CHEMICAL ENGINEERING

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.