On similar lines, considering a strip of uniform thickness dh at a depth h below the liquid surface. Discharge through the dq=area x velocity.  dQ Lxdhx Therefore discharge is given by 1 Q H 2       dq L H ha dh 0 0 3 3 2  2     L H ha ha 2   3 3 2   2 Q  C  ha   L H h 2 3   d a   Empirical Formula 3 3    (i)Francis Formula 1.84 0.1nH       Q L H ha ha 2 2   Q3.Water flows over a rectangular weir 1m wide at a depth of 15cm and afterwards passes through a right angled weir. Taking Cd for rectangular weir 0.62 and for triangular 0.59 calculate the depth over the triangular weir (July2014,July 2015) Ans: Given For Rectangular Weir Cd = 0.62, H = 15cm, L = 1m For Triangular Weir Cd = 0.59, = 90  Let the depth over triangular weir be ‘H1’. As the same discharge passes through both the weirs equating the discharge for Rectangular and triangular weir    tan H 52 C  C  3   2    8   Q H 2 3 15 Triangular  2  Re tan d c gular d 1     2  0.62 1 90 Q 8 0.59 tan   0.1532 H 52    3 15 2    0.3573m H1
On similar lines, considering a strip of uniform thickness dh at a depth h below the liquid surface. Discharge through the dq=area x velocity. dQ  Lxdhx Therefore discharge is given by 1 Q H 2  dq  L   H  ha  dh 0 0 3 3  L 2  H  ha 2  ha 2    3 3 Q  2 C  L  H  h 2  ha 2  3 d  a    Empirical Formula 3 3 (i)Francis Formula Q  1.84  L  0.1nH   H  ha  2  ha 2    Q3.Water flows over a rectangular weir 1m wide at a depth of 15cm and afterwards passes through a right angled weir. Taking Cd for rectangular weir 0.62 and for triangular 0.59 calculate the depth over the triangular weir (July2014,July 2015) Ans: Given For Rectangular Weir Cd = 0.62, H = 15cm, L = 1m For Triangular Weir Cd = 0.59,  = 90 Let the depth over triangular weir be ‘H1’. As the same discharge passes through both the weirs equating the discharge for Rectangular and triangular weir    Q  2  C  3  H 2  8  C   tan  H 52 3 d Re c tan gular 15 d Triangular  2  1    Q  2  0.62 1  0.1532  8  0.59  tan 90  H 52 3 H1  0.3573m 15  2 
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