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Ans) Apply Bernoulli equation between point 1 and 3 as shown in given figure respectively,      P1 /  + V12 / 2 g + Z1 =  P3 /  + V32 / 2 g + Z3 + HL Since, both points 1 and 2 are open to atmosphere, pressure is only atmospheric so gauge pressure P1 = P3 = 0 Velocity at surface is negligible so V1 = 0  Elevation difference, z1 - z3 = 1.22 m  HL is head loss in system Putting values, 0 + 1.22 = 0 + V2/2g + 0 + HL   => HL = 1.22 - V2/2g  Now, According to question, Q = 79.29 L/s or 0.07929 m3/s => Velocity (V) = Q / A = 0.07929 / [(/4)(0.203)2] = 2.45 m/s => HL = 1.22 - [2.452 / (2 x 9.81)] = 0.914 m  According to question, l ... See the full answer