Question

Water at 50 °Fis circulated from a large tank, through a filter, and back to the tank Calculate the volumetric flow rate through the filter for the following given information: Patm V.. Valve Pump Filter Pump power: W=200 lbf.ft/s Pump efficiency: 1p-0.9 Total pipe length: /= 200 ft Pipe diameter: d=0.1 ft Pipe surface relative roughness: /d=0.01 Elbow loss factor (each): Kelbow=1.5 Valve loss factor: Kvalve-6 Inlet and Exit type: sudden, Fig. 6.22 Filter loss factor Kriter=12 F (velocity, ft/s x diameter. (Vd) for water at 6 Valu 8000 8 10 0.1 0.2 0.4 0.6 0.8 1 40 60 80 100 200 400 600 800 1000 2 4 6 20 2000 4000 6000 10,000 Values of (Vd) for atmospheric ai t 60°F 6000 400 600 800 1000 60,000 100.000 00 6 8 10 20 40 200 2000 4000 cone 20.000 40000 0.10 TTT 0,09 Laminar-Critica HElow 0,08 zoneTransition zone f Complete turbulence, rough pipes 0.05 0.07 0.04 0.06 0.03 0.05 0.02 0.015 0,04 - 0,01 0.008 0.006 Rear Il 0.03 0.004 0,025 0,002 0.02 0,001 0.0008 0.0006 0.0004 Smooth pipes, 0.015 0.0002 0.0001 0.000,05 0.01 0.009 0.000,01 0.008 4 56 8106 2(10) 103 2(103 3 4 56 8104 2(104) 3 4 56 8105 2(105)3 4 56 8107 2107 3 4 5 6 8108 Vd Reynolds number Re = V 0,000,001 =0.000,005 Friction factor f Relative roughness- Table 6.1 Recommended Roughness Values for Commercial Ducts Material Condition ft Uncertainty, % mm Steel Sheet metal, new 0.05 ±60 0.00016 0.000007 +50 Stainless, new 0.002 Commercial, new 0.00015 0,046 ±30 Riveted 3.0 ±70 0.01 2.0 Rusted 0.007 +50 0.00085 Iron Cast, new Wrought, new 0.26 +50 0.00015 0.046 +20 0.15 Galvanized, new 0.0005 +40 Asphalted cast Drawn, new 0.12 0.0004 +50 0.000007 Brass 0.002 ±50 Drawn tubing Plastic 0.000005 0.0015 +60 Glass Smooth Smooth Concrete Smoothed 0.00013 0.04 +60 Rough Smoothed 0.007 2.0 +50 0.000033 Rubber 0.01 +60 Wood 0.0016 0.5 +40 Stave Table A.1 Viscosity and Density of Water at 1 atm , ft/s Т, € p, kg/m v, m/s Т, F A, lb - s/ft p, slug/ft A, N : s/m 1.788 E-6 0 1000 1.788 E-3 32 1.940 3.73 E-5 1.925 E-5 10 1000 1.307 E-3 1.307 E-6 50 1.940 2.73 E-5 1.407 E-5 20 998 1.003 E-3 1.005 E-6 1.937 2.09 E-5 1.082 E-5 68 0.799 E-3 0.657 E-3 30 996 0.864 E-5 0.802 E-6 86 1.932 1.67 E-5 992 40 0.662 E-6 104 1.925 1.37 E-5 0.713 E-5 50 988 0.555 E-6 122 1.917 0.548 E-3 1.14 E-5 0.597 E-5 0.511 E-5 0.467 E-3 60 983 0.475 E-6 140 1.908 0.975 E-5 70 978 0.405 E-3 0.414 E-6 158 1.897 0.846 E-5 0.446 E-5 80 972 0.355 E-3 0.365 E-6 176 0.393 E-5 1.886 0.741 E-5 90 965 0.316 E-3 0.327 E-6 194 1.873 0.660 E-5 0.352 E-5 0.283 E-3 958 0.295 E-6 212 1.859 0.591 E-5 0.318 E-5 100

Answer 1

Solution Givem Data d- 0.1 6 - 6, Kentane lofftls p- 0.9, 200t Kelbo 200 O.O Kvalve О. в 5 Rexit = At Kfilter= 12. Pe 1.94 shug1f y- .407 I0ils Pump pawer h where, Y-P3 3 = 32.26tls (0.9200) .. (0V YID1 366. 881 V .. both points at Suface equation Арpмујoag tamk -> enengy 2 + 3 + V. Y Y 29 23 & CHead lass where - 29 (51.56 6 +0.8+1t12, 2(32.2 O.1 2000f 27.3 3 64.4

s-2 Faom 0,& get we 2 20006 366.88 + 27.3 64.4 V3 23627.136 4 2000 +27.3 V (o.1) Reynolds Alumber VD Re - 5O1X ±OH'| 7107. 32 V get V= 6.0766s faam ez gefRe 43188.4 let fo Fiom ieration O.039 we eg. we feom moody_chant AeRe= 43188& €/nf get =0.01 new, f .03926 we fzom eg-y V= 6.0666 ls. -43117. S At eg-5 e Fom mood chazt iteration O.03926 =Q.03926 is Hence Solution Converged V= 6.0666 Btls AV Flow Rale T (a(6.0666) 2 : 0 = 2 0.0476 47