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# 5.2: Steady-State Energy Density Model Applied to Fluids

### Fluids at Rest

If there is no translational motion of the fluid, then Bernoulli’s Equation(5.2.1) reduces to

$\Delta P + \rho g \Delta y = 0.\tag {5.2.1}$

As we move vertically upward in a fluid at rest, the gravitational potential energy density increases, so the pressure must decrease by the same amount.  As with all gravitational potential energies, it is only the change in vertical distance that is important.  If we move in a horizontal direction, the pressure remains constant.

Another important point about Bernoulli’s equation(5.2.1) is that it is a relation involving only changes in pressure, height (and velocity squared).  Suppose we add a constant pressure to the entire fluid, by compressing it in a container, for example. Then the absolute pressure increases uniformly throughout the fluid by this same amount. Otherwise, the change in pressure due to a change in height would not be independent of the total pressure.  This result is often referred to Pascal’s Principle.

Figure 5.2.1

We can illustrate many of the features of a static fluid with the aid of the figure 5.2.1.  First, suppose there is a liquid that fills the container up to a depth d.  Above the liquid there is a gas.  Because we are only concerned here with small changes in height, we will neglect changes in gas pressure with height, since $$\rho _{gas} g∆y$$ is negligible compared to $$\rho_{liq} g \Delta y$$.

If the valve is open so the tube is open to the atmosphere, then the pressure above the fluid, $$P_5$$, will be the atmospheric pressure, $$P_A$$.  This pressure, according to Pascal’s principle, is also transmitted throughout the liquid.

Because the elevation of point (1) and point (3) are the same, there is no change in gravitational potential energy density in moving from point (1) to point (3).  Therefore, the pressures at points (1) and (3) must be the same.

At point (4), on the surface of the liquid, the pressure will be the same as at point (5), namely $$P_A$$.

What about the pressure at point (1)?  Now it is useful to “use the model” and an energy-density system diagram.  We know the pressure at point (4), so we can find the pressure at point (1) in terms of the pressure at point (4).  We let these two points be the spatial difference in the Bernoulli equation.

Figure 5.2.2

$\Delta P + \rho g \Delta y = 0 \tag{5.2.2}$

$P_4 - P_1 + \rho g(y_4 - y_1) = 0 \tag{5.2.3}$

$or$

$P_1 = P_4 + \rho g(y_4 - y_1) = P_4 + \rho gh_1 \tag{5.2.4}$

The gravitational energy density has decreased by an amount $$\rho gh_1$$ so the pressure must increase by the same amount.  Similarly, the pressure at $$P_2$$ will be greater than the pressure at $$P_1$$by the amount |$$\rho_{liq} g(y_1 - y_2)$$|.  ﻿

What happens if the tube at the top of the container is attached to a vacuum pump and the pressure is reduced to a value smaller than $$P_A$$?  The pressure differences at the various points will remain the same as before; only the absolute values of the pressure will be different.

Now suppose all of the original gas above the liquid is pumped out.  Will there be a perfect vacuum above the liquid with absolute pressure equal to zero?  No.  The liquid will begin to boil, converting liquid to vapor, and will continue to boil until the vapor pressure above the liquid surface is equal to the vapor pressure of the liquid corresponding to the temperature of the liquid.  Water at room temperature has a vapor pressure of 17.5 mm Hg (2.34 kPa), while the vapor pressure of mercury is many orders of magnitude smaller at room temperature.

### Gauge and Absolute Pressure

It is frequently the case that the absolute pressure of a system is most easily regarded as being the sum of atmospheric pressure plus an additional pressure caused by some process.  The pressure in your bicycle tires, for example, is so many lbs per square inch above atmospheric pressure, caused by the process of “pumping up your tire” which increases the pressure above atmospheric pressure.  When you use a gauge to measure your tire pressure you are measuring only the increase in pressure above one atmosphere.  This is called gauge pressure.  The absolute pressure is equal to the gauge pressure plus the pressure of one atmosphere.

Example: Calculating Height of IV Bag: Blood Pressure and Intravenous

Infusions

Intravenous infusions are usually made with the help of the gravitational force. Assuming that the density of the fluid being administered is 1.00 g/ml, at what height should the IV bag be placed above the entry point so that the fluid just enters the vein if the blood pressure in the vein is 18 mm Hg above atmospheric pressure? Assume that the IV bag is collapsible.

Strategy for (a)

For the fluid to just enter the vein, its pressure at entry must exceed the blood pressure in the vein (18 mm Hg above atmospheric pressure). We therefore need to find the height of fluid that corresponds to this gauge pressure.

Solution

We first need to convert the pressure into SI units. Since $$1.0 \space mm \space Hg = 133 \space Pa$$,

$P = 18 \space mm \space Hg \times \dfrac{133 \space Pa}{1.0 \space mm \space Hg} = 2400 \space Pa$

$= 0.24 \space Pa$

Discussion

The IV bag must be placed at 0.24 m above the entry point into the arm for the fluid to just enter the arm. Generally, IV bags are placed higher than this. You may have noticed that the bags used for blood collection are placed below the donor to allow blood to flow easily from the arm to the bag, which is the opposite direction of flow than required in the example presented here.

### Relation of Pressure to Force

Everything we have done up to this point has treated pressure as an energy density.  Pressure is also related to force, just as energy is related to force through the concept of work.

Consider a piston of area A.  The work done by an outside agent in moving the piston inward is

$dW = Fdx = - PdV. \tag{5.2.5}$

But, since dV = -Adx,  Fdx = PAdx, and we see that

$F = PA \tag{5.2.6}$

That is, the force on an area is directly proportional to the area and the pressure of the fluid at that point.  The force is perpendicular to the area.  We can imagine that the fluid exerts this force on any area, whether it is a wall of a container or just an imaginary area separating two elements of fluid.  In the latter case, the fluid exerts this same pressure on “both sides” of the area, so it is in equilibrium.

### Examples of Fluid Statics

A common example is an inverted cylinder partially or completely full of liquid and pulled part way out of a liquid open to the atmosphere as shown in the figure 5.2.3.  The interesting question is what is the pressure above the liquid in the tube?

Figure 5.2.3

The pressure at point (1) is atmospheric.  The liquid in the tube is raised a distance $$h$$ above the surface of the liquid.  Thus the pressure at the surface of the liquid in the tube must be less than the pressure at point (1) by an amount $$\rho_{liq}gh$$.  Note that the pressure in the top of the tube has no dependence on how far, $$d$$, the tube is immersed into the liquid, or on the pressure, $$P_3$$, at the open end of the tube.  It depends only on the height of the liquid in the tube above the surface.  We can understand the reduced pressure at point (2) as being due to conservation of energy.  As we go up in the tube, the gravitational potential energy density of the fluid increases, so the pressure must decrease to keep the total energy density of the fluid constant.

Whether the liquid fills the tube or does not rise entirely to the top (as shown in the figure 5.2.3) will depend on whether we allowed air to enter the tube when we inserted it into the container.

Let’s suppose we make the tube very long.  What is the maximum height to which the liquid in the tube can rise?  Thinking in terms of energy density, we know that the energy density is the same throughout the fluid.  This energy density is equal to the pressure at the free surface of the container, atmospheric pressure (putting the zero of gravitational potential energy at the surface).  If the tube has only the vapor of the liquid above the surface of the liquid (no air or other gases), then the minimum pressure above the fluid in the tube is the equilibrium vapor pressure of the fluid.  This is small compared to atmospheric pressure, so in this discussion, we will consider it to be zero.  Thus, the maximum height the liquid can rise in the tube is that height which makes the term $$\rho_{liq}gh$$ equal to atmospheric pressure.  Note that this is how a barometer measures atmospheric pressure, and is the origin of the units of pressure in terms of heights of liquids.

Another common example of static fluids is hydraulic machines.  The figure 5.2.4 shows an enclosed fluid with two pistons of different areas.  If the pistons are at the same elevation, then the pressure in the fluid is the same.  The force on each piston is simply the pressure times the area of the piston.

Figure 5.2.4

If the small piston is pressed down with a force $$F_2$$, the pressure created in the fluid is $$P = F_2/A_2$$.  The force this pressure exerts on the larger piston is then $$F_1 = PA_1 = F_2 A_1/A_2$$.  Tremendous increases in force are readily obtained in this manner, by varying the area of the pistons appropriately.  Of course, the energy or power output of a hydraulic machine, as with any machine, cannot be greater than the energy or power put in.  The distance the small piston has to move for a corresponding movement of the large piston is in exactly the inverse ratio of the forces, making the magnitudes of the work done by each piston the same.

A common example of a hydraulic system is the braking system of an automobile.  In an “old fashioned” braking system (without power assist), pushing in on the brake pedal increased the pressure throughout the system, and uniformly applied the same force to the brake shoes on each of the four wheels.  This would be tricky to do with mechanical linkages, but is “automatic” using a hydraulic system.

### Incompressible and Non-Dissipative Flow

The energy density systems for non-dissipative flow in fluids are the gravitational potential energy per unit volume, the translational kinetic energy per unit volume, and the pressure.  (Note that we can also create intensive quantities by dividing by unit mass or even by unit weight of the fluid.  In fact, soil and water scientists and civil engineers typically do divide by weight instead of volume, simply because it is more convenient for their purposes.)

In the energy-interaction model, the change in energy (or other state variable) was always from an initial to final state; that is, from a state earlier in time to a state later in time.  The interaction we were interested in occurred during the time between the initial and final states.  But in the steady-state energy model, the states are not distinguished in time; rather they are distinguished by spatial location in the physical fluid system

For static fluids in equilibrium, all locations in the connected fluid system must have the same total energy density.  (If the density of the fluid changes, we have to be careful to take that into account when we write down specific energies, but the energy density remains constant throughout the interconnected system.)

For flowing fluids, once steady-state has been reached, all locations in the connected fluid system must have the same total energy density.  We also make the restrictions that the fluid is incompressible and that all elements of the fluid move uniformly at the same speed at any particular cross-section.

With these restrictions, the expression representing conservation of energy in non-dissipative fluids is the familiar Bernoulli equation(5.2.7):

$\Delta P + \rho g \Delta y + \frac{1}{2} \rho \Delta (v^2) = 0 \tag{5.2.7}$

### Incompressible and Dissipative Flow

Whether frictional effects need to be taken into account in a flowing fluid depends on many factors, some having to do with the properties of the fluid, others having to do with the geometrical properties of the pipe or channel confining the fluid, and still others related to the rate of fluid flow and the type of flow.  In some cases the energy transferred from fluid energy-density systems to thermal energy-density systems is negligible compared to other transfers of energy.  The traditional Bernoulli equation models these cases adequately.  However, in many cases, friction effects are important or even dominant.  For example, to answer the simple question: “How much water flows out of the bottom of the vertical tube of a funnel in one second?” we have to know the details of the frictional effects.  In fact, whenever the questions we are interested in involve determination of flow rates (how much fluid is flowing), frictional effects are usually going to be important.

How do we deal with frictional effects in fluid flow?  Answer: the same way we did in the energy-interaction model.  We add a thermal energy system to our energy systems and allow for transfer of energy to the thermal system from the other systems.  For fluids, the thermal system will actually be a thermal energy per volume.  That is, we want to add a thermal energy-density system.  In this way we can extend Bernoulli’s equation to include frictional effects (5.2.8).

$\Delta P + \rho g \Delta y + \frac{1}{2} \rho \Delta (v^2) + \frac{\Delta E_{th}}{vol} = 0 \tag{5.2.8}$

#### Adding Energy from Outside Systems: Pumps

When we are analyzing dissipative fluid-flow phenomena, we still restrict ourselves to steady state phenomena.  But without an outside source of energy, the fluid energy-density systems would gradually transfer all of their energy to the thermal energy-density system and the flow would stop.  To accommodate pumps, such as the heart in our circulatory system, we add a term to the right hand side of the extended Bernoulli equation that represents the amount of energy per unit volume of fluid transferred into one or more of the fluid energy-density systems(5.2.9).  That is, we have created an open system in which energy is added to the energy-density systems from outside sources.

$\Delta P + \rho g \Delta y + \frac{1}{2} \rho \Delta (v^2) + \frac{\Delta E_{th}}{vol} = \frac{E_{pump}}{vol} \tag{5.2.9}$

### Getting a Useful Expression for $$\Delta$$Eth/vol

Although we now have a general energy conservation equation to use with many common fluid systems, we can make it much more useful by representing the rate of energy transfer to the thermal system in terms of two variables: the first is the fluid flow rate and the second is what is called the resistance of the particular section of pipe (or artery) we are dealing with.  We first give a plausibility argument for why this works.

Let’s consider how friction comes into play in a fluid flowing through a pipe.  Molecular attractions exist between the molecules of the fluid and the walls of the pipe.  This will cause the molecules closest to the pipe to be essentially stationary.  So molecules a little further away from the wall of the pipe will have to slide past the molecules nearer the wall.  But this sliding involves the momentary making and breaking of bonds as the molecules slide past each other, which leads to the creation of additional random molecular motion.  Random molecular motion, of course, is precisely what thermal energy is.  For steady streamline flow, the variation of fluid velocity appears as shown in the figure 5.2.5.  The vectors represent the velocity of the fluid in different streamlines.  This is type of flow is also called laminar flow.

Figure 5.2.5

The amount of thermal energy generated by molecules sliding past one another should be less if the average fluid velocity is less.  This will occur if the rate of flow is reduced.  It will also occur if the diameter of the pipe is increased, even if the overall amount of fluid flowing through the pipe remains the same.  The amount of thermal energy generated should also depend on the fluid itself.  Molasses molecules “don’t slide past each other” as readily as water molecules, for example.  Viscosity is the fluid property of interest here.  It turns out that it is possible to incorporate these various factors into the two parameters volumetric flow rate and resistance defined below.

Definitions:

$Volumetric ~flow~ rate \equiv \frac{Vol}{time} \equiv Current \equiv I$

$with ~units \frac{m^3}{s}$

The letter $$“I”$$ is a rather standard symbol for electric current, so we will use it here in our discussion of fluids as the symbol for the volumetric flow rate or current.  (Perhaps you have heard expressions such as “The current in the river is swift, so don’t go swimming in it!”)  For fluids, the volumetric flow rate, or current, is the volume of fluid passing through the pipe at a particular location along the pipe per second.  (Advanced texts in other disciplines might use other symbols for flow rate, such as q, for example.)

Resistance $$\equiv$$ (the parameter incorporating all factors other than current, which when multiplied by current, gives the energy transferred to the thermal system per unit volume of fluid.)

$\equiv R ~ with ~units~ of ~ \frac{Js}{m^6}$

Then, by definition,

$\frac{\Delta E_{th}}{vol} \equiv IR\tag{5.2.10}$

Our definition of the resistance, $$R$$, does not require it to be independent of the current, I.  For most fluids, however, the resistance is independent of current for currents up to a certain critical value.  Then it jumps up to a higher (usually non-constant) value as the flow becomes turbulent.  We will shortly go into greater detail as to how the resistance depends on the various fluid and pipe details.  But first, we go back and write the fully extended Bernoulli equation in a very useful way.

### Rewriting the Fully Extended Bernoulli Equation

We substitute $$IR$$ for $$\Delta E_{th}/vol$$ from equation 5.2.7 and also move it from the left side of the equation to the right side, which introduces a negative sign.

$\Delta P + \rho g \Delta y + \frac{1}{2} \rho \Delta (v^2) = \frac{E_{pump}}{vol} – IR\tag{5.2.11}$

Now we want to make sure that the algebraic signs of the two terms on the right of the above expression “make sense” to us.  Let’s check.  The net change in the three fluid energy-density systems will tend to be positive due to the action of a pump.  The pump puts energy into the velocity term, the pressure term, or the gravity potential term, or any combination of the three.  So the “+” sign is appropriate for the pump term.  The net change in the three fluid energy-density systems will tend to be negative due to the result of friction.  As we move along a streamline with the current, the term $$IR$$ will be positive, so we need a minus sign in front of this term to make the net change in the fluid terms negative.  This is an important point regarding the sign of $$“IR.”$$  We repeat:  The algebraic sign of the term $$IR$$ is positive if we move in the direction of the current.  That is, the change in energy of the thermal system, which is equal to $$IR$$, is positive as we move from position (1) to position (2) as shown in the diagram at the right.  This same situation will exist with electric circuits.

Figure 5.2.6

#### One Last Simplification

For two reasons, we will write the fully extended Bernoulli equation (5.2.11) in yet one more way.  The first reason is simply to connect with those folks who deal a lot with liquids, such as soil and water scientists and civil engineers, who use the term “head” with each of the specific terms for the fluid energy-density systems.  These terms are: “pressure head,” “gravity head,” and “velocity head.”  Together, all three terms are called the “total head.”  Rewriting the equation(5.2.11) in terms of total head, makes its similarity to electric circuits, which we discuss next, more obvious.  So, in terms of total head, the fully extended Bernoulli equation(5.2.12) can be simply written as

$\Delta (total~ head) = \frac{E_{pump}}{vol} – IR\tag{5.2.12}$

$where$

$\Delta (total ~head) \equiv \Delta(pressure ~head) + \Delta(gravity ~head) + \Delta(velocity ~head)$

$\equiv \Delta P + \rho g\Delta y + \frac{1}{2} \rho \Delta (v^2)$

### The Fluid Transport Equation

We have finally arrived at the basic fluid transport equation.  It can be applied along any continuous current path between whichever two points we specify.  The change in the fluid energy-density systems (encompassed in the total head) depends explicitly, of course, on the location of the two points along the pipe.  The pump term will be present if there is in fact a pump between the two chosen points.  The thermal energy-density term depends on the resistance between the two points.  We won’t usually write subscripts on all of these terms, but it is mandatory to have in mind the two points whenever we apply it.  This is analogous to having in mind the explicit initial and final state in our previous work with energy conservation.

From now on in these notes we will refer to the fully extended Bernoulli equation(5.2.12) as the fluid transport equation.

(Note: there is a fine point here we don’t want to belabor, but will briefly mention.  Strictly speaking, our fluid transport equation applies along streamlines.  However, if we let the velocity in the velocity head be an average velocity, then we can apply this relation to all of the fluid moving within the walls of the pipe, tube, artery, etc.)

### Power in Relation to Fluid Flow

In general, power is simply the time rate of energy transfer. With respect to an energy system, it is the change in the energy of that energy system per time.  Each term in our fluid transport equation represents either a change in an energy-density system (e.g., $$\Delta E/vol ~or~ P$$) or represents a transfer of energy per unit volume of fluid (e.g., $$IR ~or~ E_{pump}/vol$$).  If we want to determine the amount of energy change that occurs in the fluid as it passes between two points along a pipe per time, we need to multiply the energy change per volume by the volume of fluid passing through the pipe and divide by the time.  These last two factors taken together is what we have defined as the current.  So the power associated with each energy term, is simply the change in the energy of the energy-density term multiplied by the current.  In the case of transfer terms, power is simply the energy transferred per volume multiplied by the current.

$Power = \Delta (energy-density)\times current$

or

$Power = (energy~ transferred)/vol\times current$

$[energy/time] = [energy/vol] \times [vol/time]$

It is almost universal to use the symbol $$P$$ for power, and we will follow this custom.  You will need to be sensitive to the context of the equation to know whether $$“P”$$ means power or pressure.  This will not be confusing, if you always think about the meaning of the equation in which $$“P”$$ appears.

Exactly how we algebraically express power will depend, of course, on the particular context, i.e., exactly which section of the pipe or channel we are focusing on and which energy density system we are focusing on.  There is not one single expression.  Rather, we have to use the basic meaning of power and apply it appropriately to each context.  Several of these are listed here:

$P = \Delta (total ~head) I$

rate of change of all fluid energy density  systems

$P = ∆(energy-density~ sys) I$

rate of change of one fluid energy system

$P = \frac{E_{pump}}{vol} ﻿ I$

rate energy is transferred into the fluid energy density systems by a pump

$P =I^2R$

rate energy is transferred into the thermal energy density system from the fluid energy density systems

### Connection of I tovavg

Because we are dealing with an incompressible fluid that has fixed density, there is a simple relation between the velocity of the fluid and the cross-sectional area.  Since the fluid can’t pile up anywhere (what goes in at one end of the pipe must come out at the other end), the volume of fluid flowing past one point must equal the volume flowing past another point.  That is, the volume flow rate must remain constant and be the same at any cross-section.  If the cross-sectional area changes, then the velocity of the fluid must change to keep the volume constant.

Thus we are let to:

Figure 5.2.7

Fluid/second at point 1 = Fluid/second at point 2

$$dV_1/dt = dV_2/dt$$

$$A_1 dx_1 /dt = A_2 dx_2/dt$$

$$A_1 v_1 = A_2 v_2 = dV/dt = volume~ flow~ rate$$

The only change we need to make for dissipative flow, for which the speed is not constant across a given cross section, is to use the average speed.  We previously defined I to be the volume flow rate, so

$A_1 v_1 = A_2 v_2 = I \tag{5.2.13}$

for both dissipative and non-dissipative flow where we interpret $$v_1$$ and $$v_2$$ to be average velocities at those points. This has historically been referred to as the equation of continuity(5.2.13).

This last equation(5.2.13) along with the fully extended Bernoulli’s equation(5.2.12), allows us to understand all kinds of interesting phenomena involving flowing fluids. Although we derived these for incompressible fluids, they will still give the qualitative behavior for gases when the pressure changes are small in comparison to the other changes.