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.
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Open water jet against an inclined plane.
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proletariat
Civil/Environmental
The only thing I can think of is to use Bernoulli's equation after you figure out your fractions of flow up and down the inclined plane (almost 100% will go up w/ a 2% slope). One side would have z=0 and v=5.84ft/s. The other side would have v=0 and solve for z. This gives you z= 0.52', which is a no-friction best case scenario as if the water were squirting straight up out of a pipe. If you figure the water will gain .52' vertically, then it will run up the 2% (frictionless) road about 26' based on momentum alone. You would be the man (or woman) if you figured out how to get a z head loss from a mannings n to plug into your equation to give your road friction.
I may have just caused Bernoulli to roll over in his grave with my mathematical abortion, so use this info w/ caution.
I may have just caused Bernoulli to roll over in his grave with my mathematical abortion, so use this info w/ caution.
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Thanks 'proletariat' for your response. The actual velocity stated is a big 11.8 fps, the Q is 5.84 cfs. It appears that the main problem is factoring in the friction component of the road section as you acknowledged. I'm still working on this issue,any input is appreciated. Thanks again.
Not wishing to complicate matters any, but where does the water go after it has gone up the slope however far it wants to go? Does it move to one side and "get out of the way", due to a camber on the sloping surface? Or does it start to move down the slope again, as would a ball that you rolled up the slope?
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- #5
Hi Denial. The road-section is sloped upward 2% in the direction of the water jet (parallel). The road is sloped 5% perpendicular to the jet stream, therefore flow will run up the road x-section and then run downwards, perpendicular to the jet stream along the 5% slope of the road. In otherwords, the inclined plane is sloped both in the x & y directions. Hope my explanation clarifies.
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- #8
This is not a homework problem, this is realilty. I designed a detention basin that outlets into an existing street section. When the basin is working at the maximum design for a 100-year storm event the flow rate & velocity is as stated. Emperical data was provided by witness to the basin's discharge when it was plugged w/ debrie and then un-plugged, resulting in the 'water-jet' analogy. There has to be a way of determining how far a flow of water will go up an incline given the velocity of flow and the friction of the incline's plane. The challenge is mine and your's if so taken. Thanks for your time.
proletariat
Civil/Environmental
Maybe looking into some hydraulic jump equations will help you?
SlideRuleEra
Structural
I can offer a slide rule era (no pun intended) first order approximation. By that, I mean that factors that I don't know how to calculate will be converted into (approximate) equivalent factors that I can calculate. It is somewhat tedious for this problem, but not difficult - please bear with me.
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).
![[reading]](/data/assets/smilies/reading.gif "[reading] [reading]")
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).
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- #11
Hi SlideRuleEra. The 2.16' vertical distance appears consistent w/ my calcs & others of this thread. I think that in reality since this 'water-jet' is un-confined & will fan-out, that it would not shoot out into the street as far as the calcs state, added the 5% perpendicular street component, just a 'gut-feeling'. I think I've beaten this issue into the ground enough, I thank all participates for your input. Bye for now......
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