Conservative Forces

A conservative force has the following properties.

1. The work done in moving an object between two positions is independent of the path of the motion.
2. The work done when an object moves around in a closed path, finishing at its starting point, is zero.

An example of a conservative force is gravity. An example of a non-conservative force is friction since the longer the path the greater the work done by friction. The presence of non-conservative forces means that total mechanical energy (KE + PE) is not conserved since some of it is converted into heat energy.

Continuity Equation

The mass flow rate (rAv) is constant at every position along a tube that has a single entry and a single exit point for fluid flow.

r ₁A₁v₁ = r ₂A₂v₂
where r is density, A is cross-sectional area, and v is fluid speed.

If it can be assumed that r remains constant, the continuity equation reduces to
A₁v₁ = A₂v₂

Bernoulli's Principle

For a non-compressible, non-viscous fluid
P₁ + r gh₁ + ½r v₁² = P₂ + r gh₂ + ½r v₂²

where P is pressure (Pa or N/m²)
r is density (kg/m³)
h is height above an arbitrary reference point (m)
v is the velocity that a particle in the fluid is travelling at (m/s) (not to be confused with flow, which is the volume of fluid that passes a certain point in a given time interval)

Bernoulli's principle is a consequence of the law of conservation of mechanical energy: PE + KE + work done = constant. Below is a simplified explanation of Bernoulli's principle.

Consider a 1 m³ volume of water (r = 1,000 kg/m³; mass, m = r x volume = 1,000 kg) at height h.

Its PE = mgh = rgh

If it is allowed to fall height h, its PE will be converted to KE = ½mv² = ½rv²

And at any point, h, in its fall, PE + KE = rgh + ½rv² = constant

Examples of the above are waterfalls and water going down pipes of constant cross-sectional area. However, if the cross-sectional area of a pipe varies, another variable enters the equation: work done, not by gravity, but by the fliud itself.

Now consider water flowing in a horizontal tube of constant cross-sectional area. PE at all points is constant, as is KE. However, if the tube narrows at a certain point, v will increase in the narrowed portion so that the flow rate is the same in all portions of the tube. This means that KE (which equals ½rv²) increases.

So KE increases but PE remains the same. Work must have been done to accelerate the water molecules. This work is reflected in the difference in pressures between the "normal" tube and the narrowed tube.

DKE = Work done = net force x distance
Since we are dealing with a volume of 1 m³,
DKE = (net force/volume) x distance
= net force/area
= DP

It becomes apparent then that in this hypothetical tube,

P + ½rv² = constant at all points

Similarly, if there were a dilation in the tube, the opposite occurs. In the dilation, v would decrease and P increase. Work is done by the fluid on itself to slow it down when it enters the dilation. And the relationship "P + ½rv² = constant at all points" still holds.

Combining this outcome with the outcome from the previous scenario, we have Bernoulli's equation:

P₁ + r gh₁ + ½r v₁² = P₂ + r gh₂ + ½r v₂²