22 Continuity Equation

Introduction

Having established 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 cannot be created or destroyed. The resulting governing equation is the continuity equation. It is a general governing equation applicable to three-dimensional, unsteady flows, covering all types, including compressible or incompressible, viscous or inviscid, and steady or unsteady.

Learning Objectives

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

Continuity Equation – Integral Form

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 move with the fluid such that identical fluid particles are inside it, which is called a Lagrangian model. In either case, physical conservation principles must be applied to the fluid inside the control volume and any fluid crossing its boundaries.

Flow Model

Consider a fixed finite control volume \mathcal{V} bounded by a surface of area S, which is in an Eulerian frame of reference, as shown in Figure 1. The symbol S defines the area of the closed surface that bounds the control volume containing a fluid of volume {\mathcal{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 C.S., 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 elemental unit normal vector area is d\vec{S}. Also, let d{\cal{V}} be an elemental fluid volume inside the totality of the C.V.

Conservation of Mass

The fundamental principle of the conservation of mass requires that the net mass flow out of the C.V. over surface S is equal to the time rate of decrease of mass inside {\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 C.V., so the value of {\vec{V}\bigcdot d\vec{S}} will be positive for outflow and negative for inflow. Therefore, the total mass flow rate (i.e., the integral of the mass flow rate over the entire surface area) is

(3)   \begin{equation*} \text{Total mass flow rate over the C.S.} = \oiint_{\rm C.S.} \varrho \, \vec{V} \bigcdot d\vec{S} = \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} \end{equation*}

which can be physically interpreted as the net mass flow rate across the control surface. 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*} \text{The total mass inside the C.V.} = \oiiint_{\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*} \text{Time rate of decrease of mass inside the C.V.} = -\frac{\partial}{\partial t}\oiiint_{\rm C.V.} \varrho \, d {\cal{V}} = -\frac{\partial}{\partial t}\oiiint_{\cal{V}} \varrho \, d {\cal{V}} \end{equation*}

noting that the minus sign represents a 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 C.V. be equal to the time rate of decrease of mass inside the C.V., 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}\oiiint_{\cal{V}} \varrho \, d {\cal{V}} \end{equation*}

or

(7)   \begin{equation*} \boxed{\underbrace{\frac{\partial}{\partial t}\oiiint_{\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*}

The latter equation in Eq. 7 is called the continuity equation for a fluid flow, in this case, in the integral or control volume form. It is a “star” equation because it is the most general form of the governing equation, i.e., it is valid for three-dimensional, unsteady flows. It applies to all types of flows, including compressible and incompressible, as well as viscous and inviscid. Additionally, it can be used to relate aerodynamic phenomena over a finite region of the flow, such as the properties of the flow as it enters and exits the specified control volume. The unknowns in the equation 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, and these forms also allow commensurate simplifications in the overall mathematics. For example, for a steady flow, nothing changes with respect to time, i.e., {\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 other words, the mass flow entering the control volume per unit time also leaves it, so no mass accumulates within the control volume, as illustrated in Figure 2. If this assumption can be justified, eliminating time dependencies in aerodynamic problems is a significant and worthwhile simplification in most practical analyses.

 

In 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 a steady single-stream system and one-dimensional, e.g., a uniformly axisymmetric flow, then the reduced form of the continuity equation is

(9)   \begin{equation*} \oiint_S \varrho \left( \vec{V} \bigcdot d\vec{S} \right) = \sum_{S} \varrho \left( \vec{V} \bigcdot \vec{A} \right) = 0 \end{equation*}

where \vec{A} is the area vector, defined as \vec{A} = A \vec{n}, where \vec{n} is the outward unit normal to the surface. By convention, \vec{A} points outward from the C.V., as shown in Figure 3. A much easier way to express this one-dimensional result for the continuity equation is in scalar form, such that

(10)   \begin{equation*} \sum_{S} \varrho \, \vec{V} \bigcdot \vec{A} = 0 \end{equation*}

This is a statement that the algebraic sum of the mass flow rates across the control surface is zero for a steady flow. When the inlet and outlet mass flow rates are written as positive magnitudes, each stream has the same mass flow rate, denoted by \overbigdot{m}.

Single-stream system to illustrate the principle of conservation of mass for a fluid.

For the situation of a single stream inlet and outlet, then

(11)   \begin{equation*} \sum_S \varrho \, \vec{V} \bigcdot \vec{A} = \varrho_1 \left( \vec{V}_1 \bigcdot \vec{A}_1 \right) + \varrho_2 \left( \vec{V}_2 \bigcdot \vec{A}_2 \right) = 0 \end{equation*}

noting the sign on the first term because \vec{V}_1 is in the opposite direction to \vec{A}_1. This result can also be written in scalar form in terms of magnitudes as

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

or

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

which is just a statement that \overbigdot{m}_1 = \overbigdot{m}_2, i.e., mass flow is conserved.

If the flow is incompressible, i.e., \varrho = constant, then the continuity equation becomes

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

which leads to further simplification by eliminating \varrho as an unknown. Therefore, only the flow velocities need
to be related, i.e.,

(15)   \begin{equation*} \sum_{S} \vec{V} \bigcdot \vec{A} = 0 \end{equation*}

For a single inlet and a single outlet, the corresponding positive volume flow rate is denoted by Q. In this case, then

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

or

(17)   \begin{equation*} V_1 \, A_1 = V_2 \, A_2 = Q \end{equation*}

which means that Q_1 = Q_2 = Q, i.e., the volumetric flow rate is conserved.

Finally, consider a reduction to a steady, compressible, uniform, two-dimensional flow in the {x} direction, taken per unit depth. In this case, \vec{V} = u \vec{i}, and the cross-sectional area may be written as A = l \times 1, so

(18)   \begin{equation*} \oint_S \varrho \left( u \vec{i} \bigcdot d\vec{S} \right) = \sum_S \varrho \, u \, A = \sum_S \varrho \, u \, l(1) = 0 \end{equation*}

or

(19)   \begin{equation*} \varrho_1 \, u_1 \, l_1 = \varrho_2 \, u_2 \, l_2 \end{equation*}

Notice that the areas {A_1} and {A_2} have been expressed as areas per unit depth, i.e., A_1 = l_1 \times 1 and A_2 = l_2 \times 1, respectively, which is equivalent to a one-dimensional (1-D) 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 Figure 4. This is a classic problem when learning fluid dynamics. The two sectional regions are known, as evidenced by measurements. 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 or volume over which the principle of conservation of mass applies. In this case, the decision regarding the C.S. and C.V. is relatively straightforward, as the duct confines the flow. There can be no mass flow across the duct walls, a natural consequence regardless of the duct’s shape.

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, the flow is assumed to be compressible, and density must be retained as a variable. Furthermore, suppose the flow is uniformly axisymmetric. In that case, the flow velocity varies only in the {x} direction, given the adopted coordinate system, which is another significant simplification in solving this problem.

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

(20)   \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} \varrho \, \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.,

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

The mass flow coming into the C.V. through the left-hand side (face 1) is

(22)   \begin{equation*} \iint_{1} \varrho \, \vec{V} \bigcdot d\vec{S} = -\left( \varrho \, \vec{V} \bigcdot \vec{A} \right)_1 = -\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

(23)   \begin{equation*} \iint_{2} \varrho \, \vec{V} \bigcdot d\vec{S} = \left( \varrho \, \vec{V} \bigcdot \vec{A} \right)_2 = \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 C.V. per unit time must equal the net mass flow out of the C.V. per unit time, i.e.,

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

or simply that

(25)   \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

(26)   \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

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

Finally, this latter result must be examined to determine whether it reconciles expectations and is reasonable. Engineers develop the habit of asking such questions in practical problem-solving, i.e., based on the final equation(s), do the results 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 C.V., 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.

Check Your Understanding #1 – Applying continuity to a converging duct

Consider the steady flow of a gas through a converging duct, as shown in the figure. This type of control-volume model is relevant to several aerospace applications, including wind tunnel contractions, engine inlet ducts, reverse-flow diffusers, and nozzle-like flow passages. The inlet has area A_1, equivalent diameter d_1, density \varrho_1, and average velocity V_1. The outlet has area {A_2}, equivalent diameter d_2, density \varrho_2, and average velocity V_2. Assume steady, one-dimensional flow through the duct, with no mass flow through the walls.

  1. Derive the relationship between V_2 and V_1 using the continuity equation.
  2. Then calculate V_2 for d_1 = 0.22~\mbox{m}, d_2 = 0.16~\mbox{m}, \varrho_1 = 0.91~\mbox{kg m}^{-3}, \varrho_2 = 0.83~\mbox{kg m}^{-3}, and V_1 = 5.1~\mbox{m s}^{-1}.

Show solution/hide solution.

The general form of the continuity equation is

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

For steady flow, the first term is zero, so

    \[ \oiint_S \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \]

For the control volume shown, there is no mass flow through the duct walls, so only the inlet and outlet surfaces contribute to the surface integral. Therefore,

    \[ \iint_1 \varrho \, \vec{V} \bigcdot d\vec{S} + \iint_2 \varrho \, \vec{V} \bigcdot d\vec{S} =0 \]

Using the one-dimensional assumption, the inlet contribution is negative because the flow enters the control volume opposite to the outward area vector. The outlet contribution is positive because the flow leaves the control volume in the same direction as the outward area vector. Therefore,

    \[ -\varrho_1 A_1 V_1 + \varrho_2 A_2 V_2 = 0 \]

or

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

Solving for the outlet velocity gives

    \[ V_2 = \frac{\varrho_1 A_1}{\varrho_2 A_2} V_1 \]

For circular cross-sections,

    \[ A_1 = \frac{\pi d_1^2}{4}, \qquad A_2 = \frac{\pi d_2^2}{4} \]

so the area ratio becomes

    \[ \frac{A_1}{A_2} = \frac{d_1^2}{d_2^2} \]

and the velocity relation may be written as

    \[ V_2 = \frac{\varrho_1}{\varrho_2} \left( \frac{d_1}{d_2} \right)^2 V_1 \]

Substituting the numerical values gives

    \[ V_2 = \frac{0.91}{0.83} \left( \frac{0.22}{0.16} \right)^2 (5.1) \]

or

    \[ V_2 = 10.57~\mbox{m s}^{-1} \]

Therefore, the average outlet velocity is

    \[ V_2 = 10.57~\mbox{m s}^{-1} \]

If the density change had been neglected, then the incompressible estimate would be

    \[ V_2 = \left( \frac{d_1}{d_2} \right)^2 V_1 = \left( \frac{0.22}{0.16} \right)^2 (5.1) = 9.65~\mbox{m s}^{-1} \]

The compressible result is slightly higher because the outlet density is lower than the inlet density. Therefore, to conserve mass flow, the outlet velocity must increase by more than the area ratio alone would suggest.

Flow Through a Branched Pipe

Consider the flow of fluid through a branch circuit of a pipe, as shown in Figure 5. The objective, again, is to use the principles of the conservation of mass to determine a relationship between the flow properties and 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 sketch the C.V. and the C.S. and annotate them 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 velocity is constant across any cross-section.

Flow model of a branched Y-shaped pipe with a circular cross-section.

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

(28)   \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.,

(29)   \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

(30)   \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

(31)   \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. Another way of looking at this latter result for the branch flow is to write it as

(32)   \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 were no third exit, then the problem could be reduced to the one previously considered, and

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

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.,

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

For an incompressible flow, i.e., a liquid, then volume is also conserved, i.e.,

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

Check Your Understanding #2 – Converging flows into a junction

Two pipes of diameters {d_1} and {d_2} converge to form a single pipe of diameter {d}. If a liquid flows with a velocity of {V_1} and {V_2} in the two pipes, respectively, what will be the flow velocity V_3 in the third pipe? Assume steady, one-dimensional, incompressible flow with no losses.

Show solution/hide solution.

In this case, two pipes merge to form a single pipe. Conserving mass and using average flow properties, then

    \[ \overbigdot{m} = \mbox{constant} = \varrho_1 V_1 A_1 + \varrho_2 V_2 A_2 = \varrho_3 V_3 A_3 \]

If the fluid is a liquid then the assumption that \varrho = \varrho_1 = \varrho_2 = \varrho_3 = constant is justified so

    \[ Q = \mbox{constant} = V_1 A_1 + V_2 A_2 = V_3 A_3 \]

Therefore, the flow velocity in the third pipe is

    \[ V_3 = \frac{V_1 A_1 + V_2 A_2}{A_3} = \frac{V_1 (\pi \, d_1^2/4) + V_2 (\pi \, d_2^2/4)}{ (\pi \, d^2/4)} = \frac{V_1 \, d_1^2 + V_2 \, d_2^2}{d^2} \]

Flow Through Parallel Pipes

Consider the flow of fluid that splits through a branch circuit with parallel circular pipes, as shown in Figure 6, and then joins together again. The flow enters at 1 and leaves at 4. The two branches, 2 and 3, could have different cross-sectional areas. Again, it will be assumed that the fluid is incompressible, i.e., \varrho = constant, steady, and one-dimensional; recall that the one-dimensional assumption is that the flow velocity is constant across every cross-section.

Flow model of a branched parallel-pipe system with circular cross-sections.

The governing continuity equation is

(36)   \begin{equation*} \oiint_{S} \varrho \, \vec{V} \bigcdot d\vec{S} = \oiint_{1} \varrho \, \vec{V} \bigcdot d\vec{S} + \oiint_{4} \varrho \, \vec{V} \bigcdot d\vec{S} = 0 \end{equation*}

Notice also that for the two branches that

(37)   \begin{equation*} -\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}, \qquad -\oiint_{1} \varrho \, \vec{V} \bigcdot d\vec{S} = \oiint_{4} \varrho \, \vec{V} \bigcdot d\vec{S} \end{equation*}

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

(38)   \begin{equation*} - \varrho \, A_1 V_1 + \varrho \, A_4 V_4 = 0 \end{equation*}

If the density is constant, then

(39)   \begin{equation*} {-A_1 V_1 + A_4 V_4 = 0} \end{equation*}

or

(40)   \begin{equation*} A_1 V_1 = A_4 V_4 = Q \end{equation*}

where Q is the volume flow rate. Notice that for the branches, then

(41)   \begin{equation*} Q = A_1 V_1 = A_2 V_2 + A_3 V_3 = A_4 V_4 \end{equation*}

Check Your Understanding #3 – Flow split through parallel cooling ducts

Air flows through a main duct in an aircraft equipment-cooling system. The duct then splits into two parallel branches, each passing through a separate avionics bay and equipment rack, before the flow rejoins in a common exhaust duct downstream. The main duct has a diameter of d_1 = 10~\mbox{cm}, one branch has a diameter of d_2 = 6~\mbox{cm}, and the other branch has a diameter of d_3 = 4~\mbox{cm}. If the average velocity in the main duct before the split is V_1 = 2.0~\mbox{m s}^{-1}, determine what can be found about the two branch velocities V_2 and V_3 from the continuity equation alone. Assume steady, one-dimensional flow, and neglect density changes.

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For the main duct, the area is

    \[ A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \pi \left(\frac{10}{2}\right)^2 = 25\pi \, \mbox{cm}^2 \]

For the first branch, with diameter 6 cm, the area is

    \[ A_2 = \pi \left(\frac{d_2}{2}\right)^2 = \pi \left(\frac{6}{2}\right)^2 = 9\pi \, \mbox{cm}^2 \]

and for the second branch, with diameter 4 cm, the area is

    \[ A_3 = \pi \left(\frac{d_3}{2}\right)^2 = \pi \left(\frac{4}{2}\right)^2 = 4\pi \, \mbox{cm}^2 \]

For steady, one-dimensional, incompressible flow, the continuity equation requires that the volume flow rate in the main duct equal the sum of the volume flow rates in the two branches, i.e.,

    \[ Q = A_1 V_1 = A_2 V_2 + A_3 V_3 \]

Substituting the numerical values gives

    \[ 25\pi \times 2.0 = 9\pi V_2 + 4\pi V_3 \]

or

    \[ 50 = 9V_2 + 4V_3 \]

This single equation contains two unknowns, V_2 and V_3. Therefore, the two branch velocities cannot be determined from the continuity equation alone. Additional information is needed, such as the pressure drop through each branch, duct lengths, or a specified flow split.

For example, if a flow-control valve or duct design specified that 60% of the total volume flow rate goes through branch 2 and 40% goes through branch 3, then

    \[ Q_2 = 0.60 Q, \qquad Q_3 = 0.40 Q \]

The total volume flow rate is

    \[ Q = A_1 V_1 = 25\pi \times 2.0 = 50\pi~\mbox{cm}^2\,\mbox{m s}^{-1} \]

so

    \[ Q_2 = 30\pi, \qquad Q_3 = 20\pi \]

The corresponding branch velocities would then be

    \[ V_2 = \frac{Q_2}{A_2} = \frac{30\pi}{9\pi} = 3.33~\mbox{m s}^{-1} \]

and

    \[ V_3 = \frac{Q_3}{A_3} = \frac{20\pi}{4\pi} = 5.00~\mbox{m s}^{-1} \]

These numerical velocities follow only after the flow split has been specified. Continuity alone gives the constraint

    \[ 50 = 9V_2 + 4V_3 \]

but still not the individual values of V_2 and V_3.

Continuity Equation – Differential Form

To derive the continuity equation in its differential form, consider a small fluid element in the flow field defined in a Cartesian coordinate system with length dimensions {dx}, {dy}, and {dz}, as shown in Figure 7. The volume of this fluid element is d{\cal{V}} = dx \, dy \, dz, and so the mass of the fluid element is

(42)   \begin{equation*} dm = \varrho \, d\mathcal{V} = \varrho \, dx \, dy \, dz \end{equation*}

where \varrho is the fluid density at point B in three-dimensional space and time t, i.e., \varrho = \varrho(x, y, z, t).

A differential fluid-element approach is used to derive the continuity equation.

The net mass flow rate through the differential element is the sum of the mass rates through each pair of opposite faces of the element. Consider first the mass flow in the {x} direction through the left face of the element of area dy \, dz, i.e.,

(43)   \begin{equation*} \text{Mass flow rate into } d{\cal{V}} \text{ in the } x \text{ direction} = (\varrho \, u ) dy \, dz = \varrho \, u \, dy \, dz \end{equation*}

where u is the velocity component of the flow in the {x} direction. Likewise, for the right face of the element of area dy \, dz, then

(44)   \begin{equation*} \text{Mass flow rate out of } d{\cal{V}} \text{ in the } x \text{ direction} = \varrho \, u \, dy \, dz + \left(\dfrac{\partial (\varrho u)}{\partial x} \right) dx \, dy \, dz \end{equation*}

Therefore, the total mass flow rate through the element in the x direction is

(45)   \begin{equation*} \left( \varrho \, u \, dy \, dz + \left(\dfrac{\partial (\varrho u)}{\partial x} \right) dx \, dy \, dz \right) - \varrho u \, dy \, dz = \dfrac{\partial (\varrho u)}{\partial x} \, dx \, dy \, dz \end{equation*}

Similarly, for the {y} and {z} directions, then

(46)   \begin{equation*} \text{Mass flow rate out of } d{\cal{V}} \text{ in the } y \text{ direction} = \frac{\partial (\varrho v)}{\partial y} \, dx \, dy \, dz \end{equation*}

and

(47)   \begin{equation*} \text{Mass flow rate out of } d{\cal{V}} \text{ in the } z \text{ direction} = \frac{\partial (\varrho w)}{\partial z} \, dx \, dy \, dz \end{equation*}

where v and w are the velocity components in the {y} and {z} directions, respectively. Therefore, the total net mass flow rate out of the differential element is the sum of the net fluxes in all three directions, i.e.,

(48)   \begin{equation*} \text{Net mass flow rate out of } d{\cal{V}} = \left( \frac{\partial (\varrho u)}{\partial x} + \frac{\partial (\varrho v)}{\partial y} + \frac{\partial (\varrho w)}{\partial z} \right) dx \, dy \, dz \end{equation*}

Now, consider the time rate of change of mass inside the differential element (fixed volume), which will be

(49)   \begin{equation*} \frac{\partial (\varrho \, d{\cal{V}})}{\partial t} = \frac{\partial \varrho}{\partial t} \, d{\cal{V}} = \frac{\partial \varrho}{\partial t} \, dx \, dy \, dz \end{equation*}

If mass is conserved (and the volume element is fixed in space and time), then the net mass flow rate out of the fluid element is equal to the time rate of decrease of mass inside the fluid element, i.e.,

(50)   \begin{equation*} \left( \frac{\partial (\varrho u)}{\partial x} + \frac{\partial (\varrho v)}{\partial y} + \frac{\partial (\varrho w)}{\partial z} \right) dx \, dy \, dz = -\frac{\partial \varrho}{\partial t} \, dx \, dy \, dz \end{equation*}

and rearranging and simplifying gives

(51)   \begin{equation*} \frac{\partial \varrho}{\partial t} + \underbrace{ \frac{\partial (\varrho u)}{\partial x} + \frac{\partial (\varrho v)}{\partial y} + \frac{\partial (\varrho w)}{\partial z}}_{ {\nabla \bigcdot (\varrho \vec{V})} } = 0 \end{equation*}

Therefore, the differential form of the continuity equation (using vector notation) becomes

(52)   \begin{equation*} \frac{\partial \varrho}{\partial t} + \nabla \bigcdot (\varrho \vec{V}) = 0 \end{equation*}

where \vec{V} = (u, v, w). This is a mathematical statement in differential form: mass is neither created nor destroyed.

Proceeding by noticing that expanding \nabla \bigcdot (\varrho \vec{V}) using the product rule gives

(53)   \begin{equation*} \nabla \bigcdot (\varrho \vec{V}) = \varrho (\nabla \bigcdot \vec{V}) + \vec{V} \bigcdot \nabla \varrho \end{equation*}

so the continuity equation now becomes

(54)   \begin{equation*} \frac{\partial \varrho}{\partial t} + \varrho (\nabla \bigcdot \vec{V}) + \vec{V} \bigcdot \nabla \varrho = 0 \end{equation*}

The left-hand side contains the terms that make up the substantial derivative of \varrho, i.e.,

(55)   \begin{equation*} \frac{D\varrho}{Dt} = \frac{\partial \varrho}{\partial t} + \vec{V} \bigcdot \nabla \varrho \end{equation*}

so the final form of the continuity equation becomes

(56)   \begin{equation*} \frac{D\varrho}{Dt} + \varrho (\nabla \bigcdot \vec{V}) = 0 \end{equation*}

As in the case of the integral form in Eq. 7, the result in Eq. 56 applies to all types of flows, e.g., compressible or incompressible, viscous or inviscid.

Simplifications to the general form of the continuity equation include steady flow, i.e., \dfrac{\partial}{\partial t} = 0, and incompressible flow, i.e., \varrho = constant. For an incompressible flow, whether steady or unsteady, 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, this latter equation states that to satisfy the conservation of mass, the divergence of the local velocity field must be identically zero. If not, the flow would be non-physical. Notice that this condition must be satisfied regardless of whether the flow is steady or unsteady.

Check Your Understanding #4 – Physically possible flow?

Consider a steady, incompressible flow of an inviscid fluid in a two-dimensional space described by the velocity field

    \[ \vec{V} = (u, v) = (y, -x) \]

  1. Does this flow exist physically?
  2. Derive the equations of the streamlines for this velocity field.
  3. Interpret the streamlines and their directions.
Show solution/hide solution.
  1. The differential form of the continuity equation for a two-dimensional, incompressible flow is

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

    For u = y and v = -x,

        \[ \frac{\partial u}{\partial x} = 0 \quad \mbox{and} \quad \frac{\partial v}{\partial y} = 0 \]

    Therefore, the continuity equation is satisfied, i.e.,

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

    so this is a physically possible incompressible velocity field.

  2. The equation of the streamline in 2D is

        \[ \frac{dy}{dx} = \frac{v}{u} = -\frac{x}{y} \]

    Separating variables gives

        \[ y \, dy = -x \, dx \]

    Integrating both sides gives

        \[ \int y \, dy = - \int x \, dx \]

    or

        \[ \frac{1}{2}y^2 = -\frac{1}{2}x^2 + C \]

    Therefore,

        \[ x^2 + y^2 = C \]

    where C is a constant.

  3. The streamlines are circles centered at the origin. The flow direction is clockwise, which can be seen, for example, because at the point (x,y)=(1,0), the velocity is \vec{V}=(0,-1).

Continuity Equation from the RTE

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

(57)   \begin{equation*} \underbrace{\frac{d}{dt} \oiiint_{\mathrm{sys}} \beta \, \varrho \, d\mathcal{V}}_{\begin{tabular}{c} \scriptsize Time rate of \\[-3pt] \scriptsize change of $B$ \\[-3pt] \scriptsize inside the system.\end{tabular}} = \underbrace{\frac{\partial}{\partial t} \oiiint_{\mathcal{V}} \beta \, \varrho \, d\mathcal{V}}_{\begin{tabular}{c} \scriptsize Time rate \\[-3pt] \scriptsize of change of $B$ inside\\[-3pt] \scriptsize the control volume.\end{tabular}} + \underbrace{\oiint_{S} \beta \, \varrho ( \vec{V}_{\mathrm{rel}} \bigcdot d\vec{S}) }_{\begin{tabular}{c} \scriptsize Rate at which $B$ is \\[-3pt] \scriptsize leaving through the \\[-3pt] \scriptsize control surface.\end{tabular} } \end{equation*}

Integral Form

In the case of mass, \beta = 1. The mass of a material system is constant, so

(58)   \begin{equation*} \frac{d}{dt} \left( \oiiint_{\mathrm{sys}} \varrho \, d\mathcal{V} \right) = 0 \end{equation*}

Therefore, the RTE becomes

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

or

(60)   \begin{equation*} \frac{\partial }{\partial t} \oiiint_{{\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, i.e., Eq. 7.

Differential form

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

(61)   \begin{equation*} \oiiint_{{\cal{V}}} \bigg[ \frac{\partial (\beta \, \varrho)}{\partial t} + \nabla \cdot (\beta \, \varrho \vec{V}) \bigg] d{\cal{V}} = 0 \end{equation*}

The control volume is arbitrary, so the integrand must also be zero, and the differential form of the RTE becomes

(62)   \begin{equation*} \frac{\partial (\beta \, \varrho)}{\partial t} + \nabla \cdot (\beta \, \varrho \vec{V}) = 0 \end{equation*}

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

(63)   \begin{equation*} \frac{\partial \varrho}{\partial t} + \nabla \bigcdot (\varrho \vec{V}) = 0 \end{equation*}

which is the conservative differential form of the continuity equation and is equivalent to Eq. 56, as previously given, and applies at every point in the flow.

Check Your Understanding #5 – Discerning a velocity field

Consider a two-dimensional, steady velocity field where the velocity components are given by u(x, y) = -2A x and v(x, y) = A y, where A is a nonzero constant. Determine whether this velocity field satisfies the incompressible continuity equation. If it does not, explain what this implies about the proposed velocity field.

Show solution/hide solution.

The incompressible continuity equation in two dimensions is

    \[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \]

Computing the partial derivatives gives

    \[ \frac{\partial u}{\partial x} = \frac{\partial (-2Ax)}{\partial x} = -2A \]

and

    \[ \frac{\partial v}{\partial y} = \frac{\partial (Ay)}{\partial y} = A \]

so that

    \[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = -2A + A = -A \]

Because A \neq 0, the given velocity field does not satisfy the incompressible continuity equation. Therefore, it cannot represent a physically possible steady, incompressible flow field as stated. The result implies either that the flow is not incompressible or that one of the specified velocity components is incorrect.

Summary & Closure

Applying the principle of conservation of mass to fluids yields one of the fundamental governing equations of fluid dynamics, the continuity equation. In its most general integral or differential form, the continuity equation applies to all fluids, whether viscous or inviscid, compressible or incompressible, and steady or unsteady. It states that mass cannot be created or destroyed within a control volume or at a point in the flow field.

In practical applications, the continuity equation is often simplified by making assumptions appropriate to the problem, such as steady, incompressible, or one-dimensional flow. These assumptions reduce the mathematics and often lead to useful engineering relationships, such as \varrho_1 A_1 V_1 = \varrho_2 A_2 V_2 for steady one-dimensional flow or A_1 V_1 = A_2 V_2 when density changes can be neglected. However, such simplified forms are valid only when the corresponding assumptions are justified.

The continuity equation is especially useful for relating velocity, area, density, mass flow rate, and volume flow rate in ducts, nozzles, wind tunnel contractions, engine inlets, and other aerospace flow passages. However, continuity alone usually does not determine pressures, forces, power requirements, or losses. Most real fluid-flow problems require the continuity equation to be used together with the conservation principles of momentum and energy.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • Consider a scenario in which the size of the control volume varies over 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.
  • How might the continuity equation be applied to analyze blood flow in arteries and veins in the human circulatory system?
  • Explain how the continuity equation could be used to design and analyze water distribution systems, such as municipal water supply networks.
  • How does the continuity equation relate to streamlines in fluid flow? Provide an example to illustrate this relationship.

Additional Online Resources

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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.n16

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