Task 1
\nDescribe the scientific principles in respect of the following:
\nI. Laminar and turbulent flow
\nII. Uniform and non-uniform flow
\nIII. Steady and unsteady flow
\nIV. Reynolds number
\nV. Bernouli\u2019s Theorem
\nVI. Continuity equation
\nVII. Chezy\u2019s and Manning\u2019s formulae
\nVIII. Darcy\u2019s formula and HRS charts
\nIX. Channel Conveyance
\nX. The Froude Number<\/p>\n
Task 2
\na) A trapezoidal channel has a bed slope of 0.001. The base width is 10 m, the
\nside slopes are 1:3 and the discharge is 30 m3 \/ sec. Manning\u2019s n can be taken
\nas 0.015. Determine the normal depth of flow and Reynolds Number.
\nb) A symmetric trapezoidal section, 1900 mm deep and with top and bottom widths
\n3 m and 0.7 m respectively carries water at a rate of 3 m3\/sec. Manning\u2019s n may
\nbe taken as 0.012. Find:
\ni. The normal depth at a slope of 1 in 1500
\nii. The Froude number at the normal depth
\nc) In a compound channel, the flood channels are 8 m wide and have side slopes
\nof 1:2. The Manning\u2019s n is 0.035 for the banks. The base width and top width
\nfor the middle trapezoidal channel is 6 m and 18 m respectively. Manning\u2019s n
\ncan be taken as 0.015
\nIf bed slope is 0.001, calculate the Conveyance and Discharge for a flood level
\nof 4m.<\/p>\n
<\/p>\n
Task 3
\na) Describe local head losses giving examples.
\nb) A pipeline conveys water from a reservoir having a top water level of 850 m
\nabove datum, to a water treatment plant having an inlet water level of 700 m
\nabove the datum. Calculate the net head available after water goes through the
\nfollowing fittings.<\/p>\n
Fittings number head loss co-efficient (kl) Task 1 Describe the scientific principles in respect of the following: I. Laminar and turbulent flow II. Uniform and non-uniform flow III. Steady and unsteady flow IV. Reynolds number V. Bernouli\u2019s Theorem VI. Continuity equation VII. Chezy\u2019s and Manning\u2019s formulae Read More …<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-7911","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/7911","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/comments?post=7911"}],"version-history":[{"count":0,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/posts\/7911\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/media?parent=7911"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/categories?post=7911"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.benedictsol.com\/blogs\/wp-json\/wp\/v2\/tags?post=7911"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nBends 18 0.5
\nGate valves 3 0.25
\nBellmouth entry 2 0.10
\nBellmouth exit 2 0.5
\nVelocity is computed as 2.314 m\/s.
\nc) A pipeline 5 km long, 300 mm in diameter and with roughness of 0.028,
\nconveys water from a reservoir (top water level 600 m above datum) to a water
\ntreatment plant (inlet water level 450 m above datum). Assuming that the
\nreservoir remains full, estimate the discharge using the HRS charts.
\nORDER THIS ESSAY HERE NOW AND GET A DISCOUNT !!!<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"