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- 1. FLUID MECHANICSm kg = V m3V m3 = m kgW mg g KN = = = V 1000V 1000 m 3

2. Its specific gravity (relative density) is equal to the ratio of its density to that of water at standard temperature and pressure.L L SL = = W W 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.G G SG = = ah ahwhere: At standard condition 3 W = 1000 kg/m 3 W = 9.81 KN/m 3. F - 32 1.8 F =1.8C+ 32 C = KC 273RF 460F P= KPa AdF P= KPa dAwhere: F - normal force, KN A - area, m2 4. yP3 A3A P1 A1xBCzP2 A2Fx = 0 and Fy = 0 P1A1 P3A3 sin = 0 P2A2 P3A3cos = 0 From Figure: A1 = A3sin A2 = A3cos3 41 2Eq. 3 to Eq. 1 P1 = P3 Eq. 4 to Eq. 2 P2 = P3 Therefore: P1 = P2 = P3 5. 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 6. Pgage Atmospheric pressure PvacuumPabsAbsolute ZeroPabs = Pa+ Pgage Pabs = Pa - PvacuumPabs 7. moving platevv+dv dxxvFixed 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 8. = /m2/secEv = - 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 9. Where: - surface tension, N/m - specific weight of liquid, N/m3 r radius, m h capillary rise, m rhSurface Tension of Water C 0 10h0.0742202 cos r0.07560.0728300.0712400.0696600.0662800.06261000.0589 10. FREE SURFACEh1 1hh22dP = - dh Note:Negative sign is used because pressure decreases as elevation increases and pressure increases as elevation decreases. 11. Pressure Head:P h where: p - pressure in KPa - specific weight of a fluid, KN/m3 h - pressure head in meters of fluid MANOMETERS Manometer is an instrument used in measuring gage pressure in length of some liquid column. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure. 12. Open Type Manometer Fluid ADifferential Type Manometer Fluid AOpenManometer FluidFluid BManometer Fluid 13. Determination of S using a U - TubeOpenOpenFluid A x yFluid BSAx = SBy 14. Example no. 1 A building in Makati is 84.5 m high above the street level. The required static pressure of the water line at the top of the building is 2.5 kg/cm2. What must be the pressure in KPa in the main water located 4.75 m below the street level. (1120.8 KPa) Point 1: Main water line, 4.75 m below street level Point 2: 84.5 m above street level h = h2 h1 = (84.5 + 4.75) = 89.25 m P2 = 2.5 kg/cm2 = 245.2 KPaP2 P1 P1 P1P1 (h2 h1 ) P2 (h2 h1 ) 245.2 9.81(89.25) 1,120.743 KPa 15. Example No. 2 A mercury barometer at the ground floor of a high rise hotel in Makati reads 735 mm Hg. At the same time another barometer at the top of the hotel reads 590 mmHg. Assuming air density to be constant at 1.22 kg/m3, what is the approximate height of the hotel. (1608 m) Point 1: Ground floor 1.22(9.81) KN For air : 0.012 3 h1 = 0 m 1000 m P1 = 735 mm Hg = 98 Kpa kg 1.22 3 P2 - P1 - (h2 - h1 ) Point 2: Roof Top m g KN (P2 - P1 ) h2 = h (height) h2 - h1 1000 m3 P2 = 590 mm Hg = 78.7 KPa assumin g : m g 9.81 sec2h2 - h1 hh 1608.33 meters 16. Example No. 3 The reading on a pressure gage is 1.65 MPa, and the local barometer reading is 94 KPa. Calculate the absolute pressure that is being measured in kg/cm2. (17.78 kg/cm2) Example No. 4 A storage tank contains oil with a specific gravity of 0.88 and depth of 20 m. What is the hydrostatic pressure at the bottom of the tank in kg/cm2. (1.76 kg/cm2) Example No. 5 A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the mass of oil in the tank? 17. Forces Acting on Plane Surfaces Free SurfacehpShSSMMyFC.G. C.P.C.G. C.P.yp eNNF - total hydrostatic force exerted by the fluid on any plane surface MN C.G. - center of gravity C.P. - center of pressure 18. 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 19. Forces Acting on Curved Surfaces FV Free Surface DE Vertical Projection of ABFh CACLCC.G. FhC.P. BBBhp 20. Fh = hA A = BC x L A - area of the vertical projection of AB, m2 L - length of AB perpendicular to the screen, mFV = V V = AABCDEA x L, m3 2F = Fh + Fv2 21. Hoop Tension DT F Th1m DP= h F=0 2T = F T = F/2 S = T/A A = 1tT 1 F2Tt1m 22. 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)PD S KPa 2t where: S - Bursting Stress KPa P - pressure, KPa D -inside diameter, m t - thickness, m 23. Laws of Buoyancy Any 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.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 portionWVs BFW = BF W = BVB KN BF = LVs KNW = BF W = BVB BF = LVs 24. W2.Vs BF TW = BF - T W = BVB KN BF = LVs KN W = BF - T W = BVB BF = LVswhere: 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 25. 3.TWVsBFW = BF + T W = BVB KN BF = LVs KN W = BF + T W = BVB BF = LVswhere: 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 26. 4.WTVsBFVB = VsW = BF + T W = BVB KN BF = LVs KNW = BF + T W = BVB BF = LVs 27. 5.WVsBF TVB = VsW = BF - T W = BVB KN BF = LVs KNW = BF - T W = BVB BF = LVs 28. Energy and Head Bernoullis Energy equation: 2HL = U - Q Z21z1Reference Datum (Datum Line) 29. 1. Without Energy head added or given up by the fluid (No work done bythe system or on the system:P1 v12 P2 v 2 2 + + Z1 = + + Z2 + H L 2g 2g 2. With Energy head added to the Fluid: (Work done on the systemP1 v12 P2 v 2 2 + + Z1 + h t = + + Z2 + H L 2g 2g 3. With Energy head added given up by the Fluid: (Work done by the system)P1 v12 P2 v 2 2 + + Z1 + = + + Z2 + H L + h 2g 2g Where: P pressure, KPa v velocity in m/sec Z elevation, meters + if above datum - if below datum- specific weight, KN/m3 g gravitational acceleration m/sec2 H head loss, meters 30. APPLICATION OF THE BERNOULLI'S ENERGY THEOREMNozzle Base TipQ JetP1v12 2gZ1P2v22 2gZ22v 1 2 2gHL1 2 CvQAv m3 /secwhere: Cv - velocity coefficientHL 31. Venturi MeterB. Considering Head lossP1 Q' Q'1 22Meter CoefficientManometerA. Without considering Head loss 22v1 P v2 Z1 2 2g 2g A1v1 A2 v 2 actual flow2P1 v 1 P2 v 2 Z1 2g 2g Q A1v1 A2 v 2 Q theoretica flow lZ2Q' C QZ 2 HL 32. 2 Upper ReservoirSuction Gauge Discharge GaugeGate Valve1 Lower ReservoirGate Valve 33. HtP2P1 v22v1 2gQ = Asvs = Advd m3/secWP = Q Ht KWBP2TN KW 60,0002Z2Z1HLmeters 34. PWP x 100% BPmBP x 100% MPC CWP x 100% MP P m 35. MPMP where:EI(cos) KW 10003 EI(cos) KW 1000P - 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 36. HYDRO ELECTRIC POWER PLANT1 HeadracePenstockturbine2 TailraceY Gross Head 37. 1 HeadracePenstockGeneratorY Gross HeadB Draft TubeZB2B turbine inlet Tailrace 38. Fundamental Equations 1. Net Effective Head A.Impulse Type h = Y HL Y = Z1 Z 2 Y Gross Head, meters Where: Z1 head water elevation, m Z2 tail water elevation, m B. Reaction Type h = Y HL Y = Z1 Z2hPB2vB 2gZBmetersWhere: PB Pressure at turbine inlet, KPa vB velocity at inlet, m/sec ZB turbine setting, m - specific weight of water, KN/m3 39. 2. Water Power (Fluid Power) FP = Q h KW Where: Q discharge, m3/sec 3. Brake or Shaft PowerBP2 TN KW 60,000Where:T Brake torque, N-m N number of RPM 4. Turbine EfficiencyBP e x 100% FP e eh evemWhere: eh hydraulic efficiency ev volumetric efficiency em mechanical efficiency 40. 5. Generator EfficencyggGenerator Output x 100% Brake or Shaft power GP x 100% BP6. Generator SpeedN120f RPM nWhere: N speed, RPM f frequency in cps or Hertz n no. of generator poles (usually divisible by four) 41. 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).Turbine-Pump 42. 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 mm12 43. 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 1

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