Question

ENCE 4322/53222 2. Reservoir A delivers to reservoir B through two uniform pipelines AJ (3000 m long) and JB (4000 m long) of diameters 300 mm and 200 mm, respectively. The difference in elevation between the water level in the two reservoirs is 25 m. Just upstream of the change in section, which is assumed gradual (no minor losses), a controlled discharge of 30 L/s is taken off 3000 m 304 Assume kinematic viscosity - 1.14x10m/s, and a pipe effective roughness for both pipes - 0.015 mm. Find the steady discharge to reservoir B using the Colebrook-White equation for Darcy's friction factor. Use Excel Solver, and submit your Excel spreadsheet

Answer 1

Sot Glirem henath of Pipe line AT = 30 Dom dia AJ = 300mm Length of pipe line TB = 4Doom; dia JB = 200mm Elevation difference asm Discharge - 304/sec Kinematic viscosity v= H4 X 1076m/sec Roughness of pipe Kg = 0.015 mm From Colebrook - White equation for Devops friction factor Ks 3.7D Consider pipe et Reynolds number for flow in tipe 2 = V x 0.3 114 x16 25 0.015 X107 2:51 3.730. XD.30 1.14x10-6 Consider pipe - Tegnalets number for flaw in pipe 2 • * * * ser for 2 = : 7-14 X106 P ks + - Dz. - Pest 0.015 X1033 2:57 3.7% 0.2 V, X0.20 114x18 % v te, between reservoir A From energy conversation equation and reservoir B

25 = 5,4,6,2 tysi te v 29d, 29 de dengan +95 - 5, x 3000 kv," + $, X400 OULU 289-81*0: 3 3 279.87X0-2 Applying equation of mass conservation, 0.070 7m**v, = Bom mo/s + 0.0314m*x ve av, = 0.4243 +0.4441V2 ~ lung - mas z 05 +,x3000 x 10.4243+0:444102) ? $_x4000V,? 2x918) X0. 3 2 79.81*3-2 7/25= f, x5091684/0.4243 +0.444102) + 1019.3689,= from 0, 05 x10- 3be251 972 3.7(0-3) (6:424378144411x30 = -2 log 0.015 X10-3 1/4X106 From het Trial & error method z = 1 m) sec 5,= 0.016 f = 0.017

3,25 = f, * 509.684 10.4243 + 0.444112) + 1019.368 t, v,? - On sowing, V = 105 misec: with the value of above velocity, f, & fe are computed s t,> 0.016, 4, = 0.017 => velocity in pipe 2 = 1.05 ml sec steady discharge to reservoir 13 = (1-05) (5(0-2)2] = 0.032986 mesec i 1 steady discharge to reservoir B= 39.986 41 sec

C8 - X V foc =C2*509.684*(0.4243+0.4441*B2)*(0.4243+0.4441*B2)+(1019.368*E2*B2*B2) A f2 B C V2 f1 1.050515 0.016 LHS RHS wN 0.017 LHS RHS 1.00000303 1.004621852 V2 LHS RHS

X =C2*509.684 (0.4243+0.4441*B2)*(0.4243+0.4441B2)+(1019.368E2B2 B2) V2f1 1.050515 0.016 LHS RHS 0.017 LHS RHS 1.00000303 Goal Seek ? 1.004621852 x V2 LHS C8 RHS Set cell: To yalue: By changing cell OK 5852 Cancel

- X x =SQRT(E2)-2*LOG(((0.015*0.001)/(3.7*0.2))+((2.51" (1.14*10^-6))/((B2)*0.2 SQRT(E2))))) V2 f1 1.050515 0.016 LHS RHS 10.017 LHSRHS 1.00000303 Goal Seek ? 1.004621852) x V2 LHS RHS 25 24.97477 Set cell: F4 To yalue: By changing cell: SES2 OK 1 Cancel

C2 =SQRT(C2)*(-2*LOG((0.015*0.001)/(3.7*0.3))+((2.51*(1.14*10^-6))/((0.4243+0.4441*B2) *0.3*SQRT(C2))))) 1 f2 V2 1.050515 0.016 0.017 LHS RHS LHS RHS 1.00000303 Goal Seek ? 1.004621852 x Set cell: D4 V2 LHS RHS 25 24.97477 To value: By changing cell: $C$2|| OK Cancel

F4 X fac =SQRT(E2)*(-2*LOG(((0.015*0.001)/(3.7*0.2))+((2.51*(1.14*10^-6))/((B2) *0.2*SQRT(E2))))) A B C V2 f1 1.050515 0.016 LHS RHS f2 0.017 LHS RHS 1.00000303 1.004621852)

D4 - X V for =SQRT(C2)*(-2*LOG(((0.015*0.001)/(3.7*0.3))+((2.51*(1.14*10^-6))/((0.4243+0.4441*B2)*0.3*SQRT(C2))))) V2 f2 f1 1.050515 0.016 LHS RHS 0.017 LHS RHS 1.00000303 1.004621852