The velocity field of a certain two-dimensional flow is given by V(𝑥, 𝑦) = 𝑘(𝑥𝑖̂ − 𝑦𝑗̂)
where 𝑘 = 2 s^-1 . The coordinates 𝑥 and 𝑦 are in meters. Assume gravitational effects to be negligible.

If the density of the fluid is 1000 kg/m³ and the pressure at the origin is 100 kPa, the pressure at the location (2 m, 2 m) is _____________ kPa. (Answer in integer)

The correct answer is 84.

Step-by-step Explanation:

We are given the velocity field of a two-dimensional, steady, and incompressible flow:
$\mathbf{V}(x,y)=k(x\hat{i}-y\hat{j})$
where the constant is:
$k=2{\text{s}}^{-1}$

The velocity components along the x and y directions are:
$u=kx$
$v=-ky$

First, we check if the flow is irrotational by calculating the vorticity component in the z-direction:
${\omega }_z=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0-0=0$
Since the flow is steady, incompressible, and irrotational (potential flow), and gravitational effects are negligible, we can apply Bernoulli's equation between any two points in the flow field.

Bernoulli's equation states:
$p_1+\frac{1}{2}\rho V_1^{}2=p_2+\frac{1}{2}\rho V_2^{}2$

Let Point 1 be the origin (0, 0):
The pressure at the origin is:
$p_1=100\text{kPa}=100,000\text{Pa}$
The velocity magnitude at the origin is:
$V_1=0\text{m/s}$

Let Point 2 be the location (2 m, 2 m):
The velocity components at Point 2 are:
$u=2\times 2=4\text{m/s}$
$v=-2\times 2=-4\text{m/s}$
The velocity magnitude squared at Point 2 is:
$V_2^2=u^2+v^2=4^2+(-4{)}^2=16+16=32{\text{m}}^2/{\text{s}}^2$

Now, substituting the values into Bernoulli's equation:
$100,000+0=p_2+\frac{1}{2}\times 1000\times 32$
$100,000=p_2+16,000$
$p_2=100,000-16,000=84,000\text{Pa}$
Converting the pressure back to kPa:
$p_2=84\text{kPa}$