1 Dept. of Agricultural & Biological Engineering University of Illinois TSM 363 Fluid Power Systems TSM 363 Fluid Power Systems Bernoulli’s Law and Applications Tony Grift

2 Potential and kinetic energy of solid objects: calculate what it took to get them in the state they are in Energy is force through a distance: potential energy: When we drop this object the equations of motion are (assume constant gravitational pull) If the intial position and velocity are zero this reduces to:

3 Potential and kinetic energy of solid objects: calculate what it took to get them in the state they are in Bring an object up to speed by applying a constant force through a distance x At some time t, the velocity will be The distance the object has traveled is now: Substitution of the time t leads to: The total energy this took was

4 The maximum velocity an object can attain is when all potential energy is converted into kinetic energy

5 There are three forms of energy in a solid object: Potential Kinetic Internal Where is the specific heat in and T is the absolute temperature in Kelvin Since energy can neither be created nor destroyed, there is an energy balance:

6 In fluids there is another form of energy: pressure Power in an hydraulic system was pressure times flow rate or: Let’s convert energy per unit of mass as follows: So power can be translated into pressure per unit mass as follows:

7 There are four forms of energy in a fluid object, here all are given on a unit mass basis: Pressure Potential Kinetic Internal Conservation of energy now gives

8 This is the conservation of energy or Bernoulli’s law on a unit mass basis If useful work is done by the system (that is what hydraulics is all about) we need to add a work term per kg of fluid We can also convert this equation in a pressure form by multiplying by

9 Multiplying left and right gives various ‘forms of the equation’ but the energy conservation remains key There is the ‘head’ form, divide by which is the specific weight (not mass!)

10 In an hydraulic system we usually assume that the datum is constant leading to: Energy per unit mass form Pressure form ‘Head’ form

11 Example: Mechanical device (cylinder / motor) assume changes in kinetic and internal energy are negligible Assumptions

12 Example: Orifice: No change in potential nor in internal energy Assumptions

13 Example: Velocity a large tank with an opening at the bottom gives rise to a jet Assumptions

14 Example: Velocity a large tank with an opening at the bottom gives rise to a jet Assumptions

15 Example: Heat produced due to friction in a pipe: Assumptions

16 Example: Venturi tube, assume incompressible flow meaning we can apply the continuity equation Assumptions

17 Example: Venturi tube, assume incompressible flow meaning we can apply the continuity equation

18 If the pressure difference would be measured with a manometer filled with a fluid with a density the equation would be:

19 Pitot tubes are widely used to measure the velocity of a fluid (or a gas like in aircraft). The pressure difference is obtained from two ports, the dynamic (where the fluid has a velocity) and the static (drilled in the side wall) where the fluid is assumed without velocity. Assumptions When a manometer is used with fluid density

20 Example: Pressure drop through an orifice Assumptions Continuity equation

21 Example: Pressure drop through an orifice A fluid can not follow a 90 degree angle but will form a horn like jet with a reduced diameter and area. This is called the Vena contracta phenomenon characterized by a constant

22 Example: Pressure drop through an orifice

23 The flow rate is the velocity times the cross sectional area of the conduit:

24 Only valid for high Reynolds numbers (ignore viscous effects). For small Reynolds numbers is the discharge coefficient is introduced: The coefficient K is determined experimentally Orifice Equation

25 This equation can be used to estimate pressure drops across any kind of flow obstruction The value of K is a function of the Reynolds number

26 For a given flow rate The value of K can now be obtained from measured relationships as represented in the figure on the next slide for various ratios of the conduit diameter D and the orifice diameter d using the bottom scale and the vertical lines.

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28 When the flow rate q has to be determined from a certain measured head or pressure drop, the factor K can not be determined because it depends on the flow rate q itself. The slanted lines represent this relationship. The value of K can be computed using the top scale and the slanted lines. Substitution of K in now gives the flow rate

29 Example use of pressure form Bernoulli If pressure gauges are used, apply Bernoulli’s equation in pressure form and combine to obtain:

30 Example use of head form Bernoulli If pressure gauges are used, apply Bernoulli’s equation in head form and combine to obtain:

31 Assumptions in the Bernoulli derivation Fluid density is constant: fluid is incompressible In solid object case no friction was assumed: the only resistance to acceleration is inertia Bernoulli’s law only applies if no internal fluid shear friction is assumed in other words in a non-viscous fluid The flow is assume steady, that means the flow regime does not change over time

32 Bernoulli’s Law and Applications: The End