Fluid Mechanics Test 3

Concept:

Streamline: It is the line along which stream function ψ remains constant.

If ψ = f (x,y)

dψ = 0

\(d\psi = \frac{{\partial \psi }}{{\partial x}}dx + \frac{{\partial \psi }}{{\partial y}}dy\)

For streamline dψ = 0

\(\begin{array}{l} vdx - udy = 0\\ \therefore \frac{{dx}}{u} = \frac{{dy}}{v} \end{array}\)

Calculation:

Given, \(\vec V = \vec ix - \vec jy\)

So, u = x and v = -y

Streamline equation: \(\frac{{dx}}{u} = \frac{{dy}}{v}\)

\(\frac{{dx}}{x} = \frac{{dy}}{{ - y}}\)

Integrating both side

ln x = -ln y + ln c

xy = c

Given x =1 , y =2

so, c =2

so equation is xy – 2 = 0

Concept:

Continuity equation should be satisfied for incompressible, steady flow,

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0\)

Calculation:

\(\frac{\partial }{{\partial x}}\left( {ax{y^2} + 2y} \right) + \frac{\partial }{{\partial y}}\left( {3xy + {x^2}y} \right) = 0\)

(ay²) + (3x + x²) = 0

At (1, 1)

a + (3 + 1) = 0

Concept:

Flow rate between two points P & Q = Ψ_P – Ψ_Q

Calculation:

Ψ_P = 3(4)(4 + 1) + 2(4)³ = 188

Ψ_Q = 3(1 + 1) + 2(1) = 8

Ψ_P – Ψ_Q = 188 – 8

Concept:

Circulation = Vorticity × Area

where, 

Vorticity is defined as the value twice of the rotation.

ζ = 2ω

Rotation (ω) =  \({\omega _z} = \frac{1}{2}\left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right)\)

Now,

∴ Vorticity = \(\zeta = \frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} \)

Calculation:

Circulation = Vorticity × Area

\(Circulation= \left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right) \times Area\)

Circulation= (0 – 3) × π × (2)²

∴ Circulation= -12π units

Concept:

For incompressible steady flow, Laplace equation must be satisfied

\(\begin{array}{l} \frac{{{d^2}\phi }}{{d{x^2}}} + \frac{{{d^2}\phi }}{{d{y^2}}} = 0\\ \frac{{d\phi }}{{dx}} = - u = - A{e^x} \Rightarrow \frac{{{d^2}\phi }}{{d{x^2}}} = - A{e^x}\\ - A{e^x} + \frac{{{d^2}\phi }}{{d{y^2}}} = 0\\ \therefore \frac{{{d^2}\phi }}{{d{y^2}}} = A{e^x}\\ V = - \frac{{d\phi }}{{dy}} = A{e^x}y + f\left( x \right) \end{array}\)

-V = Ae^x.y + f(x)

∴ V = -Ae^x.y + f(x)

Explanation:

Given: 

The leaf is at a distance of 120 m from the center of the whirlpool.

To calculate = distance of the leaf from the center when it has moved through half a revolution

Now,

\(\begin{array}{l} \frac{{{V_\theta }}}{{{V_r}}} = - \left( {\frac{{300 \times {{10}^3}}}{{2\pi r}}} \right) \times \frac{{2\pi r}}{{\left( {60 \times {{10}^3}} \right)}} = - 5\\ \Rightarrow \frac{{{V_r}}}{{{V_\theta }}} = - \frac{1}{5} \Rightarrow {V_r} = - \frac{{{V_\theta }}}{5}\\ \Rightarrow {V_r} = \frac{{dr}}{{dt}}\;and\;{V_\theta } = r\frac{{d\theta }}{{dt}}\;\\ \Rightarrow \frac{{dr}}{{dt}} = - \frac{r}{5}\frac{{d\theta }}{{dt}} \Rightarrow \frac{{dr}}{r} = - \frac{{d\theta }}{5}\\ \mathop \smallint \limits_{120}^r \frac{{dr}}{r} = - \frac{1}{5}\mathop \smallint \limits_0^\pi d\theta \\ \Rightarrow \left| {\ln r} \right|_{120}^r = - \frac{\pi }{5}\\ \Rightarrow \ln \frac{r}{{120}} = - \frac{\pi }{5}\\ \Rightarrow \frac{r}{{120}} = {e^{ - \pi /5}} \end{array}\)

∴ r = 64 m

Concept:

Equation used: V² = u² + 2gh

Continuity equation : A₁V₁ = A₂V₂

Calculation:

Given: D = 20 mm = 0.02 m, u = 10 m/s, h = 3 m

Now,

V² = u² + 2gh

V² = 10² + 2 (-9.81) (3) [∵ Velocity of jet at height (h) = 3 m]

∴ V² = 41.14

∴ V = 6.414 m/s

Now, applying continuity equation

A₁V₁ = A₂V₂

\(\frac{\pi }{4}\left( {{D^2}} \right)\left( {10} \right) = \frac{\pi }{4}{d^2}\left( {6.414} \right)\)

d = 0.02497 m

Concept:

\({\rm{Total\;acceleration\;}}\left( a \right) = \sqrt {a_x^2 + a_y^2} \)

\({a_x} = u \cdot \frac{{\partial u}}{{\partial x}} + v \cdot \frac{{\partial u}}{{\partial y}}\)

\({a_y} = u \cdot \frac{{\partial v}}{{\partial x}} + v \cdot \frac{{\partial v}}{{\partial y}}\)

\({\rm{Also}},{\rm{\;}}u = \frac{{ - \partial \psi }}{{dy}}\;\;;\;\;v = \frac{{\partial \psi }}{{\partial x}}\)

Calculation:

Ψ = 2x²y         (Given)

\(u = \frac{{ - \partial }}{{dy}}\left( {2{x^2}y} \right) = - 2{x^2} = - 2\left( {{2^2}} \right) = - 8\)

\(v = \frac{\partial }{{\partial x}}\left( {2{x^2}y} \right) = 4xy = 4\left( 2 \right) = 8\)

\({a_x} = \left( { - 8} \right)\;\frac{\partial }{{\partial x}}\left( { - 2{x^2}} \right) + 8 \cdot \frac{\partial }{{\partial y}}\left( { - 2{x^2}} \right)\;\)

a_x = (-8)(-4x) + 0 = 32x = 64

\({a_y} = \left( { - 8} \right)\frac{\partial }{{\partial x}}\left( {4xy} \right) + \left( 8 \right)\frac{\partial }{{\partial y}}\left( {4xy} \right)\;\)

a_y  = (-8)(4y) + 8(4x)

∴ a_y = -32 + 64 = 32

Now,

\({\rm{Total\;acceleration\;}}\left( a \right) = \sqrt {a_x^2 + a_y^2} \)

\({\rm{Total\;acceleration\;}}\left( {\rm{a}} \right) = \sqrt {{{64}^2} + {{32}^2}} \)