Arectangular open channel is 20 ft wide and has a bed slope of 0.007. manning's roughness coefficient n is 0.03. it is in uniform flow condition with a discharge of 400 cfs. (a) calculate the normal depth. (b) calculate the froude number. (c) calculate the critical depth and velocity. (d) calculate the critical slope (e) determine if the flow is in the sub-critical or super-critical state.

a) normal depth, yn = 3.7ft

b) Froude number, Fr = 0.49

c) Critical depth, yc = 2.3 ft; Velocity, v = 5.4ft/s

d) Critical slope, Sc = 14.784

e) subcritical

Explanation:

Width , b= 20ft,

Bed slope, s= 0.007

Manning's roughness coefficient, n = 0.03

Discharge, Q= 400 cfs

a) normal depth, yn

Q = A(1/n x R⅔ x s½)

A = b x yn, R = b.yn/(b + 2yn)

Substituting values,

400 = 20(yn)/0.03 x (20 x yn/20 + 2yn)⅔ x (0.007)½

Simplifying further,

yn^(5/2) - 1.918yn = 19.18

Using trial and error,

yn = 3.7ft

Therefore the normal depth is 3.7 ft

b) Froude number, Fr = v/√gD

Where D is hydraulic depth given as ,

D = A/T

D = yn (for rectangular channel

Acceleration, g = 32.71 ft/s²

V = Q/A

Fr = 400/(20x3.7)x√(32.71 x 3.7)

Fr = 0.49

c) critical depth, yc = (Q²/gb²)⅓

yc = (400²/32.71*20²)⅓

yc = 2.3 ft

Velocity, v = Q/A

v = 400/(3.7*20)

v = 5.4ft/s

d) critical slope ,Sc

Sc = (Qyc = 2.3 ftn/1.49Ac(Rhc)⅔)²

Ac = b x yc, Rhc = (Ac/b + 2yc)⅔

Substituting,

Sc = (400/1.49 x 20 x 2.3 x ( 20 x 2.3 / 20 + (2 x 2.3))⅔)2

Sc = 14.784

e) since the normal depth is greater than critical depth, it is subcritical.

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explanation: