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Ann [662]
4 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]4 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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LuckyWell [14K]

Answer:

d. 1.0

Explanation:

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3 years ago
Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 1
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Answer:

- the mass flow rate of air is 7.53 kg/s

- the exit area is 0.108 m²

Explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W"_{cv / [ (h₁ - h₂ ) -\frac{V_2^2}{2} ]

Now at;

T₁ = 900K, h₁ = 932.93 k³/kg

T₂ = 500 K, h₂ = 503.02 k³/kg

so we substitute, in our given values

m" = [ 3200 kW × \frac{1\frac{k^3}{s} }{1kW} ] / [ (932.93 - 503.02  )k³/kg  -\frac{100^2\frac{m^2}{s^2} }{2}|\frac{ln}{kg\frac{m}{s^2} }||\frac{1kJ}{10^3N-m}| ]

m" = 7.53 kg/s

Therefore, the mass flow rate of air is 7.53 kg/s

now, Exit area A₂ = v₂m" / V₂

we know that; pv = RT

so

A₂ = RT₂m" / P₂V₂

so we substitute

A₂ = {[ (\frac{8.314}{28.97}\frac{k^3}{kg.K})×500 K×(7.54 kg/s) ] / [(1 bar)(100 m/s )]} |\frac{1 bar}{10N/m^2}||10^3N.m/1k^3

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Explanation:

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sveticcg [70]

Answer:

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Cheers.

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