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Examples of the Principle of Superposition

1. Source Plus a Uniform Flow:

Let us superpose a source flow at the origin and a uniform flow along the x-axis to realize how the superposition principle works.

Since we do not know what the resulting flow pattern will be we sum up stream functions of a uniform flow and a source flow at the origin first.

[ ]

To determine how superposition works, we need to visualize the flow first. A convenient way to achieve this is the determination of stagnation points in the flow, and the equation of the stagnation streamline. Remember the stagnation points are the points where fluid velocity components are zero. A streamline that passes through a stagnation point is called a stagnation streamline. The stagnation streamline can represent the shape of a solid surface since, by definition, no fluid can cross a streamline (since flow is always tangent to it). Thus, stagnation streamlines have particular importance in the study of ideal flows. For the present superposition, the stagnation points are determined from the solution of all (rs,s) values that satisfy:

+

UF Source, q

???

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If we satisfy the second equation first, we find, or,

If ,

Since “q” for a source and “U” are positive quantities, the above equation is absurd

(rs cannot be < 0). However, if , which is a possible solution.

Thus, the flow has only one stagnation point located at:

To determine the equation of the stagnation streamline, is set to a constant equal to its value at the stagnation point.

[by substitution]

Thus the plot of the stagnation streamline is given by all values of (r,) satisfying:

or,

For illustration purposes, suppose our free stream . This flow will have the stagnation point and the stagnation streamline as shown.

Note that as , the body approaches y = 3.14 asymptotically. The velocity and at the stagnation point. But as we move away from the stagnation point

U2

qrs s

3.14 m (as 0)

3.14 m (as 0)+

)2

,m57.1r(B

m1)1(2

28.6rs

s x

y1 m/s

q =6.28 m2/s

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the velocity picks up gradually on the stagnation streamline (since ideal flows slip on a solid surface). Thus, this flow is pictured as a flow over a semi-infinite body (called a Rankine-halfbody).

One interesting point to note in this analysis is that the point of singularity (origin) is trapped outside the body. Therefore although a source flow is singular, the fluid flow over the resulting body has no points of discontinuity in the flow field.

Suppose we are interested in determining what is the pressure at the point B. For this, we must first calculate the velocity at the point from the expressions of Vr and V. Then apply the inviscid Bernoulli equation between two points ( and B) on the stagnation streamline. If the pressure at “” is p, Bernoulli equation gives:

where, at the point B.

2. Doublet Plus Uniform Flow:

The ideal flow over a circular cylinder may be shown to be a superposition of a uniform flow and a doublet “” at the origin with its axis along the “-x” axis. To show this, first set up the superposition formula for .

Determination of Stagnation Points: To find (rs,s) set and solve simultaneously. If , or, . Check if both of these are valid

solutions for stagnation points. Since cannot yield because r

becomes imaginary, we solve the values of rs by setting .

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Note that the rs values cannot be negative therefore the negative root discarded.

Thus, there are 2 feasible stagnation points in the flow, and .

These points must also be used to obtain the equation for stagnation streamline.

Therefore the equation of stagnation streamline is derived when (r,) = 0. Thus

, or . Since for any arbitrary

point, or,

The boxed equation shows that r is constant ( and U are constant for this steady flow). Thus it describes the equation of a circle (or circular cylinder in 3-D). If we choose to represent the radius of the cylinder by R, for the flow. This relationship may be used for design. Suppose you want to construct a study flow over a circular cylinder of R = 1 m, in a force stream U = 1 m/s, the strength of the doublet you would use to construct this flow must be .

Evaluation of Pressure Field:

To obtain the pressure field for a steady flow over a circular cylinder, we apply Bernoulli equation between 2 points on a stagnation streamline as shown.

(neglecting z between “” and “B”)

U2r

B(r,)

x

y

“”

Stagnation streamline

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,

where [ ] and,

or,

The above expression, is the definition of pressure coefficient on a body,

which is the ratio of gauge pressure and the flow-stream’s dynamic pressure.

We notice that the plot of Cp is symmetric about the “y”-axis.

x

y

-3

1

090

180 Cp (0) = 1

Cp (90) = -3

Cp (180) = 1

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Cp starts at a value of 1 at the front stagnation point, crosses over the axis at , reaches a peak value of –3 at , crosses over the axis at

and ends at at . Since Cp is a measure of the fluid gauge pressure, the net pressure force in the x-direction (which we define as the drag force on the body)

must be zero [More formally, , which may be shown to be zero

by substituting for the pressure field and integrating]. This is the famous D’Alembert’s Paradox in fluid mechanics.

We know why the drag is zero. The real pressure curve around a circular cylinder is not symmetric about the y-axis. However, we shall discuss these implications again during discussions of flow separation in the chapter on boundary layers. For the present case, this symmetry is justifiable since the flow is inviscid.

To summarize, we have presented two super position examples. In each case, the flow stream functions are formed first. Then velocity components are computed. Next, the stagnation points and equation of stagnation streamline are found. (Note: The stagnation streamline represents the body shape over which the superposed flow results.) Then pressure is found on the body surface by the application of Bernoulli equation. Lastly, the body pressure may be integrated to evaluate forces on the body.

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