formula   0.405  0.003  Q L H 2   H    3 0.053  0.011  H  0.00112 H (iii)) Rehbock formula  0.403  Q L   Z Considering velocity of approach and End contraction, we have   3 3  2  0.1nH   Q C L H    h ha 2 2 3   d a   Considering velocity of approach and End contraction, we have 3 3   2 0.1nH Q  C L   H     h ha 2 2 3   d a   Q 11A rectangular channel 2m wide has a discharge of 250 LPS is measured by a right-angled V-notch. Find the position of the notch from the bed of the channel if maximum depthof water is not to exceed 1.3m.Take Cd = 0.62. (July 2013 , july 2015) Ans: Given: Q = 250 LPS = 0.25m3/s, Cd = 0.62, = 90,  Maximum depth in the channel not to exceed 1.3m The discharge through a right angled triangular notch is given by, The position of apex of the = Maximum Depth Height of water over – from the bed of channel of water in channel V-notch notch = 0.8m The position of apex of the from the bed of channel = (1.3 0.5) –
formula Q   0.405  0.003  L H 2  H    (iii)) Rehbock formula  Q  0.403  0.053 H  0.011 L 3  H  0.00112  Z  Considering velocity of approach and End contraction, we have 3 3 Q  2 C  L  0.1nH   H  h  2  ha 2  3 d  a    Considering velocity of approach and End contraction, we have 3 3 Q  2 C  L  0.1nH   H  h  2  ha 2  3 d  a    Q 11A rectangular channel 2m wide has a discharge of 250 LPS is measured by a right-angled V-notch. Find the position of the notch from the bed of the channel if maximum depthof water is not to exceed 1.3m.Take Cd = 0.62. (July 2013 , july 2015) Ans: Given: Q = 250 LPS = 0.25m3/s, Cd = 0.62,  = 90, Maximum depth in the channel not to exceed 1.3m The discharge through a right angled triangular notch is given by, The position of apex of the = Maximum Depth – Height of water over from the bed of channel of water in channel V-notch notch The position of apex of the from the bed of channel = (1.3 – 0.5) = 0.8m
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