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# Question No.5

#### Topic : Pipe Network Analysis (Hardy-Cross Method)

In the system shown in the following figure, a pump is installed in BC to boost the flow to C. Neglecting minor losses, determine the flow distribution and head elevations at the junctions of the pump delivers a head of 15 m. All pipes have the same roughness size of 0.03 mm. The outflows at the junctions are shown in l/sec.

Data :

Pipe |
AB |
BC |
CD |
DE |
EA |
BE |

Length (m) |
500 |
600 |
200 |
600 |
600 |
200 |

Diameter (mm) |
200 |
150 |
100 |
150 |
200 |
100 |

Pressure head elevation at A = 60 m

Given :

- $\mathit{Roughness}\mathit{height}\mathrm{:}k=0.03\mathit{mm}$
- ${Z}_{A}+{P}_{A}/\gamma =60m$
- ${H}_{\mathit{pump}}=15m$

Steps of solution :

Step (1) - Identify loops. When using hand calculations, the simplest way is to employ adjacent loops.

Loop 1 : ABEA

Loop 2 : BCDEB

The inlet flow at junction A is equal to the sum of the outlet flow at the other junctions, therefore the inlet flow at junction A is equal to :

= 60 + 50 + 50 + 40 = 200 l/sec

Step (2) - Allocate estimated flows in the pipes. Only one estimated flow in each loop is required; the remaining flows follow automatically from the continuity condition at the nodes.

Assume that water is flowing in pipe AB with a flowrate of 120 lit/sec (from junction A to junction B) and flowrate in pipe BC is 40 lit/sec (from junction B to junction C).

Therefore, the initial estimates of flowrates through the network are :

Pipe |
Q (l/sec) |

AB |
120 |

BC |
40 |

CD |
-10 |

DE |
-50 |

EA |
-80 |

BE |
+20/-20 |

Step (3) – Evaluate the head loss coefficient K for each pipe. The friction coefficient f is obtained through Moody chart or using Barr's equation. The calculations proceed in tabular form. Note that Q is written in liter/sec simply for convenience; all computations are based on Q in m3/sec.

To calculate the velocity at each pipe :

$v=\frac{Q}{A}=\frac{Q}{\frac{\pi}{4}{D}^{2}}$

To calculate Reynolds number :

${R}_{e}=\frac{vD}{\nu}$

To calculate the friction coefficient, Barr's equation is used :

$\frac{1}{\sqrt{f}}=-2\mathrm{log}\left(\frac{k/D}{3.7}+\frac{5.1286}{{R}_{e}^{0.89}}\right)$

Barr's equation is based partly on an approximation to the logarithmic smooth turbulent element in the Colebrook-White equation.

To calculate K for each pipe (using Darcy-Weisbach equation) :

$K=\frac{8fL}{{\pi}^{2}g{D}^{5}}$

To calculate the correction required in the discharge value :

$\Delta Q=\frac{-\Sigma K\u2223Q\u2223Q}{2\Sigma K\u2223Q\u2223}$

For the 1st iteration :

$\Delta {Q}_{1}=-11.948l/\mathit{sec},\Delta {Q}_{2}=13.298l/\mathit{sec}$

For the 2nd iteration :

$\Delta {Q}_{1}=-1.400l/\mathit{sec},\Delta {Q}_{2}=-0.591l/\mathit{sec}$

For the 3rd iteration :

$\Delta {Q}_{1}=-0.219l/\mathit{sec},\Delta {Q}_{2}=-0.127l/\mathit{sec}$

For the 4th iteration :

$\Delta {Q}_{1}=-0.050l/\mathit{sec},\Delta {Q}_{2}=-0.015l/\mathit{sec}$

For the 5th iteration :

$\Delta {Q}_{1}=-0.007l/\mathit{sec},\Delta {Q}_{2}=-0.007l/\mathit{sec}$

For the 6th iteration :

$\Delta {Q}_{1}=-0.002l/\mathit{sec},\Delta {Q}_{2}=0l/\mathit{sec}$

Final discharges (after 6 iterations) :

Pipe |
Q (l/sec) |
Head loss (m) |

AB |
106.373 |
21.405 |

BC |
52.558 |
13.337 |

CD |
2.558 |
0.258 |

DE |
37.442 |
14.876 |

EA |
93.627 |
20.123 |

BE |
6.185 |
1.282 |

To calculate the pressure head elevations at different junctions, use the energy equation between point A and different points ,total energy at any point can be calculated through :

$\mathit{T.E}=Z+\frac{P}{\gamma}+\frac{{v}^{2}}{2g}$

Since the minor losses can be neglected, the velocity head term is negligible (equal to zero), therefore the total energy at point A is equal to :

${\mathit{T.E}}_{A}={Z}_{A}+\frac{{P}_{A}}{\gamma}+\frac{{v}_{A}^{2}}{2g}$ $\to $ ${\mathit{T.E}}_{A}=60+0=60m$

To calculate the pressure head elevation at point B, apply Bernoulli's equation between points A & B :

${\mathit{T.E}}_{A}={\mathit{T.E}}_{B}+{h}_{lA\to B}$

${\mathit{T.E}}_{B}={Z}_{B}+\frac{{P}_{B}}{\gamma}+\frac{{v}_{B}^{2}}{2g}$ $\to $ ${\mathit{T.E}}_{B}={Z}_{B}+\frac{{P}_{B}}{\gamma}+0$

$60={Z}_{B}+\frac{{P}_{B}}{\gamma}+21.405$ $\to $ ${Z}_{B}+\frac{{P}_{B}}{\gamma}=38.595m$

Similarly, the pressure head elevations at different points is summarized in the following table :

Junction |
T.E (m) |

A |
60 |

B |
38.595 |

C |
25.258 |

D |
25.001 |

E |
39.877 |

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