advanced hydrodynamic problem - university level

[soho2025]

hi to all guys are new on this forum, and I write this post to ask you for help on a problem of mechanical engineering, hydrodynamics, advanced level that I am not able to solve and I hope that some of you definitely better than me can help me.

the problem is as follows: (I attach plumbing diagram, sorry I know that it is not accurate, I put it on the flight on the pc copying it from the book of the exercises, unfortunately it is not in scale but it should still be useful to understand the exercise)to find the entity of the force that must be applied to the hydraulic piston, placed in the blue tank below, containing 48 thousand liters of water to empty it in 25 seconds through a tube of diameter 0.9m and length 1 m that shrinks to a diameter of 0.25m and length 105 m connected up to the spherical reservoir also of 48 m3 place at the top.

density of water fluid at 20 degrees celsius = 998,21 kg / m ^ 3

dynamic viscosity of water at 20 degrees celsius = 0,001002 pa.s


calculate :

• total force to be applied to the piston to empty the tank in the specified time, considering the continuous and localized load losses, (very low-wrinkle steel tube)
• power in piston kwh

practically the piston has a 1 meter stroke and it is necessary to calculate the force and power of the latter necessary to empty the tank at the bottom completely and consequently to fill the spherical reservoir of equal volume placed at the top.

I tried to solve it, calculating the necessary force and I get about 3000 kn with a power of 1050 kwh, but I don't know where I'm wrong and the result does not lead with that of the exercise. I calculated the load losses in the 2 shrinkages, in the 3 curves, the continuous load losses and the flow force necessary from the tank, to calculate the power I calculated before the work but nothing to do.

I thank in advance whoever can solve the problem and help me.

Thanks

 

[MassiVonWeizen]

hi to all guys are new on this forum, and I write this post to ask you for help on a problem of mechanical engineering, hydrodynamics, advanced level that I am not able to solve and I hope that some of you definitely better than me can help me.


I thank in advance whoever can solve the problem and help me.

Thanks

tasks do not solve, discuss with the student.
attach your calculations, as you have developed them and your reasoning

 

[meccanicamg]

Is it that you simply wronged the units? If you place the exercise as you solved it, with so much notes and hypotheses we can understand if the error is numerical, formula or concept.
working with water in standard conditions is the application of bernoulli, concentrated and distributed load losses, water weight at altitude and little more.
is the upper tank closed or open? This makes the calculation pressure a little different.
We wait for your match.

 

[soho2025]

tasks do not solve, discuss with the student.
attach your calculations, as you have developed them and your reasoning

@massivonweizen you are right sorry I have not attached the calculations;
practically for the exercise in the calculation of force, I divided the problem in 4 points:

- point a -> calculate the force necessary to move the mass of 48 thousand liters + force of outflow from the tank to the tube. I calculated the area of the d= 0.9m tube therefore 0.635 m2 from the area I obtained the space by making volume/area then the 48 m3 of the tank / 0.635 m2 = 76 m. I then calculated the speed by making space then the 76 meters / 25 seconds = 3 m/s.
from the speed with the formula and = 1/2*m*v^2 = 0.5*48000 kg * (3m/s)^2 = 216000 joules.
dividing the energy for the space taken from the piston which is 1 meter echoes the necessary force therefore always 216000n.

- point b calculation of localized losses of shrinkage from 0.9 m to 0.25 + the 3 curves present in the pipe. with the formula of darcy ya = sa*(v)^2/2g I calculated the localized load losses, with know that it is a tabled coefficient. the velocity of course I had to recalculate it as the section of the tube shrinks therefore I used bernouilli with the formula v1*s1=v2*s2, then from 3 m/s to the exit of the tank in the tube from 0.9 becomes 227 m/s and then increases again when shrinking to 0.25 m it becomes 735 m/s. for the 3 curves that are on the 0.25 m diameter tube I always used the darcy formula. the coefficients used are sa = 4 in shrinkages and 0.6 for curves. the loss of load that I echo is in meters column water, divendo for 10,2 I get the bars then convert in pascal. knowing then the surface and the pressure echoes the applied force.

- calculation of continuous load losses, for the first stretch from 0.9 m in diameter and length 1 meter I did not calculate them as negligible given the modest length of the pipe, while in the stretch from 0.25 in diameter and length 105 meters I calculated first the number of reynolds with the formula re = (speed * diameter * density ) / viscosity I get a number of reynolds over to 300 ml because of the high speed of the flow therefore turbulent motion.
therefore with the formula j = f friction * (1/d)* density * (v^2)/2 * length tubing I calculated the continuous load loss. also here as in the localized ones I divide for 10,2 thus converting from meters column water to bar then divided 100000 for the pascals and knowing pressure and surface I get strength.

- point d minimum force to move the water mass, in this case the calculation is simple because the 48000 kg of water corresponds to 470880 n (obtained by mass * g).

in total I have:

-
- b -> 1500 kn shrinkage load loss + 400n load loss 3 curves
- c -> 414 kn continuous load losses
- d -> 470 kn minimum force to move the water mass

in total I get 2600 kn not 3000 as I had written on the comment above I was wrong before sorry, is that I am not getting anything more about this problem.

then the power actually doesn't come to me even 1050 kwh as I wrote above because I don't know how to calculate the job, my doubt is
the displacement to be considered for the calculation of the work is 1 meter that is the displacement of the piston that moves the water or the 106 meters total that the water travels to reach the tank at the top? because the 106 meters are an indirect consequence of the fact that the water being compressed by the piston in the small section tube, the pressure increases incredibly and then it arrives up going more than 100 meters but I do not think it is right to calculate it in that right way ?Sorry about the confusion in writing the post, but I just can't understand this exercise is driving me crazy all these calculations.

 

[soho2025]

Is it that you simply wronged the units? If you place the exercise as you solved it, with so much notes and hypotheses we can understand if the error is numerical, formula or concept.
working with water in standard conditions is the application of bernoulli, concentrated and distributed load losses, water weight at altitude and little more.
is the upper tank closed or open? This makes the calculation pressure a little different.
We wait for your match.

the tank at the top has closed, with regard to the units of measurement I do not know I really believe to wrong the procedure at this point, as I have done it 5 times rechecking the units of measurement

 

[soho2025]

There are other factors of force/loss of load or pressure that must be considered that I have lost ? why can't I really understand

 

[meccanicamg]

with the geometry of the pipe calculation speed in the sections and therefore the force that serves the piston.
the power to the piston is to make one meter run track that corresponds to the work to shoot the water from the piston to the high tank.

 

[meccanicamg]

how is it possible that the piston has a volume of 48m3 if it has diameter 5 meters and stroke 1 meter? the area is 19,6m2 which multiplied by 1 ago 19,6m3 per pumping cycle.
If I have to move 48m3 in 25s I would say that assuming that 1dm3=1l.... you have a range of 1920l/s and this is the constant calculation flow in the pipe....but if the cylinder does only 19,6m3/race....it is another story.

Since the incomprehensible aqua can be said in the first approximation that the volumetric flow is preserved as the mass is certainly preserved in all sections.

at each diameter corresponds its speed.


the piston will have to raise a column of water equal to the weight, but a stretch is vertical uphill, a horizontal piece.... and a descendant.... need to correctly calculate the component of geodetic pressure. then the piston will have to win the fluid dynamic water leaks in the pipe (focused and distributed leaks).
If you do not put the balance to the system with bernouls, with all the load losses you cannot know the prevalence that will have to provide the piston....so you cannot know the strength that will have to do because f=p•a.
the power, since the carrier work and speed are horizons and parallels, you can write directly as p=f•v....see qui. but careful....the speed is that of the piston, not the exit pipe.

Wherever it is to understand... If it is a piston pump, it will have to do more than one round to make the desired flow rate... otherwise there is at least one wrong figure in the text.

I'm of the idea that the reasoning you've done is not right.

 

[soho2025]

how is it possible that the piston has a volume of 48m3 if it has diameter 5 meters and stroke 1 meter? the area is 19,6m2 which multiplied by 1 ago 19,6m3 per pumping cycle.
If I have to move 48m3 in 25s I would say that assuming that 1dm3=1l.... you have a range of 1920l/s and this is the constant calculation flow in the pipe....but if the cylinder does only 19,6m3/race....it is another story.

Since the incomprehensible aqua can be said in the first approximation that the volumetric flow is preserved as the mass is certainly preserved in all sections.

at each diameter corresponds its speed.


the piston will have to raise a column of water equal to the weight, but a stretch is vertical uphill, a horizontal piece.... and a descendant.... need to correctly calculate the component of geodetic pressure. then the piston will have to win the fluid dynamic water leaks in the pipe (focused and distributed leaks).
If you do not put the balance to the system with bernouls, with all the load losses you cannot know the prevalence that will have to provide the piston....so you cannot know the strength that will have to do because f=p•a.
the power, since the carrier work and speed are horizons and parallels, you can write directly as p=f•v....see qui. but careful....the speed is that of the piston, not the exit pipe.

Wherever it is to understand... If it is a piston pump, it will have to do more than one round to make the desired flow rate... otherwise there is at least one wrong figure in the text.

I'm of the idea that the reasoning you've done is not right.

@meccanicamg Thank you for your explanation. It made me clear. with regard to the volume is 48 m3 because it is a parallelepiped practically is 5 meters wide 1 meter that is the stroke therefore the length is 9,6. 48 m3 = 9,6 *5*1.
error my excuse that I didn't specify it. regarding the characteristics of the component if it is a piston pump I do not know as on the text of the exercise is not specified, but it simply reports piston, so I do not know. the piston from the original design is connected to a panel the latter pushes water, and the panel is surface of 5 meters height for length 9,6 meters while the thickness is not reported because negligible. makes practically the piston plunger

 

[meccanicamg]

then the piston tank contains 48m3 and is squeezed into a single stroke. height 5m, stroke 1m, width 9.6m.

You have to review the approach to exercise.

You will see that you will find more sensible numbers.