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.Example Problem 8.4Consider the pipeline of Example Problem 8.1 and consider air entrainment percentagesof 0.10, 0.50, 1.0, and 2.0%.We assume the atmospheric pressure to be 14.7 lb/in2and use Case (b) restraint.For a polytropic process and an entrained air percentage of0.10%, the wave speed from Eq.8.40 is300 , 000 × 1441.94 1( − 0.001)a == 2270 ft / s241 + 3 × 1051 − 0.302()+0.001×3 × 1053 × 107 0.251.2 × 200 × 62.4 + 14.7144The following table summarizes the wave speeds for the three separate thermodynamicprocesses:% Air EntrainmentIsothermal ProcessPolytropic ProcessIsentropic Process0.12150 ft/s2270 ft/s2360 ft/s0.51160 ft/s1260 ft/s1340 ft/s1.0845 ft/s920 ft/s988 ft/s2.0610 ft/s666 ft/s777 ft/sRegardless of the process, the differences in results among the assumed processes are notgreat in view of the other uncertainties.* * *8.5 DIFFERENTIAL EQUATIONS OF UNSTEADY FLOWUp to this point we have seen for a given impulsive change in velocity ∆ V at apipeline section that we can compute the pressure head change ∆ H which will result.This ability will now be extended so the velocity and pressure head at any pipe section atany time can be determined as the result of boundary and initial conditions imposed at anysection of the system.To accomplish this, we will use Euler's equation from Chapter 7and develop another equation based on conservation of mass.© 2000 by CRC Press LLC8.5.1.CONSERVATION OF MASSWe apply conservation of mass to a control volume that coincides with the interior ofthe pipe and is of length ds:dsρ AV + ∂/ s∂ ( ρ AV) dsρ AVFigure 8.4 Control volume coinciding with the interior surface of the pipe.The result of this application isρAV − ρ AV + ∂ (ρ AV) ds(ρ Ads)(8.41)∂ s = ∂∂ tor− ∂ (ρ AV) ds = ∂ (ρ Ads)(8.42)∂ s∂ tAt this point we employ a rather unconventional form of the control volume concept inthat we require the sides of the control volume to be attached to the pipe wall.Thus thecontrol volume will elongate as the pipe stretches longitudinally.The only exception is inCase (c) where we keep the control volume at a constant length even though the pipeelongates (the total length of the pipeline remains constant even as the pipe slips in itsexpansion joints).This technique is used because the pipe stretching affects the volume ofstorage, and the relation between pipe elasticity and the available volume for the liquid isidentical to that in Section 8.2.Expanding the parentheses of Eq.8.42 yields− ρ ∂ V∂ A∂ρ∂∂ A∂ρAds + ρ Vds + AVds( ) + ρ+dsdsAds(8.43)∂ s∂ s∂ s  = ρ A ∂ t∂ t∂ tRegrouping and dividing by the control volume mass ρ Ads,1  ∂ρ + ∂ρ ∂ A∂ A∂V+ V( ds) + ∂ V = 0(8.44)ρ ∂ t∂ s  + 1 A ∂ t∂ s  + 1ds ∂ t∂ s∂ρ∂ρ∂ A∂ ARecognizing that+ V= dρ and+ V= dA , Eq.8.44 becomes∂ t∂ sdt∂ t∂ sdt1 dρ + 1 dA + ∂ V + 1 d ( ds) = 0(8.45)ρ dtA dt∂ sds dtFrom Section 8.2, K = −dp= dp so thatdV / Vdρ / ρ1 dρ = 1 dp(8.46)ρ dtK dt© 2000 by CRC Press LLCTo develop a useful expression for dA/dt in terms of p, the elastic pipe deformationsmust be considered.For the change in cross-sectional area, Eq.8.17 shows thatdA = dVc = 1 π D 2 dε2 = 1 π D 2 d( σ2 − µ dσ1)(8.47)dL22E1 dA = 2 d( σ2 − µ dσ1)(8.48)AEIn evaluating these stresses we will again examine Case (b) restraint; hencedσ2 = D dp and dσ1 = µ dσ2(8.49)2 esodpdσ2 − µ dσ1 = 1 − µ2( ) dσ2 = 1 − µ2() D(8.50)2 e dtFinally,1 dA =dp1 − µ2() D(8.51)A dteE dtConsidering longitudinal expansion,d( ds) = dε1 ds(8.52)which is zero for Case (b).Thus1 d ( ds) = 0(8.53)ds dtCombining all these results in Eq.8.45 gives1 dp +dp1 − µ2() D + ∂ V = 0(8.54)K dteE dt∂ sdp1+1 − µ2= 0(8.55)() DdtKeE  + ∂ V∂ s1From Eq.8.31 it is clear that the term in the brackets is.This statement is alsoa 2ρcorrect for Case (a) and Case (c) pipe restraint.Making this substitution for the terms inbrackets leads to1 dp + a 2 ∂ V = 0(8.56)ρ dt∂ sWhen we combine this result with the Euler equation of motion, Eq.7.19, we have twoindependent partial differential equations for p( s,t) and V( s,t):dV + 1 ∂ p + dzg+ f V V = 0(8.57)dtρ ∂ sds2 Ddpa 2 ∂ V + 1= 0(8.58)∂ sρ dt© 2000 by CRC Press LLC8.5.2.INTERPRETATION OF THE DIFFERENTIAL EQUATIONSBefore moving to the solution of these equations in Chapter 9, we can learn about thenature of these solutions by looking at a linearized subset of the full equations.If we firstexpress the pressure p in terms of the piezometric head H via the relation p = ρ g( H - z), then Eqs.8.57 and 8.58 becomedV + ∂ Hg+ f V V = 0(8.59)dt∂ s2 DanddHa 2 ∂ V + g= 0(8 [ Pobierz całość w formacie PDF ]

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