Conservation of Energy: Energy Equation & Bernoulli’s Equation

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.n18

19 Conservation of Energy: Energy Equation & Bernoulli’s Equation

Introduction

In developing the energy equation for a fluid flow, the applicable physical principle is a thermodynamic one in that energy cannot be created or destroyed but only converted from one form to another. This latter principle is formally embodied in the first law of thermodynamics applied to a system of a given (fixed) mass, i.e.,

(1) $\begin{equation*} \Delta Q - \Delta W = \Delta E \end{equation*}$

where $\Delta Q$ is the net heat added to the system (such as from a thermal source), $\Delta W$ is the net work done by the system (such as mechanical work), and $\Delta E$ is the change in energy of the mass inside that system, the principle being illustrated in the schematic below. Internal energy can be thought of as the sum of the macroscopic kinetic and potential energy of the molecules comprising the fluid, as well as their microscopic vibrational energy, e.g., temperature effects. Therefore, another way of writing the first law of thermodynamics for a fixed system is

(2) $\begin{equation*} ( Q_{\rm in} - Q_{\rm out} ) - ( W_{\rm in} - W_{\rm out} ) = \Delta E \end{equation*}$

The first law of thermodynamics states that the change in internal energy of a fixed system is equal net heat added less the net work done on that system.

If the system does no external work, all the heat added will manifest as an increase in internal energy, e.g., an increase in temperature. Of course, heat and work can be added to or subtracted from an engineering system, i.e., heat and work can be transferred across the system’s boundaries. The system’s energy can be viewed as being associated with that system’s specific amount of mass. Therefore, the energy terms in the resulting equations are usually written as an energy per unit mass, i.e., per “ $\rho \, d{\cal{V}}$ ” in the spirit of the differential calculus when also following the previous derivations of the equations that apply to the conservation of mass and momentum. If $q$ is the heat added per unit mass and $w$ is the work done per unit mass, then the conservation of energy in this form is

(3) $\begin{equation*} dq - dw = de \end{equation*}$

Objectives of this Lesson

Set up the most general form of the energy equation in integral form.
Know how to simplify the energy equation into various, more practical forms.
Understand how to derive a surrogate for the energy equation for steady, incompressible, inviscid flow, called the Bernoulli equation.

Setting Up the Energy Equation

As before, consider a fixed finite control volume ${\cal V}$ bounded by a surface of area $S$ , as shown in the figure below. At any point inside the control volume or on the control surface, the flow velocity is denoted as $\vec{V}$ . At a point on the control surface, the unit normal $\vec{n}$ or the unit normal vector area $d\vec{S}$ can be defined. Also, let $d{\cal V}$ be an elemental fluid volume inside the control volume, which will contain a fluid of elemental mass $\rho \, d{\cal V}$ .

A finite control volume, fixed in space, as used to set up the energy equation in its most general form, i.e., as a governing equation.

The objective is to develop an analogous equation to the continuity and momentum equations for energy conservation as it applies to a fluid flow. As will be shown, the resulting energy equation describes a power balance equation for the fluid system, i.e., a balance of the rate of work done on the flow versus the rate of energy added to the flow and/or conversion from one form of energy in the flow to another.

Heat Addition

An increment in energy, $de$ can be added to the elemental fluid volume $d{\cal{V}}$ in the form of heat, $dq$ , so that $de = dq$ . Heat energy can be added (or subtracted) in the form of conduction, convection, or radiation, e.g., because of a temperature difference. Heat energy is often added to a system through combustion, e.g., the liberation of heat energy from stored chemical energy in a fuel. Therefore, the rate of volumetric heat addition will be

(4) $\begin{equation*} \mbox{Rate of volumetric heat addition} = \iiint_{\cal{V}} \dot{q} \, \rho \, d{\cal{V}} \end{equation*}$

If the flow is viscous, then frictional heating can also add heat. This latter effect can be important on a high-speed (supersonic) aircraft or a spacecraft during re-entry into the Earth’s atmosphere. While viscous heating effects must be recognized as a complicated process involving the consideration of shear stresses and thermal effects from shock wave formation in the high-speed fluid flow, for now, the effects can be cumulatively denoted by $\dot{Q}_{\mu}$ . Therefore, for the volume of fluid, then

(5) $\begin{equation*} \mbox{Total rate of heat addition} = \iiint_{\cal{V}} \dot{q} \, \rho d{\cal{V}} + \dot{Q}_{\mu} \end{equation*}$

Work

If some work is done on the system, i.e., $dw$ per unit mass, the system’s energy will increase further. Work can be done by the effects of forces resulting from the action of pressure and/or body forces and/or viscous shear and/or mechanical work. Remember that work is equivalent to a force times a distance, and power is the rate of doing work; therefore, power is equivalent to the product of a force and a velocity.

In the case of pressure effects, a pressure force arises from the pressure acting over an area, i.e., for a point on the control surface then $-p \, d\vec{S}$ , the minus sign indicating that the pressure force acts inward for an outward-pointing $d\vec{S}$ (by convention). Therefore, following the same type of process used for the derivation of the momentum equation, then the work done by the pressure forces can be expressed as

(6) $\begin{equation*} \mbox{Rate of doing work by pressure forces} = -\oiint_{S} (p \, d\vec{S}) \cdot \vec{V} \end{equation*}$

Similarly, the effects of body forces can be written as

(7) $\begin{equation*} \mbox{Rate of doing work by body forces} = \iiint_{\cal{V}} ( \rho \vec{f}_b d{\cal{V}} ) \cdot \vec{V} \end{equation*}$

In the case of viscous work, this contribution can be written as $\dot{W}_{\mu}$ for the same reason as heat addition from viscous effects. This term is the rate of work done by viscous stresses acting on the control volume over the control surface.

Finally, some mechanical work could be added to the system, say $\dot{W}_{\rm mech}$ , an example being the work done by a fan or propeller. Work can also be extracted from a fluid flow, such as with a turbine. Therefore, adding all the contributions together gives

(8) $\begin{equation*} \mbox{Total rate of work} = -\oiint_{S} (p \, d\vec{S}) \cdot \vec{V} + \iiint_{\cal{V}} ( \rho \vec{f}_b d{\cal{V}} ) \cdot \vec{V} + \dot{W}_{\mu} + \dot{W}_{\rm mech} \end{equation*}$

Internal Energy

The next step is to examine the internal energy inside the control volume. This form of energy will comprise some “temperature” energy or the microscopic kinetic energy per unit mass of the fluid molecules, $e$ , some macroscopic kinetic energy per unit mass of the fluid, $V^2/2$ , and some macroscopic potential energy per unit mass, $g z$ . In general, internal energy can be written as

(9) $\begin{equation*} \mbox{Energy per unit mass inside $d\cal{V}$} = e + \frac{V^2}{2} + g z \end{equation*}$

where the resultant velocity $V$ is obtained from $V^2 = u^2 + v^2 + w^2$ .

Notice that the microscopic energy kinetic energy per unit mass of the fluid molecules, $e$ , can be considered as the sum of their translational, rotational, and vibrational energy, as shown in the schematic below. The temperature of the fluid is proportional to the average kinetic energy of the molecules, so a thermal gradient, for example, will cause the molecules to travel faster in the direction of increasing temperature.

The microscopic energy kinetic energy of the fluid molecules, $e$ , can be considered as the sum of their translational energy, rotational energy, and vibrational energy,

The macroscopic kinetic energy per unit mass of the fluid per unit mass, $V^2/2$ , and the macroscopic potential energy per unit mass, $g z$ , can be interpreted within the framework of a continuum model, as shown in the schematic below. Increasing the velocity of the fluid will increase its kinetic energy, and increasing the height of the fluid or its pressure “head” will increase its potential energy. Some work on the fluid must be done to increase either.

The macroscopic kinetic and potential energy of the fluid can be interpreted within the framework of a continuum model.

The total energy will then be obtained by integration over the entire mass of fluid contained within the control volume, and remembering that mass can also flow across the control surface from the control volume. Across the element $dS$ , the mass flow rate will be $\rho \, \vec{V} \cdot d\vec{S}$ . Therefore, the total energy flow rate across the entire surface of the control volume then follows by integration, i.e.,

(10) $\begin{equation*} \mbox{Net flow of energy across $S$} = \oiint_{S} (\rho \, \vec{V} \cdot d\vec{S}) \left( e + \frac{V^2}{2} + g z\right) \end{equation*}$

In addition, unsteady effects may be present so that the energy in the system can change because of temporal variations of the flow field properties inside $\cal{V}$ . For this contribution then, the total energy inside the control volume is

(11) $\begin{equation*} \mbox{Total energy inside $\cal{V}$} = \iiint_{\cal{V}} \rho \left( e + \frac{V^2}{2} + g z\right) d{\cal{V}} \end{equation*}$

and the so the time rate of change of energy inside the control volume will be

(12) $\begin{equation*} \mbox{Time rate of change of energy inside $\cal{V}$} = \frac{\partial}{\partial t} \iiint_{\cal{V}} \rho \left( e + \frac{V^2}{2} + g z\right) d{\cal{V}} \end{equation*}$

Finally, adding these two latter terms together (Eqs. 10 and 12) gives

(13) $\begin{eqnarray*} \mbox{Total rate of change in energy} & = & \frac{\partial}{\partial t} \iiint_{\cal{V}} \rho \left( e + \frac{V^2}{2} + g z \right) d{\cal{V}} \\ && \nonumber + \oiint_{S} (\rho \, \vec{V} \cdot d\vec{S}) \left( e + \frac{V^2}{2} + g z \right) \end{eqnarray*}$

Final Form of Energy Equation

Now all of the parts that make up the final form of the energy equation are available in its integral form. To this end, the principle of conservation of energy has been applied to a fluid flowing through a finite control volume that may have heat added or removed and/or mechanical work is done on the fluid or extracted from it.

Because energy is conserved, then in words, it can be stated that: “The rate of heat added plus the rate of doing work to the flow will be equal to the total rate of change of energy of the flow,” i.e.,

(14) $\begin{eqnarray*} \iiint_{\cal{V}} \dot{q} \, \rho d{\cal{V}} + \dot{Q}_{\mu} -\oiint_{S} (p \, d\vec{S}) \cdot \vec{V} + \iiint_{\cal{V}} ( \rho \vec{f}_b d{\cal{V}}) \cdot \vec{V} + \dot{W}_{\mu} + \dot{W}_{\rm mech} = & & \nonumber \\ \frac{\partial}{\partial t} \iiint_{\cal{V}} \rho \left( e + \frac{V^2}{2} + g z\right) d{\cal{V}} + \oiint_{S} (\rho \, \vec{V} \cdot d\vec{S}) \left( e + \frac{V^2}{2} + g z\right) \end{eqnarray*}$

which is mathematically the most general form of the energy equation. Identifying the components again gives

(15) $\begin{eqnarray*} \underbrace{\iiint_{\cal{V}} \dot{q} \, \rho d{\cal{V}}}_{\begin{tabular}{c} Net flow rate\\ of heat\\ into C.V. \end{tabular}} & + & \underbrace{\dot{Q}_{\mu}}_{\begin{tabular}{c} Rate of heat\\ added into C.V. \\ from viscous effects \end{tabular}} -\underbrace{ \oiint_{S} (p \, d\vec{S}) \cdot \vec{V}}_{\begin{tabular}{c} Work done \\ on the C.S. by \\ pressure forces \end{tabular}} + \underbrace{\iiint_{\cal{V}} ( \rho \vec{f}_b d{\cal{V}}) \cdot \vec{V}}_{\begin{tabular}{c} Work done \\ on the C.V. by \\ body forces \end{tabular}} \\[12pt] & & + \underbrace{\dot{W}_{\mu}}_{\begin{tabular}{c} Work done \\ on the C.V. \\ by viscous forces \end{tabular}} + \underbrace{\dot{W}_{\rm mech}}_{\begin{tabular}{c} Work done \\ on the C.V. by \\ mechanical sources \end{tabular}} \\[18pt] = & & \underbrace{\frac{\partial}{\partial t} \iiint_{\cal{V}} \rho \left( e + \frac{V^2}{2} + g z\right) d{\cal{V}}}_{\begin{tabular}{c} Time rate of \\ change of energy \\ inside C.V. \end{tabular}} \quad + \quad \underbrace{\oiint_{S} (\rho \, \vec{V} \cdot d\vec{S}) \left( e + \frac{V^2}{2} + g z\right) }_{\begin{tabular}{c} Net flow of energy \\ out of C.V. per \\ unit time \end{tabular}} \end{eqnarray*}$

This latter equation (Eq. 15) now completes the set of three conservation equations, i.e., mass (continuity), momentum, and energy. Recall for completeness that the continuity equation is

(16) $\begin{equation*} \underbrace{\frac{\partial}{\partial t}\iiint_{\cal{V}} \rho d {\cal{V}}}_{\begin{tabular}{c} Time rate of \\ change of mass \\ inside C.V. \end{tabular}} + \underbrace{ \oiint_S \rho \, \vec{V} \cdot d\vec{S}}_{\begin{tabular}{c} Net mass \\ flow rate \\ out of C.V. \end{tabular}} = 0 \end{equation*}$

and the momentum equation is

(17) $\begin{eqnarray*} \vec{F} & = & \underbrace{\iiint_{\cal{V}} \rho \vec{F} _b d{\cal{V}}}_{\begin{tabular}{c} Sum of \\ body forces \\ acting on C.V. \end{tabular}} -\underbrace{\oiint_S p \, d\vec{S}}_{\begin{tabular}{c} Sum of \\ pressure forces \\ acting on C.S. \end{tabular}} + \underbrace{ \vec{F}_{\mu}}_{\begin{tabular}{c} Sum of \\ viscous forces \\ acting on C.V. \end{tabular}} \\[12pt] & & = \underbrace{ \frac{\partial}{\partial t}\iiint_{\cal{V}} \rho \, \vec{V} d {\cal{V}}}_{\begin{tabular}{c} Time rate of \\ change of momentum \\ inside C.V. \end{tabular}} + \underbrace{\oiint_S (\rho \, \vec{V} \cdot d\vec{S}) \vec{V}}_{\begin{tabular}{c} Net flow of \\ momentum \\ out of C.V. \\ per unit time \end{tabular}} \end{eqnarray*}$

In such collective equations, the unknown parameters may include the velocities, pressures, densities, and temperatures, although not all quantities may be unknown or required in any given problem. A fourth equation is also available to round out the set, namely the equation of state $p = \rho R T$ , which is “handy” because it reduces the number of unknowns.

Simplifications of Energy Equation

The general form of the energy equation in Eq. 14 may look somewhat forbidding. Nevertheless, as with the other conservation equations, the general form can be written in various simplified forms depending on the assumptions that might be made and justified, e.g., steady flow, no body forces, no heat added to the system, no mechanical work, no viscous forces, one-dimensional flow, etc.

However, because thermodynamic principles are being used here, caution must be applied in using the energy equation to ensure that the needed terms are retained in all potential simplification processes. Naturally, in real engineering problem solving, any over-simplification of any governing equations without careful justification is likely to meet with disastrous results.

By way of illustrating the steps that could be used in the simplifications of the energy equation, many practical engineering problems involve fluid systems with just one inlet and one outlet, the mass flow rate through such a system being constant, which is sometimes referred to as single-stream systems. A general depiction of such a single-stream system is shown in the figure below, where the flow comes in at one side of the system and exits on the other. Notice that, in general, there could be a difference in the heights of the inlet and the outlet, so the gravitational potential energy or “head” terms must be retained.

The idea of a single-stream device to which heat is added as well as mechanical work.

For such a single-stream system, the general form of the energy equation can be reduced to the form

(18) $\begin{equation*} \dot{Q} + \dot{W}_{\rm shaft} - \dot{m} \left( \frac{p_2}{\rho_2} - \frac{p_1}{\rho_1} \right) = \dot{m} \left( e_2 - e_1 + \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) \right) \end{equation*}$

where the subscripts 1 and 2 refer to the inlet and outlet conditions, respectively.

Look carefully at Eq. 14 and consider how Eq. 18 is obtained. As you do that, remember that a mass flow rate or the “ $\dot{m}$ term” is always viewed as the flow velocity times the density of the flow times the area through which the flow passes. Notice also the form of the pressure term. The rate of doing work by the pressure forces is equivalent to the product of the pressure times an area times a velocity, so the pressure term in Eq. 18 becomes equivalent to a net pressure times the mass flow rate divided by density.

The left-hand side of the preceding equation represents the energy input, and the right-hand side represents the energy output. The energy input, in this case, comes from the rate of heat transfer to the fluid $\dot{Q}$ , the rate of mechanical work being done on the fluid $\dot{W}_{\rm mech}$ (such as through a shaft driven from outside of the system by an engine or other powerplant), and any work done by the pressure forces.
In such a system, the mass flow through the system is conserved (mass inside the system is constant if the flow is steady), so on the basis of per unit mass then Eq. 18 becomes

(19) $\begin{equation*} q + w_{\rm shaft} = e_2 - e_1 + \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1) + \left( \frac{p_2}{\rho_2} - \frac{p_1}{\rho_1} \right) \end{equation*}$

which is a common form of the steady flow energy balance in thermodynamics. Rearranging this equation gives

(20) $\begin{equation*} w_{\rm shaft} = \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) + \left( \frac{p_2}{\rho_2} - \frac{p_1}{\rho_1} \right) + (e_2 - e_1 - q) \end{equation*}$

The latter term in Eq. 20 can be viewed as the sum of the frictional losses, i.e., the value of $e_2 - e_1 - q \ne 0$ . Now, if the flow is ideal with no irreversible processes such as friction, the total energy must be conserved, and the energy term $e_2 - e_1 - q = 0$ , i.e., the internal energy does not change if there is no heat transfer from frictional effects. Now an even simpler form of the energy conservation equation is available, i.e.,

(21) $\begin{equation*} w_{\rm shaft} = \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) + \left( \frac{p_2}{\rho_2} - \frac{p_1}{\rho_1} \right) \end{equation*}$

Remember that adding or subtracting mechanical work to a flow is common in aerospace systems, such as utilizing a turbine or propeller. In the case of the latter equation, the addition of mechanical work to the system may appear as a change in the kinetic energy of the flow and/or to the potential energy of the flow and/or to the internal temperature and/or pressure in the system or indeed any combination of these effects.

Bernoulli’s Equation

Proceeding further by assuming that no mechanical work is introduced into the system, i.e., $w_{\rm shaft} = 0$ , then $w_{\rm mech} = 0$ , then in this case

(22) $\begin{equation*} 0 = \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) + \left( \frac{p_2}{\rho_2} - \frac{p_1}{\rho_1} \right) \end{equation*}$

Now, if it is also assumed that the flow is incompressible, i.e., $\rho_1 = \rho_2 = \rho = \mbox{constant}$ , then the latter equation can be rearranged into the form

(23) $\begin{equation*} p_1 + \frac{1}{2} \rho V_1^2 + \rho g z_1 = p_2 + \frac{1}{2} \rho V_2^2 + \rho g z_2 \end{equation*}$

or simply that

(24) $\begin{equation*} p + \frac{1}{2} \rho V^2 + \rho g z = \mbox{constant} \end{equation*}$

which is known as the Bernoulli equation or Bernoulli’s principle after Daniel Bernoulli. It will be apparent that it is a statement of energy conservation in that a fluid exchanges its specific kinetic energy for pressure, either static or potential, the specific kinetic energy being the kinetic energy per unit volume.

A more general form of the Bernoulli equation is to leave it in an unintegrated form, i.e.

(25) $\begin{equation*} \frac{dp}{\rho} + \frac{1}{2} d( V^2 ) + g z = 0 \end{equation*}$

After integration then

(26) $\begin{equation*} \int \frac{dp}{\rho} + \frac{1}{2} V^2 + g z = \mbox{constant} \end{equation*}$

and assuming incompressible flow gives

(27) $\begin{equation*} \frac{p}{\rho} + \frac{1}{2} V^2 + g z = \mbox{constant} \end{equation*}$

or

(28) $\begin{equation*} p + \frac{1}{2} \rho V^2 + \rho g z = \mbox{constant} \end{equation*}$

as has been written down previously.

The Bernoulli equation is one of the most famous fluid mechanics equations and is used to solve many practical problems. It has been derived here as a particular degenerate case of the general energy equation, but it can also be derived in several other ways.

Consider the figure below, which represents a streamtube flow of an ideal fluid that is steady, incompressible, and inviscid. At the inlet to the streamtube, the cross-sectional area is $A_1$ with flow velocity $V_1$ and at a height $z_1$ from a reference level. At the outlet, the area is $A_2$ , and the velocity is $V_2$ at height $Z_2$ . In a given time, a mass of fluid $dm$ can be considered to have moved in the streamtube from the inlet to the outlet.

Streamtube flow model used for an alternative derivation of the Bernoulli equation.

Let $p_1$ be the pressure acting uniformly at the inlet at station 1. Because pressure acts inward then the pressure here creates a force that pushes the fluid downstream towards outlet. The work done by the pressure force here will be

(29) $\begin{equation*} W_1 = \left( p_1 \, A_1 \right) dx_1 \end{equation*}$

remembering that work is equal to force times distance. Let $p_2$ be the pressure acting uniformly at the outlet at station 2. In this case, the work done by the pressure force will be

(30) $\begin{equation*} W_2 = -\left( p_2 \, A_2 \right) dx_2 \end{equation*}$

the minus sign indicating that the pressure force here acts upstream the opposite direction to the fluid flow. In addition, there will be work done under gravity, which will be

(31) $\begin{equation*} W_g = - dm \, g \left( z_2 - z_1 \right) = - \rho \, d{\cal{V}} \, g \left( z_2 - z_1 \right) \end{equation*}$

where the mass $dm$ is $\rho \, d{\cal{V}$ and the minus sign here reminding us that work must be put into the fluid system to increase its potential energy. Therefore, the total external work is

(32) $\begin{equation*} W_1 + W_2 + W_g = p_1 \, A_1 \, dx_1 - p_2 \, A_2 \, dx_2 - \rho \, d{\cal{V}} \, g \left( z_2 - z_1 \right) \end{equation*}$

The kinetic energy of the flow must now be considered. The change in kinetic energy of the fluid as it moves through the streamtube is

(33) $\begin{equation*} \Delta KE = \frac{1}{2} dm V_2^2 - \frac{1}{2} dm V_1^2 = \frac{1}{2} \, \rho d{\cal{V}} \, V_2^2 - \frac{1}{2} \, \rho d{\cal{V}} \, V_1^2 \end{equation*}$

The application of the principle of conservation of energy requires that

(34) $\begin{equation*} W_1 + W_2 + W_g = \Delta KE \end{equation*}$

so that

(35) $\begin{equation*} p_1 \, A_1 \, dx_1 - p_2 \, A_2 \, dx_2 - \rho \, d{\cal{V}} \, g \left( z_2 - z_1 \right) = \frac{1}{2} \, \, \rho d{\cal{V}} \, V_2^2 - \frac{1}{2} \, \rho d{\cal{V}} \, V_1^2 \end{equation*}$

Notice also that conservation of mass (continuity equation) requires that

(36) $\begin{equation*} \rho \, d{\cal{V}} = \rho \left( A_1 dx_1 \right) = \rho \left( A_2 dx_2 \right) \end{equation*}$

or just

(37) $\begin{equation*} A_1 \, dx_1 = \, A_2 \, dx_2 \end{equation*}$

so

(38) $\begin{equation*} \left( p_1 - p_2 \right) d{\cal{V}} - \rho \, d{\cal{V}} \, g \, \left( z_2 - z_1 \right) = \frac{1}{2} \, \rho \, d{\cal{V}} \, \left( V_2^2 - V_1^2 \right) \end{equation*}$

Cancelling out the volume $d{\cal{V}}$ and rearranging this latter equation gives

(39) $\begin{equation*} p_1 + \frac{1}{2} \, \rho V_1^2 + \rho g z_1 = p_2 + \frac{1}{2} \, \rho V_2^2 + \rho g z_2 \end{equation*}$

or

(40) $\begin{equation*} p + \frac{1}{2} \, \rho V^2 + \rho g z = \mbox{constant} \end{equation*}$

which, again, is the Bernoulli equation.

Remember that in the derivation of the Bernoulli equation, the flow has been assumed to be steady and incompressible with negligible frictional losses (i.e., an ideal fluid) and where there is no mechanical work added or subtracted from the flow. While the Bernoulli equation is found to have many practical applications, it should be remembered that it has been derived based on these prior assumptions, so its practical use requires considerable caution in actual application.

Worked Example #1 – Water Flow Out of a Nozzle

Water ( $\rho$ = 1,000 kg/m $^3$ ) is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 300 kPa. At the nozzle the pressure decreases at the outlet (discharge) to atmospheric pressure (101.3 kPa). There is no change in height. Use the Bernoulli equation to calculate the velocity of the water exiting the nozzle.

Using the Bernoulli equation ( $\rho =$ constant) then

$p_1 + \frac{1}{2} \rho V_1^2 = p_2 + \frac{1}{2} \rho V_2^2$

Rearranging to solve for $V_2$ gives

$V_2 = \sqrt{ \frac{2 (p_1 - p_2)}{\rho} + V_1^2} = \sqrt{ \frac{2 (300.0 - 101.3) \times 10^3}{1,000} + 1.0^2} = 19.96~\mbox{m/s}$

Worked Example #2- Pressure Drop Along a Pipe

Air flows at low speed through a pipe with a volume flow rate of 0.135 m $^3$ /s. The pipe has an inlet section diameter of 21 cm and and outlet section diameter of 9 cm. A water manometer measures the pressure difference between the inlet and outlet sections. Assuming an ideal fluid then determine the differential height $\Delta h$ shown on the manometer. Take the density of air to be 1.21 kg/m $^3$ and the density of water to be 1,000 kg/m $^3$ . Make any assumptions you feel are justified.

It is reasonable from the information given to assume one-dimensional, steady, incompressible, inviscid flow. The flow rates, flow velocities, and Mach numbers are low enough that compressibility effects can be neglected. Let the inlet be condition 1 and the outlet condition 2. The continuity equation allows us to relate the inlet and outlet conditions, i.e.,

$\dot{m} = \rho A_1 V_1 = \rho A_2 V_2 = \mbox{constant}$

or because the flow is assumed incompressible then just

$\dot{Q} = A_1 V_1 = A_2 V_2 = \mbox{constant} = 0.135~\mbox{m$^3$/s}$

It follows that

$V_1 = \frac{\dot{Q} }{A_1}$

and

$V_2 = \frac{\dot{Q} }{A_2}$

so it is possible to calculate $A_1$ , i.e.,

$A_1 = \frac{\pi d_1^2}{4} = \frac{\pi \, 0.21^2}{4} = 0.0346~\mbox{m$^2$}$

and also $A_2$ , i.e.,

$A_2 = \frac{\pi d_2^2}{4} = \frac{\pi \, 0.09^2}{4} = 0.00636~\mbox{m$^2$}$

so that

$V_1 = \frac{\dot{Q} }{A_1} = \frac{0.135}{0.0346} = 3.90~\mbox{m/s}$

and

$V_2 = \frac{\dot{Q} }{A_2} = \frac{0.135}{0.00636} = 21.22~\mbox{m/s}$

It has been confirmed that the flow velocities are very low, much lower than a Mach number of 0.3, and so the assumption of incompressible flow is justified.

The pressure difference between points 1 and 2 is needed so the use of the Bernoulli equation is justified in that it is an incompressible flow. The Bernoulli equation gives

$p_1 + \frac{1}{2} \rho V_1^2 = p_2 + \frac{1}{2} \rho V_2^2$

so

$p_1 - p_2 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 = \frac{1}{2} \rho \left( V_2^2 - V_1^2 \right)$

Inserting the values for $\rho$ , $V_1$ and $V_2$ gives

$p_1 - p_2 = \frac{1}{2} \rho \left( V_2^2 - V_1^2 \right) = \frac{1}{2} \times 1.21 \left( 21.22^2 - 3.9^2 \right) = 263.25~\mbox{Pa}$

In terms of $h$ then

$p_1 - p_2 = \rho_{\rm H_2 0} \, g \, h = 263.25~\mbox{Pa}$

so solving for $h$ , which is the differential height on the manometer, gives

$h = \frac{p_1 - p_2 }{\rho_{\rm H_2 0} \, g} = \frac{263.25}{1,000 \times 9.81} = 0.0268~\mbox{m} = 2.68~\mbox{cm}$

Compressible Form of Bernoulli Equation

A practical case of a compressible flow is the isentropic (i.e., reversible and adiabatic) high-speed flow of gases through nozzles, diffusers, engines, etc. An isentropic flow is characterized by the relationship

(41) $\begin{equation*} \frac{p}{\rho^{\gamma}} = C = \mbox{constant} \end{equation*}$

where $\gamma$ is the ratio of specific heats. Solving for the density $\rho$ gives

(42) $\begin{equation*} \rho = C^{-1/\gamma} p^{1/\gamma} \end{equation*}$

As previously derived, the general form of the Bernoulli equation can be written as

(43) $\begin{equation*} \int \frac{dp}{\rho} + \frac{1}{2} V^2 + g z = \mbox{constant} \end{equation*}$

Substituting for $\rho$ into the first term gives

(44) $\begin{equation*} \int \frac{dp}{\rho} = \int C^{1/\gamma} p^{-1/\gamma} dp = \left( \frac{\gamma}{\gamma - 1} \right) \frac{p}{\rho} \end{equation*}$

Therefore, a form of the Bernoulli equation for steady, isentropic, compressible flow of an ideal gas becomes

(45) $\begin{equation*} \left( \frac{\gamma}{\gamma - 1} \right) \frac{p}{\rho} + \frac{1}{2} V^2 + g z = \mbox{constant} \end{equation*}$

Summary & Closure

Applying the principle of energy conservation to fluid flow results in a rather formidable-looking equation, at least in its more general form. The energy equation is developed from thermodynamic principles, i.e., the first law of thermodynamics, and uses the concepts of heat, work, and power. The types of energy are internal, potential, and kinetic energy, all of which are involved in most fluid problems. The energy equation is a power equation based on its units, i.e., the rate of doing work. When thermodynamic principles are involved in fluid flow problems, the equation of state is also helpful in establishing the relationships between the known and unknown quantities.

The most common application of the energy equation is to so-called single-stream systems, in which a certain amount of fluid energy comes into a system where work is added or extracted. Then energy comes out of the system. Further simplifying the energy equation to incompressible, inviscid flows without energy addition leads to the Bernoulli equation, which can be used successfully in many practical problems to relate pressures and flow velocities. The Bernoulli equation is also a statement of energy conservation, i.e., fluid exchanges its specific kinetic energy for either static or potential pressure. Nevertheless, the Bernoulli equation must be used carefully and always applied within the assumptions and limitations of its derivation.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

The energy equation is often referred to as a redundant equation for analyzing incompressible, inviscid flows. Why?
What are the three main assumptions used in the derivation of the Bernoulli equation?
The flow through the turbine blades of a jet engine can be modeled using the incompressible form of the Bernoulli equation. True or false? Explain.
In the analysis of the flow through a propeller operating at low flow speeds, then the Bernoulli equation can be used. Explain how to do that.

Additional Online Resources

To learn more about the energy equation, as well as the Bernoulli equation and its uses, check out some of these online resources:

A good video on the first law of thermodynamics and internal energy.
A video lecture describing the derivation of the energy equation for a fluid.
Video on conservation of energy in control volume form.
A great video on better understanding the Bernoulli equation.
The use of graphics in this video reinforce the meaning of the Bernoulli equation.
Another good video describing the applications of the Bernoulli equation.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n18