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# NCEES 2008 Water Resources Question

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Hello -

I have two questions from the NCEES 2008 Book, Water Resources Afternoon Questions.

1) For Problem 508 - The question is "A rain event has an intensity of 1.5 in/hour for the first hour followed by 0.7 in/hr for the second hour. The 1 hour unit hydrograph for the watershed is given as Q = 0.5 (T = 1 hour), 1.2 (T=2 hours) and 0.4 (T= 3 hours). Neglecting infiltration, the discharge (CFS) from the watershed during the second hour is most nearly: 2.64, 2.15, 1.44 or 1.20." The books answer has the solution as 1.5 in/hour x 0.5 cf/in + 0.7 in/hr x 1.2 cf/in. = 2.15 cfs. I do not understand how they go this answer, especially because the units do not add up.

2) For Problem 509 - The question is "Assume that you are evaluating an agricultural watershed. The soil is classified as clay with a high swelling potential. The watershed is a pasture that has 65% ground cover and is not heavily grazed. The potential maximum retention (in) after runoff begins (also called the soil storage capacity) is most nearly." How can you tell from this problem whether you use straight row, contoured, or contoured and terraced in Table 20.5 of the Lindenburg book?

If anyone can help with these questions I would greatly appreciate it.

Thanks.

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Question 1 - Problem 508

Units not shown, I hate it when they do that.

A Unit Hydrograph is a hydrograph of of dropping 1 inch evenly on the watershed. You get it by dividing every point on the hydrograph by the average excess precipitation. So, where they hydrograph Q has units of cf/s, a unit hydrograph has units of cf/(s x in).

Draw out the unit hydrograph with the 3 points given. It goes up and down over a course of 3 hours. The values are:

Q1=0.5cf/(s x in)

Q2=1.2cf/(s x in)

Q3=0.4cf/(s x in)

Now, the first hour you are given values in inch/hour. The rain events last 1 hour.

0.5 in/hr x 1 hour = 1.5 in

0.7 in/hr x 1 hour = 0.7 in

Now multiply for the first two hours, as asked for in the problem:

0.5cf/(s x in) x 1.5 in + 1.2cf/(s x in) x 0.7 in =

Now that should help you with your units, as you can see you'd wind up with cf/s. However, to get the answer they did you would need to multiply 1.5x1.2 + .7x.5 = 2.15. Are you sure you wrote the question up correctly?

Question 1 - Problem 509

Read through Step 3, page 20-16. It appears that you need 'additional tables' for pasture land. Table 20-5 is specified for cultivated land.

I'm taking Transportation, so am only learning enough to get through the morning breadth. Hopefully won't need to worry about the extra tables.

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Hello -

I have two questions from the NCEES 2008 Book, Water Resources Afternoon Questions.

1) For Problem 508 - The question is "A rain event has an intensity of 1.5 in/hour for the first hour followed by 0.7 in/hr for the second hour. The 1 hour unit hydrograph for the watershed is given as Q = 0.5 (T = 1 hour), 1.2 (T=2 hours) and 0.4 (T= 3 hours). Neglecting infiltration, the discharge (CFS) from the watershed during the second hour is most nearly: 2.64, 2.15, 1.44 or 1.20." The books answer has the solution as 1.5 in/hour x 0.5 cf/in + 0.7 in/hr x 1.2 cf/in. = 2.15 cfs. I do not understand how they go this answer, especially because the units do not add up.

2) For Problem 509 - The question is "Assume that you are evaluating an agricultural watershed. The soil is classified as clay with a high swelling potential. The watershed is a pasture that has 65% ground cover and is not heavily grazed. The potential maximum retention (in) after runoff begins (also called the soil storage capacity) is most nearly." How can you tell from this problem whether you use straight row, contoured, or contoured and terraced in Table 20.5 of the Lindenburg book?

If anyone can help with these questions I would greatly appreciate it.

Thanks.

Hi,

For question 508 there is errata for this question that explains the answer a little better, you should porbably get for this sample exam.

This might help:

Rainfall 1 is from time, t=0 to t=1 at an intensity of 1.5 in/hr

Rainfall 1 is from time, t=1 to t=2 at an intensity of 0.7 in/hr

The discharge in the watershed during the second hour consists of the 2nd hr of rainfall 1 plus the 1st hr of rainfall 2

Q= (1.2 cfs/in)(1.5 in/hr)(1 hr) + (0.5 cfs/in)(0.7in/hr)(1 hr) = 1.8 +0.35 =2.15 cfs

For question 509 I believe you use the NRCS Method S = (1000/CN) - 10; CN = curve number and S = soil storage capacity.

Since the question reference a pasture that is not heavily grazed, so that implies grass cover. I have an older CERM so my table # maybe different.

From Table 20.4 P. 20-17

CN for open space

poor condition (grass cover <50%) 89

fair condition (grass cover 50-75%) 84

good condition (grass cover >75%) 80 so you can use CN= 84 and S=1.90 or NCEES used an average CN = (89+80)/2=84.5 and S=1.83

FYI

straight row, contoured, or contoured and terraced refer to row crops such as corn planted in straight rows, or crops like dryland wheat where the ground plowed and planted in the direction of the contours for soil conservation.

I believe you can get the erratta for the 2008 PE sample exam at PPI's site.

Good Luck

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Can someone please link the NCEES errata? All I could find is this, which is only 1 problem (#116):