6 (i) What are the conditions necessary to form a hydraulic jump? (5 marks) (ii)

Question

6 (i) What are the conditions necessary to form a hydraulic jump? (5 marks) (ii) The flow in an open channel of 0.25 m2 cross sectional area goes over a sharp crested rectangular weir 500 mm wide with a head of 150 mm. Calculate the discharge assuming that (a) the approach velocity in the channel is negligible, and, (b) the approach velocity in the channel is NOT negligible. (7 marks) (ii) If the same flow subsequently goes over a sharp crested V-notch with a subtended angle of 90, what would be the head assuming the approach velocity is negligible Assume that the coefficient of discharge for the rectangular weir is 0.65 and that for the triangular weir is 0.60. (8 marks)

Explanation / Answer

Answer 6 (i)

Hydraulic jump is a natural phenomenon that occurs whenever flow changes from supercritical to subcritical flow. A flow is said to be supercritical when froude number of flow Fr>1 and when Fr <1 then the flow is said to be subcritical. Thus the condition of transition from supercritical to subcritical flow is most necessary to form hydraulic jump. Also specific energy is lost in form of heat but law of conservation of momentum is applicable.

Answer 6 (ii)

a) discharge without considering velocity of approach

Q= (2/3)Cd B sqr(2g) H^(3/2) = (2/3)*0.65*0.5*sqrt(2*9.81)*(0.15^(3/2)) = 0.055 m3/s

b) considering velocity of approach

Va= 0.055/0.25 = 0.22 m/s

Considering Kinetic energy correction factor = 1

Ha= 1*0.22*0.22/(2*9.81) = 2.467*10^-3 m

Thus Q= (2/3)Cd B sqrt(2*9.81)((H+Ha)^1.5 - Ha^1.5)

Q= 0.057 m3/s

Answer 6(iii)

Q= (8/15) Cd( H^2.5 ) sqrt(2*9.81) tan(90/2) =

0.055= 1.417 H^2.5 thus H = 0.2726 m

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