Find normal depth and Froude number?

[mhanna632]

A symmetric trapezoidal section, 1900 mm deep and with top and bottom widths 3 m and 0.7 m respectively carries water at a rate of 3 m3/sec. Manning’s n may be taken as 0.012. Find:

i. The normal depth at a slope of 1 in 1500

ii. The Froude number at the normal depth

Another past paper question

 

[Happy]

This is similar to your last questions:

Q=(1/n)(A)(R^.66666)(S)^.5

n=.012

Find Angle@ (see CERM Table 19.2)

Given the T=3, b=.7 and d(full)=1.9, You have 1.15H for 1.9V

Therefore Angle@=58.81 deg

A=(b+d/tan@)d

R=(bdsin@ + d^2cos@)/(bsin@ + 2d)

S=1/1500

S=.000666

Plug into equation above to find d,

Q=(1/n)(A)(R^.66666)(S)^.5

Q=(1/n) [(b+d/tan@)d] [(bdsin@ + d^2cos@)/(bsin@ + 2d)] [s^.5]

3=(1/.012) [(.7+d/tan 58.81)d] [(.7dsin58.81 + d^2cos58.81)/(.7sin58.81 +2d)] [.000666^.5]

d=1.375

 

[Happy]

Froude #, section 19-27 CERM

Fr=v/(gL)^.5

For trapezoidal sections, L is A/T

from above, A=2.1069 m^2

L=2.1069/3

L=.7023

v=Q/A

v=3/2.1069

v=1.42m/s

Fr=1.42 / [ (9.81)(.7023) ]^.5

Fr=.541