ENERGY is given by:\(E = y + \frac{{{V^2}}}{{2g}} = y + \frac{{{Q^2}}}{{2g{A^2}}}\)where,y = Depth of flowQ = Discharge of flowA = Area of cross-sectionV = Velocity of flowand, Area of cross section of trapezoidal section is given by:A = By + my2where, B = Bottom WIDTH of channelm = Side slope of channel Calculation:Given Data:y = 1.5 mBottom width, B = 6 mQ = 15 m3/secSide slope, m = 1Let's FIND area of the cross section from the above formula A = (6 × 1.5) + (1 × 1.52) = 11.25 m2Now, Putting all these data in Specific Energy formula, we get,\(E = 1.5 + \frac{{{{15}^2}}}{{2 \TIMES 9.81 \times {{11.25}^2}}} = 1.591\;m = 1.6\;m\)

"> ENERGY is given by:\(E = y + \frac{{{V^2}}}{{2g}} = y + \frac{{{Q^2}}}{{2g{A^2}}}\)where,y = Depth of flowQ = Discharge of flowA = Area of cross-sectionV = Velocity of flowand, Area of cross section of trapezoidal section is given by:A = By + my2where, B = Bottom WIDTH of channelm = Side slope of channel Calculation:Given Data:y = 1.5 mBottom width, B = 6 mQ = 15 m3/secSide slope, m = 1Let's FIND area of the cross section from the above formula A = (6 × 1.5) + (1 × 1.52) = 11.25 m2Now, Putting all these data in Specific Energy formula, we get,\(E = 1.5 + \frac{{{{15}^2}}}{{2 \TIMES 9.81 \times {{11.25}^2}}} = 1.591\;m = 1.6\;m\)

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Calculate the approximate specific energy of a trapezoidal channel having a bottom width of 6 metres, ratio of side slopes is 1: 1 and the depth of flow at a discharge speed of 15 cubic metres per second is 1.5 metres.

Fluid Mechanics Open Channel Flow in Fluid Mechanics . 6 months ago

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Concept:Specific ENERGY is given by:\(E = y + \frac{{{V^2}}}{{2g}} = y + \frac{{{Q^2}}}{{2g{A^2}}}\)where,y = Depth of flowQ = Discharge of flowA = Area of cross-sectionV = Velocity of flowand, Area of cross section of trapezoidal section is given by:A = By + my2where, B = Bottom WIDTH of channelm = Side slope of channel Calculation:Given Data:y = 1.5 mBottom width, B = 6 mQ = 15 m3/secSide slope, m = 1Let's FIND area of the cross section from the above formula A = (6 × 1.5) + (1 × 1.52) = 11.25 m2Now, Putting all these data in Specific Energy formula, we get,\(E = 1.5 + \frac{{{{15}^2}}}{{2 \TIMES 9.81 \times {{11.25}^2}}} = 1.591\;m = 1.6\;m\)

Posted on 30 Nov 2024, this text provides information on Fluid Mechanics related to Open Channel Flow in Fluid Mechanics. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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