NCEES Sample exam WR&E #508

[Lisa]

NCEES Sample exam - problem 508 - Water Resources and Environmental

I'm having some trouble understanding the solution to this problem. Can anyone help?

A rain event has an intensity of 1.5 in/hr for the first hour followed by 0.7in/hr for the second hour. The 1-hour unit hydrograph for the watershed is given as:

Q (cfs/in) 0.5, 1.2, 0.4

T (hours) 1, 2, 3

Neglecting infiltration, the discharge (cfs) from the watershed during the second hour is most nearly:

(A) 2.64

(B) 2.15

© 1.44

(D) 1.20

From the solution:

Q = effective rainfall x unit hydrograph ordinate (runoff/inch of rainfall)

Flow during the second hour: Q = 1.5 x 1.2 + 0.7 x 0.5 = 1.8 + 0.35 = 2.15 cfs

 

[Casey]

The answer will be more obvious if you arrange the data into columns

Unit Hyrdograh

Time------Q--------Q@hour1(1.5in/hr)--------Q@hour2(0.7in/hr)

1_______0.5_______1.5*0.5=0.75___________-----

2_______1.2_______1.5*1.2=1.8____________0.7*0.5=0.35

3_______0.4_______1.5*0.4=0.6____________0.7*1.2=0.84

4_______________________________________0.7*0.4=0.28

So looking at hour 2 the flow rates add up like this

1.8cfs + 0.35cfs = 2.15cfs

Let me know if the above doesn't help....

Regards,

Casey

 

[Tido]

The answer will be more obvious if you arrange the data into columns

Unit Hyrdograh

Time------Q--------Q@hour1(1.5in/hr)--------Q@hour2(0.7in/hr)

1_______0.5_______1.5*0.5=0.75___________-----

2_______1.2_______1.5*1.2=1.8____________0.7*0.5=0.35

3_______0.4_______1.5*0.4=0.6____________0.7*1.2=0.84

4_______________________________________0.7*0.4=0.28

So looking at hour 2 the flow rates add up like this

1.8cfs + 0.35cfs = 2.15cfs

Let me know if the above doesn't help....

Regards,

Casey

Correct.

Now this reminds me that I should write my note on this one:

(please ignore if you are not going to take the WR depth)

How about if I give you 1.1 inches (averaging 1.7 and 0.5in from the above) rain that happens in 2 hours, and ask the runoff at the 2 hours.

obviously the asnwer is not

0.55*0.5 + 0.55*1.2

By comparing the answer of this and the Q508, it gives you an idea on the effect of intensity on runoff generation

Best

 

[Guest]

The answer will be more obvious if you arrange the data into columns

That's a boss answer Casey! Nice explanation!

JR

 

[squishles10]

I skipped this one after numerous tries. I hate it when someone comes along and makes it seem so obvious. I doubt of get of still bit at least that makes sense.

 

[Casey]

I skipped this one after numerous tries. I hate it when someone comes along and makes it seem so obvious. I doubt of get of still bit at least that makes sense.

Don't feel so bad... I have no idea how many hours it took me to figure out what to do with these silly hydrographs...

 

[IlPadrino]

Don't feel so bad... I have no idea how many hours it took me to figure out what to do with these silly hydrographs...

I felt like I threw ago some easy points on the exam because I didn't understand the hydrology stuff nearly as well as I could have. Unit hydrographs still confuse me. I think many people would find it really helpful if someone started adding to PE Notes Wiki - Hydrology

 

[Casey]

I felt like I threw ago some easy points on the exam because I didn't understand the hydrology stuff nearly as well as I could have. Unit hydrographs still confuse me. I think many people would find it really helpful if someone started adding to PE Notes Wiki - Hydrology

I told you that you could harass me after the exam!

 

[squishles10]

unit hydrographs don't bother me! But that question is just too much...

 

[Tido]

I skipped this one after numerous tries. I hate it when someone comes along and makes it seem so obvious. I doubt of get of still bit at least that makes sense.

I am sorry if I cause any problem. I thought I am just helping without making it so abvious. I will stick to what is being asked nex time.

Here is how to do it. Since what is given is a 1-hr unit hydrograph one first need to change it to a 2hr unit hydrograph. Since 2 is a multiplier 1hr this can be done using the so called S-curve

g(t) = dt [h(t) + h(t-dt) + h(t-2dt) + ...]

then shiftin the g(t) by the required hour to find g'(t)

g'(t) = g(t - 2dt)

the 2hr unit hydrograph is then

h'(t) = (1/2dt)[g'(t) - g(t)]

in the table form it is as followes

time(hr).............h(t) ................... g(t)........g'(t)...........h'(t)

1............................0.5.......................0.5..........0.0.........

..0.25

2............................1.2.......................1.7.........0.0..........

..0.85

3............................0.4.......................2.1.........0.5..........

..1.60

4............................0.0.......................2.1..........1.7.........

..0.20

5 ...........................0.0.......................2.1.........2.1............

0.00

The last column is the 2hr unit hydrograph. Now after 2hr the flow will be

1.1 * 0.25 = 0.275 cfs

This is significantly different from the one from the above question. Again, sorry if I make a confusion here.

Best

 

[Casey]

I am sorry if I cause any problem. I thought I am just helping without making it so abvious. I will stick to what is being asked nex time.
Here is how to do it. Since what is given is a 1-hr unit hydrograph one first need to change it to a 2hr unit hydrograph. Since 2 is a multiplier 1hr this can be done using the so called S-curve

I think you are making something that is relatively simple into somthing quite complex.

The way I saw it was that there were two 1-hour stores that occurred one right after the other....

This is assuming that your post is talking about the original quesiton, #508... if not, disregard this post.

Thanks...

 

[Tido]

I think you are making something that is relatively simple into somthing quite complex.
The way I saw it was that there were two 1-hour stores that occurred one right after the other....

This is assuming that your post is talking about the original quesiton, #508... if not, disregard this post.

Thanks...

Ok, I will disregard the post! arty-smiley-048:

 

[Tad]

For a visual of the 1.5 in/hr runoff hydrograph + the 0.7 in/hr hydrograph...well, maybe not, I can't seem to paste a graph in here. Maybe I can attach one...looks as if I can, wonder if you can see this graph. Anyway...

Is there some best way to solve this?

I set up four equations in four unknowns.

If we are looking for the one-hr unit hydrograph (i.e., the runoff resulting from one inch per hour of rain for one hour),

and we know the runoff of 0.75 cfs occuring at time 1 hr is the hydrograph value at time one hour of a 1.5 in/hr hydrograph;

then let's assign the ordinates of a unit (1 inch of direct runoff) hydrograph:

if Y1 is the one hour unit hydrograph value at time = 1 hr

Y2 is the one hour unit hydrograph value at time = 2 hr

Y3 is the one hour unit hydrograph value at time = 3 hr

Y4 is the one hour unit hydrograph value at time = 4 hr

then

1.5*Y1 = 0.75 cfs at one hr

the discharge of 2.15 cfs occuring at time 2 hrs is the addition of the hydrograph value at time one hour of a 0.7 in/hr hydrograph plus the hydrograph value at time two hours of the 1.5 in/hr hydrograph; so

0.7*Y1 + 1.5*Y2 = 2.15 cfs at 2 hr

and for the discharge of 1.44 at time = three hours...

0.7*Y2 + 1.5*Y3 = 1.44 cfs

and for the discharge of 0.28 cfs at time = four hours

0.7*Y3 + 1.5*Y4 = 0.28 cfs

The first of the equations solves quickly, and the other equations each then in turn become one equation in one unknown and also solve quickly by hand:

Y1 = 0.5 cfs

Y2 = 1.2 cfs

Y3 = 0.4 cfs

Y4 = 0

 

[IlPadrino]

I told you that you could harass me after the exam!

Uh... the exam is OVER! You PASSED. Now please help some that haven't gotten to their goal!

 

[Casey]

Uh... the exam is OVER! You PASSED. Now please help some that haven't gotten to their goal!

Would you have mercy on me if I told you I was studying for the SEII exam?

 

[Tad]

Correct.
Now this reminds me that I should write my note on this one:

(please ignore if you are not going to take the WR depth)

How about if I give you 1.1 inches (averaging 1.7 and 0.5in from the above) rain that happens in 2 hours, and ask the runoff at the 2 hours.

obviously the asnwer is not

0.55*0.5 + 0.55*1.2

By comparing the answer of this and the Q508, it gives you an idea on the effect of intensity on runoff generation

Best

If you mean 1.1 inches/hr for two hours,

then the answer is...

1st hr rainfall hydrograph, at time 2 hr = 1.2*1.1 cfs

plus

2nd hr rainfall hydrograph, after 1 hour = 0.5*1.1 cfs

1.2*1.1 + 0.5*1.1 = 1.87 cfs.

If you mean 1.1 inches over two hour period,

then then answer would be

0.55*0.5 + 0.55*1.2,

wouldn't it?

Tad

 

[maximus808]

Ok guys, here's a similar problem. Question 119 of page 37 of the NCEES Problems.

Given: 1-hr unit hydrograph figure with following data:

Time 0 1 2 3 4 5 6 (hours)

0 75 200 100 50 25 0 (cfs)

A 2-hour storm of intensity 0.5in/hr in this watershed produces a total excess volume of water (acre-ft) of most nearly.

a)0.01

b)37

c)75

d)450

I'm trying to follow the solution and here's what they did,

They set up a table with columns of Time, Qu, Q1, Q2, and Qtotal (Q1+Q2)

They sum the total of Q total and convert to acre-ft to get the answer of 37 acre ft (B.)

My questions are what does the Qu represent although I know it's the given Q values in the given table. Q1 is found by multiplying the Qu x the 2 hour storm intensity of 0.5 in/hr. But how is Q2 found? Q2 seems to follow the value after Q1 table? Any clarification would be greatly appreciated. Thanks.

 

[maximus808]

anyone? Thanks.

 

[sac_engineer]

anyone? Thanks.

Here we go... I hope this makes sense without diagrams:

The table that was provided is a unit-hydrograph. The CFS values are based on a per inch of rainfall (cfs/in) for a one-hour rainfall intensity.

Since the problem states that a 2-hour rainfall had occurred, then you need to add up the Q/in values (0+75+200+100+50+25+0=450) and multiply by 2 for the unit CFS (900 cfs/in) and multiply again by 0.5" for the actual CFS (450 cfs).

However, to properly calculate the total volume of water, we need to graph the actual CFS at each time interval and calculate the area below the line.

Here are the data domains to develop the volume by graph:

T (hours) = {0,1,2,3,4,5,6}

Qu = {0,75,200,100,50,25,0}

Q1 = {0,37.5,100,50,25,12.5,0}

Q2 = {0,0,37.5,100,50,25,12.5} (Q2 = Q1 shifted 1 hour)

QTOT = {0,37.5,137.5,150,75,37.5,12.5}

If you graph QTOT (cfs) vs. T (hr), the area under the line = 450 cfsh (units are cfs x h = volume of water).

Therefore 450 cfsh = 450 cfsh x 3600 sec/hr x 1hr / 43560 cf/AF = 37.2 AF

I think if you do a few of these problems, you might find that volume of water is the same result as the sum of QTOT, but I can't verify at this time.

Good luck!

 

[Caseyuconn]

anyone? Thanks.

Here we go... I hope this makes sense without diagrams:

The table that was provided is a unit-hydrograph. The CFS values are based on a per inch of rainfall (cfs/in) for a one-hour rainfall intensity.

Since the problem states that a 2-hour rainfall had occurred, then you need to add up the Q/in values (0+75+200+100+50+25+0=450) and multiply by 2 for the unit CFS (900 cfs/in) and multiply again by 0.5" for the actual CFS (450 cfs).

However, to properly calculate the total volume of water, we need to graph the actual CFS at each time interval and calculate the area below the line.

Here are the data domains to develop the volume by graph:

T (hours) = {0,1,2,3,4,5,6}

Qu = {0,75,200,100,50,25,0}

Q1 = {0,37.5,100,50,25,12.5,0}

Q2 = {0,0,37.5,100,50,25,12.5} (Q2 = Q1 shifted 1 hour)

QTOT = {0,37.5,137.5,150,75,37.5,12.5}

If you graph QTOT (cfs) vs. T (hr), the area under the line = 450 cfsh (units are cfs x h = volume of water).

Therefore 450 cfsh = 450 cfsh x 3600 sec/hr x 1hr / 43560 cf/AF = 37.2 AF

I think if you do a few of these problems, you might find that volume of water is the same result as the sum of QTOT, but I can't verify at this time.

Good luck!

My question with this problem is how come it does not follow the rules of a lagging hydrograph. The duration is greater than 25% bigger than the original duration. According to the lagging hydrograph method the Q1+Q2 should be divided by 'n' which is two in this case and then multiplied by the storm precipitation which would be P=It or 1inch.

This gives a total Q of 487.5 and a V of 40.3 ac-ft.

Any insight into why this was not done?