Fluid Mechanics Formula

Fluid mechanics could be defined as the division of engineering science which deals with the fluids behavior in both motion and rest situations.

Fluid mechanics is one of the four divisions of mechanics namely quantum mechanics, relative mechanism, fluid mechanics and elastic body mechanics.

Fluid mechanics is separated into three distinct classes. These are statics, kinematics, and the dynamics out of which static and dynamics are then divided into compressible and incompressible flow. These are furthermore split into turbulent and laminar forms.

Fluid mechanics is grounded upon five great principles of physics and they are as follows:

1. Conservation of linear momentum
2. Conservation of angular momentum
3. Conservation of energy
4. Conservation of mass
5. Conservation of thermodynamics

In fluid mechanics if the speed of flow of a liquid if not too huge then it generally flows in coats with regular gradation in their respective velocities and therefore called streamline flow. While if the rate of flow is too high then numerous irregularities begin to show and do not flow in layers which we term as turbulent flow. The turbulence commences after the critical velocity which is given by a relation

$V_{c}=\frac{k\eta}{\rho&space;r}$

Solved Example

Some Examples on Fluid Mechanics Formula are:

Problem: The distance amid two pistons is 0.015 mm and the viscous fluid flowing through produces a force of 1.2 N per square meter to keep these two plates move at a speed 35 cm/s. Calculate the fluid viscosity in the middle of the plates?

Transforming every unit into meters.

Distance in the middle of the pistons = 0.015 × 10-3 m or 0.015 mm

The speed of the pistons at which they travel = 0.35 m/s or 35 cm/s

Force in the middle of the two pistons = 1.2 N / m2

$Force\;per\;unit\;area=\frac{viscosity(change\;in\;velocity)}{distance\;between\;pistons}$

$viscosity=\frac{force\;per\;unit\;area\times&space;distance\;between\;two\;pistons}{change\;in\;velocity}$

$viscosity=(\frac{1.2\times&space;0.015\times&space;10^{-3}}{0.35})=0.0514\times&space;10^{-3}=5.14\times&space;10^{-3}Ns/m^{2}$