Chapter 9

David Wagner 2007/06/29 15:40

Problem 1.

  • Hose Pressure Drop [tbl 9-6]: 1.5/50*80=2.4 psig
  • Manifold Pressure Drop: 2.5 psig
  • Pressure at Outlet: 110-(90+2.4+2.5)=15.1 psig
  • Loss/1000': 15.1/1400 → 10.8 psig/1000'
  • Flow: 3*110*1.07=353 cfm
  • Drop of 2” Pipe: 9.4 < 10.8
2” Diameter Pipe

Problem 2.

Submersible pumps are usually used when a great deal of suction lift would otherwise be required. Jet pumps and air-lift pumps may also be used.

Problem 3.

9'

Problem 4.

  • 100' of 3” hose; Q=200 gpm
  • SH=10+20=30' [tbl 9-8] → 9.6/100' * 100' = 9.6 ft=FH
  • TH=SH+FH = 30+9.6=39.6' [tbl 9-9]→
Model 15M Pump:40/10/241

Problem 5.

  • SH: 50' [fig 9-4]; Elbows: 3*11'=33'; CV: 1*25'; FV: 70'
  • Pipe: 300'+33+25+70=468'; 4”@280gpm [tbl 9-9]: 5.4'/100'*400'=25.3'
  • Hose: 4.3'/100'*20'=0.8'
  • FH: 25.3+0.8=26.1'; TH=50+26.1=76.1'
Model 40M Pump

Problem 6.

  • 4*200/.95=842 cfm
  • 4” Pipe: 1500'
  • Gate Valve: 2*2.3'=4.6'
  • Elbows: 6*11'=66

Equivalent Length of 4” Pipe: 1570.6'

  • Pipe: -1570.6'*1.6/1000'= -2.5 psig
  • Manifold: -3 psig

Pressure at Manifold: 100-5.5= 94.5 psig

  • 1.25” hose: -1.048*100'*25.9/1000'= -2.7 psig

Pressure at each dirll=94.5-2.7=91.8

92 psig

Problem 7.

  • 2*800+1*1100+5*100=3200 cfm
  • {1.21*3200}/{0.9*.92}=4376
4376 cfm

Problem 8.

Air compressor capacity is rated as as the STP volume of air the compressor will deliver at as specified pressure.

Problem 9.

The pressure loss in a pipeline decreases as the initial pressure increases.


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