whenthesteady-statedesignfails!

whenthesteady-statedesignfails!内容摘要：

at lake surface Period of Oscillations pAAgLT p itw e t 2 222 24/317 02mmsmmT plan view area of wet pit (m2) 24 pipeline length (m) 3170 inner diameter of pipe (m) gravity (m/s2) T = 424 s 2 1 0120 200 400 600 800 1000 1200t i m e ( s )Q (m3/s)432101234z (m)Q zPendulum Period? 2 LT gTransients In previous example we assumed that the velocity was the same everywhere in the pipe We did not consider pressibility of water or elasticity of the pipe In the next example water pressibility and pipe elasticity will be central V V2Valve Closure in Pipeline Sudden valve closure at t = 0 causes change in discharge at the valve What will make the fluid slow down?____ Instantaneous change would require __________ Impossible to stop all the fluid instantaneously infinite force What do you think happens? ↑p at valve Transients: Distributed System Tools Conservation of mass Conservation of momentum Conservation of energy We‟d like to know pressure change rigid walls elastic walls propagation speed of pressure wave time history of transient Pressure change due to velocity change velocity density pressure unsteady flow steady flow P00V0 VV 0P0  P0  P00P0  P0  aV0 V0  VHGL V0 a V0  V  aMomentum Equation 2121 ppxx FFMM x 12111 AVM x  22222 AVM x   221112111 ApApVVAV aV0 V0  VHGL 222111 AVAV  1 2 Mass conservation A1  A2 p = p2 p1 pVV 11sspp FFFWMM  2121Neglect head loss! Magnitude of Pressure Wave pVV 11aV0 V0  V1 2 1V aV 0Vap  aVHgDD=0Va pHgD = DDecrease in V causes a(n) _______ in HGL. increase Propagation Speed: Rigid Walls Conservation of mass aV0 V0 V0 0  1)(000aVV00 )( aVVSolve for V ))(()( 0000   VaVAaVAPropagation Speed: Rigid Walls aV0 V0 V0 0 momentum VaVp  )( 00aV 0 0  2ap00 )( aVV mass 0200 )( aVpNeed a relationship between pressure and density! Propagation Speed: Rigid Walls  pK pa 2Ka definition of bulk modulus of elasticity Example: Find the speed of a pressure wave in a water pipeline assuming rigid walls. G Pa 2 .2K3K g /m 1 0 0 0m / s 1480100010 x 9 aspeed of sound in water (for water) Propagation Speed: Elastic Walls aV0 V0 V0 0 0Ka  D t = thickness of thin walled pipe E = bulk modulus of elasticity for pipe Additional parameters D = diameter of pipe tDEKKa10effect of water pressibility effect of pipe elasticity solution Propagation Speed: Elastic Walls Example: How long does it take for a pressure wave to travel 500 m after a rapid valve closure in a 1 m diam。