78 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Substituting forwin the equation for dQand integrating to give the discharge rate over
the notchQthen:
QD 2
√
2 g k
∫H
0
hnh^0.^5 dh
D 2
√
2 g k
∫H
0
hnC^0.^5 dh
D 2
√
2 g k[1/nC 1. 5 ]HnC^1.^5
Since it is required thatQ/H:
nC 1. 5 D 1
and: nD 0. 5
Thus: QD 2
√
2 g kH
SinceQD 0 .1m^3 /s whenHD 0 .15 m:
kD 0. 1 / 0. 15 [1/ 2
√
2 g]D 0 .0753 m^1.^5
Thus, withwandhin m: wD 0. 0753 h^0.^5
and, withwandhin mm: wD 2374 h^0.^5
and using this equation, the profile is plotted as shown in Figure 6a.
300 200 100 0 100
100
200
300
400
500
200 300
Distance from centre line, w (mm)
h (mm)
Figure 6a.