JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III COMPUTATIONAL AERODYNAMICS Previous Question Paper (February/March - 2022 - R15)

B.Tech | III | II | R15 | February/March - 2022

Answer PaperPaper Code: 126YWMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

Derive the Momentum equation for a control volume moving with fluid with neat diagram.

15 marks

Answer

The momentum equation for a fluid flow is derived from Newton's Second Law of Motion, which states that the net force acting on a fluid element is equal to the rate of change of momentum of that element. For a control volume moving with the fluid (a Lagrangian approach), the system consists of a fixed mass of fluid particles.

Consider a fluid element of mass m moving with velocity V. The fundamental law is:

ΣF = d(mV)/dt

For a control volume moving with the fluid, the mass m is constant. Thus, the equation simplifies to:

ΣF = m(dV/dt) = ma

In a fluid continuum, we consider an infinitesimal control volume of volume dV and density ρ. The forces acting on this volume are categorized into body forces (such as gravity) and surface forces (pressure and viscous stresses).

1. Body Forces: The force due to gravity is ρg dV.

2. Surface Forces: These arise from the pressure distribution p and the viscous stress tensor τ. The net surface force is the integral of the stress vector over the control surface, which by the divergence theorem is expressed as the divergence of the stress tensor: ∇ · σ, where σ = -pI + τ.

Applying Newton's Second Law to the fluid element:

ρ(DV/Dt) = ρg + ∇ · σ

Substituting the stress tensor components:

ρ(DV/Dt) = ρg - ∇p + ∇ · τ

Here, DV/Dt is the material derivative, representing the acceleration of the fluid particle: ∂V/∂t + (V · ∇)V. This equation is known as the Cauchy momentum equation. For a Newtonian fluid, the viscous stress tensor τ is related to the velocity gradients via the coefficient of viscosity μ, leading to the Navier-Stokes equations:

ρ(∂V/∂t + (V · ∇)V) = -∇p + μ∇²V + (λ + μ)∇(∇ · V) + ρg

This equation accounts for the convective acceleration, pressure gradients, viscous diffusion, and external body forces acting on the moving control volume.

2

Derive the equation for substantial derivative with neat sketch.

15 marks

Answer

The substantial derivative, also known as the material or particle derivative, represents the time rate of change of a fluid property (such as velocity, pressure, or temperature) experienced by a specific fluid element as it moves through a flow field. It accounts for both the local change at a fixed point and the convective change due to the motion of the fluid element.

Consider a fluid property f(x, y, z, t). As a fluid element moves from position r at time t to r + dr at time t + dt, the change in the property df is given by the total differential:

df = (∂f/∂t)dt + (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz

Dividing by the time increment dt, we obtain the rate of change following the fluid element:

Df/Dt = ∂f/∂t + (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt) + (∂f/∂z)(dz/dt)

Since the velocity components of the fluid element are u = dx/dt, v = dy/dt, and w = dz/dt, the equation becomes:

Df/Dt = ∂f/∂t + u(∂f/∂x) + v(∂f/∂y) + w(∂f/∂z)

In vector notation, this is expressed as:

Df/Dt = ∂f/∂t + (V ⋅ ∇)f

Where ∂f/∂t is the local derivative (change at a fixed point) and (V ⋅ ∇)f is the convective derivative (change due to movement through spatial gradients). A sketch would show a fluid particle at position (x, y, z) at time t moving along a streamline to (x+dx, y+dy, z+dz) at time t+dt, illustrating that the total change is the sum of the temporal variation and the spatial displacement effect.

3

Explain the procedure to find the mathematical behaviour of partial differential equation using eigenvalue method with an example of linearized velocity potential equation for subsonic and supersonic case and its effect on CFD.

15 marks

Answer

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4

Give the classification of partial differential equations, explaining the characteristics of each type of PDE.

15 marks

Answer

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5

Derive finite difference schemes for the following partial differential equations and indicate their order of accuracy:

15 marks
a

(∂u/∂t) + a (∂u / ∂x) = 0.

5 marks

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b

(∂u/∂t) = a (∂2u / ∂x2).

5 marks

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c

(∂2u / ∂x2) + (∂2u / ∂y2) = 0.

5 marks

Answer

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6

Using the Taylor’s series approximation, derive the finite difference expressions of first order and second order differential terms of variable, φ for forward, backward and central methods.

15 marks

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7

Explain cell-centered and cell-vertex discretization methodologies with their constraints in selection. Discuss general formulation of a numerical scheme for 2D problem based on finite volume method.

15 marks

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8

Describe the philosophy of Pressure Correction technique. Explain how boundary conditions are specified consistent with the philosophy of Pressure Correction method.

15 marks

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About this paper

This page contains the solved question paper with answers of COMPUTATIONAL AERODYNAMICS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech AE III II, February/March - 2022, R15.

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