Question

Using EPANET2.0 Software, determine the flow rate in each pipe for the simple network shown in the Figure. Assume the fully turbulent flow exists for all pipes. The pipe lengths, diameters and the Darcy-Weisbachand Hazen-William roughness factor f for each pipe are given in the Table.

60U/s 40L/ Question-1 Using EPANET 2.0 Software, determine the flow rate in each pipe for the simple network shown in the Figure. Assume the fully turbulent flow exists for all pipes. The pipe lengths, diameters and the Darcy-Weisbach and Hazen-William roughness factor f for each pipe are given in the Table. 20L/S LOOPI 30U/s 40L/S 250/s 251/5 20/s sus 154/s LOOPII LOOP III 10 150's 201/s OL/s 151/s Pipe Length (m) Diameter (mm) 1 2 3 4 250 100 125 125 100 100 100 125 100 125 5 6 7 Roughness Factor D-W H-W 0.025 120 0.025 120 0.025 120 0.025 120 0.025 120 0.025 120 0.025 120 0.025 120 0.025 120 0.025 120 300 200 200 200 200 150 150 250 150 250 8 9 10

Answer 1

SOLUTION IS AS BELOW-

LOOPS ARE CONSIDERED AS THE BELOW DIAGRAMME-

60 L/s 1 40 L/s 20 LIS 40 LIS 5 30 L/s 25 L/s 25 L/s 6. 15 L/s 15 L/s V20 LIS 7 9 20 L/s OLIS 8 10 15 L/S 15 L/s

Consider loop-(1):-

Pipe	Qo	f	f*Qo^2	2*f*Qo
1	40	0.025	40	2
2	40	0.025	40	2
3	25	0.025	15.625	1.25
4	25	0.025	-15.625	1.25
5	20	0.025	-10	1
Sum			70	7.5

ΔQ =-Σf02/Σ2fQο

$\Delta$Q=-70/7.5

$\Delta$Q=-9.333 L/s , [note:-$\Delta$Q add in clockwise direction flow with sign and subtract in anticlockwise direction flow with sign]

correction in flow-

pipe-1=40+(-9.333)=30.667 L/s

pipe-2=40+(-9.333)=30.667 L/s

pipe-3=25+(-9.333)=15.667 L/s

pipe-4=25-(-9.33)=34.333L/s

pipe-5=20-(-9.333)=29.333 L/s    

Consider loop-(II):-

Pipe	Qo	f	f*Qo^2	2*f*Qo
4	25	0.025	15.625	1.25
7	20	0.025	10	1
8	20	0.025	10	1
6	5	0.025	0.625	0.25
Sum			36.25	3.5

ΔQ =-Σf02/Σ2fQο

$\Delta$Q=-(36.25)/3.5

$\Delta$Q=-10.3571 L/s

Correction in flow-

pipe-4=corrected value in loop-(1)+$\Delta$Q

pipe-4=34.333+(-10.3571)=23.9759 L/s

pipe-7=20+(-10.3571)=9.6429 L/s

pipe-8=20+(-10.3571)=9.6429 L/s

pipe-6=5+(-10.3571)=-5.3571 L/s

Consider Loop-(III):-

Pipe	Qo	f	f*Qo^2	2*f*Qo
3	25	0.025	-15.625	1.25
9	15	0.025	5.625	0.75
10	0	0.025	0	0
7	20	0.025	-10	1
Sum			-20	3

ΔQ =-Σf02/Σ2fQο

$\Delta$Q=-(-20)/3

$\Delta$Q=+6.667 L/s

Correction in flow-

pipe-3=corrected value in Loop-(1)-$\Delta$Q

pipe-3=15.667-6.667=9 L/s

pipe-9=15+6.667=21.667 L/s

pipe-10=0+6.667=6.667 L/s

pipe-7=corrected value of pipe-7 in loop-(2)-$\Delta$Q

pipe-7=9.6429-6.667=2.9759 L/s

correct flow in pipes-

flow in pipe-1=30.667 L/s

flow in pipe-2=30.667 L/s

flow in pipe-3=9 L/s

flow in pipe-4=23.9759 L/s

​​​​​​​flow in pipe-5=29.33 L/s

​​​​​​​flow in pipe-6=-5.3571 L/s

​​​​​​​flow in pipe-7=2.97559 L/s

​​​​​​​flow in pipe-8=9.6429 L/s

​​​​​​​flow in pipe-9=21.667 L/s

​​​​​​​flow in pipe-10=6.667 L/s

FINAL FLOW IN EACH PIPE IS AS BELOW-

60 L/S 1 30.667 LIS 30.667 LIS 5 29.333 LIS 2 30 L/s 23.9759 L/s 9LIS 15.3571 Lis 6 2.9759 L/s 21.667 LIS 7 9 9..6429 LIS 6.667 Lis 8 10 15 LIS 15 L/s