Flow in a Partially Filled Drain Line
Problem solution was updated to correctly calculate angle of flow level in partially filled pipe, theta.
Given the following parameters for a partially filled cast iron pipe drain line, determine the approximate flow in cubic feet per second.
Slope: \(S=1 \%\)
Radius: \(r=6 \ in\)
Depth of Flow: \(y=8 \ in\)
A) 3.8 cfs
B) 7.4 cfs
C) 8.3 cfs
D) 10.0 cfs
Solution
Answer: A) 3.8 cfs
Manning’s Roughness Coefficient for a Cast Iron Pipe:
\[n=0.013\]
Depth of Flow:
\[y=8 \ in\]
Radius:
\[r=6 \ in\]
Unfilled Segment Height:
\[h=2*6 \ in-8 \ in=4 \ in\]
\[θ=\sout{2*cos^{-1}(\frac{r-h}{h})}\]
\[θ=2*cos^{-1}(\frac{r-h}{r})\]\[\sout{θ=2*cos^{-1}(\frac{6 \ in- 4 \ in}{4 \ in})}\]
\[θ=2*cos^{-1}(\frac{6 \ in- 4 \ in}{6 \ in})\]\[\sout{θ=2.09 \ rad}\]\[θ=2.46 \ rad\]
Calculate Area of Flow:
\[A=πr^2- \frac{r^2 (θ-sin \ θ)}{2}\]\[A=\sout{π(6 \ in)^2}- \frac{\sout{(6 \ in)^2 (2.09 -sin \ 2.09)}}{\sout{2}}\]\[A=π(6 \ in)^2- \frac{(6 \ in)^2 (2.46 -sin \ 2.46)}{2}\]
\[A=0.63 \ ft^2\]
\[A=0.56 \ ft^2\]Calculate Wetted Perimeter:
\[P=rθ\]
\[\sout{P=0.5 \ ft*2.09 \ rad=1.05 \ ft}\]\[P=0.5 \ ft*2.46 \ rad=1.23 \ ft\]
Calculate Hydraulic Radius:
\[R=\frac AP\]
\[\sout{R= \frac{0.63 \ ft^2}{1.05 \ ft}=0.6 \ ft}\]
\[R= \frac{0.56 \ ft^2}{1.23 \ ft}=0.45 \ ft\]Calculate Flow Based on Manning’s Equation:
\[Q=\frac{1.49}{n} AR^{2/3} S^{1/2}\]\[\sout{Q=\frac{1.49}{0.013} (0.63 \ ft^2 )(0.6 \ ft)^{2/3} (0.01)^{1/2}}\]\[Q=\frac{1.49}{0.013} (0.56 \ ft^2 )(0.45 \ ft)^{2/3} (0.01)^{1/2}\]
\[\sout{Q=5.1 \ cfs}\]\[Q=3.77 \ cfs\]