All Problems quiz5

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Oct 30, 2023

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N Question 1 of 1 0/4 = View Policies Show Attempt History Your Answer Correct Answer X Your answer is incorrect. Find the pressure drop for a 70 °F water flow rate of 500 gal/min flowing upward in a vertical 6-in. schedule 80 (actual I.D. of 5.761- in.) commercial steel pipe. The pipe is 351 ft long. p1-p2= 3.426817 psi eTextbook and Media Solution Find the pressure drop for a 70 °F water flow rate of 500 gal/min flowing upward in a vertical 6-in. schedule 80 (actual I.D. of 5.761-in.) commercial steel pipe. The pipe is 351 ft long. Solution B £t/ 0 4 4 (500 gal/min) < T g:]/mm ) V_X= D2 5761 ¢\ 2 =6'15? & (353+1t) 0.00015£t =0.00031 ek Using Table 8.1,& =0.00015 ftso 7 = 6.158) (381) ft Re = Qz M:Z,Slxlos v 1.052x107° L& The Moody chart gives f = 0.0171. Note that the friction factor could be obtained from the Moody chart or the Haaland equation to approximately the same precision. The mechanical energy equation between the bottom B and the top T of the pipe is G o bl —+ 5 tgm-—wy=—+ + gzt + ghr
Solution Find the pressure drop for a 70 °F water flow rate of 500 gal/min flowing upward in a vertical 6-in. schedule 80 (actual I.D. of 5.761-in.) commercial steel pipe. The pipe is 351 ft long. Solution B £t/ 0 4 4 (500 gal/min) < T g:]/mm ) V_X= D2 5761 ¢\ 2 =6'15? & (353+1t) 0.00015£t =0.00031 ek Using Table 8.1,& =0.00015 ftso 7 = 6.158) (381) ft Re = Qz %:2,81“05 v 1.052x107° L& The Moody chart gives f = 0.0171. Note that the friction factor could be obtained from the Moody chart or the Haaland equation to approximately the same precision. The mechanical energy equation between the bottom B and the top T of the pipe is V3 Vi P54 7B gy —w =L Ty or gl P 2 P 2 ws = 0 and Vg = V1 from the continuity equation. LV? Also ghy, :fBT So s Py —pr=pgGr ) +pf 54 2 = (623%) (3511) (1) Ibm s\ (6158) Ib-s? i + (62'3F) 0.0171) ( (if—;l—)fl) 2 32.2ftIbm 144in> = 151.86psi + 3.18 psi ~ [155psi
View Policies Show Attempt History Your Answer Correct Answer Assume that laminar flow of 60°F water has been maintained in a 1-in. schedule 160 (actual I.D. of 0.815-in.), horizontal, 100-ft-long commercial steel pipe at a Reynolds number of 21,000. (Yes, laminar flow is possible at a Reynolds number that is this high under very controlled conditions.) Compare the pressure drop for this flow and that of turbulent flow for the same Reynolds number. Laminar flow p;-py= 0.423106954177 psi Turbulentflow pq-p,= 4.113731658621 psi
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Solution The mechanical energy equation from the inlet 1 to the outlet 2 is 2 VZ PV P pi A pga —we =y +ast Assuming fully developed flow with z; = z5, w, = 0, and constant velocity (from the continuity equation) +pgza + pgh P1—py=pght or LV? P =P _prT The velocity is found from Reynolds number —5 ft2 JRe (1,21 x 10 ‘—_)(21,000) i V=== (T =374— For laminar flow: 64 64 f= Re = 31,000 =0.00305 So slug (3.742)° Py —pa = (1 94—) (0.00305) ('}fifl)fl T( 1b 2 - = 60 9"2 (144m2 ) - For turbulent flow: Table 8.1 gives e = 0.00015 so - = 200 600 ables.1gives e = V. SOD— (%)fl_ A The Moody chart gives f = 0.030 so 2 1006 (3.74%) (ZE)n 2 Cegp b (1 ) _ p =592 8 (L) = [411psi by - (1 94 30ue ) (0.030) 2 Turbulent pressure drop is about 10 times that of laminar flow, because there is the added kinetic energy in the turbulent eddies that is dissipated by viscosity.
view Folicies Show Attempt History Your Answer Correct Answer Assume that laminar flow of 60°F water has been maintained in a 1-in. schedule 160 (actual I.D. of 0.815-in.), horizontal, 100-ft-long commercial steel pipe at a Reynolds number of 20,000. (Yes, laminar flow is possible at a Reynolds number that is this high under very controlled conditions.) Compare the pressure drop for this flow and that of turbulent flow for the same Reynolds number. Laminarflow pi-pp= 0.402959003978 psi Turbulent flow p1-po= 3.758899300307 psi
Solution The mechanical energy equation from the inlet 1 to the outlet 2 is Vi Vi 41 +"Tl +pgz1 —ws =p, +aTz +pgza + pgh Assuming fully developed flow with z; = z5, w, = 0, and constant velocity (from the continuity equation) P1—py=pght or LV? P =P _prT The velocity is found from Reynolds number -5 £ S Re _ (121x 1072 ) 20,000 sl b (521 T For laminar flow: 64 64 =R - 20,000 =0.00320 So 100f (356%)° (S 2 6] pL-py = (194‘;‘#) (0.00320) b i _ f = 58,0“—1( l44in2) = [0.403 psi For turbulent flow: .00015 fi Table 8.1 gives e = 0.00015 so £ M =0.00221 CRNEEaT The Moody chart gives f = 0.030 so 5 lug 100t (356 - = (19422) 0030 AT e ft 5 - I (i _ f =345 ( 144in’ ) = [376psi Turbulent pressure drop is about 10 times that of laminar flow, because there is the added kinetic energy in the turbulent eddies that is dissipated by viscosity.
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