What am I doing wrong?
Find the quantity and velocity of flow in an 18" dia conc pipe flowing 94% full. The pipe has a slope of 1%.
First have to figure the central angle for which I have been using the following:
theta = 2 ACos(r-h/r)...i come up with 303.2847dd --->303-17-05dms
2
2
plug that into a = r(theta*pi -sin theta) .75 (303.2847*pi -303.2847sin)
2 ( 180 ) 2 (180 )
I get a = 1.25 book says I should have 1.72
I didn't have to size a pipe. Take the erosion control manual and know how to figure trap/v-ditches.
That is an absolutely unfair question. When they say 94% full do they mean 94% of the diameter as depth of flow or do they mean 94% of the cross section area.
Maybe that is where your issue with the math is.
Great example of a pathetic test question.
:good:
I paraphrased b/c I couldnt do sketch here, but they mean 94% depth. 16.92" tall water to bottom of pipe.
> What am I doing wrong?
>
> Find the quantity and velocity of flow in an 18" dia conc pipe flowing 94% full. The pipe has a slope of 1%.
>
> First have to figure the central angle for which I have been using the following:
>
>
> theta = 2 ACos(r-h/r)...i come up with 303.2847dd --->303-17-05dms
>
> 2
> 2
> plug that into a = r(theta*pi -sin theta) .75 (303.2847*pi -303.2847sin)
> 2 ( 180 ) 2 (180 )
>
>
> I get a = 1.25 book says I should have 1.72
by my calcs, from a pipe chart; 18" rcp at 1%
v=6.6 fps
q=12 cfs
my presumption is 94% full puts the flow at very close to the critical flow curve, so i used that.
interesting that you are not being tested my a more practical approach, such as pipe performance or pipe sizing in a sd circuit, or hgl
Open channel flow
Have fun figuring the cross-sectional area without using a computer to do it for you.
some worthwhile reading:
http://www.cedengineering.com/upload/Partially%20Full%20Pipe%20Flow%20Calculations.pdf
http://www.wsdot.wa.gov/publications/manuals/fulltext/M22-24/FieldFormulas.pdf