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Fluid Mechanics - Bernoulli’s Equation

Bernoulli’s equation is an expression of the Law of Energy Conservation for an in viscid fluid stream, named after Daniel Bernoulli1. It states that the sum total energy at any point in a passive fluid stream (i.e. no pumps or other energy-imparting machines in the flow path) must be constant. Two versions of the equation are shown here:



  z = Height of fluid (from a common reference point, usually ground level)

  ρ = Mass density of fluid

  γ = Weight density of fluid (γ = ρg)

  g = Acceleration of gravity

  v = Velocity of fluid

  P = Pressure of fluid


Each of the three terms in Bernoulli’s equation is an expression of a different kind of energy, commonly referred to as head:


Elevation and Pressure heads are potential forms of energy, while Velocity head is a kinetic form of energy. Note how the elevation and velocity head terms so closely resemble the formulae for potential and kinetic energy of solid objects:



The only real differences between the solid-object and fluid formulae for energies is the use of mass density (ρ) for fluids instead of mass (m) for solids, and the arbitrary use of the variable z for height instead of h. In essence, the elevation and velocity head terms within Bernoulli’s equation come from the assumption of individual fluid molecules behaving as miniscule solid masses.

It is very important to maintain consistent units of measurement when using Bernoulli’s equation! Each of the three energy terms (elevation, velocity, and pressure) must possess the exact same units if they are to add appropriately2. Here is an example of dimensional analysis applied to the first version of Bernoulli’s equation (using British units):



As you can see, both the first and second terms of the equation (elevation and velocity heads) bear the same unit of slugs per foot-second squared after all the “feet” are canceled. The third term (pressure head) does not appear as though its units agree with the other two terms, until you realize that the unit definition of a “pound” is a slug of mass multiplied by the acceleration of gravity in feet per second squared, following Newton’s Second Law of motion (F = ma):



Once we make this substitution into the pressure head term, the units are revealed to be the same as the other two terms, slugs per foot-second squared:



In order for our British units to be consistent here, we must use feet for elevation, slugs per cubic foot for mass density, feet per second squared for acceleration, feet per second for velocity, and pounds per square foot for pressure. If one wished to use the more common pressure unit of PSI (pounds per square inch) with Bernoulli’s equation instead of PSF (pounds per square foot), all the other units would have to change accordingly: elevation in inches, mass density in slugs per cubic inch, acceleration in inches per second squared, and velocity in inches per second.

Just for fun, we can try dimensional analysis on the second version of Bernoulli’s equation, this time using metric units:



Here, we see that all three terms end up being cast in simple units of meters. That is, the fluid’s elevation, velocity, and pressure heads are all expressed as simple elevations. In order for our metric units to be consistent here, we must use meters for elevation, meters per second for velocity, meters per second squared for acceleration, pascals (newtons per square meter) for pressure, and newtons per cubic meter for weight density.

The following example shows how we would apply Bernoulli’s equation to the solution of pressure at a point in a water piping system, assuming no frictional losses at any point:



Water has a nominal density of 62.4 pounds per cubic foot, but this is weight density (γ) and not mass density (ρ). If we wish to use the form of Bernoulli’s equation where all terms are in units of pressure (), we must have a value of ρ for water.

The relationship between weight density γ and mass density ρ is the exact same relationship between weight (FW) and mass (m) in a gravitational field (g). Newton’s Second Law equation relating force to mass and acceleration (F = ma) works well to relate weight to mass and gravitational acceleration:



Dividing both sides of this equation by volumetric units (V ) (e.g. cubic feet) gives us our relationship between γ and ρ:



Water has a weight density of 62.4 pounds per cubic foot in Earth gravity (32.2 feet per second squared), so:



Now we are ready to begin our Bernoulli’s equation calculations. Since we have the freedom to choose any arbitrary point in the piping system as our reference elevation (z = 0), we will set the location of the first pressure gauge as this reference height, so the second pressure gauge will have a positive elevation value of 3 feet. First, calculating the values of all terms (elevation, velocity, and pressure) at the first point, near the discharge of the pump:


  z1ρg    (0 ft) (1.94 slugs/ft3) (32.2 ft/s2
  0 lb/ft
  (11 ft/s)2 (1.94 slugs/ft3) / 2 
  117.4 lb/ft
  P     (46 lb/in2) (144 in2/1 ft2
  6624 lb/ft
  0 lb/ft2 + 6.56 lb/ft2 + 6624 lb/ft2  
  6741.4 lb/ft


Note the absolutely consistent use of units: all units of distance are feet. All units of time are seconds. Failure to maintain consistency of units will result in (often severely) incorrect results!3

Next, we will calculate the values of the elevation and velocity heads at the location of the second pressure gauge. Here, the pressure is unknown, but the elevation is given as 3 feet higher than the first gauge, and the velocity may be calculated by pipe size. We know that the pipe here is 6 inches in diameter, while it is 10 inches in diameter where the velocity is 11 feet per second. Since area varies with the square of the diameter, and velocity varies inversely with area, we can tell that the velocity at the second pressure gauge will be 2.78 times greater than at the first pressure gauge:


Tabulating our calculations and results:

  Head     Calculation 
  (3 ft) (1.94 slugs/ft3) (32.2 ft/s2)  
  187.4 lb/ft
  (30.56 ft/s)2 (1.94 slugs/ft3) / 2 
  905.6 lb/ft3 
  187.4 lb/ft2 + 905.6 lb/ft2 + P
  1093 lb/ft2 + P

Knowing that the total head calculated at the first location was 6741.4 lb/ft2, and the Conservation of Energy requires total heads at both locations be equal (assuming no energy lost to fluid friction along the way), P2 must be equal to:



Converting pounds per square foot into the more customary unit of pounds per square inch (PSI):



Note how much lower the pressure is at the second gauge than at the first: 39.2 PSI compared to 46 PSI: almost a 7 PSI decrease in pressure. Note also how little vertical distance separates the two gauges: only 3 feet. Clearly, the change in elevation between those two points in insufficient to account for the large loss in pressure4. Given a 3 foot difference in elevation, one would expect a pressure reduction of about 1.3 PSI for a static column of water, but what we’re seeing in this piping system is a pressure drop of nearly 7 PSI. The difference is due to an exchange of energy from potential to kinetic form, as the fluid enters a much narrower pipe (6 inches instead of 10) and must increase velocity.

Furthermore, if we were to increase the flow rate discharged from the pump, resulting in even more velocity through the narrow pipe, pressure at P2 might even drop lower than atmospheric. In other words, Bernoulli’s equation tells us we can actually produce a vacuum by accelerating a fluid through a constriction. This principle is widely used in industry with devices known as eductors: tapered tubes through which fluid flows at high velocity to produce a vacuum at the throat.

Some eductors use high-velocity steam as the working fluid to produce significant vacuums. Other eductors use process liquid flow, such as the eductor shown in this next photograph where wastewater flow creates a vacuum to draw gaseous chlorine into the stream for biological disinfection:



Here, the eductor helps fulfill an important safety function. By creating a vacuum to draw dangerous chlorine gas from the supply tank into the water stream, the chlorine gas piping may be continuously maintained at a slightly negative pressure throughout. If ever a leak were to develop in the chlorine system, this vacuum would cause ambient air to enter the chlorine pipe rather than toxic chlorine gas to exit the pipe, making a leak far less dangerous than if the chlorine gas piping were maintained in a pressurized state.


1According to Ven Te Chow in Open Channel Hydraulics, who quotes from Hunter Rouse and Simon Ince’s workHistory of Hydraulics, Bernoulli’s equation was first formulated by the great mathematician Leonhard Euler and made popular by Julius Weisbach, not by Daniel Bernoulli himself.

2Surely you’ve heard the expression, “Apples and Oranges don’t add up.” Well, pounds per square inch and pounds per square foot don’t add up either!

3It is entirely possible to perform all our calculations using inches and/or minutes as the primary units instead of feet and seconds. The only caveat is that all units throughout all terms of Bernoulli’s equation must be consistent. This means we would also have to express mass density in units of slugs per cubic inch, the acceleration of gravity in inches per second squared (or inches per minute squared), and velocity in units of inches per second (or inches per minute). The only real benefit of doing this is that pressure would remain in the more customary units of pounds per square inch. My personal preference is to do all calculations using units of feet and seconds, then convert pressures in units of PSF to units of PSI at the very end.

4A simple approximation for pressure loss due to elevation gain is approximately 1 PSI for every 2 vertical feet of water (1 PSI for every 27.68 inches to be more exact).

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Comments (1)Add Comment
Thank you
written by Aditya, March 24, 2016
Such a nice explanation of a very important equation! The dimensional analysis and the discussion surrounding it was very informative and useful. Thanks for the post.

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