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Equation of Continuity

At a Glance

Why This Chapter Matters

The equation of continuity is the mathematical expression of mass conservation in fluid flow — the first equation any fluid mechanics problem must satisfy. UPSC 2018 tested it computationally: verify that a given velocity field satisfies the continuity equation or find a stream function for a given velocity component. The computation is straightforward if you know the two forms (compressible and incompressible) and the stream function relations.

Minimum Theory

Physical Statement

Mass is neither created nor destroyed in fluid flow. If ρ\rho is the fluid density and v=(u,v,w)\mathbf{v} = (u, v, w) is the velocity field, the rate of change of mass in any fixed control volume equals the net mass flux into it.

General (Compressible) Form

ρt+(ρv)=0,\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0,

or in Cartesian components:

ρt+(ρu)x+(ρv)y+(ρw)z=0.\frac{\partial \rho}{\partial t} + \frac{\partial(\rho u)}{\partial x} + \frac{\partial(\rho v)}{\partial y} + \frac{\partial(\rho w)}{\partial z} = 0.

Incompressible Form

For incompressible flow, ρ=const\rho = \text{const} (density is constant in time and space), so ρ/t=0\partial\rho/\partial t = 0 and ρ\rho factors out:

v=0,i.e.,ux+vy+wz=0.\nabla \cdot \mathbf{v} = 0, \quad \text{i.e.,} \quad \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0.

In 2D incompressible flow (no zz-dependence, w=0w = 0):

ux+vy=0.\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0.

The Stream Function ψ\psi

For 2D incompressible flow, the continuity equation is automatically satisfied by introducing the stream function ψ(x,y)\psi(x, y) defined by:

u=ψy,v=ψx.u = \frac{\partial \psi}{\partial y}, \qquad v = -\frac{\partial \psi}{\partial x}.

Verification: ux+vy=2ψxy2ψyx=0\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y} = \dfrac{\partial^2 \psi}{\partial x\,\partial y} - \dfrac{\partial^2 \psi}{\partial y\,\partial x} = 0. \checkmark

Properties of ψ\psi:

Finding ψ\psi from a Given Velocity Component

Given u(x,y)u(x, y) and the incompressibility condition, find vv and ψ\psi:

  1. From u/x+v/y=0\partial u/\partial x + \partial v/\partial y = 0, integrate to get v=(u/x)dy+g(x)v = -\int (\partial u/\partial x)\,dy + g(x).
  2. Determine g(x)g(x) from any additional condition (e.g., a boundary condition or a given expression for vv).
  3. Find ψ\psi by integrating: ψ=udy\psi = \int u\,dy (treating xx as constant), then check ψ/x=v-\partial\psi/\partial x = v.

Question Archetypes

ArchetypeRecognition
verify-continuity”Verify that v=(u,v)\mathbf{v} = (u,v) satisfies the equation of continuity”
find-stream-function”Given uu, find vv and the stream function ψ\psi

verify-continuity (1 question; 2018)

Recognition Cues

Solution Template

  1. State the continuity equation for incompressible flow: u/x+v/y=0\partial u/\partial x + \partial v/\partial y = 0 (2D) or u/x+v/y+w/z=0\partial u/\partial x + \partial v/\partial y + \partial w/\partial z = 0 (3D).
  2. Compute each partial derivative from the given expressions.
  3. Sum the partial derivatives and show the sum is zero.
  4. State the conclusion: the velocity field satisfies continuity, hence represents an incompressible flow.
  5. If stream function is asked: integrate u=ψ/yu = \partial\psi/\partial y to find ψ\psi, verify v=ψ/xv = -\partial\psi/\partial x.

Worked Example

2018 Paper 2, 2018-P2-QMF (10 marks)

For the 2D velocity field u=x2y2u = x^2 - y^2, v=2xyv = -2xy, (a) verify that the equation of continuity is satisfied, and (b) find the stream function ψ\psi.

Part (a): Verify continuity.

For 2D incompressible flow, the continuity equation is:

ux+vy=0.\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0.

Compute:

ux=x(x2y2)=2x,\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x,

vy=y(2xy)=2x.\frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(-2xy) = -2x.

Sum:

ux+vy=2x+(2x)=0.\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 2x + (-2x) = 0. \quad \checkmark

The velocity field satisfies the equation of continuity. \blacksquare

Part (b): Find the stream function.

Use the defining relations u=ψ/yu = \partial\psi/\partial y and v=ψ/xv = -\partial\psi/\partial x.

From ψ/y=u=x2y2\partial\psi/\partial y = u = x^2 - y^2, integrate with respect to yy (treating xx as constant):

ψ=x2yy33+f(x),\psi = x^2 y - \frac{y^3}{3} + f(x),

where f(x)f(x) is an arbitrary function of xx (the “constant” of integration with respect to yy).

Now use v=ψ/xv = -\partial\psi/\partial x:

ψx=(2xy+f(x))=v=2xy.-\frac{\partial\psi}{\partial x} = -(2xy + f'(x)) = v = -2xy.

Therefore:

(2xy+f(x))=2xy    f(x)=0    f(x)=C (constant).-(2xy + f'(x)) = -2xy \implies f'(x) = 0 \implies f(x) = C \text{ (constant)}.

The stream function is:

ψ=x2yy33+C.\boxed{\psi = x^2 y - \frac{y^3}{3} + C.}

Remark. Note that ψ=x2yy3/3=Im(x+iy)3=Im(z3)\psi = x^2 y - y^3/3 = \text{Im}(x+iy)^3 = \text{Im}(z^3), confirming that w=z3w = z^3 is the complex potential — the velocity field corresponds to the cubic flow. \blacksquare

Common Traps

Marks-Aware Writing

This is a 10-mark Section A question, likely split as 5+5 for parts (a) and (b).

For part (a): state the continuity equation, compute partials, show they sum to zero — 3 lines is sufficient.

For part (b): show the integration step clearly, include f(x)f(x), and determine it from the second relation — do not just write the answer; the method carries the marks.

Practice Set

Only one historical question on this atom (shown above).

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