By J. A. Fox (auth.)
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Additional resources for An Introduction to Engineering Fluid Mechanics
X 9810 x 1 x (10 + 1 + 5)2 = 1·26 meganewtons Vertical component of the force V = weight of contained fluid = area x w = (6 x 10 + 1 x 5 +! x 10 x 10 + ~ X 12) x 9810 = 1 ·14 meganewtons Resultant force R = y(H2 + V2) = y(1·26 2 + 1·14 2)meganewtons = 1·69 meganewtons 24 An Introduction to Engineering Fluid Mechanics 6m Fig. 20 Direction of resultant: Let Q be the angle the resultant makes to the horizontal. Then Q = arctan (H/V) = arctan (1·26/1·14) = 47° 53' Position of resultant: It is first necessary to locate the centroid of the mass of water contained between a vertical through A and the dam profile.
Y x Fig. 9 Between these streamlines a fluid flux occurs and as no flow can cross a streamline (by defmition) this flux ~Q must be made up of flow across AB which in turn must be the same as the sum of the flow across CB and that across AC. The flow across CB is -v 8x (8x is negative). Thus 8Q But and =- vax + u 8y Hydrodynamics 43 8if; = - v8x +u 8y so 8Q == 8 if; Thus streamlines, which have been defmed as lines of constant value of the stream function, must also be lines of constant flux; the flux between a streamline of stream function if; and one having a value for its stream function of zero is numerically equal to if;.
IO) A bridge across a river has the profile illustrated in Fig. 24. The width of the bridge is 30 ft [10 m] . Calculate the upthrust on the bridge when the river 15'[4 5mJ 15'[45mJ 0
An Introduction to Engineering Fluid Mechanics by J. A. Fox (auth.)