last minute questions

[zxu]

a 4-foot diameter concrete interceptor sewer (n = 0.013, s = 0.02) carries wastewater at a depth of 1.5 feet. what is the velocity of flow (assume n varies with depth).

the given answer

step 1:

d/D= 1.5/4

use App. 16A, r=0.2061*4=.83

v using manning equation = 14.3cfs

step 2

then using App 19.C

d/D=1.5/4 get v/vfull=0.68

step 3

then the final v=0.68*14.3=9.8cfs

my confusion is why not calculate the velocity when the piple is full (4ft) and then use the Apendix 19.3 for adjustment.

thanks

 

[cantaloup]

a 4-foot diameter concrete interceptor sewer (n = 0.013, s = 0.02) carries wastewater at a depth of 1.5 feet. what is the velocity of flow (assume n varies with depth).

I think you got incorrect answer. The correct answer is 14.1 fps

Here is my 2 solutions:

Sol 1: V(full flow) = 16.17 fps

d/D = 1.5/4 = 0.375

from App 19C (assume constant n) with d/D = 0.375, V/V(full) = 0.875

then V = 16.17 * 0.875 = 14.15 fps

Sol 2:

from App 16A, with d/D = 0.375, r(h)/D = 0. 2040 (approx. interpolated)

then r(h) = 0.2040 * 4 = 0.816

Use Manning equation with r(h) then calculate V = 14.11 fps

 

[jartgo]

a 4-foot diameter concrete interceptor sewer (n = 0.013, s = 0.02) carries wastewater at a depth of 1.5 feet. what is the velocity of flow (assume n varies with depth).
the given answer

step 1:

d/D= 1.5/4

use App. 16A, r=0.2061*4=.83

v using manning equation = 14.3cfs

step 2

then using App 19.C

d/D=1.5/4 get v/vfull=0.68

step 3

then the final v=0.68*14.3=9.8cfs

my confusion is why not calculate the velocity when the piple is full (4ft) and then use the Apendix 19.3 for adjustment.

thanks

I believe the solution that is confusing you, is incorrect. In step 1, you need to calculate velocity at full flow, then v/vfull is about 0.68 as you suggest, then actual v would be about 11fps.

 

[jartgo]

I think you got incorrect answer. The correct answer is 14.1 fps
Here is my 2 solutions:

Sol 1: V(full flow) = 16.17 fps

d/D = 1.5/4 = 0.375

from App 19C (assume constant n) with d/D = 0.375, V/V(full) = 0.875

then V = 16.17 * 0.875 = 14.15 fps

Sol 2:

from App 16A, with d/D = 0.375, r(h)/D = 0. 2040 (approx. interpolated)

then r(h) = 0.2040 * 4 = 0.816

Use Manning equation with r(h) then calculate V = 14.11 fps

cantaloup, the question asked to assume n varies with depth...your solution would be correct for a constant n (as you pointed out in your assumptions). For varying n, v/vfull is 0.68. I still stand behind the answer of V is about 11fps.