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WorksheetsDay 44-Nov 4-worksheet-Diploma-OCH-Channel types&Chezys equation
Total questions: 15
Worksheet time: 10mins
A non-prismatic channel differs from a prismatic one because —
Bed slope and cross-section vary
Velocity distribution is uniform
It is always rectangular
Flow is always steady
In a mobile-boundary channel, boundaries —
Remain fixed under flow
Are non-deformable
Change shape due to fluid movement
Are always concrete-lined
If the hydraulic radius (R) of a channel doubles for the same slope and Chezy’s C, the velocity —
Halves
Becomes √2 times
Doubles
Increases four-fold
The unit of Chezy’s constant (C) is —
m/s
m¹ᐟ²/s
m²/s
Dimensionless
For uniform flow, bed slope (S₀) equals the water-surface slope (Sw).
False
Rigid-boundary channels deform under the action of flowing water.
True
False
For a triangular channel with θ = 45°, the side slope ratio is (a) .
In Chezy’s analysis, the average shear stress on bed is given by τ = (a) .
The force causing flow in an open channel is the component of (a) along the bed slope.
The symbol γ represents the (a) of water.
According to Chezy’s formula, discharge is given by Q = (a) .
The line connecting points of equal velocity in open-channel flow is called (a) .
If R = 0.8 m, S₀ = 0.0004, and C = 50, find velocity using Chezy’s formula.
(a)
Calculate the discharge through a channel having a bed slope 1 in 1000, area 12m2, hydraulic radius of 1.2m and Chezy’s constant being equal to 50.
(a)
The perimeter of a circular channel section is 18.84m, calculate the discharge through the channel when it is running full having a bed slope of 1 in 1500 and C = 60.
(a)
