CPENG78
Civil/Environmental
- Sep 2, 2008
- 186
I need a little push with this calculation. Even though my initial reaction is that this is a simple calculation, I think the lack of sleep is catching up to me (crazy deadline for a different project)and not letting me visualize the approach.
So I have a 54 inch pipe (193 lf) which outfalls into a sump situation. Water builds up in the pipe and is allowed to bubble up over a ridge (top of ridge elevation = 7.6 ft) to eventually end up in open space (wetlands). The invert of the pipe at the outfall (sump) is 2.62. The 1st manhole upstream is at invert 3.94 and the rim is at 9.5, at which point pipe changes in size to a 42 inch line. There are 243 lf of this 42 inch pipe before it reaches the 2nd manhole upstream at invert elevation of 4.96 and rim elevation of 10.6.
Assuming the 54 inch pipe is completely submerged, what is the required head you would need to push that water so that it will continue to bubble up over the ridge?
I started with Bernoulli's equation, from the point of interest of unknown elevation to the ridge at 7.6.
pressue head at either end is 0 since I'm assuming both ends open to the atmosphere (assuming water surface will not exceed the rim elevation creating pressurized flow)
velocity in both the 54 inch and 42 inch pipes is determined as 7.2 fps and 12 fps respectively based on assumption that pipes are running full and the Q = 115 cfs. Velocity head at 54 inch determined to be 2.24 and the veolocity head at the 42 inch is 0.80.
I'm ignoring friction losses for now.
Therefore am I oversimplifying this and determiing the head required to push the water over the ridge to be 0.80-2.24 or 1.44 ft? Just doesn't sound right to me. Any input on what I'm missing is greatly appreciated.
So I have a 54 inch pipe (193 lf) which outfalls into a sump situation. Water builds up in the pipe and is allowed to bubble up over a ridge (top of ridge elevation = 7.6 ft) to eventually end up in open space (wetlands). The invert of the pipe at the outfall (sump) is 2.62. The 1st manhole upstream is at invert 3.94 and the rim is at 9.5, at which point pipe changes in size to a 42 inch line. There are 243 lf of this 42 inch pipe before it reaches the 2nd manhole upstream at invert elevation of 4.96 and rim elevation of 10.6.
Assuming the 54 inch pipe is completely submerged, what is the required head you would need to push that water so that it will continue to bubble up over the ridge?
I started with Bernoulli's equation, from the point of interest of unknown elevation to the ridge at 7.6.
pressue head at either end is 0 since I'm assuming both ends open to the atmosphere (assuming water surface will not exceed the rim elevation creating pressurized flow)
velocity in both the 54 inch and 42 inch pipes is determined as 7.2 fps and 12 fps respectively based on assumption that pipes are running full and the Q = 115 cfs. Velocity head at 54 inch determined to be 2.24 and the veolocity head at the 42 inch is 0.80.
I'm ignoring friction losses for now.
Therefore am I oversimplifying this and determiing the head required to push the water over the ridge to be 0.80-2.24 or 1.44 ft? Just doesn't sound right to me. Any input on what I'm missing is greatly appreciated.
