Show that the two-dimensional flow described (in metre-second units) by the equation \(\psi=x+2 x^{2}-2 y^{2}\) is irrotational. What is the velocity potential of the flow? If the density of the fluid is \(1.12 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) and the piezometric pressure at the point \((1,-2)\) is \(4.8 \mathrm{kPa}\), what is the piezometric pressure at the point \((9,6)\) ?

In fluid dynamics, the stream function is a very useful concept to describe the flow of incompressible fluids. It is denoted by \( \psi \), and in two-dimensional flow, the stream function is a scalar function whose partial derivatives give us the velocity components of the flow. For the given flow \( \psi = x + 2x^2 - 2y^2\), the velocity components are determined as follows:
\( u = \frac{\partial \psi}{\partial y} = -4y \)
\( v = -\frac{\partial \psi}{\partial x} = -(1 + 4x) \)
These components help in understanding the direction and magnitude of the flow at different points in the fluid field.

The velocity potential, denoted by \( \phi \), is another important scalar function in fluid dynamics. It is especially useful in describing irrotational flows. The potential function is connected to the velocity components of the flow via partial derivatives:
\( u = \frac{\partial \phi}{\partial x} \), and \( v = \frac{\partial \phi}{\partial y} \.\)
To determine the velocity potential \( \phi \) for the given flow, integrate the velocity components:
\( u = -4y \to \phi = -2y^2 + f(x) \)
Substitute this into the expression for \( v \) and solve for \( f(x) \):
\( -(1 + 4x) = \frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x}(-2y^2 + f(x)) \),
which simplifies to \( f(x) = x + 2x^2 \).
Hence, the velocity potential \( \phi = -2y^2 + x + 2x^2 \).

Bernoulli’s equation is a fundamental principle in fluid dynamics that describes the conservation of energy in a flowing fluid. For incompressible flows, the equation is expressed as:
\( p + \frac{1}{2} \rho (u^2 + v^2) = \text{constant} \)
Here, \( p \) denotes the piezometric pressure, \( \rho \) is the fluid density, and \( u \) and \( v \) are the velocity components. This equation states that the sum of the pressure energy, kinetic energy, and potential energy remains constant along a stream line. To find the piezometric pressure at different points, substitute the velocity components and known values into Bernoulli’s equation and solve for the unknown pressure. For example, at point \( (1, -2) \), the piezometric pressure is 4.8 kPa, leading to:
\( 4.8 \text{kPa} + \frac{1}{2} \times 1.12 \times (8^2 + (-5)^2) = p_{(9, 6)} + \frac{1}{2} \times 1.12 \times (24^2 + 37^2)\), solving to \( p_{(9, 6)} = -36226.4 \text{Pa} \).

Piezometric pressure is a combination of the pressure energy and the gravitational potential energy per unit volume within a fluid. It is used in Bernoulli’s equation to account for both static and dynamic pressures. It’s given as:
\( p = \text{static pressure} + \frac{1}{2} \rho v^2 \)
This quantity remains constant along a streamline for incompressible flows. For the given problem, the piezometric pressure at one point is used to determine it at another point using the flow velocities derived from the stream function:
\( p_{(1, -2)} + \frac{1}{2} \rho (u^2 + v^2) = p_{(9, 6)} + \frac{1}{2} \rho (u^2 + v^2) \). Substitute and solve for the piezometric pressure at the new point. Understanding piezometric pressure helps in evaluating how pressure changes within a flowing fluid.