Answer:
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.
