pump system 0000 - Dr. Syahrullail Samion · 2021. 1. 2. · PUMP SYSTEM #01 Two identical pumps...
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Theorectical and tutorial
Transcript of pump system 0000 - Dr. Syahrullail Samion · 2021. 1. 2. · PUMP SYSTEM #01 Two identical pumps...
Theorectical and tutorial
PUMPSYSTEM#01
TwoidenticalpumpsareconnectedtoapipesystemasshowninFigureQ4.Thepumpsareusedtodeliverwaterintothehigh-levelreservoir.Theheaddischargecharacteristicofthepumpsisgivenbyequation(1),whereQisthepumpdischargeinliterpersecondandHistheheaddevelopedbythepump
𝐻 = 70 + 0.16𝑄 − 0.001𝑄!(1)ThepipeABis150mlongwithadiameterof300mm.ThepipeBCis250mlongwithadiameter400mm.Thefrictionfactorforbothpipesis0.03.TheregulatingvalveVisusedtoregulateflowinpipeBC.Thestaticliftsagainstwhichthepumpsareworkingare40mand70m.Foragivenofvalve,thepumpoperatesagainstastaticliftof70manddevelopsadischargeof100literpersecond.Determinethedischargeofthesecondpumpandthelossofcoefficientofthevalve.NeglectallminorlossesandpipelossfromBtotank2.
SolutionTank(1)toTank(3)
0 + 0 + 0 + 𝐻"# = 0 + 0 + 70 +(0.03)(150)
0.3𝑣$!
2𝑔+(0.03)(250)
0.4𝑣%!
2𝑔+ 𝐾&'(&)
𝑣%!
2𝑔
𝑄 = 0.1𝑚*/𝑠
𝑣$ =𝑄𝐴$
=0.1
𝜋4 (0.3)
!= 1.41𝑚/𝑠
𝐻"# = 70 +(0.03)(150)
0.31.41!
2𝑔+ 0.96𝑣%! + 0.05𝐾𝑣%! (1)
𝐻" = 70 + 0.16𝑄 − 0.001𝑄!(Thisequationisgiveninthequestion)
𝐻"# = 70 + 0.16(100) − 0.001(100)! = 76𝑚 (2)
Tank(2)toTank(3)
0 + 0 + 0 + 𝐻"! = 0 + 0 + 40 +(0.03)(250)
0.4𝑣%!
2𝑔+ 𝐾
𝑣%!
2𝑔
𝐻"! = 40 + 0.96𝑣%! + 0.05𝐾𝑣%! (3)
𝐻"! − 40 = 0.96𝑣%! + 0.05𝐾𝑣%! (4)
Fromequation(1)and(2)
𝐻"# = 76 = 70 +(0.03)(150)
0.31.41!
2𝑔+ 0.96𝑣%! + 0.05𝐾𝑣%!
76 = 70 + 1.52 + 0.96𝑣%! + 0.05𝐾𝑣%!
4.48 = 0.96𝑣%! + 0.05𝐾𝑣%!
Fromequation(4) 𝐻"! − 40 = 0.96𝑣%! + 0.05𝐾𝑣%!
𝐻"! − 40 = 4.48𝑚
𝐻"! = 44.48𝑚
Then,
𝐻"! = 44.48 = 70 + 0.16𝑄 − 0.001𝑄!
0 = 25.52 + 0.16𝑄 − 0.001𝑄!
𝑄 = 258.5𝑜𝑟 − 98.5(Wecanignorethenegativevalue)
Dischargeforpump2is258.5liter/sor0.2585m3/s
𝑄 = 𝐴%𝑣%
𝑣% =𝑄𝐴%
=0.2585𝜋4 (0.4)
!= 2.06𝑚/𝑠
Fromequation(4)
𝐻"! − 40 = 0.96𝑣%! + 0.05𝐾𝑣%!
44.48 − 40 = 0.96(2.06)! + 0.05𝐾(2.06)!
4.48 = 4.07 + 0.212𝐾
𝐾 = 1.93
PUMPSYSTEM#2
𝑓 = 0.05𝑙 = 1𝑘𝑚𝐷 = 0.15𝑚𝑝𝑢𝑚𝑝𝑝𝑜𝑤𝑒𝑟 = 80𝑘𝑊Calculatethemaximumflowrateinthepump
𝑃𝑜𝑤𝑒𝑟, 𝑃 = 𝜌𝑔𝑄𝐻
Bernoulliequation:
0 + 0 + 0 + 𝐻! = 0 + 0 + 0 +;ℎ"
𝐻! = ;ℎ" = 𝑓 >𝑙𝐷AB𝑣#
2𝑔E
𝑄 = 𝐴𝑣
𝑣 =𝑄𝐴=4𝑄𝜋𝑑#
𝐻! = 𝑓 >𝑙𝐷AB16𝑄#
𝜋#𝐷$E= 𝑓 >
𝑙𝐷%A B16𝑄#
2𝑔𝜋#E= 0.05 >
10000.15%
A B16𝑄#
2𝑔𝜋#E
𝐻! = 54390𝑄#
𝑃𝑜𝑤𝑒𝑟, 𝑃 = 𝜌𝑔𝑄𝐻!
= 9810𝑄(54390𝑄#)
= 533565900𝑄&
If𝑃 = 80𝑘𝑊
80000 = 533565900𝑄&
𝑄& = 6669.57
𝑄 = 18.82𝑚&/𝑠
Ifwaterlevelwasdifferent:
Bernoulliequationbecomes:
0 + 0 + 0 + 𝐻! = 0 + 0 + 10 +Yℎ"
