U r explaining very concisely at the same time from very basic making it visualise, I have to admit it's even better than some of the nptel videos, I have seen all ur hydrograph videos. Thank u so much sir
Thanks Ravi Pratap, have you checked our site learn.apseduverse.com/ ? Video lectures covering entire syllabus with study material, test series and most importantly, doubt support! Starting from Rs.1399/-
Thank you so much brother for your videos... It is very much helpful... One thing I wanted to say is please name the video in serial number order so that linked videos will be easy to find....
Sir tarika bhut bdiya smjhate ho But yaar aap systematic way m nhi likhte Kuch mbhi kahin pr bhi likhne lg jate ho Or nahi ques. No. 1 wagreh daalte Jab mrzi chalu kr dete ho kahi bhi asa mt kro Way of writing thik kriye Baaki bhut bdiya smjhate ho Great
A 3 hour unit hydrograph for a basin is assumed to have a shape of a triangle of base 24 hour and peak of 50 cumec, occurring after 08 hours from the beginning. A 3 hr storm occurred over the basin with an intensity of 2.6 cm/hr. Assuming an average loss rate of 6 mm/hr and a constant base flow of 6 cumec, determine the stream flow at 4, 8 and 16 hours from the beginning. Also calculate the area of the basin and its coefficient of runoff. Can u please solve this for me, i m totally frustrated plz help me
sir please help me with gate AG 2010 question no. 55. question:the peak of a flood hydrograph due to a 1h duration isolated storm in a catchment of area 13.45 sq km is 135 m3/s. the total depth of rainfall is 54 mm. assume a constant base flow of 10 m3/s and phi index equal to 4mm. Q54: the peak of 1h unit hydrograph for the catchment in m3/s is: a) 15 b) 20 c) 25 d) 30 Q55: assuming the above 1h unit hydrograph to be trangular in shape with the time to peak as 1 hour,the peak of the 2 h unit hydrograph for the catchment in m3/s is: a) 13.25 b) 18.75 c) 21.25 d) 26.75
q.54 answer is ((135-10)m^3/s/(5.4-0.4)cm)=25m^3/s q55:first calculate base period using: area of UHG =(depth of runoff*catchment area), i got base period 3 hrs here. then plot 1hr UHG and another UHG lagging 1hr. Then at time =2; we get max ordinate of DRH actually I got. I got the value as 37.5. after we need to divide the value with 2 to get peak of 2hr UHG. the value i've got is 18.75. correct me if im wrong.
U r explaining very concisely at the same time from very basic making it visualise, I have to admit it's even better than some of the nptel videos, I have seen all ur hydrograph videos.
Thank u so much sir
That's great to hear Suraj. Keep watching :)
Thanx a ton man, i wish my teachers were as good as you! Keep up the good work!
Thanks a lot sir 😍 i am going to skip hydrology for GATE but now i can do better in it and again thank you so much.
One of the few good water resource engineering professor out there
like magic, now everything makes sense. Thank you sir!!!!
GATE 2025 attendence sir, your video lecture is really helpful
"So i think you have understood, whatever i have told here." Beautiful.
Sir u r great ur effort is too good for student like us
Love u sir
Thanks Ravi Pratap, have you checked our site learn.apseduverse.com/ ? Video lectures covering entire syllabus with study material, test series and most importantly, doubt support! Starting from Rs.1399/-
Thank you so much brother for your videos... It is very much helpful... One thing I wanted to say is please name the video in serial number order so that linked videos will be easy to find....
In playlists you can find them in order. :)
Awesome effort thank you sir
Thanks sir for this videos....i hope it will help me in gujrat engineering services exam...
All the best Jay :)
THANK YOU So Much......Specially for S Hydrograph
Very nice 👍👍 so wonderful
second video derivation of 2 hr uh from 4 hr uh hydrograph sir???
Extra ordinary work
impressive work
Great video and very helpful, thank you!
Sir tarika bhut bdiya smjhate ho
But yaar aap systematic way m nhi likhte
Kuch mbhi kahin pr bhi likhne lg jate ho
Or nahi ques. No. 1 wagreh daalte
Jab mrzi chalu kr dete ho kahi bhi asa mt kro
Way of writing thik kriye
Baaki bhut bdiya smjhate ho
Great
sahi bola vai
Thanks man, great help ☺️
Hope you have checked our site too!
Keep going bro
Har trade ka app banva do⌚😇😇😇😇 sab asaan ho jayega
Excellent explanation.
Thanks Sangamesh!
Tq for ur content
How did you get the third column? The purple one?
A 3 hour unit hydrograph for a basin is assumed to have a shape of a triangle of base
24 hour and peak of 50 cumec, occurring after 08 hours from the beginning. A 3 hr storm
occurred over the basin with an intensity of 2.6 cm/hr. Assuming an average loss rate of
6 mm/hr and a constant base flow of 6 cumec, determine the stream flow at 4, 8 and 16
hours from the beginning. Also calculate the area of the basin and its coefficient of runoff.
Can u please solve this for me, i m totally frustrated plz help me
Which app u r using sir??
One Note
Can we derive a 3 hrs unit hydrograph from given 5hrs unit hydrograph?
Yes we can derive it. The data should be available in 1hr interval for it.
1 hr? Means the interval should be one hour!!?
thank you
Thank you sir
precisely explain..
Good explanation
Thanks Bharath , hope you have checked our site too : learn.apseduverse.com
The way of explanation is very good.
Thanks Bhaskar! Do check our website (learn.apseduverse.com/) if you like our explanations :)
Can I give more than 1 like!? It deserves so!
Ha ha, thanks man ! :) Do watch our other subject videos too.
i owe you one sir
This is for gate or ESE??
Sir plz upload from 4hr duration to 2hr duration hydrograph....
Or suggest me if there is already..
did you find one
At the end, shouldn't the Qs = A * D
Instead of Qs = A/D ?
Actually area(cm2)*intensity(cm/hr) gives discharge(cm3/hr)
Where intensity is 1/D(cm/hr)
That's why it must be A*(1/D)=A/D
Thanks! Since it is a unit hydrograph, thus the corresponding excess rainfall depth is 1 cm... I had forgotten that.
Qs = A(km^2) * 1cm / D (hrs)
Qs = A *10^6 m^2 * 0.01 m / D hrs
Qs= A *10^4 m^3 / D (hrs)
Qs = 2.778*A/D (m^3/s)
thank u sir...
sir please help me with gate AG 2010 question no. 55.
question:the peak of a flood hydrograph due to a 1h duration isolated storm in a catchment of area 13.45 sq km is 135 m3/s. the total depth of rainfall is 54 mm. assume a constant base flow of 10 m3/s and phi index equal to 4mm.
Q54: the peak of 1h unit hydrograph for the catchment in m3/s is:
a) 15
b) 20
c) 25
d) 30
Q55: assuming the above 1h unit hydrograph to be trangular in shape with the time to peak as 1 hour,the peak of the 2 h unit hydrograph for the catchment in m3/s is:
a) 13.25
b) 18.75
c) 21.25
d) 26.75
q.54 answer is ((135-10)m^3/s/(5.4-0.4)cm)=25m^3/s
q55:first calculate base period using: area of UHG =(depth of runoff*catchment area), i got base period 3 hrs here.
then plot 1hr UHG and another UHG lagging 1hr. Then at time =2; we get max ordinate of DRH actually I got. I got the value as 37.5. after we need to divide the value with 2 to get peak of 2hr UHG. the value i've got is 18.75.
correct me if im wrong.
Thanku sir
Good
Ye to simple waale hai, 2hr sei 7hr waale karwao sir
How 2 convert 4hr UH to 2hr plss sir jaldi btado ppr h
S curve in last column its 50.why its 60
for it, you have to make a column for time duration lagged by 10 hours and then add the entities.
What is the full form of your channel #apsed