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# Question No.3

#### Topic : Pump-Pipeline System Analysis & Design

Calculate the steady discharge of water between the tanks in the system shown in the following figure and the power consumption. Pipe diameter Ds = Dd = 200 mm; length = 2,000 m; k = 0.03 mm (uPVC). Losses in valves, bends plus the velocity head amount to 6.2 v2 / 2g. Static lift = 10 m.

Pump Characteristics :

Discharge (l/sec) |
0 |
10 |
20 |
30 |
40 |
50 |

Total Head (m) |
25 |
23.2 |
20.8 |
16.5 |
12.4 |
7.3 |

Efficiency (per cent) |
- |
45 |
65 |
71 |
65 |
45 |

The efficiencies given are the overall efficiencies of the pump and motor combined.

Given :

- ${D}_{s}={D}_{d}=200\mathit{mm}=0.2m$
- $L=\mathrm{2,000}m$
- $k=0.03m$
- ${h}_{\mathit{minor}}=6.2\frac{{v}^{2}}{2g}$
- ${H}_{s}=10m$

Steps of solution :

The solution to such problems is basically to solve simultaneously the head-discharge relationships for the pump and pipeline :

For the pump, head delivered at discharge Q may be expressed by :

${H}_{m}=A{Q}^{2}+BQ+C$

For the pipeline, the head required to produce a discharge Q is given by :

${H}_{m}={H}_{s}+{h}_{f}+{h}_{\mathit{minor}}$

${H}_{m}={H}_{s}+\frac{8fL}{{\pi}^{2}g{D}^{5}}{Q}^{2}+\frac{{K}_{m}}{2g{A}^{2}}{Q}^{2}$

A graphical solution is the simplest method and also gives the engineer a visual interpretation of the matching of the pump & pipeline.

First of all, we shall start with the system curve. Values of H corresponding to a range of Q values will be calculated.

To summarize the procedure for completing the previous table :

Flow velocity is calculated through :

$v=\frac{Q}{A}=\frac{Q}{\frac{\pi}{4}{D}^{2}}$ $\to $ $v=\frac{Q}{\frac{\pi}{4}\ast {0.2}^{2}}$

The kinematic viscosity for water is taken as :

$\nu =1.13\ast {10}^{-6}{m}^{2}/\mathit{sec}$

Reynolds number is calculated through :

${R}_{e}=\frac{vD}{\nu}=\frac{v\ast 0.2}{1.13\ast {10}^{-6}}$

The roughness height for uPVC pipes is :

$k=0.03\mathit{mm}$

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

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

The friction losses (using Darcy-Weisbach equation) occurring in the pipeline is calculated through :

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

To calculate the minor/secondary losses through the pipeline :

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

Finally, the total manometric head required is :

${H}_{m}={H}_{s}+{h}_{f}+{h}_{\mathit{minor}}$

The following graph shows the system curve for the previously tabulated values :

The following graph shows the pump characteristic curve (Head vs Discharge curve ) :

The computed system curve data and pump characteristic curve data are now plotted on the same graph as shown in the following figure :

The intersection point gives the operating conditions; in this case :

${Q}_{o}=28l/\mathit{sec},{H}_{m}=17.3m$

The operating efficiency is :

${\eta}_{\mathit{total}}=0.70$

To calculate the power consumption :

$\mathit{Output}\mathit{Power}={\gamma}_{w}{Q}_{o}{H}_{m}=\mathrm{9,810}\ast \frac{28}{1000}\ast 17.3=\mathrm{4,751.964}\mathit{Watt}$

${\eta}_{\mathit{total}}=\frac{\mathit{Output}\mathit{Power}}{\mathit{Input}\mathit{Power}}$ $\to $ $0.70=\frac{\mathrm{4,751.964}}{\mathit{Input}\mathit{Power}}$

$\mathit{Input}\mathit{Power}=\mathrm{6,788.52}\mathit{Watt}=6.8\mathit{kW}$

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