This document discusses key concepts in hydraulics and fluid mechanics. It defines important fluid properties like density, specific volume, viscosity, and surface tension. It describes Pascal's law and factors that influence pressure like elevation and atmospheric pressure. Key concepts in fluid flow are also summarized like Bernoulli's equation, venturi meters, orifices, and pumps. The document provides equations for calculating forces, pressure, discharge, and efficiency in hydraulic systems.

1. Hydraulics By: Engr. Yuri G. Melliza

2. FLUID MECHANICS Properties of Fluids 1. Density () 2.Specific Volume () 3.Specific Weight ()

3. 4. Specific Gravity or Relative Density For Liquids: Its specific gravity (relative density) is equal to the ratio of its density to that of water at standard temperature and pressure. For Gases:Its specific gravity (relative density) is equal to the ratio of its density to that of either air or hydrogen at some specified temperature and pressure. where: At standard condition W = 1000 kg/m3 W = 9.81 KN/m3

4. 5. Temperature 6. Pressure If a force dF acts on an infinitesimal area dA, the intensity of pressure is, where: F - normal force, KN A - area, m2

5. PASCAL’S LAW: At any point in a homogeneous fluid at rest the pressures are the same in all directions. y P3 A3 A P1 A1 x  B C z P2 A2 Eq. 3 to Eq. 1 P1 = P3 Eq. 4 to Eq. 2 P2 = P3 Therefore: P1 = P2 = P3 Fx = 0 and Fy = 0 P1A1 – P3A3 sin  = 0  1 P2A2 – P3A3cos = 0  2 From Figure: A1 = A3sin   3 A2 = A3cos   4

6. Atmospheric pressure:The pressure exerted by the atmosphere. At sea level condition: Pa = 101.325 KPa = .101325 Mpa = 1.01325Bar = 760 mm Hg = 10.33 m H2O = 1.133 kg/cm2 = 14.7 psi = 29.921 in Hg = 33.878 ft H2O Absolute and Gage Pressure Absolute Pressure:is the pressure measured referred to absolute zero and using absolute zero as the base. Gage Pressure:is the pressure measured referred to atmospheric pressure, and using atmospheric pressure as the base

7. Pgage Atmospheric pressure Pvacuum Pabs Pabs Absolute Zero Pabs = Pa+ Pgage Pabs = Pa - Pvacuum

8. 7. Viscosity: A property that determines the amount of its resistance to shearing stress. moving plate v v+dv dx x v Fixed plate S dv/dx S = (dv/dx) S = (v/x)  = S/(v/x) where:  - absolute or dynamic viscosity in Pa-sec S - shearing stress in Pascal v - velocity in m/sec x -distance in meters

9. 8. Kinematic Viscosity: It is the ratio of the absolute or dynamic viscosity to mass density.  = / m2/sec 9. Elasticity: If the pressure is applied to a fluid, it contracts,ifthe pressure is released it expands, the elasticity of a fluid is related to the amount of deformat- ion (contraction or expansion) for a given pressure change. Quantitatively, the degree of elasticity is equal to; Ev = - dP/(dV/V) Where negative sign is used because dV/V is negative for a positive dP. Ev = dP/(d/) because -dV/V = d/ where: Ev - bulk modulus of elasticity, KPa dV - is the incremental volume change V - is the original volume dP - is the incremental pressure change

10.     r h 10. Surface Tension: Capillarity Where:  - surface tension, N/m  - specific weight of liquid, N/m3 r – radius, m h – capillary rise, m Surface Tension of Water

11. FREE SURFACE h1 1• h2 h 2• Variation of Pressure with Elevation dP = - dh Note:Negative sign is used because pressure decreases as elevation increases and pressure increases as elevation decreases.

13. Fluid A Fluid A Open Fluid B Manometer Fluid Manometer Fluid Open Type Manometer Differential Type Manometer

14. Open Open Fluid A x y Fluid B Determination of S using a U - Tube SAx = SBy

15. S S S Free Surface  M M hp F •C.G. •C.G. •C.P. •C.P. yp e N N Forces Acting on Plane Surfaces F - total hydrostatic force exerted by the fluid on any plane surface MN C.G. - center of gravity C.P. - center of pressure

16. where: Ig- moment of inertia of any plane surface MN with respect to the axis at its centroids Ss - statical moment of inertia of any plane surface MN with respect to the axis SS not lying on its plane e - perpendicular distance between CG and CP

17. Forces Acting on Curved Surfaces FV Free Surface D E Vertical Projection of AB F C’ L C C A C.G. Fh C.P. B B B’

18. A = BC x L A - area of the vertical projection of AB, m2 L - length of AB perpendicular to the screen, m V = AABCDEA x L, m3

19. D h 1 m D P = h T T F F 1 m T t T Hoop Tension F = 0 2T = F T = F/2  1 S = T/A A = 1t  2

20. S = F/2(1t)  3 From figure, on the vertical projection the pressure P; P = F/A A = 1D F = P(1D)  4 substituting eq, 4 to eq. 3 S = P(1D)/2(1t) where: S - Bursting Stress KPa P - pressure, KPa D -inside diameter, m t - thickness, m

21. Laws of BuoyancyAny body partly or wholly submerged in a liquid is subjected to a buoyant or upward force which is equal to the weight of the liquid displaced. 1. W where: W - weight of body, kg, KN BF - buoyant force, kg, KN  - specific weight, KN/m3  - density, kg/m3 V - volume, m3 Subscript: B - refers to the body L - refers to the liquid s - submerged portion Vs BF W = BF W = BVB BF = LVs W = BF W = BVB KN BF = LVs KN

22. 2. BF T Vs W where: W - weight of body, kg, KN BF - buoyant force, kg, KN T - external force T, kg, KN  - specific weight, KN/m3  - density, kg/m3 V - volume, m3 Subscript: B - refers to the body L - refers to the liquid s - submerged portion W = BF - T W = BVB KN BF = LVs KN W = BF - T W = BVB BF = LVs

23. T W BF Vs 3. where: W - weight of body, kg, KN BF - buoyant force, kg, KN T - external force T, kg, KN  - specific weight, KN/m3  - density, kg/m3 V - volume, m3 Subscript: B - refers to the body L - refers to the liquid s - submerged portion W = BF + T W = BVB g BF = LVs g W = BF + T W = BVB g BF = LVs g

24. W T Vs BF 4. VB = Vs W = BF + T W = BVB g BF = LVs g W = BF + T W = BVB BF = LVs

25. W Vs BF T 5. VB = Vs W = BF - T W = BVB g BF = LVs g W = BF - T W = BVB BF = LVs

26. Energy and Head Bernoullis Energy equation: 2 HL = U - Q Z2 1 z1 Reference Datum (Datum Line)

27. 1. Without Energy head added or given up by the fluid (No work done by the system or on the system: 2. With Energy head added to the Fluid: (Work done on the system) 3. With Energy head added given up by the Fluid: (Work done by the system) Where: P – pressure, KPa - specific weight, KN/m3 v – velocity in m/sec g – gravitational acceleration Z – elevation, meters m/sec2 + if above datum H – head loss, meters - if below datum

28. APPLICATION OF THE BERNOULLI'S ENERGY THEOREM Nozzle Base Tip Q Jet where: Cv - velocity coefficient

29. Venturi Meter B. Considering Head loss inlet 1 throat exit 2 Meter Coefficient Manometer A. Without considering Head loss

30. Orifice: An orifice is an any opening with a closed perimeter Without considering Head Loss and from figure: Z1 - Z2 = h, therefore 1 a h Vena Contracta By applying Bernoulli's Energy theorem: Let v2 = vt 2 a where: vt - theoretical velocity, m/sec h - head producing the flow, meters g - gravitational acceleration, m/sec2 But P1 = P2 = Pa and v1is negligible, then

31. COEFFICIENT OF DISCHARGE(Cd) COEFFICIENT OF VELOCITY (Cv) COEFFICIENT OF CONTRACTION (Cc) where: v' - actual velocity vt - theoretical velocity a - area of jet at vena contracta A - area of orifice Q' - actual flow Q - theoretical flow Cv - coefficient of velocity Cc - coefficient of contraction Cd - coefficient of discharge

32. Jet Trajctory 2 d v sin v  1 3 v cos R = v cos (2t) If the jet is flowing from a vertical orifice and the jet is initially horizontal where vx = v. v = vx y x

33. Upper Reservoir Suction Gauge Discharge Gauge Lower Reservoir Gate Valve Gate Valve PUMPS: It is a steady-state, steady-flow machine in which mechanical work is added to the fluid in order to transport the liquid from one point to another point of higher pressure.

34. 1. TOTAL DYNAMIC HEAD 4. BRAKE or SHAFT POWER FUNDAMENTAL EQUATIONS 2. DISCHARGE or CAPACITY Q = Asvs = Advdm3/sec 3. WATER POWER or FLUID POWER WP = QHtKW

35. 5. PUMP EFFICIENCY 6. MOTOR EFFICIENCY 7. COMBINED PUMP-MOTOR EFFICIENCY

37. For 3 Phase Motorwhere: P - pressure in KPa T - brake torque, N-m v - velocity, m/sec N - no. of RPM  - specific weight of liquid, KN/m3 WP - fluid power, KW Z - elevation, meters BP - brake power, KW g - gravitational acceleration, m/sec2 MP - power input to HL - total head loss, meters motor, KW E - energy, Volts I - current, amperes (cos) - power factor

38. Penstock Headrace turbine Y – Gross Head Tailrace HYDRO ELECTRIC POWER PLANT A. Impulse Type turbine (Pelton Type) 1 2

39. 1 Headrace Generator Penstock Y – Gross Head B ZB Draft Tube 2 B – turbine inlet Tailrace B. Reaction Type turbine (Francis Type)

40. Fundamental Equations Where: PB – Pressure at turbine inlet, KPa vB – velocity at inlet, m/sec ZB – turbine setting, m  - specific weight of water, KN/m3 1. Net Effective Head Impulse Type h = Y – HL Y = Z1 – Z2 Y – Gross Head, meters Where: Z1 – head water elevation, m Z2 – tail water elevation, m B. Reaction Type h = Y – HL Y = Z1 –Z2

41. 2. Water Power (Fluid Power) FP = Qh KW Where: Q – discharge, m3/sec 3. Brake or Shaft Power Where: T – Brake torque, N-m N – number of RPM Where: eh – hydraulic efficiency ev – volumetric efficiency em – mechanical efficiency 4. Turbine Efficiency

42. 5. Generator Efficency 6. Generator Speed Where: N – speed, RPM f – frequency in cps or Hertz n – no. of generator poles (usually divisible by four)

43. Turbine-Pump Pump-Storage Hydroelectric power plant: During power generation the turbine-pump acts as a turbine and during off-peak period it acts as a pump, pumping water from the lower pool (tailrace) back to the upper pool (headrace).

44. A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the same elevation. The pressure at point 1 is 200 KPa. Q = 30 L/sec flowing from 1 to 2, and the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at 2 if the liquid is oil with S = 0.80. (174.2 KPa) 300 mm 100 mm 1 2

45. A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm water main. In a differential gage partly filled with mercury (the remainder of the tube being filled with water) and connected with the meter at the inlet and at the throat, what would be the difference in level of the mercury columns if the discharge is 150 L/sec? Neglect loss of head. (h=273 mm)

46. The liquid in the figure has a specific gravity of 1.5. The gas pressure PA is 35 KPa and PB is -15 KPa. The orifice is 100 mm in diameter with Cd = Cv = 0.95. Determine the velocity in the jet and the discharge when h = 1.2. (9.025 m/sec; 0.071 m3/sec) PA 1.2 m PB