surface of the reservoir (1) and the underwater pipe exit (2), and let atmospheric pressure be our datum pressure point, and h=0 at sea level, hx is the distance from the pipe outlet to sea level (ie underwater depth of outlet). We'll ignore the difference in density of fresh and salt water, it's only a few percent anyway.

P1 + ρgh1 + ½ρv12= P2 + ρgh2 + ½ρv22

We assume the reservoir is large - that is, the surface has no velocity (v1 = 0 ), Let's use hres for the sea-reservoir height difference.

0 + ρghres + 0 = ρghx + (- ρghx ) + ½ρv22

(note that P2 = ρgh2 pipe outlet depth doesn't matter, as expected)

ρghres = ½ρv22

v = √(2gh)

if we let hres = 3m as mentioned earlier we have:

√ (2*9.81m/s2*3 m ) = 7.7 m/s

> the flow velocity at the pipe outlet is 7.7 m/s (ignoring all losses)
This can be used to find the dynamic pressure for all points in the pipe.

14. Nov 9, 2015

### morrobay

Ok thanks .Then the dynamic pressure at outfall is .3 atm.
It is counter intuitive that there would be outflow against 6atm static pressure at 52 m depth.

15. Nov 9, 2015

### insightful

From pipe sizing tables, I get that the friction head loss in 300m of 2m diameter pipe flowing at 7.7 m/s is about 6m.

16. Nov 9, 2015

### insightful

If the outfall pipe is everywhere under water, is not the static pressure at the discharge of the pipe irrelevant?

17. Nov 9, 2015

### insightful

As I understand your system, with a 3m deep reservoir, I would predict an outflow of about 12 m3/s. What is the design flow rate?

18. Nov 10, 2015

### billy_joule

Dynamic pressures is the kinetic energy per unit volume of a fluid.

https://en.wikipedia.org/wiki/Dynamic_pressure

I'm not sure where you got the idea that dynamic pressure must be greater than static pressure for flow to occur but it's not right.

In this case, the frictional head loss cannot be calculated from the frictionless flow velocity as the flow velocity is dependant on the frictional head loss and vice versa.
This is why the answer makes no sense; a head loss of 6 m when the total head is only 3 m. Which, if taken literally, means the water will flow uphill into the reservoir.

To find the flow velocity with friction accounted for, the Darcy Weisbach equation can be used:
https://en.wikipedia.org/wiki/Darcy–Weisbach_equation
https://en.wikipedia.org/wiki/Darcy_friction_factor_formulae