Chapter10_SM

# Chapter10_SM - Chapter 10 Open Channel Flow 10.1 The...

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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 . ________________________________________________________________________ P10.2 Water at 20 ° C flows in a 30-cm-wide rectangular channel at a depth of 10 cm and a flow rate of 80,000 cm 3 /s. Estimate ( a ) the Froude number; and ( b ) the Reynolds number. Solution : For water, take ρ = 998 kg/m and 3 μ = 0.001 kg/m-s. The surface wave speed is 0.08 m 3 /s 10 cm 30 cm s m m s m gy c o / 99 . 0 ) 1 . 0 )( / 81 . 9 ( 2 = = = The average velocity is determined from the given flow rate and area: The Reynolds number should be based upon hydraulic radius: ) .( ) (sup 69 . 2 / 99 . 0 / 67 . 2 : 67 . 2 267 ) 10 )( 30 ( ) / 000 , 80 ( 3 a Ans ercritical s m s m c V Fr number Froude s m s cm cm cm s cm A Q V o = = = = = = = (0.3 )(0.1 ) 0.06 (0.3 0.1 0.1 ) (998)(2.67)(0.06) : Re 160,000 ( ) .( ) (0.001) h h h R Am m Rm Pm VR Reynolds number turbulent Ans b == = ++ =

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734 Solutions Manual Fluid Mechanics, Fifth Edition 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 Solution: If it is a simple shallow-water wave phenomenon, the time delay would be 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. o L (21 mi)(5280 ft/mi) t 3015 s ??? c 32.2(42) Ans. Δ Δ= = 50 min ocess, dependent on estuary shape, bottom friction, and tidal period. This doesn’t agree with the measured Δ t 30 min. In reality, tidal propagation in estuaries is a dynamic pr 10.4 The water-channel flow in Fig. P10.4 has a fr qualify as an ee surface in three places. Does it open-channel flow? Explain. What does the dashed line represent? Solution: No , this is not an open-channel Fig. P10.4 low in them. The dashed line represents the ulic grade line” (HGL). flow. The open tubes are merely piezometer or pressure-measuring devices, there is no f pressure distribution in the tube, or the “hydra 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. Fig. P10.5 number is analogous to Mach number here: Solution: This is a “supercritical” flow, analogous to supersonic gas flow. The Froude θ = = = 1 sin sin(19 ) Sol c Fr V = == 2 (9.81 m/s )(0.25 m) , ve . (Fr 3.1) gy 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, s in Fig. P10.6. From this information,

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## This note was uploaded on 04/24/2011 for the course MECH 122 at San Jose State University .

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Chapter10_SM - Chapter 10 Open Channel Flow 10.1 The...

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