hydrographs?

Aurora09 · Mar 10, 2019

Does anyone have a better explanation for the hydrograph problem in Example 20.4, Part B from the CERM? I'm not understanding their solution so if someone can please explain that'd be great!

Jbone27 PE · Mar 12, 2019

I don't have that exact book but if you would like to explain your problem and question I bet me or someone else could help you out.

Aurora09 · Mar 18, 2019

@Jbone27 PEhey! sorry for the late response but here is the problem:

A 6 hr storm rains on a 25 mi2 (65 km2) drainage

watershed. Records from a stream gaging station draining

the watershed are shown. (a) Construct the unit

hydrograph for the 6 hr storm. (b) Find the runoff rate

at t = 15 hr from a two-storm system if the first storm

drops 2 in (5 cm) starting at t = 0 and the second storm

drops 5 in (12 cm) starting at t = 12 hr.

So the average precipitation was calculated as P = V/Area which gives you 4.37 cm. The rest of the solution for Part B was:

(b) To find the flow at 15 h, add the contributions from

each storm. For the 5 cm storm, the contribution is the

15 h runoff multiplied by its scaling factors; for the 12 cm

storm, the contribution is the 15 h - 12 h = 3 h runoff

multiplied by its scaling factors:

let me know if you understand the solution

Jbone27 PE · Mar 19, 2019

Aurora09 said:
@Jbone27 PEhey! sorry for the late response but here is the problem:

A 6 hr storm rains on a 25 mi2 (65 km2) drainage

watershed. Records from a stream gaging station draining

the watershed are shown. (a) Construct the unit

hydrograph for the 6 hr storm. (b) Find the runoff rate

at t = 15 hr from a two-storm system if the first storm

drops 2 in (5 cm) starting at t = 0 and the second storm

drops 5 in (12 cm) starting at t = 12 hr.

So the average precipitation was calculated as P = V/Area which gives you 4.37 cm. The rest of the solution for Part B was:

(b) To find the flow at 15 h, add the contributions from

each storm. For the 5 cm storm, the contribution is the

15 h runoff multiplied by its scaling factors; for the 12 cm

storm, the contribution is the 15 h - 12 h = 3 h runoff

multiplied by its scaling factors:View attachment 12699

let me know if you understand the solution

Okay so I'm assuming you are just looking for an explanation on the scaling? If not please let me know.

So to get the solution they went to the hydrograph constructed based on the 4.37 cm precip and found Q = 35 m^3/s at 15 hrs and 10 m^3/s at 3 hrs.

Keeping in mind these values are for 4.37 cm you have to scale each value so for the 5 cm event 5/4.37 = 1.14 or 114% ( I like to think of them in percentages). Replicated for the 12 cm event 12/4.37 = 2.75 or 275%.

Q = 35*1.14 + 10*2.75

Q = 67.5

I feel like I just rewrote their solution but just a little more in my mindset. Point being you have runoff values based on the 4.37 event but the events you are calculating are some degree different so you have to scale for the variance.

Vice versa you could say @ 15 hrs you get 35/4.37 = 8 so you get (8 m^3/s) per cm and @ 3 hrs you get 10/4.37 = 2.29 (m^3/s) per cm then multiplied those values by your corresponding storm events.

Hope this helps. Please let me know if you have any further questions.

Aurora09 · Mar 19, 2019

Jbone27 PE said:
Okay so I'm assuming you are just looking for an explanation on the scaling? If not please let me know.

So to get the solution they went to the hydrograph constructed based on the 4.37 cm precip and found Q = 35 m^3/s at 15 hrs and 10 m^3/s at 3 hrs.

Keeping in mind these values are for 4.37 cm you have to scale each value so for the 5 cm event 5/4.37 = 1.14 or 114% ( I like to think of them in percentages). Replicated for the 12 cm event 12/4.37 = 2.75 or 275%.

Q = 35*1.14 + 10*2.75

Q = 67.5

I feel like I just rewrote their solution but just a little more in my mindset. Point being you have runoff values based on the 4.37 event but the events you are calculating are some degree different so you have to scale for the variance.

Vice versa you could say @ 15 hrs you get 35/4.37 = 8 so you get (8 m^3/s) per cm and @ 3 hrs you get 10/4.37 = 2.29 (m^3/s) per cm then multiplied those values by your corresponding storm events.

Hope this helps. Please let me know if you have any further questions.

Hi @Jbone27 PE. Thanks for taking your time to reply! I should have been more specific in my original post as to what part of the solution confused me, but anyways I was stumped on why they subtracted 15 hrs - 12 hrs = 3 hr. I would have taken the Q value at t = 12 hr. I guess I'm not understanding the concept behind what is actually happening with the rainfall.

Jbone27 PE · Mar 20, 2019

Oh okay. To better visualize this assume you have 2 identical unit hydrographs in the same plot. The first begins at T=0. The second doesn't begin until T=12. Therefore when when the second hydrograph reaches T=15 it has only been producing runoff for 3 hours.