In an ideal fluid flow along a tube of uniform cross-section, located in a horizontal plane and bent, the Bernoulli's principle can be applied. According to Bernoulli's principle, the total energy per unit mass of a fluid particle remains constant along a streamline.
The Bernoulli's equation for steady, incompressible flow is given by:
\[ P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} \]
where:
- \(P\) is the pressure,
- \(\rho\) is the density of the fluid,
- \(v\) is the velocity of the fluid,
- \(g\) is the acceleration due to gravity,
- \(h\) is the height above a reference point.
In your case, considering two points, 1 and 2, along the bent tube:
\[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \]
Since the fluid elements have the same net energy density, you can cancel out the terms involving the height (\(h\)) and rearrange the equation:
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]
This equation represents the conservation of energy for the fluid particles between points 1 and 2 in the bent tube. It shows that the sum of pressure energy and kinetic energy per unit mass at point 1 is equal to the sum of pressure energy and kinetic energy per unit mass at point 2.
