This was clear, precise, and at the same time descriptive. Brilliant work. I have been trying to get a grasp of the concept for a whole day now. Your explanation was the best. Thank you. Cheers!
Nice video! But for the people who are watching this, there is something wrong. The discharge of the unit hydrograph isn't simply Q of DHR divided by total excess rainfall. That's a incorrect concept. You got to know also the data of excess rainfall so that you can obtain the number of intervals in the UH and calculate the all the discharge of the UH separately using sets of equation.
1. Over a period of 50 years, a particular catchment underwent urbanization. Before urbanization, the catchment was mostly having agricultural cultivation and the storm loss rate was estimated as 0.5cm/h. However, after urbanization, this value dropped to 0.3cm/h. The 1-h unit hydrographs for this catchment before and after urbanization were both estimated to be triangular in shape with a time base of 12 hours and a peak magnitude of 100 m3/s. However, the time to peak changed from 4 hours to 3 hours. Find the increase in peak flow magnitude after urbanization for a 2-h storm in which 2 cm of rain fell during the first hour and 1.2 cm during the second hour. Assume base flow to be negligible. Answer for this qn plz 🙏
Here I agree with you using trapezoidal formula but can u plz explain me why Simpson rule is used in SK Garg book although it cannot be used for even ordinates.
A waterbody having a surface area of 2000 hectares. 10.8 litres of water was added to bring the water surface in the evaporation pan to the stipulated level ad the nearby rain gauge measured a 3.6 mm of rainfall. Diameter of the pan 122 cm and the pan coefficient is 0.7. The evaporation from the water body can be estimated as ________________Mm3.(Answer should be round off to 2 decimal places). please tell me sir answer this question
Even after rainfall ends there would be discharge at the outlet, if we see the hydrograph the falling limb does not depend on the rainfall pattern and intensity, it will depends on the topographical features of catchment. So to make clear or easy understanding, Discharge is measuring at the outlet of the catchment which is very far from the actual rainfall occurred area so it takes time to travel the runoff from the rain gauge point to the discharge measuring point..
in second question, rainfall excess comes in 'cm' and for UH discharge you divide discharge ordinate by 4.32 cm but discharge ordinate is in 'cum per sec. please clear the issue. Thank you
This is because of unitary method. Discharge variations for rainfall excess of 1cm. We have to obtain unit hydrograph. If you are confusing with the units. Then think again. Q(m^3/s)/x gives the Q for 1cm ie unit hydrograph in m^3/s
@10:22 How come he did not divide by the 6-UH 1cm/unit? so in this case I think it is 1cm/unit which would give you still the same values because dividing by 1 so I'm wondering if that's what were supposed to but it's just not shown here
Welcome Shashank. All topics are covered in our site, hope you have checked our site learn.apseduverse.com/ where you can register for the full package which contains all our lectures, study material and test series with doubt support (All your doubts will be answered within 24 hours!)
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Hello Anurag, glad to hear from you. Thanks for your feedback, we will look into it. Currently, we do not have courses in Hindi. But we do offer courses in English in TH-cam and in our app. In this you get access to all topics in Civil Engineering with solved examples and study material.
Thank You Your Video Scored me 10 marks in semester Exam 😇
Most welcome 😊
This was clear, precise, and at the same time descriptive. Brilliant work. I have been trying to get a grasp of the concept for a whole day now. Your explanation was the best. Thank you. Cheers!
Thanks Saif, glad to hear that. Continue watching :)
but not very clear in calculation the area of the unit hydrograph
This is pretty cool.you are elaborate enough.please continue
Best ever video I've ever watched in my life
Thanks Aman!
Keep up the good work. This was the best step-by-step video I have ever seen on this topic.
Thanks Mike! :)
Nice video! But for the people who are watching this, there is something wrong.
The discharge of the unit hydrograph isn't simply Q of DHR divided by total excess rainfall. That's a incorrect concept.
You got to know also the data of excess rainfall so that you can obtain the number of intervals in the UH and calculate the all the discharge of the UH separately using sets of equation.
1. Over a period of 50 years, a particular catchment underwent urbanization. Before urbanization, the catchment was mostly having agricultural cultivation and the storm loss rate was estimated as 0.5cm/h. However, after urbanization, this value dropped to 0.3cm/h. The 1-h unit hydrographs for this catchment before and after urbanization were both estimated to be triangular in shape with a time base of 12 hours and a peak magnitude of 100 m3/s. However, the time to peak changed from 4 hours to 3 hours. Find the increase in peak flow magnitude after urbanization for a 2-h storm in which 2 cm of rain fell during the first hour and 1.2 cm during the second hour. Assume base flow to be negligible.
Answer for this qn plz 🙏
You are the best, please never stop
your online courses are truly superior...
Perfect explanation sir
12:30
Here I agree with you using trapezoidal formula but can u plz explain me why Simpson rule is used in SK Garg book although it cannot be used for even ordinates.
nice lectures, thank you
I am not from India, and it is stopping me from signing up in the website
Thank you Shreshtha. Just hit us an email to support@apsed.in. We will help you with it.
Such a clear explanation. Thank You very much.
After converting 6 hrs to secs, why was it not divided by 2 according to the trapezoidal formula?
Ordinates of 6H Unit hydro graph are in m^3/sec and rainfall in cm, can we multiply them directly isn't the units inconsistent?
No. Because ordinate of UH=(Ordinate of drh*1cm)/R cm
Where R is rainfall excess
while dividing discharge with rainfall excess we should convert the units or not
A waterbody having a surface area of 2000 hectares. 10.8 litres of water was added to bring the water surface in the evaporation pan to the stipulated level ad the nearby rain gauge measured a 3.6 mm of rainfall. Diameter of the pan 122 cm and the pan coefficient is 0.7. The evaporation from the water body can be estimated as ________________Mm3.(Answer should be round off to 2 decimal places).
please tell me sir answer this question
nice video .......content is very good but try to give some soft copy of notes
Sir @9.20.. it starts at 12hr so shouldn't it be multiplied as 2.8×600 instead of 2.8×250 ?
sir i have a doubt in first question..... rainfall is given for 3 successive intervels and ordinates for 66 hours
Even after rainfall ends there would be discharge at the outlet, if we see the hydrograph the falling limb does not depend on the rainfall pattern and intensity, it will depends on the topographical features of catchment. So to make clear or easy understanding, Discharge is measuring at the outlet of the catchment which is very far from the actual rainfall occurred area so it takes time to travel the runoff from the rain gauge point to the discharge measuring point..
At 18:56 why dividing metre by centimetre sir??
why cant they explain it how u did?...thank you sir so informative, i liked this
At 7.16 ,, why you have multiplied m3/s and cm?
Can we take interpolated value or not?
Many thanks, it was very helpful and clear.
Thank you very much sir! My lecturer explained this horribly.
Nice video sir😊
Excess rainfall=precipitation - infiltration, but here u considered phi*duration
Why is need infiltration lose that pi-index substrate from rainfall magnitude??
Why do we multiply UH with rainfall magnitude????
DRH is In cm and base flow in metres.. Then how added?
great video, but there are so many ads -________-
in second question, rainfall excess comes in 'cm' and for UH discharge you divide discharge ordinate by 4.32 cm but discharge ordinate is in 'cum per sec. please clear the issue.
Thank you
While dividing 4.32 it is assumed as a unitless quantity.
check the definition of Unit HG its a DRH from 1 cm of effective rainfall that means 'discharge per 1 cm' here 23.15 cumec per 1 cm..
thanks
Thank You. Awesome explanations
If the area of catchment is not given in question then what shall we take to find the unit hydrograph ordinates..
Use tripodizal method
What if initial loss is given in mm..what needs to be done
If phi is not given in the question then what will do
thanks a lot..this is very comprehensive...thanks again
q(m^3/s)/4.32(cm)
This is because of unitary method. Discharge variations for rainfall excess of 1cm. We have to obtain unit hydrograph.
If you are confusing with the units. Then think again. Q(m^3/s)/x gives the Q for 1cm ie unit hydrograph in m^3/s
@10:22 How come he did not divide by the 6-UH 1cm/unit? so in this case I think it is 1cm/unit which would give you still the same values because dividing by 1 so I'm wondering if that's what were supposed to but it's just not shown here
in the 2nd question unit of discharge is m^3/s and we are dividing it with cm..is it ok>>??
Uh is in m3/sec but you are directly multiplying with cm
pls solve some more GATE questions related to hydrograph
How do you get 6×3600 I dont understand
u guys are saviors!
Great helpful sir
i m very much thankful to you sir.........thank you so much please upload also another lectures please
Welcome Shashank. All topics are covered in our site, hope you have checked our site learn.apseduverse.com/ where you can register for the full package which contains all our lectures, study material and test series with doubt support (All your doubts will be answered within 24 hours!)
Thank you for doing this video. Very helpful.
how flood Hydrograph changed to Effective rainfall hydrograph?
Video was helpful. Thanks sir
Thanks Dakshat. Hope you have checked our website where we offer more things : learn.apseduverse.com
Nice thank you Mr
A good job brother
your video is helpfull..tnx
Sir 3. 4. 5. cm wale question me DRH kitne hr ka hai
12 hr ya fir 18 hr ka
Sir plz clear kriye ❓❔⁉️
I also had a same confusion, I think it should be of 18 hrs
how to calculate river flow if a 2hr storm of 4cm excess rainfall occur...for the same unit hydrograph ??
Ask your father!
Nyc video.very helpful...thank u.
Thanku so much , its really helpful
Dear there is a question on u ..that help me only I want today ...
How to remove base flow?
Thankyou
You’re welcome 😊
Yaa Really helpful ur video
Plzz keep it up
It was very helpful....
Thanks Palash :)
thanks a lot ...
Thanku sir
how can you determine the baseflow of 30m^3/s?
dont do research on that bro. simply take the given value.
very helpful
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I am unable to understand your handwriting
ThaNks a lot ☺
Supposed phi index not given then...only raillfall multiple
best one
Best
nice video
thank you sir
Thank you..
Thanku sir .
super
❤❤❤❤
plz upload the next part
8:37 how is 0 multiplied by 3.8 is 190 sir....plzzz explain
That I send on email...
Thank u sir....
I love you
منو من العراق جاي يباوعله
Hindi me video banaao
Devru guru ninu
Please improve your writing
Send me yours emial o
hindi ?????????
Hello Anurag, glad to hear from you. Thanks for your feedback, we will look into it. Currently, we do not have courses in Hindi. But we do offer courses in English in TH-cam and in our app. In this you get access to all topics in Civil Engineering with solved examples and study material.
very bad handwriting
Terrible Indian accent...
Thankyou
Thanks a lot Sir
Are you student of civil
thank you sir
Thank you sir