Sammanfattning hydraulik Bernoullis ekvation Rörelsemängdsekvationen Energiekvation applikationer Rörströmning Friktionskoefficient, Moody s diagram Pumpsystem
BERNOULLI S EQUATION 2 p V z H const. Quantity Name Measure of 2g Bernoulli s equation is a useful relationship between pressure, p, velocity, V, and geometric height, z, above a reference plane (datum). H Energinivå Total energi P/γ Tyckhöjd tryckenergi Z Geometrisk höjd Lägesenergi H: energy head (m) z: elevation head above datum (m) V: velocity (m/s) g: gravity acceleration (m/s 2 ) p: pressure (Pa) γ: weight density for the flowing fluid (N/m 3 ) V 2 /(2g) Hastighetshöjd Rörelseenergi p z piezometri c head or H. G. L Hydraulic Grade Line =>Trycknivå γ = w = ρ g
Grundläggande ekvationer Pitotrör (manometri, peizometer) Ångtryck, kavitation Energi- och trycklinjer Strömning med energiförlust Bernoullis ekvation (K.E.; B.E.; E.E.)
Momentum Equation (RRM ekv.) F Q( V V ) ( x direction) X XOUT, X, IN F Q( V V ) ( y direction) Y Y, OUT Y, IN F Q ( V V ) ( z direction) Z Z, OUT Z, IN F : Sum of all external forces acting on the control volume (like the streamtube). : Density of fluid Q: Flowrate V OUT : Velocity out of the control volume V IN : Velocity in to the control volume : Correction coefficient for momentum (non-uniform velocity)
Methodology Using The Momentum Equation For A Fluid Flow Problem 1. Define an appropriate control volume (Draw picture) 2. Define coordinate axes (For and ; find out + and -) 3. Determine all forces (magnitude and direction) acting on the control volume (Draw picture) 4. Determine flowrate, outflow and inflow velocities to the control volume (if not given, use continuity equation and Energy / Bernoulli equation) 5. Solve momentum equation: ΣF = ρq(v 2 - V 1 ) = Total forces acting on CV F V
Rörströmning I--IV Energiförluster vid rörströmning Lokala energiförluster Seriekopplade rörledningar Parallellkopplade rörledningar Trereservoirproblemet Allmänfriktionslag Turbulent rörströmning Friktionskoefficient, Moody s diagram Icke-cirkulära rör
(total energi) (trycknivå) Bell-mouth = Trattformig
Pipe systems branched pipe systems J Solution 3 Possible flow situations: 1) From reservoir 1 and 2 to reservoir 3 2) From reservoir 1 to reservoir 2 and 3 3) From reservoir 1 to 3 (Q 2 = 0) For the situation as shown: Energy equation H J = P J /w + z J + V 2 J/2g h f1 + h local,1 = z 1 H J h f2 + h local,2 = z 2 H J h f3 + h local,3 = H J z 3 Continuity equation Q 3 = Q 1 + Q 2 As H J (H J is total head at J) is initially unknown, a method of solution is as follows: 1) Guess H J 2) Calculate Q 1, Q 2, and Q 3 3) If Q 1 + Q 2 = Q 3, then the solution is correct 4) If Q 1 + Q 2 Q 3, then return to 1).
Ett vattenflöde på 60 l/s strömmar genom rörledningen i Figur 2.5 a) Om vattnet stiger 3.0 m ovan rörets centrum i Pitot-röret vid B, hur högt stiger vattnet i piezometern vid A? a) Hur stort är det statiska trycket (i Pa) vid rörcentrum i B? Försumma alla energiförluster. + 3.0 m Figur 2.5
Vad är skillnad mellan A o B? p w 2 V z H konst. g 2 A
I figuren visas en Venturimeter (flödesmätare). Vilket vattenflöde går genom rörledningen för det aktuella differentialmanometerutslaget i ledningen? Relativa densiteten för kvicksilver är S Hg = 13.56. Försumma eventuella förluster mellan 1 och 2.
2.5 En vattenstråle avbördas med ett konstant flöde av 35 l/s från den övre tanken, se figuren nedan. Om jetdiametern vid sektion 1 är 10 cm, vilka krafter mäts av vågarna vid A och B. Anta att en tom tank väger 135 kg och att tankens ytarea är 0.4 m 2. H = 3 m och h = 0.3 m.
1 z V t g t 2 z0 2 (Based on Cast Parable) V V g t z z0
I37: This corrugated ramp is used as an energy dissipator in a twodimensional open channel flow. For a flowrate of 5.4 m 3 /(s m) calculate the head lost, the power dissipated, and the horizontal component of force exerted by the water on the ramp. 2 X 1
The hydraulic characteristics for the pipe system, H syst, is obtained from the energy equation L Q 2 H z h z f syst losses D 2 ga 2 (local losses neglected in this case) H syst states how much energy that is needed to transport 1 kg of water from A to B H p states how much energy the pump can provide to the water When the pump is introduced in the pipe system the flowrate and pump head will adjust so that H syst = H p H syst : Systemkurvan består av skillnad i nivå + friktionsförluster
PARALLEL PUMPING To deal with cases where you have pumps operating in parallel you can consider them as being replaced by a fictive equivalent pump with a pump curve obtained by horizontal addition of the single pumps pump curves Horisontal addition
PUMPS IN SERIES To deal with cases where you have pumps operating in series you can consider them as being replaced by a fictive equivalent pump with a pump curve obtained by vertical addition of the single pumps pump curves Vertical addition
J30: Water (20 C) is pumped between two reservoirs through two identical, parallel pipes each with a diameter of 0.2 m, length 1000 m, and equivalent sand roughness of 4 10-4 m. a) What flow is expected through the pump? b) How much energy (kwh) is needed to pump 1 m 3 of water? The efficiency of the pump η = 0.75