# Physics with Calculus/Mechanics/Fluid Mechanics

## Introduction to Fluid Mechanics

What is Fluid Mechanics?

Fluid mechanics is the study of the forces on and between, and the energy and motion of, fluids. The solution to essentially any problem involving fluids--whether the design of a sprinkler system, a dam, or common fresh water and sewage house piping, for instance--requires knowledge of fluid mechanics. Fluid mechanics is composed of two subfields: fluid statics, which is the study of fluids at rest; and fluid dynamics, which is the study of fluids in motion. Before we discuss these two fields in detail, however, it is obligatory for us to delve into several rudimentary yet imperative concepts and definitions.

Definition of a Fluid

A fluid is defined as a collection of molecules where the molecules conform to the shape of their container. Thus, both liquids and gases can be defined as fluids. In contrast, solids tend to retain their shape and hence do not conform easily upon changes in container shape. Obviously, water, oil, and mercury are all examples of fluids; by contrast, steel piping, and concrete are all obvious examples of solids.

Useful properties of fluids

There are several properties of fluids which are integral in the study of fluid mechanics, the most important of which is specific weight (γ). γ is defined as the weight of the fluid divided by its volume; hence, it has the units, in SI, ${\begin{matrix}{\frac {N}{m^{3}}}\end{matrix}}$ ; in English units, γ is in ${\begin{matrix}{\frac {lb}{ft^{3}}}\end{matrix}}$ . The weight of a fluid (or of a solid, for that matter) is

$\mathbf {w} =Mg$

where w is weight, M is mass, and g is the gravitation constant. The units of g in SI are ${\begin{matrix}{\frac {m}{s^{2}}}\end{matrix}}$ , and in English units g is in ${\begin{matrix}{\frac {ft}{s^{2}}}\end{matrix}}$ . Hence, γ is equal to

${\boldsymbol {\gamma }}={\begin{matrix}{\frac {w}{V}}\end{matrix}}$

where V is the volume, which is not to be confused with velocity, which we will define later as v. Another important fluid property is density (ρ), which is defined as the mass of a fluid per unit volume.

${\boldsymbol {\rho }}={\begin{matrix}{\frac {M}{V}}\end{matrix}}$

Its units in SI and English are ${\begin{matrix}{\frac {kg}{m^{3}}}\end{matrix}}$  and ${\begin{matrix}{\frac {slugs}{ft^{3}}}\end{matrix}}$ , respectively. The relationship between density and specific weight is a rather obvious one which the scrupulous reader will no doubt discover quickly:

${\boldsymbol {\gamma }}={\boldsymbol {\rho }}g={\begin{matrix}{\frac {Mg}{V}}\end{matrix}}$

since, obviously, the weight divided by the volume is equivalent to the mass times g divided by the same volume of fluid.

Pressure

The pressure on or due to a fluid is defined as the force exerted divided by the area on which it is exerted. Mathematically, we define pressure to be :

$P={\begin{matrix}{\frac {\mathbf {F} }{A}}\end{matrix}}$

The units of pressure are pascals (Pa); one pascal is equal to ${\begin{matrix}{\frac {1N}{m^{2}}}\end{matrix}}$ . Pressure will form the basis of our first detailed study of fluid mechanics, fluid statics.

## Fluid Statics

Fluid statics, as mentioned previously, is the study of fluids at rest. The pressures being incurred by the sides of a dam holding back non-moving water, for example, could be derived via fluid statics concepts.

There are a few aspects of pressure which we will discuss in this section, which all relate to the question of how pressure changes and acts in a static pool of fluid. For example, we wish to be able to answer questions such as: how many pascals of pressure would there be at a depth of 5 meters of water? Additionally, we would like to be able to make conclusions about the directions in which the fluid pressures acts; for instance, is the pressure on a diver at 5 meters acting upwards equivalent to the pressure on the said diver acting downwards? Furthermore, does the amount of water above the diver affect the pressures the diver will experience? In other words, does the pressure at the bottom of an 8 cm glass with a radius of 1 cm equal the pressure at the bottom of an 8 cm glass with a radius of 8 cm? We begin our discussion with a review of elementary physics.

(As an aside, it is generally assumed that a 'boldface' symbol or letter indicates a vector. In this case, however, boldface is used to emphasize and is not necessarily indicative of a vector.)

Newton's Laws, revisited

Isaac Newton's first and second laws are shown below in mathematical form.

For all objects:

$\mathbf {F} =M\mathbf {a}$

where $\mathbf {a}$  is the acceleration of the object and M is the mass of the object. Therefore, when an object is under no external force, $\mathbf {a}$  must be zero. Hence, in that particular case:

$\sum \mathbf {F_{x}} =\sum \mathbf {F_{y}} =\sum \mathbf {F_{z}} =0$

In fluid statics, of course, we assume that the fluid in question is at rest with respect to its surroundings--and hence $\mathbf {a}$  is zero and the sum of all the forces in each direction must be, as shown above, zero.

We imagine an arbitrary cube of fluid in the middle of a static pool of fluid as shown in Figure 1. We label the unknown pressures acting upon this cube as shown in Figure 1. Let us set the pressures labeled with a "1" to be in the 'positive' directions, and those labeled with a "2" we'll let act in the 'negative' directions. Note that these signs are arbitrary and if switched would produce exactly the same result. We will assume that the fluid does not compress due to the pressures acting on it.

Thus, we have six pressures: $\mathbf {P_{1x}} ,\mathbf {P_{2x}} ,\mathbf {P_{1y}} ,\mathbf {P_{2y}} ,\mathbf {P_{1z}}$ , and $\mathbf {P_{2z}}$  (only some are shown in the figure). Each of these pressures act upon a face of the cube. Of course, the depth of this cube is arbitrary, so we must redefine each of these pressures to some arbitrary pressure at the center of the cube, which we will label $\mathbf {P_{A}}$ .

Each of these pressures can be related to $\mathbf {P_{A}}$  via the change in pressure from the center point (A) to each of the surfaces. For $\mathbf {P_{1x}}$ , for example, the change in pressure is equal to the partial derivative of the pressure with respect to the x direction times the distance between A and the surface on which $\mathbf {P_{1x}}$  acts:

${\boldsymbol {\delta P_{X}}}={\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)$

where ${\boldsymbol {\delta P_{X}}}$  is the change in pressure between point A and the surface one which $\mathbf {P_{1x}}$  and $\mathbf {P_{2x}}$  act. ${\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)$  is the distance from A to either surface. Changes in pressure in the Y and Z directions are similarly found:

${\boldsymbol {\delta P_{Y}}}={\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)$

${\boldsymbol {\delta P_{Z}}}={\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta Z}}\left({\frac {1}{2}}\right)$

Keeping with the usual convention of assuming 'right' is positive, $\mathbf {P_{2x}}$  is equal to $\mathbf {P_{A}}$  plus the change in pressure:

$\mathbf {P_{2x}} ={\boldsymbol {\delta P_{X}}}+P_{A}={\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A}$

Similarly, the change in pressure is negative for $\mathbf {P_{1x}}$ :

$\mathbf {P_{1x}} =-{\boldsymbol {\delta P_{X}}}+P_{A}=-{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A}$

Of course, all of the other pressures can be derived in the same way. Here we assume--remember, though it's arbitrary the opposite sign assignment would produce the same result--that "up" in the Y direction and "out" from the monitor in the Z direction are both positive:

$\mathbf {P_{1y}} =-{\boldsymbol {\delta P_{Y}}}+P_{A}=-{\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A}$

$\mathbf {P_{2y}} ={\boldsymbol {\delta P_{Y}}}+P_{A}={\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A}$

$\mathbf {P_{1Z}} =-{\boldsymbol {\delta P_{Z}}}+P_{A}=-{\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta Z}}\left({\frac {1}{2}}\right)+P_{A}$

$\mathbf {P_{2Z}} ={\boldsymbol {\delta P_{Z}}}+P_{A}={\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta Z}}\left({\frac {1}{2}}\right)+P_{A}$

Now that we have derived the pressures on each face of the cube, we seek to use Newton's Laws to solve for pressure. We know that the total force on the cube must be equal to its mass times its acceleration. However, we began this derivation assuming that the cube was static. Hence,

$\sum \mathbf {F_{T}} =0$

Since we now know the pressures, we can find the forces acting on each face. The force acting in the positive X direction, for instance, is equal to

$\mathbf {F_{1x}} =({\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}})\mathbf {P_{1X}} ={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {-\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A})$

Since, remember,

$\mathbf {F} =PA$

Here, of course, taking a look at Figure 1 it becomes apparent that

${\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}=A$

Using the same method, we state that

$\mathbf {F_{2x}} =({\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}})\mathbf {P_{2X}} ={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A})$

Therefore, we can sum the forces in the X direction to get

$\sum \mathbf {F_{X}} =0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A})-{\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {-\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A})$

$0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+P_{A}+{\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)-P_{A}$

$0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)+{\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta X}}\left({\frac {1}{2}}\right)$

$0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\frac {\partial {\boldsymbol {P}}}{\partial x}}={\frac {\partial {\boldsymbol {P}}}{\partial x}}{\boldsymbol {\delta V}}$

where the ${\boldsymbol {\delta }}$  symbol means 'a small amount', and hence ${\boldsymbol {\delta V}}$  means 'a little volume'.

Note that in this simple equation lies a partial answer to one of our initial questions. We sought to determine the way in which pressure changes. The above equation tells us that the product of the volume of a cube of fluid and the change in pressure in the X-direction is always zero. Assuming that the volume of our arbitrary cube is nonzero, this means that the change in pressure in the X-direction must be zero! The significance of this is greater than you may think. Personal experience tells us that the pressure we feel when diving in a pool depends upon our depth, not our position in the pool. We know from experience that the pressure at the bottom of a pool at one end doesn't feel any different from the pressure at the bottom of the same pool at the other end. Yet, without such experience we may never come to realize this aspect of our world. Using fairly simple logic and mathematics, however, we can not only prove such hypotheses, we can predict them accurately even without direct experience or data.

Moving on to finishing our proof, we can sum the forces in the Z direction in exactly the same way we did for the X-direction. This is because the only two forces acting in the Z-direction are opposing forces due to pressure. As in the X-direction, $\mathbf {P_{A}}$  cancels out and we are left with a similar equation:

$0={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\frac {\partial {\boldsymbol {P}}}{\partial z}}={\frac {\partial {\boldsymbol {P}}}{\partial z}}{\boldsymbol {\delta V}}$

Hence, the change of pressure in the Z-direction in a pool of water must be zero. I suggest to the reader to derive the above if she is not fully convinced of it.

At first we may be tempted to claim that the derivation of pressure for the Y-direction would be essentially the same as in the X- and Z-directions. However, the careful reader will have noticed that in Figure 1 the weight of the fluid has not been neglected. The weight of the fluid is equal to the specific weight times the volume:

${\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\boldsymbol {\gamma }}={\boldsymbol {\delta V\gamma }}={\boldsymbol {W}}$

This is because specific volume, you will remember, is equal to the weight divided by the volume. Therefore, multiplying the specific weight by the volume gives weight.

Therefore, the sum of the forces in the Y direction reveals the following equality:

$\sum \mathbf {F_{Y}} =W={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\boldsymbol {\gamma }}={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A})-{\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}({\frac {-\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta Y}}\left({\frac {1}{2}}\right)+P_{A})$

The pressure at A once again cancels out, and we have

${\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\boldsymbol {\gamma }}={\boldsymbol {\delta Z}}{\boldsymbol {\delta Y}}{\boldsymbol {\delta X}}{\frac {\partial {\boldsymbol {P}}}{\partial y}}={\frac {\partial {\boldsymbol {P}}}{\partial y}}{\boldsymbol {\delta V}}$

Hence, we can cancel out the small volume and we have

${\boldsymbol {\gamma }}={\frac {\partial {\boldsymbol {P}}}{\partial y}}$

Now (finally!!) we can derive the equation for pressure. Multiplying by the derivative of y and integrating, we have

$\int _{P_{1}}^{P_{2}}dP=\int _{y_{1}}^{y_{2}}{\boldsymbol {\gamma }}dy$

${\boldsymbol {\Delta P}}=(y_{2}-y_{1}){\boldsymbol {\gamma }}$

Therefore, the change in pressure between two point in a static fluid is equal to the product of the specific weight of the fluid and the change in depth of the fluid. There are, therefore, two laws we can state regarding fluid statics:

1. Pressure only changes with respect to the depth of a fluid; no other dimensional change affects pressure.

2. Since a point in a static fluid must have no net force acting upon it, pressure acts equally from all angles.

Generally, fluid mechanics problems assume a reference pressure which is equivalent to atmospheric pressure. This is because, since pressure is always measured as a change, it becomes more practical to let atmospheric pressure equal zero, and hence measure all other pressures with respect to this reference pressure. This is called the 'gauge pressure'. For example, if one wants to measure the pressure at the bottom of a lake, it becomes easier to let the atmospheric pressure on the surface of the lake be zero, and therefore the pressure at some depth of the lake is

$\mathbf {P} =(h){\boldsymbol {\gamma }}$

Here, $\mathbf {h}$  is the distance from the reference line of the system, which in the case of a lake, for instance, would be the distance from the surface of the water. The 'absolute pressure', on the other hand, is a pressure which is measured from a reference value of zero. In other words, in this case it would include the atmospheric pressure being imparted on the surface of the lake.

Example Problem 1:Finding the pressure in a tube of water.

Consider the tank and tube setup to the right in Figure 2. Unless we state otherwise, let us assume that every pressure is a gauge pressure. Find the following:

${\boldsymbol {i)}}$  The pressure at point A, $\mathbf {P_{A}}$ .

Solution:

$\mathbf {P_{A}}$  = $\mathbf {h_{A}}$ $\mathbf {\gamma }$ , where $\mathbf {\gamma }$  is the specific weight of the fluid--in this case, water. Hence,

$\mathbf {P_{A}}$  = $2(9.8)\left({\frac {KN}{m^{3}}}\right)m$

$\mathbf {P_{A}}$  = $19.6\left({\frac {KN}{m^{2}}}\right)$

$\mathbf {P_{A}}$  = $19.6\mathbf {KPa}$

${\boldsymbol {ii)}}$  The pressure at point B, $\mathbf {P_{B}}$ .

Solution:

$\mathbf {P_{B}}$  = $[\mathbf {h_{A}} +\mathbf {h_{B}} ]\mathbf {\gamma }$

$\mathbf {P_{B}}$  = $3(9.8)\left({\frac {KN}{m^{3}}}\right)m$

$\mathbf {P_{B}}$  = $29.4\left({\frac {KN}{m^{2}}}\right)$

$\mathbf {P_{B}}$  = $29.4\mathbf {KPa}$

${\boldsymbol {iii)}}$  The absolute pressure at point C, $\mathbf {P_{c}}$ .

Solution:

The absolute pressure at point C is the atmospheric pressure, ${\boldsymbol {P_{0}}}$

${\boldsymbol {iv)}}$  The absolute pressure at point B, $\mathbf {P_{b}}$ .

Solution:

$\mathbf {P_{b}}$  = $[\mathbf {h_{A}} +\mathbf {h_{B}} ]\mathbf {\gamma } +\mathbf {P_{0}}$

$\mathbf {P_{b}}$  = $3(9.8)+101\mathbf {KPa} \left({\frac {KN}{m^{3}}}\right)m$

$\mathbf {P_{b}}$  = $130.4\left({\frac {KN}{m^{2}}}\right)$

$\mathbf {P_{b}}$  = $130.4\mathbf {KPa}$

Example Problem 2:Finding the minimum length of the pipe 'r' needed if no water is to leak from the system.

Figure 3: Finding the length of pipe 'r' needed to ensure that no water leaks from the end of the pipe.

A designer of some type of canister wants to know the minimum length of the pipe 'r', as shown in Figure 3 to the right, that she will need to buy in order to ensure that no water leaks from the system. In addition, she tells you that the pressure at the bottom of the canister is a constant $466\mathbf {KPa}$ . Find the least amount of piping 'r' that she must buy.