If the velocity components in a two-dimensional flow field are expressed as

$u = \frac{y^{3}}{3} + 2 x - x^{2} y$

$v = x y^{2} - 2 y - \frac{x^{3}}{3}$

These functions represent a possible case of:

Question

Question

This question was previously asked in

JKPSC Lecturer Paper I Civil 2022 Official Paper

Answer 1

Concept:

Continutiy equation in Three-Dimension

$\frac{\partial ρ}{\partial t} + \frac{\partial (ρ U)}{\partial X} + \frac{\partial (ρ V)}{\partial Y} + \frac{\partial (ρ W)}{\partial Z} = 0$

Where,

U, V, and W are components of velocity in X, Y and Z direction respectively

When the flow is steady $, \frac{\partial ρ}{\partial t} = 0$

$∴ \frac{\partial (ρ U)}{\partial X} + \frac{\partial (ρ V)}{\partial Y} + \frac{\partial (ρ W)}{\partial Z} = 0$

When flow is steady and incompressible, ρ = constant

$∴ \frac{\partial U}{\partial X} + \frac{\partial V}{\partial Y} + \frac{\partial W}{\partial Z} = 0$

When the flow is steady, incompressible and 2-D, $\frac{\partial W}{\partial Z} = 0$

$∴ \frac{\partial U}{\partial X} + \frac{\partial V}{\partial Y} = 0$

For flow being irrotational

$\frac{\partial V}{\partial X} - \frac{\partial U}{\partial Y} = 0$

Calculation:

Given:

$\begin{array}{l} U = \frac{y^{3}}{3} + 2 x - x^{2} y \\ V = x y^{2} - 2 y - \frac{x^{3}}{3} \end{array}$

As per continuity equation, for flow to occur

$∴ \frac{\partial U}{\partial X} + \frac{\partial V}{\partial Y} = 0$

$\begin{array}{l} \frac{\partial U}{\partial X} = 2 - 2 x y \\ \frac{\partial V}{\partial Y} = 2 x y - 2 \end{array}$

2 - 2xy + 2xy - 2 = 0

Since, $\frac{\partial U}{\partial X} + \frac{\partial V}{\partial Y} = 0$ , then flow is incompressible.

For irrational flow

$\begin{array}{l} \frac{\partial V}{\partial X} = y^{2} - x^{2} \\ \frac{\partial U}{\partial Y} = y^{2} - x^{2} \end{array}$

$\frac{\partial V}{\partial X} - \frac{\partial U}{\partial Y} = 0$

So, flow is irrotational also.