← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q5c — Step-by-Step Solution

10 marks · Section B

Equation of Continuity · Mechanics & Fluid Dynamics · Read the full method →

Question

For an incompressible fluid flow, two components of velocity (u,v,w)(u,v,w) are given by u=x2+2y2+3z2, v=x2y−y2z+zxu=x^2+2y^2+3z^2,\ v=x^2y-y^2z+zx. Determine the third component ww so that they satisfy the equation of continuity. Also, find the zz-component of acceleration.

Technique

Continuity ∇ ⁣⋅q⃗=0\nabla\!\cdot\vec q=0 to get wzw_z, integrate in zz; then zz-acceleration az=(q⃗⋅∇)wa_z=(\vec q\cdot\nabla)w for steady flow.

Solution

Setup. For an incompressible flow the equation of continuity is

∂u∂x+∂v∂y+∂w∂z=0.\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0.

Step 1 — Compute the known divergence terms

u=x2+2y2+3z2 ⇒ ∂u∂x=2x.u=x^2+2y^2+3z^2\ \Rightarrow\ \frac{\partial u}{\partial x}=2x. v=x2y−y2z+zx ⇒ ∂v∂y=x2−2yz.v=x^2y-y^2z+zx\ \Rightarrow\ \frac{\partial v}{\partial y}=x^2-2yz.

Hence

∂w∂z=−(∂u∂x+∂v∂y)=−(2x+x2−2yz)=−x2−2x+2yz.\frac{\partial w}{\partial z}=-\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=-(2x+x^2-2yz)=-x^2-2x+2yz.

Step 2 — Integrate w.r.t. zz

w=∫(−x2−2x+2yz) dz=−x2z−2xz+yz2+f(x,y).w=\int(-x^2-2x+2yz)\,dz=-x^2z-2xz+yz^2+f(x,y).

Taking the simplest determination f(x,y)=0f(x,y)=0 (no arbitrary tangential field required),

  w=−x2z−2xz+yz2.  \boxed{\;w=-x^2z-2xz+yz^2.\;}

Step 3 — zz-component of acceleration

The acceleration is the material derivative Dq⃗Dt\dfrac{D\vec q}{Dt}. For steady flow (∂/∂t=0\partial/\partial t=0) its zz-component is

az=u∂w∂x+v∂w∂y+w∂w∂z.a_z=u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}.

With

∂w∂x=−2xz−2z,∂w∂y=z2,∂w∂z=−x2−2x+2yz,\frac{\partial w}{\partial x}=-2xz-2z,\qquad \frac{\partial w}{\partial y}=z^2,\qquad \frac{\partial w}{\partial z}=-x^2-2x+2yz,

substitute:

az=(x2+2y2+3z2)(−2xz−2z)+(x2y−y2z+zx)(z2)+(−x2z−2xz+yz2)(−x2−2x+2yz).a_z=(x^2+2y^2+3z^2)(-2xz-2z)+(x^2y-y^2z+zx)(z^2)+(-x^2z-2xz+yz^2)(-x^2-2x+2yz).

Expanding and collecting (a common factor zz appears),

Answer

  az=z(x4+2x3−2x2yz+2x2−4xy2−6xyz−5xz2+y2z2−4y2−6z2).  \boxed{\;a_z=z\big(x^4+2x^3-2x^2yz+2x^2-4xy^2-6xyz-5xz^2+y^2z^2-4y^2-6z^2\big).\;}
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