Question

Hydro Mechanics

Please solve the question in an instructive way (step by step), also with a legible (readable) handwriting, ty.

Q1) Water is flowing with a discharge rate of Q=96 m²/sec in a rectangular channel section of 4 m. in depth and 6 m. in width. There is a smooth downward step of 0.25 m. and a reduction of 3 m. in the channel width. a)Calculate the water depths before and after the step. b)If choking occurs, calculate the change in the step height to prevent the formation of choking and show on the figure the step height and water depths before and after the step. SIDE VIEW y=4 m Az= 0.25 m B=6 m. B=3 m. TOP VIEW

Answer 1

NOTE- The channel bed has been lower so it will be a negative hump case.
Solution - leiven that. Pog е ие — 1 - side view - 19, 4m oz=0.25m ! TH L1 B = 6m - 3m TOP VIEW TI TYTV - Discharge in the channel, is = 96 m3/sec Width of Meetangulere channel in section , Depth of watve in el estang ulon channel at reition 0. 7,34m B, om Downward step height Az - 0.25m width of channel in section ③, B₂ = 3m - specific energy of channel sehion & 4+ 96² E = y + y² egy, Bi? = 4.815m 2x9.81 X 4²x6² ". Specifie energy in section E: E, -AZ = 4.815 - (-0.25) = 5.065 Now, for this specific energy E2, het us calulate the

Page no-2 minimum width in section @ for flow to be posible. can be done by anuming Ez as the curred specific energy for a This width Bei where Benthe minimum possible width - 3 ye where, de = critical depth 13 V2 Where 92 Dischange per unit width in erection - Accelaration due to grovity. 9 43 196 Yc = 9.794 BC 9.81 B 2 ye 5.065 9.794 Be -> B = 4,940m But given width in section 5 is B₂ = 3m. So, there will be choking condition in section and flow will Mestablish after some time at britical clefth in section ③

Page no-3 . Critical flow depth in section ole where / = 4 96 = 32 m3/slm 3 V3 2. 32 - 4.708m 9.81 Y. - 4.708m This specific energy in rection will be - E, 3 2 Ye Critical specific energy, Ę = 3 yc, where ye is the critical depth <3 * 4.908 = 7.063m 7.065m So energy in rection ① will be E = Ein Az € = Etoz =4.063 - 0.25 = 6.813 m Let us now calulate the flow depth in ration ④ where y = new flow depth. 6.813 = y at 96² 249.81 xy2x62 y = 6.505m, 1.579m

bage no-4 subcritical higher depth. lyi -6.505m flow is So, take New flow depths after step 1. = 6.505m y, - 4.708m Joz=0.25m As we have seen that choking occurs in section due to reduction in width to 3m in rection ② Let us lower the channel bed additionally by im to prevent choking. Az'=-0.25-h Now, for B₂ = 3m, £₂ = 7.063m [Already calculated in (a) E-A2 = 7.063m 4.815- (-0-26-1) 7.063 n = 1.998 am h-2m Thus, we need to lower the bed additionally by 2m Change in step height = 2m Now, Let us calculate the flow depth in reation and . in rection ②, E = 7.063m => Y t. +7-063 specific energy 29 42.B

Pagenoeg = 4.741m & specific energy in section ® will be E = E + oz' <7.063+ (-0.25=2) E = 4.813., which is the same specific energy as in the begining & flow depth will not change & thus y, = 4m Step height ,02'=-0.25 -2 = -2.25m ly, = 4m y = 4.741m A2=-2.25m Note- Negative sign in the step height indicates the nature of hump. As the bed has been lowered to the sign of hump will be will be negative.