King Fahd University of Petroleum & Minerals
Civil Engineering Dept.
CE 331 Cover Sheet
Spring 07
Instructor: Rashid .I. Allayla
King Fahd University of Petroleum & Minerals
Civil Engineering Dept.
CE 331 Engineering Hydrology
Dr. Rashid Allayla
HW No. 1
Problem 1.
The annual evaporation from a lake is found to be 125 cm. If the lake’s surface area is 12 km2, what is the daily evaporation rate in cm/d?
Solution:
Daily evaporation = 125/365=0.34 cm
Problem 2.
The storage in a reach of river is 20,000 m3 at a given time. Determine the storage 1 hr later if the average rates of inflow and outflow during the hour are 20 and 18 m3/s, respectively.
Solution:
Total inflow after one hour = (20 m3/s)(3600 s/hr) = 72000 m2
Total outflow after one hour = (18 m3/s)(3600 s/hr) = 64800 m2
Gain = 7200-64800 = 57600 m3
Storage = 20000+57600= 77600 m3
Problem 3.
The chart shown below represents a record of rainfall vs. time. Find the average rainfall intensity between 6 am and noon. Also find the total rainfall on the day.
Average Intensity between 6 & noon = 1.5 inch
Tota rain = area under the curve = ½ (0.2 x 12) + ½ [(0.2+0.15)x6+ ½ [(0.15+0.25)x6] = 1.2 + 1.05 + 1.2 = 3.45 inches
Problem 5:
Rain gage X was out of operation for a month during a storm. The rainfall amount for the adjacent three stations A, B and C were 10.2, 11, and 15 mm. The annual precipitation for all gages are: A = 200 mm, B = 1150 mm, C = 140 mm, and X = 120 mm, Estimate the amount of rainfall at the missing month at station X.
Px = 1/3 [(Nx / NA) PA + (Nx / NB) PB + (Nx / NC) PC]
Px=1/3[(120/200)(10.2)+(120/1150)(11)+(120/140)(15)] = 2.04+0.38+4.29=6.71 mm
HW No. 2
Problem 1:
Compute the daily evaporation from Class A pan if the amount of water required to bring the level to the fixed point are as follows:
Day 1 2 3 4 5
Rainfall (in) 0 0.65 012 0 0.01
Water added 0.29 0.55 0.07 0.28 0.10
Evaporation ??
Solution:
Day 1 2 3 4 5 Rainfall (in) 0 0.65 012 0 0.0
Water added 0.29 0.55 0.07 0.28 0.10
Evaporation 0.29 1.2 0.19 0.28 0.11
Problem No. 2:
Given initial infiltration capacity f0 of 3.0 in/hr and a time constant k of 0.30 hr-1 derive infiltration capacity curve vs. time if the final infiltration capacity is 0.5 in/hr. What is the amount of water infiltrated (in inches) after 9 hours?
Solution:
f = 0.5 + 2.5 e-0.3k
F = tf∞ + (f0 - f∞)/k [1-e-10K] = 9 (0.5) + (2.5 / 0.3) e-0.3x9
Problem No. 3:
The infiltration rate for excess rain on an area was 4.5 in/hr at the beginning of rain, and it decreased exponentially to equilibrium of 0.5 in/hr. A total of 30 in. of water infiltrated during the 10 hr-hr interval. Determine the value of k in Horton’s equation
HW No. 3
Problem 1:
Use Meyer equation to find the daily evaporation from a lake during which the following data were obtained : air temperature = 90o F, water temperature = 60o F, wind speed = 20 mph and relative humidity = 20%.
Solution:
-Using Meyer formula: From table (A.2) for vapor pressure, es (@60oF) = 0.52 in. Hg
es (@90oF) = 1.42 in. Hg
e = 1.42 x 0.20 = 0.284 in Hg
For C = 0.36 then E = 0.36 ( 0.52 – 0.284) [1+20/10]
= 0.08496 x 3 = 0.255 in
Problem 2:
Use coaxial chart to find lake evaporation in cm/day for the following conditions: The water temperature of Pan “A” is 60oF, wind movement is 100 mi/day, elevation above sea level is 1000 ft, the difference of temperature between class “A” pan surface and air is +5 degrees Fahrenheit and the pan evaporation is 0.3 in/day.
Solution:
From the chart @ pan temp. of 60oF and wind speed of 100 mi/day αP = 0.6
From the coaxial chart:
Problem 3:
The infiltration rate for small area was observed to be 4.5 in/hr at the beginning of the rain, and it decreased exponentially to an equilibrium of 0.5 in/hr after 10 hrs. a total of 30 inches of water infiltrated during the 10 hr interval. Determine the value of k in Horton equation.
Solution:
f = f∞ + (f0 - f∞) e-Kt
f∞ = 0.5 in/ hr and ∫0 to 10 f (t) dt = 30 in
f = 4.5 in / hr then ∫0 to 10 [ f∞ + (f0 - f∞) e-Kt ] dt = 30
then: 10 f∞ + (f0 - f∞)/k [1-e-10K] = 30
Then: (10)(0.5) + (4.5-0.5) / k (1-e-10K) = 30
then K = 0.1027 hr-1
______
Problem 4:
Tabulated below are total rainfall intensities during each hour of a storm over a drainage basin.
a) Plot rainfall hyetograph (intensity vs. Time)
b) Determine total storm precipitation in inches.
c) If the net storm rain is 2.00 in, determine the Φ-Index in in/hr for the drainage basin.
d) Determine the area of the basin in acres if the net rain is 2 inches and the measured volume of runoff is 2015 cfs-hr
e) Determine the volume of direct runoff in ft3 that would result from the following storm:
Hour: 1 2 3 4
IntensityLin/hr): 0.4 0.05 0.30 0.2
Hour rainfall Intensity
1 0.41
2 0.49
3 0.22
4 0.31
5 0.22
6 0.08
7 0.07
8 0.09
9 0.08
10 0.06
11 0.11
12 0.12
13 0.15
14 0.23
15 0.28
16 0.26
17 0.21
18 0.09
19 0.07
20 0.06
21 0.03
22 0.02
23 0.01
24 0.01
25
Solution:
a) The total rain is the area under the hyetograph = 3.68 in.
b) try 0.08 < Φ > 0.9 then (0.41- Φ) (1) + (0.49 – Φ) + … = 2 in
3.19 – 14 Φ = 2 then Φ = 0.085 in/ hr OK
Volume of direct R.O. = 2015 cfs-hr
d) Area = (2015-hr) /(2 in/hr x 1/12 in/ft) = 12090 ft2
f) Volume of direct R.O. = (0.4-0.085)(1) + (0.3 - 0.085)(1) + (0.2 - 0.085)(1) = 0.615 in
______
Problem 5:
Given a rain pattern and a simplified Horton curve. Correct the Horton curve and find the mount of runoff.
Solution:
The equation f = 10 – 1.714 t
F = 10 t – 0.857 t2 = 9
Then: 0.857t2 - 10t +9 = 0
Then: t [@ rain = 9] = { -(-10) +/- SQRT [(10)2 – (4)(0.857)(9)] / 2} = 0.99 hr
Result: Shift the curve @ t=0.99 to t=2hr and find the area above the curve.