Question

10) A two dimensional flow field is
described by the velocity
components u = 5x3 and V = -15x2y
The stream function at point P(1.2)
would be numerically equal to
a) 5 units​

Answer 1

SOLUTION

A two dimensional flow field is described by the velocity components u = 5x³ and v = - 15x²y

TO DETERMINE

The stream function at point P(1,2)

EVALUATION

Here it is given that a two dimensional flow field is described by the velocity components u = 5x³ and v = - 15x²y

Now

$\displaystyle\sf{ \frac{ \partial u}{ \partial x} = 15 {x}^{2} }$

$\displaystyle\sf{ \frac{ \partial v}{ \partial y} = - 15 {x}^{2} }$

$\displaystyle\sf{ \therefore \: \: \frac{ \partial u}{ \partial x} = - \frac{ \partial v}{ \partial y}}$

Thus the flow is possible

$\sf{Let \: stream \: function = \Psi}$

$\displaystyle\sf{ \therefore \: \:u = \frac{ \partial \Psi}{ \partial y} \: \: and \: \: v = - \frac{ \partial \Psi}{ \partial x}}$

$\displaystyle\sf{ u = \frac{ \partial \Psi}{ \partial y} \: \: gives}$

$\displaystyle\sf{ \frac{ \partial \Psi}{ \partial y} = 5 {x}^{3} }$

Again

$\displaystyle\sf{ v = - \frac{ \partial \Psi}{ \partial x} \: \: \: gives}$

$\displaystyle\sf{ \frac{ \partial \Psi}{ \partial x} = 15 {x}^{2}y }$

$\displaystyle\sf{ \therefore \: \:d \Psi}\displaystyle\sf{ = \frac{ \partial \Psi}{ \partial x} \: dx + \frac{ \partial \Psi}{ \partial y} dy }$

$\displaystyle\sf{ \implies \: \:d \Psi}\displaystyle\sf{ = 15 {x}^{2} y \: dx + 5 {x}^{3} dy }$

$\displaystyle\sf{ \implies \: \: \Psi}\displaystyle\sf{ =5 {x}^{3} y }$

Thus the required stream function

$\boxed{ \: \:\displaystyle\sf{ \: \Psi}\displaystyle\sf{ =5 {x}^{3} y \: \: }}$

Now the stream function at point P(1,2)

$= \sf{5 \times {(1)}^{3} \times 2}$

$= \sf{5 }$

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