The x component of velocity in a two dimensional incompressible flow is given by $u = 1.5 x$ . At the point (x, y) = (1, 0), the y component of velocity v = 0. The equation for the y component of velocity is

v = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = 1.5 y

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = - 1.5 x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

v = - 1.5 y

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $v = - 1.5 y$
For a two-dimensional incompressible flow, the continuity equation is given by
$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$
$\Rightarrow 1.5 + \frac{\partial v}{\partial y} = 0$
$\Rightarrow \frac{\partial v}{\partial y} = - 1.5$
$\Rightarrow v = - 1.5 y + C$
But at y = 0, v = 0
$∴$ $0 = - 1.5 \times 0 + C$
$\Rightarrow C = 0$
$T h u s v = - 1.5 y$

Suggest Corrections

Similar questions

u = 2 y^{2} + 6 x y

u = y^{2} + 4 x y

u = \frac{x}{T_{1}}, v = \frac{- y}{T_{2}} a n d w = 0

T_{1} (\neq 0) a n d T_{2} (\neq 0)

T h e c o n t i n u i t y e q u a t i o n i s \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}

T h e a c c e l e r a t i o n i n x - d i r e c t i o n i s a_{x} = \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}

T h e c o n d i t i o n o f i r r o t a t i o n a l i t y i s \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x}

u = (λ x y^{2} - x^{3} y^{2}) a n d v = (x^{2} y^{3} - 3 y^{3})

λ

Join BYJU'S Learning Program