# Reynolds number and inertial force

The Reynolds number is defined as the ratio of ´inertial´ forces to viscous forces.

$$Re = \frac{\text{Inertial Forces}}{\text{Viscous Forces}}$$

Now, viscous forces make sense to me. They are frictional shear forces that come about due to the relative motion of the different layers in a flowing fluid, resulting in different amount of friction, hence, different viscosity values.

However, I am not really sure how to think about the 'inertial force'. This, to me, is somewhat of a dynamic effect since large Re numbers indicate turbulence in most cases, where there is a lot of motion, vortices and eddies. But what exactly is the inertial force and how can it be explained physically?

Inertial force, as the name implies is the force due to the momentum of the fluid. This is usually expressed in the momentum equation by the term $$(\rho v)v$$. So, the denser a fluid is, and the higher its velocity, the more momentum (inertia) it has. As in classical mechanics, a force that can counteract or counterbalance this inertial force is the force of friction (shear stress). In the case of fluid flow, this is represented by Newtons law, $$\tau_x = \mu \frac{dv}{dy}$$. This is only dependent on the viscosity and gradient of velocity. Then, $$Re = \frac{\rho v L}{\mu}$$, is a measure of which force dominates for a particular flow condition.

The inertial forces are what gives rise to the dynamic pressure. Another way to look at the Reynolds Number is by the ratio of dynamic pressure $$\rho u^2$$ and shearing stress $$μ v/ L$$ and can be expressed as $$Re =\frac{\rho u^2} {μ v/ L} = \frac{ u L} {\nu}$$

At very high Reynolds numbers, the motion of the fluid causes eddies to form and give rise to the phenomena of turbulence.

• I think your answer can be improved by putting more emphasis on the dynamic pressure to explain it in the context of forces. – Bernhard Oct 8 '13 at 20:40
• I'm not sure if I'm correct, but in the last latex equation, shouldn't it be vL/(v kinematic viscosity) rather than uL/v? Or could you explain why its written as rho.v.L/mu before? – midnightBlue Nov 25 '14 at 20:43
• @mcodesmart in your answer, $v$ is velocity, $\nu$ is kinematic viscosity, and $u$ is what? – Armadillo Dec 12 '17 at 19:57
• In en.wikipedia.org/wiki/Characteristic_length they suggest setting L to system volume divided by system boundary area do you agree this is a good choice for L? – Emil Sep 28 '18 at 16:23

The ratio of the inertial force to the viscous forces of the fluid is known as the Reynolds number. Now the viscous forces are the forces due to to the friction between the the layers of any real fluid. In fluid mechanics we take the fluid as in the continuum condition, which means fluid particles are very closely packed so necessarily there is friction between layers of fluid. The inertial forces are the forces which are due to due the particles of fluid resisting any change in momentum.

Newton’s second law can be written as

$$m a = f_1 + f_2 + \cdots + f_n$$

The term $$m a$$ having dimensions of force and being proportional to the mass of the body, which is a measure of its inertia, is often named inertia force. So, according to the previous equation, the inertia force is just the resultant force acting on the body under analysis. In Fluid Mechanics it is advantageous to use mass per unit volume of the body (fluid in this case), that is its density, so that Newton’s law (or rather, the Navier-Stokes equation) is written with the terms having dimensions of force per unit volume of fluid.

When fluids flow, different types of forces act on the fluid. These are represented in the previous equation by $$f_1$$, $$f_2$$,$$\cdots$$ , $$f_n$$. Suppose viscous forces are represented by $$f_2$$. Back to the original question, the Reynolds number (Re) associated with the fluid flow would be, in this case, $$m a / f_2$$. So, in fluid flow, Re is a measure of the ratio between the resultant force (or inertia force) and viscous force acting on the fluid. Notice that the viscous force is part of the inertia force. In other words, Re is the ratio between the resultant force acting on the fluid and one of its components.

Our intuition regarding effects of inertia forces is quite good because in daily life our muscles overcame inertia of static bodies all the time. Inertia of moving bodies are also easily perceptible when our velocity (for instance in a moving car) changes magnitude or direction, that is, when we are accelerated relative to the ground (notice that a is a factor in the inertia force). Effects of viscous forces are much more subtle and require specific experiments.

• "Re is the ratio between the resultant force acting on the fluid and one of its components." That would mean that Re>=1, which is obviously false. – Ratbert Jan 24 at 15:20
• @Ratbert , only if the sign is the same for every force – user1420303 Mar 28 at 2:13

In Fluid Mechanics, fluid is considered as a continuous medium i.e the atoms of the fluid are stacked closely. Consider an atom stack of a fluid as a group of people standing in a queue. By their nature of state, they remain standing/ stagnant. When a force is provided at any one end of the qeue, the people in the queue moves in the corresponding direction. The force which has to be applied at the end of queue should be greater than the force given by the people to stay stagnant in the line (inertial force which is due to the mass of the people acted upon by the gravity). This force is called as inertial force. Now, when there are two or more queues standing parallel with each member of one qeue holding their hands with the adjacent queue members, then when the force applied at any one end of the queue is greater, then there exist a relative velocity between the adjacent queue members and because of this (also that the queue members are holding hands with adjacent queue members-connected-concept of continuum) , they tend to rotate and hinder with the motion of the adjacent queue members. This phenomenon is equivalent to eddy formation and turbulence

• Knowing enough fluid mechanics, it is possible to have some feeling about the precise physical content of your answer. However, it would be better to add some more formal detail, in order to provide a key to understand the reason Reynolds number is defined that way. – GiorgioP Jan 23 at 22:52

## protected by AccidentalFourierTransformSep 19 at 0:27

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