4. Potential Flow Theory

(From Aerodynamics for Students Website)

We consider only inviscid flows in this chapter. This leads to a simple analysis. In fact we will be solving only the continuity equation for mass to calculate velocity components. Pressure is obtained from Bernoulli equation. Of course, various assumptions will have to made to make the analysis easier. We discuss these in course of the text.

We first derive the continuity equation which is a statement of the fact that mass is conserved. Then we introduce the stream function which is a powerful concept in fluid dynamics. By analyzing the kinematics of fluid motion we proceed to introduce concepts of Circulation and Irrotationality. Definition of Velocity Potential follows. We then write down the stream functions and velocity potentials for some of the simple flows like a uniform flow, source and sink flow and vortex flow. These flows are then superposed to arrive at solutions for complicated flows. Flow about a circular cylinder is then analyzed in some detail.

Conservation of Mass

Figure 4.1: Differential Control Volume

We derive the equation for mass conservation by considering a differential control volume at P(x,y,z) as shown in Fig.4.1. Let the dimensions of the volume be dx, dy and dz and velocity components at P be u, v and w. Assuming that the mass flow rate is continuous across the volume we can calculate the mass flow rates at the various faces of the cell by a Taylor Series expansion as we had done previously (Eqn. 2.5). Accordingly we have,

$\displaystyle ML =\left[ \rho u - {\partial \over {\partial x}}(\rho u) 
 dx\ri...
...z; MR =\left[ \rho u + {\partial \over {\partial x}}(\rho u) 
 dx \right] dy dz$    
$\displaystyle MB =\left[ \rho v - {\partial \over {\partial y}}(\rho v) 
 dy\ri...
...z; MT =\left[ \rho v + {\partial \over {\partial y}}(\rho v) 
 dy \right] dx dz$    
$\displaystyle MA =\left[ \rho w - {\partial \over {\partial z}}(\rho w) 
 dz\ri...
...y; MF =\left[ \rho w + {\partial \over {\partial z}}(\rho w) 
 dz \right] dx dy$ (4.1)

The net mass flow rate into the control volume as a consequence is given by,
$\displaystyle MNET =  MR - ML + MT - MB + MA - MF$ (4.2)

Applying the Reynolds transport theorem for mass (Eqn. 3.30) will give,

$\displaystyle {\int_{CV} {{\partial \rho} \over {\partial t}} d
 \forall}  +  MNET  =  0$ (4.3)

From Eqn.4.1 and 4.2 we have,

$\displaystyle MNET = \left( {\partial \over {\partial x}}(\rho u) + {\partial ...
...\partial y}}(\rho v)
 + {\partial \over {\partial z}}(\rho w) \right) dx dy dz$ (4.4)

Further in Eqn.4.3 noting that the control volume is tiny, the integral can be approximated as

$\displaystyle {\int_{CV} {{\partial \rho} \over {\partial t}} d
 \forall}  \approx  {{\partial \rho} \over {\partial t}} dx dy dz$ (4.5)

The Reynolds Transport Theorem thus gives,

$\displaystyle {{\partial \rho} \over {\partial t}} dx dy dz + \left( {\partial ...
...tial y}}(\rho v)
 + {\partial \over {\partial z}}(\rho w) \right) dx dy dz = 0$ (4.6)

Cancelling out dx dy dz, we have,

$\displaystyle {{\partial \rho} \over {\partial t}}  + {\partial \over {\partial...
...rtial \over {\partial y}}(\rho v)
 + {\partial \over {\partial z}}(\rho w) = 0$ (4.7)

Eqn. 4.7 is known as the Continuity Equation. Note that it is a very general equation with hardly any assumption except that density and velocities vary continually across the element we have considered.

If we now bring in the gradient operator, namely,

$\displaystyle \nabla = \hat{i}{\partial \over {\partial x}} + \hat{j}{\partial \over {\partial y}}
 + \hat{k}{\partial \over {\partial z}} $ (4.8)

and represent velocity as a vector,

$\displaystyle \vec{V} = \hat{i}u + \hat{j}v + \hat{k}w$ (4.9)

Then the Continuity Equation can be written in a compact manner as
$\displaystyle {{\partial \rho} \over {\partial t}}  + \nabla \cdot (\rho \vec{V}) = 0$ (4.10)

Written in this form it enables one to consider any other system of coordinates with ease.

Continuity Equation in Cylindrical Polar Coordinates

We have derived the Continuity Equation, 4.10 using Cartesian Coordinates. It is possible to use the same system for all flows. But sometimes the equations may become cumbersome. So depending upon the flow geometry it is better to choose an appropriate system. Many flows which involve rotation or radial motion are best described in Cylindrical Polar Coordinates. Let us now write equations for such a system. In this system coordinates for a point P are $ r, \theta$ and $ z$, which are indicated in Fig.4.2. The velocity components in these directions respectively are $ v_r, v_\theta$ and $ v_z$. Transformation between the Cartesian and the polar systems is provided by the relations,

$\displaystyle r = \sqrt{x^2 + y^2},   \theta = \tan^{-1}{y \over x},  z=z$ (4.11)

The gradient operator is given by,

$\displaystyle \nabla P = {1 \over r} {\partial \over \partial r} (rP) + {1 \over r} {\partial \over \partial
 \theta}(P) + {\partial \over \partial z}(P)$ (4.12)

As a consequence the continuity equation becomes,
$\displaystyle {{\partial \rho}\over {\partial t}} + {1 \over r} {\partial \over...
...over \partial
 \theta}(\rho v_\theta) + {\partial \over \partial z}(\rho v_z)=0$ (4.13)

Figure 4.2: Cylindrical Polar Coordinate System

@

Continuity Equation for steady flow

For a steady flow the time derivative vanishes. As a result 4.7 becomes,

$\displaystyle {\partial \over {\partial x}}(\rho u) + {\partial \over {\partial y}}(\rho v)
 + {\partial \over {\partial z}}(\rho w) = 0$ (4.14)

The equation in polar coordinates also undergoes the same simplification.
$\displaystyle {1 \over r} {\partial \over \partial r} (r \rho v_r) + {1 \over ...
...over \partial
 \theta}(\rho v_\theta) + {\partial \over \partial z}(\rho v_z)=0$ (4.15)

These equations are the ones that are to be used for a compressible flow as we have kept density, $ \rho$ still variable.

Continuity Equation for an Incompressible flow

For an incompressible flow density is a constant. Accordingly we have

$\displaystyle {\partial u\over {\partial x}} + {\partial v\over {\partial
 y}}
 + {\partial w \over {\partial z}} = 0$ (4.16)

and in polar coordinates we have,
$\displaystyle {1 \over r} {\partial \over \partial r} (r v_r) + {1 \over r} {\partial \over \partial
 \theta}( v_\theta) + {\partial \over \partial z}( v_z)=0$ (4.17)

As noticed for the control volume analysis the continuity equation for an incompressible flow is the same whether the flow is steady or unsteady.

Stream function

Stream function is a very useful device in the study of fluid dynamics and was arrived at by the French mathematician Joseph Louis Lagrange in 1781. Of course, it is related to the streamlines of flow, a relationship which we will bring out later. We can define stream functions for both two and three dimensional flows. The latter one is quite complicated and not necessary for our purposes. We restrict ourselves to two-dimensional flows.

Consider a two-dimensional incompressible flow for which the continuity equation is given by,

$\displaystyle {\partial u \over {\partial x}} + {\partial v \over {\partial y}} = 0$ (4.18)

A stream function $ \psi$ is one which satisfies

$\displaystyle u = {\partial \psi \over {\partial y}} ,    v = -{\partial \psi \over {\partial
 x}}$ (4.19)

Substituting these into Eqn.4.18, we have,

$\displaystyle {\partial \over {\partial x}} \left(\partial \psi \over {\partial...
...l \over {\partial y}}\left(-{\partial \psi \over {\partial x}} \right) \equiv 0$ (4.20)

Thus the continuity equation is automatically satisfied. Thus if we can find a stream function $ \psi$ that meets with the eqn.4.19 the continuity equation need not be solved. For the rest of the chapter we will be invariably describing flows with a stream function.

Line = constant is a streamline

Let us consider a line given by $ \psi = C$, a constant as shown in Fig. 4.3. We have

$\displaystyle \psi = C$
   
i.e.,
   
giving
   
  -vdx + udy = 0   after substituting for etc.    
(4.21)

Going back to Eqn. 4.21, this is the equation to a streamline. What we have proved then is that =constant line is a streamline of the flow. Alternately equation to a streamline is given by = constant.

Figure 4.3 : A Stream Line in A Flow

@

$ d\psi$ between two streamlines is proportional to the Volumetric Flow

Figure 4.4: Flow between two Stream Lines.

Consider the volumetric flow through a small element of thickness ds placed on a streamline as shown in Fig. 4.4. The volumetric flow through the element is given by

$\displaystyle dQ $ $\displaystyle = u dy - vdx$    
@ $\displaystyle ={{\partial \psi} \over {\partial y}} dx - {(-{{\partial \psi}
 \over {\partial x}}}) dy$    
@ $\displaystyle ={{\partial \psi} \over {\partial y}} dx + {{{\partial \psi}
 \over {\partial x}}} dy$    
@ $\displaystyle = d\psi$ (4.22)

which indicates that the volumetric flow rate is proportional to the difference between stream functions. If we now integrate the Eqn.4.22 between two stream lines $ \psi=C$ and $ \psi=D$, we have,

$\displaystyle Q = \int_1^2 dQ = \int_{\psi_1} ^{\psi_2}
 d\psi = \psi_2 - \psi_1$ (4.23)

Stream Function in Polar Coordinates

The velocity components in polar coordinates are related to the stream function by,

$\displaystyle v_r = {1 \over r} {{\partial \psi} \over {\partial \theta}},
    v_\theta = - {{\partial \psi} \over {\partial r}}$ (4.24)

Kinematics of Fluid Motion

Consider a two-dimensional fluid element, a square ABCD for simplicity. when the fluid flows this element is subject to various forces and as a result undergoes a complex motion and a possible deformation as indicated in Fig.4.5 and assumes a shape like A`B`C`D`. It appears that the complex deformation of the element can be split into four basic constituents -

  1. Translation
  2. Linear Deformation
  3. Rotation
  4. Angular Deformation

Figure 4.5 : Deformation of A fluid Element

Figure 4.6 : Basic Deformation of A fluid Element

These elemental deformations have been sketched in Fig.4.6 . Let us now consider each of these separately.


Translation

Translation is the type of motion where the element retains its shape. Its side do not undergo any change in length and the four angles do remain square. The square element ABCD, we have considered is bodily shifted from its position to a new one A`B`C`D`. If the motion is without acceleration in a steady, uniform flow it is easy to calculate the position of any particle in the fluid at different instants of time. But it should be noted that the fluid particles may undergo acceleration.

Figure 4.7: Translation of A fluid Element

Considering now a particle in the fluid element we can write down an expression for its acceleration. At time t, the particle is at (x,y,z) and its velocity is given by,

$\displaystyle \overrightarrow{V}_p(t) = \overrightarrow{V}(x,y,z,t)$ (4.25)

where $ \overrightarrow{V}(x,y,z,t)$ is the fluid velocity. At time (t+dt) the particle will find itself at a different location and its velocity now is given by,
$\displaystyle \overrightarrow{V}_p(t+dt) = \overrightarrow{V}(x+dx,y+dy,z+dz,t+dt)$ (4.26)

Consequently the change in velocity is given by,
$\displaystyle d
 \overrightarrow{V}_p = {\partial \overrightarrow{V} \over {\pa...
...over {\partial
 z}} dz_p + {\partial \overrightarrow{V} \over {\partial
 t}} dt$ (4.27)

The acceleration of the particle is obtained by dividing throughout by dt. Accordingly,

$\displaystyle {d\overrightarrow{V}_p \over {\partial t}} = {\partial \overright...
...tial
 z}} {{dz_p}\over{dt}} + {\partial \overrightarrow{V} \over {\partial
 t}}$ (4.28)

Now denoting the speed in x-direction, dx/dt by u, speed in y-direction dy/dt by v and speed in z-direction dz/dt by w, we have,

$\displaystyle a_p= {d\overrightarrow{V}_p \over {\partial t}} = {\partial \over...
...w{V} \over {\partial
 z}} w + {\partial \overrightarrow{V} \over {\partial
 t}}$ (4.29)

where ap is the particle acceleration. The derivative dVp/dt is usually denoted by DV/Dt and is called Material Derivative or the Particle Derivative or the Total Derivative.

Linear Deformation

Consider the same element ABCD again. If the element has to undergo a linear deformation it is necessary that u velocity change in the x-direction and v velocity in the y-direction. Let the velocities at A be u and v. Then at B the u-velocity will be $ u+{\partial u \over \partial x}dx$ and at D it will be $ v+{\partial v \over \partial y}dy$. As a result the element stretches both in x and y directions and assumes a shape A`B`C`D` shown in Fig. 4.8.


Figure 4.8: Linear Deformation of a Fluid Element

Now, the stretch in x-direction is given by distance BB' which is equal to

$\displaystyle BB' = {\partial u \over \partial x} dx  dt$ (4.30)

Corresponding change in the volume of the element (for a unit depth normal to the paper) is given by
$\displaystyle d\forall_x = {\partial u \over \partial x} dx  dy dt$ (4.31)

Similarly the change in the volume of the element in the y-direction is given by,
$\displaystyle d\forall_y = {\partial v \over \partial y} dy  dx dt$ (4.32)

Neglecting the change in volume due to CC```C`C``, The total change in volume is

$\displaystyle d\forall = \left \{{{\partial u \over \partial x} + {\partial v \over \partial y}}\right\} dx  dy dt$ (4.33)

Thus the rate of change of volume expressed as a fraction of the initial volume is given by,

$\displaystyle {1 \over \forall}{d\forall \over {dt}} = {{\partial u \over \partial x} + {\partial v \over \partial y}}$ (4.34)

The left hand side is called the Volume Dilitation rate of the element. We have seen that the Right hand side of the equation is zero for all incompressible flows.

Rotation

Considering the same element ABCD again, we notice that any rotation of AB or AD is brought about by a change in u velocity along y-direction and that of v velocity along x-direction. Let $ d\alpha$ and $ d\beta$ be the angles through which sides AB and AD rotate.


Figure 4.9 : Rotation of a fluid Element

Now it is required that

$\displaystyle BB' $ $\displaystyle = {\partial v \over \partial x} dx  dt$    
$\displaystyle DD' $ $\displaystyle = {\partial u \over \partial y} dy  dt$ (4.35)

Since angles $ d\alpha$ and $ d\beta$ are small we have,

$\displaystyle d\alpha $ $\displaystyle ={{BB'} \over {AB}}= {{{{\partial v \over \partial x} dx  dt}} \over {dx} }
 = {\partial v \over \partial x} dt$    
$\displaystyle d\beta $ $\displaystyle ={{DD'} \over {AD}}= {{{{\partial u \over \partial y} dy 
 dt}} \over {dy} }
 = {\partial u \over \partial y} dt$ (4.36)

Rotation or angular velocity of the element (about the z-axis) is defined as

$\displaystyle \omega_z = {1 \over 2} \left( {d\alpha \over {dt}} - {d\beta \over
 {dt}}\right)$ (4.37)

Combining eqns. 4.36 and 4.37 we have,
$\displaystyle \omega_z = {1 \over 2} \left( {\partial v\over {\partial x}} - {\partial u \over
 {\partial y}}\right)$ (4.38)

Similarly we have for rotation about the other axes,
$\displaystyle \omega_x = {1 \over 2} \left( {\partial w\over {\partial y}} - {\...
...\left( {\partial u\over {\partial z}} - {\partial w \over
 {\partial x}}\right)$ (4.39)

We see that $ \omega$ is a vector given by $ \omega = \hat{i}\omega_x + \hat{j}\omega_y + \hat{k}\omega_z$, which can be written in vector notation as,

$\displaystyle \omega =  {1 \over 2} (curl \vec{V}) = {1 \over 2}\left\vert\begi...
...ial y} & \partial \over {\partial z} \ 
 u & v & w \ 
 \end{array}\right\vert$ (4.40)

We now introduce another term Vorticity which is defined as twice rotation or

$\displaystyle \zeta = 2\omega = curl \vec{V}$ (4.41)

This brings us to a class of flows for which vorticity (i.e., rotation) is zero. These are known as Irrotational Flows. These are governed by the equation,

curl    
which for a two-dimensional flow becomes                         
(4.42)

Meaning of Irrotationality

It should be noted that the concept of Irrotationality applies to a fluid element in a given flow than to the flow itself. The main flow may be a vortex where the streamlines are circles. But the individual elements of fluid may not rotate or distort making the flow irrotational. This is shown in Fig. 4.10 where an irrotational flow about an aerofoil is sketched. Note that even though the main follows a path which seems to indicate distortion, the fluid elements are simply translated.

Figure 4.10: Meaning of Irrotationality.

Angular Deformation

Our previous discussion of rotation also leads to the definition of angular deformation or rate of shear strain. For a two-dimensional flow this is given by,

$\displaystyle \dot \gamma = {d\alpha \over {dt}} + {d\beta \over
 {dt}} = {\partial v\over {\partial x}} + {\partial u \over {\partial
 y}}$ (4.43)

Shear stress for the element is thus given by

$\displaystyle \tau_{xy}= \mu \dot\gamma =  \mu \left( {\partial v\over {\partial x}} + {\partial u \over {\partial
 y}}\right)$ (4.44)

Circulation

We now discuss one other property of flows, that of Circulation. Consider any closed curve C in a flow as shown. Circulation is defined as the line integral around the curve of the arc length ds times the tangential component of velocity. Shear stress for the element is thus given by

Figure 4.11 : Definition of Circulation

$\displaystyle \Gamma = \oint_C \vec{V}.ds = \oint_C  V  cos\alpha  ds = \oint_C
 (udx + vdy + wdz)$ (4.45)

An expression for circulation can be derived by considering a small differential area in the curve, which is shown enlarged in Fig.4.12.

Figure 4.12: Definition of Circulation, continued

If the velocity components at A are (u,v) then we have

Velocity along AB: $\displaystyle (u,v)$    
Velocity along BC: $\displaystyle (u + {\partial u \over \partial x} dx,v + {\partial v \over \partial x} dx))$    
Velocity along DC: $\displaystyle (u + {\partial u \over \partial y} dy,v + {\partial v \over \partial y} dy)$    
Velocity along AD: $\displaystyle (u,v)$ (4.46)

Circulation along the boundary of the differential element is given by

$\displaystyle d\Gamma = (udx+vdy)_{AB} +  (udx+vdy)_{BC} +  (udx+vdy)_{CD} + 
 (udx+vdy)_{DA}$ (4.47)

Noting that along AB, dy = 0 etc, the above equation reduces to

$\displaystyle d\Gamma =  (udx)_{AB} +  (vdy)_{BC} + (udx)_{CD} + 
 (vdy)_{DA}$ (4.48)

Upon substituting for velocity components from Eqn.4.46, we have

$\displaystyle d\Gamma = (udx) +  (v + {\partial v \over \partial x} dx)dy -
  (u + {\partial u \over \partial y} dy)dx -  (vdy)$ (4.49)

which on simplification gives,

$\displaystyle d\Gamma = \left({\partial v \over \partial x} - {\partial u \over \partial
 y}\right) dx dy$ (4.50)

Integrating this for the entire region C gives,

$\displaystyle \Gamma = \int_C d\Gamma = \int \int \left({\partial v \over \partial x} - {\partial u \over \partial
 y}\right) dx dy$ (4.51)

In other words we have,

$\displaystyle \Gamma = \oint_C (u dx + v dy)= \int \int \left({\partial v \over \partial x} - {\partial u \over \partial
 y}\right) dx dy$ (4.52)

Thus we see that a complicated area integral (also a double integral) is reduced to a single integral along the curve.

The other important observation to make is that for an irrotational flow, circulation is zero.

Velocity Potential

We have seen that for an irrotational flow $ curl(\vec{V}) =0  
or   \nabla \times \vec{V}=0$. It follows from vector algebra that there should be a potential such that

$\displaystyle \vec{V} = \nabla \phi$ (4.53)

$ \phi$ is called the Velocity Potential. The velocity components are related to $ \phi$ through the following relations.

$\displaystyle u = {{\partial \phi} \over {\partial x}},  v = {{\partial \phi} \over {\partial
 y}},  w = {{\partial \phi} \over {\partial z}}$ (4.54)

Velocity potential is a powerful tool in analysing irrotational flows. First of all it meets with the irrotationality condition readily. In fact, it follows from that condition. As a check we substitute the velocity potential in the irrotationality condition, thus,

$\displaystyle {\partial v\over {\partial x}} - {\partial u \over {\partial
 y}}...
... - {\partial \over {\partial y}}{ {\partial \phi} \over {\partial
 x}} \equiv 0$ (4.55)

The next question we ask is does the velocity potential satisfy the continuity equation? To find out we consider the continuity equation for incompressible flows and substitute the expressions for velocity coordinates in them. Accordingly,

$\displaystyle {\partial u\over {\partial x}} + {\partial v \over {\partial
 y}}...
...artial^2 \phi} \over {\partial x^2}} + {{\partial^2 \phi} \over {\partial y^2}}$ (4.56)

It is clear that to meet with the continuity requirements the velocity potential has to satisfy the equation,

$\displaystyle {{\partial^2 \phi} \over {\partial x^2}} + {{\partial^2 \phi}
 \over {\partial y^2}} = 0$ (4.57)

In vector notation it is

$\displaystyle \nabla^2  \phi = 0$ (4.58)

As with stream functions we can have lines along which potential $ \phi$ is constant. These are called Equipotential Lines of the flow. Thus along a potential line $ \phi = C$.

The equation 4.58 is called the Laplace Equation and is encountered in many branches of physics and engineering. A flow governed by this equation is called a Potential Flow. Further the Laplace equation is linear and is easily solved by many available standard techniques, of course, subject to boundary conditions at the boundaries.

Note that in terms of velocity potential expression for circulation(Eqn. 4.45, see Circulation) assumes a simple form.

$\displaystyle \Gamma $ $\displaystyle = \oint_C \vec{V}\cdot d\vec{s}$    
and
   
@ $\displaystyle \vec{V} \cdot ds = \nabla \phi \cdot ds = d \phi$ (4.59)
giving
   
$\displaystyle \Gamma $ $\displaystyle = \oint_C d \phi$ (4.60)

Relationship between $ \phi$ and $ \psi$

We notice that velocity potential $ \phi$ and stream function $ \psi$ are connected with velocity components. It is necessary to bring out the similarities and differences between them.

Stream function is defined in order that it satisfies the continuity equation readily (eqn. 4.20 see Stream function). We do not know yet if it satisfies the irrotationality condition. So we test out below. Recall that the velocity components are given by

$\displaystyle u = {\partial \psi \over {\partial y}} ,    v = -{\partial \psi \over {\partial
 x}}$    

Substituting these in the irrotationality condition, we have
$\displaystyle {\partial v\over {\partial x}} - {\partial u \over {\partial
 y}}...
...artial^2 \psi} \over {\partial x^2}} - {{\partial^2 \psi} \over {\partial y^2}}$ (4.61)

Which leads to the condition that $ \nabla^2 \psi = 0$ for irrotationality.

Thus we see that the velocity potential $ \phi$ automatically complies with the irrtotatioanlity condition, but to satisfy the continuity equation it has to obey that $ \nabla^2 \phi = 0$. On the other hand the stream function readily satisfies the continuity condition, but to meet with the irrotationality condition it has to obey $ \nabla^2 \psi = 0$.

Thus we see that the streamlines too follow the Laplace Equation. So it is possible to solve for a potential flow in terms of stream function.

Property $ \psi$ $ \phi$
Continuity Equation Automatically Satisfied satisfied if $ \nabla^2 \phi$=0
Irrotationality Condition satisfied if $ \nabla^2 \psi$=0 Automatically Satisfied

Table 4.1 : Properties of stream function and velocity potential

Streamlines and equipotential lines are orthogonal to each other. We have seen that the velocity components of the flow are given in terms of velocity potential and stream function by the equations,

$\displaystyle u $ $\displaystyle = {{\partial \phi} \over {\partial x}} = {{\partial \psi} \over
 {\partial y}}$    
$\displaystyle v $ $\displaystyle = {{\partial \phi} \over {\partial y}} = -{{\partial \psi}
 \over {\partial x}}$ (4.62)

Those familiar with Complex Variables theory will recognise that these are the Cauchy-Riemann equations and that $ \phi = C$ and $ \psi = D$ are orthogonal and that both $ \phi$ and $ \psi$ obey Laplace Equation. However, we will prove the orthogonality condition by other means.

Figure 4.13 : Orthogonality of Stream lines and Equi-potential lines

Since $ \phi = C$, it follows that

$\displaystyle d\phi$ $\displaystyle = {{\partial \phi} \over {\partial x}}dx + {{\partial \phi}
 \over
 {\partial y}} dy$    
@ $\displaystyle =0$    
@ $\displaystyle =u dx + v dy$ (4.63)

The gradient of the equipotential line is hence given by
$\displaystyle \left( {dy} \over {dx} \right)_{\phi = C} = -{u \over v}$ (4.64)

On the other hand the gradient of a stream line is given by
$\displaystyle \left( {dy} \over {dx} \right)_{\psi = D} = {v \over u}$ (4.65)

Thus we find that

$\displaystyle \left( {dy} \over {dx} \right)_{\phi = C}\left( {dy} \over {dx}
 \right)_{\psi = D} = -1$ (4.66)

showing that equipotential lines and streamlines are orthogonal to each other. This enables one to calculate the stream function when the velocity potential is given and vice versa.

Fig. 4.14 shows the flow through a bend where the streamlines and the equipotential lines have been plotted. The two form an orthogonal network.

Figure 4.14 : Stream lines and Equi-potential lines for flow through a bend

Occurrence of Irrotational Flows

A question that naturally arises is "Where do we find irrotational flows?". A uniform flow is definitely irrotational. But one hardly finds a uniform flow in nature. Further, there is hardly anything to calculate for a uniform flow.

The other region where we can expect an irrotational flow is away from any solid body. Recall the "Thought Experiment" with two parallel plates (What is a Fluid?) when the space in between is filled with a fluid. Here once the top plate starts moving we have seen that a velocity gradient is set up in the flow normal direction. This gives rise to $ \partial u / \partial y$ which contributes directly to vorticity or rotation. As such this flow is NOT irrotational. A similar velocity gradient is set up when a fluid flows past a solid body as shown in Fig.4.15. The velocity right on the body surface is zero and it build up gradually we move in a normal direction away from the body. This region is highly rotational and is called the Boundary Layer. But at some distance form the body this velocity gradient flattens out and the velocity becomes constant in the flow normal direction. This is one of the irrotational regions of flow. As indicated in the figure the flow in the wake of the body is also NOT irrotational.

Figure 4.15: Occurrence of irrotational and rotational regions for flow past a body

Figure 4.16: Occurrence of irrotational and rotational regions for flow through a pipe.

At the entrance to a pipe as shown in Fig.4.16 one has a uniform flow. As the flow enters the pipe, velocity components are forced to be zero on the surface of the pipe. A boundary layer develops and starts to grow. At the beginning one sees a inviscid core encircled by a boundary layer. The flow in the inviscid core is irrotational. However, as we move downstream the boundary layers grow and merge to give a fully developed flow when the entire flow is NOT irrotational.

It is also worth noting that the flow is irrotational wherever Bernoulli equation is valid.

We could foresee from this that an inviscid flow is likely to be irrotational. In fact it is broadly true except in case of High Speed flows where shocks could occur. As indicated in Fig. the region behind a shock in a high speed flow has severe gradients of velocity making $ \omega$ not negligible.

Figure 4.17 : Rotational flow behind a shock wave in a high speed flow.

Simple Examples of Plane Potential Flows

In this section we consider some of the simple potential flows. The examples considered are such that there is an analytical expression for $ \phi, \psi, u,,$ for each of them. While calculating such flow a good coordinate system is important. Of course, it is true that any example could be handled with the Cartesian coordinates. But depending on the problem, the expressions may become too complicated. Further, the geometry of the flow itself indicates the coordinates to be chosen. Accordingly it is necessary to write down the important formulas involving stream function and velocity potential in the Cartesian and Polar coordinate systems.

Equations in Cartesian Coordinates

The velocity components are given by

$\displaystyle u $ $\displaystyle = {{\partial \phi} \over {\partial x}} = {{\partial \psi} \over
 {\partial y}}$    
$\displaystyle v $ $\displaystyle = {{\partial \phi} \over {\partial y}} = -{{\partial \psi}
 \over {\partial x}}$ (4.67)

The Laplacian is given by,

$\displaystyle {{\partial^2 \phi} \over {\partial x^2}} + {{\partial^2 \phi}
 \o...
...rtial^2 \psi} \over {\partial
 x^2}} + {{\partial^2 \psi} \over {\partial y^2}}$ (4.68)

Equations in Polar Coordinates

Velocity Components:

$\displaystyle v_r $ $\displaystyle = {{\partial \phi} \over {\partial r}} = {1 \over r} {{\partial \psi} \over {\partial \theta}} $    
$\displaystyle    v_\theta $ $\displaystyle = {1 \over r} {{\partial \phi} \over {\partial \theta}} = - {{\partial \psi} \over {\partial r}}$ (4.69)

Laplacian:

Uniform Flow

The simplest possible potential flow is a uniform flow, $ U_\infty=C$. The velocity potential is given by

Figure 4.18 : Uniform Flow

$\displaystyle u = U_\infty = {{\partial \phi} \over {\partial x}},   v = {{\partial \phi} \over {\partial
 y}} = 0$ (4.71)

On integration we have
$\displaystyle \phi = U_\infty x + C$ (4.72)

where C is the constant of integration, which can be chosen arbitrarily.

The stream function for uniform flow can be easily calculated and is given by,

$\displaystyle U_\infty = {{\partial \psi} \over {\partial
 y}},   v = {{\partial \psi} \over {\partial
 x}} = 0$ (4.73)

As an exercise the student is asked to write the velocity potential and stream function for flows for which (a) u = 0, v = V and (b) $ u = U_\infty cos \alpha$.

Source and Sink

Consider a radial flow going away from the origin at a velocity $ v_r$ as shown in Fig.4.19. This constitutes a Source Flow. This is a purely radial flow with no component of velocity in the tangential direction, i.e., $ v_\theta = 0$. If m is the volumetric flow rate we have

Figure 4.19 : Source Flow and Sink Flow.

$\displaystyle (2 \pi r) v_r = m$    
i.e., (4.74)

We can now write down velocity potential and stream function for this flow :

$\displaystyle \phi $ $\displaystyle = {m \over {2\pi}} \ln r$    
$\displaystyle \psi $ $\displaystyle = {m \over {2\pi}} \theta$ (4.75)

It is easily verified that $ v_\theta = 0$ for this flow. Further, the equation we started out with , namely, Eqn.4.74 is the continuity equation for the source flow. It states that the Volumetric flow rate (mass flow rate when multiplied by density) is constant in a radial direction and is equal to m, which is called the Strength of the source.

Another point to make is that the radial velocity $ v_r$ becomes infinite at r = 0. So the origin is a singularity of the flow.

If m is negative we have a flow which flows inwards and is called a Sink flow, which again has a singularity at the origin.

Vortex

We now consider flows which go in a circumferential direction with no radial flow. These are Vortex flows as shown in Fig. 4.20.

Figure 4.20: A Vortex Flow

The velocity potential and stream function are given by,

$\displaystyle \phi $ $\displaystyle = K \theta$    
$\displaystyle \psi $ $\displaystyle = -K \ln r$ (4.76)

The velocity components are given by

$\displaystyle v_\theta $ $\displaystyle = {1 \over r}{\partial \phi \over {\partial \theta}} = {K
 \over r}$    
$\displaystyle v_r $ $\displaystyle = 0$ (4.77)

It is seen that $ v_\theta$ is infinite at the origin and decreases as r increases and becomes zero as r approaches infinity.

A question arises now as to whether we are contradicting ourselves? How is it that a vortex flow is irrotational? We should note that the term "Irrotational" refers to the behaviour of a fluid element and not to the path taken by it. At an elemental level the flow is still irrotational. Such a vortex is called a Free Vortex. A good and familiar example is that of a bath tub vortex. Contrary to this we have a Forced Vortex which behaves like a solid body. These have their velocity given by $ v_\theta = K r$, with a zero velocity at the origin. The velocity increases as one moves away from the origin. A water filled tank is a good example.

Circulation around a Vortex

Let us now calculate the circulation (see Circulation) around a free vortex. We have

$\displaystyle \Gamma = \oint_C d \phi = \int_0 ^{2 \pi} {K \over r}
 ds = \int_0 ^ {2 \pi} {K \over r} (r d\theta) = 2 \pi K$ (4.78)

which is non-zero. Where is the flaw in our integration then? It is a simple matter to find out. We have a singularity in our region, namely, r = 0! If we exclude the singularity by making a small cut around the origin, we will in fact get the result that circulation around the vortex is zero.

It is usual to write the equation for velocity potential and stream function in terms of circulation $ \Gamma$, thus

$\displaystyle \phi = {\Gamma \over {2 \pi}} \theta,   \psi = - {\Gamma \over {2
 \pi}} \ln r$ (4.79)

A Source-Sink Pair

Figure 4.21 : Source-Sink Pair

Consider a source and a sink placed at (-a,0) and (a,0) respectively as shown in Fig.4.21. By combining their stream functions we have a stream function for the combination given by,

$\displaystyle \psi = -{m \over {2\pi}}(\theta_1 - \theta_2)$ (4.80)

By taking tangents of the two sides (after manipulation) we have

$\displaystyle \tan \left( -{2 \pi \psi} \over m \right) = {\tan
 (\theta_1 - \t...
...} = {{{\tan \theta_1 - \tan\theta_2}} \over
 {1 + \tan \theta_1 \tan \theta_2}}$ (4.81)

From geometry it follows that
$\displaystyle \tan \theta_1 = {{r \sin \theta} \over {r \cos \theta - a }},   
 \tan \theta_2 = {{r \sin \theta} \over {r \cos \theta + a }}$ (4.82)

Upon substituting these into Eqn. 4.81, one gets,

$\displaystyle \tan \left( -{2 \pi \psi} \over m \right) = {{2ar \sin
 \theta}\over {r^2 - a^2}}$ (4.83)

so that
$\displaystyle \psi = -{m \over{2\pi} } \tan^{-1} \left( {2ar \sin \theta} \over
 {r^2 - a^2}\right)$ (4.84)

Figure 4.22 : Flow about a Source-Sink Pair

When the distance between the source and the sink becomes smaller, i.e., a is small we have,

$\displaystyle \psi = -{{m a r \sin \theta } \over {\pi (r^2 - a^2)}}$ (4.85)

The streamline pattern for the source-sink flow is sketched in the Fig.4.22.

Doublet

A Doublet is formed when the source and sink approach each other,i.e., $ a\rightarrow
0$ and at the same time $ m \rightarrow \infty$ such that $ ma
/\pi$ is constant, we see that
$\displaystyle {r \over {r^2 - a^2}}\rightarrow {1 \over r}$ (4.86)

As a consequence the stream function becomes
$\displaystyle \psi = -{{K \sin \theta} \over r}$ (4.87)

The velocity potential is

$\displaystyle \phi = {{K \cos \theta} \over r}$ (4.88)

Figure 4.23: Doublet Flow

The streamlines and the equipotential lines for a doublet are sketched in Fig.4.23. It is seen that the streamlines are circles which are tangential to the x-axis while the equipotential lines are also circles but tangential to y-axis.

Superposition of Elementary Flows

One question that naturally arises is "Why are we discussing these flows such as uniform flow, source flow vortex and doublet flows when they do not actually exist. The answer is "Yes, they do not exist. But conceptually they are useful.". In fact, they serve as alphabets of potential flow. By combining these flows we can build up more complicated flows, which are meaningful.

We note that the flows we have discussed are linear flows. By linearity it is meant that if A and B are two solutions, even $ mA\pm nB$ (where m and n are numbers) is also a solution. Therefore we are allowed to superpose the elementary flows one above the other to obtain a different flow. If the constituent flows are irrotational, the combined flow too is irrotational.

The student is advised to run the following programs which plot streamlines and velocity potential lines for various cases of superposition.

streamf.exe

velpot.exe

Uniform Flow and a Source

Let us now place a source in the path of a uniform flow. The stream function and the velocity potential for the resulting flow are given by adding the two stream functions and velocity potentials as follows,

$\displaystyle \psi $ $\displaystyle = \psi_{UF} + \psi_{S}$    
@ $\displaystyle =U_\infty r\sin \theta + {m \over {2\pi}} \theta$ (4.89)
$\displaystyle \phi $ $\displaystyle = \phi_{UF} + \phi_{S}$    
@ $\displaystyle =U_\infty r\cos \theta + {m \over {2\pi}} \ln r$ (4.90)

One of the interesting features to determine for the resulting force is the stagnation point of the flow, i.e., where the velocity goes to zero.

One could calculate this from the equations. It is clear that for this flow the stagnation point will occur on the x-axis. The location can be arrived at purely intuitionally. The source produces a radial flow of magnitude

$\displaystyle v_r = {m \over {2r}}$    

while the uniform flow produces a velocity of U in the positive x-direction. When these two cancel out at a point we have the stagnation point. A negative radial flow that can cancel the uniform flow is possible only to the left of the x-axis, say at x = -b. Hence,

$\displaystyle U_\infty = {m \over {2 \pi b}}$    
leading to
   
$\displaystyle b = {m \over {2 \pi U_\infty}}$ (4.91)

At x = -b, we have $ \theta=\pi$ and r = b. Substituting these values in the expression for $ \psi$, i.e., Eqn. 4.89 we get the value of $ \psi$ at the stagnation point to be
$\displaystyle \psi_{stagn} = {m \over 2}$ (4.92)

An equation to the streamline passing through the stagnation point, i.e., stagnation streamline is obtained as follows,

$\displaystyle {m \over 2} = U_\infty r \sin \theta + {b U_\infty \theta}$    
But hence    
$\displaystyle U_\infty \pi b = U_\infty r \sin \theta + {b U_\infty \theta}$    
leading to
   
$\displaystyle r = {{b(\pi - \theta)} \over {\sin
 \theta}}  or  y = {{b(\pi - \theta)}}$ (4.93)

Figure 4.24 : Flow about Rankine Half Body

The streamlines for this flow are sketched in Fig.4.24. It is clear that we can make the stagnation streamline the solid body. In fact any streamline of a flow can be treated as a solid body since there is no flow across it. In the present example if we ignore the streamlines inside the "body" we have described the flow about a solid body given by Eqn. 4.25. This body is referred to as a Rankine Half Body as it is "open" at the right hand end.

Limits of $ \theta$ for this body are 0 and $ 2 \pi$. At these values we have y approaching $ \pm \pi b$, which is called the Half Width of the body.

Figure 4.25 : Sketch of Flow about Rankine Half Body

The velocity components for this flow are given by

$\displaystyle v_r = U_\infty \cos \theta + {m \over {2 \pi r}}$ (4.94)
$\displaystyle v_\theta = -U_\infty \sin \theta $ (4.95)
The square of velocity reduces to
   
$\displaystyle V^2 = U_\infty ^2 \left( 1 + 2 {b \over r} \cos \theta + {b^2
 \over r^2} \right)$ (4.96)

If the pressure in the free stream is $ p_\infty$ it follows from Bernoulli Equation that,

which enables us to calculate the pressure. Usually in aerodynamic applications involving significant velocities and pressures any contribution due to elevation changes is negligible. The equation for pressure assumes a simple form,

It is left as an exercise for the student to show that the maximum velocity over the surface of the body occurs at the location $ \theta = 63^0$ and is approximately equal to $ 1.26 U_\infty$.

Rankine Oval

We saw that the previous example defined a half body open at one end. Can we come up with a closed body by a suitable combination? An inspection of the streamlines suggests that by placing a sink in addition to the source one should be able to define a closed body. In other words a uniform flow past a source-sink combination is what we are after. The stream function for this is given by

$\displaystyle \psi $ $\displaystyle = U_\infty r \sin \theta -{m \over{2\pi} } \tan^{-1} \left(
 {2ar \sin \theta} \over {r^2 - a^2}\right)$ (4.99)
or in Cartesian coordinates
   
$\displaystyle \psi $ $\displaystyle = U_\infty y -{m \over{2\pi} } \tan^{-1} \left( {2ay} \over
 {x^2 + y^2 - a^2}\right)$ (4.100)

Figure 4.26 : Flow past a Rankine Oval

When the streamlines for this flow are plotted (Fig.4.26) one discovers that the one given by $ \psi = 0$ (shown in red) forms a closed curve. This obviously forms the "body", i.e., the stream function we have written describes the flow about this body. Shapes such as this are called Rankine Ovals. The distance to the stagnation points from the origin or the Half Body Length is given by

$\displaystyle {l \over a} = \sqrt{{m \over {\pi U_\infty a}}+1}$ (4.101)

Figure 4.27 : Rankine Oval

The other feature of interest, Half Width is found by determining the point of intersection of y-axis with the body, i.e., $ \psi = 0$ line. An expression for h is,

$\displaystyle {h \over a} = {1 \over 2} \left\{{h \over a}^2 - 1 \right\}\tan
 \left\{ 2 \left({ {\pi U_\infty a} \over m}\right) {h \over a}
 \right\}$ (4.102)

the solution for which is to be obtained by iteration. Rankine ovals include a wide range of bodies which can be obtained by varying the value of the parameter $ \pi U_\infty a /m$. These could be bodies stretched in any of the two directions. When stretched in x-direction one obtains elliptic bodies with a small half width compared to the span. The solution obtained could be a good approximation to the flow especially if viscous effects are small. On the other hand a considerable half width would indicate a bluff body prone to effects like separation. The solution obtained can hardly be accepted in this case.

Flow Around a Circular Cylinder

Flow around a circular cylinder can be approached from the previous example by bringing the source and the sink closer. Then we are considering a uniform flow in combination with a doublet. The stream function and the velocity potential for this flow are given by,

Figure 4.28 : Schematic for Flow past a Circular Cylinder

$\displaystyle \psi $ $\displaystyle = U_\infty r \sin \theta - {{K \sin \theta} \over r}$ (4.103)
$\displaystyle \phi $ $\displaystyle = U_\infty r \cos \theta + {{K \cos \theta} \over r}$ (4.104)

Streamlines for this flow are plotted in Fig. 4.29.


Figure 4.29
: Flow past a Circular Cylinder


Figure 4.30 : Stagnation Points for Flow about a Circular Cylinder

The velocity components are given by,

$\displaystyle v_r $ $\displaystyle = {1 \over r} {{\partial \psi} \over \partial \theta} = \cos \theta \left(U_\infty - {K \over r^2}\right)$ (4.105)
$\displaystyle v_\theta $ $\displaystyle = -{{\partial \psi} \over { \partial r}} = -\sin \theta
 \left( U_\infty + {K \over r^2} \right)$ (4.106)

It is seen that the radial velocity is zero when

If we recognise this particular streamline as the surface of the circular cylinder then the radius of the cylinder a is given by,

The equations for the streamline, velocity potential and the velocity components are replaced by,

$\displaystyle \psi $ $\displaystyle = U_\infty r \left( 1 -  {a^2 \over r^2} \right)\sin \theta$ (4.109)
$\displaystyle \phi $ $\displaystyle = U_\infty r \left( 1 +  {a^2 \over r^2} \right)\cos \theta$ (4.110)
$\displaystyle v_r $ $\displaystyle = U_\infty \left( 1 -  {a^2 \over r^2} \right)\cos \theta$ (4.111)
$\displaystyle v_\theta $ $\displaystyle = -U_\infty \left( 1 +  {a^2 \over r^2} \right)\sin
 \theta$ (4.112)

The velocity components on the surface of the cylinder are obtained by putting r = a in the above expressions. Accordingly,

$ \sin \theta$ has a zero at 0 and 1800 and a maximum of 1 at = 900 and 2700. The former set denotes the stagnation points of the flow and the later one denotes the points of maximum surface velocity (of magnitude $ 2U_\infty$). Thus the velocity decreases from a value of $ 2U_\infty$ at $ \theta$ equals 900 to $ U_\infty$ as one moves away in a normal direction s shown in Fig 4.30.

The surface pressure distribution is calculated from Bernoulli equation. If we denote the free stream speed and pressure as $ U_\infty$ and $ p_\infty$ we have

Substituting for $ v_{\theta s} = -2 U_\infty \sin \theta$, we have

We can also express pressure in terms of pressure coefficient, Cp,

$\displaystyle C_p =1 - {\left( {v_s \over U_\infty}\right)^2}$ (4.116)
leading to
   
$\displaystyle C_p = 1 - 4 \sin^2 \theta$ (4.117)

Fig. 4.31 shows Cp plotted as a function of $ \theta$. A symmetry about y -axis is apparent. When compared to the experimentally observed Cp distribution we see that there is some agreement in the region between$ \theta$= 00 and $ \theta$= 900 . But any agreement is lost in the other regions. The reasons for this are obvious. Viscous forces dominate the flow in the region to the right of the centreline giving rise to separation. The pressure tends to plateau out in a separated region, the level depending on whether it is a laminar separation or a turbulent one.


Figure 4.31 (a) : Cp distribution for flow past a circular cylinder.


Figure 4.31 (b) : Cp distribution for flow past a circular cylinder plotted around the cylinder.

Symmetry in the theoretical Cp distribution about both y-axis and x-axis shows that drag and lift forces about the cylinder are each zero. This may also be proved by integrating pressure around the cylinder, thus,

Drag, $\displaystyle Drag, D = -\int_0 ^{2 \pi} p_s  \cos \theta  a  d \theta $ (4.118)
Lift, $\displaystyle Lift, L = -\int_0 ^{2 \pi} p_s  \sin \theta a d \theta $ (4.119)

By substituting for the surface pressure, ps from Eqn.4.115 we find,

$\displaystyle D$ $\displaystyle = - \int_0 ^ {2 \pi} p_\infty a \cos \theta d \theta -
 \frac{1}{...
...0 ^ {2 \pi} \left( \cos \theta
 - 4 \sin^2 \theta \cos \theta \right ) d \theta$ (4.120)
@ $\displaystyle =  -p_\infty a \left [\sin \theta \right ]_0 ^{2
 \pi}-\frac{1}{...
...1}{2} \rho U_\infty
 ^2 a \left [ \frac{4}{3} \sin^3 \theta \right ]_0 ^{2 \pi}$ (4.121)
@ = -0 -0 + 0 (4.122)
$\displaystyle L$ $\displaystyle = - \int_0 ^ {2 \pi} p_\infty a \sin \theta d \theta -
 \frac{1}{...
...y ^2 a \int_0 ^ {2 \pi} \left( \sin \theta
 - 4 \sin^3 \theta \right ) d \theta$ (4.123)
@ $\displaystyle =  -p_\infty a \left [ \cos \theta \right ]_0 ^{2
 \pi}-\frac{1}...
...fty ^2 a \left [ \frac{4}{3} \cos^3
 \theta - 4 \cos \theta \right ]_0 ^{2 \pi}$ (4.124)
@ = -0 -0 + 0 (4.125)

What we have just calculated is in contrast to the experimental results which do predict a significant drag for the flow about a circular cylinder. This seems to have caused in what is called D'Alembert's Paradox in honour of Jean le Rond D'Alembert (1717-1783). Now it is no more a paradox. As we discussed above we calculate a zero drag because we have not taken viscosity into account.

Flow about a Lifting Cylinder

A lifting flow can be generated by adding a free vortex to the flow about a circular cylinder just described. The stream function and the velocity potential now become,

$\displaystyle \psi $ $\displaystyle = U_\infty r \left( 1 -  {a^2 \over r^2} \right)\sin \theta - {\Gamma \over {2 \pi}} \ln r$ (4.126)
$\displaystyle \phi $ $\displaystyle = U_\infty r \left( 1 +  {a^2 \over r^2} \right)\cos \theta + {\Gamma \over {2 \pi}} \theta$ (4.127)

Fig. 4.32 shows the streamlines for this flow.

Figure 4.32 : Flow past a Lifting Cylinder

Consequently the velocity components will be,

$\displaystyle v_r $ $\displaystyle = U_\infty \left( 1 -  {a^2 \over r^2} \right)\cos \theta$ (4.128)
$\displaystyle v_\theta $ $\displaystyle = -U_\infty \left( 1 +  {a^2 \over r^2} \right)\sin
 \theta + {\Gamma \over {2 \pi r}}$ (4.129)

At r = a, the radial velocity is still zero allowing us to consider the same circular cylinder as the "body". The tangential velocity on the surface of the cylinder is given by,

The streamline pattern for this flow depend upon the location of the stagnation points given by,

The surface velocity can now be written as

Figure 4.33 : Stagnation Points for a Lifting Cylinder

The location of the stagnation points and the resulting streamline pattern are shown in Fig 4.33 and Fig. 4.32 respectively.

Stagnation Points for a lifting circular cylinder

The stagnation points we saw in Fig. 4.33 are for the case when the circulation imposed on the cylinder was such that $ \Gamma < 4 \pi U_\infty a$. But from Eqn. 4.131 it is evident that angle $ \beta$, hence the position of the stagnation points is a strong function of circulation, $ \Gamma$. This is illustrated in Fig.4.34. With zero circulation the stagnation points lie at $ \theta$ = 0, $ 2 \pi$. As circulation $ \Gamma$ increases the stagnation points move (upwards or downwards depending upon the direction of rotation). When $ \Gamma
= 4 \pi U_\infty a$ they coincide at $ \theta$ = $ \frac{\pi}{2}$ or $ \theta$ = - $ \frac{\pi}{2}$. If circulation is further increased the stagnation point will no longer be found on the cylinder surface, but will appear in the flow as shown in (d) in Fig. 4.34.

Figure 4.34 : Effect of circulation on flow about a cylinder

Surface Pressure Distribution and Lift

The surface pressure is calculated from the Bernoulli equation as

$\displaystyle p_s = p_\infty + {1 \over 2} \rho U_\infty ^2 - 2 U_\infty ^2
 \left( \sin \theta
 -  \sin \beta \right )^2$ (4.133)
or
   
$\displaystyle C_p = 1 - {\left( {v_\theta \over U_\infty}
 \right)^2} = 1 - 4 \left(\sin \theta - \sin \beta \right)^2$ (4.134)


Figure 4.35 (a) : Cp distribution for a lifting cylinder,
$ \beta$=-150.


Figure 4.35 (b) : Cp distribution for a lifting cylinder plottd around the cylinder.$ \beta$=-150

The Cp distribution is plotted in Fig.4.35 (a) and is also shown plotted along the cylinder surface in Fig.4.35 (b). Asymmetry about x-axis is evident indicating the generation of lift. Drag however is zero. Magnitude of the lift force is calculated by integration as in Eqn. 4.118 and 4.119 .

We have from 4.133,

$\displaystyle p_s = p_\infty + {1 \over 2} \rho U_\infty ^2 \left ( 1 - 4 \left(
 \sin \theta
 -  \sin \beta \right )^2 \right)$ (4.135)
$\displaystyle \noalign{i.e.,}$    
$\displaystyle p_s = p_\infty + {1 \over 2} \rho U_\infty ^2 \left (1 - 4 \sin^2 \theta
 -4 \sin^2 \beta + 8 \sin \theta \sin \beta \right)$ (4.136)

Lift is now given by

$\displaystyle L = -\int_0 ^ {2 \pi} a \sin \theta \left[ p_\infty + {1 \over 2}...
...1 - 4 \sin^2 \theta
 -4 \sin^2 \beta + 8 \sin \theta \sin \beta \right) \right]$ (4.137)

giving,

$\displaystyle L = -\int_0 ^ {2 \pi} a \left[ p_\infty + {1 \over 2} \rho
 U_\in...
...theta
 -4 \sin^2 \beta \sin \theta + 8 \sin^2 \theta \sin \beta \right) \right]$ (4.138)

This reduces to

$\displaystyle L $ $\displaystyle = -\int_0 ^ {2 \pi} a p_\infty \sin \theta - {1 \over 2} \rho
 U_\infty ^2 a \int_0 ^ {2 \pi} 4 \sin^3 \theta$ (4.139)
@ $\displaystyle + {1 \over 2} \rho U_\infty ^2 a \int_0 ^ {2 \pi}\sin^2 \beta
 \s...
...eta +{1 \over 2} \rho U_\infty ^2 a \int_0 ^{2 \pi}
 8 \sin^2 \theta \sin \beta$ (4.140)

As we had in Eqn. 4.121 and 4.124 , the first three integrals in 4.139 and 4.140 are each zero. It follows that Lift is now given by,

$\displaystyle L $ $\displaystyle ={1 \over 2} \rho U_\infty ^2 a \int_0 ^{2 \pi}
 8 \sin^2 \theta \sin \beta$ (4.141)
@ $\displaystyle =4 \rho U_\infty ^2 \sin \beta \left[ \frac{\theta}{2} -
 \frac{\sin^2 \theta}{4} \right ]_0 ^{2 \pi}$ (4.142)
@ $\displaystyle = 4 \pi \rho U_\infty ^2 a \sin \beta $ (4.143)

Upon substituting for $ \sin \beta $ we have

Thus lift developed by a rotating circular cylinder is equal to the product of density, freestream speed and circulation.

Kutta-Joukowsky Theorem

The result derived above, namely, $ L=-\rho U_\infty \Gamma$ is a very general one and is valid for any closed body placed in a uniform stream. It is named the Kutta-Joukowsky theorem in honour of Kutta and Joukowsky who proved it independently in 1902 and 1906 respectively. The theorem finds considerable application in calculating lift around aerofoils. See Fig.4.37.

@

Figure 4.37 : Kutta-Joukowski Theorem

Magnus Effect

We have shown in Eqn.4.144 that a force is produced when circulation is imposed upon a cylinder placed in uniform flow (see Fig. 4.38) . This force is nothing but the lift. This effect is called Magnus Effect in honour of the scholar Heinrich Magnus (1802 - 1870). Sports involving balls, such as golf, baseball, tennis see this effect in action. A spinning ball when hit in a horizontal direction follows a curved trajectory because of this effect.

Figure 4.38 : Magnus Effect