2 0.2612 a  ha 2   0.0035m 19.62  g 2    Q C L H H 32 0.1  n   h  h 3 d a a Q = m3/s 2 0.63   Q 3 0.4 0.225  0.1 2   0.00350.225  0.003532  0.07186 Q New Va=  0.266 / m s 0.6 A  0.45 2  h a  0.266  2  19.62 a g New Q = 0.0719 m3/s Q.7It is required to establish the throat diameter of a venturimeter in an installation of 100mm diameter pipe conveying water. The maximum range available in differential manometer gauge is 50cm of mercury deflection. Find the maximum throat diameter which will indicate the fill gauge deflection when the flow rate is 20 LPS assuming coefficient of venturimeter as 0.984. (July 2014) Given D1 = 100mm = 0.1m, Q = 20 LPS = 0.02 m3/s, Cd = 0.984 Ans: Deflection of mercury manometer deflection (x) = 50cm = 0.5m The discharge through Orifice is given by,               2    0.984   g h 0.031416  Q  Cd     2 2 9.81  h     A1 1        A           2 D1 = 0.2m,
ha  2 a 2  g 0.2612  19.62  0.0035m Q  2  C  L  0.1 n  H  h H  h 32 3 d a a Q  2  0.63 3 Q = m3/s 0.4  0.1 2  0.225  0.00350.225  0.003532 New Va= Q  0.07186 A 0.6  0.45  0.266 m / s 2 h  a  0.266  a 2  g 19.62 New Q = 0.0719 m3/s Q.7It is required to establish the throat diameter of a venturimeter in an installation of 100mm diameter pipe conveying water. The maximum range available in differential manometer gauge is 50cm of mercury deflection. Find the maximum throat diameter which will indicate the fill gauge deflection when the flow rate is 20 LPS assuming coefficient of venturimeter as 0.984. (July 2014) Ans: Given D1 = 100mm = 0.1m, Q = 20 LPS = 0.02 m3/s, Cd = 0.984 Deflection of mercury manometer deflection (x) = 50cm = 0.5m The discharge through Orifice is given by,             Q  Cd   2   2  g  h  0.984     0.031416    2  9.81 h   A1   1      A       2     D1 = 0.2m,