A steady incompressible flow field is given by u = 2x² + y²and v = -4xy. The convective acceleration, along x - direction at point (1, 2) is

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ESE Mechanical 2015 Paper 1: Official Paper

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Answer 1

Concept:

Convective acceleration is given by,

$a_{x} = u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z}$

$a_{y} = u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} + w \frac{\partial v}{\partial z}$

$a_{z} = u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y} + w \frac{\partial w}{\partial z}$

Calculation:

Given:

u = 2x2 + y2 and v = -4xy

To get, $a_{x} = u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z}$

At point (1, 2)

$u \frac{\partial u}{\partial x} = (2 x^{2} + y^{2}) \times \frac{\partial (2 x^{2} + y^{2})}{\partial x} = (2 x^{2} + y^{2}) \times (4 x)$

Now putting the coordinates of x and y in the above equation (x = 1, y = 2)

$u \frac{\partial u}{\partial x} = (2 x^{2} + y^{2}) \times (4 x) = (2 \times 1^{2} + 2^{2}) \times (4 \times 1) = 24 u n i t s$

Same

$v \frac{\partial u}{\partial y} = - 4 x y \times \frac{\partial (2 x^{2} + y^{2})}{\partial y} = - 8 x y^{2}$

Now putting the coordinates of x and y in the above equation (x = 1, y = 2)

$v \frac{\partial u}{\partial y} = - 8 x y^{2} = - 8 \times 1 \times 2^{2} = - 32$

$w \frac{\partial u}{\partial z} = 0 S i n c e, u i s i n d e p e n d e n t o f z$

Therefore,

a_{x} = u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = 24 - 32 = - 8 u n i t s