Stream Function and Velocity Potential

Both the stream function and the velocity potential are scalar functions of the velocity-field coordinates. The derivatives of both of these functions are related to the velocity components of the flow, so that if the function can be found, the velocity field is determined. The existence of the two functions does not depend on the same condition. Thus, one function may exist and be useful in a problem whereas the other function may not exist at all. These functions make the analysis of certain flow prob­lems much easier, as shown herein, so that it is important to know when they may be used and when they do not apply.

Stream Function. A unique stream function exists for two-dimensional planar flows and for axisymmetric flows. For a three-dimensional flow, a stream function may be defined for each of two intersecting surfaces; the intersection of these two surfaces determines a streamline. The two – or three-dimensional flows may be com­pressible or incompressible, viscous or inviscid. If a stream function is defined for a flow, the flow must satisfy the continuity equation (i. e., the flow must be physically possible). Only two-dimensional planar-incompressible flows are considered here.

Consider the continuity equation, Eq. 3.52, for incompressible, two-dimensional planar flow; namely:

du + dv _ 0 Эх dy

Then ask: Is it possible to define a scalar function of the coordinates (x, y) that iden­tically satisfy this equation? If so, call this function y(x, y). After some thought, we conclude that there is indeed such a scalar function and that in its simplest form, it must be a function such that:

u _d¥ and v _-^¥. (4.11)

dy dx

If Eq. 4.11 is true, then the continuity equation is identically satisfied because:

d2¥ d2¥ _ 0

dxdy dydx

(The order of differentiation is unimportant because there are no discontinuities in the flow.) The minus sign is introduced with the v term rather than the u term in Eq. 4.11 because it is observed that the derivative of the stream function yields a velocity component that is orthogonal to the derivative direction. Thus, a nega­tive Ax with a minus sign yields a positive v component. The student should not be uneasy about Eq. 4.11 being derived here by inspection; the same equation may be derived in a fully rigorous mathematical manner if desired.

It may be shown that the derivative of the stream function in any direction n yields the magnitude of the velocity in a direction perpendicular to n. What is impor­tant is that if y(x, y) can be determined or specified, then the velocity components of the flow field follow directly by differentiation (this is important later).

It has been shown that if a two-dimensional flow satisfies the continuity equation (i. e., is physically possible), then a stream function exists. What can be said about the properties of such a stream function?

Recall from vector analysis that the gradient of a scalar is a vector and, further, that the gradient vector is at right angles to isolines, which are lines along which the scalar is a constant. With this in mind, form the following:

Подпись: V • Vy = (иі +vj) •"эу i+эу

ydx dy,

Now, if the dot product of two vectors is zero, then the two vectors must be every­where at right angles to one another. It follows that the velocity vector is everywhere perpendicular to the vector gradient of the stream function. However, the gradient vector is everywhere perpendicular to isolines y = constant. However, if the velocity vector is perpendicular to a vector that is perpendicular to isolines, it follows that the velocity vector must be everywhere tangent to the isolines y = constant. This means that lines y = constant are streamlines; if a stream function can be found, then the streamlines of the flow field are determined as well.

Because no flow can cross a streamline, the volume flow (or mass flow for a compressible flow field) in a stream tube defined by any two streamlines and a unit height out of the paper must be constant. By choosing appropriate values of the constant for y = constant, streamlines may be labeled by the volumetric flow rate between them and an arbitrarily designated “base” streamline, as shown in Fig. 4.5.

Подпись: V Stream Function and Velocity Potential Stream Function and Velocity Potential Подпись: = 0, Подпись: (4.12)

A stream function also may be defined for polar coordinates in the same manner as for Cartesian coordinates. Write the incompressible flow continuity equation, Eq. 3.11, in polar form by suitably expressing the divergence of the velocity vector in polar coordinates—namely:

Подпись: ur Подпись: 1 dy r de Подпись: and Stream Function and Velocity Potential Подпись: (4.13)

where the polar-coordinate notation is defined in Fig. 4.3. It may be determined from inspection that a stream function in this coordinate system must be defined as:

if the requirement of continuity as expressed in Eq. 4.12 is to be satisfied. To summarize:


Stream Function and Velocity Potential Подпись: y = V2 y = VI y = VO

A stream function exists for any steady flow that satisfies the continuity equation (i. e., physically possible).

2. The derivative of the stream function in any direction yields the velocity com­ponents of the flow field at right angles to that direction.

3. Explicit expressions for velocity components in terms of derivatives of the stream function were developed for two-dimensional planar flows in Eqs. 4.11 and 4.13.

4. Lines of у = constant are streamlines.

EXAMPLE 4.5 Given: The equation of a family of streamlines in the physically possible flow of a perfect fluid is given by:

Подпись:Подпись: 23

— = constant. 3

Find the slope of the streamline passing through the point (1,2) and verify that the expression for the stream function as given by Eq. 4.11 is consistent with this answer.

Approach: Appeal to Eq. 4.1 for the definition of a streamline; then, find the slope from this result and also by evaluating Eq. 4.11 for this flow field.

Подпись: v u Stream Function and Velocity Potential

Solution: From Eq. 4.1,

Differentiating the given equation for the streamline, y2dx + 2xydy – x2dx = 0. Solving,

Подпись: dy dx * y and at(1.2),

2 xy v ‘

dy = v = 3 dx u 4

This is the required streamline slope at (1,2). Now, because the stream function is constant along a streamline, the equation of the streamline also must be the equation of the stream function—namely,

2 *3 у= xy – y.

Applying Eq. 4.11 at the point (1,2), u = 4 and v = -3 so that the ratio of the velo­city components is v/u = -3/4, as before. Hence, the slope of the streamline agrees with the prior result.

Appraisal: This example reinforces the validity of the derivation leading to Eq. 4.11. Because the problem was framed as a “physically possible flow field,” no check using the continuity equation was performed (neither was it necessary) before the problem was begun.

Подпись: and u0 = (0 -2r).
Stream Function and Velocity Potential

example 4.6 Given: A velocity field is given in polar coordinates for a perfect fluid flow as:

Required: Find the stream function for this flow.

Approach: Use Eq. 4.13 to link velocity components with the stream function in polar coordinates.

Solution: Using Eq. 4.13,

102 _2

ur = —- =—— 1 ^–!- = 02 – r.

r r d0 r d0

Integrating this partial derivative,


¥ = ^ – r0 + f(r)’

where f(r) is an arbitrary function of r (recall that there is not simply a constant of integration here because a partial derivative is being integrated). Now, using this result for ¥, form:

u0=-j¥ = 0 +0-f ‘(r) = 0-2r

from the given information. Solving for f'(r) and integrating, f ‘(r) = 2r ^ f (r) = r2 + constant.

From this, it follows that:


¥ = — – r0 + r2 + constant, which is the required stream function.

Appraisal: The resulting stream function also is the equation for the streamlines for this flow, with different values of the integration constant corresponding to different streamlines. If it is only required to find velocity components from a given stream function, the value of this constant is not important because it drops out on differentiation for the velocity components. Note that it is well to check the given velocity field to see that it satisfied the continuity equation (i. e., Eq. 3.52) before the stream function was sought.

Velocity Potential: This function of the field coordinates exists only for irrotational flows; the flows may be compressible or incompressible, two – or three-dimensional. Only inviscid, incompressible flows are treated here. As discussed, an inviscid sub­sonic flow is irrotational unless vorticity is created in the flow upstream of the region being investigated.

From Section 4.2, if a flow is irrotational, then this implies that the vorticity is zero and that the curl of the velocity vector is also zero—or, in vector notation, V x V = 0 . Now, there is a vector identity that says V x (Уф) = 0, where ф is any scalar function. From this identity, it follows that the velocity vector may be defined in terms of the gradient of a scalar function of the field coordinates if the flow is irro – tational. This function is called the velocity potential and is usually given the symbol ф(х, у). Thus,

V = У{ф(х, у)} + vj = £ і + дф j,

and, equating vector components:

Подпись: (4.14)Эф Эф U“ЭХ’ v“Эу■

Because the density is not involved in the derivation, the same expression holds true for an inviscid compressible flow. If the compressible flow is supersonic, there may be regions of the flow that are not irrotational by reason of entropy gradients (as behind curved shock waves), even though the flow is assumed to be inviscid. A third Cartesian dimension simply adds a third term in the development and the expression for the third velocity component follows directly.

Подпись: Ur Подпись: Эф U = 1 Эф dr ’ 9 r Э9' Подпись: (4.15)

A parallel development using the vector expressions proper for polar coordi­nates leads to the result:

Notice that the derivative of the velocity potential with respect to a coordinate direction yields the velocity component in the same direction.

The velocity potential has the same role as the stream function in that if the velocity potential can be determined or specified, then the velocity components of the flow field follow by differentiation. The velocity potential often is useful as well in the treatment of simultaneous partial-differential equations for a flow field. The introduction of the velocity potential allows the equations to be combined in terms of a single dependent variable (i. e., the velocity potential) although the single equation is of higher order than the simultaneous equations.

Lines of constant velocity potential are not streamlines. However, they have a certain property with respect to streamlines that may be observed by forming the vector-dot product [У{ф(х, у)}] • [V|/{(x, y)}]. Evaluation of this product yields zero for any flow field for which the two scalar functions exist. This states that the two gradient vectors are perpendicular to one another; however, these two vec­tors, in turn, are each perpendicular to isolines of their particular function. The conclusion, then, is that isolines of constant-velocity potential (i. e., equipotential lines) are everywhere perpendicular to isolines of constant-stream function (i. e., streamlines). Thus, equipotential lines form an orthogonal mesh with streamlines everywhere in a flow field—provided, of course, that the potential exists (i. e., the flow is irrotational).

Henceforth, it is assumed that any flow field under consideration satisfies the continuity equation so that a stream function exists. However, a particular physically possible flow may or may not be irrotational, depending on the circumstances.

To summarize:

1. A velocity potential exists for any flow that is irrotational.

2. The derivative of the velocity potential in any direction yields the velocity com­ponent of the flow field in that same direction.

3. Explicit expressions for velocity components in terms of derivatives of the velo­city potential were developed for two-dimensional planar flows in Eqs. 4.14 and 4.15.

4. Equipotential lines are everywhere orthogonal to streamlines.

EXAMPLE 4.7 Given: A two-dimensional flow field is described by:

У = (2Х+У+1) і + (-2y) j.

Required: Determine the velocity potential for this flow.

Approach: Before trying to find the velocity potential, check first whether one exists.

Solution: If the velocity potential exists, then the flow must be irrotational and the curl of the velocity vector must be zero. For this two-dimensional flow, this means that the coefficient of the unit vector k in Eq. 4.6 must be zero. Evaluating,

I =№) – a), o.

Thus, the flow is not irrotational and no velocity potential exists.

Appraisal: Care must be taken to establish the existence of the velocity potential.

EXAMPLE 4.8 Given: A physically possible irrotational flow is y = (2x + 1) i – (2y) j.

Required: Find the velocity potential for this flow.

Approach: Use Eq. 4.14 and integrate to find the velocity potential.

Solution: Using Eq. 4.14,

u = дФ = 2x +1 ^ф = x2 + x + f(y) дФ = f'(y).

dx dy

Also, however, v = ^ = -2y => f'(y) = -2y ^ f (y) = – y2 + constant.


Substituting for f(y) in the previous expression for the velocity potential, ф = x2 + x – y2 + constant, where again the evaluation of the constant of integration is of no practical importance because the primary concern is with the derivatives of the velocity potential—that is, with the velocity components.

Appraisal: The determination of the velocity potential knowing the velocity components of a flow field is straightforward and parallels the finding of the stream function from given velocity components.