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Fluid Mechanics 07

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**Hydraulic and Energy Grade Lines**

EGL:- Energy Grade line indicate the total head at any point in the system. πΈπΊπΏ= πππππππ‘π¦ βπππ + ππππ π π’ππ βπππ + πΈπππ£ππ‘πππ π»πππ πΈπΊπΏ= π£ 2 2βπ + π Ι£ +π HGL:- Hydraulic Grade Line indicate the piezometric head at any point in the system π»πΊπΏ= ππππ π π’ππ βπππ + πΈπππ£ππ‘πππ βπππ π»πΊπΏ= π Ι£ +π§

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**Hydraulic and Energy Grade Lines**

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**Pump Add head to the System**

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Turbine

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Nozzle Nozzle increase the velocity and if discharge to atmospheric the term of pressure head will be zero

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**Change in pipe diameter**

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Negative Pressure

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Example A pump draws water (50Β°F) from a reservoir, where the water-surface elevation is 520 ft, and forces the water through a pipe 5000 ft long and 1 ft in diameter. This pipe then discharges the water into a reservoir with water-surface elevation of 620 ft. The flow rate is 7.85 cfs, and the head loss in the pipe is given by

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Determine the head supplied by the pump, hp, and the power supplied to the flow, and draw the HGL and EGL for the system. Assume that the pipe is horizontal and is 510 ft in elevation.

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**Solution π1 Ι£ +π§1+ π£1 2 2βπ +βπ= π2 Ι£ +π§2+ π£2 2 2βπ +βπ‘+βπΏ Where**

π1 Ι£ +π§1+ π£1 2 2βπ +βπ= π2 Ι£ +π§2+ π£2 2 2βπ +βπ‘+βπΏ Where P1=P2=Patm=zero V1=v2=zero Ht=zero, z1=520 ft, z2=620 ft π£= π π΄ = 7.85 (Ξ /4)β( 1 2 ) =10ππ‘/π hL= 0.01β πΏ π· β( π£ 2 2βπ )=0.01β β( β32.2 ) =77.6 ft

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βπ= π§2βπ§1 +βπΏ= 620β =178 ft πππ€ππ=Ι£βπββπ=62.4β7.85β178=159βπ

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