Fluid Mechanics Test 2

Concept:

Any velocity filed which represent the fluid flow; it must satisfy the equation of continuity. 

The equation of continuity for a 2 D, steady and incompressible flow is given as:

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

Where, u and v are the x and y components of the velocity.

Calculation:

V = (ax + by) i + (cx + dy) j

 The x component of velocity, u = ax + by ⇒ \(\frac{{\partial u}}{{\partial x}} = a\)

The x component of velocity, v = cx + dy ⇒ \(\frac{{\partial v}}{{\partial y}} = d\)

If it represents the fluid flow then it must satisfy the continuity equation i.e.

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

⇒ a + d = 0

Concept:

Stream function (ψ) is the function of space and time in 2 D defined in such a way that the continuity equation is satisfied and flow is possible. It is given as:

ψ = f^{^} (x, y)

\(u = - \frac{{\partial \psi }}{{\partial y}}\;and\;v = \;\frac{{\partial \psi }}{{\partial x}}\)

Where u and v are the x and y component of velocity field respectively.

Discharge passing through between stream lines is given as:

Q = ψ₁ – ψ₂

Where ψ₁ is the stream line at point 1 and ψ₂ is the stream line at point 2.

Calculation:

Given:

u = 3y² and v = -3x²

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

Continuity equation : \(\frac{{du}}{{dx}} + \frac{{dv}}{{dy}} = 0\)

This implies stream function exists.

Now:

\(u = - \frac{{\partial \psi }}{{\partial y}} = 3{y^2}\)

Integrating above w.r.t. ‘y’

ψ = -y³ + δ(x)

\(\frac{{\partial \psi }}{{\partial x}} = \delta '\left( x \right) = v\)

⇒ \(\delta '\left( x \right) = - 3{x^2}\) ⇒ δ(x) = -x³ + C

∴ ψ = -y³ – x³ + C

There is no need to evaluate the constant as it will cancelled out later.   

ψ₁ = ψ (1, 2) = - (1)³ – (2)³ + C = -9 + C

ψ₂ = ψ (3, 4) = -(3)³ – (4)³ + C = -91 + C

Concept:

Both velocity potential function (ϕ) and stream function (ψ) must satisfy Cauchy-Reimann equations:

\(\frac{{\partial \phi }}{{\partial x}} = \frac{{\partial \psi }}{{\partial y}}\)

\(\frac{{\partial \phi }}{{\partial y}} = - \frac{{\partial \psi }}{{\partial x}}\)

Calculation:

ϕ = 4x (3y - 4)

ϕ = 12xy – 16x

\(\Rightarrow \frac{{\partial \phi }}{{\partial x}} = 12y - 16\)

\(\frac{{\partial \psi }}{{\partial y}} =\frac{{\partial \phi }}{{\partial x}} = 12y - 16\)

Integrating w.r.t y keeping x constant,

ψ = 6y² – 16y + c(x)

ϕ = 12xy – 16x

\(\frac{{\partial \phi }}{{\partial y}} = 12x = - \left[ {c'\left( x \right)} \right]\)

\( \frac{{\partial \psi }}{{\partial x}}=-\frac{{\partial \phi }}{{\partial y}} \Rightarrow c'(x)=-12x\)

c(x) = -6x²

ψ = 6y² – 16y – 6x² = 6(y² – x²) – 16y

ψ = 6(y² – x²) – 16y

At x = 2, y = 3

ψ = 6{9 - 4} – (16)(3)

ψ = 30 – 48

ψ = -18

Concept:

It is given that mass of the flow is conserved, this implies that equation of continuity is valid.

The equation of continuity in vector form can be written as:

\(\frac{{\partial \rho }}{{\partial t}} + \;\rho \left( {\;\overrightarrow {{\rm{\Delta }}.} {\rm{\vec V}}} \right) = {\rm{\;}}0{\rm{\;}}\)

Where, \(\overrightarrow {{\rm{\Delta }}.} {\rm{\vec V}}\) is the divergence of velocity filed of flow in sec^-1.

Calculation:

Given : ρ = 1 + 0.5t

\(\Rightarrow \frac{{\partial \rho }}{{\partial t}} = 0.5\;kg\;/{m^3}/sec\)

\(\frac{{\partial \rho }}{{\partial t}} + \rho \left( {\vec \nabla \cdot \vec U} \right) = 0\)

⇒ \(0.5 + 2 \times \left( {\vec \nabla \cdot \vec U} \right) = 0\)

\( \Rightarrow \vec \nabla \cdot \vec U = \frac{{ - 0.5}}{2} = - 0.25\;se{c^{ - 1}}\)