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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
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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:
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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
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4 The maximum velocity an object can attain is when all potential energy is converted into kinetic energy
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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:
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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:
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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
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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
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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!)
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10 In an hydraulic system we usually assume that the datum is constant leading to: Energy per unit mass form Pressure form ‘Head’ form
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11 Example: Mechanical device (cylinder / motor) assume changes in kinetic and internal energy are negligible Assumptions
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12 Example: Orifice: No change in potential nor in internal energy Assumptions
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13 Example: Velocity a large tank with an opening at the bottom gives rise to a jet Assumptions
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14 Example: Velocity a large tank with an opening at the bottom gives rise to a jet Assumptions
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15 Example: Heat produced due to friction in a pipe: Assumptions
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16 Example: Venturi tube, assume incompressible flow meaning we can apply the continuity equation Assumptions
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17 Example: Venturi tube, assume incompressible flow meaning we can apply the continuity equation
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18 If the pressure difference would be measured with a manometer filled with a fluid with a density the equation would be:
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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
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20 Example: Pressure drop through an orifice Assumptions Continuity equation
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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
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22 Example: Pressure drop through an orifice
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23 The flow rate is the velocity times the cross sectional area of the conduit:
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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
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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
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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
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29 Example use of pressure form Bernoulli If pressure gauges are used, apply Bernoulli’s equation in pressure form and combine to obtain:
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30 Example use of head form Bernoulli If pressure gauges are used, apply Bernoulli’s equation in head form and combine to obtain:
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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
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32 Bernoulli’s Law and Applications: The End
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