Advanced Hydrodynamic Problem - University level

[erobz]

ok thank you very much now I try to calculate with this formula, just a last thing sorry View attachment 316536
the y in the image here attached (sorry but on the keyboard i don't know how to write it) what it is ? and Z it is the piezometric height right ? in this case would be 100 m the maximum height that the water reaches ? Am I correct ?
sorry to ask so many questions but as you can see I am really bad at physics :(

View attachment 316537

$\gamma$ (gamma) is the specific weight of the fluid. $\gamma = \rho g$

$z$ is the static elevation head for the point of reference. I would take $z_1 = 0$.

 

[Hydro2022]

@erobz
I tried to calculate the pressure with the equation but I can't solve it as unfortunately I have two unknowns both P1 that of the tank but also P2 that is the pressure inside the 250 mm diameter pipe. Do you think there is anothe way to calculate the pressure ?

thanks in advance

luckily I have another 2 weeks before I have to submit the exercise

 

[erobz]

$P_2$ is 0 gauge. You should be taking point 1 as the entrance of the 0.9 m diameter pipe. Point 2 should be the exit of the 0.25 m pipe. We are assuming the upper tank is at atmospheric pressure (0 gauge).

The only issue that you would need to split it up for, is if the flow in the large 0.9 m pipe is laminar. We know the flow in the 0.25 m pipe is turbulent.

 

[Hydro2022]

$P_2$ is 0 gauge. You should be taking point 1 as the entrance of the 0.9 m diameter pipe. Point 2 should be the exit of the 0.25 m pipe. We are assuming the upper tank is at atmospheric pressure (0 gauge).

The only issue that you would need to split it up for, is if the flow in the large 0.9 m pipe is laminar. We know the flow in the 0.25 m pipe is turbul

$P_2$ is 0 gauge. You should be taking point 1 as the entrance of the 0.9 m diameter pipe. Point 2 should be the exit of the 0.25 m pipe. We are assuming the upper tank is at atmospheric pressure (0 gauge).

The only issue that you would need to split it up for, is if the flow in the large 0.9 m pipe is laminar. We know the flow in the 0.25 m pipe is turbulent.

@erobz
ok I tried again with the calculation, I eliminated from the equation the factors that are not present in the problem, so z1 / hp / ht / P2 as it is 0.

V1 = 3.02 m / s (speed in the section of 0.9 m) calculated with the flow rate of 1.92 m3 / s
V2 = 39.2 m / s (speed in the section of 0,25m) calculated from the flow rate of 1.92 m3 / s
With these speeds I calculated the continuous and localized pressure drops for a total of 500 meters of water column

in the end the formula comes to me like this

where x = P1

in the end the result is P1 = 6 670 000 Pascal

 

[Hydro2022]

@erobz
this pressure now to get the Force I have to multiply it by which surface? that of the tank therefore 48 m2 or that of the 0.9 m section where the water comes out of the tank?
thank you

 

[Hydro2022]

because if I multiply it by the area with a diameter of 0.9 meters, the result is a force of 4235 kN
while if I multiply it by the surface of the tank comes a monstrous force of 320 000 kN

the first solution is the one that comes closest to the true result of 11,000 kN

 

[Hydro2022]

maybe we are getting closer to the solution I hope so

 

[erobz]

@erobz
this pressure now to get the Force I have to multiply it by which surface? that of the tank therefore 48 m2 or that of the 0.9 m section where the water comes out of the tank?
thank you

View attachment 316768

It should be multiplied by the piston surface area. We are ignoring head losses in the tank.

 

[erobz]

If you completely ignore head losses it comes out to about $\approx 85 \cdot 10^3 \, \rm{kN}$. Clearly there is a serious issue.

The static elevation head alone is well over the supposed answer.

$$F_{z} = z_2 \cdot \rho \cdot g \cdot A = 100 \rm{m} \cdot 997 \rm{\frac{kg}{m^3}} \cdot 9.81 \rm{\frac{m}{s^2}} \cdot 48\rm{m^2} \approx 47 \cdot 10^3 \rm{kN}$$

 

[Hydro2022]

If you completely ignore head losses it comes out to about $\approx 85 \cdot 10^3 \, \rm{kN}$. Clearly there is a serious issue.

The static elevation head alone is well over the supposed answer.

$$F_{z} = z_2 \cdot \rho \cdot g \cdot A = 100 \rm{m} \cdot 997 \rm{\frac{kg}{m^3}} \cdot 9.81 \rm{\frac{m}{s^2}} \cdot 48\rm{m^2} \approx 47 \cdot 10^3 \rm{kN}$$

thank you very much for your answer I really appreciate your help, I will try to understand why the result is much lower than the monstrous value we get

 

[erobz]

thank you very much for your answer I really appreciate your help, I will try to understand why the result is much lower than the monstrous value we get

View attachment 316781

I don't think you are going to be able too. Are you sure it's not $110,000 \, \rm{kN}$? What I showed was simply the force on the cylinder to simply hold the $100 \rm{m}$ column statically was practically 5 times the quoted result of $11,000 \rm{kN}$. That is as low as the force could possibly get. The result of $11,000 \rm{kN}$ is unattainable.

 

[Hydro2022]

If you completely ignore head losses it comes out to about $\approx 85 \cdot 10^3 \, \rm{kN}$. Clearly there is a serious issue.

The static elevation head alone is well over the supposed answer.

$$F_{z} = z_2 \cdot \rho \cdot g \cdot A = 100 \rm{m} \cdot 997 \rm{\frac{kg}{m^3}} \cdot 9.81 \rm{\frac{m}{s^2}} \cdot 48\rm{m^2} \approx 47 \cdot 10^3 \rm{kN}$$

only a doubt, in the calculation of the hydrostatic force of 47 000 kN, the force is vertical, why the surface of the 0.25 m is not considered but is multiplied by the entire surface of the tank? the force does not act vertically at the base of the tube of 0,25 m?

 

[erobz]

only a doubt, in the calculation of the hydrostatic force of 47 000 kN, the force is vertical, why the surface of the 0.25 m is not considered but is multiplied by the entire surface of the tank? the force does not act vertically at the base of the tube of 0,25 m?View attachment 316785

You need to review hydrostatic pressure if you are having conceptual difficulty with that part. You should be well beyond that in coursework if you are expected to solve these problems.

 

[Lnewqban]

only a doubt, in the calculation of the hydrostatic force of 47 000 kN, the force is vertical, why the surface of the 0.25 m is not considered but is multiplied by the entire surface of the tank? the force does not act vertically at the base of the tube of 0,25 m?View attachment 316785

The diameter of the pipes carrying the flowing water are only relevant when estimating the resistance that those present to that flow.
Only the height of the vertical column of water matters, regarding hydrostatic (non-moving fluid) pressure.

Please, see:
https://en.wikipedia.org/wiki/Torricelli's_law

https://en.wikipedia.org/wiki/Pascal's_law

Not considering the final pressure of the compressed air in the spherical top tank (assuming that the top tank is open to the atmosphere), the value of the static pressure along the horizontal line that runs through the centers of the blue tank, the 0.9 m diameter pipe and the horizontal section and elbow of the 0.25 m diameter pipe would be the same: $980.64~kPa$ or $980.64~N/m^2$.

Since $P=F/A$, that pressure would push the $48~m^2$ piston with a force of $47040~kN$.

Yes, the walls of the 0.25 m diameter pipe will be "feeling" that same pressure, as well as the masses of fluid located directly forward and after that section.

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/14-1-fluids-density-and-pressure/

 

[Lnewqban]

Perhaps the length of the vertical pipe could be 10 meters rather than 100?
A 10-meter water column would induce a static pressure of $98.06 ~kN/m^2$ on the piston.

Please, see attached PDF drawing at scale to appreciate the real dimensions of the problem as shown at post #1.