Differential-Continuity Equation

A derivation of the differential form of the continuity equation is carried out next. As explained, differential equations provide a tool for a more complete description of the flow field. In particular, a detailed description of the complete field behavior at any point in the flow then can be accomplished.

Recall from advanced calculus that a form of Green’s Theorem, the divergence theorem (also called Gauss’s Theorem), relates surface integrals to volume integrals —namely, if A is a vector, then:

j] (A ■ n)dS = JJJ (V • A)dE = JJJ divAdV. (3.47)

s v V

In what follows, we use the gradient-operator notation (V ^ gradient, V – ^ diver­gence, Vx ^ curl) instead of grad, div, and curl. Now, by identifying the vector A in Eq. 3.47 with the vector (pV) in Eq. 3.24, the surface integral in Eq. 3.24 may be changed to a volume integral by using the divergence theorem. Also, because the control volume that we are using is fixed in space, the limits on the volume integral in Eq. 3.24 do not vary with time. That being the case, the time derivative may be brought through the volume integral such that:

Then, using Eqs. 3.47 and 3.48 and putting both terms in Eq. 3.24 under the same volume integral:

Now, Eq. 3.49 must hold for any arbitrary control volume; therefore, the integrand must be zero. The integrand then represents the differential form of the continuity requirement.

The only way that Eq. 3.49 can be satisfied, then, is for the integrand to be zero. Thus,

dp+v • (pv) = o.

This is the general continuity equation in differential-equation form, valid at any field point in an unsteady compressible flow. Again, as in the control-volume conti­nuity formulation, there are special forms of Eq. 3.50 that fit cases of interest and simplify the equation. For steady compressible flow, for example, Eq. 3.50 becomes:

V • (pV) = 0.

For incompressible flow, Eq. 3.50 becomes:

V • V = 0,

which is valid for both steady and unsteady flow because in an incompressible flow, the density is constant and all of its derivatives therefore are zero. Table 3.5 summa­rizes the special cases of the continuity equation.

These differential equations can be written for any coordinate system by suit­ably expanding the vector operations. For example, in a three-dimensional Cartesian coordinate system with unit vectors i, j, and k in the coordinate directions x, y, and z and with the velocity vector having components V = ui + vj + wk, Eq. 3.50 becomes:

The first of the conservation laws now is expressed in terms of both an inte­gral and a differential equation. The equivalent equations in other coordinates (e. g., cylindrical and spherical) are found by using the vector operators written in that system. For example, in three-dimensional cylindrical coordinates with unit vectors, er, ee, and ez, the velocity vector can be written as V = urer + ueee + uzez, and the divergence of the velocity vector is:

„ 1 д. .1 due duz

V = – d-(rur) + ~~дЄе + ~дГ. r dr r r de dz

This expression is set to zero for continuity for an incompressible flow.

In the case of a compressible flow, the density multiplies each of the scalar velo­city components as before. It is important to understand the process described here, to apply the equations in the coordinate system that best fits the geometry of the problem to be solved.

example 3.8 Given: A velocity field for an incompressible flow is described by:

V = (2xy2) i + (2x2y) j.

Required: Determine if this flow field is physically possible.

Approach: To be physically possible, the flow field must satisfy the conservation

of mass.

Solution: Test the flow field by checking whether it satisfies the continuity equation for an incompressible flow, Eq. 3.13. Then,

du dv „2 „2 r

— + — = 2y2 + 2jt Ф 0,

dx dy

except at the origin. This flow field is not physically possible.

Appraisal: A flow field that is described as a given mathematical function or that is to be solved for any problem should always be checked to see whether it is physically possible—that is, whether it satisfies the continuity equation.

The Differential Form of the Momentum Equation

As in the case of the continuity equation, the starting point for the derivation of the differential form of the momentum equation is the integral form, Eq. 3.32:

d – JJJ p WE + || pV(V. n)dS = F = JJJ pbdV + JJ TdS. (3.55)

1 v S V S

The surface force is decomposed into two parts: one due to normal pressure force only, Tp; and the other containing both normal and shear components due to vis­cosity, tv, because there can be a viscous correction to the normal force due to the pressure. Thus, Eq. 3.55 becomes:

I JJJpVdV+Я pvV • ")dS=ffl pbdV+Л * pdS+JJ (356)

V S V s S

In following the same procedure used for the continuity equation, we must assemble all of the terms under a single-volume integral. Work with the force components first; the body-force term is already in the required form. Because Tp = – pndS, the surface integral can be converted from a surface to a volume integral by using the following form of the divergence theorem*:

JJ GndS = JJJVGdV. (3.57)

s V

where G is any scalar (e. g., p in the present case) and the gradient of a scalar is a vector. The second term on the right side of Eq. 3.56 then becomes:

JJt pdS = JJ-P«dS = – JJJVPdV’. (3.58)

S S v

* The derivation is shown here for convenience. Starting with the divergence theorem,

JJ (A n)dS = JJJ (V ■ A)dV.

S V

Put A = GB, where B = constant and G = scalar variable.

JJ^. ndS = JJJV (GB)dV,

S V

but V-(GB) = GV-B + B-VG = B-VG

JJ ^. nd^? = JJJb. VGdt>

Si V

B * JJ GndS = B * JJJVGdV^, because B = constant.

Si E

Thus, JJ GndS = JJJVGdK

S V2

The last term involving the viscous stresses may be rewritten in terms of a volume integral after first expressing tv in terms of its normal and tangential viscous stress components. Because details of the viscous-force modeling are the subject of Chapter 8, the volume integral corresponding to the viscous term is represented symbolically for the present as:

JM – Ш (3-59)

S V

and is carried along as a placeholder. Regardless of the details, this term is zero if the flow is assumed to be inviscid—a simplification of which we often take advantage.

Working with the first acceleration term on the left side of Eq. 3.56, the time derivative may be passed through the integral sign because the limits of integration are independent of time; the control volume is fixed. Thus,

fjJJp™"-JJJfw^- <3-60>

ut V i?

The second term on the left side:

convective acceleration – JJpV(V • n)dS, (3.61)

S

requires special consideration. We again attempt to convert the surface integral of Eq. 3.61 to a volume integral by means of the divergence theorem,

JJ(A. n)dS – JJJ(V. A)dV,

s V

where A is any vector. Notice that there is an apparent difficulty here in converting the surface integral to a volume integral by using the divergence theorem in this form because pV(V • n) in Eq. 3.61 is not of the form (A • n). In fact, pV(V • n) is a vector and (A • n) is a scalar. The divergence theorem is not directly applicable to the conversion of Eq. 3.61 into a volume integral. Notice, however, that the problem is circumvented if we focus on any one scalar component of the velocity vector V. Then, the procedure we follow is to work first with one component and then gener­alize the results to full vector form.

For simplicity, use Cartesian coordinates and focus on the x-component, u, of the velocity vector:

V = ui + vj + wk. (3.62)

Then, the x-component of the vector surface integral in Eq. 3.61 is converted readily to a volume integral by means of the divergence theorem. We find:

JJ pu(V ■ n) dS = JJJ V (puV)dV

S V

Inserting this and the x-components of the other terms into Eq. 3.56:

Jfl|(Pu)dV + JJJV (PuV)dV – i. JJJpbdV-lllVPW +JJJ t. dV,

V v L V v v _

where the x-components of the vectors on the right side are found by dotting with the unit vector, i. Collecting all of the terms under the same volume integral results in:

Thus, for an arbitrary control volume, we require that the integrand vanishes with the result that for the x-component,

I(pu) + V (puV)-i.(pb-Vp + ^) = 0. (3.63)

This is the so-called conservation form of the differential-momentum equation (x-component). It is clear that the other two components can be found by a similar analysis. The three momentum-equation components are used commonly in conser­vation form to set up numerical solutions in computational fluid dynamics (CFD) procedures, discussed later in the book.

It is useful to expand the first term,

Э, . du dp

st(p u)=p¥+u 17 •

and the second term by means of the vector identity V^A) = A • Vф + фV• A. Then,

V • (puV) = pV • Vu + uV • (pV).

The combination of the first and second terms yields:

d(pu) + V – (puV) = p^u + u^r + pV – Vu + uV. (pV) = dt dt dt

du „

— + V ■ Vu

dt where the part multiplied by u is the continuity equation and must be set to zero so that mass is conserved. Noting that the pressure gradient in Cartesian form is:

„ 3p. 3p . dp,

Vp=¥i +3pJ +¥ k

we can write the momentum equation in Cartesian component form as:

This suggests that in compact vector form, the differential-momentum equation is:

Г 3v ^

pl^V + V-VVj = – Vp + pb + fv,

which is sometimes called the primitive-variable form of the momentum equation. In effect, we split off the continuity equation from the conservation form (i. e., Eq. 3.63
and the other two components) to arrive at the primitive form. Expressed in words, this equation states that “the mass per unit volume times the total acceleration equals the force per unit volume” at any point in the field of motion. In the usual particle-mechanics problem, the time derivative of the velocity vector is the accel­eration. Because of our use of Eulerian-field point of view, the derivative of V is the local acceleration term; the second term, V • VV, is the convective acceleration term that corrects for the rate of change of fluid properties due to motion in the field.

Although the derivation was carried out for Cartesian coordinates, Eq. 3.66 is the correct result for any coordinate system provided that we properly define the convective accleration V • VV and correctly interpret the implied mathematics. Note that unlike the Cartesian scalar-component form in Eq. 3.65, this term cannot be interpreted as simply V dotted with the gradient of V (the gradient of a vector is not defined!). A careful discussion and interpretation of this seemingly pathological term is required. Many errors in the application of the differential-momentum equation can be traced to an insufficient understanding of this term and its exten­sion to other than simple rectangular Cartesian coordinate systems. Observe also that this term is nonlinear. That is, it involves products of the velocity components whereas, for comparison, the local acceleration term is linear in the sense that it con­tains no products (or transcendental functions) of the variables.

The operator V • V is sometimes called the convective operator. In Cartesian form, it is correct to interpret this as an operator:

„ d d d V • V = u— + v— + w— dx dy dz

and then apply it to each scalar component of the vector on which it acts. For example, for the x-component of Eq. 3.65, we find that taking the dot product of the velocity vector with the gradient of u:

du. du. du, 1 du du du —i + —j + — k = u — + v — + w —

dx dy dz J dx dy dz

is the same result found by treating V V as an operator acting on u; thus,

The convective element of the momentum balance must be treated with great care. It is best to treat the form, V • V, as a shorthand notation that can be evaluated cor­rectly and explicitly as demonstrated for Cartesian coordinates. To find the correct form for any coordinate system, it is necessary to use vector analysis.[9] The form that is correct in any coordinate system is:

Then, whenever it is necessary to write the momentum equation for cylindrical or spherical coordinates, we use Eq. 3.67 to arrive at the correct equations. This expression is demonstrated herein when we evaluate the momentum equation in component form for cylindrical coordinates (see Eq. 3.71). As is typical, mathemat­ical results such as Eq. 3.67 often have important physical implications. For example, the first term contains the square of the local flow speed, which suggests a connection to the kinetic energy of the flow. The second term involves the curl of the velocity vector, V x V, which is associated in subsequent chapters with important properties of the flow field, such as rotation and vorticity.

Now that the general form for the momentum balance, Eq. 3.66, is deduced, we can proceed as usual to write reduced forms that suit situations of special interest. In the complete form, the viscous-force term, fv, is replaced by a suitable model that relates the components of the viscous force to the geometry of the flow. The Newtonian model for viscous force introduced in Chapter 2 (see Eq. 2.4) is general­ized in Chapter 8 for this purpose. The resulting equation (or equations, if written in component form) is called the Navier-Stokes Equation after Navier (French) and Stokes (English), who developed the equations independently in the mid-1800s. Famous names are associated with other special forms of the momentum balance. For example, an important case is Euler’s equations, in which the viscous forces are dropped. The inviscid approximation has many applications in aerodynamics problems. Table 3.6 summarizes important special forms of the differential – momentum equation.

The second set of equations (Eq. 3.68) in Table 3.6 represents the Cartesian form of the Navier-Stokes equations for an incompressible fluid with a Newtonian model for the viscous forces. The viscosity coefficient, p, is assumed to be constant (see Chapter 2). The details of the viscous model used here are elaborated on in Chapter 8. Notice that in the other cases shown in the table, there is no assumption of incompressibility. Equations 3.69-3.71 are valid for compressible flow because after splitting the continuity equation off of the original momentum balance, there are no derivatives of the density remaining in the general equation. However, if vis­cous forces are retained, density gradients may have important consequences.

Equation 3.70, the two-dimensional Euler’s equations for Cartesian coordinates, is used frequently and should be compared to the set shown in Eq. 3.71 in Table 3.6, written in polar coordinates. Students should test their understanding of these results by reducing the complete vector-momentum equation (Eq. 3.66) to the special cases shown in the table. It is important to see that treatment of the convective acceleration by using the V* V operator approach (correct only for the Cartesian case) does not work for cylindrical coordinates. It is necessary to use the full form of the convective accleration in vector form (Eq. 3.67) to derive the result shown in Eq. 3.71.

Because the momentum equation is a vector equation, the result is that three more of the required six equations needed for the general fluid-mechanics problem now are established. If the flow is assumed to be incompressible, then the continuity and momentum equations are sufficient to define the problem. This simplification is used to advantage in subsequent chapters.

example 3.9 Given: Consider the following flow field. It is steady and inviscid with no body forces. The flow field is described by V = (2x)i + (3x-2y)j. Assume the density (constant) to be p = 0.2 slug/ft3.

Required: Find the pressure gradient in this flow field in the x-direction at a cer­tain point specified by (x = 3, y = 2).

Approach: Because the x-component of the pressure gradient (i. e., dp/dx) is required, use the x-component of the differential-momentum equation (Eq. 3.70) with constant density.

Solution: First, is this flow physically possible? Check using Eq. 3.52: Yes (verify this). Next, substitute Eq. 3.70 and evaluate:

u ^ + v ^ = 4x + 0 = _1 dp, and solving, dp = _2.4 lbf/ft2/ ft. dx dy p dx dx

Appraisal: For this particular flow field, the pressure gradient in the x-direction is independent of y. This, of course, is not a general result. Note that this example does not represent a solution of the differential-momentum equation but rather is an exercise in applying the equation.

EXAMPLE 3.10 Required: Derive Euler’s equation in two-dimensional polar coordinates by direct evaluation of the general momentum equation (Eq. 3.66) and Eq. 3.67.

Approach: Assume steady, inviscid flow without body forces and insert Eq. 3.67 for the convective acceleration so that the momentum equation is valid for any coordinate system. Then, use the vector expressions for the gradient and curl in polar coordinates.