SOE2156: Fluids Lecture 7

Transcription

1 Weirs and SOE2156: Fluids Lecture 7 Vee

2 Vee Last lecture { assumed the channel was uniform (constant depth, shape, slope etc.) { steady uniform Found that : location of free surface to be determined 2 possible solutions { subcritical and supercritical. Solution can jump between these (hydraulic jump) Froude number important can also be critical { Fr = 1 Single solution for this case : maximum for a given specic energy What if the channel is? { steady?

3 Froude number still indicates whether the is sub- or super-critical. As the cross-section changes, the conditions will too. The free surface may pass through the critical depth. Points where this happens are a limiting factor in the design of the channel, and are knn as control sections. Vee d 1 d c Hydraulic jump s<s c s>s c d 2 Critical depth s<s c Examples of this behaviour include and umes, used (principally) to measure rates of water. Weirs are vertical obstructions { can be thin or broad-crested. are narrings in the channel.

4 Typical arangement : Total energy h E d c Vee Suciently broad that the free surface is parallel to the crest Fl upstream tranquil { Fr < 1 Fl dnstream unimpeded { free jet { discharge over the weir will be the maximum possible Thus over weir at critical depth

5 Vee Fl rate over weir given by Q = CbE 3=2 in terms of the upstream specic energy E. Usually assume V = 0 upstream, so E ' H, and thus Q = CbH 3=2 { general weir equation. A single measurement of this value gives the discharge. From theoretical analysis C = 1:705 but this is usually determined from experiment. N.B. The depth over the crest of the weir is xed. Raising the weir crest will not alter this, but will alter the overall depth upstream.

6 Thin Weirs Vee Alternative arangement { weir is thin (at plate) with specied opening. Assumptions : Fl behind weir V = 0 Pressure in nappe atmospheric No energy losses Streamlines horizontal over crest

7 H 1 H δh Vee Start from Q = AV Consider the thin strip marked A = bh

8 Vee Applying Bernoulli gives p V = 2gh so Q = bhp 2gh and thus Z H Q = dq 0

9 Vee Applying Bernoulli gives p V = 2gh so Q = bhp 2gh and thus Z H Z H Q = dq = 0 0 bp 2gh dh

10 Vee Applying Bernoulli gives so and thus Q = Z H 0 dq = V = p 2gh Q = bhp 2gh Z H 0 b p 2gh dh = b p 2g 2 3 H 3=2

11 Vee N.B. This is wrong : we usually measure H 1, not H A more detailed analysis can be carried out, with ' H 1 =3, giving 2 p2g 3=2 Q = 0:81 bh 1 3 This is still inaccurate : we have neglected the velocity head upstream (important for small channels). In reality, just use the formula 2 p 3=2 Q = C d 2g bh 3 where we have assumed H 1 ' H C d is a coecient to be determined for each weir. Generally C d ' 0:62

12 Vee notch wier Vee A similar analysis for a weir of angle is possible. In this case the width of the strip and the simple analysis gives b = 2(H h) tan =2 Q = 8 15 p 2g tan 2 H 5=2

13 b 1 V 1 b 2 V 2 Vee : : : plan view Width of channel reduced ) speeds up ) discharge per unit width increases. Specic energy constant ) for subcritical, depth decreases ) for supercritical, depth increases This arrangement can be used for measurement.

14 Vee If the free surface does not pass through the critical depth, we call this a venturi ume h d 1 d 2 Total energy Critical depth If the free surface passes through the critical depth in the throat, then we have a standing wave ume Total energy h d 1 d c hydraulic jump