3897first exam solution

College of Graduate StudiesHydrological Processes and Systems−Fall 2010First Exam (100 points) −12/10/2010 − Solution

Instructor: Dr. Sameer Shadeed

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Q1. Estimate ETo using Jensen-Haise method during August for Nablus area located 32oN and elevated 600 m (amsl). [10 points]

Parameter June July AugustMax. Temp. (oC) 29.0 30.0 29.5Min. Temp. (oC) 17.0 20.0 19.5Sunshine (n/N)% 70 75 83Solar radiation, Ra (mm/day) 17 16.8 15.6ETo (mm/day)

Solution:The warmest month is Julyes max (for T = 30) = 42.43mbes min (for T = 20) = 23.38 mbRs = 0.77[0.25+0.5(n/N)]Ra = 0.77[0.25+0.5(0.83)]*15.6 = 7.988 mm/day

CT = 1/[45-(600/137)+(365/42.43-23.38) = 0.017Tx = -2.5 – 0.14(42.43-23.38)-(600/500) = -6.367

Tavg. (July) (30+20)/2 = 25

ETo (July) = 0.0165[25-(-6.367)](7.988) = 4.13 mm/day



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Q2. A 5-hour rainfall event took place where the intensity in the second period was unrecorded. However, the runoff generated from this event equaled 18 mm while the Φ-index value was 5 mm/hr. Find out the missing intensity value. [10 points]

Solution:There are two possibilities. Either to have the missing rainfall event intensity at a value higher than Φ-index or less. If less, then Runoff = (6 – 5) + (0) + (10 – 5) + (12 – 5) + (7 –5) = 15 mm. Therefore, the unrecorded rainfall is higher than Φ-index ⇒ Runoff = (6 – 5) + (i-5) + (10 – 5) + (12 – 5) + (7 – 5) = 18 mm. This gives i + 10 =18 or i= 8 mm occurring over a duration of 1 hour which means an intensity of 8 mm/hr.



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Q3. A lake with a surface area of 3 km2 was monitored over a period of time. During a one week period the inflow was 1.4 m3/s, the outflow was 1.1 m3/s, and a 1.6 cm seepage loss was measured. During the same week, the total precipitation was 5 cm. Evaporation loss was estimated at 4 cm. Estimate the storage change for this lake during the week. [10 points]

Solution:I-O = ΔS(1.4 m3/s + 5 cm) - (1.1 m3/s + 4 cm + 1.6 cm) = ΔSΔS = (0.3 m3/s + 1 cm)

= {[0.3 (m3/s)(3600s/hr)(24hr/day)(7day/week)(1week)]/[3x106 m2]}* (100 cm/1 m) – 0.6 cm= 6.05 cm + – 0.6 cm = 5.45 cm



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Q4. You have the following four cases of runoff hydrographs for four situations. Justify the different shape-looking hydrographs. Consider “a” versus “b” and “c” versus “d”. [10 points]

Solution: There are two issues to consider here; the peak amount and the time to peak. In case “a”, we need more time for runoff to arrive at the outlet thus compared to “b”

it will need more time and thus delay in peak. In case “c”, the outlet drains larger are at a shorter time, which leads to a less time to

peak.



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Q5. For the storm given in table 1, convert the resulting direct runoff hydrograph (DRH) summarized in table 2 to a 3h-unit hydrograph if the Φ-index for the storm was 1 cm/hr. [20points]

Table 1: Storm distribution

T (hr) Intensity (cm/hr)1 12 33 34 1

Table 2: Direct runoff hydrographT (hr) DRH (m3/s)

0 01 1002 2403 2004 1005 406 0

Solution: First we find the excess rainfall by subtracting Φ-index from the rainfall

hyetograph. The excess rainfall represents 4.0 cm of rainfall (2 cm/hr for 2 hours)T (hr) Intensity (cm/hr) Φ-index (mm/hr) Excess Rainfall (mm/hr)

1 1 1 02 3 1 23 3 1 24 1 1 0

We need to convert the hydrograph to 1 cm of direct runoff over the catchment. Dividing the DRH ordinates by 4, 2hr-unit hydrograph obtained as follows

T (hr) DRH (m3/s) 2hr-UH (m3/s)

0 0 01 100 252 240 603 200 504 100 255 40 106 0 0



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Given the 2h-UH, use S-hydrograph procedure to construct a 3h-UH as followsT

(hr)DRH (m3/s)

2h-UH (m3/s)

S-Hydrograph(m3/s)

S-Hydrograph Lagged by 3h (m3/s)

Difference(m3/s)

3h-UH(m3/s)

0 0 0 0 0 0.001 100 25 25 25 16.672 240 60 60 60 40.003 200 50 75 0 75 50.004 100 25 85 25 60 40.005 40 10 85 60 25 16.676 0 0 85 75 10 6.677 85 85 0 0.00



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Q6: The upper part of the Faria catchment is composed of two sub-catchments which are Al-Faria sub-catchment and Al-Badan sub-catchment as shown in Figure 1. The drainage areas for these sub-catchments are 56 km2 and 83 km2 respectively. A 1mm-unit hydrographs (UHs) for these two sub-catchments are tabulated in Table 1. A major storm event was recorded on February 14, 2004. The event brought 40 mm of rainfall and lasted 16 hours. The event was averaged over the Al-Badan sub-catchment as provided in Table 2. For this event, the Φ-index value was 4.6 mm.

1. Estimate the volume of direct runoff that occurred in million cubic meters over Al-Badan sub-catchment. [5 points]

2. Estimate the volume of infiltration in million cubic meters for this storm event based on the measured rainfall. [5 points]

3. Using the 1-mm IUH of Al-Badan, estimate the storm hydrograph. [10 points]

4. Using the simulated storm hydrograph, estimate the volume of direct runoff that occurred in million cubic meters over Al-Badan sub-catchment and compare with result of part 1. [5 points]

5. A dam is to be constructed 5 km downstream of Al-Malaqi Bridge, where Al-Badan and Al-Faria streams meet, combine the unit hydrographs of the both sub-catchments at Al- Malaqi Bridge and route the combined unit hydrographs to the dam location using Muskingum channel-routing method (use x= 0.2 and K = 2 hr). [15 points]

0 10 Kilometers

Faria sub-catchmentsAl-Badan SubAl-Faria SubLower Sub

Main StreamFaria outline

N

#

Al-Malaqi Bridge

#

Proposed Dam Site

Figure 1: The Three Sub-catchments of the Faria Catchment



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Table 1: 1-mm IUHs for Al-Badan and Al-Faria sub-catchments

Time (hr)Q (m3/s)

Al-Badan Al-Faria0 0 01 2.91 2.472 4.26 3.213 3.35 2.514 2.64 1.975 2.09 1.566 1.66 1.237 1.32 0.988 1.06 0.789 0.85 0.6210 0.68 0.4911 0.54 0.3912 0.44 0.3113 0.35 0.2514 0.28 0.215 0.23 0.1616 0.19 0.1317 0.15 0.118 0.12 0.0819 0.1 0.0620 0.08 0.0521 0.07 0.0422 0.05 0.0323 0.04 0.0324 0.04 0.0225 0.03 0.0226 0.02 0.0127 0.02 0.0128 0.02 0.0129 0.01 0.0130 0.01 0.0131 0.01 032 0.0133 0.0134 0.0135 0

Table 2: Hourly rainfall data averaged over Al-Badan sub-catchment

T (hr) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16RD (mm) 0 3.2 3.8 2.5 3.5 5.5 1.5 0 0 0 0 0 4.5 8 5 2.5



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Solution:1. In order to calculate the volume of direct runoff, we do the following:

i. Estimate the excess rainfall by subtracting the value of Φ-index from the rainfall data T (hr) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16RD (mm) 0 3.2 3.8 2.5 3.5 5.5 1.5 0 0 0 0 0 4.5 8 5 2.5Φ-index 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6 4.6ERD (mm) 0 0 0 0 0 0.9 0 0 0 0 0 0 0 3.4 0.4 0

ii. Depth of direct runoff = excess rainfall depth = 0.9 + 3.4 + 0.4 = 4.7 mmiii. Volume of direct runoff = Depth of direct runoff multiply by the catchment area

= (4.7X10-3 m)(83X106 m2) = 390100 m3 = 0.39 MCM

2. The volume of infiltration = (gross rainfall – excess rainfall) multiply by the catchment area

= (40X10-3 - 4.7X10-3)(83X106 m2) = 2929900 m3 = 2.93 MCM

3. The storm hydrograph of Al-Badan sub-catchment is estimated using Excel (see the provided excel sheet) and the result is shown in the following figure

4. From the simulated storm hydrograph, the volume of direct runoff = Δt∑Q = 1hr (111.16 m3/s) (3600s/hr) = 400176 m3 = 0.4 MCM As a comparison with the estimated volume from the excess rainfall depth (0.39 MCM), the rounding error can be calculated as 1-(0.39/0.4) = 0.025 = 2.5%

5. The combined unit hydrograph for both Al-Faria and Al-Badan sub-catchments are estimated using Excel (see the provided excel sheet) and the routed unit hydrograph of the combined unit hydrographs is shown in the following figure



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Good Luck

3897first exam solution

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