Two reservoir maintain a constant difference of water levels of 11.25 m and are connected by a 10 cm diameter pipeline of 294.3 m length. The total of all head losses, by friction valve losses, bend losses, inlet and exit losses, and velocity head can be taken as $98.1 v^{2} / 2 g$ (in m) where v is the flow velocity through the pipe (in m/sec). Assuming that the valve at the downstream end is suddenly opened so that there is no pressure wave, what will be the time taken for the velocity of flow in the pipe to attain 95% of the steady terminal velocity? Take
$\frac{1}{9.81} = 0.102$ .

2.25 l o g_{e} 19

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2 l o g_{e} 19

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2.25 l o g_{e} 39

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2 l o g_{e} 39

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Solution

The correct option is D $2 l o g_{e} 39$
$H = \frac{V_{0}^{2}}{2 g} + k . \frac{V_{0}^{2}}{2 g}$

where $V_{0} \Rightarrow$ steady flow velocity

$11.25 = \frac{V_{0}^{2}}{2 g} [1 + k]$

$11.25 = \frac{V_{0}^{2}}{2 g} [1 + 98.1]$

$11.25 = V_{0}^{2} \times 5.051 = 1.49 m / s e c$

Time take for establishment of flow

$t = \frac{L}{(1 + k) V_{0}} l n \frac{V_{0} + V}{V_{0} - V}$

$= \frac{294.3}{(99.1) (1.49)} I n (\frac{V_{0} + V}{V_{0} - V})$

$= 2 I n (\frac{V_{0} + V}{V_{0} - V}) . . . . (1)$

$w h e r e \frac{Q}{Q_{0}} = 0.95$

$T h e n \frac{V}{V_{0}} = 0.95$

Put the value in equation

$t = 2 I n (\frac{1 + 0.95}{1 - 0.95}) = 2 l o g_{e} 39$

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