Question

The multi-pipe system shown below aims to distribute water (p = 1000 kg/m3 and u = 0.00089 Pa.s) to three tanks at different elevations ZA ZB, and ZC of 101,30, and 55, respectively by means of gravity. All of the tanks are exposed to atmospheric pressure. Your task is to determine the elevation required in tank A (ZA) to deliver 1 m3/s (in pipe 1). In this project, the minor losses can be neglected and the pipe can be considered to be smooth. The pipe diameters are 30 cm for pipe 1 and 20 cm for pipes 2, 3, 4, and 5 ZB Tank A Pipe 2 L2 Tank B Pipe 23 L1 = 200 m L2 = 40 m L3 = 60 m L4 = 50 m L5 = 40 m ZD Pipe 1544 Hz Tank D Pipe 5 Pipe 4 Tanko Input your answer (ZA) in the box below. Please ignore if you get "wrong" note as the answer must be checked manually Submit your EES script and solution as PDF or DOC in the "Exam 03 Part B - Work" folder and your answers for flowrates in tanks B, C, and D

Please post EEs code and handwork

Answer 1

The equation for flow distribution and head losses are written. The friction factor is computed for given flow rate assuming smooth wall using Prandtl relation for turbulent flows. The equations for flow rate and head losses are simultaneous equations which are solved iteratively bu assuming the intial solution and updating till the convergence is achieved. The solution of simultaneous equation gives us the Za, v2, v3, v4, v5. Once we know the velocity, we can find the flow rate to the respective tanks. Note: There is a typo in the problem. The datum given should be zB = 101, zC = 55 and zD =30 but in the first line zA, zB and zC are given.

LO 200 14/ 11 -0.2 . 2x9.81 (307 100] 200 Q = 8 + 8 5 (30) ² U = 15 (26) ² 0 + 1 (20) a 9v, Ilvtvg 4 - ① G = Q & Q 21 (20) og = 1 (20) ² + 2 (20) ² » Ug = y + - • 2 F 24-105(0.002), )* ***60**) - **5); ADA ***E SCO ALEXA 2-3 0-009 «{) Q = 1 = 1 1 10.3)” U = 0,5 14,15 ml S - -x14. O 0.009 2x9.81 0.3 0.009 2x9.81 F = 2 by (he f') -0.8 f = 0.009 (Smouth wall Pipe Re - pu, d, 1000) * 14.15* 0. 3 4708687.433 (Turbodent) (8.9x104)

gv = 40 + 40 - 127-35/4 = V2 + V V₂ + ₂ = 31.8 - 6 y = y + vs - . za - 101 = 61:23 +(0.09) 2-3 34-55 = 61-23 + (0-1376) y2 + (0.946), -6 ZA - 30 = 61.23 + (0-1370) vx2 +(0:09) - © Equating 4 & €1-23 +55 + (0-1376) x (0+)+(0.1146) 4,2 - 61-23 +36 +601376) (4 * vs)+(0109) v" 25+ (0.1146) v = 0.09 v ². per 101461.23 + 0.09% (31-8 –(+v)]* =*p55+61-23 + 04346 (,+v)*+ (0-11416)? 46 +2.862 7 0.09 ( +v) = 0.134C (4,+V_)* +(0140) y, 48-862 = (0-2276) (+V_)*+601146),? Solving the above sen sinut écouleurly iteratively will give Q=Bdku x 02 x 12 = 0.376 mils u = 5.84 m/s exy = lex - x 5.84 = 0.1835 mils V = 14:15 mlo ZA = 175.2m " x 0.2 x 14.15 = 0.445 m3/s. 2 x м o.. = 12 Rater Flow X 0.2 V = 12 m/s V = 20ls C Dų