Water Pressure in Pipe: 85L/s at 35m Depth

[Nuke787]

Homework Statement

For this problem use: ρwater = 1000 kg/m3

A cylindrical water-storage tank has a height of 7.9 m and a radius of 6.2 m. The storage tank is full of water but is vented to the atmosphere. The bottom of the tank is 35 m off the ground. A 12 cm diameter pipe runs vertically down from the tank and goes 1.7 m underground before turning horizontal. The water flow in the horizontal pipe is 85 L/s.

What is the water pressure in the horizontal pipe underground? (kPa)

Homework Equations

p=F/A
pascal (Pa) = 1N/m²
p=ρgd
Atmospheric Pressure = 101.3kPa

The Attempt at a Solution

I know that gauge pressure is ρ*gravity*depth. I am assuming my problem is that I am not using the appropriate height as rho and gravity are given. For heights I have tried numerous combinations of heights. I have tried 1.7, 1.7+35, 1.7+35+7.9. This is a hydrodynamics problem, so my method may be incorrect in trying to solve for the pressure.

 

[supratim1]

as it is vented out to atmosphere, add atmospheric pressure to it.

 

[supratim1]

and height is 1.7+35+7.9

 

[Nuke787]

This still yields an incorrect answer. I've got 14 tries on the problem thus far, that happens to be one of them. Is my methodology correct in attempting to find the pressure. I have more equations, but the only one I can clearly relate to pressure is Bernoulli's principle which the question does not provide all the variables for to my understanding.

 

[supratim1]

I think your approach is right. I don't see any other way. may be the answer is mistaken in the book you are referring. or may be there is something missing in the question.

 

[Nuke787]

It's an online homework for engineering. I don't know the answer, all I know is whether I get it right or wrong. Would it make a difference if the ground was used as a datum instead of the pipe?

 

[Nuke787]

The question nicely informed me that I made a common mistake and that water pressure is measured in gauge pressure. I got an answer of 505.472 after atmospheric pressure was added and I used the ground as a datum. I did 35+7.9 as one height and used -1.7 as the other. I'm assuming since it said common mistake that I am headed in the right direction with previous experiences with their random hints.

 

[supratim1]

gauge pressure is applicable underground too. i don't think you should subtract 1.7

 

[Nuke787]

If you use 1.7 as positive or negative it still yields an incorrect answer. I've tried numerous combinations of the heights and can't seem to get it right. As far as the gauge pressure is concerned, it is most definitely applicable underground. Adding atmospheric pressure gives you the absolute pressure. I believe the question wants an answer solely based off of the gauge pressure which is the depth ( or height) multiplied by gravity and rho.

 

[tiny-tim]

Welcome to PF!

Hi Nuke787! Welcome to PF!

A cylindrical water-storage tank has a height of 7.9 m and a radius of 6.2 m. The storage tank is full of water but is vented to the atmosphere. The bottom of the tank is 35 m off the ground. A 12 cm diameter pipe runs vertically down from the tank and goes 1.7 m underground before turning horizontal. The water flow in the horizontal pipe is 85 L/s.
…
I know that gauge pressure is ρ*gravity*depth. I am assuming my problem is that I am not using the appropriate height as rho and gravity are given. For heights I have tried numerous combinations of heights. I have tried 1.7, 1.7+35, 1.7+35+7.9. This is a hydrodynamics problem, so my method may be incorrect in trying to solve for the pressure.

You haven't used the flow rate.

(This is a Bernoulli's equation problem)

 

[supratim1]

oh yes,sorry i didnt notice that! use that 85 L/s

 

[Nuke787]

Well played sir. Using Q= vA and the continuity equation v_1A1=v₂A₂ to solve for the velocities I'm assuming. I was hesitant to use Bernouilli's as it involves an initial pressure on the right side P₁ and on the left side a P₂. I'm assuming I'm looking for P₂. For P₁ would I find it by using my afore mentioned method of height*gravity*rho?

 

[supratim1]

yes, that is P1

 

[tiny-tim]

Hi Nuke787!

_{I was hesitant to use Bernouilli's as it involves an initial pressure on the right side P1 and on the left side a P2. I'm assuming I'm looking for P2. For P1 would I find it by using my afore mentioned method of height*gravity*rho?}

Bernoulli's equation is along a streamline.

You compare P etc at one end of the streamline to P etc at the other end.

In this case, your streamline starts at the water surface, so P₁ is P_atm

 

[Nuke787]

Alright so this is where I'm at.
75 L/s= .075 m³/s
Q = vA
.075 m³/s=v₂((.06m)²*pi)
v₁A₁=v₂A₂
v₁*(pi*(6.2m)²*7.9m)
P₁+(1/2)$$\rho$$v₁²+$$\rho$$gh₁ = P₂+(1/2)$$\rho$$v₂²+$$\rho$$gh₂

(101.3kPa+(1/2)1000kg/m³(7.861418572*10^-5m/s)²+1000kg/m³(9.81m/s²)(35m+7.9m)=P₂+(1/2)(1000kg/m³)(6.631455m/s)²+1000kg/m³(9.81m/s²)(-1.7m)

How does that look, and as far as the last term, assuming that I use the ground as my datum, am I correct in making it negative?

 

[tiny-tim]

Hi Nuke787!

How does that look, and as far as the last term, assuming that I use the ground as my datum, am I correct in making it negative?

(I haven't checked your arithmetic, but …)

Yes that looks right,

except isn't it 85 L/s, not 75?

and if you're using gauge pressure, then P₁ = 0.

And yes, the 1.7 is negative (or you could use the water surface as your zero-level, and put the whole 1.7 + 7.9 on the same side).

 

[Nuke787]

That might explain why I got it wrong, I'll crunch the numbers again. It will be a miracle from god if I pass this engineering assignment. Thanks for all the help in advance.

 

[Nuke787]

Got it on try 26. Thanks a ton for the help guys.

 

[uros]

Hydrodynamic pressure (P):
P=ρv²/2

Volumetric flow rate (Q):
Q=Av
(A → area; v → velocity)

Attempt to solve:

Q=85L/s→0.085m³/s

A=πr²
A=0.0036πm²

v=Q/A
v=23.61m/s

P=ρv²/2
P=278742.284Pa