Inviscid Fluid Flow

A

n important area in momentum transfer is inviscidflow, in which, by virtue of the absence of shear stress, analytical solutions to the differential equations offluidflow are possible.

The subject of inviscidflow has particular application in aerodynamics and hydrodynamics and general application toflow about bodies—the so-called external flows. In this chapter, we shall introduce the fundamentals of inviscidflow analysis.

▶ 10.1

FLUID ROTATION AT A POINT

Consider the element offluid shown in Figure 10.1a. In timeDtthe element will move in thexyplane as shown. In addition to translation, the element may also deform and rotate.

We have discussed the deformation previously in Chapter 7. Now let us focus our attention on the rotation of the element. Although the element may deform, the orientation will be given by the average rotation of the line segmentsOBandOAor by denoting the rotation by

wzˆ d dt

a‡b 2

where the counterclockwise sense is positive. From Figure 10.1a, we see that

O y

x t A

B

O

A B

Figure 10.1a Rotation of afluid element.

124

w^zˆ lim

Dx;Dy;Dz;Dt®0

1 2

arctanf‰…vyj_x‡Dx vyj_x†DtŠ=Dxg Dt

‡arctanf ‰…vxj_y‡Dy vxj_y†DtŠ=Dyg Dt

which becomes, in the limit,

w^zˆ1 2

@vy

@x

@vx

@y

(10-1) The subscript zindicates that the rotation is about thezaxis.

In the xzand yzplanes the rotation at a point is given by wyˆ1

2

@vx

@z

@vz

@x

(10-2) and

w^xˆ1 2

@vz

@y

@vy

@z

(10-3) The rotation at a point is related to the vector cross product of the velocity. As the student may verifyÑ

Ñvˆ @vz

@y

@vy

@z

ex‡ @vx

@z

@vz

@x

ey‡ @vy

@x

@vx

@y

ez

and thus

Ñvˆ2w (10-4)

The vectorÑvis also known as thevorticity.When the rotation at a point is zero the flow is said to be irrotational.For irrotational flow Ñvˆ0, as can be seen from equation (10-4). The significance offluid rotation at a point may be examined by a different approach. The Navier–Stokes equation for incompressibleflow, equation (9–29), may also be written in the form

rDv

Dtˆ ÑP‡rg m‰Ñ …Ñv†Š (9-29)

It may be observed from the above equation that if viscous forces act on afluid, theflow must be rotational.

Figure 10.1b illustrates the effect of vorticity, and whether thefluid element is rotational or irrotational. The vorticity is often described as the measure of the moment of momentum of a small differential area about its own center of mass. The vorticity vector is characterized mathematically as the curl of the velocity vector. In Figure 10.1b, a velocity gradient is

Figure 10.1b Rotation of afluid element as the result of a velocity gradient.

10.1 Fluid Rotation at a Point ◀ 125

moving along aflat plate. Near the plate, as a result of the no-slip boundary condition, the velocity of thefluid is zero at that wall and increases into the bulkfluid. Near the wall, a small differential area offluid is acted upon by the velocity gradient such that at the bottom of thefluid element, the velocity will be less than at the top of the element resulting in a net rotation or spin of the element in the clockwise direction as shown by the arrow in the center of the differential area. The result of this is the rotation of thefluid element and rotational flow. Out in the bulk, the velocity is the same at the top and bottom of thefluid element, resulting in the differential element having no net rotation—thus the term irrotationalflow.

Vorticity has many applications in weather patterns, such as in tropical cyclones¹and in flow of blood in the body.

The kinematic condition Ñvˆ0 is not the first time we have encountered a kinematic relation that satisfies one of the fundamental physical laws offluid mechanics.

The law of conservation of mass for an incompressibleflow,Ñ

^vˆ0, is also expressed as a kinematic relation. The use of this relation is the subject of the next section.

Example 1

A two-dimensional velocity is given by the equation

vˆ …6y†ex‡ …6x†ey

Is theflow rotational, or irrotational?

To be irrotational, theflow must satisfy the equationr vˆ0, and as a result, Equation 10.1 must be equal to zero.

wzˆ1 2

@vy

@x

@vx

@y

ˆ0

First, we mustfind values for@vy/@xand@vx/@yto see if the condition of (equation (10-1)) is satisfied.

From the given equation, wefind the velocity components:

vxˆ6y and vyˆ6x Next we take the necessary partial derivatives dictated by equation (10-1):

@vx

@y ˆ6

@vy

@x ˆ6 And thus, plugging in the values,

wzˆ1

2…6 6† ˆ0 So the condition of irrotationalflow is satisfied, and this particle will not rotate.

Let’s also examine another different, but similar, condition. A two-dimensional velocity is given by the equation vˆ …6y†ex …6x†ey

Is thisflow rotational, or irrotational? We must again see if equation (10-1) is satisfied.

Finding values for@vy/@xand@vx/@yas we did previously, vxˆ6y and vyˆ 6x

1J. C. L. Chan,Ann. Rev. Fluid Mech.,37, 99 (2005).

Next we take the necessary partial derivatives dictated by equation (10-1):

@vx

@y ˆ6

@vy

@x ˆ 6 giving

wzˆ1

2…6 … 6†† ˆ6≠0 So the condition is not satisfied, and this is an example of rotationalflow.

▶ 10.2

THE STREAM FUNCTION

For a two-dimensional, incompressible flow, the continuity equation is

Ñ

^vˆ@v_@x^x‡@v_@y^{yˆ0 (9-3)}

Equation (9-3) indicates that vx and vy are related in some way so that @vx/@xˆ (@vy/@y). Perhaps the easiest way to express this relation is by having v_x and v_yboth related to the same function. Consider the function F(x,y), ifvxˆF(x,y), then

@vy

@y ˆ @F

@x or vyˆ

Z @F

@xdy

Unfortunately, the selection of vxˆF(x, y) results in an integral for vy. We can easily remove the integral sign if we make the original F(x,y) equal to the derivative of some function with respect to y.For example, ifF(x,y)ˆ(@Y(x,y)/@y], then

vxˆ@Y

@y As @vx/@xˆ (@vy/@y), we may write

@vy

@y ˆ @

@x

@Y

@y or @

@y vy‡@Y

@x

ˆ0

for this to be true in general:

vyˆ @Y

@x

Instead of having two unknowns, vx and vy, we now have only one unknown, Y. The unknown, Y, is called the stream function. The physical significance of Y can be seen from the following considerations. As YˆY(x,y), the total derivative is

dYˆ@Y

@xdx‡@Y

@ydy

10.2 The Stream Function ◀ 127

Also

@Y

@xˆ vy and @Y

@y ˆvx

and thus

dYˆ vydx‡vxdy (10-5)

Consider a path in thexyplane such thatYˆconstant. Along this path,dYˆ0, and thus equation (10-5) becomes

dy

dxj_Yˆconstantˆvy

vx

(10-6) The slope of the pathYˆconstant is seen to be the same as the slope of a streamline as discussed in Chapter 3. The functionY(x,y) thus represents the streamlines. Figure 10.2 illustrates the streamlines and velocity components forflow about an airfoil.

The differential equation that governs Y is obtained by consideration of the fluid rotation,w, at a point. In a two-dimensionalflow,w^zˆ¹₂‰…@vy=@x† …@vx=@y†Š;and thus, if the velocity componentsvyandvxare expressed in terms of the stream functionY, we obtain, for an incompressible, steadyflow,

2w^zˆ@²Y

@x² ‡@²Y

@y² (10-7)

When theflow is irrotational, equation (10-7) becomes Laplace’s equation:

ѲYˆ@²Y

@x² ‡@²Y

@y² ˆ0 (10-8)

Example 2

The stream function for a particularflow is given by the equationYˆ6x² 6y². We wish to determine the velocity components for thisflow, andfind out whether theflow is rotational or irrotational.

We defined the stream function as

@Y

@xˆ vy

Figure 10.2 Streamlines and the stream function.

and

@Y

@yˆvx

Thus,

vxˆ@Y

@yˆ 12y vyˆ @Y

@xˆ 12x The equations for the velocity components arevxˆ 12yandvyˆ 12x.

Next, we want to determine whether theflow is rotational or irrotational. To do this, we must satisfy equation (10-1).

wzˆ1 2

@vy

@x

@vx

@y

ˆ0 Solving for the necessary partial derivatives,

@vy

@x ˆ 12

@vx

@y ˆ 12 Thus,

wzˆ1

2… 12 … 12†† ˆ0 and thisflow is seen to be irrotational.

▶ 10.3

INVISCID, IRROTATIONAL FLOW ABOUT AN INFINITE CYLINDER In order to illustrate the use of the stream function, the inviscid, irrotational flow pattern about a cylinder of infinite length will be examined. The physical situation is illustrated in Figure 10.3. A stationary circular cylinder of radiusais situated in uniform, parallelflow in the xdirection.

As there is cylindrical symmetry, polar coordinates are employed. In polar coordi- nates,²equation (10-8) becomes

@²Y

@r² ‡1 r

@Y

@r‡1 r²

@²Y

@q² ˆ0 (10-9)

where the velocity componentsvrandv_q are given by vr ˆ1

r

@Y

@q vqˆ @Y

@r (10-10)

2The operatorѲin cylindrical coordinates is developed in Appendix A.

10.3 Inviscid, Irrotational Flow about an Infinite Cylinder ◀ 129

The solution for this case must meet four boundary conditions. These are as follows:

1. The circlerˆamust be a streamline. As the velocity normal to a streamline is zero, vrjrˆaˆ0 or@Y=@qjrˆaˆ0:

2. From symmetry, the line qˆ0 must also be a streamline. Hence, vqj_qˆ0ˆ0 or

@Y/@r|_qˆ0ˆ0.

3. Asr®

∞

the velocity must befinite.

4. The magnitude of the velocity asr®

∞

^isv∞, a constant.

The solution to equation (10-9) for this case is

Y…r;q† ˆv_∞rsinq 1 a² r²

(10-11) The velocity componentsvrand vqare obtained from equation (10-10):

vr ˆ1 r

@Y

@q ˆv_∞cosq 1 a² r²

(10-12) and

v_qˆ @Y

@r ˆ v_∞sinq 1‡a² r²

(10-13) By settingrˆain the above equations, the velocity at the surface of the cylinder may be determined. This results in

vrˆ0 and

vqˆ 2v_∞sinq (10-14)

The velocity in the radial direction is, of course, zero, as the cylinder surface is a streamline.

The velocity along the surface is seen to be zero atqˆ0 andqˆ180°. These points of zero velocity are known asstagnation points.The forward stagnation point is atqˆ180°, and the aft or rearward stagnation point is at qˆ0°. The student may verify that each of the boundary conditions for this case are satisfied.

r y

a x

Figure 10.3 Cylinder in a uniformflow.

▶ 10.4

IRROTATIONAL FLOW, THE VELOCITY POTENTIAL

In a two-dimensional irrotational flow Ñvˆ0, and thus @vx/@yˆ@vy, /@x. The similarity of this equation to the continuity equation suggests that the type of relation used to obtain the stream function may be used again. Note, however, that the order of differentiation is reversed from the continuity equation. If we let vxˆ@f(x,y)/@x, we observe that

@vx

@y ˆ @²f

@x@yˆ@vy

@x or

@

@x

@f

@y vy

ˆ0 and for the general case

vyˆ@f

@y

The functionfis called thevelocity potential. In order forfto exist, the flow must be irrotational. As the condition of irrotationality is the only condition required, the velocity potential can also exist for compressible, unsteadyflows. The velocity potential is commonly used in compressibleflow analysis. Additionally, the velocity potential,f, exists for three-dimensionalflows, whereas the stream function does not.

The velocity vector is given by

vˆvxex‡vyey‡vzezˆ@f

@xex‡@f

@yey‡@f

@zez

and thus, in vector notation,

vˆÑf (10-15)

The differential equation definingfis obtained from the continuity equation. Considering a steady incompressible flow, we have Ñ

^vˆ0; thus, using equation (10-15) for v, we obtain

Ñ

^{ÑfˆÑ2fˆ0 (10-16)}

which is again Laplace’s equation; this time the dependent variable isf. Clearly,Yandf must be related. This relation may be illustrated by a consideration of isolines ofYandf. An isoline of Y is, of course, a streamline. Along the isolines

dYˆ@Y

@xdx‡@Y

@ydy or

dy dx

Yˆconstantˆvy

vx

10.4 Irrotational Flow, the Velocity Potential ◀ 131

and

dfˆ@f

@xdx‡@f

@ydy dy dx

dfˆ0

ˆ vx

vy

Accordingly,

dy=dxj_fˆconstantˆ 1 dy=dx

Yˆconstant

(10-17) and thusYand forthogonal. The orthogonality of the stream function and the velocity potential is a useful property, particularly when graphical solutions to equations (10-8) and (10-16) are employed.

Figure 10.4 illustrates the inviscid, irrotational, steady incompressible flow about an infinite circular cylinder. Both the streamlines and constant-velocity potential lines are shown.

Example 3

The steady, incompressibleflowfield for two-dimensionalflow is given by the following velocity components:vxˆ16y xand vyˆ16x‡y. Determine the equation for the stream function and the velocity potential.

First, let’s check to make sure continuity is satisfied:

@vx

@x‡@vy

@y ˆ @

@x…16y x† ‡@

@y…16x‡y† ˆ 1‡1ˆ0 (1)

So continuity is satisfied, which is a necessary condition for us to proceed.

We defined the stream function as

@Y

@yˆvx (2)

and

@Y

@xˆvy (3)

Figure 10.4 Streamlines and constant- velocity potential lines for steady, incompressible, irrotational, inviscidflow about a cylinder.

Thus,

vxˆ@Y

@yˆ16y x (4)

vyˆ @Y

@xˆ16x‡y (5)

We can begin by integrating equation (4) or equation (5). Either will result in the same answer. (Problem 10.24 will let you verify this.) We will choose to begin by integrating equation (4) partially with respect toy:

Yˆ8y² xy‡f₁…x† (6)

wheref₁…x†is an arbitrary function ofx.

Next, we take the other part of the definition of the stream function, equation (3), and differentiate equation (6) with respect tox:

vyˆ @Y

@xˆy f₂…x† (7)

Here,f₂…x†isdf

dx, sincefis a function of the variablex.

The result is that we now have two equations forvy, equations (5) and (7). We can now equate these and solve for f₂…x†:

vyˆy f₂…x† ˆ16x‡y Solving forf₂…x†,

f₂…x† ˆ 16x So

f₁…x† ˆ 16x²

2 ˆ 8x²‡C

The integration constant C is added to the above equation sincefis a function ofxonly. Thefinal equation for the stream function is

Yˆ8y² xy 8x²‡C (6)

The constant C is generally dropped from the equation because the value of a constant in this equation is of no significance.

Thefinal equation for the stream function is

Yˆ8y² xy 8x²

One interesting point is that the difference in the value of one stream line in theflow to another is the volumeflow rate per unit width between the two streamlines.

Next we want tofind the equation for the velocity potential. Since a condition for the velocity potential to exist is irrotational flow, we mustfirst determine whether theflow in this example is irrotational.

To do this, we must satisfy equation (10-1).

wzˆ1 2

@vy

@x

@vx

@y

ˆ1

2…16 16† ˆ0 So theflow is irrotational, as required.

We now want to determine the equation for the velocity potential. The velocity potential is defined by equation (10-15):

vˆ rf

or @f

@xˆvxˆ16y x

10.4 Irrotational Flow, the Velocity Potential ◀ 133

fˆ16xy x² 2‡f…y†

Differentiating with respect toyand equating tovx,

@f

@yˆ16x‡ d

dyf…y† ˆ16x‡y Thus,

d

dyf…y† ˆy and

f…y† ˆy² 2 so that thefinal equation for the velocity potential is

fˆ16xy x² 2‡y²

2

▶ 10.5

TOTAL HEAD IN IRROTATIONAL FLOW

The condition of irrotationality has been shown to be of aid in obtaining analytical solutions influidflow. The physical meaning of irrotationalflow can be illustrated by the relation between the rotation or vorticity,Ñv, and the total head,P/r‡v²/2‡gy. For an inviscid flow, we may write

Dv

Dtˆg ÑP

r …Euler’s equation†

and

Dv Dt ˆ@v

@t ‡Ñ v²

2 v …Ñv† …Vector identity†

As the gradient of the potential energy is g, Euler’s equation becomes, for incompressible flow,

Ñ P r‡v²

2 ‡gy

ˆv …Ñv† @v

@t: (10-18)

If theflow is steady, it is seen from equation (10-18) that the gradient of the total head depends upon the vorticity, Ñv. The vector (Ñv) is perpendicular to the velocity vector; hence, the gradient of the total head has no component along a streamline. Thus, along a streamline in an incompressible, inviscid, steadyflow,

P r‡v²

2 ‡gyˆconstant (10-19)

This is, of course, Bernoulli’s equation, which was discussed in Chapters 6 and 9. If the flow is irrotational and steady, equation (10-18) yields the result that Bernoulli’s equation is

valid throughout theflowfield. An irrotational, steady, incompressibleflow, therefore, has a constant total head throughout theflowfield.³

▶ 10.6

UTILIZATION OF POTENTIAL FLOW

Potentialflow has great utility in engineering for the prediction of pressurefields, forces, and flow rates. In thefield of aerodynamics, for example, potentialflow solutions are used to predict force and moment distributions on wings and other bodies.

An illustration of the determination of the pressure distribution from a potentialflow solution may be obtained from the solution for theflow about a circular cylinder presented in Section 10.3. From the Bernoulli equation

P r‡v²

2 ˆconstant (10-20)

We have deleted the potential energy term in accordance with the original assumption of uniform velocity in thexdirection. At a great distance from the cylinder the pressure isP_∞, and the velocity is v_∞, so equation (10-20) becomes⁴

P‡rv²

2 ˆP_∞‡rv²_∞

2 ˆP₀ (10-21)

where P0is designated the stagnation pressure(i.e., the pressure at which the velocity is zero). In accordance with equation (10-19), the stagnation pressure is constant throughout thefield in an irrotationalflow. The velocity at the surface of the body isvqˆ 2v_∞sinq, thus the surface pressure is

PˆP0 2rv²_∞sin²q (10-22)

A plot of the potentialflow pressure distribution about a cylinder is shown in Figure 10.5.

0 30 60 90 120 150 180

–3 –2 –1 0 1 2

1 2

Figure 10.5 Pressure distribution on a cylinder inan inviscid, incompressible, steadyflow.

3A more general result, Crocco’s theorem, relates the vorticity to the entropy. Thus, it can be shown that a steady, inviscid, irrotationalflow, either compressible or incompressible, is isentropic.

4The stagnation pressure as given in equation (10-21) applies to incompressibleflow only.

10.6 Utilization of Potential Flow ◀ 135

▶ 10.7

POTENTIAL FLOW ANALYSIS—SIMPLE PLANE FLOW CASES

In this section, a number of cases will be considered in which solutions are achieved for two- dimensional, incompressible irrotationalflow. We begin with some very straightforward flow situations.

Case 1.Uniformflow in thexdirection.

For a uniform flow parallel to the x axis, with velocity v_∞ˆconstant, the stream function and velocity potential relationships are

vxˆv_∞ @Y

@y ˆ@f

@x vrˆ0 @Y

@xˆ@f

@y which integrate to yield

Yˆv_∞y fˆv_∞x Case 2.A line source or sink.

A line source, in two dimensions, is aflow that is radially outward from the source, which is the origin in this example. The reverse, or sinkflow, has theflow directed inward.

The source strength is the volumeflow rate per unit depth,Qˆ2prvr:The radial velocity associated with a source is

vrˆ Q 2pr

and the azimuthal velocity is given byv_qˆ0. The stream function and velocity potential are evaluated from the expressions

vrˆ Q 2prˆ1

r

@Y

@q ˆ@f

@r vqˆ0ˆ @Y

@r ˆ1 r

@f

@q Integrating these expressions, we obtain for the line source

Yˆ Q 2pq fˆ Q

2plnr

For sinkflow, the sign of the radial velocity is negative (inward), and thus,Qis negative.

The expressions for a line source or sink present a problem atrˆ0, the origin, which is a singular point. At rˆ0, the radial velocity approaches infinity. Physically this is unrealistic, and we use only the concept of line source or sink flow under conditions where the singularity is excluded from consideration.

Case 3. A line vortex.

Vortexflow is that which occurs in a circular fashion around a central point, such as a whirlpool. Afree vortexis one wherefluid particles are irrotational, i.e., they do not rotate as they move in concentric circles about the axis of the vortex. This would be analogous to people sitting in cabins on a ferris wheel. For an irrotationalflow in polar coordinates (see Appendix B), the productrvqmust be constant. The stream function and velocity potential can be written directly,

vrˆ0ˆ1 r

@Y

@q ˆ@f

@r vqˆ K

2prˆ @Y

@r ˆ1 r

@f

@q which, upon integration, become

Yˆ K 2plnr fˆ K

2pq

whereKis referred to as thevortex strength. WhenKis positive, theflow is observed to be counterclockwise about the vortex center.

▶ 10.8

POTENTIAL FLOW ANALYSIS—SUPERPOSITION

It was shown earlier that both the stream function and the velocity potential satisfy Laplace’s equation for two-dimensional, irrotational, incompressible flow. As Laplace’s equation is linear we can use known solutions to achieve expressions for both Y and f for more complex situations using the principle ofsuperposition. Superposition, simply put, is the process of adding known solutions to achieve another—i.e., ifY1andY2are solutions to ѲYˆ0, then so isY3ˆY1‡Y2a solution.

The reader is reminded that the solutions obtained for these very specialized flow conditions are idealizations. They apply for inviscidflow, which is a reasonable approxi- mation for conditions outside the region, near a solid body, where viscous effects are manifested. This region, the boundary layer, will be considered in some depth in Chapter 12.

Some cases will now be considered where the elementary planeflows of the previous section give some interesting and useful results through the process of superposition.

Case 4. The doublet.

A useful case is achieved from considering a source-sink pair on the x axis as the separation distance,2a,approaches zero. Geometrically, we can note that the streamlines and velocity potential lines are circles with centers on theyandxaxes, but with all circles passing through the origin that is a singular point.

The strength of a doublet, designatedl, is defined as thefinite limit of the quantity 2aQ asa®0. For our case, the source is placed on thexaxis at aand the sink is placed on the xaxis at‡a. The resulting expressions forY andfin polar coordinates are

10.8 Potential Flow Analysis—Superposition ◀ 137

Yˆlsinq r fˆlcosq

r

Case 5.Flow past a half body—superposition of uniformflow and a source.

The stream function and velocity potentials for uniformflow in thexdirection and for a line source are added together, yielding

YˆYuniform flow‡Ysource

ˆv_∞y Q

2pqˆv_∞rsinq‡ Q 2pq fˆfuniform flow‡fsource

ˆv_∞x‡ Q

2plnrˆv_∞rcosq‡Q 2plnr

Case 6.Flow past a cylinder—superposition of uniformflow and a doublet.

As a final illustration of the superposition method, we will consider a case of considerable utility. When the solutions for uniformflow and the doublet are superposed, the result, similar to the past case, defines a streamline pattern inside and around the outside surface of a body. In this case the body is closed and the exteriorflow pattern is that of ideal flow over a cylinder. The expressions forYand fare

YˆYuniform flow‡Ydoublet

ˆv_∞y lsinq

r ˆv_∞rsinq lsinq r

ˆ v_∞r l r

sinq fˆfuniform flow‡fdoublet

ˆv_∞x‡lcosq

r ˆv_∞rcosq‡lcosq r

ˆ v_∞r‡l r

cosq

It is useful, at this point, to examine the above expressions in more detail. First, for the stream function

Yˆ v_∞r l r

sinq

ˆv_∞r 1 l=v_∞ r²

sinq

where, as we recall,lis the doublet strength. If we chooselsuch that l

v_∞ˆa² wherea is the radius of our cylinder, we obtain

Y…r;q† ˆv_∞rsinq 1 a² r²

which is the expression used earlier, designated as equation (10-11).