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.13.3.The air chamber must be sufficiently large to supply the needs of the pipeline withoutemptying and permitting air to enter the pipeline.Also, the initial air volume must belarge enough to prevent the rate of pressure drop from being excessively high.An initialair volume that is too small will cause the pump discharge pressure to behave as if the airchamber is absent, thereby giving little or no assistance in preventing low pressures.When the flow finally reverses and begins to move back toward the pumps, the check valvecloses (actually it usually is already closed), and flow occurs into the chamber.To providesome damping for the system, the losses for flow into the chamber are deliberately madehigher than the losses for flow from the chamber.This can be done by using a nozzlesimilar to the one shown in Fig.13.6 or by having two connections to the chamber, onewith a low loss for outflow and one with a higher loss for inflow.Generally, gooddamping can be accomplished without causing high pressures during the backflow phase.Occasionally the air chamber alone is not adequate to prevent column separation.Lowpressures can occur at local summits along the pipeline where the effect of the air chamberat the pump is inadequate.In these cases a one-way surge tank at each summit can be usedto "drape" the EL-HGL above the pipeline on both sides of the summit.Figure 13.8illustrates this technique.The set of equations describing the behavior of an air chamber is rather complex.Wewill assume that the air chamber is at the upstream end of the pipeline so the boundarycondition at this point will consist of the relations for both the pumps and the air chamber.We will assume that all of the pumps fail simultaneously.Steady state EL - HGLAir chamberFigure 13.8 Propagation of a negative wave after power failure with an air chamber and a one-way surge tank.The appropriate equations are the following:C– VP = C 1 + C 2 HP(13.11)Conservation of massN puQ + Qc = VP A(13.12)© 2000 by CRC Press LLCPump work-energyHsump + hp = HP(13.13)Chamber work-energyQ()c = Cout Aout2 g Hc − HP(13.14)QPump head increasehp = N 2 Nst C 7+ C 8N(13.15)In these equations Hc is the head in the chamber, Cout is the outflow coefficient, Aoutis the outflow cross-sectional area, and Qc is the discharge from the chamber.We have six unknowns in these five equations, so another relation is required.Thisequation will describe the thermodynamic process that the air in the chamber undergoes.The most commonly used process is the polytropic processpγ η = poγ η(13.16)oin which po and γ o are the absolute pressure and specific weight of the air in thechamber under steady-flow conditions, p and γ are those values at a later time, and η isthe polytropic exponent, generally chosen to be 1.2.There is some disagreement over theappropriateness of this value.Graze (1972) and Graze et al.(1976) have shown that thisprocess does not describe precisely the thermodynamic behavior of the air.One compli-cating feature during the air expansion process is that the freezing temperature of the liquidis often reached.The latent heat released by the freezing of condensed liquid vaporcomplicates the thermodynamic process beyond the simplicity of the polytropic model.However, in light of the many other uncertainties in the analysis, we will continue to usethe polytropic equation until a better model that is reasonably easy to use appears.With Hatm as the atmospheric pressure head, the polytropic equation can be writtenH()H+ H()γc + Hatm − zP γ wcoatm − zPwγ η=γ η(13.17)oIf the initial air volume is Vc and the volume at a later time is Voc , the equation can bewritten as ηH() Vcoc = zP − Hatm + Hc − z(13.18)oP + Hatm Vc in which the air volume at any time can be calculated in a way that is similar to the fluidvolume computations for the one-way surge tank:V ()( )( )c t + ∆ t= Vc t + Qc t ∆ t(13.19)Now we solve the six equations simultaneously.For the most general case we assume flow continues both through the pumps and fromthe chamber.Knowing Vc from Eq.13.19, we can calculate Hc from Eq.13.18,reducing the number of unknowns to five.Solving the five equations for Qc produces© 2000 by CRC Press LLCQc = 0.5 C 5 1 − 1 − 4 C 6 (13.20)C 25 in which2 A 2 N NCoutst C 75 = 2 gCout(13.21)N pu − N Nst AC 7 C 2andH()()sump + N Nst /N puC 1 C 7 A + N NpuC 8C226 = 2 gCout Aout − Hc +(13.22)1 − N N()st AC 2 C 7 /N puAfter calculating Qc, Q is found from2Q =1 AC 1 + AC 2 Hc −Qc(13.23)N22pu 2 gCout Aout − QcIf Q ≥ 0, then the solution is acceptable, and the remaining unknowns may be calculatedfrom Eqs.13.11 through 13.15.If Q < 0, then the pump check valves must be closed, and Q must be set to zero.Now Qc must be calculated fromQc = 0.5 C 5 − 1 + 1 + 4 C 6 (13.24)C 25 where2 A 2Cout5 = 2 gCout(13.25)C 2 AandC226 = 2 gCout Aout Hc + C 1C 2 (13
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