Example 6.5. The lowest level of an impounding reservoir, situated in hills and being used as a source of supply, is higher than the lepel of the service woservoir by 150 metres and the distance...


Example 6.5. The lowest level of an impounding reservoir, situated in hills<br>and being used as a source of supply, is higher than the lepel of the service<br>woservoir by 150 metres and the distance between them is 50 km. The storgge<br>eruoir and the service reservoir were originally connected by a single pipe<br>line designed to carry a max. daily. demand' of 24 million litres per day<br>However, it was later found necessary to increase the supply to 35 million litres<br>per day. It was, therefore, decided to lay another pipe of the same diameter<br>alongside the first one in a certain reqd. length and cross-connected. Calculate<br>the dia. of the pipes and the length of second pipe required to be laid. Make use<br>of William Hazen's equation. Also derive an expression expressing H, in terms<br>of L and d from the original William Hazen's formula. Assume CH = 120.<br>

Extracted text: Example 6.5. The lowest level of an impounding reservoir, situated in hills and being used as a source of supply, is higher than the lepel of the service woservoir by 150 metres and the distance between them is 50 km. The storgge eruoir and the service reservoir were originally connected by a single pipe line designed to carry a max. daily. demand' of 24 million litres per day However, it was later found necessary to increase the supply to 35 million litres per day. It was, therefore, decided to lay another pipe of the same diameter alongside the first one in a certain reqd. length and cross-connected. Calculate the dia. of the pipes and the length of second pipe required to be laid. Make use of William Hazen's equation. Also derive an expression expressing H, in terms of L and d from the original William Hazen's formula. Assume CH = 120.

Jun 11, 2022
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