The steady velocity field in an inviscid fluid of density 1.5 is given to be $\overrightarrow{V}=(y^2-x^2)\hat{i}+(2xy)\hat{j}$ . Neglecting body forces, the pressure gradient at (x =1, y = 1) is ______.

The correct option is -6î - 6ĵ.

1. Understanding the Governing Equation
For a steady, inviscid flow neglecting body forces, the motion of the fluid is governed by Euler's equation of motion:

$\rho \overrightarrow{a}=-\nabla P$

where:
• $\rho$ is the density of the fluid (given as 1.5),
• $\overrightarrow{a}$ is the acceleration vector, and
• $\nabla P$ is the pressure gradient vector.
Rearranging the equation, we can write the pressure gradient as:

$\nabla P=-\rho \overrightarrow{a}$

2. Determining the Velocity Components and Derivatives
The given velocity field is:

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

From this, the horizontal velocity component $u$ and vertical velocity component $v$ are:

$u=y^2-x^2$

$v=2xy$

Now, let's calculate the required partial derivatives of the velocity components with respect to $x$ and $y$:

$\frac{\partial u}{\partial x}=-2x$

$\frac{\partial u}{\partial y}=2y$

$\frac{\partial v}{\partial x}=2y$

$\frac{\partial v}{\partial y}=2x$

3. Calculating Acceleration Components at (x = 1, y = 1)
First, evaluate the velocity components $u$ and $v$ at the point $(1,1)$:

$u=1^2-1^2=0$

$v=2\left(1\right)\left(1\right)=2$

For steady flow, the local acceleration is zero, so the acceleration components are solely convective:

$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\left(0\right)(-2)+\left(2\right)\left(2\right)=4$

$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=\left(0\right)\left(2\right)+\left(2\right)\left(2\right)=4$

Thus, the acceleration vector at the point $(1,1)$ is:

$\overrightarrow{a}=4\hat{i}+4\hat{j}$

4. Computing the Pressure Gradient
Substitute the fluid density $\rho =1.5$ and the computed acceleration vector $\overrightarrow{a}$ back into Euler's relation:

$\nabla P=-1.5(4\hat{i}+4\hat{j})=-6\hat{i}-6\hat{j}$

Therefore, the pressure gradient at $(x=1,y=1)$ is indeed -6î - 6ĵ.