wow ... that moment when you realize that a 13 mins vidoe is 1000 times better than a 2 hours long lecture at the university. thank you so much Phil keep it up!
Thank you so much Phil! I am enrolled in an intro course to Geotech right now and my professor goes way too fast when explaining concepts. Your explanation is spot on and clear. Thank you for sharing!
It is the circles, guys. You have to have 4 tangent lines surrounding a circle. 1. You start with the equipotential lines at the ground level and impermeable layer 2. Then you draw flow lines than are parabolas that end perpendicular to the ground equipotential line. You can make how many you feel is right. (The fewer lines the bigger the circles you'll imagine later are going to be, the more lines, the more circles you'll have to draw but it'll be more accurate) 3. (The most difficult part.)Draw parabolic equipotential lines that are (1) perpendicular to all flow lines they cross and (2) a circle can be drawn in the resulting inner area that all sides touch the surrounding lines at one point (tangent lines to the circles) 4. Count-out flow channels and drops. I'm just a student btw, this just what I got from the video.
Im having a hard time concentrating on your lecture wondering how you wrote it backwards and explain it at the same time 😅😅.. That's just wowwwwwww👍👍👍👍👍👍
An example of unconfined flow with a sheet pile driven halfway into an aquifer is shown in Fig. 7.19(which is same is the figure you have drawn.). The flow net is horizontally symmetrical, providing that the top and bottom flow lines (borders) are parallel and the penetration of the sheet pile is exactly one-half the depth of the aquifer. Note that the number of flow channels Nf is 3 and the number of equipotential drops N dis 6. Regardless of how many flow channels you use, the Nf!Nd ratio, or the shape factor, will always be 1/2 for the geometry of this example! any comment on this?
Very helpful video. I have a question about the units. I realize that flow is expressed in volume per time, but if K is in (m/s), and H is in (m), should Q be in (m2/s)?
Phil how do you know how many flow lines and equipotential lines you need? I've read everywhere and no one gives me a straight answer or an equation to follow. Thank you!
Hi Ryan, good question. It doesn't matter for the maths as the number of equipotential lines will also increase. If you stick to the rules you should end up with the same Nf/Nh no matter how many flow lines are used. In practice, a flow net with too many lines is tricky to draw. Remember, you'll be rubbing things out and redrawing as you go along.
Question sir. How do you find the pore pressure of a certain point if it is located at the very center of equipotential line? We know that pore pressure = 9.81(H + h - mH/nd) Where H = head of water h = height of saturated soil above the point of interest m = number of equipotential drops from zero to point of interest nd = total number of equipotential drops. So how would you know the value of "m" in that case?
I often see values of 5.6,6.5,4.5, etc. when calculating Nf. How on earth is this possible if calculating the #of flow paths? Shouldn't this always be a whole number? Or does this only apply to anisthropic materials?
Wouldn't (4x10^-7 m^3/s) = (0.03456 m^3/day) = (12.61m^3/yr) = (12,600 L/yr) ? I think the value you calculated was the daily value and not the yearly. Although it seems like too large a value
wow ... that moment when you realize that a 13 mins vidoe is 1000 times better than a 2 hours long lecture at the university. thank you so much Phil keep it up!
right! Better than my post post post doc professor
Mate you are a legend. Very well explained. Thank you so much.
Thank you Phil..... Lots of love for you .....
Thank you so much Phil! I am enrolled in an intro course to Geotech right now and my professor goes way too fast when explaining concepts. Your explanation is spot on and clear. Thank you for sharing!
Thanks a lot , you saved a lot of time of us . We appreciate it!
you're great, i could easily grab this topic onto my head
Thank you so much for clear explanation.
Massive help, thanks!
Thanks from India...
Very helpful... Demonstration
Thank you from Saudi Arabia
You are awesome!
It is the circles, guys. You have to have 4 tangent lines surrounding a circle.
1. You start with the equipotential lines at the ground level and impermeable layer
2. Then you draw flow lines than are parabolas that end perpendicular to the ground equipotential line. You can make how many you feel is right. (The fewer lines the bigger the circles you'll imagine later are going to be, the more lines, the more circles you'll have to draw but it'll be more accurate)
3. (The most difficult part.)Draw parabolic equipotential lines that are (1) perpendicular to all flow lines they cross and (2) a circle can be drawn in the resulting inner area that all sides touch the surrounding lines at one point (tangent lines to the circles)
4. Count-out flow channels and drops.
I'm just a student btw, this just what I got from the video.
Thank alot for sharing about soil mechanics
Much love from RIPOSE LECTURES
Thank you man, much appreciated from Kenya. (student)
Watch here
thanks, this helped me :)
Im having a hard time concentrating on your lecture wondering how you wrote it backwards and explain it at the same time 😅😅.. That's just wowwwwwww👍👍👍👍👍👍
thank you sir for clearing our concept
Good explanation sir👍👍
Thanks from Bangladesh
Are you writing backwards?? My mind is blown!
I can write backward. With a little practice it's not too hard.
They probably reflected the footage after recording - my guess
They? If only I had a production team.
Hi Phil, why don't you make a video about hardening soil material behaviour?
He is just mirroring the video
Thanks.
Thank you
thank you :D
Thank you
I love you
Step by step video solutions of civil engineering questions
Thank you so much this helped a lot
thanks
Beside of anisotropic condition can flow net change due to other reasons like head, permeability etc
An example of unconfined flow with a sheet pile driven halfway into an aquifer is shown in
Fig. 7.19(which is same is the figure you have drawn.). The flow net is horizontally symmetrical, providing that the top and bottom flow lines (borders) are parallel and the penetration of the sheet pile is exactly one-half the depth of the aquifer. Note
that the number of flow channels Nf is 3 and the number of equipotential drops N dis 6. Regardless of
how many flow channels you use, the Nf!Nd ratio, or the shape factor, will always be 1/2 for the geometry of this example!
any comment on this?
cool
tq 🙏
Very helpful video. I have a question about the units. I realize that flow is expressed in volume per time, but if K is in (m/s), and H is in (m), should Q be in (m2/s)?
Great spot with the units! It's really m3/s per m of hydraulic structure (if you imagine the structure coming out of /going into the screen).
Oh, I see. Thanks a lot.
Phil how do you know how many flow lines and equipotential lines you need? I've read everywhere and no one gives me a straight answer or an equation to follow. Thank you!
Hi Ryan, good question. It doesn't matter for the maths as the number of equipotential lines will also increase. If you stick to the rules you should end up with the same Nf/Nh no matter how many flow lines are used. In practice, a flow net with too many lines is tricky to draw. Remember, you'll be rubbing things out and redrawing as you go along.
We have the same unanswered question
It is actually 12,622.78 liters per year sir. That 35 liters is per day.
Question sir. How do you find the pore pressure of a certain point if it is located at the very center of equipotential line? We know that pore pressure = 9.81(H + h - mH/nd)
Where H = head of water
h = height of saturated soil above the point of interest
m = number of equipotential drops from zero to point of interest
nd = total number of equipotential drops.
So how would you know the value of "m" in that case?
How does he write upside down?
I often see values of 5.6,6.5,4.5, etc. when calculating Nf. How on earth is this possible if calculating the #of flow paths? Shouldn't this always be a whole number? Or does this only apply to anisthropic materials?
Thanks! I'd say that it would be better if you wore dark clothes so that we could see the light lines and words on the board
Yeah, maybe I will next time. I was worried about becoming a floating head, which may have been even more troubling
hahahahaha!
@@philrenforth9569 😂
Yo, isn't the differernce between the 2 hights 7m, on the diagram.
no the heights are 2m and 1m. But I understand the 2 looks like an 8
@@AB-gu9ui thanks bro 😫
pls consider wearing dark shaded shirt
Wouldn't (4x10^-7 m^3/s) = (0.03456 m^3/day) = (12.61m^3/yr) = (12,600 L/yr) ?
I think the value you calculated was the daily value and not the yearly. Although it seems like too large a value
Hi
I i thought H was 8m -1m??
Not sure where you got your values, but H (or really 'delta' H) is the pressure difference across the system. In this case 1m.
Is the left water height 2m? Because it looks like 8 also.
The result should be 35 liters per day not per year
should't the H be 7 m instead of 1
its a 2 not a 8, i made the same mistake
Nice video, thanks sir!