Question

Construct expressions for the stream function and velocity potential of a Half Body. This is a source in a uniform stream. 1. Visualize the Flow Net (the streamlines and velocity potential lines) 2. Determine an expression for the velocity field. Note that the book uses cylindrical coordinates here

Answer 1

A circular cylinder is one of the basic geometrical shapes and the flow passing over it can be simulated by combination of a uniform flow and doublet. When the distance between source-sink pair approaches zero, the shape Rankine oval becomes more blunt and approaches a circular shape.

Consider the superposition of a uniform flow of velocity
V.
and a doublet of strength
A
as shown in Fig. The direction of the doublet is upstream, facing into uniform flow.

= 0 r= R Stagnation streamline A B V. Uniform flow Flow over a cylinder Doublet

Fig. : Superposition of a uniform flow and doublet.

The stream function for the combined flow is,

Wunomsouble: = V ^ sin 9-A sin Vrsine1-- 27T 2TV Vrsin1- .2 2nV

1

The velocity field is obtained as,

R2 1-Vcos v aw 1+ sine 2 r r 2e ar

2

In order to locate the stagnation point, assign the velocity components in Eq.2 to zero value and simultaneously solve for
r and e
. There are two stagnation points, located at
(r,e) (R.0) and (R.T)
and denoted by points A and B, respectively. The equation of streamlines that passes through the stagnation points A and B, is given by the following expression;

(3)

This equation is satisfied by for all values of θ. Since R is a constant, Eq. (3) may be interpreted as the equation of a circle with radius R with center at the origin. It is satisfied by
e=0 and T
for all values of R. Different values of R may be obtained by varying the uniform velocity and/or doublet strength. Hence, entire horizontal axis through the points A and B, extending infinitely far upstream and downstream, is a part of stagnation streamline. The above discussions can be summarized as follows;

• The dividing streamline is a circle of radius R. The family of circles can be obtained by assigning different values of R with various doublet strength and free stream velocity.

• The flow inside the circle is generated from the doublet whereas flow outside the circle comes from the uniform flow. So, the flow inside the circle may be replaced by solid body and the external flow will not feel the difference.

• Thus, the inviscid, irrotational, incompressible flow over a circular cylinder of radius R can be simulated by adding a uniform flow of velocity
V.
and a doublet of strength
A
and R is related to
V.
AND
A
.

(4)

• • The dividing streamline that passes through the stagnation points A and B as shown in Fig. 1.

Referring to the Fig. 1, it is seen that the entire flow field is symmetrical about both horizontal and vertical axes through the center of the cylinder. It means the pressure distribution is also symmetrical about both the axes. When the pressure distribution over the top part of the cylinder is exactly balanced by the bottom part, there is no lift . Similarly, when the pressure distribution on the front part of the cylinder is exactly balanced by rear portion, then there is no drag. This is in contrast to the realistic situation i.e. a generic body placed in a flow field will experience finite dragand zero lift may be possible. This paradox between the theoretical result of zero drag in an inviscid flow and the knowledge of finite drag in real flow situation is known as d' Alembert's paradox .

Pressure Coefficient

In general, pressure is a dimensional quantity. Many a times, it is expressed in a non-dimensional form with respect to free stream flow and the ‘pressure coefficient' is defined as follows;

(5)

Here, are the free stream density and free stream velocity, respectively. The term is called as dynamic pressure. For incompressible flow, if a body is immersed in the free stream, then Bernoulli's equation can be written at any arbitrary point in the flow field as,

(6)

On the surface of the cylinder shown in Fig. 1, the velocity distribution can be obtained from Eq. (2) i.e.

v, 0 v =-2V sin e

(7)

Fig. 2: Maximum velocity in the flow over a circular cylinder.

Here, is geometrically normal to the surface and is tangential to the surface of the cylinder as shown in Fig. 2(a). The negative sign signifies that is positive in the direction of increasing θ. It may be observed that the velocity at the surface reaches to maximum value of at the top and bottom of the cylinder as shown in Fig. 2(b). Eqs. (6) and (7) can be combined to obtain the surface pressure coefficient as,

(8)

The pressure distribution over the cylinder is plotted in Fig. 3. Here, varies from 1 at the stagnation point to -3 at the points of maximum velocity. It is also clear that the pressure distribution at the top half the cylinder is equal to the bottom half and hence the lift is zero. Similarly, the pressure distribution on the front part of the cylinder is exactly balanced by rear portion and there is no drag. Both, normal force and axial force coefficients are same as lift and drag coefficients. They are calculated from as given below;

(9)

Here, LE and TE stands for leading edge and trailing edge, respectively. The subscripts u and lrefers to upper and lower surface of the cylinder. The chord c is the diameter of the cylinder (R).

Fig. 3: Surface pressure coefficient for a circular cylinder.

When the surface pressure matches with free stream pressure, then Eq. (3.7.8) reduces to,

(10)

These points as well as the stagnation points and location of minimum pressure are illustrated in Fig. 4.

Fig. 4: Pressure values at various locations on the surface of the cylinder.