Question

Solve Problem 12.3, using the Cross technique. Problem 12.3 In the branched pipe system shown in Fig. 12.8, 850 /min of water at 10°C is flowing in a DN 100 Schedule 40 pipe at A. The flow splits into two DN 50 Schedule 40 pipes as shown and then rejoins at B. Calculate (a) the flow rate in each of the branches and (b) the pressure difference pA-pB. Include the effect of the minor losses in the lower branch of the system. The total length of pipe in the lower branch is 60 m. The elbows are standard. FIGURE 12.8 ON 100 Schedule 40 DN 100 Schedule 40 30m DN 50 Schedule 40 Angle val ve fully open Step-by-step solution There is no solution to this problem yet Get help from a Chegg subject expert. Ask an expert

Answer 1

Head loss in the upper branch L-30m DN50 Diameter nominal 5cm-2in 2.54 Inside diameter of pipe for 2 in schedule 4o steel pipe D-0.0525m hr = 001 9 2gD 2g×0.0525 , -7カ7' 2g× 0.0525 (A), = [1 142A+ 4.56] Assume fa 0.02 and h, -h, 142-001.2 11 .42 -27.4階 850x10-m / s 60 4--x 0,0525-2.168x 10-3㎡ 4 0,01416x850 b_ 2.554, -255x2. 168x103x60 Vo-1.55 -3.97m /s v-kinematic viscosity-1.3x106 v. × D. 3.97×0.0525 D 0.0525 -1141·-0.021 1.3x1016xI0 D & 4.6x105 0.0525-1.03x10,e ) 0.0525 ε 4.6x105 ï,×D, 32.56 × 0,0525 1.3x106 5-1141-ל.. :0.0215 Recompute ( with f, -0.0215

571x 0.02 1x2.11 142 x 0.02 1+4.56]× 2g y,:·29.11/11.9 ly, v = 1.56 0 2.564, 850x 103 2.56x2.168×101x60 b= = 2.55 m/s = 1 .56y, = 3.98m /s no change ins, and f, a) 60000 L/min - 518L/ min -8628x10-х Im3 s = 4v, = 2 168x10 "m2 × 2.55×60000-332L / min b) h-5 Im 9.81kN 571x0,021x -3.98-950kPa 571x0.021x--m9