By Zoeb Hussian, Zulkifly Abdullah, Zanial Alimuddin
Following a concise review of fluid mechanics knowledgeable via a number of engineering purposes and examples, this reference provides and analyzes significant varieties of fluid equipment and the main periods of generators, in addition to pump expertise. It bargains execs and scholars in hydraulic engineering with history thoughts in addition to functional assurance of recent turbine applied sciences, totally explaining some great benefits of either steam and fuel generators. Description, layout, and operational details for the Pelton, Francis, Propeller, and Kaplan generators are supplied, as are outlines of assorted varieties of strength crops. It offers solved examples, bankruptcy difficulties, and a radical case study.
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Extra resources for Basic Fluid Mechanics and Hydraulic Machines
With the help of second law of Newton force acting by a jet on stationary and moving plate is obtained. The impact ofjet on vanes has direct application on hydraulic turbines. 2 Scope of Fluid Mechnanics The dimensional analysis deals with the units of measurement in SI units both fundamental and derived units, and non-dimentional quatities. The properties offluids deal with measurement of mass, density, specific weight, specific gravity, compressibility offluids, surface tension, capillary action.
Determine power available to the turbine when the flow rate is 30 ml/s. Nozzle 20m -+ . ~ / 7;-/;; ~ 0 CD + + 6 r -- /7777}/);r Fig. 19 Schematic of hydro power plant. L 2g - I where HT = represents head developed by turbine PI = P2 = atmospheric pressure = 0 VI = 0 because of large reservoir. 7 A pitot-static tube is used to measure air velocity. 2 kg/m 3 • Solution A Pitot-static tube measures stagnation pressure and static pressure. 8 The wind blows at a speed of 100 km/hr in a storm. Calculate the force acting on a I m x 2 m window which is facing the storm.
Applying Bernoulli equation between 1 and 2 points, vi. 2 PI V P2 - + -l = +- y vi. For continuous flow Substituting V 2 in eq. 23 2g 2 VI Y = 2g 2g [(PI - P2 )] pg ..... 23) Fluid EJpw V = , 37 a2 -,---=-------::- (a~ - a~)112 Volume flow rate Q = a,V, where H = PI - P2 If m = area ratio pg al = -a 2 or 2(PI - P2 ) ..... 24) P[I-:: J In practice, some loss of energy occurs between points 1 and 2 and therefore actual discharge (Q) will be less than theoretical discharge (Q) and therefore a coefficient of discharge Cd is introduced Cd = Actual discharge Theoretical discharge = Qa Q .....
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