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# Question No.5

#### Topic : Solving Pipe Networks using Quantity Balance Method

Determine the flows in the network illustrated in the below figure; minor losses are given by Cm v2/2g.

Data :

Pipe |
Length (m) |
Diameter (mm) |
Roughness (mm) |
Minor Loss Coefficient Cm |

AB |
20,000 |
500 |
0.3 |
20 |

BC |
5,000 |
350 |
0.3 |
10 |

BD1 |
6,000 |
300 |
0.3 |
10 |

BD2 |
6,000 |
250 |
0.06 |
10 |

Given :

- ${Z}_{A}=100m,{Z}_{C}=70m,{Z}_{D}=60m$

Steps of solution :

Step (1) - Estimate ZB (pressure head elevation at B) = 80 m (Note that the elevation of the pipe junction itself does not affect the solution.)

${Z}_{B}=80m$

Step (2) - Allocate estimated flow velocities in the pipes.

$v=2m/\mathit{sec}\mathit{for}\mathit{all}\mathit{pipes}$

The friction factor f may be obtained from the Moody diagram, or using Barr's equation, using an initially estimated velocity in each pipe. Subsequently f can be based on the previously calculated discharges. However, unless there is a serious error in the initial velocity estimates, much effort is saved by retaining the initial f values until perhaps the final correction.

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 :

${R}_{e}=\frac{vD}{\nu},\nu =1.011\ast {10}^{-6}{m}^{2}/\mathit{sec}$

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 friction head loss coefficient Kf for each pipe (using Darcy-Weisbach equation) :

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

To calculate minor head loss coefficient Km for each pipe (representing minor losses) :

${h}_{l\mathit{minor}}={C}_{m}\frac{{v}^{2}}{2g}={C}_{m}\frac{{Q}^{2}}{2g{A}^{2}}$ $\to $ ${K}_{m}=\frac{{C}_{m}}{2g{A}^{2}}$

To calculate head loss coefficient K for each pipe (representing both friction & minor losses) :

$K={K}_{f}+{K}_{m}=\frac{8fL}{{\pi}^{2}g{D}^{5}}+\frac{{C}_{m}}{2g{A}^{2}}$

To calculate the correction required in ZB value :

$\Delta {Z}_{B}=\frac{2(\Sigma {Q}_{\mathit{IB}}-{F}_{B})}{\Sigma \frac{{Q}_{\mathit{IB}}}{{h}_{L,\mathit{IB}}}}$

For this problem, no outflow occurs at junction B, therefore :

${F}_{B}=0$

Therefore :

$\Delta {Z}_{B}=\frac{2\Sigma {Q}_{\mathit{IB}}}{\Sigma \frac{{Q}_{\mathit{IB}}}{{h}_{L,\mathit{IB}}}}$

For the first iteration :

Initial estimate for velocity is 2 m/sec for all pipes !

Correction to ZB :

$\Delta {Z}_{B}=\frac{2\ast (-0.054)}{0.0210}=-5.146m\to {Z}_{B}=80-5.146=74.854m$

For the second iteration :

The initial estimates for velocity is updated according to the previously calculated value !

Correction to ZB :

$\Delta {Z}_{B}=\frac{2\ast (0.004)}{0.0242}=0.328m\to {Z}_{B}=74.854+0.328=75.183m$

For the third iteration :

The initial estimates for velocity is updated according to the previously calculated value !

Correction to ZB :

$\Delta {Z}_{B}=\frac{2\ast (0.001)}{0.0237}=0.076m\to {Z}_{B}=75.183+0.076=75.258m$

For the fourth iteration :

Correction to ZB :

${Z}_{B}=75.259m$

Final Discharges in the given pipe network :

Final discharges (after 4 iterations) :

Pipe |
Q (l/sec) |
Flow Direction |

AB |
156.513 |
A to B |

BC |
56.518 |
B to C |

BD1 |
58.979 |
B to D |

BD2 |
41.016 |
B to D |

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