FluidMechWhite5eCh10 - Chapter 10 Open Channel Flow 10.1...

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Chapter 10 Open Channel Flow 10.1 The formula for shallow-water wave propagation speed, Eq. (10.9) or (10.10), is independent of the physical properties of the liquid, i.e., density, viscosity, or surface tension. Does this mean that waves propagate at the same speed in water, mercury, gasoline, and glycerin? Explain. Solution: The shallow-water wave formula, o c( g y ) , = is valid for any fluid except for viscosity and surface tension effects . If the wave is very small, or “capillary” in size, its propagation may be influenced by surface tension and Weber number [Ref. 5 7]. If the fluid is very viscous, its speed may be influenced by Reynolds number. The formula is accurate for water, mercury, and gasoline but would be inaccurate for glycerin . 10.2 A shallow-water wave 12 cm high propagates into still water of depth 1.1 m. Compute (a) the wave speed; and (b) the induced velocity δ V . Solution: The wave is high enough to include the y terms in Eq. (10.9): (1 / )(1 /2 ) 9.81(1.1)(1 0.12/1.1)[1 0.12/{2(1.1)}] / . (a) cg y y y y y Ans δδ =+ + + = 3.55 m s == = ++ (3.55 m/s)(0.12 m) . (b) 1.1 0.12 m cy VA n s yy 0.35 m/s 10.3 Narragansett Bay is approximately 21 (statute) mi long and has an average depth of 42 ft. Tidal charts for the area indicate a time delay of 30 min between high tide at the mouth of the bay (Newport, Rhode Island) and its head (Providence, Rhode Island). Is this delay correlated with the propagation of a shallow-water tidal-crest wave through the bay? Explain. Solution: If it is a simple shallow-water wave phenomenon, the time delay would be o L (21 mi)(5280 ft/mi) t 3015 s ??? c 32.2(42) Ans. ∆= = 50 min This doesn’t agree with the measured t 30 min. In reality, tidal propagation in estuaries is a dynamic process, dependent on estuary shape, bottom friction, and tidal period.
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734 Solutions Manual Fluid Mechanics, Fifth Edition 10.4 The water-channel flow in Fig. P10.4 has a free surface in three places. Does it qualify as an open-channel flow? Explain. What does the dashed line represent? Solution: No , this is not an open-channel flow. The open tubes are merely piezometer Fig. P10.4 or pressure-measuring devices, there is no flow in them. The dashed line represents the pressure distribution in the tube, or the “hydraulic grade line” (HGL). 10.5 Water flows rapidly in a channel 25 cm deep. Piercing the surface with a pencil point creates a wedge-like wave of included angle 38 ° , as shown. Estimate the velocity V of the water flow. Solution: This is a “supercritical” flow, Fig. P10.5 analogous to supersonic gas flow. The Froude number is analogous to Mach number here: θ = = = = == 2 (9.81 m/s )(0.25 m) 1 sin sin(19 ) , Solve . (Fr 3.1) gy c Fr V V V VA n s m 4.81 s 10.6 Pebbles dropped successively at the same point, into a water-channel flow of depth 42 cm, create two circular ripples, as in Fig. P10.6. From this information, estimate (a) the Froude number; and (b) the stream velocity.
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This note was uploaded on 09/07/2009 for the course CBME Kinetics & taught by Professor Lobban during the Spring '09 term at The University of Oklahoma.

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FluidMechWhite5eCh10 - Chapter 10 Open Channel Flow 10.1...

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