Lesson 5:
Bernoulli's Theorem

 

 

Lecture

Introduction

One of the most useful laws of physics is the Law of Conservation of Energy. It simply states that energy can neither be created nor destroyed, but it can be converted from one form of energy to another. Energy is always needed to move water from one point to another and Bernoulli's Theorem describes this in a simple equation.

 

 

Piezometric Surface and Hydraulic Grade Line

The water surface in an open vertical tube attached to the bottom of a water tank will be at the same level as the water in the tank.  This is called the piezometric surface.  Now consider the arrangement of pipes in the figure below.  The pipeline is under constant pressure and with a valve closed at the end.  This is considered to be a static system since the water is not moving.  The water will rise in each of the tubes to the same height as in the water tank.  If you were to draw a line connecting each piezometric surfaces you would have what is called the hydraulic grade line.  The hydraulic grade line is always horizontal in a static system.

 

 

Now consider the next figure where we have opened the valve.  This we will call a dynamic system since the water is flowing.  The piezometric surfaces experience a drop as water travels farther along the pipeline.  This drop occurs due to energy losses resulting from the rough interior surfaces of the pipe.  These head losses are symbolized as HL in the energy equation.

 

 

Forms of Energy in Hydraulic Systems

Pressure Head

From the previous lesson on pressure we derived an equation that says that pressure at any point is determined by the height or depth of the water above it. 

This pressure is referred to as pressure head or simply head.  It is the height above some reference point (typically the ground) that gives water energy and causes it to flow.    If a tank of water is 60 ft deep, the pressure head is 60 ft.  To determine the pressure in psi we simply divide 60 ft by 2.31 psi/ft to get 26 psi.  As long as the system is static, the pressure in all horizontal points will be the same.  However, if the water is flowing (dynamic) pressure will continue to decrease along the length of the pipe.  Solving this equation for h we have h = P/γ. Pressure head is symbolized as P/γ in the energy equation which we will look at later. 

 

 

 

 

Elevation Head

Let’s now say that we have a tank that contains 60 ft of water but it is located on a hill of 100 ft above our reference point.  The 100 ft above the reference point is called the elevation head.  The pressure would still be created from our horizontal hydraulic gradeline because our system is static.  If we were to read a pressure gage at our point of reference it would show the pressure created by the water from the top of the water in the tank plus the elevation above the reference point.

 

 

This is precisely the reason many water storage tanks are placed at high elevations.  Sometimes cities also choose to draw water from distance sources to take advantage of elevation head.  More commonly, however, is the use of elevated storage tanks to provide the necessary pressure to customers.  Elevation head is symbolized as Z in the energy equation.

 

 

 

Velocity Head

Moving water possesses energy from its motion just as a locomotive.  This energy due to its motion is called kinetic energy.  In hydraulics we use the term velocity head to describe water’s kinetic energy and like the previous two forms of energy it is expressed in feet.  The formula for velocity head is



Where v is the velocity of the water and g is the acceleration due to gravity.  It is important to note that velocity head values are not used in static systems only dynamic systems. 

Velocity head becomes an important value in the calculating the energy of systems where water freely exits the end of a pipe. There is no is pressure head left at this point and it is the velocity head that continues the water’s motion.  Open channel flow is another situation where velocity head and elevation head are important factors.  Since the channel is not under pressure, only the slope of the channel and its velocity contribute to the water’s energy.         

Velocity head is also useful in calculating flow.  Pressure head and elevation head are measured by a pressure gage whereas velocity head is not.  Recall that by Pascal’s law that pressure is transmitted undiminished in all directions.  Water pressure is transferred to the pressure gage commonly installed above the pipe in question and perpendicular to flow.  However, velocity head can only be measured in the direction that the water is moving. A special gage called a Pitot tube measures both the pressure head and velocity head.  By reading the pressure measurement from this device the flow of the fluid can be calculated. It becomes a tool to go from units of pressure to feet to velocity to flow.   Another important fact is that velocity head only changes from point to point only if the diameter of the pipe changes.    

 

 

Total Head

Total energy at any point in a hydraulic system is the sum of the pressure head, elevation head, and velocity head and is called the total head.  Here it is in equation form:

or

 

If energy losses due to friction are taken into consideration the equation becomes:

 – Head Loss

or

 

Once again these energy losses come from the fluid rubbing against the pipes, valves, etc. and also the internal friction that exists between the water molecules themselves as they rub against each other. 

 

 

 

Bernoulli's Theorem

We have learned that at any point in a hydraulic system the total energy is the same as long as friction losses are ignored (Law of Conservation of Energy).  Recall, that the total energy at any point is the referred to as the Total Head which is the sum of Pressure Head, Elevation Head, and Velocity Head.  Bernoulli’s Theorem says that we can equate the energy at two different points in a steady state system. Steady state refers to the fact that no fluid has been withdrawn between the two points.  In equation form it looks like this:

 

If energy losses due to friction are taken into consideration the equation becomes:

 

Where P1 = pressure at point 1, P2 = pressure at point 2, v1 = velocity at point 1, v2=velocity at point 2, Z1=height at point 1, and Z2 = Height at point two. 

 

Examples

  1. A flow of 2,000 gpm takes place in a pipe with an inside diameter of 12 in. What is the velocity head?

 

To solve this we must first find the velocity of the fluid using the equation for flow.

Data                                                     Q = 2,000 gpm, d = 12 in                                                                  

Basic Equation                                   

Working Equation                             

Values plugged in and solved                      

 

Note: Two conversions were used, gal to feet and min to seconds.

Using the equation for velocity head we have:




 

  1. From the figure below determine the total energy at point 1 and point 2 and calculate the total energy lost in moving 2000 gpm from point 1 to point 2. Assume the fluid is water.



 

To solve this problem we will the energy equation.

Point 1

Data                                        Q = 2000 gpm, P1 = 20 psig, Z1=20 ft, P2 = 10 psig, Z2 = 4 ft
                                                Velocity = 5.7 ft/s, Velocity head = 0.5 ft (from example 1)

 

Basic Equation                       

Plug in values                          

 

 

Point 2

Basic Equation                              

Plug in values                         

 

Difference in H1 and H2            





 

  1. In the figure below a pipeline enlarges from 24 in at point 1 to 36 in at point 2.  The velocity at point 1 is 5 ft/s and the average pressure at that point is 50 psig.  If all losses sum to 2 ft, what will the pressure be at point 2 if the pipe slopes 15 downward from point 1 to point 2? 





    Data                                        P1 = 50 psig, v1= 5 ft/s, Z1 = 15 ft, Z2 = 0, HL= 2 ft

    Since there is no value for v2 we will need to solve for it first using the continuity equation.

    Basic Equation                       

    Working Equation                 

    Plug in values                         

     

    We can now solve for P2 using Bernoulli’s Equation.

     

    Basic Equation                       

    Working Equation                 

     

    Plug in Values                        

    Convert ft to psig