Question

An unfinished concrete semicircular channel with diameter of 3 ft is used to transport water over a distance of 1 mile. If flow is 90 ft3/s when the channel is full, calculate the minimum elevation difference in ft.

Answer 1

448.8 ft

Step-by-step explanation:

Area of semi-circle=

Hydraulic radius is given by

`R_h=\frac {A_c}{\pi R}` where
`\pi R` is wetted perimeter,
`A_c` is cross-sectional flow area and
`R_h` is hydraulic radius

`R_h=\frac {3.5325}{\pi (1.5)}=0.74 ft`

From Manning’s equation, flow rate is given as

`V=\frac {a A_c R_h^(2/3)S_o^(0.5)}{n}`

Here,
`S_o` is bottom slope,
`a=1.486 ft^(1/3)/s`

Manning’s coefficient is taken as n=0.014 for open unfinished concrete

Substituting 90 for v, 1.486 for a, 3.5325 for
`A_c`, 0.74 for
`R_h` and 0.014 for n

`90=\frac {1.486* 3.5325* 0.74^(2/3)* S_o^(0.5)}{0.014}`

`4.29S_o^(0.5)=1.26`

`S_0^(0.5)=0.293`

`S_o=0.085`

Change in elevation is given by

`\triangle z=S_o L` where L is distance, here given as 1 mile

1 mile=5280 ft

`\triangle z=0.085* 5280=448.8 ft`