3.5

3.5 Bernoulli Equation

A non-turbulent, perfect, compressible, and barotropic fluid undergoing steady motion is governed by the Bernoulli Equation:

where g is the gravity acceleration constant (9.81 m/s2; 32.2 ft/s2), V is the velocity of the fluid, and z is the height above an arbitrary datum. C remains constant along any streamline in the flow, but varies from streamline to streamline. If the flow is irrotational, then C has the same value for all streamlines. The function  is the "pressure per density" in the fluid, and follows from the barotropic equation of state, p= p(r). For an incompressible fluid, the function  simplifies to p/r, and the incompressible Bernoulli Equation becomes:

Derivation from Navier-Stokes

The Navier-Stokes equation for a perfect fluid reduce to the Euler Equation:

Rearranging, and assuming that the body force b is due to gravity only, we can eventually integrate over space to remove any vector derivatives,

If the fluid motion is also steady (implying that all derivatives with respect to time are zero), then we arrive at the Bernoulli equation after dividing out by the gravity constant (and absorbing it into the constant C),

Note that the fluid's barotropic nature allowed the following chain rule application,

with the "pressure per density" function  defined as,

Examples:

1. Water (density = 1000 kg/m³) flows through a hose with a velocity of 1 m/sec.  As it leaves the nozzle the constricted area increases the velocity to 20 m/sec.  The pressure on the water as it leaves is atmospheric pressure (1 Atm = 100,000 N/m²).  What is the pressure on the water in the hose?  Express the answer in N/m² and Atm.
   
   First let us list the information we are given.
   
   Inside:  v₁ = 1 m/sec
   P₁ = ?
   Outside:  v₂ = 20 m/sec
   P₂ = 100,000 N/m²
   Density of water = d = 1000 kg/m³
   
   Notice that care has been taken to put all values in mks units.  We can now use Eq. 9.5  to find the unknown pressure P₁.

               

Express the result in atmospheres.

2. Bernoulli’s equation can also be used to show how the design of an airplane wing results in an upward lift.  The flow of air around an airplane wing is illustrated below.  In this case you will notice that the air is traveling faster on the upper side of the wing than on the lower.

 

As a result the pressure will be greater on the bottom of the wing, and the wing will be forced upward.  Let us consider an airplane wing where the flow of air (density = 1.3 kg/m³) is 250 m/sec over the top of the wing and 220 m/sec over the bottom.

1. Calculate the pressure difference (P₁ – P₂) between the bottom and the top wing.
   
   Begin by listing the information given and checking that the units are consistent.
   Air velocity on bottom = v₁ = 220 m/sec
   Air velocity on top = v₂ = 250 m/sec
   Air density = 1.3 kg/m³
   Pressure difference P₁ – P₂ = ?
   Now use Bernoulli’s equation 

          
           

2. If the area of the two wings of the airplane is 10 m² what is the upward force?
   
   The result of the first part of this example told us that the net upward pressure is 9165 N/m².  Using the definition of pressure, P = F/A, we can write
   
              F = PA = 9165 N/m² x 10 m² = 91,650 N
   
   If we convert this force to pounds we obtain an upward force of approximately 20,000 lbs.  This means that this air foil in question is capable of supporting a 10 ton airplane in level flight.