Answer to Question #288562 in Classical Mechanics for joseph

Question

A flow of 834 cfs is conveyed in a 10-ft-wide rectangular channel at a depth of 6.0 ft. What is the specific energy of the channel? Is this flow subcritical or supercritical? If n = 0.014, what channel slope must be provided to produce a uniform flow at this depth? Check your solution using appropriate computer software.

Answer 1

Solution

Given data are

Flow rate Q=834cfs=23.35m³/sec.

Width w=10ft=3.048m

n=0.0014

Depth d=6ft=1.828m

Now

Specific energy

$E=\frac{Q^2}{2gA^2}+d$

Putting all values

$E=\frac{(23.35)^2}{2\times 9.8\times (3.048\times1.828)^2}+1.828$

E=2.723J

Now

For checking the flow rate

$F_r=\frac{\frac{Q}{A}}{\sqrt{gd}}$

$F_r=\frac{\frac{23.35}{3.048\times1.828}}{\sqrt{9.8\times1.828}}$

So $F_r=0.889<1$

So this is subcritical.

So

Slope for uniform flow

S=$\frac{1.49AR^{0.66}D^{0.5}}{n}$

Putting all values we get

S=$1.298\times10^{-3}$