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Ann [662]
3 years ago
7

A flow field is characterized by the stream function ψ= 3x2y−y3. Demonstrate thatthe flow field represents a two-dimensional inc

ompressible flow. Show that the magnitude ofthe velocity at a point depends only on the distance from the origin of the coordinates.
Plot the strim ψ = 2.

Engineering
1 answer:
boyakko [2]3 years ago
8 0

Answer:

δu/δx+δu/δy = 6x-6x =0

9r^2

Explanation:

The flow is obviously two-dimensional, since the stream function depends only on the x and y coordinate. We can find the x and y velocity components by using the following relations:  

u =δψ/δy = 3x^2-3y^2

v =-δψ/δx = -6xy

Now, since:  

δu/δx+δu/δy = 6x-6x =0

we conclude that this flow satisfies the continuity equation for a 2D incompressible flow. Therefore, the flow is indeed a two-dimensional incompressible one.  

The magnitude of velocity is given by:  

|V| = u^2+v^2

    =(3x^2-3y^2)^2+(-6xy)^2

    =9x^4+18x^2y^2+9y^2

    =(3x^2+3y^2)^2

    =9r^2

where r is the distance from the origin of the coordinates, and we have used that r^2 = x^2 + y^2.  

The streamline ψ = 2 is given by the following equation:  

3x^2y — y^3 = 2,

which is most easily plotted by solving it for x:  

x =±√2-y^3/y

Plot of the streamline is given in the graph below.  

Explanation for the plot: the two x(y) functions (with minus and plus signs) given in the equation above were plotted as functions of y, after which the graph was rotated to obtain a standard coordinate diagram. The "+" and "-" parts are given in different colors, but keep in mind that these are actually "parts" of the same streamline.  

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See the attached file for the calculation.

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How to find the voltage(B Aab) in series parallel circuit? ​
Sindrei [870]

Answer:

  Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

  R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

  Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

  Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

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That forms a divider with R1 to give V1:

  V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

  Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

  (Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

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The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

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