Velocity of water out of reservoir.

In summary, the conversation discusses ways to algebraically determine the constant, k, in the equation u=k sqrt(w) for a cylindrical tank where the efflux speed is proportional to the square root of the depth of the hole from the surface. Suggestions include finding the acceleration in i and j components and integrating for velocity, as well as taking into account the dynamic and static pressure inside the hole. However, some argue that taking pressure into account is not necessary for the problem at hand.
  • #1
John09
2
0
I have a cylindrical tank and I know that the efflux speed is proportional to the square root of the depth of the hole from the surface. So u=k sqrt(w). I need to algebraically determine the constant or k in that situation. Has anyone got any ideas as to how I should approach this? I was thinking that I could try and find the acceleration in i and j components and integrate it for velocity but didn't get far.

Thanks for any help.
 
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  • #2
If you're ignoring the viscous effects, the efflux speed will be such that the dynamic pressure is equal to the static pressure just inside the hole. Dynamic pressure is 1/2*rho*v2, so rearranging for v, we can get that v = sqrt(2*p/rho). Since the pressure in a tank is simply from hydrostatic equilibrium (P = rho*g*h), we can plug in for P:

v = sqrt(2*rho*g*h/rho) = sqrt(2*g*h).

So, your constant is sqrt(2g).
 
  • #3
Hm I don't think taking pressure into account is necessary as it is not part of our coursework.
 
  • #4
If you want to know the velocity in fluid dynamics you need to know 2 things. Volume and pressure to find velocity. Unless you can invent some kinda new math cjl is right.
 
  • #5


I would approach this problem by first understanding the underlying principles and equations governing fluid dynamics. The efflux speed of water from a reservoir can be described by Bernoulli's principle, which states that the total energy of a fluid (in this case, water) remains constant throughout its flow. This means that the kinetic energy of the water as it exits the reservoir is equal to its potential energy at the surface of the reservoir.

In order to algebraically determine the constant k in the given situation, we can use the equation for Bernoulli's principle, which is:

P + 1/2ρv^2 + ρgh = constant

Where P is the pressure, ρ is the density of water, v is the velocity, g is the acceleration due to gravity, and h is the height of the water in the tank.

We can rewrite this equation as:

v = √(2gh + 2k)

Where k is the constant we are trying to determine.

To solve for k, we can use the given information that the efflux speed is proportional to the square root of the depth of the hole from the surface. This means that as the depth increases, the efflux speed also increases.

We can set up a proportion using two different depths and their corresponding efflux speeds:

√(2gh1 + 2k) / √(2gh2 + 2k) = √(h1) / √(h2)

Simplifying this equation, we get:

√(2gh1 + 2k) / √(2gh2 + 2k) = √(h1/h2)

Squaring both sides, we get:

(2gh1 + 2k) / (2gh2 + 2k) = h1/h2

Solving for k, we get:

k = (h1 - h2)(gh1h2) / (h1 + h2)

This is the algebraic expression for the constant k in terms of the two depths and the height of the water in the reservoir.

In conclusion, by using Bernoulli's principle and setting up a proportion, we can algebraically determine the constant k for the given situation of water efflux from a reservoir.
 

FAQ: Velocity of water out of reservoir.

What is the formula for calculating the velocity of water out of a reservoir?

The formula for calculating the velocity of water out of a reservoir is V = √(2gh), where V is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the water above the outlet.

How does the height of the water affect the velocity of water out of a reservoir?

The height of the water has a direct impact on the velocity of water out of a reservoir. The higher the water level, the greater the potential energy, resulting in a higher velocity of water out of the reservoir.

What factors can affect the velocity of water out of a reservoir?

Aside from the height of the water, other factors that can affect the velocity of water out of a reservoir include the size and shape of the outlet, the volume of water in the reservoir, and any obstructions or restrictions in the outlet.

How is the velocity of water out of a reservoir measured?

The velocity of water out of a reservoir can be measured using a flow meter, which can calculate the rate of flow at the outlet. Alternatively, it can be measured manually by timing how long it takes for a known volume of water to pass through the outlet.

What are some practical applications of knowing the velocity of water out of a reservoir?

Knowing the velocity of water out of a reservoir is important in various fields such as civil engineering, hydrology, and agriculture. It can help in designing and maintaining efficient water systems, predicting flood risks, and managing irrigation and drainage systems.

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