NCEES 2008 Morning Problem 119

[maximus808]

Can someone please clarify the following problem:

1. Given the following data:

Time (hrs) vs. Discharge (cfs)

0,0

1,75

2,200

3,100

4,50

5,25

6,0

A two hour storm even of intensity 0.5 in.hr.

How much EXCESS VOLUME of water (acre-ft) is produced?

The answer sets up a table of Time, Qu, Q1 + Q2 = Qtotal. It adds the Qtotal then x 3600s/hr x 1ac/43560 ft^2 = 37.19 ac. ft.

What does Qu, Q1, and Q2 represent and how are these factors determined? thanks.

 

[sac_engineer]

Can someone please clarify the following problem:
1. Given the following data:

Time (hrs) vs. Discharge (cfs)

A two hour storm even of intensity in.hr.

How much EXCESS VOLUME of water (acre-ft) is produced?

The answer sets up a table of Time, Qu, Q1 + Q2 = Qtotal. It adds the Qtotal then x 3600s/hr x 1ac/43560 ft^2 = ac. ft.

What does Qu, Q1, and Q2 represent and how are these factors determined? thanks.

I answered this in your other post.

Other Post

Qu = unit discharge (cfs per inch of rain for 1-hour storm)

The Qu values represent the discharge of water after 1" of rainfall in the first hour only. Thus, based on the given table, if 1" of rain falls in the first hour only, the discharge at t=4hours is 50 cfs.

Q1 = actual discharge for 1st hour of storm (Qu x 0.5)

Q2 = actual discharge for 2nd hour of storm (Q1 shifted 1 hour later)

Note: if the storm was 3 hours, then a Q3 set of values (Q1 shifted 2 hours later) would be required

Qtotal = Q1+Q2 = total discharge because runoff from 1st hour must be added to runoff from 2nd hour of rainfall

Good luck!

 

[PEin2010]

Sac_engineer,

Is it true across the board that a unit hydrograph represents the discharge of water after 1" of rainfall only?

Also, you state that "....Thus, based on the given table, if 1" of rain falls in the first hour only, the discharge at t=4hours is 50 cfs." Would this be a case of a 4 hr storm with an intensity of 1 inch/hr?

Thank you so much for your help!

Can someone please clarify the following problem:
1. Given the following data:

Time (hrs) vs. Discharge (cfs)

0,0

- -

A two hour storm even of intensity - in.hr.

How much EXCESS VOLUME of water (acre-ft) is produced?

The answer sets up a table of Time, Qu, Q1 + Q2 = Qtotal. It adds the Qtotal then x 3600s/hr x 1ac/43560 ft^2 = ac. ft.

What does Qu, Q1, and Q2 represent and how are these factors determined? thanks.

I answered this in your other post.

Other Post

Qu = unit discharge (cfs per inch of rain for 1-hour storm)

The Qu values represent the discharge of water after 1" of rainfall in the first hour only. Thus, based on the given table, if 1" of rain falls in the first hour only, the discharge at t=4hours is 50 cfs.

Q1 = actual discharge for 1st hour of storm (Qu x 0.5)

Q2 = actual discharge for 2nd hour of storm (Q1 shifted 1 hour later)

Note: if the storm was 3 hours, then a Q3 set of values (Q1 shifted 2 hours later) would be required

Qtotal = Q1+Q2 = total discharge because runoff from 1st hour must be added to runoff from 2nd hour of rainfall

Good luck!

 

[Marie925]

From the solution given:

(- ft^-/sec) X (3,600 sec/hr) X (1 hr) X (1 acre/ 43,560 ft^2) = - acre-ft

In this last step of the solution where - cfs is converted to acres-ft, is the (1hr) the interval of the unit hydrograph? IE is the flow multiplied by the interval to get volume?

At first I thought this was part of the unit conversion..... Then my first thought was that it should be multiplied by 2 hrs since it is a 2 hour storm.

Thanks,

Marie925

 

[sac_engineer]

The solution appears more simple than it looks. The fact that the storm intensity is constant at 0.5 in/hr for a two hours makes the calculation easier, but you need to understand how the total volume is calculated in case you're presented with a question where the intensity for the first hour is different than the second.

Given the unit hydrograph, you can calculate the volume of water at each hour, as follows:

Assume rain starts at 12pm

First Hour of Rain:

hour volume

12pm 0

1 pm 75 cfs x 0.5 in / 1 in x 1 hr = 37.5 cfs/hr (this is a volume, but we'll convert units at the end)

2 pm 200 x 0.5 = 100 cfs/hr

3 pm 100 x 0.5 = 50 cfs/hr

etc...

Second Hour of Rain:

- same intensity, but you need to shift the calcs by one hour, as follows:

hour volume

12 pm 0

1 pm 0 (this is when the 2nd hour begins/continues)

2 pm 75 cfs x 0.5 in/ 1 in x 1 hr = 37.5 cfs/hr

3 pm 200 cfs x 0.5 in/hr = 100 cfs/hr

etc.

You must sum the volume at each time interval:

12 pm: 0 + 0 cfs x in/hr

1 pm: 0 + 37.5

2 pm: 100 + 37.5

3 pm: 50 + 100

etc.

The total volume from all time intervals will be 450 cfs/hr = 37.2 ac.ft.

It's easier to do these calcs in a table format so you can shift the volume and sum each row that represents a time interval.

Good luck!

 

[Marie925]

The total volume from all time intervals will be 450 cfs/hr = 37.2 ac.ft.

Thanks, I understand the Unit Hydrograph and the lagging method.

I am JUST asking about the last step to converst the Flow to Volume. Above you say 450 cfrs/Hr but that is not what is given in the solution. A Flow is calculated. 450 cfs and it is converted and an extra term is used to get it to volume. It is multiplied by (1hr).

(450 ft^3/sec) X (3,600 sec/hr) X (1 hr) X (1 acre/ 43,560 ft^2) = 37.19 acre-ft

(solution straight from the book)

I am trying to find out what this term is. I believe it to be the interval of the hydrograph, because the flow was already calculated and totaled over the entire 2 hour storm.

But I want to verify this incase there are hydrographs that have intervals different from 1 hour.

Thanks,

Marie

 

[Marie925]

NEVERMIND I found the answer to my questions in another thread.

It is the Interval that you multiply by when converting the Total Flow Q that you calculated to Volume.

See the last post in the other thread:

<deleted>

"Yeah, sum the values to get the total direct runoff -->450cfs. Multiply that by the time interval between hydrograph ordinates (1hr=3600s) to get the volume of runoff. 450cfs x 3600s = 1,620,000ft^3 convert to acre-ft --> 1620000/43560 = 37.19 acre-ft "

I was pretty sure it was the interval, but wanted to verify.

Thanks!

 

[sac_engineer]

My units were wrong. Instead of cfs/hr, it should be cfs x hr to give you a volume. The time (i.e. 1 hour) was already included for each time interval calculation.