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.,HRESDN=1240.,ZEND=1100.,HATM=30.,QTRY=0.,QACC=0.50,TMAX=60.,PFILE=T,HVPRNT=T,PPLOT=T,GRAPH=T,RERUN=F/1 30.15000.013 3590.800.2 30.30000.013 3590.800.&PUMPS NPUMPS=4,NSTAGE=3,IPUMP=1,RPM=1775.,WRSQ=475.,© 2000 by CRC Press LLCQN=0.,1000.,2000.,3000.,4000.,4500.,HNSQ=129.,127.5,121.,103.5,67.5,0.,TNSQ=50.,58.,78.,92.,97.,80./&GRAF NSAVE=2,IOUTSA=1,PIPE=1,2,0,0,NODE=999,1,0,0/The booster pump power failure program PROG4 is used to analyze the problem.The source and executable programs are on the CD.The following plot of extreme pressure values along the pipeline is one of the primary results to come from this analysis.We observe that high heads occur on the suction side of the pump, as well as low heads onthe discharge side.No column separation occurs in this case.Maximum head1300'Steady State EL - HGL1200'head1100'MaximumheadumSteady State1000'EL-HGLMinim- 30"Minimum head30,000' 30.01900'15,000' - 30"800'0.013* * *11.1.4.SETTING UP THE EQUATIONS FOR SOURCE PUMPSWe will follow the same general procedure as for booster pumps.There is a check valveon each pump discharge line.There is also a low-friction, essentially frictionless, bypassline with a check valve around the pump station to supply the pipeline if the pump headshould drop to zero during the transient.We again assume all pumps fail simul-taneously.Four equations are needed to model the pump behavior:Discharge side C-VP = C(11.20)d3 + C 4 HPdConservation of massN pu Q = VP Add(11.21)Work-energyHsump + hp = HP(11.22)dhpQPump characteristic= Nst C 7+ C 8N 2N(11.23)Here Npu is the number of pumps in parallel, and Hsump is the pump sump elevation.© 2000 by CRC Press LLCWhen N is found as in the previous case, we have only four unknowns here.WithEqs.11.20-11.23 we can proceed with a solution.In this case we findHsump + Nst N C 7 C 3 Ad + Nst N 2 C 8N puHP =(11.24)1 − Nst N C 7 C 4 AdN puIf HP is less than Hsump, then the pump bypass is open, and we must then equate HPto Hsump and use Eq.11.20 to compute the velocity.If the velocity were found to benegative, then we would set VP = 0 and compute HP from Eq.11.20.dThe next example problem applies this analysis to a source pump configurationsubjected to a power failure.Example Problem 11.2Four pumps in parallel are used to pump approximately 12,000 gal/min from areservoir at elevation 395 ft to a storage reservoir at elevation 840 ft, as shown below(not to scale).The pump discharge lines have check valves and lead into a manifold whichin turn supplies the 30-in welded steel pipeline.El.840'2 mi.- 30"- El.810'El.700'3 mi.- 30"ƒ = 0.019ƒ = 0.0132000' - 30"- El.415'ƒ = 0.013El.395'The pipeline extends 2000 ft horizontally from the pump station at an elevation of415 ft.It then slopes upward for three miles to elevation 700 ft.The remaining twomiles of pipe are reinforced concrete and slope gradually upward to enter the storagereservoir at elevation 810 ft.The friction factor and the wave speed for the steel pipe are0.013 and 3590 ft/s, respectively.For the concrete pipe these values are 0.019 and3490 ft/s.The pumps are the same Ingersoll-Dresser 15H277 turbine pumps used inExample Problem 11.1, except they now have five stages.Refer to the previous ExampleProblem for the pump characteristics.The 11.83-in impeller will be used.The totalrotary moment of inertia of each pump and motor unit is 475 lb-ft2.Find the consequences of pump power failure.This is a source pump configuration, so PROG3 will be used to determine the effect ofpump power failure.The input data file for this program is shown below.This programalso uses the subroutine PGRAPH which makes it possible to generate additional tablesof output data, printer plots, and data files for external plot programs.© 2000 by CRC Press LLCDEMONSTRATION OF PROGRAM NO.3, INPUT DATA FILE "EP112.DAT"SOURCE PUMP FAILURE, 4 INGERSOLL-DRESSER 15H277 5-STAGE PUMPS&SPECS NPIPES=3,NPARTS=3,IOUT=5,HRES=840.,HSUMP=395.,ZEND=810.,HATM=33.,QACC=0.50,TMAX=10.,DTNEW=0.,PFILE=T,HVPRNT=T,PPLOT=T,GRAPH=T,RERUN=F/1 30.2000.013 3590.415.2 30.15840.013 3590.415.3 30.10560.019 3490.700.&PUMPS NPUMPS=4,NSTAGE=5,RPM=1775.,WRSQ=475.,QN=0.,1000.,2000.,3000.,4000.,4500.,HNSQ=129.,127.5,121.,103.5,67.5,0.,TNSQ=50.,58.,78.,92.,97.,80./&GRAF NSAVE=3,IOUTSA=1,PIPE=1,2,3,0,NODE=1,1,1,0/The results show that column separation occurs about 5 sec after power failure.At thattime the program execution ends because this program is not prepared to analyze vaporcavities.A plot of the EL-HGL for times prior to column separation is presented.1000'St900'eady State EL - HGLsec.3.9sec.800'5sec.88t=0.7sec.t=11sec.74t=2.6sec.9t=3700't=45.1=t600'500'400'* * *© 2000 by CRC Press LLC11.2 PUMP STARTUPPressure surges caused by pump startup can be difficult to predict, particularly if air isinitially in the system.This air may be in the pump discharge column or located at highspots along the pipeline.Whatever the case, the air-exhaustion process must be simulatedin order to approximate the pressures which could occur during startup.On the other hand, if the pump startup is associated with a recent power failureshutdown, then the restart sequence is important in controlling the pressures
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