A flow has velocity potential function given by $\phi=x^3-3xy^2$. Verify whether it represents valid flow fluid. If it does, then determine the stream function and also calculate the velocity and pressure at (1,-3) given that the pressure at (4,1) is 14 kPa and the fluid is water.

Data:-

$\phi = x^3-3xy^2,$ $P_{(4,1)}=14kPa,$ $w=9810\frac{N}{m^3}$.........(water)

To find:-

i) To verify valid fluid flow

ii) $\psi=?$

iii) Velocity and pressure at (1,-3)=?

Solution:-

i) In velocity potential function,

velocity component in x-direction 'u' is given by

$u=-\frac{\partial \phi}{\partial x}$

and

velocity component in y-direction 'v' is given by

$v=-\frac{\partial \phi}{\partial y}$

$u=-\frac{\partial \phi}{\partial x}=-\frac{\partial}{\partial x}(x^3-3xy^2)=-(3x^2-2y^2)$

$\therefore u=3y^2-3x^2$

& $v=-\frac{\partial \phi}{\partial y}=-\frac{\partial}{\partial y}(x^3-3xy^2)=-(0-6xy)$

$\therefore u=6xy$

For possible case of flow it should satisfy continuity equation

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

$\frac{\partial u}{\partial x}=-6x$ & $\frac{\partial v}{\partial y}=6x$

$\therefore \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=-6x+6x=0$..........satisfy

Therefore, it is possible case of flow.

To check whether potential function exist or not it should satisfy case of irrational flow

$W_t=\frac{1}{2}(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})$

$\frac{\partial v}{\partial x}=6y$ and $\frac{\partial u}{\partial y}=6y$

Here, rotational component is absent, therefore it is valid velocity potential function

ii) Stream function $(\psi)$

$\frac{\partial \psi}{\partial x}=-v$ and $\frac{\partial \psi}{\partial y}=u$

Consider $\frac{\partial \psi}{\partial y}=u$

$\therefore \partial \psi=u.dy$

$\therefore \partial \psi=(3y^2-3x^2)dy$

Integrating above equation w.r.t y

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

$\psi y^3-3x^2y+f(x)$..............(1)

Differentiating above equation w.r.t x

$\frac{\partial \psi}{\partial x}=0-6xy+f'(x)$............(2)

But $\frac{\partial \psi}{\partial x}=-v=-6xy$

Equation (2) becomes

$-6xy=-6xy+f'(x)$

$\therefore f'(x)=0$

Integrating w.r.t x

$\therefore f(x)=x$............put in equation (1)

$\psi=y^3-3x^2y+x$

(iii)Velocity and pressure at (1,-3)

Velocity at (1,-3) i.e., at x=1 & y=-3

$u=3y^2-3x^2=3(-3)^2-3(1)^2$

$\therefore u=24 units$

and $v=6xy=6\times 1\times -3=-18 units$

$v_{(1,-3)}=\sqrt{u^2+v^2}=\sqrt{(24)^2+(18)^2}$

$=30 units$

Velocity at (4,1)

$u=3y^2-3x^2=-45 units$

$v=6xy=24 units$

$v_(4,1)=\sqrt{u^2+v^2}=51 units$

Applying Bernoulis equation between (1,-3) & (4,1)

$\frac{P_{(1,-3)}}{W_w}+\frac{V_{(1,-3)}^2}{2\times 9}+t_{(1,-3)}=\frac{P_{(4,1)}}{W_w}+\frac{V_{(4,1)}^2}{2\times 9}+t_{(4,1)}$

$\frac{P_{(1,-3)}}{9810}+\frac{30^2}{2\times 9.81}=\frac{14\times 10^3}{9810}+\frac{51^2}{2\times 9.81}$

$P_{(1,-3)}=64.5\times 10^3N/m^3$