18 Conservation of Mass: Continuity Equation


Having laid down the fundamental forms of the flow models used in aerodynamic analyses, the governing equations can now be formulated to describe fluid dynamic or aerodynamic flows. The approach uses the three physical conservation principles: mass, momentum, and energy. In this first case, the applicable physical principle is that mass can neither be created nor destroyed. The resulting governing equation is called the continuity equation. It is a general equation valid for three-dimensional, unsteady flows and applies to all types of flows, e.g., compressible or incompressible, viscous or inviscid.

Learning Objectives

  • Know how to derive the most general form of the continuity equation in its control volume or integral form.
  • Learn how to use the continuity equation to solve simple flow problems.
  • See how the continuity equation can also be derived from the Reynolds Transport Equation.

Flow Model

As previously discussed, the flow model is a control volume that may either be fixed in space with the fluid moving through it (the most common application), which is called an Eulerian description of the flow, or the volume can move with the fluid such that the identical fluid particles are inside it, which is called a Lagrangian model. In either case, the physical conservation principles must be applied to the fluid inside the control volume and any fluid crossing its boundaries.

Consider a fixed finite control volume {\cal V} bounded by a surface of area S, as shown in the figure below. The symbol S defines the area of the closed surface that bounds the control volume containing a fluid of volume \cal{V}. The control volume is abbreviated to “C.V.” and the control surface to “C.S.” All properties are allowed to vary with spatial location (i.e., with respect to x, y, and z) and in time t so that

(1)   \begin{eqnarray*} \varrho & = & \varrho ( x, y, z, t ) \\[6pt] \vec{V} & = & \vec{V} (x, y, z, t) \end{eqnarray*}

A finite control volume, fixed in space, with the fluid flowing in and out across a control surface.

At any point on the control surface, the velocity is \vec{V}, which is given in terms of the Cartesian components as

(2)   \begin{equation*} \vec{V} = u \, \vec{i} + v \, \vec{j} + w \, \vec{k} \end{equation*}

At the same point, the unit normal vector area is d\vec{S}. Also, let d{\cal V} be an elemental fluid volume inside the total control volume.

Conservation of Mass

The fundamental principle of the conservation of mass requires that the net mass flow out of the control volume over surface S is equal to the time rate of decrease of mass inside the control volume {\cal V}. Now that physical statement must be translated into mathematics.

Following the concept of mass flow and mass flux discussed previously, the elemental mass flow across area dS is \varrho \, \vec{V} \bigcdot d\vec{S}. Remember that by convention, d\vec{S} always points out of the control volume, so the value of \vec{V}\bigcdot d\vec{S} will be positive. Therefore, the total mass flow rate (i.e., the integral of the mass flow rate over the entire surface area) is

(3)   \begin{equation*} \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} \end{equation*}

which can be physically interpreted as a net outflow leaving the control volume. Notice that the double integral here means the summation over the surface S, i.e., an area integral.

The small fluid mass contained within the elemental volume inside the C.V. is \varrho d{\cal V}. Hence, the total mass inside the C.V. is

(4)   \begin{equation*} \iiint_{\cal{V}} \varrho \, d {\cal{V}} \end{equation*}

where the triple integral means a volume integral. So, the time rate of decrease of mass inside the C.V. is

(5)   \begin{equation*} -\frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, d {\cal{V}} \end{equation*}

noting that the minus sign represents the decrease of mass, i.e., what is leaving the C.V.

Because the principle of conservation of mass requires that the net mass flow out of the control volume be zero, then Eq. 3 must equal Eq. 5, i.e.,

(6)   \begin{equation*} \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = -\frac{\partial}{\partial t}\iiint_{\cal{V}} \varrho \, d {\cal{V}} \end{equation*}


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

This latter equation is called the continuity equation for a fluid flow, in this case, in the integral or control volume form. It is a general equation valid for three-dimensional, unsteady flows and applies to all types of flows, e.g., compressible or incompressible, viscous or inviscid. In addition, it can be used to relate aerodynamic phenomena over a finite region of the flow, e.g., the properties of the flow as it comes into and leaves the specified control volume. The unknowns in the equation may include the flow velocities and the flow density.

Simplifications of the Continuity Equation

Various reductions or simplifications of the continuity equation can be used to solve practical problems, these forms also allowing commensurate simplifications in the overall mathematics. For example, for a steady flow, nothing changes with respect to time and \partial/\partial t \equiv 0. This simplification means that the continuity equation reduces to

(8)   \begin{equation*} \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

In words, the mass flow that comes into the control volume per unit of time then leaves the control volume simultaneously, i.e., no mass accumulates inside the control volume. The elimination of time dependencies in aerodynamic problems, if this assumption can be justified, is a significant and worthwhile simplification in most forms of practical analysis.

In the case of a steady flow, the mass flow into the control volume equals the mass flow out of the control volume.

Proceeding further by assuming that the flow is both steady (\partial/\partial t \equiv 0) and incompressible (\varrho = constant), then in this case, the governing continuity equation becomes

(9)   \begin{equation*} \oiint_S \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

which leads to a further significant reduction in the complexity of the governing equation, i.e., the elimination of \varrho as an unknown so that only the flow velocities need to be related.

Finally, consider a reduction to a steady but compressible, one-dimensional flow in the x direction, e.g., a uniformly axisymmetric flow. In this case, \vec{V} = u and so

(10)   \begin{equation*} \oint_S \varrho \, u \left( \, \vec{i} \bigcdot d\vec{S} \right) = 0 \end{equation*}


(11)   \begin{equation*} \varrho \, u \, l = \mbox{constant} \end{equation*}

Note that the original surface integral becomes a length l or an area per unit depth.

A much easier way to express this latter one-dimensional result is that

(12)   \begin{equation*} \varrho \,  A \, V = \mbox{constant} \end{equation*}

This is a statement that the mass flow rate through the control volume is constant for a steady flow.

Flow Through a Converging/Diverging Duct

Consider the steady, uniform flow through a converging/diverging duct with a circular cross-section with inlet area A_1 and outlet area A_2, as shown in the figure below. This is a classic problem when learning fluid dynamics. The two areas are known, such as by measurement. If the flow is assumed to be steady and uniformly axisymmetric (i.e., one-dimensional), then determine the relevant form of the continuity equation to relate the flow conditions at the outlet to those at the inlet.

Flow model of a converging/diverging duct with a circular cross-section.

The first step in the solution is to define a coordinate system and the control surface/volume over which to apply the principle of conservation of mass. In this case, the decision on the control volume is relatively easy as the duct itself bounds the flow, and there can be no mass flow over the walls of the duct, and this naturally applies no matter what the duct’s shape is.

The flow is steady, so \partial/\partial t \equiv 0, and no further justification is needed in this case. However, nothing is mentioned about whether the flow is compressible or incompressible. Because air is a gas, it must be assumed that the flow is compressible and that density must be retained as a variable. Furthermore, suppose the flow is uniformly axisymmetric. In that case, the flow velocity changes only in one direction, i.e., in the x direction based on the adopted coordinate system, another significant simplification toward the solution of this problem.

In light of the preceding assumptions, therefore, in this case, then

(13)   \begin{equation*} \oiint_{S} \varrho \, \vec{V} \bigcdot d\vec{S} = \iint_{1} \varrho \, \vec{V} \bigcdot d\vec{S} + \iint_{2} \varrho \, \vec{V} \bigcdot d\vec{S} + \iint_{\rm walls} \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

where the latter term is zero because there is no mass flow over the walls, i.e.,

(14)   \begin{equation*} \iint_{\rm walls} \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

The mass flow coming into the control volume through the left-hand side (face 1) is

(15)   \begin{equation*} \iint_{1} \varrho \, \vec{V} \bigcdot d\vec{S} = -\varrho_1 A_1 V_1 \end{equation*}

the one-dimensional assumption being used and the minus sign on the first term indicating that the flow is in the opposite direction to d\vec{S}. Similarly, the flow coming out of the right-hand side (face 2) is then

(16)   \begin{equation*} \iint_{2} \varrho \, \vec{V} \bigcdot d\vec{S} = \varrho_2 A_2 V_2 \end{equation*}

which is positive in this case because the flow is now in the direction of d\vec{S}. Therefore, because the flow is steady, the principle of conservation of mass states that the net mass flow is zero, so what mass flow comes into the control volume per unit time must equal the net mass flow out of the control volume per unit time, i.e.,

(17)   \begin{equation*} -\varrho_1 A_1 V_1 + \varrho_2 A_2 V_2 = 0 \end{equation*}

or simply that

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

Rearranging the latter equation gives the outlet conditions

(19)   \begin{equation*} \varrho_2 V_2 = \left( \frac{A_1}{A_2} \right) \varrho_1 V_1 \end{equation*}

i.e., the mass fluxes are related by the area ratio A_1/A_2. If the flow was further assumed to be incompressible, then \varrho = constant, and so

(20)   \begin{equation*} V_2 = \left( \frac{A_1}{A_2} \right) V_1 \end{equation*}

Finally, this latter result must be examined to see if it reconciles expectations and makes sense. Engineers get into the habit of asking such questions in practical problem solving, i.e., based on the final equation(s), does (do) the result(s) make physical sense?

For example, if the outlet area were to be smaller than the inlet area (i.e., A_2 < A_1), then the expectation is that the flow velocity will increase as it flows into and out of the control volume, which it does according to the equations because A_1/A_2 > 1. Notice that while this particular problem may appear easy, and indeed it is in this case, it provides an excellent example of how the conservation laws, in the integral form, can be applied to a fluid dynamics or aerodynamics problem.

Worked Example #1 – Calculating flow velocities in a converging duct

Consider the steady flow of a particular gas through a horizontal, converging pipe with an inlet diameter d_1 of 0.22 m and an outlet diameter d_2  of 0.16 m. The density of the gas is known to change from \varrho_1 = 0.91 kg/m^3 at the inlet to \varrho_2 = 0.83 kg/m^3 at the outlet. If the inlet flow velocity of the gas V_1 is 5.1 m/s, what is its exit velocity V_2? Assume one-dimensional flow.

The general form of the continuity equation is

    \[ \frac{\partial}{\partial t}\iiint \varrho \, d{\cal{V}} + \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \]

In this case, for steady, one-dimensional flow, then

(21)   \begin{equation*} -\varrho_1 \, A_1 \, V_1 + \varrho_2 \, A_2 \, V_2 = 0 \end{equation*}


    \[ \overbigdot{m} = \mbox{constant} = \varrho_1 V_1 A_1 = \varrho_2 V_2 A_2 \]

where the density must be retained as a variable. In terms of diameters, then

    \[ \varrho_1 V_1 d_1^2 = \varrho_2 V_2 d_2^2 \]

And solving for V_2 gives

    \[ V_2 = \frac{\varrho_1 V_1 d_1^2}{\varrho_2 d_2^2} \]

Substituting the numerical values gives

    \[ V_2 = \frac{0.91 \times 5.1 \times 0.22^2}{0.83 \times 0.16^2} = 10.57~\mbox{m s$^{-1}$} \]

Flow Through a Branched Duct

Consider the flow of hydraulic fluid through a branch circuit of a pipe, as shown below. The objective again is to use the principles of the conservation of mass to determine a relationship between the flow properties between the inlet and outlet conditions. The inlet and outlet areas of the pipe are assumed to be known. Remember that the first step in the analysis is to think about a sketch of the control volume and the control surface and annotate it appropriately. It will be assumed that the fluid is incompressible, i.e., \varrho = constant, steady, and one-dimensional; the one-dimensional assumption is that the flow velocities are constant over every cross-section.

The governing equation for continuity of flow, in this case, becomes

(22)   \begin{equation*} \oiint_{S} \varrho \, \vec{V} \bigcdot d\vec{S} = \oiint_{1} \varrho \, \vec{V} \bigcdot d\vec{S} + \oiint_{2} \varrho \, \vec{V} \bigcdot d\vec{S} + \oiint_{3} \varrho\vec{V} \bigcdot d\vec{S} + \oiint_{\rm walls} \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

where the mass flow over the solid walls would be zero, i.e.,

(23)   \begin{equation*} \oiint_{\rm walls} \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

If the flow velocities are constant over their respective areas (the one-dimensional assumption), then

(24)   \begin{equation*} -\varrho A_1 V_1 + \varrho A_2 V_2 + \varrho A_3 V_3 = 0 \end{equation*}

And if the density is constant (which it is for a liquid), then

(25)   \begin{equation*} -A_1 V_1 + A_2 V_2 + A_3 V_3 = 0 \end{equation*}

noting the negative sign on the first term and its significance. In general, considering the flow out of the junction as positive and the flow into the junction as negative, then for steady flow at any junction, the algebraic sum of all the mass flows must be zero, i.e.,

(26)   \begin{equation*} \sum \varrho \, A \, V = 0 \end{equation*}

Another way of looking at this latter result is to write it as

(27)   \begin{equation*} A_1 V_1 = Q = A_2 V_2 + A_3 V_3 \end{equation*}

where Q is the volume flow rate.  Of course, if there was no third exit, then the problem could be reduced to the one previously considered, and

(28)   \begin{equation*} A_1 V_1 = A_2 V_2 = Q = \mbox{constant} \end{equation*}

In general, with N branches, then

(29)   \begin{equation*} \sum_{i=1}^{N}  \, A_i V_i = 0 \end{equation*}

which is a statement that the net algebraic sum of the mass flow inputs and outputs to the control volume is zero.

Time-Dependent Flows

Time-dependent flows are often more challenging to deal with in fluid dynamics. However, by way of introduction, consider a relatively simple problem of a tank of circular cross-section that lets a fluid leave through a drain valve at the bottom, as shown below. The discharge velocity, V_d, of the flow from the drain varies with the height, h, of the fluid level above the drain according to V_d = \sqrt{ 2 g h}. Assume that the fluid is a liquid, so \varrho = constant. Notice that the height, h, will decrease with time.


Illustration of a time-dependent flow problem where a fluid continuously discharges from a drain.

The general form of the continuity equation is

    \[ \frac{\partial}{\partial t}\iiint \varrho \, d{\cal{V}} + \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \]

In this case, \varrho = constant so that it can be reduced to

(30)   \begin{equation*} -\frac{d{\cal{V}}}{dt} +  a \, V_d  = 0 \end{equation*}

where the volume of fluid \cal{V} in the tank for any height h is

(31)   \begin{equation*} {\cal{V}} = A h \end{equation*}

and where A is the cross-sectional area of the tank and where a is the area of the jet of liquid from the drain.

It is given that

(32)   \begin{equation*} V_d = \sqrt{ 2 g h} \end{equation*}

Notice that this flow rate depends on the instantaneous height of the liquid in the tank, h. This result is called Torricelli’s theorem, and it applies to a liquid that flows out of an orifice based under hydrostatic pressure.

In a short time interval dt, then the decrease in the volume \cal{V} of fluid in the tank will be

(33)   \begin{equation*} \frac{d \cal{V}}{dt} = - A \frac{dh}{dt} \end{equation*}

The minus sign denotes that h decreases as the tank empties.  Notice that the units in the preceding equation are in terms of a volume flow rate.

By continuity considerations (conservation of mass), this flow rate must be equal to the flow rate of fluid coming out of the drain, i.e.

(34)   \begin{equation*} \frac{d \cal{V}}{dt} = Q \end{equation*}

where Q is the flow rate of the liquid from the tank through the drain, then

(35)   \begin{equation*} \frac{d {\cal{V}}}{dt} = Q \end{equation*}

and so

(36)   \begin{equation*} - A \frac{dh}{dt} = a V_d = a \sqrt{ 2 g h} \end{equation*}

Rearranging gives

(37)   \begin{equation*} \frac{dh}{dt} = - \frac{a}{A} \sqrt{ 2 g h} \end{equation*}

which is an ordinary differential equation relating the height of the fluid level to time. Separating the variables and integrating them gives

(38)   \begin{equation*} t = \int_{0}^{t} dt = - \frac{A}{a} \int_{h_0}^0 \frac{dh}{\sqrt{ 2 g h}} = - \frac{A}{a \sqrt{ 2 g}} \int_{h_0}^0 \frac{dh}{\sqrt{h}} = \frac{A}{a \sqrt{ 2 g}} \int_0^{h_0} \frac{dh}{\sqrt{h}} \end{equation*}

where the limits of integration are noted, i.e., at time t = 0, then h = h_0, and when h = 0, the tank is considered empty. Performing the integration gives

(39)   \begin{equation*} t = \frac{2A}{a \sqrt{ 2 g}} \sqrt{h_0} \end{equation*}

In terms of the diameter of the tank D and the outlet diameter d of the drain, then

(40)   \begin{equation*} \frac{A}{a} = \frac{D^2}{d^2} \end{equation*}

Therefore, the time to empty the tank will be

(41)   \begin{equation*} t = \frac{2D^2}{d^2} \sqrt{\frac{h_0}{2 g}} \end{equation*}

where h_0 is the initial height of the liquid in the tank when the drain is opened.

Worked Example #2 – Discharge of propellant mass from a spacecraft

An orbiting satellite is propelled using a monopropellant thruster, with the propellant stored in a pressurized storage tank. The satellite has an initial mass of 5,000 kg, including its propellant mass. To give a slight corrective boost to its orbit, it opens the control valve and ejects the propellant at a constant rate of 0.1 kg/s at an effective exit velocity of 2,000 m/s. Assume that the satellite operates in a vacuum and no gravity forces are acting on it. Determine the change in velocity and new mass of the satellite after 90 seconds from the start of the propulsive burn.

This is a time-varying flow problem because a mass of fluid (propellant) is continuously discharged from a tank, so the mass of propellant inside is decreasing. The general form of the continuity equation is

    \[ \frac{\partial}{\partial t}\iiint \varrho \, d {\cal{V}} + \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \]

and in this case, it can be reduced to

    \[ - \frac{d M_p}{d t} + \overbigdot{m}_p  = 0 \]

where M_p is the mass of the propellant and \overbigdot{m}_p is the propellant flow rate. Notice that the minus sign occurs because the propellant mass is decreasing. Remember that a mass flow rate will be in units of mass per unit time, i.e., kg/s or kg s-1.

Let the initial mass of the spacecraft be M_0. Therefore, conservation of mass gives the current mass of the spacecraft, M, at time t later as the propellant is discharged as

    \[ M  = M_0 - \left( \frac{dM_p}{dt} \right)  t = M_0 -\overbigdot{m}_p \, t \]

where \overbigdot{m}_p = constant.

The thrust force from the propulsion system is given by Newton’s second law (force equals time rate of change of momentum), i.e.,

    \[ T = \overbigdot{m}_p V_e \]

and so the acceleration of the spacecraft, which is not constant because of its continuously decreasing mass, is given by

    \[ a = \frac{T}{M} = \frac{\overbigdot{m}_p \, V_e}{M} = \frac{dV}{dt} \]


    \[ \frac{dV}{dt} = \frac{\overbigdot{m}_p \, V_e}{M_0 -\overbigdot{m}_p \, t} \]

Integrating this previous equation gives

    \[ \Delta V = V_e \ln \left( \frac{M_0}{M_0 -\overbigdot{m}_p \, t} \right) \]

This previous equation has a special name called the Rocket Equation. After 90 seconds of thrusting, i.e., t = 90, then

    \[ \Delta V = 2,000 \ln \left( \frac{5,000}{5,000 - 0.1 \times 90.0 } \right) = 3.6~\mbox{m/s} \]

and the final mass will be

    \[ M  = M_0 -\overbigdot{m}_p \, t = 5,000 - 0.1 \times 90 = 4,991~\mbox{kg} \]

Derivation of the Continuity Equation using the RTE

Recall that the Reynolds Transport Equation (RTE) can be expressed as

(42)   \begin{equation*} \underbrace{\frac{D}{Dt} \iiint_{\mathrm{sys}} \beta \, \varrho \, d\mathcal{V}}_{\begin{tabular}{c} \footnotesize  Time rate of \\ \footnotesize  change of $B$ \\ \footnotesize  inside the system.\end{tabular}} = \underbrace{\frac{\partial}{\partial t} \iiint_{\mathcal{V}} \beta \, \varrho \, d\mathcal{V}}_{\begin{tabular}{c}  \footnotesize  Time rate \\ \footnotesize  of change of  $B$ inside\\ \footnotesize  the control volume.\end{tabular}} + \underbrace{\oiint_{S} \beta \, \varrho ( \vec{V}_{\mathrm{rel}} \bigcdot d\vec{S}) }_{\begin{tabular}{c} \footnotesize  Rate at which $B$ is \\ \footnotesize  leaving through the \\ \footnotesize  control surface.\end{tabular} } \end{equation*}

In the case of mass, then \beta = 1, and for a fixed C.V., the mass of the system and the C.V. are the same, so

(43)   \begin{equation*} \frac{D}{Dt} \iiint_{\mathrm{sys}} \beta \, \varrho \, d\mathcal{V} = 0 \end{equation*}

Therefore, the RTE becomes

(44)   \begin{equation*} 0  = \frac{\partial }{\partial t} \iiint_{{\cal{V}}} (1) \varrho \, d {\cal{V}} + \oiint_{S} (1) \varrho ( \vec{V} \bigcdot d\vec{S} ) \end{equation*}


(45)   \begin{equation*} \frac{\partial }{\partial t} \iiint_{{\cal{V}}} \varrho \, d {\cal{V}} + \oiint_{S} \varrho \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

which will be recognized as the continuity equation in its integral form.

Differential form of the continuity equation

The differential form of the continuity equation can also be derived from the RTE. Recall that using the divergence theorem, the RTE for a fixed control volume becomes

    \[ \iiint_{{\cal{V}}} \bigg[ \frac{\partial (\beta \, \varrho)}{\partial t} + \beta \, \varrho \,\left( \nabla  \bigcdot \vec{V} \right) \bigg] d{\cal{V}} = 0 \]

so a differential form of the RTE becomes

    \[ \frac{\partial (\beta \, \varrho)}{\partial t} + \beta \, \varrho \,\left( \nabla  \bigcdot \vec{V} \right) = 0 \]

where in this case, \beta = 1 and the mass of the system and the C.V. are the same, so

    \[ \frac{\partial \varrho}{\partial t} + \varrho \,\left( \nabla  \bigcdot \vec{V} \right) = 0 \]

which will apply at every point in the flow. If the flow is incompressible, then \varrho = constant, so the continuity equation becomes

    \[ \nabla  \bigcdot \vec{V} = 0 \]

or in terms of the scalar components when \vec{V} = (u, v, w), then

    \[ \nabla  \bigcdot \vec{V}  = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 \]

Therefore, in words, this equation states that the conservation of mass requires that the divergence of the velocity field must be identically zero. If it is not, then the flow would be non-physical.

Summary & Closure

Applying the principle of the conservation of mass to fluids results in a governing “star” equation called the continuity equation. This equation applies to all fluids, viscous or inviscid, compressible or incompressible, steady or unsteady. In application, the continuity equation can be simplified from its most general form by making various assumptions as they may apply to the problem of interest. However, it should be remembered that all assumptions must be justified, which in some cases may be challenging to establish a priori, so caution should be used. The most common simplification is to write the continuity equation in one-dimensional form, i.e., in terms of fluid properties that change in only one direction. While the continuity equation can help solve certain simple classes of fluid flow problems, the solution of most real problems will generally need to invoke conservation principles of momentum and energy to elicit the required information.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • Think about a situation where the size of the control volume changes with time. What form of the continuity equation would be needed in this case?
  • Think of some fluid flow problems where a Lagrangian flow model might be preferable to solve the problem.

Additional Online Resources

  • A good video on some basics of the continuity equation.
  • Another video on the use of the continuity equation in fluid mechanics.


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Introduction to Aerospace Flight Vehicles Copyright © 2022 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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