Tutorial for CIVL252 Hydraulics

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Presentation transcript:

Tutorial for CIVL252 Hydraulics By DUAN Huanfeng ceduan@ust.hk

Example 1 Problem rectangular channel with 6ft width. y0, y1 are 65ft and 1ft, respectively horsepower lost =? in the hydraulic jump y0 y1

Solution From section 0 to 1 From section 1 to 2 Head loss (Method 1)

head loss (Method 2) For section 1 and 2: V1=64.2ft/s; y1=1ft; V2=?; y2=15.5ft; According to continuity EQ: V2=64.2*1/15.5=4.142ft/s According to Energy EQ: HL = (y1+V12/2g) - (y2+V22/2g)=49.2ft

Example 2 Problem Water flows at a steady rate of 12 cfs per foot of width in the wide rectangular concrete channel shown. Pls determine the water surface profile from section 1 to section 2. 1 q=12 cfs/(s*ft) Slope = 0.04 Rectangular weir 2 3ft

Solution First, determine type of the flow regime. Assume n =0.015; the depth is h1.

Second, determine whether a water jump occurs upstream of the weir.

Furthermore, determine the distance from the weir to where the hydraulic jump occurs.

Complete the water surface profile 0.74ft 3.13ft 5.26ft 57.0ft