# Bae 4314 finalexam fall 2021

BAE 4314 – HYDROLOGY – SPRING 2021
FINAL EXAM (TAKE-HOME)- TOTAL POINTS: 100
DUE: BY 11:59 PM, DECEMBER 7, 2021
NOTE: The exam is open book open note but you must work individually. No
communication is allowed with other individuals. Submit spreadsheets for questions that
calculation questions) under each question. Use as much space as you need.
Name: ______________________________________
1. Use the Muskingum routing method to find outflow at time t2 when inflow is 100 cms. Assume the
following conditions: (i) Δt= 1 hr, k= 2.3 hr, x= 0.1516; (ii) at t1 inflow=outflow= 50 cms. (20 Pts).
∆t = 1 hour
K = 9.3 hour
x = 0.1516
∆t the ‘𝑡1 ’ inflow (I𝐼𝑛− ) = Outflow Q-) = 50 cm
∆t ‘𝑡2 ’, inflow 𝐼𝑛+ = 100 cm
Outflow φn =?
outflow ∆t time interval t in given by
φn= Co Int
C. In – 1 + φ Q n-1)
outflow Tome Inflow
(I) 50 – Int 50 = φui On? 100 In are Mushkingum constanta Co, C. and C -kn+0.5
∆t K-ka+0.5 ∆t -(2.3)
(0.1516) +0.5 (1) .3-6.3)
(0.1516) +0.5 (1)
CO 2=-0.34868 +0.5 . 2.3-0.34868 +0.5
co =0.15132 2.45132 [co= 0.06173]
Ka+ 0.5.At K-kx+0.5
∆t c. 0.84868 2.45132
= 0.346211 Cotcit Ca = 1
= 1-cora (C=0.59206)
en = 0.06473) (100) +(0.34621)(50)+0.59 206)(50) – 6.173 + 17.3105 +29.603 1∆ an =
53.0865 cano
The outflow of a waterhed at time 𝑡2
i∆= 53.086 CMs
2. An aquifer test is performed on a well in an aquifer of practical infinite aerial extent (no
hydrogeologic boundaries in the nearby vicinity). The aquifer is confined with no vertical leakage.
The pumped well is pumped at a constant rate of 400 gpm. The drawdown is measured in an
observation well located 35 feet from the pumping well. The data from the aquifer test are given in
the table below. Analyze the data below and determine T (30 Pts).
Time (min)
1
3
5
7
10
12
15
18
20
25
30
40
50
70
100
Drawdown(ft)
0.1
0.2
0.6
1.1
1.4
1.6
2.0
2.3
2.5
2.8
3.2
3.5
4.0
4.6
5.3
Time (min)
120
150
180
200
300
500
700
1000
2000
5000
7500
10000
20000
35000
50000
From table consider
φ=400 gpm
π=35ft
S1 = 17.2 ft 𝑡1 = 10,000 min
S2 = 20.6 ft 𝑡2 = 35,000 min
2.3 𝜑
𝑡
S2-S1= 4𝜋𝑡 log10 𝑡1
2
20.6–17.2 =
2.3 x 400×192.5ft3/day
4𝜋𝑡
35000
log 10 10000
T= 42.72 T = 96354.45
T = 2255.488 ft2/day
𝜑
47 𝑡
S= 4𝜋𝑡 [loge 𝜋2 𝐴-0.5772
S=21.5 ft
t= 50000 min
= 34.72 Day
400∗192.5
By 121.5 = 4𝜋∗2255.488 [loge
4X2255.480×34.72
7.914 – loge (255.7)A – 0.5772
352 A
– 0.5772
Drawdown (ft)
5.8
6.2
6.8
7.1
8.1
9.5
10.4
11.3
13.1
15.5
16.6
17.2
19.1
20.6
21.5
Storage coefficient 1A = 0.05248
3. A well of radius 0.2 m produces 550 m3/d from an unconfined aquifer with K = 10 m/d. The well is
located 250 m from a river, and the saturated thickness of the unconfined aquifer is 30 m. Calculate
and plot the steady-state drawdown at various distances between the well and the river? (20 Pts)
φ
2π kh (hω−h1)
𝜋
𝜋𝜔
2.3𝑙𝑜𝑔10 2
hω − h1 = drowdown
φ = 550m3/day
K= l0 m/day
H= 30m .
We Need to find drawdown at. 200, 150 I00 m distances
55o =
55o =
55o =
2π ∗10∗30 (hω−h1)
200
0.2
2.3𝑙𝑜𝑔10
2π ∗10∗30 (hω−h2)
100
0.2
2.3𝑙𝑜𝑔10
2π ∗10∗30 (hω−h3)
150
0.2
2.3𝑙𝑜𝑔10
=2.01m
=1.929 m
=1.811 m
drawdown at
200 m = 2.
150m=1.919m
100m=1.811m
4. A detention basin (small flood control reservoir) is characterized by the following table of water
surface elevation versus storage (S) and outflow rate (O). The initial water surface in the reservoir is
at an elevation of 120 ft.
Elevation
(ft)
120
125
130
Storage
(ft3)
0
27,000
135,000
Outflow
(ft3/s)
0
20
50
135
288,000
80
An inflow hydrograph to the reservoir is provided below.
Time
(hours)
0
1
2
3
4
5
6
7
8
Discharge
(ft3/s)
0
10
50
110
70
50
30
10
0
What is the maximum water elevation in the reservoir during passage of the inflow hydrograph? (30 Pts)
110
100
90
80
70
60
50
110ft3/sec
Q 40
ft/sec
30
20
10
0
time (hr)
rule
1
2
3
4
5
6
7
8
9
10

Volume=2[(𝜑𝑜+ 𝜑𝑛) +2(𝜑1+ 𝜑2………….𝜑𝑛−1 )]
1∗60∗60
=
2
-[(o+o)+2(10+50+110+70+50+30+10)]
Volume =1188000 ft3
elevation = 120 ft
storage = o(ft3)
elevation = x
storage = 1188000
elevation – 130ft
storage = 135000
𝑥−120
10
=
1188000 135000
X=208 ft
Max. W elevation – 208 ft.

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