Deriving Bernoulli's Equation from a Few Simple Concepts

Among the classes I took as undergraduate was fluid mechanics, and among the most useful equations I learned, or at least to which I was reintroduced, was Bernoulli's equation. This equation is useful for understanding the flow of fluid in pipes, and as way of remembering the equation, I like to use a few concepts I easily remember to help derive the equations.

The way I typically start is by thinking of the equation as a kind energy conversation equation. This ultimately leads to a kind of height conservation equation (this will become clearer as we reach our desired equation). To be more clear, all the terms in the equation have a set of units that evaluate to length. Additionally, there are three terms: one term for gravitational potential energy, another for kinetic energy, and one for pressure energy. Finally, each side of the equation will represent the energy (or height) state of the fluid a points A and B, respectively, and will have all three aforementioned terms (See the image below). Knowing these things, I can derive the equation.

The first term to derive is that associated with gravitational potential energy. When we think about this form of energy, we have to remember how an object of some mass m, when suspended in the air, has a potential energy associated with it, mgh, where g is the acceleration due to gravity and h is the elevation of the object relative to an established datum or coordinate system. Remember the following: although we have the energy of an object of mass m, we are dealing with fluid flowing in a pipe, so we are thinking of a changing mass (but not the quantity of mass over an interval of time for a fixed volume). To fix this, we can obtain the energy per unit mass by dividing by m; this is acceptable and can be useful, but we can go further. We want to only have height, so we can divide by g to obtain simply h, or the elevation of the fluid. This is what we are after for the first term. As a side note, we can think of this term as a kind of energy per unit mass of the fluid (it isn't exactly that because we also divided by the constant g).

The second term to consider is that of kinetic energy. From mechanics, we remember that the translational energy associated with some object of mass m moving at speed v is 1/2*m*v^2. This term is in units of energy, but once again we can divide by m to get energy per unit mass. However, we are still striving for a term with units that evaluate to length. Since we remember that the potential energy term was divided by m and g, we can divide this kinetic energy term by g as well to obtain v^2/(2*g). If you check the units, it evaluates to length.

Finally, the pressure energy term is derived. Now we begin with pressure P, but this is not an energy term. To get energy, we would need to multiply by a volume, but that is not useful here. Instead we can think of the factors associated with the pressure of a static fluid at some depth h (we are not considering the pressure exerted on the surface of the fluid), ρ*g*h, where ρ is the density. When the pressure of a fluid expressed this way is divided by ρ and g, we obtain h, which gives the desired units; dividing P by these factors gives h.

When we consider all of this, we can put the terms together, while considering that points A and B will have these terms, to obtain the following:

[P/(ρg) + v^2/(2g) + h]A = [P/(ρg) + v^2/(2g) + h]B

This states mathematically that for the energy we considered (or in terms of units the sum of heights), the energy at point A will equal the energy at B.

Note that we did not state the assumptions at the beginning. An important thing to keep in mind is that this equation assumes there is nothing going on that decreases (or increases) the energy between A and B. In reality, there are energy losses, assuming there is no addition of energy somewhere in between. The fluid flow in this ideal case is said to be inviscid, irrotational, and steady.