Answer :

Solution :

$\overrightarrow{V}=(y^2-x^2)\hat{i}+(2xy)\hat{j}$

u = (y² - x²) , v = 2xy

x =1, y = 1, ρ = 1.5 kg/m³

for inviscid flow $\mu ({▽}^{2}u)=0$, body force (ρg) = 0

So the equation for inviscid flow and negligible body force is:

$\rho [u\frac{\delta u}{\delta x}+v\frac{\delta u}{\delta y}]=-\frac{\delta p}{\delta x}$

$\frac{\delta p}{\delta x}=$- 1.5 [ (y2 - x2) (-2x) + 2xy (2y)]

at x = 1 y = 1

δp / δx = - 1.5 [ 0 + 2 × 2 ] = - 6i

Similarly for y direction :

$\rho [u\frac{\delta v}{\delta x}+v\frac{\delta v}{\delta y}]=-\frac{\delta p}{\delta y}$

δp / δy = - 1.5 [ (y2 - x2) (2y) + 2xy (2x)]

at x = 1 y = 1

δp / δy = - 1.5 [ 0 + 2 × 2] = - 6j

Hence, Pressure gradient is = - 6i - 6j