A straight 100 m long raw water gravity main is to carry water from an intake structure to the jack well of a water treatment plant. The required flow through this water main is 0.21 m $^{3}$ . Allowable velocity through the main is 0.75 m/s. Assume f = 0.01, g = 9.81 m/s $^{2}$ . The minimum gradient (in cm/100 m length) to be given to this gravity main so that the required amount of water flows without any difficulty is
4.8

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Solution

The correct option is A 4.8
$Q = 0.21 m^{3} / s$
Allowable velocity = 0.75 m/s
$f = 0.01$
$g = 9.81 m / s^{2}$

$\frac{π d^{2}}{4} = \frac{Q}{V} = \frac{0.21}{0.75} = 0.28 m^{2}$

$\Rightarrow d = 0.597 m$

$\Rightarrow \frac{f l V^{2}}{2 g d} = h_{f} = \frac{0.01 \times 100 \times (0.75)^{2}}{2 \times 9.81 \times 0.597} m$

$= 4.8 c m$
$= 0.048 m$

$\Rightarrow$ Minimum gradient

$= \frac{h_{f}}{l} = \frac{4.8 c m}{100 m}$

Hence, answer is 4.8

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The correct option is A 4.8 Q=0.21 m3/s Allowable velocity = 0.75 m/s f=0.01 g=9.81 m/s2 πd24=QV=0.210.75=0.28 m2 ⇒ d=0.597 m ⇒ flV22gd=hf=0.01×100×(0.75)22×9.81×0.597m =4.8 cm =0.048 m ⇒ Minimum gradient =hfl=4.8cm100m Hence, answer is 4.8