11.11  Sources and Sinks

    If the two-dimensional motion of an ideal fluid consists of an outward radial flow from a point and is symmetrical in all directions, then the point is called a simple source.  A source at the origin can be considered as a line perpendicular to the z plane along which fluid is being emitted.  If the rate of emission of volume of fluid per unit length  ,  then the origin is said to be a source of strength  m,  the complex potential for the flow is  

            ,  

and the velocity  V  at the point is given by  

            .  

For fluid flows, a sink is a negative source and is a point of inward radial flow at which the fluid is considered to be absorbed or annihilated.  Sources and sinks for flows are illustrated in Figure 11.93.

 

            Figure 11.93  Sources and sinks for an ideal fluid.

11.11.1  Source: A Charged Line

    In the case of electrostatics a source will correspond to a uniformly charged line perpendicular to the z-plane at the point .  We will show that if the line L is located at    and carries a charge density of    coulombs per unit length, then the magnitude of the electric field is  .  Hence  E  is given by  

(11-41)        ,  

and the complex potential is  

            ,  and  
            
            .  

A sink for electrostatics is a negatively charged line perpendicular to the z-plane.  The electrostatic field for electrostatic problems corresponds to the velocity field for fluid flow problems, except that their corresponding potentials differ by a sign change.

    To establish Equation (11-41), we start with Coulomb's law, which states that two particles with charges q and Q exert a force on one another with magnitude   ,  where r is the distance between particles and C is a constant that depends on the scientific units.  For simplicity, we assume that and the test particle at the point z has charge .  

    The contribution induced by the element of charge    along the segment of length situated at a height h above the plane has magnitude    given by  

            .  

    It has the same magnitude as induced by the element    located a distance -h  below the plane.  From the vertical symmetry involved their sum,  ,  lies parallel to the plane along the ray from the origin, as shown in Figure 11.94.

            Figure 11.94  Contributions to from the elements of charge situated at , above and below the z plane.

    By the principal of superposition, we add all contributions from the elements of charge along L to obtain  .  By vertical symmetry,   lies parallel to the complex plane along the ray from the origin through the point z.  Hence the magnitude of   is the sum of all components    that are parallel to the complex plane, where t is the angle between   and the plane.  Letting    in this summation process produces the definite integral   

            .  

    Next, we use the change of variable    and the trigonometric identity    to obtain the equivalent integral:  

            .  

    Multiplying this magnitude    by the unit vector    establishes Formula (11-41).  If , then the field is directed away from and, if , then it is directed toward .  An electrical field located at is given by   

            ,  

and the corresponding complex potential is  

            .

Example 11.30.  (Source and sink of equal strength)  Let a source and sink of unit strength be located at the points and , respectively.  The complex potential for a fluid flowing from the source at    to the sink at    is  

            .  

          

            Figure 11.95 (a)  Source and sink of unit strength.

The velocity potential and stream function are  

            ,   and  

            ,  

respectively.  Solving for the streamline  ,  we start with   

              

and obtain the equation

            .

A straightforward calculation shows that points on the streamline must satisfy the equation  

            ,  

which is the equation of a circle with center at    that passes through the points  .  Several streamlines are indicated in Figure 11.95(a).

Explore Solution 11.30.

Example 11.31.  (Two sources of equal strength)  Let two sources of unit strength be located at the points  .  The resulting complex potential for a fluid flow is  

            .  

                 Figure 11.95 (b)  Two sources of unit strength.

The velocity potential and stream function are


            ,   and  

            ,  

respectively.  Solving for the streamline  ,  we start with  

            

and obtain the equation  

            .  

     If we express this equation in the form    then we can get  

            ,   or  
            
            .

Now use the rotation of axes  

              

then the streamlines must satisfy the equation   

              

and are rectangular hyperbolas with centers at the origin that pass through the points  .  Several streamlines are indicated in Figure 11.95(b).

Explore Solution 11.31.

Ideal Fluid Flow

    Let an ideal fluid flow in a domain in the z plane be effected by a source located at the point .  Then the flow at points z, which lie in a small neighborhood of the point , is approximated by that of a source with the complex potential  

            .  

If    is a conformal mapping and  ,  then S(z) has a nonzero derivative at and   

            

where    as  .  Taking logarithms yields

            .

Because  ,  the term    approaches the constant value    as  .  As    is the complex potential for a source located at the point , the image of a source under a conformal mapping is a source.

    We can use the technique of conformal mapping to determine the fluid flow in a domain D in the z plane that is produced by sources and sinks.  If we can construct a conformal mapping    so that the image of sources, sinks, and boundary curves for the flow in D are mapped onto sources, sinks, and boundary curves in a domain G where the complex potential is known to be , then the complex potential in D is given by .

Example 11.32.  Suppose that the lines    are considered as walls of a containing vessel for a fluid flow produced by a single source of unit strength located at the origin.  The conformal mapping     maps the infinite strip bounded by the lines    onto the w plane slit along the boundary rays    and  ,  and the image of the source at    is a source located at  .  The complex potential  

              

determines a fluid flow in the w plane past the boundary curves    and  , which lie along streamlines of the flow.  Therefore the complex potential for the fluid flow in the infinite strip in the z plane is  

            .  

Several streamlines for the flow are illustrated in Figure 11.96. 

          

                 Figure 11.96  A source in the center of a strip.

Explore Solution 11.32.

Example 11.33.  Suppose that the lines    are considered as walls of a containing vessel for the fluid flow produced by a single source of unit strength located at the point    and a sink of unit strength located at the point  .  The conformal mapping    maps the infinite strip bounded by the lines    onto the w plane slit along the boundary rays    and  .  The image of the source at is a source at , and the image of the sink at is a sink at .  The potential  

              

determines a fluid flow in the w plane past the boundary curves , which lie along streamlines of the flow.  Therefore the complex potential for the fluid flow in the infinite strip in the z plane is

            .  

Several streamlines for the flow are illustrated in Figure 11.97. 

          

                 Figure 11.97  A source and a sink on the edges of a strip.

Explore Solution 11.33.

    We can use the technique of transformation of a source to determine the effluence from a channel extending from infinity.  In this case, we construct a conformal mapping    from the upper half-plane so that the single source located at is mapped to the point at infinity that lies along the channel.  The streamlines emanating from in the upper half-plane are mapped onto streamlines issuing from the channel.

Example 11.34.  Consider the conformal mapping   

            ,  

which maps the upper half-plane onto the domain consisting of the upper half-plane joined to the channel  .  The point is mapped onto the point and is mapped onto along the channel.  Images of the rays  , where is a constant, are streamlines issuing from the channel as indicated in Figure 11.98. 

          

                 Figure 11.98  Effluence from a channel into a half-plane.

Explore Solution 11.34.

Extra Example 1.  Use a Schwarz-Christoffel transformation, (see Section 11.9), to show that the conformal mapping  

              

will map the flow in the upper half-plane from a source at    onto the flow from a channel with    into the first quadrant, as shown in Figure 11.105.
Hint: Set   and .   

     

                 Figure 11.105  Flow from a channel into a quadrant.

Solution.  Using the Schwarz-Christoffel transformation method, (see Section 11.9), the exterior angles are  ,  and the formula for the derivative is  

               
integrate

            
            
            
            
            

    
use the conditions  ,    and solve the resulting system for A and B.  The desired result is

            .

Explore Extra Solution 1.

Extra Example 2.  Use a Schwarz-Christoffel transformation, (see Section 11.9), to show that the conformal mapping  

            
            

will map the flow in the upper half-plane from a source at    onto the flow from a channel with    into a sector, as shown in Figure 11.106.
Hint: Set   and .  Use the change of variable    in the resulting integral. 

     

                 Figure 11.106  Flow from a channel into a sector in the first quadrant.

Solution.  Using the Schwarz-Christoffel transformation method, (see Section 11.9), the exterior angles are  ,  and the formula for the derivative is  

               
integrate

            
            
use the substitutions

              
and get

              

Use the identity    and write the integral as

            

Now use the substitution and get

            
            
use the conditions  ,    and solve the resulting system for A and B.  The desired result is

            .

Explore Extra Solution 2.

Extra Example 3.  Use a Schwarz-Christoffel transformation, (see Section 11.9), to show that the conformal mapping  

              

will map the flow in the upper half-plane from a source at    onto the flow in a right-angled channel with  , as shown in Figure 11.107. 

     

                 Figure 11.107  Flow in a right-angled channel.

Solution.  Using the Schwarz-Christoffel transformation method, (see Section 11.9), if we choose    and  ,  then the formula  

            

will determine a mapping of the upper half-plane onto the domain indicated in Figure 11.74(a).  With  ,  we let ,  then    and  .  

    The limiting formula for the derivative becomes   

            

where  , which will determine a mapping from the upper half plane onto the channel as indicated in Figure 11.74(b) and Figure 11.107.

    Using integrals in Table 11.2, we obtain

           

    If we use the principal branch of the inverse sine function, then the boundary values    lead to the system  

               and   ,  

which we can solve to obtain .  Hence the required solution is

              

Explore Extra Solution 3.

Extra Example 4.  Use a Schwarz-Christoffel transformation, (see Section 11.9), to show that the conformal mapping  

              

will map the flow in the upper half-plane onto the flow from a channel back into a quadrant, as shown in Figure 11.108,
where    and  . 

     

                 Figure 11.108  Flow from a channel  back into the first quadrant.

Explore Extra Solution 4.