# Bernoulli equation / fluid flow / finding height

• Polarbear10
In summary: Thank you, Sir.Your question has made me realize that my example about pumps may not be the best, as different types of pumps react differently to increasing pressure in the systems they feed.
Polarbear10
Homework Statement
Bernoulli equation / fluid flow
find height using Bernoulli equation
Relevant Equations
Bernoulli equation
Qv = V/t = A*V
Hi, everyone! Doing some fluid flow/Bernoulli tasks.

Ok, so the task is:
«A hose with a radius of 0,035m is connected to a nozzle, which reduces the radius of the hose to 0,018m (2). The hose carries (qv) 0,0075 m3/sec and the totalpressure (3) in the wide section (1) is 211 kPA. The density of the water is set at 1000 kg/m3.””

• 1: Task one is to find the dynamic pressure in the wide section. I found the velocity to be 1,95 m/s, using V1 = qv/A1. Then calculated the dynamic pressure using Pd = ½ * ρ * v^2 and got 1,89 kPa.
• 2: Second task was to calculate the static pressure. Using Pt = Pd – Ps, and found that PS = 209kPa.
• 3: Then the third task was to find the pressure in the pressure gauge (4). I then calculated the velocity2, using V2=A1*V1/A2, and got 7,37 m/s. I then used Bernouli-equation to calculate the pressure that would be in the gauge (4), and found P2 to be 186 kPa.

4: But the last task: “The nozzle is raised to a higher level. The pressure gauge (3) has a pressure on 241,2 kPa. How high was the nozzle raised”?
I'm struggling to see where I would start. What happens to the pressure in the pressure gauge (4)? Does it change when the nozzle is raised, or will it be the same as in task 3, with P2 being 186 kPa. Also, does the velocity change?

Here is a drawing of the situation, before the nozzle gets lifted.
c

Chestermiller
The velocity can’ change, as the flowmass must be constant.
You should accont for the additional pressure differential induce by the different height new locations of the gauges.

Lnewqban said:
The velocity can’ change, as the flowmass must be constant.
You should accont for the additional pressure differential induce by the different height new locations of the gauges.
Hi! Did you mean that the velocity can or can't change since the flowmass must be constant?

But would it be correct to say that P1; is the new pressure on the pressure gauge (3) = 241,2kPa?
And P2 being what I calculated in task 3?

Polarbear10 said:
Hi! Did you mean that the velocity can or can't change since the flowmass must be constant?

But would it be correct to say that P1; is the new pressure on the pressure gauge (3) = 241,2kPa?
And P2 being what I calculated in task 3?
I meant can’t; sorry.
It seems that the pump making the fluid move could compensate for the additional pressure of lifting the hose, by increasing P1 and P3 and keeping the same mass flowing through the pipe.
I seems that we can assume that the original values of P2 and P4 have remained constant.

Lnewqban said:
I meant can’t; sorry.
It seems that the pump making the fluid move could compensate for the additional pressure of lifting the hose, by increasing P1 and P3 and keeping the same mass flowing through the pipe.
I seems that we can assume that the original values of P2 and P4 have remained constant.
Thanks! That's what I reckon too, so I set up the equation using P1 as 241,2kPa and P2 being 186kPa.
Ended up getting a height (h2) of 3 meters (rounded)..

Lnewqban
Lnewqban said:
It seems that the pump making the fluid move could compensate for the additional pressure of lifting the hose, by increasing P1 and P3 and keeping the same mass flowing through the pipe.
Isn't it more likely that the motor maintains constant power, i.e. volumetric flow rate times static pressure?

haruspex said:
Isn't it more likely that the motor maintains constant power, i.e. volumetric flow rate times static pressure?
Thank you, Sir.
Your question has made me realize that my example about pumps may not be the best, as different types of pumps react differently to increasing pressure in the systems they feed.
Perhaps lifting an imaginary supply tank in the same amount as the end of the hose of our problem has been relocated could be a better example.

## 1. What is the Bernoulli equation and how is it used in fluid flow?

The Bernoulli equation is a fundamental equation in fluid dynamics that describes the conservation of energy in a fluid system. It states that the sum of the pressure, kinetic energy, and potential energy of a fluid remains constant along a streamline. This equation is used to analyze the behavior of fluids in motion, such as in pipes, pumps, and airfoils.

## 2. How do you find the height of a fluid using the Bernoulli equation?

The Bernoulli equation can be rearranged to solve for the height of a fluid. The equation is written as P + 1/2ρv^2 + ρgh = constant, where P is pressure, ρ is density, v is velocity, g is the acceleration due to gravity, and h is the height of the fluid. To find the height, simply rearrange the equation to solve for h: h = (constant - P - 1/2ρv^2)/ρg. This will give you the height at a specific point in the fluid system.

## 3. How does the Bernoulli equation explain the relationship between fluid speed and pressure?

The Bernoulli equation states that as the speed of a fluid increases, the pressure decreases. This is because the total energy of the fluid must remain constant, and an increase in velocity means an increase in kinetic energy. Therefore, the pressure must decrease to maintain the constant energy. This is known as the Bernoulli principle.

## 4. Can the Bernoulli equation be applied to all types of fluids?

The Bernoulli equation can be applied to any fluid, as long as it is incompressible and there are no external forces acting on the fluid. Incompressible fluids are those that do not change in density when subjected to pressure, such as water or air at low speeds. However, the equation may need to be modified for compressible fluids, such as air at high speeds, to account for changes in density.

## 5. How does the Bernoulli equation relate to the conservation of mass in fluid flow?

The Bernoulli equation is based on the principle of conservation of mass, which states that mass cannot be created or destroyed, only transferred or converted into different forms. In fluid flow, this means that the mass of the fluid entering a system must be equal to the mass leaving the system. The Bernoulli equation takes this into account by including the density of the fluid in its calculations.

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