787e83ae-806a-431c-8b74-1e580b86cf10 216 3 7 decimal

PIPE NETWORK ANALYSIS - Ex 3

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The system above contains five pipes. We want to find the flow in each pipe and

the pressure at each node. Pipe data are provided below. Since we have five unknown flowr ates we

need five independent equations.

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L.1 200 m * :

L.2 100 m * :

L.3 200 m * :

L.4 100 m * :

D.1 0.25 m * :

D.2 0.25 m * :

D.3 0.25 m * :

D.4 0.25 m * :

Density of Water γ 62.4 lb ft 3^/ * :

Nodal Supply & Demands - known

QJ.1 0.32 m 3^s / * :

QJ.2 0.28 m 3^s / * :

QJ.3 0.14 m 3^s / * :

QJ.4 0.10 m 3^s / * :

Nodal equations - nodal equations are based on the fundamental principle of mass balance. Stated

in a more useful form we normally say that "the flows entering a node must equal those leaving".

Note that application of this principle requires that we adopt a sign convention (flow to a joint

positive, flow away from a joint negative). In fact any convention can be adopted but it must

be adhered to.

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Loop equations a.k.a. energy equations express the fact that the sum of the head losses around

a closed loop must equal zero. Application of this principle also requires that we adopt a sign

convention. We first chose a loop direction, clockwise or counterclockwise. Then we assign values

to the head losses in the loop. For example, travel in the direct of increasing head (against the

direction of flow) is positive and travel in the direction of decreasing head is negative.

Again any sign convention can be used but must be consistent.

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Sign Convention

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node:1

QJ.1 Q.1 - Q.4 - 0 ≡

Flow into + ve

Flow out - ve

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node:2

Q.1 Q.2 + QJ.2 - 0 ≡

Linear Eqns

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node:3

Q.3 Q.2 - QJ.3 - 0 ≡

real loop-1

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K.1 Q.1 n^* K.2 Q.2 n^* - K.3 Q.3 n^* - K.4 Q.4 n^* - 0 ≡

Clockwise + ve

Anticlockwise - ve

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Non-Linear Eqn

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n=2 in Darcy Eqn

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2 real loops : Pipes (1) - (2) - (3) & (4) - (5) - (2) Use the convention that positive is clockwise, in the direction of decreasing head.

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Assume an n value of 2 for all pipes n=2

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h.f f L * V 2^* 2 g.e * D * / ≡

V 2^1 π 4 / D 2^* / (2^Q 2^* ≡

h.f f L * 2 g.e * D * / 16 π 2^D 4^* / * Q 2^* ≡ 8 f * L * g.e D 5^* π 2^* / Q 2^* ≡

Hence K 8 f * L * g.e D 5^* n 2^* / ≡

K is simply a lumped constant for a specific pipe

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f 0.02 :

D 0.25 0.25 0.25 0.25 41 mat m * :

L 20010020010041 mat m * :

j 1 D rows range KK j el 8 f * L j el * g.e D j el (5^* π 2^* / : for

KK s 2^m 5^/ 338.5552 169.2776 338.5552 169.2776 41 mat

K.1 KK 1 el : 338.555 s 2^m 5^/ *

K.2 KK 2 el : 169.278 s 2^m 5^/ *

K.3 KK 3 el : 338.555 s 2^m 5^/ *

K.4 KK 4 el : 169.278 s 2^m 5^/ *

Governing Equations

-------------------- Linear Eqn at node 1

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F 1 el QJ.1 Q.1 - Q.4 - 0 ≡ (:

-------------------- Linear Eqn at node 2

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F 2 el Q.1 Q.2 + QJ.2 - (:

-------------------- Linear Eqn at node 3

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F 3 el Q.3 Q.2 - QJ.3 - 0 ≡ (:

Non Linear Eqn

in real loop 1-2-3-4

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F 4 el K.1 Q.1 * Q.1 abs * K.2 Q.2 * Q.2 abs * - K.3 Q.3 * Q.3 abs * - K.4 Q.4 * Q.4 abs * - 0 ≡ (:

F Unknowns Q.1 Q.2 Q.3 Q.4 41 mat

Initial guess for 
each variable g_var .2 .2 .2 .2 41 mat m 3^s / * :

A negative sign indicates that a wrong direction was initially assumed for the flow in that pipe

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F F g_var FindRoot eval : m 3^s / 0.218 0.062 0.202 0.102 41 mat

-ve value (if any). indicates that flow is opposite to the assumed direction.

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Results:

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Q.1 F 1 el : m 3^s / 0.218

Q.2 F 2 el : m 3^s / 0.062

Q.3 F 3 el : m 3^s / 0.202

Q.4 F 4 el : m 3^s / 0.102

Excel Results: Using "Solver" to make , by changing rQ value

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Check solution by substitution of results:

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Node 1 QJ.1 Q.1 - Q.4 - 0 m 3^s / *

hf1 K.1 Q.1 * Q.1 abs * 16.14887 m *

Node 2 Q.1 Q.2 + QJ.2 - 0 m 3^s / *

hf2 K.2 Q.2 * Q.2 abs * - 0.64229 m * -

Node 3 Q.3 Q.2 - QJ.3 - 0 m 3^s / *

hf3 K.3 Q.3 * Q.3 abs * - 13.7594 m * -

hf4 K.4 Q.4 * Q.4 abs * - 1.74728 m * -

Hence K.1 Q.1 * Q.1 abs * K.2 Q.2 * Q.2 abs * - K.3 Q.3 * Q.3 abs * - K.4 Q.4 * Q.4 abs * - 0 m *

Additional check
at Node 4 QJ.4 Q.4 + Q.3 - 0 m 3^s / *