MrsDrPoe: Conservation of Momentum, Part V

It's Thesis Tuesday on the blog, and today we'll be continuing our examination of the Conservation of Momentum equation, this time focusing on the inviscid reduction of its differential form.

If you recall, the difficulty with the differential momentum equation is that we need to define the stress terms. Previously, we obtained the following results:

d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) = -d/dx(p) + mu*(d/dx(d/dx(u))+d/dy(d/dy(u)) + d/dz(d/dz(u)))

d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) = -d/dy(p) + mu*(d/dx(d/dx(v))+d/dy(d/dy(v)) + d/dz(d/dz(v)))

d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) = -d/dz(p) + mu*(d/dx(d/dx(w))+d/dy(d/dy(w)) + d/dz(d/dz(w)))

If we examine the case of inviscid flow, this means we are looking at an area where viscous effects are negligible (mu = 0). Generally this occurs in the far-field, i.e. in a region of the fluid far away from any walls or boundaries. This leaves us with:

d/dx(Sxx) + d/dy(Txy) + d/dz(Txz) = -d/dx(p)

d/dx(Tyx) + d/dy(Syy) + d/dz(Tyz) = -d/dy(p)

d/dx(Tzx) + d/dy(Tzy) + d/dz(Szz) = -d/dz(p)

So our stress tensor terms are ONLY related to changes in pressure. If we replace these new equalities in the differential form of the equation, we end up with a set of equations known as Euler's Equations. It is important to note, however, that this set is only valid for inviscid flows.

If we apply the assumptions of steady, incompressible flow with no outside forces acting on the fluid to Euler's Equations, we get:

u*d/dx(u) + v*d/dy(u) + w*d/dz(u) = gx - (1/rho)*d/dx(p)


u*d/dx(v) + v*d/dy(v) + w*d/dz(v) = gy - (1/rho)*d/dy(p)


u*d/dx(w) + v*d/dy(w) + w*d/dz(w) = gz - (1/rho)*d/dz(p)

Using a vector identity (which we won't discuss in further detail), and applying along a streamline (lines always parallel or tangent to the flow), we can further reduce this equation to:

0.5*V*V + (p/rho) + g*y = constant (along a streamline)

or: 0.5*V1*V1 + (p1/rho) + g*y1 = 0.5*V2*V2 + (p2/rho) + g*y2

This is Bernoulli's Equation, and it is written for two specific points along a streamline.

Restrictions:

Remember that this equation is only applicable to flows of fluids where the assumptions we made to derive the equation (steady, incompressible, inviscid, along a streamline, no outside forces) hold. It is CRUCIAL to realize that if the flow does not meet these qualifications, this simplified equation CANNOT be used!

If our flow cannot be treated as incompressible, but the other assumptions hold, we can modify the Bernoulli Equation so that it can still be used. For compressible gases, we can apply the ideal gas law to obtain:

0.5*V*V + y*g + R*T*ln(p) = constant (along a streamline)

For unsteady flows, we can modify the equation by the addition of a single term integrating the acceleration over a streamline:

0.5*rho*V1*V1 + p1 + rho*g*y1 = rho*int(d/dt(V))ds + 0.5*rho*V2*V2 + p2 + rho*g*y2

Physical Meaning:

The Bernoulli Equation is simply the mathematical statement of the principle "the work done on a particle by all forces acting on the particle is equal to the change of the kinetic energy of the particle." We can better see that if we look at each of the terms in the equation. As the particle moves, we see forces of pressure and gravity acting on it (the z*g and p/rho terms). This is directly connected to the amount of kinetic energy that the particle has (0.5*V*V).

If we multiply each term in the Bernoulli Equation by the density, rho:

0.5*rho*V*V + p + g*rho*y = constant (along a streamline)

we obtain the pressure form of the equation. Here p represents the thermodynamic pressure of the fluid as it flows (this pressure determines the state of a fluid, i.e. whether it is a gas or a liquid). This value is typically termed the static pressure of the fluid, since it could be measured by moving along with the fluid or being static in relation to the fluid. The third term is the hydrostatic pressure, which is associated with the hydrostatic pressure condition. The second term is called the dynamic pressure since it is associated with the velocity of the fluid.

The combination of the first two terms in the equation is the stagnation pressure. This pressure represents the conversion of all the kinetic energy into a pressure rise, which occurs at a stagnation point where the velocity is equal to zero. The streamline leading to a stagnation point is called a stagnation streamline. The sum of all three terms in this form of the Bernoulli Equation is known as the total pressure, which is constant along a streamline.

Often in Mechanical Engineering, the Bernoulli Equation will be written in head form; this form of the equation is particularly useful for pipe flow and can be derived by dividing each term in the original equation by the gravitational constant, g:

(p/gamma) + (0.5*V*V/g) + y = constant (along a streamline)

In this form, the pressure term is called the pressure head, which represents the height of a column of fluid needed to produce the pressure p. The velocity term is called the velocity head, which represents the vertical distance needed for the fluid to fall freely to reach the velocity V from rest. The height term is the elevation head, and it represents the potential energy of the fluid.


Next week we'll look at some very simple examples of how to employ the Bernoulli Equation. Stay tuned!

MrsDrPoe: Hydrostatic Pressure Forces

Last week we talked about how pressure changes in a static body of fluid. If you recall, we noted that pressure increases as the fluid depth increases and that this pressure variation is linearly related to depth, or in mathematical terms: pressure = specific weight * depth. We call this pressure hydrostatic pressure. Today we will be briefly discussing one result of its existence.

If anything is submerged in a static body of fluid, the fluid applies a force on the object due to the hydrostatic pressure. This hydrostatic pressure force is equal to the pressure acting on the surface of the object times the surface area of the object; it always acts normally (perpendicularly) into the surface.

If we examine a swimming pool for instance, the water in the pool causes a hydrostatic pressure force on the bottom and the sides of the pool. Since the bottom of the pool is all at the same depth, the pressure acting at each point on this surface is equal. This causes a rectangular pressure distribution like the top image in the figure below. Each side of pool varies in depth, so the pressure acting at each point on the surface is not the same. This causes a triangular distribution like the one in the bottom image of the figure below.


We can calculate the resulting pressure force on these surfaces by employing the pressure prism method. This means that we will find the volume of the imaginary prism created by these pressure distributions. For the pool bottom, a rectangular prism is formed; the force is equal to the width of the pool into the page times the length of the pool times the pressure on the surface. For one side of the pool, a triangular prism is formed; the force is equal to the width of the pool into the page times 0.5 times the submerged length of the pool side times the pressure acting on the deepest point on the side of the pool (the bottom of the side).

To find the location of these resulting forces on the pool surfaces, we must find the centroids of the pressure prisms. For a rectangular prism, this would be in the very center of the surface (1/2 width, 1/2 length) and directed downward (normal to the surface). For a triangular prism, this would be at 1/2 of the width and 2/3 of the submerged length of the pool side from the top of the water.

The process for finding hydrostatic pressure forces on curved surfaces is similar but a bit more complicated, so we won't talk about that at this time.

MrsDrPoe: Pressure Variation in Fluids

Good morning and welcome to "Thesis Tuesday" on the blog! Today we're going to be investigating pressure variation in a static (or motionless) body of fluid with as little math as possible. If you do happen to be interested though, this relation can be derived by examining Newton's Second Law, which states that the sum of all the forces acting on an object is equal to the object's mass times its acceleration.

In a static fluid, pressure is a function of depth ONLY. This means that if we look at two points in a body of fluid (such as the ocean) that are right beside each other like this:

* *

the pressure at each point is the same as that at the other point because they are at the same depth. But, if we look at two points like this:

_______________*_________________________________*
_______________________________or this:
_______________*___________________________________________*

the pressure at each point is NOT the same as the one above or below it, respectively.

We also know that pressure is not just a function of depth, but that it actually INCREASES with increasing depth. If we think about our experiences, this physical reality should make sense to us. Why was it so much trouble to fix the Deepwater Horizon oil spill? "Every day" tools and equipment could not be used in this instance because the water pressure at the leak depth was SO great!

The pressure at a given depth in a given (incompressible) fluid can be calculated by multiplying the specific weight of a fluid by the depth. For instance, the specific weight of salt water is 64 pounds/cubic foot, and the depth of the oil leak was 5,000 feet, so the pressure at the leak was 320,000 pounds/square foot. The average surface area of a human is 18.6 square feet, which means that a person located at the oil spill would experience a force of 5,952,000 pounds on his/her body. The 2010 Mississippi State Football team weighed 25,357 pounds collectively...this means that the pressure force that a person located at the leak would feel is equivalent to taking this entire team:


cloning them approximately 233 times and setting ALL of these guys on top of him/her. Pretty crazy huh?

MrsDrPoe: Fluid Pressure Introduction

The next major concepts that we will discuss as far as fluid mechanics goes are related to pressure in fluids. So as a brief introduction and my final post for today, I leave you with a few terms more pressure-related terms to remember for our future discussions on the topic.

pressure - in a stagnant fluid, pressure is defined as the normal force (force perpendicular to the fluid surface) divided by the area over which the force is acting on (similar to the definition of shear stress)

atmospheric pressure - pressure of the atmosphere in a given location

absolute pressure - exact pressure of a fluid in a given location relative to a vacuum

gage pressure - relative pressure of a fluid in a given location; the pressure that a pressure gage inserted into the fluid would read at a given location

absolute pressure = gage pressure + atmospheric pressure