# Bernoulli equation for calculation of water flow

• gamecult
In summary, Bernoulli's equation is used to calculate the pressure, velocity, and flow rate of water in a pipe. The problem presented involves calculating these values at the outlet of a pipe that is 900 meters long and located on the seafloor. The calculation assumes no head loss or friction in the pipe and a differential head of 0.87 meters between the inlet and outlet points.
gamecult
Homework Statement
A seawater collection system consists of two tanks connected by pipe 900m long and 31.5cm in diameter
The pipe inlet is on the seabed at a level of -11.5m and the pipe outlet to Tank 2 is on the beach at a level of -0.875m.
Calculate in detail the water velocity and water flow at the outlet
Both point 1 and 3 are in open aire (Only under atmospheric pressure).
See the drawing for more explanation.
Relevant Equations
P_1+1/2 ρv_1^2 + ρgh_1 = P_3+1/2 ρv_3^2 + ρgh_3
Hello everyone;
Please need some help to check if my calculation are correct (and if possible some explantation)

Bernoulli's equation between point 1 and 3 is given by:
P_1+1/2 ρv_1^2 + ρgh_1 = P_3+1/2 ρv_3^2 + ρgh_3​
 P_1 = P_(atm ) v_1= 0 m/s h_1= 0.875 m P_3 = P_(atm ) v_3= ? m/s h_3= 0 m

P_(atm )+1/2 ρ(0 m/s)_1^2 + ρg(0.875 m) = P_(atm )+1/2 ρv_3^2 + ρg(0 m)
ρg(0.875 m)=1/2 ρv_3^2
(9,81 m/s^2 )(0.875 m)=1/2 v_3^2
v_3=4.10 m/s
In our case, the pipe diameter is d = 0.315 m (31.5 cm), so the section is:
A = π × (0.315/2)^2 = 0.0779 m^2​
The volume flow Q is then:
Q = A×v_3 = 0.0779 m^2 × 4.10 m/s = 0.320 m^3/s
To convert flow from m^3/s to m^3/s
Q = 0.320 m^3/s * 3600 s/h = 1.152 m^3/h

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Do you have any specific inquiries, doubts, etc...?

gamecult said:
Q = 0.320 m^3/s * 3600 s/h = 1.152 m^3/h
I can tell you immediately that this incorrect. 0.320*3600 = 1152.
Also, why is v1 = 0? Isn't the level of the tank dropping as the water flows out the conduit?
More importantly, when the problem says "Calculate in detail the water velocity and water flow at the outlet" which outlet do you think this is? I would say it's the one at the very bottom.

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kuruman said:
I can tell you immediately that this incorrect. 0.320*3600 = 1152.
Also, why is v1 = 0? Isn't the level of the tank dropping as the water flows out the conduit?
But the water is flowing up the pipe into tank 3, filling it up. It's a sea water collection system. There is a differential head between points 1 and 3 of ##0.87 \rm{m}##. Point 1 is taken to be the surface of the ocean, the kinetic head at 1 is assumed to be vanishingly small.

In real life there is head loss between 1 and 3 which will decrease the flow rate in the pipe, but we are apparently working under the assumption of inviscid flow.

kuruman
Oops, I reversed the flow in my head. That makes more sense.

erobz
erobz said:
Do you have any specific inquiries, doubts, etc...?
That's almost all inquiries I have I made the assumption that there is no head loss or friction inside the HDPE tube.
I wanted to know the water flow at the outlet to see if it was enough or not.
I also wanted to know if my calculation are correct or not.

kuruman said:
I can tell you immediately that this incorrect. 0.320*3600 = 1152.
Also, why is v1 = 0? Isn't the level of the tank dropping as the water flows out the conduit?
More importantly, when the problem says "Calculate in detail the water velocity and water flow at the outlet" which outlet do you think this is? I would say it's the one at the very bottom.
0.320*3600 = 1152
Sorry, when I copy-pasted from Word it changed the thousands separator (,) into a period (.)
In addition, the water flows from the lower part to the upper part. I assumed that the speed at the top of sea level is 0 because the level will never drops

gamecult said:
I made the assumption that there is no head loss or friction inside the HDPE tube.
So it's not a homework problem? If not, no losses in a ##900^+ ~\rm{m}## long pipe is a poor assumption.

Why is the tube 900m long? Is the collection tank 1km away from the edge of the water? And why is the pipe inlet on the seafloor? Is it more convenient than just floating it near the surface near the beach?

Lnewqban
gamecult said:
when I copy-pasted from Word it changed the thousands separator (,) into a period (.)
Interesting… I would guess you have Word set up for French conventions, so normally you would type period for the thousands delimiter. But, aware of your anglophone audience, you used a comma. So I wonder if the code point Word used was "decimal introducer", rather than simply "comma", and pasting that in made it translate to a period.

berkeman said:
Why is the tube 900m long? Is the collection tank 1km away from the edge of the water? And why is the pipe inlet on the seafloor? Is it more convenient than just floating it near the surface near the beach?
Thanks for the question :
The tube is 900m aways to unsure that the quality of Seawater is taken from a point and at a level protected from pollution. Also, the tube is laid at the bottom of the sea so as not to interfere with maritime traffic and for the water quality.

haruspex said:
Interesting… I would guess you have Word set up for French conventions, so normally you would type period for the thousands delimiter. But, aware of your anglophone audience, you used a comma. So I wonder if the code point Word used was "decimal introducer", rather than simply "comma", and pasting that in made it translate to a period.
You guessed very well, i use a French version of word. I usually write numbers like this : 1,152.0 m3/h

Just an FYI. Including major head loss term from the pipe, the volumetric flowrate is ## \approx 170 ~\rm{ \frac{m^3}{hr}}##.

If you're interested in the details I'll share them, if it's a non-issue that's ok too.

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erobz said:
Just an FYI. Including major head loss term from the pipe, the volumetric flowrate is ## \approx 170 ~\rm{ \frac{m^3}{hr}}##.

If you're interested in the details I'll share them, if it's a non-issue that's ok too.
I would like to have more informations and details about head loss and friction loss. I dont know how to do it trying to learn.

gamecult said:
I would like to have more informations and details about head loss and friction loss. I dont know how to do it trying to learn.
Cliff notes version:

The Energy Equation in fluid mechanics is derived from applying Reynolds Transport Theorem for a control volume. It accounts for energy being added by a pump, extracted by a turbine, or lost due to viscous dissipation along a path the fluid takes in going from one point to another. It's going to look like Bernoulli's with some additional terms:

$$\frac{P_1}{\rho g} + z_1 + \frac{V_1^2}{2g}+h_p = \frac{P_3}{\rho g} + z_3+ \frac{V_3^2}{2g}+h_t+ \sum_{1 \to 3}h_l \tag{Energy Equation}$$

Between points (1) and (3)
##h_t## is head extracted by a turbine
##h_l## head loss - viscous dissipation

The last term is typically broken up into two components. Major head loss - from pipes, and minor head loss from plumbing components, elbows, valves, couplings, etc..

$$\sum_{1 \to 3}h_l = \sum_{pipes} h_l + \sum_{components}h_l$$

The sum over the pipes is represent by the Darcy-Weisbach Equation applied to each section of pipe:

$$\sum_{pipes}h_l = \sum f_i\frac{L_i}{D_i} \frac{V_i^2}{2g}$$

##f## is the friction factor
##L## is the length of pipe
##D## is the pipe diameter

The sum of the components is usually represented by a factor ##k## that is found in tables for a particular component:

$$\sum_{comp.}h_l =\sum k_i\frac{V_i^2}{2g}$$

The Energy Equation then becomes:

\begin{aligned}\frac{P_1}{\rho g} + z_1 + \frac{V_1^2}{2g}+h_p =& \\ & \frac{P_3}{\rho g} + z_3 + \frac{V_3^2}{2g}+h_t+ \overbrace{ \sum f_i\frac{L_i}{D_i} \frac{V_i^2}{2g} }^{\text{pipes}}+ \overbrace{ \sum k_i\frac{V_i^2}{2g}}^{\text{components}} \end{aligned}

$$\text{(Energy Equation)}$$

Any questions so far?

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gamecult said:
Homework Statement: A seawater collection system consists of two tanks connected by pipe 900m long and 31.5cm in diameter
The pipe inlet is on the seabed at a level of -11.5m and the pipe outlet to Tank 2 is on the beach at a level of -0.875m.
Calculate in detail the water velocity and water flow at the outlet
Both point 1 and 3 are in open aire (Only under atmospheric pressure).
See the drawing for more explanation.
Relevant Equations: P_1+1/2 ρv_1^2 + ρgh_1 = P_3+1/2 ρv_3^2 + ρgh_3

Bernoulli's equation between point 1 and 3 is given by:
P_1+1/2 ρv_1^2 + ρgh_1 = P_3+1/2 ρv_3^2 + ρgh_3​

I must be missing something. Is there a flowpath between point 3 and point 1?? The sketch shows the pipe between P2 and P1

gmax137 said:
I must be missing something. Is there a flowpath between point 3 and point 1?? The sketch shows the pipe between P2 and P1

Seems odd, but this is a streamline.

erobz said:
View attachment 325890

Seems odd, but this is a streamline.
I do not understand why P1 enters into the equations. The flow width between P1 and P2 is vast, meaning the flow rate is minuscule and head loss can be ignored.
Just consider the stream from P2 to P3.

haruspex said:
I do not understand why P1 enters into the equations. The flow width between P1 and P2 is vast, meaning the flow rate is minuscule and head loss can be ignored.
Just consider the stream from P2 to P3.
It really doesn’t enter in, it’s just an efficient place along a streamline. P1 = 0 gauge, and the kinetic head at 1 is effectively 0. If we go directly from 2 to 3 you end up having to solve for P2 ( sure you could do this by inspection but why bother, it’s not pertinent info). It’s a waste of a step. Losses aren’t being considered between 1 and 2. The indices on the summation refer to portions of pipe.

Edit: I see what was confusing you( I think). I had my summation of losses going from 1 to 2 in post 15. My bad. I was thinking in a general sense( between any two points “1 and 2”) to explain it. I changed them to go from 1 to 3 to match the context of this problem.

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erobz said:
Seems odd, but this is a streamline.

Thanks @erobz I wasn't seeing it this way.

haruspex said:
I do not understand why P1 enters into the equations. The flow width between P1 and P2 is vast, meaning the flow rate is minuscule and head loss can be ignored.
Just consider the stream from P2 to P3.
This is more what I was seeing. It doesn't really matter, though, since the elevation difference is still 0.87 m. EDIT: I mean the difference in the (pressure plus elevation head).
erobz said:
So it's not a homework problem? If not, no losses in a ##900^+ ~\rm{m}## long pipe is a poor assumption.
The ##f \frac {L} {d}## is ~40 (assuming turbulent flow; f~0.013). This is not insignificant.

I get a Re of 1.3 million using 4.1 m/s through 0.3 m pipe (check my arithmetic). So turbulent.

So, now you need to iterate, using the Bernoulli with head loss approach outlined by @erobz in post #15.

There is an assumption here, that the water is being drawn off the upper tank to maintain the elevation of the liquid surface at the -0.87 meter. If the water level in the tank rises, the flow will slow down, and if the level rises to sea level the flow would just stop, right?

erobz
gmax137 said:
The ##f \frac {L} {d}## is ~40 (assuming turbulent flow; f~0.013). This is not insignificant.

I get a Re of 1.3 million using 4.1 m/s through 0.3 m pipe (check my arithmetic). So turbulent.

So, now you need to iterate, using the Bernoulli with head loss approach outlined by @erobz in post #15.
It ends up settling in around ##f \approx 0.016##
gmax137 said:
There is an assumption here, that the water is being drawn off the upper tank to maintain the elevation of the liquid surface at the -0.87 meter. If the water level in the tank rises, the flow will slow down, and if the level rises to sea level the flow would just stop, right?
Energy Equation is valid only for steady flow. If the water is accumulating above the elevation of 3 we will have transient flow from then on. But yes, if it makes it to sea level, flow stops in either case.

gmax137
erobz said:
P1 = 0 gauge, and the kinetic head at 1 is effectively 0. If we go directly from 2 to 3 you end up having to solve for P2
Well, ##V_2## is also zero. ##\frac{P_1}{\rho g}+z_1=\frac{P_2}{\rho g}+z_2##.

Btw, @gmax137, a Borda mouthpiece at the pipe entrance should help. See 4.10 of https://egyankosh.ac.in/bitstream/123456789/29678/1/Unit-4.pdf

berkeman and gmax137
haruspex said:
Well, ##V_2## is also zero. ##\frac{P_1}{\rho g}+z_1=\frac{P_2}{\rho g}+z_2##.

Btw, @gmax137, a Borda mouthpiece at the pipe entrance should help. See 4.10 of https://egyankosh.ac.in/bitstream/123456789/29678/1/Unit-4.pdf
Usually we are taking 2 at the pipe entrance, and ##V_2## will not be zero. In going from (2) to (3) For a straight pipe ##V_2 = V_3##, and the kinetic head terms cancel on each side of the equation. We can throw a wrench into that by having the outlet a different diameter, they will not cancel and there will be an additional variable to contend with.

In going from (1) to (2) we still have ( with (3) as the elevation datum ##z_3 = 0##):

$$\cancel{\frac{P_1}{\gamma}}^{0} + z_1 + \cancel{\frac{V_1^2}{2g}}^{\approx 0} =\frac{P_2}{\gamma}- z_2 + \frac{V_2^2}{2g}$$

$$\implies \frac{P_2}{\gamma} = z_1 + z_2 - \frac{V_2^2}{2g}$$

Then going from (2) to (3):

$$\frac{P_2}{\gamma}- z_2 + \cancel{\frac{V_2^2}{2g}}^{V_2 = V_3} = \cancel{\frac{P_3}{\gamma}}^{0}+ \cancel{z_3}^{0} + \cancel{\frac{V_3^2}{2g}}^{V_2 = V_3}$$

$$\implies \frac{P_2}{\gamma}- z_2 = 0$$

Substituting for ##\frac{P_2}{\gamma}##, with ##V_2 = V_3##:

$$\implies z_1 = \frac{V_2^2}{2g} = \frac{V_3^2}{2g}$$

For this reason going directly from (1) to (3) is the more efficient choice ( i.e. the solution is completed in a single application of Bernoulli's).

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erobz said:
Usually we are taking 2 at the pipe entrance, and ##V_2## will not be zero. In going from (2) to (3) For a straight pipe ##V_2 = V_3##, and the kinetic head terms cancel on each side of the equation. We can throw a wrench into that by having the outlet a different diameter, they will not cancel and there will be an additional variable to contend with.

In going from (1) to (2) we still have ( with (3) as the elevation datum ##z_3 = 0##):

$$\cancel{\frac{P_1}{\gamma}}^{0} + z_1 + \cancel{\frac{V_1^2}{2g}}^{\approx 0} =\frac{P_2}{\gamma}- z_2 + \frac{V_2^2}{2g}$$

$$\implies \frac{P_2}{\gamma} = z_1 + z_2 - \frac{V_2^2}{2g}$$

Then going from (2) to (3):

$$\frac{P_2}{\gamma}- z_2 + \cancel{\frac{V_2^2}{2g}}^{V_2 = V_3} = \cancel{\frac{P_3}{\gamma}}^{0}+ \cancel{z_3}^{0} + \cancel{\frac{V_3^2}{2g}}^{V_2 = V_3}$$

$$\implies \frac{P_2}{\gamma}- z_2 = 0$$

Substituting for ##\frac{P_2}{\gamma}##, with ##V_2 = V_3##:

$$\implies z_1 = \frac{V_2^2}{2g} = \frac{V_3^2}{2g}$$

For this reason going directly from (1) to (3) is the more efficient choice ( i.e. the solution is completed in a single application of Bernoulli's).
There may be some confusion re P2. I am taking it as a point a short distance in front of the pipe entrance, whereas you seem to mean a point within the pipe.

If I rename my P2 as P2a, the velocity is zero there and ##\frac{P_1}{\rho g}+z_1=\frac{P_{2a }}{\rho g}+z_2##.
Going from P2a to your P2 the velocity jumps to v and the pressure drops correspondingly. But we don't need to find the pressure there since we can compare P2a with P3.

But I agree there is no difference in effort between using P1 and using P2a. It's only a question of which view is conceptually more comfortable.

erobz said:
Cliff notes version:

The Energy Equation in fluid mechanics is derived from applying Reynolds Transport Theorem for a control volume. It accounts for energy being added by a pump, extracted by a turbine, or lost due to viscous dissipation along a path the fluid takes in going from one point to another. It's going to look like Bernoulli's with some additional terms:

$$\frac{P_1}{\rho g} + z_1 + \frac{V_1^2}{2g}+h_p = \frac{P_3}{\rho g} + z_3+ \frac{V_3^2}{2g}+h_t+ \sum_{1 \to 3}h_l \tag{Energy Equation}$$

Between points (1) and (3)
##h_t## is head extracted by a turbine
##h_l## head loss - viscous dissipation

The last term is typically broken up into two components. Major head loss - from pipes, and minor head loss from plumbing components, elbows, valves, couplings, etc..

$$\sum_{1 \to 3}h_l = \sum_{pipes} h_l + \sum_{components}h_l$$

The sum over the pipes is represent by the Darcy-Weisbach Equation applied to each section of pipe:

$$\sum_{pipes}h_l = \sum f_i\frac{L_i}{D_i} \frac{V_i^2}{2g}$$

##f## is the friction factor
##L## is the length of pipe
##D## is the pipe diameter

The sum of the components is usually represented by a factor ##k## that is found in tables for a particular component:

$$\sum_{comp.}h_l =\sum k_i\frac{V_i^2}{2g}$$

The Energy Equation then becomes:

\begin{aligned}\frac{P_1}{\rho g} + z_1 + \frac{V_1^2}{2g}+h_p =& \\ & \frac{P_3}{\rho g} + z_3 + \frac{V_3^2}{2g}+h_t+ \overbrace{ \sum f_i\frac{L_i}{D_i} \frac{V_i^2}{2g} }^{\text{pipes}}+ \overbrace{ \sum k_i\frac{V_i^2}{2g}}^{\text{components}} \end{aligned}

$$\text{(Energy Equation)}$$

Any questions so far?
Thank you so much it is more clear for me to understand about the energy loss. I will try to do the calculation including the energy loss

gmax137 said:
I must be missing something. Is there a flowpath between point 3 and point 1?? The sketch shows the pipe between P2 and P1

This a more clear (I hope) Illustration

gamecult said:
Thank you so much it is more clear for me to understand about the energy loss. I will try to do the calculation including the energy loss
Well, the last piece to consider is the friction factor. See Moody Diagram. You will calculate a Reynolds Number ##Re##, and a Relative Roughness ##\frac{k_s}{D}## ( HDPE will be considered "smooth" on the diagram ##\frac{k_s}{D} \approx 0##), calulate flowrate using ##f##. That procedure will usually need to be iterated until ##f## settles.

Alternatively there is an empirical equation you can use that ( reported ##\approx 3\%## difference from Moody) does well for ##4 \text{e3} < Re < 1\text{e8}## and ## 1\text{e-5} < \frac{k_s}{D} < 2\text{e-2}##:

$$f = \frac{1}{4 \left[ \log_{10} \left( \frac{1}{3.7} \frac{k_s}{D}+ \frac{5.74}{Re^{0.9}} \right) \right]^2}$$

And to calculate the Reynolds number, you will need density and viscosity; so you will need to pick a temperature for your seawater. I used to have property tables for seawater but since I retired I don't have them anymore. For a calculation like this, using regular water properties is probably ok.

erobz
Alternatively kinematic viscosity ##\nu##, instead of density and viscosity:

$$Re = \frac{VD}{\nu}$$

I don't have a complete table for seawater, but there is table in my textbook that contains Sea Water @##10°C, 3.3 \% ~\rm{salinity}## which gives ##\nu = 1.4\text{e-6} ~\rm{\frac{m^2}{s}}## (which is the value I used for the computation)

gmax137
gmax137 said:
And to calculate the Reynolds number, you will need density and viscosity; so you will need to pick a temperature for your seawater. I used to have property tables for seawater but since I retired I don't have them anymore. For a calculation like this, using regular water properties is probably ok.
https://www.engineeringtoolbox.com/sea-water-properties-d_840.html

## What is the Bernoulli equation and how is it used to calculate water flow?

The Bernoulli equation is a principle in fluid dynamics that describes the behavior of a fluid under varying conditions of flow and height. It states that an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. The equation is often used to calculate water flow by relating the pressure, velocity, and height at different points along a streamline.

## What are the key components of the Bernoulli equation?

The key components of the Bernoulli equation are pressure energy, kinetic energy, and potential energy. Mathematically, it is expressed as: P + 0.5 * ρ * v^2 + ρ * g * h = constant, where P is the pressure, ρ is the fluid density, v is the flow velocity, g is the acceleration due to gravity, and h is the height above a reference point.

## How does the Bernoulli equation apply to real-world water flow scenarios?

In real-world scenarios, the Bernoulli equation helps engineers and scientists predict how water will behave in different systems, such as pipes, rivers, and airfoils. By understanding the relationship between pressure, velocity, and height, they can design more efficient systems for water distribution, irrigation, and even flood control.

## What assumptions are made when using the Bernoulli equation?

The Bernoulli equation assumes that the fluid is incompressible and non-viscous, and that the flow is steady and along a streamline. These assumptions simplify the complex nature of fluid flow, making it easier to apply the equation to practical problems, though they may not always perfectly represent real-world conditions.

## Can the Bernoulli equation be used for all types of fluid flow?

No, the Bernoulli equation is best suited for incompressible, non-viscous fluids in steady, streamline flow. It does not accurately describe compressible fluids (like gases at high velocities) or situations with significant viscosity (like oil in some industrial applications). For such cases, more complex models and equations are required.

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