Today we continue with Physics, and more specifically the branch of "Classical Mechanics", in order to get into Exercises on Fluid Dynamics.
So, without further ado, let's get straight into it!
Volume Flow Rate Example
Consider a water pipe with varying radius that's filled completely with water. The volume flow rate is 0.8 m3 / sec. Supposing ideal fluid, calculate:
the flow speed at a cross-section of radius 0.2 m, and
the radius of the pipe at a section where the flow speed is 3 m / sec
The flow rate is constant throughout the pipe, and so for the first calculation, we simply have to solve Q = Av for the velocity v. The area of a circle is given by πr2 and thus the result is:
Similarly, for the second case, the missing quantity is now the area A:
And so, the radius is approximately:
Flow Continuity Example
Consider another water pipe with varying radius that's filled completely with ideal fluid. At some cross-section of area 0.1 m2 the flow speed is 2 m / sec. Let's calculate:
the flow speed at cross-sections of 0.05 m2 and 0.15 m2 respectively, and
the volume of fluid that flows through the pipe each hour
Let's skip the volume flow rate calculation, and directly calculate the missing flow speed from the flow continuity equation. As such, the flow speed at the two mentioned cross-sections is:
The volume that flows through the pipe during an hour, which equals 3600 sec, equals the volume flow rate times the duration, V = Qt. The volume flow rate can be calculated at any given cross-section because it remains constant. Choosing the initially given information, the volume that passes in an hour is:
Bernoulli’s Equation Example
Consider a closed fluid tank filled with water up to a height of 10 m, and air on top with a gauge pressure of 5 atm. Water flows out of a small cross-section of some cm at the bottom. What's the velocity of the water flowing out of that cross-section?
The density of water is 1 x 103 Kg / m3 and atmospheric pressure is 1 atm = 1.013 x 105 Pa.
The difference between the areas A1 and A2 is enormous, and so v1 << v2, allowing us to consider the smaller velocity v1 = 0.
In a similar manner, the height at which the water flows out of the cross-section the bottom is much less than the height h1, allowing us to consider y2 = 0.
As such, two terms of Bernoulli's Equation can be easily neglected:
The pressure at the second point equals the atmospheric pressure, p2 = pa, as it's open to the atmosphere. With that everything is known and what remains to be done is solving for v2.
Viscosity Example
Let's lastly also get into an example on real fluids...
Consider water of 20° C that flows through a pipe of 30 cm radius. The viscosity of water at the mentioned conditions is about 1 centipoise (1 cP = 0.01 P). If the velocity of the water at the center of the pipe is 3 m / s and the flow is laminar, calculate:
the velocity 15 cm from the center and pipe walls (at a median distance),
the velocity 1 cm from the pipe wall
the pressure difference at a length of 1 m
The given velocity of 3 m / sec at the center is also the maximum velocity of the flow. As we move further and further from the center the velocity decreases, until it reaches 0 at the pipe walls.
More specifically, if R is the radius of the pipe and r the distance from the center, it can be proven that the velocity is given by the following equation:
As such, in the median distance, the velocity drops to:
Whilst, the velocity at a distance of 30 - 1 = 29 cm is:
Additionally, it can be proven as well that the average velocity of the flow is exactly half the maximum velocity, and so 1.5 m / sec in the case of this example.
The volume flow rate is equal to the area times that average velocity and so:
Now we can apply Poiseuille’s Law in order to determine the pressure difference:
