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&[DATE] &[TIME] - &[FILENAME]

PIPE NETWORK - Ex 3 Improved

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The system above contains 4 pipes, and therefore, we need 4 independent equations to solve.

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1. DATA:

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Nodal Supplies & demands QJ 0.32 0.28 0.14 0.10 41 mat m 3^s / * :

Pipe diameters D 0.25 0.25 0.25 0.25 41 mat m * :

Pipe lengths L 20010020010041 mat m * :

Friction factor f 0.02 :

Density of Water γ 1000 kg m 3^/ * :

Elevation at 
Node 1 WS.1 L 2 el : 100 m *

Exponent for friction head loss. 
"n" may be,say 2.0 or 1.85 or ... n 2 :

n=2 in Darcy Eqn

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Assume pipe flow directions as shown in the diagram -------------------------------------------------------->

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node:1

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QJ.1 Q.1 - Q.4 - 0 ≡

Sign Convention

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Flow into + ve

Flow out - ve

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node:2

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Q.1 Q.2 + QJ.2 - 0 ≡

Linear Eqns

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node:3

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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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One real loop : Pipes 1-2-3-4. Use the convention that positive is clockwise, in the direction of decreasing head.

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2. CALCULATIONS:

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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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Alternatively, using 
"Vectorize" function KK 8 f * L * g.e D (5^* π 2^* / vectorize : s 2^m 5^/ 338.5552 169.2776 338.5552 169.2776 41 mat

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

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

----------->

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Assumed directions of flows

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Note that if n = 2, Smath computes correct units with no problem. However if n is not equal to 2, (say if n = 1.85), then we have to balance the units in the equations for h_f as shown below.

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uQ QJ 1 el UnitsOf : 1 m 3^s / *

uK K 1 el UnitsOf : 1 s 2^m 5^/ *

Define --------->

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General eqn for head loss hf K Q n^* ≡ K Q * Q abs (n 1 -^* ≡

Pipe friction losses, 
with units adjusted. 
Symbolic optimization r 1 K rows range hf r el K r el uK / (Q r el uQ / (* Q r el uQ / abs (n 1 -^* m (* : 11 line for

Symbolic Optimization hf 655360000 s * Q 1 el * s Q 1 el * m 3^/ abs * 196133 m 2^* π 2^* / 327680000 s * Q 2 el * s Q 2 el * m 3^/ abs * 196133 m 2^* π 2^* / 655360000 s * Q 3 el * s Q 3 el * m 3^/ abs * 196133 m 2^* π 2^* / 327680000 s * Q 4 el * s Q 4 el * m 3^/ abs * 196133 m 2^* π 2^* / 41 mat

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

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Using n≡1.85 in h.f=K*Q^n@#

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Governing Equations required for solution

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F QJ 1 el Q 1 el - Q 4 el - 0 ≡ Q 1 el Q 2 el + QJ 2 el - 0 ≡ Q 3 el Q 2 el - QJ 3 el - 0 ≡ hf 1 el hf 2 el - hf 3 el - hf 4 el - 0 ≡ 41 mat :

ΔE 10 :

Symbolic optimization F Unknowns Q 11 mat

Initial guess for 
each variable g_var 0.2 0.2 0.2 0.2 41 mat m 3^s / * :

Using n≡2 in h.f=K*Q^n

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Q F g_var FindRoot : 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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3. RESULTS:

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Pipe flows Q m 3^s / 0.218 0.062 0.202 0.102 41 mat

Friction losses hf m 16.14887 0.64229 13.7594 1.74728 41 mat

Assumed 'n' n 2

Pressure at node 1 P.1 γ WS.1 * g.e * : psi 142.233

Pressure at Node 2 P.2 P.1 hf 1 el γ * g.e * - : psi 119.264

Pressure at node 3 P.3 P.2 hf 2 el γ * g.e * + : psi 120.178

Alternatively, pressure at node 3 P.1 hf 4 el γ * g.e * - hf 3 el γ * g.e * - psi 120.178

------->

Pressure at node 4 P.4 P.1 hf 4 el γ * g.e * - : psi 139.748

Average sea-level pressure is 101.325 kPa or 29.92 inches (in Hg) or 760 millimetres of mercury (mm Hg).

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P.atm 101.325 kPa * :

------->

ie.

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P.atm psi 14.7

------->

P.atm γ g.e * / ft 33.9

Sanity Check

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QJ 1 el Q 1 el - Q 4 el - Q 1 el Q 2 el + QJ 2 el - Q 3 el Q 2 el - QJ 3 el - QJ 4 el Q 4 el + Q 3 el - hf 1 el hf 2 el - hf 3 el - hf 4 el - 51 mat 0 m 3^s / * 0 m 3^s / * 0 m 3^s / * 0 m 3^s / * 0 m * 51 mat

F 000041 mat

Mercury γ.Hg 846 lbf ft 3^/ * : (Density ≡ sg.Hg 13.56 : (Sp. gravity ≡ 21 line

P.atm kPa 101.325

Equivalent height of water P.atm γ.Hg / sg.Hg * ft 33.92