# 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

you choose to use a spreadsheet for. Type your answers or a summary of your answer (for

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

Answer :

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

ℎ

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.

…