24 Energy Equation & Bernoulli’s Equation

Introduction

Energy is the capacity to do work, a foundational physical concept in science and engineering. Many forms of energy are recognized, i.e., potential, kinetic, chemical, electrical, and nuclear. For thermodynamic analyses, it is helpful to classify energy into two categories: macroscopic and microscopic energy. Macroscopic energy is the energy a system possesses as a whole with respect to a fixed external reference, including both kinetic and potential energies. Microscopic energy refers to the internal modes of a molecule (translation, rotation, vibration, and chemical bonding). A “system” in this context is a collection of matter of fixed identity.

When developing the energy equation for fluid flow, the governing physical principle is thermodynamic, namely that energy cannot be created or destroyed; it can only be converted from one form to another. This is embodied in the first law of thermodynamics applied to a system of fixed mass, i.e.,

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

Here Q is the net heat transferred into the system, W is the net work done by the system, and {\Delta E} is the change in the system’s total energy. By convention, Q > 0 when heat enters the system and Q < 0 when heat leaves the system. Likewise, W > 0 when work is done by the system on its surroundings, and W < 0 when work is done on the system. An equivalent bookkeeping form that separates the inlets and outlets is

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

as illustrated in Figure 1. Heat and work cross the system boundary and can be written in rate form as

(3)   \begin{equation*} \underbrace{\frac{dQ}{dt}}_{\begin{tabular}{c}\scriptsize Time rate of\\[-3pt]\scriptsize heat into system\end{tabular}} \;-\; \underbrace{\frac{dW}{dt}}_{\begin{tabular}{c}\scriptsize Time rate of\\[-3pt]\scriptsize work by system\end{tabular}} \;=\; \underbrace{\frac{dE}{dt}}_{\begin{tabular}{c}\scriptsize Time rate of change\\[-3pt]\scriptsize of system energy\end{tabular}} \end{equation*}

 

The first law of thermodynamics states that the change in energy of a fixed system equals the net heat added minus the net work done by that system.

It is often convenient to express energy on a per-unit-mass basis. The total energy per unit mass of the fluid is

(4)   \begin{equation*} e = u + \frac{V^{2}}{2} + g\,z \end{equation*}

where u is the internal energy per unit mass, representing the microscopic energy of the fluid arising from molecular motion and interactions (such as temperature), while V^{2}/2 and g\,z are the macroscopic kinetic and potential energies associated with the bulk motion and position of the fluid. For many aerospace applications, the potential energy term is small compared with u and V^{2}/2, but it must be retained when changes in height or altitude are significant.

Many aerospace components are also open systems with mass flow from an inlet to an outlet, as shown in Figure 2. The application of the first law of thermodynamics to such open systems is common in aerospace engineering. For a control volume, the First Law becomes a power balance, i.e., the rate of heat added to the control volume, minus the rate of work done by the control volume, is equal to the rate of change of energy within the control volume plus the net rate at which the mass flow carries energy.

The application of the first law of thermodynamics to an open system with a mass flow is more common in aerospace engineering.

This balance may first be written as

(5)   \begin{equation*} \overbigdot{Q} - \overbigdot{W} = \frac{d}{dt}\left(\Delta E\right) + \sum_{\rm out}\overbigdot{m}\,e - \sum_{\rm in}\overbigdot{m}\,e \end{equation*}

where e is the total energy per unit mass of the fluid, i.e.,

(6)   \begin{equation*} e = u + \frac{V^2}{2} + g\,z \end{equation*}

Notice again that Eq.~5 is really a power balance because it relates the rates of change of energy.

For flowing fluids, pressure forces also do work as the fluid enters and leaves the control volume. The pressure acting on an elemental area produces a force p\,d\vec{S}, and because power is the product of force and velocity, the corresponding rate of work is proportional to (p\,d\vec{S}) \bigcdot \vec{V}. The associated mass flow rate across the same element is \varrho\,\vec{V} \bigcdot d\vec{S}. Therefore, the pressure work per unit mass of the fluid is p/\varrho, which represents the flow work required to push fluid into or out of the control volume. This flow work contributes an additional term to the transported energy, giving

(7)   \begin{equation*} \overbigdot{Q} - \overbigdot{W}_{\rm shaft} = \frac{d}{dt}\left(\Delta E\right) + \sum_{\rm out}\overbigdot{m}\left(e + \frac{p}{\varrho}\right) - \sum_{\rm in}\overbigdot{m}\left(e + \frac{p}{\varrho}\right) \end{equation*}

where \overbigdot{W}_{\rm shaft} is the net rate of shaft or other mechanical work done by the control volume. Because e = u + V^2/2 + g\,z, the pressure-work term may be combined with the internal energy to determine the enthalpy, i.e.,

(8)   \begin{equation*} h = u + \frac{p}{\varrho} \end{equation*}

so that the transported energy may also be written as

(9)   \begin{equation*} h + \frac{V^2}{2} + g\,z \end{equation*}

Notice that under steady operation, the storage term inside the control volume, d(\Delta E)/dt, vanishes.

Learning Objectives

  • 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.
  • Learn how to solve fundamental engineering problems using the energy equation.

Setting Up the Energy Equation

As before, in developing the equations governing the conservation of mass and momentum, consider a fluid flow through a fixed finite control volume (C.V.) of volume {\cal {V}}, bounded by a control surface (C.S.) of area S, as shown in Figure 3. The flow velocity is {\vec{V}} at any point inside the C.V. or on the C.S. At a point on the C.S., the unit outward normal, {\vec{n}}, and the differential vector area, d\vec{S}, are defined. Also, let d{\cal {V}} be an elemental fluid volume inside the C.V., which contains a fluid of elemental mass \varrho \, d{\cal {V}}.

A finite control volume fixed in space is used to establish the energy equation in its most general form.

The objective is to develop an equation expressing the conservation of energy for a fluid flowing through a control volume. In general, the energy of a fluid consists of three primary components: internal energy associated with molecular motion, kinetic energy associated with the bulk velocity of the flow, and potential energy associated with position in a gravitational field. In addition, energy may be transferred into or out of the control volume by heat transfer and work interactions. Accordingly, the energy equation represents a balance among the rate of heat addition to the flow, the rate of work done on or by the flow, and the rates of change and transport of these energy components within the fluid.

In this chapter, the internal energy per unit mass is denoted by u, the total energy per unit mass by

(10)   \begin{equation*} e = u + \frac{V^2}{2} + g\,z \end{equation*}

and the enthalpy per unit mass by

(11)   \begin{equation*} h = u + \frac{p}{\varrho}. \end{equation*}

The flow speed is denoted by V = |\vec{V}|.

Heat Addition

Energy may be added to a fluid by heat transfer. In general, changes in the energy of a fluid element arise from both heat transfer and work interactions, so that the differential relation du = dq applies only when no work is done by or on the element. Heat transfer may occur by conduction, convection, or radiation, resulting from temperature differences within the fluid or between the fluid and its surroundings. In many practical systems, heat is also added through combustion processes, where chemical energy is converted into thermal energy.

If \overbigdot{q} denotes the rate of heat addition per unit mass, then the total rate of heat transfer into the control volume is

(12)   \begin{equation*} \overbigdot{Q} = \iiint_{\cal{V}} \varrho \, \overbigdot{q} \, d{\cal{V}} \end{equation*}

In viscous flows, mechanical energy may be dissipated into internal energy through viscous effects. This process, referred to as viscous dissipation, can be significant in high-speed flows, such as those encountered on supersonic aircraft or spacecraft during atmospheric re-entry. While the detailed modeling of viscous dissipation involves shear stresses and, in compressible flows, shock-wave effects, its net contribution may be represented by an additional term, \overbigdot{Q}_{\mu}.

If an internal-energy equation is being written, the effect of viscous dissipation may be represented as an internal conversion term, so that the thermal contribution may be written schematically as

(13)   \begin{equation*} \overbigdot{Q}_{\rm total} = \iiint_{\cal{V}} \varrho \, \overbigdot{q} \, d{\cal{V}} + \overbigdot{Q}_{\mu} \end{equation*}

Alert: Symbol conflict!

Symbol conflict is common between branches of engineering. In the situation just examined, Q represents heat, and \overbigdot{Q} denotes the rate of heat addition. In other contexts, Q may also denote the volume flow rate. It is essential to avoid such conflicts by redefining a new symbol. For example, the volumetric flow rate can be denoted by {\overbigdot{{\cal{V}}}} to avoid ambiguity.

Work

Energy may also be transferred to or from a fluid by work interactions. In general, work is done by forces acting on the fluid, and because power is the rate of doing work, it may be expressed as the product of a force and a velocity.

For a control volume, work interactions arise from pressure forces acting on the control surface, body forces acting throughout the fluid volume, viscous stresses, and any external mechanical devices such as pumps or turbines.

In the case of pressure forces, the pressure acting on an elemental area produces a force p\,d\vec{S}. The corresponding rate of work is therefore

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

where the negative sign reflects that the pressure force acts inward, opposite to the outward-pointing surface normal.

If body forces other than gravity act on the fluid, they contribute a rate of work given by

(15)   \begin{equation*} \mbox{Rate of work by body forces} = \iiint_{\cal{V}} \varrho \, \vec{f}_{b,\rm ng} \bigcdot \vec{V} \, d{\cal{V}} \end{equation*}

In the present formulation, gravitational effects are already included through the potential-energy term g\,z in e, so gravity should not also be included in \vec{f}_b. Viscous stresses acting on the control surface also perform work, which may be represented collectively as \overbigdot{W}_{\mu}.

Finally, mechanical work may be added to or extracted from the flow by devices such as pumps, turbines, or propellers, which is denoted by \overbigdot{W}_{\rm mech}.

Therefore, the total rate of work interaction is

(16)   \begin{equation*} \overbigdot{W} = -\oiint_{S} (p\,d\vec{S}) \bigcdot \vec{V} + \iiint_{\cal{V}} \varrho \, \vec{f}_{b,\rm ng} \bigcdot \vec{V} \, d{\cal{V}} + \overbigdot{W}_{\mu} + \overbigdot{W}_{\rm mech} \end{equation*}

Internal Energy

The next step is to examine the internal energy within the control volume. Internal energy represents the microscopic energy of the fluid associated with molecular motion and intermolecular interactions. The total energy per unit mass of the fluid is written as

(17)   \begin{equation*} e = u + \frac{V^2}{2} + g \, z \end{equation*}

where u is the internal energy per unit mass, V^2/2 is the macroscopic kinetic energy per unit mass, g\,z is the macroscopic potential energy per unit mass, and V = |\vec{V}| is the velocity magnitude. The internal energy u arises from molecular translation, rotation, vibration, and intermolecular forces; see Figure 4. For most engineering applications, it is sufficient to regard the internal energy as a function of temperature.

The internal energy per unit mass, u, arises from the microscopic motion and interactions of molecules.

At the molecular level, the average energy per molecule is proportional to temperature. For a monatomic gas, this relationship is

(18)   \begin{equation*} E = \frac{3}{2} k_B \, T \end{equation*}

where k_B is Boltzmann’s constant. Diatomic gases, such as nitrogen and oxygen, also possess rotational degrees of freedom, giving

(19)   \begin{equation*} E = \frac{5}{2} k_B \, T \end{equation*}

at moderate temperatures. At higher temperatures, vibrational modes become active, further increasing the internal energy.

The macroscopic kinetic energy per unit mass, V^2/2, and potential energy per unit mass, g \, z, are associated with the bulk motion and position of the fluid. These forms of energy arise from the continuum motion of the fluid and are distinct from the microscopic internal energy, as shown in Figure 5. In this figure, V_1 and V_2 denote the parcel speeds at two locations, while z_1 and z_2 denote the corresponding elevations measured from a common datum.

The macroscopic kinetic and potential energy of the fluid arise from its bulk motion and position.

Energy In & Out of Control Volume

The total energy within the control volume is obtained by integrating over the mass of fluid contained within the C.V. In addition, energy may be transported across the control surface by the flow of mass into and out of the control volume. Across an elemental surface area dS with outward unit normal \vec{n}, the mass flow rate is \varrho \, \vec{V} \bigcdot \vec{n} \, dS, which is equivalently written as \varrho \, \vec{V} \bigcdot d\vec{S}, where d\vec{S} = \vec{n}\, dS is the elemental unit normal area. Because the specific total energy of the fluid is

(20)   \begin{equation*} e = u + \frac{V^2}{2} + g \, z \end{equation*}

the net rate of energy transport across the control surface is

(21)   \begin{equation*} \mbox{Net convective transport of energy across the C.S.} = \oiint_{S} (\varrho \, \vec{V} \bigcdot d\vec{S}) \, e \end{equation*}

If the flow is unsteady, then the total energy within the control volume may vary with time. The total energy inside the control volume is

(22)   \begin{equation*} \mbox{Total energy inside the C.V.} = \iiint_{\cal{V}} \varrho \, e \, d{\cal{V}} \end{equation*}

and the corresponding rate of change is

(23)   \begin{equation*} \mbox{Time rate of change of energy inside C.V.} = \frac{\partial}{\partial t} \iiint_{\cal{V}} \varrho \, e \, d{\cal{V}} \end{equation*}

Combining the storage and transport terms (Eqs. 21 and 23) gives

(24)   \begin{equation*} \mbox{Total rate of change of energy} = \frac{\partial}{\partial t} \iiint_{\cal{V}} \varrho \, e \, d{\cal{V}} + \oiint_{S} (\varrho \, \vec{V} \bigcdot d\vec{S}) \, e \end{equation*}

Final Form of the Energy Equation

Now, all of the parts that make up the final form of the energy equation are available in integral form. To this end, the principle of conservation of energy is applied to a fluid flowing through a finite C.V. that may have heat added or removed and may have mechanical work added to or extracted from it, i.e.,

(25)   \begin{eqnarray*} \underbrace{\iiint_{\cal{V}} \overbigdot{q} \, \varrho \, d{\cal{V}}}_{\begin{tabular}{c} \scriptsize Rate of heat\\[-3pt] \scriptsize added to C.V. \end{tabular}} &-& \underbrace{\oiint_{S} (p \, d\vec{S}) \bigcdot \vec{V}}_{\begin{tabular}{c} \scriptsize Pressure\\[-3pt] \scriptsize work \end{tabular}} + \underbrace{\iiint_{\cal{V}} \varrho \, \vec{f}_{b,\rm ng} \bigcdot \vec{V} \, d{\cal{V}}}_{\begin{tabular}{c} \scriptsize Non-gravitational\\[-3pt] \scriptsize body-force work \end{tabular}} \\[12pt] &+& \underbrace{\oiint_{S} \vec{\tau} \bigcdot \vec{V} \, dS}_{\begin{tabular}{c} \scriptsize Viscous\\[-3pt] \scriptsize work \end{tabular}} + \underbrace{\overbigdot{W}_{\rm mech}}_{\begin{tabular}{c} \scriptsize Mechanical\\[-3pt] \scriptsize work \end{tabular}} \\[18pt] &=& \underbrace{\frac{\partial}{\partial t} \iiint_{\cal{V}} \varrho \, e \, d{\cal{V}}}_{\begin{tabular}{c} \scriptsize Time rate of\\[-3pt] \scriptsize change of energy\\[-3pt] \scriptsize inside C.V. \end{tabular}} + \underbrace{\oiint_{S} \left( \varrho \, \vec{V} \bigcdot d\vec{S} \right) e}_{\begin{tabular}{c} \scriptsize Net convective\\[-3pt] \scriptsize transport of energy\\[-3pt] \scriptsize out of C.V. \end{tabular}} \end{eqnarray*}

where the specific total energy is

(26)   \begin{equation*} e = u + \frac{V^2}{2} + g\,z \end{equation*}

with u being the internal energy per unit mass, V^2/2 the macroscopic kinetic energy per unit mass, and g\,z the macroscopic potential energy per unit mass.

Because energy is conserved, the equation may be stated in words as: “The rate of heat added to the control volume minus the rate of work done by the control volume is equal to the rate of increase of energy within the control volume plus the net rate of energy transport across its boundaries.” This statement is consistent with the first law of thermodynamics, i.e.,

(27)   \begin{equation*} \overbigdot{Q} - \overbigdot{W} = \overbigdot{E} \end{equation*}

Rewriting the energy equation again, without the components being identified, gives

(28)   \begin{eqnarray*} \iiint_{\cal{V}} \overbigdot{q} \, \varrho \, d{\cal{V}} -\oiint_{S} (p \, d\vec{S}) \bigcdot \vec{V} + \iiint_{\cal{V}} \varrho \, \vec{f}_{b,\rm ng} \bigcdot \vec{V} \, d{\cal{V}} + \oiint_{S} \vec{\tau} \bigcdot \vec{V} \, dS + \overbigdot{W}_{\rm mech} &=& \frac{\partial}{\partial t} \iiint_{\cal{V}} \varrho \, e \, d{\cal{V}} + \oiint_{S} \left( \varrho \, \vec{V} \bigcdot d\vec{S} \right) e \end{eqnarray*}

where \vec{f}_{b,\rm ng} denotes any non-gravitational body force per unit mass. Gravity has already been accounted for by the g\,z term in the specific total energy.

This latter equation (Eq. 28) completes the set of three conservation equations: mass (continuity), momentum, and energy. Recall, for completeness, that the continuity equation is

(29)   \begin{equation*} \underbrace{\frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, d{\cal{V}}}_{\begin{tabular}{c} \scriptsize Time rate of \\[-3pt] \scriptsize change of mass \\[-3pt] \scriptsize inside C.V. \end{tabular}} + \underbrace{\oiint_S \varrho \, \vec{V} \bigcdot d\vec{S}}_{\begin{tabular}{c} \scriptsize Net mass \\[-3pt] \scriptsize flow rate \\[-3pt] \scriptsize out of C.V. \end{tabular}} = 0 \end{equation*}

and the momentum equation is

(30)   \begin{flalign*} \!\!\!\!\!\!\!\! \vec{F} = \underbrace{\iiint_{\cal{V}} \varrho \, \vec{f}_b \, d{\cal{V}}}_{\begin{tabular}{c} \scriptsize Sum of \\[-3pt] \scriptsize body forces \\[-3pt] \scriptsize acting on C.V. \end{tabular}} - \underbrace{\oiint_S p \, d\vec{S}}_{\begin{tabular}{c} \scriptsize Sum of \\[-3pt] \scriptsize pressure forces \\[-3pt] \scriptsize acting on C.S. \end{tabular}} + \underbrace{\vec{F}_{\mu}}_{\begin{tabular}{c} \scriptsize Sum of \\[-3pt] \scriptsize viscous forces \\[-3pt] \scriptsize acting on C.V. \end{tabular}} = \underbrace{\frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, \vec{V} \, d{\cal{V}}}_{\begin{tabular}{c} \scriptsize Time rate of \\[-3pt] \scriptsize change of momentum \\[-3pt] \scriptsize inside C.V. \end{tabular}} + \underbrace{\oiint_S \left( \varrho \, \vec{V} \bigcdot d\vec{S} \right) \vec{V}}_{\begin{tabular}{c} \scriptsize Net flow of \\[-3pt] \scriptsize momentum out of C.V. \\[-3pt] \scriptsize per unit time \end{tabular}} \end{flalign*}

In such collective equations, the unknown parameters may include velocities, pressures, densities, and temperatures, although not all quantities are unknown or required in a given problem. A fourth equation is also available to round out the set, namely the equation of state p = \varrho \, R \, T, which is useful because it reduces the number of unknown thermodynamic properties by one.

Simplifications of the Energy Equation

The general form of the energy equation in Eq. 28 may look somewhat forbidding when written out in its entirety. Nevertheless, as with the other conservation equations, it can be simplified by making assumptions (and justifying them), such as steady flow, absence of body forces, no heat transfer, no mechanical work, negligible viscous effects, or one-dimensional flow. However, because thermodynamic principles are involved, caution is required when simplifying the energy equation to ensure that no essential terms are inadvertently neglected. In engineering analysis, unjustified simplifications can lead to erroneous or misleading results.

Single Stream System

To illustrate the steps involved in simplifying the energy equation, it is useful to recognize that many practical engineering problems involve fluid systems with a single inlet and a single outlet, in which the mass flow rate is constant. Such systems are often referred to as single-stream systems. A general depiction is shown in Figure 6. In such cases, there may be differences in velocity, pressure, and elevation between the inlet and outlet, so the corresponding energy terms must be retained.

A single-stream system has a single inlet and a single outlet, with a constant mass flow rate.

For such a system, assuming steady flow, the energy equation reduces to

(31)   \begin{equation*} \overbigdot{Q} - \overbigdot{W}_{\rm shaft} = \overbigdot{m} \left[ (u_2 - u_1) + \left( \frac{p_2}{\varrho_2} - \frac{p_1}{\varrho_1} \right) + \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) \right] \end{equation*}

Using the definition of enthalpy, h = u + \dfrac{p}{\varrho}, this equation may be written more compactly as

(32)   \begin{equation*} \overbigdot{Q} - \overbigdot{W}_{\rm shaft} = \overbigdot{m} \left[ (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) \right] \end{equation*}

For steady flow, the mass flow rate is conserved, so

(33)   \begin{equation*} \overbigdot{m} = \varrho_1 A_1 V_1 = \varrho_2 A_2 V_2 \end{equation*}

Dividing through by \overbigdot{m}, the equation per unit mass becomes

(34)   \begin{equation*} q - w_{\rm shaft} = (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) \end{equation*}

Rearranging gives

(35)   \begin{equation*} w_{\rm shaft} = q - \left[ (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g ( z_2 - z_1 ) \right] \end{equation*}

In practical systems, irreversible effects such as viscous dissipation, turbulence, and heat transfer losses reduce the useful mechanical energy available in the flow. These effects are often grouped into a loss term. It is also useful at this point to introduce the engineering convention that work added to the fluid, such as by a pump, fan, or propeller, is written as w_p, while work extracted from the fluid, such as by a turbine, is written as w_t. With this convention, the mechanical-energy equation may be written as

(36)   \begin{equation*} \frac{p_1}{\varrho_1} + \frac{V_1^2}{2} + g \, z_1 + w_p = \frac{p_2}{\varrho_2} + \frac{V_2^2}{2} + g \, z_2 + w_t + \mbox{losses} \end{equation*}

where w_p is the specific mechanical work added to the fluid, w_t is the specific mechanical work extracted from the fluid, and the loss term represents irreversible conversion of mechanical energy into internal energy.

What are the dimensions of Eq. 36? For this equation to be dimensionally homogeneous, all terms must have the same dimensions and units, i.e.,

    \[ \left[ \frac{p}{\varrho} \right] = \frac{M L^{-1} T^{-2}}{M L^{-3}} = L^{2} T^{-2} \]

and

    \[ \left[ \frac{V^2}{2} \right] = (L T^{-1})^2 = L^{2} T^{-2} \]

and

    \[ \left[ g \, z \right] = (L T^{-2})(L) = L^{2} T^{-2} \]

Therefore, the equation is dimensionally homogeneous, with dimensions of L^{2} T^{-2}, corresponding to energy per unit mass, with units of \mathrm{m^2\,s^{-2}} (SI) or \mathrm{ft^2\,s^{-2}} (USC).

Particular Case: No Losses

The total mechanical energy is conserved in an ideal flow with no irreversible processes such as turbulence, friction, or viscosity. In this case, the loss term is zero, and Eq. 36 reduces to

(37)   \begin{equation*} \frac{p_1}{\varrho_1} + \frac{V_1^2}{2} + g \, z_1 + w_p = \frac{p_2}{\varrho_2} + \frac{V_2^2}{2} + g \, z_2 + w_t \end{equation*}

or, in terms of power, after multiplying through by the mass flow rate,

(38)   \begin{equation*} \overbigdot{m} \bigg( \frac{p_1}{\varrho_1} + \frac{V_1^2}{2} + g \, z_1 \bigg) + P_p = \overbigdot{m} \bigg( \frac{p_2}{\varrho_2} + \frac{V_2^2}{2} + g \, z_2 \bigg) + P_t \end{equation*}

Mechanical Work

When mechanical work is added to or extracted from a flow, such as by a pump, fan, turbine, or propeller, it is useful to keep the added and extracted powers as separate positive quantities. Let P_p denote the power added to the fluid by a pump, fan, or propeller, and let P_t denote the power extracted from the fluid by a turbine. This convention avoids ambiguity in the sign of the shaft-work term. Therefore, the mechanical energy equation may be written as

(39)   \begin{equation*} \underbrace{ \overbigdot{m} \bigg( \frac{p_1}{\varrho_1} + \frac{V_1^2}{2} + g \, z_1 \bigg) }_{\begin{tabular}{c} \scriptsize Rate of energy \\[-3pt] \scriptsize entering C.V. \end{tabular}} + \underbrace{ P_p}_{\begin{tabular}{c} \scriptsize Power added \\[-3pt] \scriptsize to the fluid \end{tabular}} = \underbrace{\overbigdot{m} \bigg( \frac{p_2}{\varrho_2} + \frac{V_2^2}{2} + g \, z_2 \bigg) }_{\begin{tabular}{c} \scriptsize Rate of energy\\[-3pt] \scriptsize leaving C.V. \end{tabular}} + \underbrace{ P_t}_{\begin{tabular}{c} \scriptsize Power extracted \\[-3pt] \scriptsize from the fluid \end{tabular}} \end{equation*}

Finally, if the fluid is assumed to be incompressible, i.e., \varrho_1 = \varrho_2 = \varrho, then

(40)   \begin{equation*} \overbigdot{m} \bigg( \frac{p_1}{\varrho} + \frac{V_1^2}{2} + g \, z_1 \bigg) + P_p = \overbigdot{m} \bigg( \frac{p_2}{\varrho} + \frac{V_2^2}{2} + g \, z_2 \bigg) + P_t \end{equation*}

or, in terms of energy per unit mass,

(41)   \begin{equation*} \frac{p_1}{\varrho} + \frac{V_1^2}{2} + g \, z_1 + w_p = \frac{p_2}{\varrho} + \frac{V_2^2}{2} + g \, z_2 + w_t \end{equation*}

which is called the extended Bernoulli equation.

Energy Equation in Terms of “Head”

Civil and hydraulic engineers, who deal primarily with liquids rather than gases, often express the energy equation in terms of pressure head and specific weight. The pressure head, h or H, is defined as the height of a liquid column corresponding to a given pressure, so that

(42)   \begin{equation*} h \equiv H = \frac{p}{\varrho \, g} \end{equation*}

and has units of length (meters or feet). This pressure-head representation is often convenient for analyzing incompressible flows.

Caution: Symbol conflict! In this section, the symbol h denotes hydraulic head, which has units of length. Earlier in the chapter, h was used to denote enthalpy per unit mass, with units of energy per unit mass. Therefore, the meaning of h must be inferred from the context.

Specific Weight

The specific weight is the weight per unit volume of a fluid. It is commonly denoted by \gamma, but here the symbol w is used to avoid confusion with the ratio of specific heats. Therefore,

(43)   \begin{equation*} w = \frac{\mbox{\small Weight of fluid}}{\mbox{\small Volume of fluid}} = \varrho \, g \end{equation*}

or equivalently,

(44)   \begin{equation*} \varrho = \frac{w}{g} \end{equation*}

Using this definition, the incompressible energy equation (Eq. 41) can be written in head form as

(45)   \begin{equation*} \frac{p_1}{w} + \frac{V_1^2}{2g} + z_1 + h_p = \frac{p_2}{w} + \frac{V_2^2}{2g} + z_2 + h_t \end{equation*}

where all terms have units of length.

If frictional losses are included, represented by a head loss h_l, then

(46)   \begin{equation*} \frac{p_1}{w} + \frac{V_1^2}{2g} + z_1 + h_p = \frac{p_2}{w} + \frac{V_2^2}{2g} + z_2 + h_t + h_l \end{equation*}

Pressure Head of a Pump or Turbine

The power associated with a fluid stream may be written as

(47)   \begin{equation*} P = \overbigdot{m} \, g \, h \end{equation*}

Using \overbigdot{m} = \varrho \, \overbigdot{\cal V}, this becomes

(48)   \begin{equation*} P = w \, \overbigdot{\cal V} \, h \end{equation*}

For a pump, the required input power exceeds the useful fluid power, so

(49)   \begin{equation*} P_p = \frac{w \, \overbigdot{\cal V} \, h_p}{\eta_p} \end{equation*}

and so

(50)   \begin{equation*} h_p = \frac{\eta_p \, P_p}{w \, \overbigdot{\cal V}} \end{equation*}

For a turbine, the output power is less than the available fluid power, so

(51)   \begin{equation*} P_t = \eta_t \, w \, \overbigdot{\cal V} \, h_t \end{equation*}

giving

(52)   \begin{equation*} h_t = \frac{P_t}{\eta_t \, w \, \overbigdot{\cal V}} \end{equation*}

Caution: Units versus variables! Do not confuse the pressure head added by a pump, h_p, which is a variable with units of length, with horsepower, denoted by “hp.” Conventionally, units are never italicized, whereas variables are italicized. The notation is similar, but the meanings are entirely different.

Check Your Understanding #1 – Using the energy equation to calculate pumping power

Apply conservation of energy principles to calculate the power required by a hydraulic pump to deliver fluid (oil) from one location to another. The pump is 75% efficient in converting mechanical input work into useful fluid energy. The inlet pressure, p_1, is 100 kPa, and the outlet pressure, p_2, is 500 kPa. The diameter of the inlet pipe, {d_1}, is 10 cm, and the outlet pipe, {d_2}, is 5 cm. The flow rate, {\overbigdot{{\cal{V}}}}, is 60 m3/hr. The height difference, {z_2 - z_1}, is 3.2 m. The density of the hydraulic fluid is 850 kg/m3. Assume the internal fluid losses are equivalent to a head loss of h_l = 1.4~\mbox{m}.

Show solution/hide solution.

First, it is necessary to find the inlet and outlet velocities, {V_1} and {V_2}, respectively, i.e.,

    \[ V_1 = \frac{\overbigdot{{\cal{V}}}}{A_1} = \frac{4 {\overbigdot{{\cal{V}}}}}{\pi d_1^2} = \frac{4 \times 60.0/3,600}{\pi \times 0.1^2} =2.12~\mbox{m/s} \]

and for {V_2}, then

    \[ V_2 = \frac{\overbigdot{{\cal{V}}}}{A_2} = \frac{4 {\overbigdot{{\cal{V}}}}}{\pi d_2^2} = \frac{4 \times 60.0/3,600}{\pi \times 0.05^2} =8.49~\mbox{m/s} \]

The relevant form of the energy equation is

    \[ \frac{p_1}{\varrho \, g} + \frac{V_1^2}{2g} + z_1 + {h}_{p} = \frac{p_2}{\varrho \, g} + \frac{V_2^2}{2g} + z_2 + h_l \]

So, solving for h_p gives

    \[ h_{p} = \frac{p_2 - p_1}{\varrho \, g} + \frac{V_2^2 - V_1^2}{2g} + (z_2 - z_1) + h_l \]

Inserting the known values gives

    \[ h_{p} = \frac{(500.0 - 100.0) \times 10^3}{850.0 \times 9.81} + \frac{8.49^2 - 2.12^2}{2.0 \times 9.81} + 3.2 + 1.4 = 56.15~\mbox{m} \]

The power required from the pump is

    \[ P_p = \frac{ \varrho \, g \, {\overbigdot{{\cal{V}}}} \, h_p}{\eta_p} \]

Inserting the numerical values gives

    \[ P_p = \frac{ 850.0 \times 9.81 \times (60.0/3,600) \times 56.15}{0.75} = 10.41 ~\mbox{kW} \]

Bernoulli’s Equation

For a steady, incompressible, and inviscid flow with no mechanical work added to or extracted from the fluid system, i.e., w_p = w_t = 0 and no energy losses, Eq. 41 reduces to

(53)   \begin{equation*} \frac{p_1}{\varrho} + \frac{V_1^2}{2} + g \, z_1 = \frac{p_2}{\varrho} + \frac{V_2^2}{2} + g \, z_2 \end{equation*}

This equation may be rearranged into the form

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

or, more generally,

(55)   \begin{equation*} \boxed{ \underbrace{p}_{\begin{tabular}{c}\scriptsize Static\\[-3pt]\scriptsize pressure\end{tabular}} + \underbrace{\frac{1}{2}\,\varrho\,V^2}_{\begin{tabular}{c}\scriptsize Dynamic\\[-3pt]\scriptsize pressure\end{tabular}} + \underbrace{\varrho\,g\,z}_{\begin{tabular}{c}\scriptsize Hydrostatic\\[-3pt]\scriptsize pressure\end{tabular}} = \underbrace{\mbox{constant}}_{\begin{tabular}{c}\scriptsize Total mechanical\\[-3pt]\scriptsize pressure along streamline\end{tabular}} }^{~*} \end{equation*}

which is known as Bernoulli’s equation. Note that this form of the Bernoulli equation has units of pressure (force per unit area), not energy per unit mass. Therefore, it may be interpreted as a surrogate form of the energy equation expressed per unit volume. The equation shows how pressure, kinetic energy, and potential energy are interconverted in a flowing fluid, while their sum remains constant along a streamline.

Bernoulli’s equation is one of the most widely used relations in fluid mechanics and can be applied to many practical problems. It has been derived here as a particular case of the general energy equation for steady, inviscid, incompressible flow with no shaft work or energy losses. Although it is not expressed in terms of energy per unit mass, it is entirely consistent with the conservation of energy.

Bernoulli’s equation may also be written in several equivalent forms depending on the chosen normalization. In terms of energy per unit mass, then

(56)   \begin{equation*} \frac{p}{\varrho} + \frac{V^2}{2} + g \, z = \text{constant} \end{equation*}

Multiplying by \varrho gives the form in terms of energy per unit volume (pressure form), i.e.,

(57)   \begin{equation*} p + \frac{1}{2} \, \varrho \, V^2 + \varrho \, g \, z = \text{constant} \end{equation*}

Dividing by g gives the head form, i.e.,

(58)   \begin{equation*} \frac{p}{\varrho \, g} + \frac{V^2}{2g} + z = \text{constant} \end{equation*}

All three forms are entirely equivalent and differ only by normalization.

Bernoulli’s Equation from Euler’s Equation

In the previous chapter, the streamline form of the Euler equations was obtained as

(59)   \begin{equation*} V\, dV = -\frac{1}{\varrho}\,dp - g\,dz \end{equation*}

which expresses the balance of forces acting on a fluid particle moving along a streamline in steady, inviscid flow. Integrating this equation along a streamline gives

(60)   \begin{equation*} \int V\,dV + \int \frac{1}{\varrho}\,dp + \int g\,dz = 0 \end{equation*}

For an incompressible flow, where \varrho is constant, this becomes

(61)   \begin{equation*} \frac{V^2}{2} + \frac{p}{\varrho} + g \, z = \text{constant} \end{equation*}

or equivalently

(62)   \begin{equation*} \frac{1}{2} \, \varrho \, V^2 + p + \varrho \, g \, z = \text{constant} \end{equation*}

which is Bernoulli’s equation. Therefore, Bernoulli’s equation can be viewed as a direct consequence of the momentum equation and provides a convenient relationship between velocity, pressure, and elevation in a fluid flow.

Another Derivation of Bernoulli’s Equation

A more physically intuitive derivation of Bernoulli’s equation can be obtained by considering the streamtube flow shown in Figure 7. The flow is assumed to be steady, incompressible, and inviscid, with no heat transfer and no mechanical work added to or extracted from the fluid.

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

Consider a fluid element of mass dm moving through the streamtube from an inlet section (1) to an outlet section (2). At the inlet, the area is A_1, the velocity is V_1, and the elevation is z_1. At the outlet, the corresponding values are {A_2}, V_2, and z_2. The pressure at the inlet, p_1, exerts a force on the fluid element that does work in the direction of motion:

(63)   \begin{equation*} W_1 = (p_1 \, A_1)\,dx_1 \end{equation*}

At the outlet, the pressure p_2 exerts a force opposing the motion, so that

(64)   \begin{equation*} W_2 = -(p_2 \, A_2)\,dx_2 \end{equation*}

The work done by gravity on the fluid element is

(65)   \begin{equation*} W_g = -dm \, g (z_2 - z_1) \end{equation*}

which is negative when the fluid moves to a higher elevation. Therefore, the total work done on the fluid element is

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

Because the flow is inviscid and adiabatic, with no shaft work, the work done on the fluid results in changes to its kinetic and potential energy, i.e.,

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

where the change in kinetic energy is

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

Using continuity,

(69)   \begin{equation*} dm = \varrho \, d{\cal V} = \varrho \, A_1 \, dx_1 = \varrho \, A_2 \, dx_2 \end{equation*}

Substituting these relations gives

(70)   \begin{equation*} (p_1 - p_2)\, d{\cal V} - \varrho \, d{\cal V} \, g (z_2 - z_1) = \frac{1}{2} \, \varrho \, d{\cal V} (V_2^2 - V_1^2) \end{equation*}

Canceling d{\cal V} and rearranging yields

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

or

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

which is the famous Bernoulli’s equation.

Caution: The derivation of the Bernoulli equation assumes steady, incompressible, inviscid flow with no heat transfer and no shaft work. Under these conditions, there is no change in internal energy, and all energy exchanges occur between pressure, kinetic, and potential forms. Application of the Bernoulli equation to real flows requires careful consideration of these assumptions.

Check Your Understanding #2 – Pressure change in a contraction

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 an outlet section diameter of 9 cm. A water manometer measures the pressure difference between the inlet and outlet sections. Assuming no frictional losses, 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.

Show solution/hide solution.

Based on the information, assuming a one-dimensional, steady, incompressible, and inviscid flow seems reasonable. The flow rates, 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 relates the inlet and outlet conditions, i.e.,

    \[ \overbigdot{m} = \varrho \, A_1 \, V_1 = \varrho \, A_2 \, V_2 = \mbox{constant} \]

Or because the flow is assumed incompressible, then just

    \[ {\overbigdot{{\cal{V}}}} = A_1 \, V_1 = A_2 \, V_2 = \mbox{constant} = 0.135~\mbox{m${^3}$/s} \]

It follows that

    \[ V_1 = \frac{{\overbigdot{{\cal{V}}}} }{A_1} \]

and

    \[ V_2 = \frac{{\overbigdot{{\cal{V}}}} }{A_2} \]

Calculating the areas gives

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

and

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

Now the velocities can be calculated, i.e.,

    \[ V_1 = \frac{{\overbigdot{{\cal{V}}}} }{A_1} = \frac{0.135}{0.0346} = 3.90~\mbox{m/s} \]

and

    \[ V_2 = \frac{{\overbigdot{{\cal{V}}}} }{A_2} = \frac{0.135}{0.00636} = 21.22~\mbox{m/s} \]

which are much lower than a Mach number of 0.3, so the assumption of incompressible flow is justified.

The pressure difference between points 1 and 2 is required, which necessitates the use of some form of the energy equation. Using the Bernoulli equation is justified because the given information pertains to an incompressible, frictionless flow. The Bernoulli equation is

    \[ p_1 + \frac{1}{2} \, \varrho \, V_1^2 = p_2 + \frac{1}{2} \, \varrho \, V_2^2 \]

so

    \[ p_1 - p_2 = \frac{1}{2} \, \varrho \, V_2^2 - \frac{1}{2} \varrho \, V_1^2 = \frac{1}{2} \, \varrho \left( V_2^2 - V_1^2 \right) \]

Inserting the values for \varrho, {V_1} and {V_2} gives

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

In terms of \Delta h, which is the differential height on the manometer, then

    \[ p_1 - p_2 = \varrho_{\rm H_2O} \, g \, \Delta h = 263.25~\mbox{Pa} \]

so solving for \Delta h gives

    \[ \Delta h = \frac{p_1 - p_2 }{\varrho_{\rm H_2O} \, g} = \frac{263.25}{1,000 \times 9.81} = 0.0268~\mbox{m} = 2.68~\mbox{cm} \]

Other Forms of the Bernoulli Equation

Other forms of the Bernoulli equation may be obtained by evaluating the integral

(73)   \begin{equation*} \int \frac{dp}{\varrho} + \frac{V^2}{2} + g \, z = \mbox{constant} \end{equation*}

along a streamline. The form of the first term depends on the thermodynamic relationship between pressure and density.

For an incompressible flow, where \varrho = \text{constant}, then

(74)   \begin{equation*} \int \frac{dp}{\varrho} = \frac{p}{\varrho} \end{equation*}

so the classic Bernoulli equation is recovered, i.e.,

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

Isothermal Process

For an isothermal flow of an ideal gas, T = constant, so from the equation of state, i.e.,

(76)   \begin{equation*} p = \varrho \, R \, T \end{equation*}

rearranging gives

(77)   \begin{equation*} \varrho = \frac{p}{R \, T} \end{equation*}

Therefore,

(78)   \begin{equation*} \int \frac{dp}{\varrho} = \int \frac{R \, T}{p} \, dp = R \, T \ln\left(\frac{p}{p_{\rm ref}}\right) \end{equation*}

and the Bernoulli equation becomes

(79)   \begin{equation*} R \, T \, \ln\left(\frac{p}{p_{\rm ref}}\right) + \frac{1}{2} \, V^2 + g \, z = \mbox{constant} \end{equation*}

Adiabatic Process

For a reversible adiabatic (isentropic) process, then

(80)   \begin{equation*} \frac{p}{\varrho^{\gamma}} = \text{constant} \end{equation*}

which gives

(81)   \begin{equation*} \varrho = C \, p^{1/\gamma} \end{equation*}

Substituting into the integral gives

(82)   \begin{equation*} \int \frac{dp}{\varrho} = \int \frac{1}{C} p^{-1/\gamma} dp = \frac{\gamma}{\gamma - 1} \frac{p}{\varrho} \end{equation*}

so that the Bernoulli equation becomes

(83)   \begin{equation*} \frac{\gamma}{\gamma - 1} \frac{p}{\varrho} + \frac{1}{2} \, V^2 + g \, z = \mbox{constant} \end{equation*}

Unsteady Bernoulli Effects

The Bernoulli equation may be extended to certain inviscid flows in which the velocity field changes with time. In such cases, the local-acceleration term from Euler’s equation must be retained. The differential form becomes

(84)   \begin{equation*} \varrho \frac{\partial V}{\partial t} + \frac{dp}{ds} + \varrho \, V \, \frac{dV}{ds} + \varrho \, g \, \frac{dz}{ds} = 0 \end{equation*}

Integrating along a streamline from point 1 to point 2 gives

(85)   \begin{equation*} p_1 + \frac{1}{2} \, \varrho \, V_1^2 + \varrho \, g \, z_1 = p_2 + \frac{1}{2} \, \varrho \, V_2^2 + \varrho \, g \, z_2 + \varrho \int_{s_1}^{s_2} \frac{\partial V}{\partial t} \, ds \end{equation*}

The additional term, i.e.,

(86)   \begin{equation*} \varrho \int_{s_1}^{s_2} \frac{\partial V}{\partial t} \, ds \end{equation*}

accounts for unsteady acceleration along the streamline. For example, if the velocity varies linearly with time, such as

(87)   \begin{equation*} V(t) = V_0 + a \,t \end{equation*}

then

(88)   \begin{equation*} \frac{\partial V}{\partial t} = a \end{equation*}

and the unsteady contribution becomes

(89)   \begin{equation*} \varrho \int_{s_1}^{s_2} a \, ds = \varrho \, a \,(s_2 - s_1) \end{equation*}

which represents the pressure-equivalent contribution associated with the local acceleration of the fluid along the streamline. Its sign depends on whether the flow is accelerating or decelerating in time and on the direction in which the streamline integral is evaluated.

Check Your Understanding #3 – Fuel transfer pump power in an aircraft

Jet fuel is transferred within an aircraft from one tank to another by a pump and piping system at a volume flow rate of 0.080~\mbox{m}^3/\mbox{s}. At station 1, the pipe diameter is 0.20~\mbox{m}, the pressure is 180~\mbox{kPa}, and the elevation is z_1 = 2.0~\mbox{m}. At station 2, the pipe diameter is 0.12~\mbox{m}, the pressure is 300~\mbox{kPa}, and the elevation is z_2 = 14.0~\mbox{m}. Losses in the system are equivalent to a head loss of h_l = 3.5~\mbox{m}. Determine the useful pumping power added to the fuel. Assume the fuel density is \varrho = 800~\mbox{kg/m}^3.

Show solution/hide solution.

For steady incompressible flow, the energy equation in specific form is

    \[ \frac{p_1}{\varrho} + \frac{V_1^2}{2} + g \, z_1 + w_p = \frac{p_2}{\varrho} + \frac{V_2^2}{2} + g \, z_2 + w_l \]

where w_p is the specific energy added by the pump and w_l represents the specific energy loss. The inlet and outlet areas are

    \[ A_1 = \frac{\pi d_1^2}{4} = 0.0314~\mbox{m}^2 \qquad \text{and} \qquad A_2 = \frac{\pi d_2^2}{4} = 0.0113~\mbox{m}^2 \]

so the velocities are

    \[ V_1 = \frac{0.080}{0.0314} = 2.55~\mbox{m/s} \qquad \text{and} \qquad V_2 = \frac{0.080}{0.0113} = 7.07~\mbox{m/s} \]

The head loss is converted to a specific energy loss using

    \[ w_l = g \, h_l = 9.81 \times 3.5 = 34.3~\mbox{m}^2/\mbox{s}^2 \]

Solving for the specific pump work gives

    \[ w_p = \left( \frac{p_2 - p_1}{\varrho} \right) + \frac{V_2^2 - V_1^2}{2} + g (z_2 - z_1) + w_l \]

Substituting the numerical values gives

    \[ w_p = \frac{(300 - 180)\times 10^3}{800} + \frac{7.07^2 - 2.55^2}{2} + 9.81(14.0 - 2.0) + 34.3 \]

or

    \[ w_p = 150.0 + 21.8 + 117.7 + 34.3 = 323.8~\mbox{m}^2/\mbox{s}^2 \]

The useful pumping power added to the flow is

    \[ P_p = \varrho \, \overbigdot{{\cal V}} \, w_p \]

so

    \[ P_p = 800 \times 0.080 \times 323.8 = 20.7~\mbox{kW} \]

This result shows that the pump must supply useful energy to the fuel to overcome the pressure rise, the increase in flow speed, the elevation change, and the losses in the fuel-transfer system.

Energy Equation from the RTE

The Reynolds Transport Theorem and the Reynolds Transport Equation (RTE), when applied to total energy, provide a general framework for deriving the energy equation for a control volume. For a system, the First Law of Thermodynamics gives

(90)   \begin{equation*} \frac{D}{Dt} \iiint_{\mathrm{sys}} \varrho \, e \, d\mathcal{V} = \overbigdot{Q} - \overbigdot{W} \end{equation*}

Using the RTE, the control-volume form is

(91)   \begin{equation*} \frac{\partial}{\partial t} \iiint_{\mathcal{V}} \varrho \, e \, d\mathcal{V} + \oiint_{S} \varrho \, e \, \vec{V} \bigcdot d\vec{S} = \overbigdot{Q} - \overbigdot{W} \end{equation*}

Using the divergence theorem, the surface integral may be written as a volume integral, i.e.,

(92)   \begin{equation*} \oiint_{S} \varrho \, e \, \vec{V} \bigcdot d\vec{S} = \iiint_{\mathcal{V}} \nabla \bigcdot (\varrho \, e \vec{V}) \, d\mathcal{V} \end{equation*}

Therefore, the differential form may be written schematically as

(93)   \begin{equation*} \frac{\partial}{\partial t} (\varrho \, e) + \nabla \bigcdot (\varrho \, e \vec{V}) = \varrho \, \overbigdot{q} - \nabla \bigcdot (p \vec{V}) + \text{(viscous work terms)} \end{equation*}

where \overbigdot{q} is the rate of heat addition per unit mass, so the corresponding heat-addition rate per unit volume is \varrho \, \overbigdot{q}. The term -\nabla \bigcdot (p \vec{V}) represents pressure work, i.e., the local rate at which pressure forces do work on the flow. Using the product rule together with the continuity equation gives

(94)   \begin{equation*} \varrho \frac{De}{Dt} = \varrho \, \overbigdot{q} - \nabla \bigcdot (p \vec{V}) + \text{(viscous work terms)} \end{equation*}

where

(95)   \begin{equation*} e = u + \frac{V^2}{2} + g \, z \end{equation*}

is the total energy per unit mass. This equation shows that the change in total energy is governed by heat addition, pressure work, and viscous work.

For inviscid, adiabatic flow with no shaft work and negligible viscous effects, the pressure-work term must still be retained. Therefore, the differential energy equation should not be interpreted as simply giving

(96)   \begin{equation*} \frac{De}{Dt}=0 \end{equation*}

unless the pressure-work contribution has first been treated consistently. A more useful thermodynamic form for an inviscid, adiabatic flow with no shaft work is

(97)   \begin{equation*} \frac{D}{Dt}\left(u+\frac{V^2}{2}+g z\right) + p\,\frac{D}{Dt}\left(\frac{1}{\varrho}\right) = 0 \end{equation*}

which shows explicitly how pressure-volume work enters the energy balance.

For incompressible flow, D(1/\varrho)/Dt=0, so this thermodynamic form alone does not produce the pressure term in Bernoulli’s equation. The pressure term enters through the mechanical work done by pressure forces, or equivalently, through the momentum equation. For a steady inviscid flow, Euler’s equation gives

(98)   \begin{equation*} V\,dV + \frac{dp}{\varrho} + g\,dz = 0 \end{equation*}

Integrating along a streamline for an incompressible flow, where \varrho = \text{constant}, yields

(99)   \begin{equation*} \frac{p}{\varrho} + \frac{V^2}{2} + g \, z = \text{constant} \end{equation*}

which is Bernoulli’s equation.

Therefore, Bernoulli’s equation may be interpreted as a special mechanical-energy relation derived from the energy principle together with the inviscid momentum equation, under the assumptions of steady, incompressible, inviscid, and adiabatic flow with no shaft work or losses.

Check Your Understanding #4 – Differential energy equation and Bernoulli’s equation

Starting with the differential energy equation for an inviscid, adiabatic flow with no shaft work, i.e.,

    \[ \frac{D}{Dt}\left(u+\frac{V^2}{2}+gz\right) + p\,\frac{D}{Dt}\left(\frac{1}{\varrho}\right) =0 \]

show that, for steady incompressible flow, this equation reduces to Bernoulli’s equation along a streamline, i.e.,

    \[ \frac{p}{\varrho}+\frac{V^2}{2}+gz=\mbox{constant} \]

Explain clearly where the pressure term enters the result.

Show solution/hide solution.

For an incompressible flow, the density is constant, so

    \[ \frac{D}{Dt}\left(\frac{1}{\varrho}\right) = 0 \]

The internal energy of an incompressible, adiabatic, inviscid flow does not change through compression work. However, pressure forces still do mechanical work on the fluid as it accelerates or decelerates along a streamline. Therefore, the mechanical-energy form is obtained using the scalar product of the inviscid form of the momentum equation (Euler’s equation) with the velocity vector. For steady inviscid flow, Euler’s equation is

    \[ \varrho \left(\vec{V}\bigcdot\nabla\right)\vec{V} = -\nabla p - \varrho \, g \, \nabla z \]

Taking the scalar product with d\vec{s}, where d\vec{s} is directed along a streamline, gives

    \[ \varrho \, V \, dV = -dp - \varrho \, g\,dz \]

Dividing by \varrho gives

    \[ V\,dV+\frac{dp}{\varrho} + g\,dz = 0 \]

For incompressible flow, \varrho is constant so that this equation may be integrated along the streamline, i.e.,

    \[ \int V\,dV+\int\frac{dp}{\varrho}+\int g\, dz = 0 \]

Therefore,

    \[ \frac{V^2}{2}+\frac{p}{\varrho}+g \, z=\mbox{constant} \]

which is Bernoulli’s equation.

The pressure term enters through the pressure-force work, not by identifying pressure as internal energy. In the control-volume energy equation, this same effect appears as the flow-work contribution p/\varrho, which combines with the internal energy to form the enthalpy h=u+p/\varrho. For incompressible, inviscid, adiabatic flow with no shaft work or losses, the remaining mechanical-energy balance is therefore

    \[ \frac{p}{\varrho}+\frac{V^2}{2}+gz=\mbox{constant} \]

Summary & Closure

Applying the principle of energy conservation to fluid flow yields a general equation derived from the first law of thermodynamics that incorporates heat transfer, work, and the fluid’s internal, kinetic, and potential energies. The equation is most naturally expressed as a power balance that accounts for the rates at which energy is transferred into, out of, and stored within a control volume. In many practical aerospace applications, this equation is applied to single-stream systems, in which fluid enters and leaves at a constant mass flow rate. Energy may be added to or removed from the flow through heat transfer or mechanical work, thereby changing the form of the energy carried by the flow.

Under the simplifying assumptions of steady, incompressible, inviscid flow with no heat transfer, shaft work, or losses, the energy principle together with the inviscid momentum equation leads to Bernoulli’s equation. This relation connects pressure, velocity, and elevation along a streamline and is widely used in engineering analysis. Although often written in terms of pressure or energy per unit volume, Bernoulli’s equation is consistent with energy conservation. It describes the exchange among pressure, kinetic, and potential energy along a streamline. However, its application requires care, as real flows may involve viscosity, turbulence, compressibility, heat transfer, and mechanical work, necessitating the use of the more general energy equation.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • The energy equation is often considered redundant for the analysis of incompressible, inviscid flows. Why?
  • What are the three main assumptions used in deriving 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 conjunction with conservation of mass and momentum, the Bernoulli equation can be used to analyze flow through a propeller operating at low speeds. Could you 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.
  • Video sequence from Dr. Biddle’s YouTube channel on fluid mechanics.
  • A video lecture describing the derivation of the energy equation for a fluid.
  • Video on the conservation of energy in control-volume form.
  • This is an excellent video for a better understanding of the Bernoulli equation.
  • The use of graphics in this video reinforces 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 Copyright © 2022–2026 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

Share This Book