A 600-mm-radius sewer pipe is laid on a slope of 0.001 and has a 0.012, was found to be 7/8 full. Determine the discharge roughness coefficient n = through the pipe.
A 600-mm-radius sewer pipe is laid on a slope of 0.001 and has a 0.012, was found to be 7/8 full. Determine the discharge roughness coefficient n = through the pipe.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Explain every step of the solution thank you.
![A 600-mm-radius sewer pipe is laid on a slope of 0.001 and has a
coefficient n = 0.012, was found to be 7/8 full. Determine the
through the pipe.
roughness
discharge
Problem 8-5 (CE Board May 1993)
Solution
A
r 0.6 m
r= 0.6 m
4.
360-0
0.6 m
Q = A - R2/351/2
A=D 좋Atotal
T (0.6)2
%3D
A = 0.9896 m2
A1 = Atwtal - A = a 1²
A1 = Asector Atriangle
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9524bfbe-478a-4502-b535-498e402c3883%2Fd1621f7a-7a96-417e-a46d-00e9ea7c61f8%2Fzxztjdl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 600-mm-radius sewer pipe is laid on a slope of 0.001 and has a
coefficient n = 0.012, was found to be 7/8 full. Determine the
through the pipe.
roughness
discharge
Problem 8-5 (CE Board May 1993)
Solution
A
r 0.6 m
r= 0.6 m
4.
360-0
0.6 m
Q = A - R2/351/2
A=D 좋Atotal
T (0.6)2
%3D
A = 0.9896 m2
A1 = Atwtal - A = a 1²
A1 = Asector Atriangle
%3D
![FLUID MECHANICS
&HYDRAULICS
CHAPTER EIGHT
Open Cnannel
a2= 0,-r² sin 0
0,-sin 0 0.785
Solve 0 by trial and error:
0 = 101.185°
%3D
101.185°(T/180°) – sin (101.185°) = 0.785 (OK)
Then;
T((0.6)(360° - 101.185°)
P =
180°
P 2.71 m
R = A/P = 0.9896/2.71
R 0.365 m.
Q= (0.9896) dz (0.365)/(0.001)/2
Q = 1.332 m/s
Problem 8- 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9524bfbe-478a-4502-b535-498e402c3883%2Fd1621f7a-7a96-417e-a46d-00e9ea7c61f8%2F5po53yo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:FLUID MECHANICS
&HYDRAULICS
CHAPTER EIGHT
Open Cnannel
a2= 0,-r² sin 0
0,-sin 0 0.785
Solve 0 by trial and error:
0 = 101.185°
%3D
101.185°(T/180°) – sin (101.185°) = 0.785 (OK)
Then;
T((0.6)(360° - 101.185°)
P =
180°
P 2.71 m
R = A/P = 0.9896/2.71
R 0.365 m.
Q= (0.9896) dz (0.365)/(0.001)/2
Q = 1.332 m/s
Problem 8- 6
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