Question

8-C) As shown below water is pumped and sprayed through a nozzle into the atmosphere at 20 m/s at an elevation 10 m above the reservoir surface. The mass flow rate is 40 kg/s and the volumetric flow rate is 0.040 m3/s. The large diameter inlet pipe total length is 50 m. The pump effidency is 65%. The longer smaller diameter pipe leaving the pump is 200 m in total length. The pipes have different diameters as indicated below. The pipe material is smooth plastic pipe with insignificant roughness. Minor losses for the inlet, filter and valve are shown, neglect losses in other fittings. The nozzle tip diameter is not known. Given: Water density-1000 kg/m3 and viscosity 0.00112 N-s/m2 Pipe (1): Total L1 = 50 m, D, = 0.20 m, A: 0.0314 m2 and Pipe (2): L2 = 200 m D2 = 0.10 m, A2 = 0.00785 m2 a) Solve for the minimum system pressure at the pump inlet section (1), p,- kPagauge (negative) b) Solve for the maximum system pressure at the pump exit section (2), p2 kPa,gauge. c) Solve for the actual pump shaft power (including the Δke), (2) = kW (rough approx, 50 kW) NOTE: Carefully graph the averaged velocity profile (u vs. r) inside the long pipe (2) with Reynolds #- Indicate VBAR and Vmax 5 10 5 20 (m/s) Free Jet Ps PATM Vs-20 m/s Nozzle tip Filter KL = 20 40 m Valve K-10 Drawing is not to scale 5 m 2) D-0.20 m 6 m Pump 5m D2 0.10m Long pipe (2) ← L2 = 200 m total 10 m nlet KL-0.8

Answer 1

Please refer the following images for the solution.

Soluion Given data V3 20 m/s H lom 040 m3/s L1:50m L2: 2007 D2 O.10m (KL)に2.0 (KL) 10 9) Minimum system pressure ㅓhe pump inlet section (1) Applying Bennoul lis elm betu un inld ofHe pipe and sechond) の .PL: Cu) a ta press une huada/lke inlof cィthe Pipe) wh one , 4m 2. 60 0412732 m/s final Rynolds number x 1 2732 x 0.26 VD Re, 227357. 1429

R624000 As the ftoa)、ın sucion line istabulent, 15 fU ) Fo tur bulent flow Re) (227357.1429 2 r9.8 P, 1.2732 +10+2-0177 4+0+0 2x9.8 oooxg-8) 7946415 P

s) Apling Bengullis equahion od the inled ond outlef o te pipe line a 20 2 D4 23,02- - 0.040 2 8 -5.0 g 29 m/s 62 G-= - o 0791 s O.oo30 46 CRez C4s4723.2143 28Dn 2 0177 2 x 3.8 hL3=2,0171+ 32 . 2144+13.21g g ー hL347.452m] Hp77.839 m

Py 2 he pump & Comstaiy 7461,273 ooox 38) 2xg.8 Sa me lerel 7. 3 jooox98厂 2x3.8) -8-100+0.08262477, 83s-- +1-321939 9810 2. ()Actu pump shaft powen p = 46.990