CLOSED NETWORK WITH ACCESSORIES SOLVED WITH EXCEL SOLVER

 CLOSED NETWORK WITH ACCESSORIES SOLVED WITH THE EXCEL SOLVER TOOL

Example: In the network shown in the figure, the valve in pipe 2-3 is partially closed and produces a head loss equal to 10V2-3^2/2g. The pressure head at node 1 is 40 m of column of water. For the indicated data, determine the flow rate in the system pipes, as well as the pressure head at node 4. The network is arranged in a horizontal plane. Kinematic viscosity =10^-6 m2/s.

 Excel template write to iyoba14@hotmail.com

Solution :

QT= 60+40+30+30+40 = 200 l/s

Step 01: Assume the flows that circulate through pipe 1, 2, 3, 4, 5, 6 and 7 in l/s, complying with the continuity equation at each node.

Step 03: We determine the area for each of the pipes 

A=3.1416D^2/4

Step 04: We calculate the speed for each of the pipes in m/s 
V = Q/A

Step 05: We calculate the Reynolds number for each of the pipes.

Re=V x D / kinematic viscosity

Step 06: From Barr's formula to determine the coefficient of friction:
1/raiz(f) = -2 log(K/(3.7*D) + 5.1286/Re^0.89)
RHS=-2 log(K/(3.7*D) + 5.1286/Re^0.89) 

Step 07: We calculate the friction coefficient for each of the pipes.

f=1/RHS^2

Step 08: We calculate the friction losses in pipes 1,2, 3, 4, 5,6 and 7 with the Darcy equation.

hf = f L/D V^2/(2g)

To pipe 2 the total pressure loss = hf + the local pressure loss, produced by the valve.

RESULTS:

Pressure head at node 1 = 40 m
Pressure height at node 4 = 42.765 m

 Excel template write to iyoba14@hotmail.com