Open water jet against an inclined plane.
Open water jet against an inclined plane.
(OP)
I'm trying to figure out how far up an asphalt road section a jet of water would go. The flow rate is 5.84 cfs at a velocity of 11.8 fps. The Manning's n-value is 0.016 for the road section which is inclined upward at 2%. My references have equations that assume no friction so I'm coming up with some pretty large and unrealistic numbers.





RE: Open water jet against an inclined plane.
I may have just caused Bernoulli to roll over in his grave with my mathematical abortion, so use this info w/ caution.
RE: Open water jet against an inclined plane.
RE: Open water jet against an inclined plane.
RE: Open water jet against an inclined plane.
RE: Open water jet against an inclined plane.
RE: Open water jet against an inclined plane.
Just Curious
Rik
RE: Open water jet against an inclined plane.
RE: Open water jet against an inclined plane.
RE: Open water jet against an inclined plane.
Primary energy losses will result from these factors:
1. Friction of the water flowing over asphalt.
2. Water flowing up the 2% slope (stored as potential energy)
Friction: Water flowing a on level, frictionless surface would move horizontally "forever" (think of this as positive drainage). On the real asphalt surface described what would it take to establish (minimal) positive drainage? Using engineering judgment, I will say a slope of 1.5% (sloped in the direction of the flowing water)
Therefore friction can be accounted for as the equivalent of water flowing UP a hypothetical 1.5% slope.
Water flowing up a 2% slope: From the discussion above, looks like all of you accept this concept as a valid energy loss.
As Denial astutely observed, there is another important factor; gravity is continuously diverting water down the (sideways) 5% slope, but I'll leave that out, for now.
So much for the assumptions, now for the math:
The solvable equivalent approximation has the the 11.8 fps water jet flowing up a theoretical 3.5% slope (2% true slope plus 1.5% hypothetical friction equivalent slope)
proletariat's use of Bernoulli's equation should give the same answer, but I like to use Newtons Laws (assume that all energy from the 11.8 fps jet is absorbed by converting it into potential energy):
Velocity = Acceleration x Time
11.8 fps = (32.2 ft/sec^2) x Time
Time = 0.366 seconds
Vertical Distance = (1/2) x Acceleration x Time^2
Vertical Distance = (1/2) x (32.2 ft/sec^2) x (0.366 Seconds)^2
Vertical Distance = 2.16 ft
Horizontal Distance = (2.16 ft) / (3.5% theoretical slope)
Horizontal Distance = 62 ft
Because of other losses, such as the sideways 5% slope, impact of the water jet on the pavement, etc., the distance moved uphill will be less than 62 ft. If you ask me for my "best estimate", I would say 50 feet (assumed one significant figure accuracy).
I know this "solution" leaves a lot out, but that is how this sort of thing was approximated in the past (pre-computer).
www.SlideRuleEra.net
RE: Open water jet against an inclined plane.