Question

Q3. Water flows along the rectangular channel such that after it falls to the lower elevation, the depth h=0.3 m. Find ), ), ), 0 ) h=0.3 m. ) ) 092425, 26, Zp, Zą. Zr. YZ 29 2g' 28' 28' 29 Determine the volumetric discharge through the channel. The channel has a width of 1.5 m. 0.5m JA 0.25m I 0.25m h = 0.3m To.15m 0.15m

Answer 1

Sol-

Water flows along Rectangular channel such that after it falls from higher elevation to lower elevation.

given that h=0.3m

Section 2 Soli 0.5m 0.25% 0.25m / Section 2 4 0.3m {м - 10:15h - 10.15m Note: A point should be remember that when A laminar flow flows through circuler Pipe. the velocity distribution into the pipe can be describe) defined as Chraph ---C (contre I line) of Centralous According to the above graph the velocity sich flowing water inside the pike at point AC, P, R will be zero' (VA = Ve a Vp = Vo = 0] so. line of the pipe as Datum let Assume bottom So:

ZA = 1+0.5 1.5m ZB = 1 +0.25 = 1.25m. Z Z Im Ze & 0.3m Zu= 0.15m Ri 0 Using Bernaullie equation b/w porsect & section & ( given h el to get the best you + (0.15 –1.25) + ces using continuity Equation cethlwsection ① & sectima AB VB = ARVO I x 0.5² xVB = 1 x 0.32 x a. ve = 278 VB ] - so that his prezometric head = ( 12 +23) - (1 120)

B - Pen 0.3-7R + 20 = 0.3-185+0.15 Vw substituting the value of Eq(3) into Eq11) -0.8 & VB²-vo? =-11. 28 Vg2 V6 2 -0.2 25 now substituting the value of eq12) in above equation NB² - (2-18val 2013 29 6.73 Vg2 = 0.3 29 Vg2 2 0.874 FVR = 0.93 ms) & Vo = 278x08 mit 0.93 - 2.6mis. 0.93 = 0.1325m 26.0414m VB - 28 2x9.81 Now volumetric discharge through the channal

is a = - = ABVB 1 x 0.52 x 0.93 0.183 m3/ Note: - I hope you understand the concept very well. I am veryporalittle a sorry change. circuler. is given I assume the here in the rectanguler. pipe as Question It Sonry Los Now I hope you can make it.