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HELP PLEASE
- Thread starter NJPHIL3
- Start date
D
Deleted member 4993
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Can you post a photocopy of the assignment? The problem sounds "incomplete"!IF IT TAKES 33 MINS FOR THE WATER TO FALL 1 INCH....HOW FAR DOES IT FALL AFTER 10 MINS
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem.
i REALLY HAVE NO IDEA WHERE TO START. i AM TRYING TO CALCULATE THE PERC RATE IN MY YARD FOR A SEPTIC. IT TOOK 198 MINS FOR THE WATER TO FALL 6" SO THAT IS 33 MINS AN INCH BUT I NEED TO FIGURE OUT HOW FAR IT FELL IN TENTHS OF AN INCH AFTER 10 MINS TO FILL OUT A FORM FOR THE TOWN
D
Deleted member 4993
Guest
Please do not design a real septic tank out of these totally approximate data.IS IT 0.303" EVERY TEN MINS????
You might be inviting whole host of nasty problems - and it will take lot of resources to correct the flaws.
Please go to your local municipal office and start a discussion with an engineer.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
If you are talking about "percolation" then "fall" is the wrong word. After reading your first post, I started thinking about water falling under gravity where the acceleration is constant but downward velocity is increases.
In percolation the rate is constant so if "the level of water decreases by 1 inch in 33 minutes" then, since 10 minutes is \(\displaystyle \frac{10}{33}\) of 33 minutes, the level of water dercreases \(\displaystyle \frac{10}{33}(1)=\frac{10}{33}\) inches.
In percolation the rate is constant so if "the level of water decreases by 1 inch in 33 minutes" then, since 10 minutes is \(\displaystyle \frac{10}{33}\) of 33 minutes, the level of water dercreases \(\displaystyle \frac{10}{33}(1)=\frac{10}{33}\) inches.
D
Deleted member 4993
Guest
It could also be "seepage" - where the rate decreases logarithmically because the ground gets saturated. OP was not willing to investigate - instead threw a temper tantrum.If you are talking about "percolation" then "fall" is the wrong word. After reading your first post, I started thinking about water falling under gravity where the acceleration is constant but downward velocity is increases.
In percolation the rate is constant so if "the level of water decreases by 1 inch in 33 minutes" then, since 10 minutes is \(\displaystyle \frac{10}{33}\) of 33 minutes, the level of water decreases \(\displaystyle \frac{10}{33}(1)=\frac{10}{33}\) inches.