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.7.3.1.THE EULER EQUATIONThe Euler equation is derived by applying Newton's second law to a small cylindricalcontrol volume of fluid at the pipe centerline, as shown in Fig.7.4.The resultingequation will apply to one-dimensional flow along the pipeline when we disregardvariations in fluid or flow properties across the cross section.Further, the equation willapply to flows of both constant and variable density, so it is valid for both rigid-columnand water hammer flows.( p + ∂p/∂s ds) dAstreamlinesdsτ π d ds.dvθdWp dAFigure 7.4 A cylindrical fluid element with all forces shown.Along the streamline direction s, Newton's second law givesdvFs∑ = mas = m(7.14)dtwhere m is the fluid mass in the cylindrical fluid parcel.The term dv/dt is in general thetotal or substantial derivative of the fluid velocity.Substituting the applied forces intothis equation and writing the mass in terms of density and volume results indvpdA − p + ∂ p ds( ) = dW(7.15)∂ s dA − dW sinθ − τ π d dsgdtIf we divide by dW to produce a non-dimensional equation that is written per unit weight,and the local pipe slope is expressed in terms of distance and elevation along the pipe, wearrive at the one-dimensional Euler equation− 1 ∂ p − ∂ z − 4τ = 1 dv(7.16)γ ∂ s∂ sγ dg dt© 2000 by CRC Press LLCIf we now expand the cross-sectional area of the parcel to fill the pipe cross section andintroduce the average velocity V, we obtain a more useful equation:− 1 ∂ p − ∂ z − 4τ o = 1 dV(7.17)γ ∂ s∂ sγ Dg dtHere τ o is the shear stress at the wall.Because the wall shear stress is usually not ofprimary interest and because we will be working almost entirely with cylindrical pipes, weprefer to express the shear stress in terms of the Darcy-Weisbach friction factor f asτ o = 1 f ρ V V(7.18)8The form of the velocity representation in this equation is desirable because it preserves theproper direction of the shear force whenever the flow reverses direction.With the substitution of Eq.7.18 into Eq.7.17 and the assumption that the localelevation of the pipe can be described solely as a function of location s, we obtain theEuler equation of motion1 dV + 1 ∂ p + dz + f V V = 0(7.19)g dtγ ∂ sdsD 2 gIf we also introduce the piezometric head H via p = ρ g( H - z) and expand the total derivative in the formdV = ∂ V + ∂ VV(7.20)dt∂ t∂ sthen the Euler equation can be written in the alternative form1 ∂ V+ ∂V VH + V 2 + f= 0(7.21)g ∂ t∂ s2 gD 2 gsince V∂ V/∂ s = ∂ (V2/2)/∂ s.The sum of piezometric head and velocity head that appears in the middle term is the sum that is usually displayed in diagrams as the Energy Line.7.3.2.RIGID-COLUMN FLOW IN CONSTANT-DIAMETER PIPESThe neglect of the elasticity of the pipe and fluid in a pipe of constant diameter forcesany change in velocity, in theory, to occur instantaneously throughout the entire pipe.Inaddition, the steady form of the mass conservation equation applies throughout the pipe sothat the velocity everywhere in the pipe is the same at any given time.This section willonly examine such flow in single pipes in order to emphasize basic principles; similartransient flows in pipe networks are studied in Chapter 12.There are relatively few closed-form solutions of Eqs.7.19 or 7.21, even with theserestrictions.One of these solutions describes the development or establishment of flowfrom rest through a horizontal pipe from a constant-head reservoir, as shown in Fig.7.5.In the simpler version of the flow establishment problem we assume that the fluid isinviscid, i.e.without friction, so the last term in Eq.7.21 is dropped and only the inertiaof the fluid is important.First we must choose the limits of integration, with respect tothe distance s, to begin the solution of Eq.7.21; we select section 1 at the upstreamend of the pipe (even though the pressure is changing rapidly and the velocity is indeednonuniform over this section; see Street et al., 1996, pp.362, 367) and section 2 to be© 2000 by CRC Press LLCHRLValveReservoir12Figure 7.5 Constant-head reservoir with valve at downstream end of horizontal pipe.just upstream from the valve.Now we formally integrate with respect to the distance sfrom the reservoir, point 1, to the valve, point 2:1∂ V∫ds = −∂ H + V 2∫ ds(7.22)g 1−2 ∂ t1−2 ∂ s2 gSince continuity assures us that all of the fluid in the pipe must undergo the sameacceleration, this leads toL ∂ V= −H + V 2 + H + V 2(7.23)g ∂ t2 g2 g21From Eq.7.23 we see that the acceleration term is the difference in fluid energy per unitweight, which is also the difference in energy grade line values, between the two end pointsof the integration.The valve is instantaneously opened fully at t = 0, and flow developsthereafter.When t = 0+ at section 2, H2 drops to zero and remains so, and V2 = Vwill grow with time from 0 to the steady state velocity V 0 [ Pobierz całość w formacie PDF ] |
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