Fluid Mechanics-Frank M White Solution Ch10

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Fluid Mechanics-Frank M White Solution Ch10에 대한 보고서 자료입니다.

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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, 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. 3−7]. If the fluid is very viscous, its speed may be influenced by Reynolds number. The formula is accurate for water, mercury, and gasoline, in larger channels, but would be inaccurate for glycerin.

P10.2 Explain whether or not a disturbance to an open-channel flow can be detected upstream when the flow is subcritical or supercritical. Solution: Subcritical flow Fr<1 ⇒ C= gh > V
Therefore ←–C + V < 0 C + V > 0→ waves propagate both upstream and downstream

Supercritical flow Fr>1 ⇒ C= gh < V –C + V > 0→ C + V > 0→ both waves propagate downstream Disturbance to flow can be detected both upstream and downstream for subcritical flow, but only downstream for supercritical flow.

P10.3 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 cm3/s. Estimate (a) the Froude number; and (b) the Reynolds number. Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m-s. The surface wave speed is co = gy = (9.81 m/s2)(0.1 m) = 0.99m/s
The average velocity is determined from the given flow rate and area:


V = Q A = (80,000 cm3 /s) (30 cm)(10 cm) = 267 cm s = 2.67 m s Froude number: Fr = V co = 2.67m/s 0.99m/s = 2.69 (supercritical) Ans.(a)

10 cm
30 cm 0.08 m3/s
2
The Reynolds number should be, for this writer, based upon hydraulic radius:


Rh = A P = (0.3m)(0.1m) (0.3+0.1+0.1m) = 0.06m Reynolds number: ReRh = ρ VRh µ = (998)(2.67)(0.06) (0.001) = 160,000(turbulent) Ans.(b)


10.4 Narragansett Bay is approximately 33.8 km long and has an average depth of 12.8 m. 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 Δt= ΔL co = (33.8×103) 9.81(12.8) ≈3016 s≈50 min Ans.??? This doesn’t agree with the measured Δt ≈ 30 min. In reality, tidal propagation in estuaries is a dynamic process, dependent on estuary shape, bottom friction, and tidal period.


10.5 The water-channel flow in Fig. P10.5 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. P10.5 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).

P10.6 Water flows down a rectangular channel that is 1.2 m wide and 1 m deep. The flow rate is 0.95 m3/s. Estimate the Froude number of the flow.

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