Hydraulics for engineers

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  • 1. FLUID MECHANICS m kg ρ= V m3 V m3 υ= m kg W mg ρg KN γ= = = 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. ρL γ L SL = = ρW γW 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. ρG γG SG = = ρ ah γ ah where: At standard condition 3 W = 1000 kg/m 3 W = 9.81 KN/m
  • 3. °F - 32 1.8 °F =1.8°C+ 32 °C = K C 273 R F 460 F P= KPa A dF P= KPa dA where: F - normal force, KN A - area, m2
  • 4. y P3 A3 A P1 A1 x B C z P2 A2 Fx = 0 and Fy = 0 P1A1 – P3A3 sin = 0 P2A2 – P3A3cos = 0 From Figure: A1 = A3sin A2 = A3cos 3 4 1 2 Eq. 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 Pvacuum Pabs Absolute Zero Pabs = Pa+ Pgage Pabs = Pa - Pvacuum Pabs
  • 7. 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
  • 8. = / m2/sec 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
  • 9. Where: - surface tension, N/m - specific weight of liquid, N/m3 r – radius, m h – capillary rise, m r h Surface Tension of Water C 0 10 h 0.0742 20 2σ cos θ γr 0.0756 0.0728 30 0.0712 40 0.0696 60 0.0662 80 0.0626 100 0.0589
  • 10. FREE SURFACE h1 1• h h2 2• dP = - dh Note:Negative sign is used because pressure decreases as elevation increases and pressure increases as elevation decreases.
  • 11. Pressure Head: P h γ 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 A Differential Type Manometer Fluid A Open Manometer Fluid Fluid B Manometer Fluid
  • 13. Determination of S using a U - Tube Open Open Fluid A x y Fluid B SAx = 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 KPa P2 P1 P1 P1 P1 (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 sec2 h2 - h1 h h 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 Surface hp S h S S M M y F •C.G. •C.P. •C.G. •C.P. yp e N N F - 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 D E Vertical Projection of AB F h C A C’ L C C.G. Fh C.P. B B’ B hp
  • 20. Fh = γhA A = BC x L A - area of the vertical projection of AB, m2 L - length of AB perpendicular to the screen, m FV = γV V = AABCDEA x L, m3 2 F = Fh + Fv 2
  • 21. Hoop Tension D T F T h 1m D P= h F=0 2T = F T = F/2 S = T/A A = 1t T 1 F 2 T t 1m
  • 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 portion W Vs BF W = BF W = BVB KN BF = LVs KN W = BF W = BVB BF = LVs
  • 24. W 2. Vs BF T W = BF - T W = BVB KN BF = LVs KN W = BF - T W = BVB BF = LVs 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
  • 25. 3. T W Vs BF W = BF + T W = BVB KN BF = LVs KN W = BF + T W = BVB BF = LVs 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
  • 26. 4. W T Vs BF VB = Vs W = BF + T W = BVB KN BF = LVs KN W = BF + T W = BVB BF = LVs
  • 27. 5. W Vs BF T VB = Vs W = BF - T W = BVB KN BF = LVs KN W = BF - T W = BVB BF = LVs
  • 28. Energy and Head Bernoullis Energy equation: 2 HL = U - Q Z2 1 z1 Reference 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 γ 2g γ 2g 2. With Energy head added to the Fluid: (Work done on the system P1 v12 P2 v 2 2 + + Z1 + h t = + + Z2 + H L γ 2g γ 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 γ 2g γ 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 THEOREM Nozzle Base Tip Q Jet P1 v12 2g Z1 P2 v22 2g Z2 2 v 1 2 2g HL 1 2 Cv Q Av m3 /sec where: Cv - velocity coefficient HL
  • 31. Venturi Meter B. Considering Head loss P1 γ Q' Q' 1 2 2 Meter Coefficient Manometer A. Without considering Head loss 2 2 v1 P v2 Z1 2 2g γ 2g A1v1 A2 v 2 actual flow 2 P1 v 1 P2 v 2 Z1 γ 2g γ 2g Q A1v1 A2 v 2 Q theoretica flow l Z2 Q' C Q Z 2 HL
  • 32. 2 Upper Reservoir Suction Gauge Discharge Gauge Gate Valve 1 Lower Reservoir Gate Valve
  • 33. Ht P2 P1 γ v2 2 v1 2g Q = Asvs = Advd m3/sec WP = Q Ht KW BP 2πTN KW 60,000 2 Z2 Z1 HL meters
  • 34. ηP WP x 100% BP ηm BP x 100% MP ηC ηC WP x 100% MP ηP ηm
  • 35. MP MP where: EI(cosθ) KW 1000 3 EI(cosθ) KW 1000 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
  • 36. HYDRO ELECTRIC POWER PLANT 1 Headrace Penstock turbine 2 Tailrace Y – Gross Head
  • 37. 1 Headrace Penstock Generator Y – Gross Head B Draft Tube ZB 2 B – 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 –Z2 h PB 2 vB 2g ZB meters Where: 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 Power BP 2 TN KW 60,000 Where: T – Brake torque, N-m N – number of RPM 4. Turbine Efficiency BP e x 100% FP e eh evem Where: eh – hydraulic efficiency ev – volumetric efficiency em – mechanical efficiency
  • 40. 5. Generator Efficency g g Generator Output x 100% Brake or Shaft power GP x 100% BP 6. Generator Speed N 120f RPM n Where: 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 mm 1 2
  • 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 150 L/sec? Neglect loss of head. (h=273 mm)
  • 44. 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