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Home work solution MONO LAKE problem(Hydrology)...

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Homework 1 Solution

1.

2006 Hydrogeology Due 3/14/2006

At a water elevation of 6391 ft, Mono Lake has a volume of 2,939,000 ac-ft, and a surface area of 48,100 ac. Annual inputs to the lake include 8 in. of direct  precipitation, runoff from gauged streams of 150,000 ac-ft per year, and ungauged runoff and groundwater inflow of 37,000 ac-ft per year. Evaporation is 45 in. per year. (A) Make a water budget showing inputs, in ac-ft per year and outputs in ac-ft  per year. Does Does the input balance the output? (B) Will the average lake level rise or fall from the 6391-ft elevation over the long term? (C) What would be the lake surface area when the inputs balance the outputs?(Assume that the volume of gauged and ungauged runoff and ground-water inflows remain constant with a change in lake surface area. ) (D) What is the residence time* for water in Mono Lake when the water surface is at 6391 ft? *The residence time of a body of water is the average time that is would take for the volume of water to be exchanged once.

Solution: First convert all measurements to units of acre-feet per year Precipitation = 8 in/yr × 0.083 ft/in = 0.67 ft/yr = 0.67 ft/yr × 48,100 ac = 32,067 ac-ft/yr Evaporation

= 45 in/yr × 0.083 ft/in = 3.75 ft/yr

= 3.75 ft/yr × 48,100 ac = 180,375 ac-ft/yr (A) INPUTS(ac-ft/yr) Precipitation

32,000

Streamflow

150,000

Ground water inflow Total

37,000 219,000

OUTFLOWS (ac-ft/yr) Evaporation

180,000

Total

180,000

Input is greater than output (B) The lake level will rise over the long term, as inputs are greater than the only output, which is evaporation. (C) -1-

Homework 1 Solution

2006 Hydrogeology Due 3/14/2006

In order for the inputs and outputs to be equal, the lake will rise so that the surface area is greater and more water evaporates as evaporation as well as the precipitation input is a function of lake surface area. The excess input at an area of 48,100 acre surface area is 219,000 ac-ft/yr – 180,000 ac-ft/yr or 39,000 ac-ft/yr The net evaporation is total evaporation less precipitation  Net evaporation = 45 in/yr – 8 in/yr = 37 in/yr = 3.08 ft/yr Increase in area = (39,000 ac-ft/yr) / (3.08 ft/yr) = 12,649 ac Total area of lake = 12, 662 ac + 48,100 ac = 60,762 ac = 61,000 ac (D) The residence time = (2,939,000 ac-ft) / (180,000 ac-ft/yr) = 16.3 yr 2.

Figure 2.28 is the hydrograph of a river with a long summer baseflow recession. Compute the volume of annual recharge that occurs between runoff year 1 and runoff year 2.

Solution: 3

Qo in year 1 is 280 ft /s 0.1 Qo in

year 1 is reached in 5.5 months -2-

Homework 1 Solution

2006 Hydrogeology Due 3/14/2006 4

7

t1 = 5.5 mo × 8.64× 10  sec/day× 30 day/mo = 1.4256 × 10  s V tp = Qo t 1 / 2.3026 7

V tp = (280× 1.4256 × 10 ) / 2.3026 = 1.73

×10

9

3

 ft

Total recession in year 1, t , is 6.4 months. (t/t1)

V t  = V tp / 10t 

9 3 (6.4/5/5) V t  = 1.73 × 10  ft  / 10 8 3 V t  = 1.19 × 10  ft  Where Vt is the volume of water that could contribute to  baseflow at the end of the first year’s recession. 3 Qo in year 2 is 320 ft /s

0.1 Qo is reached in 5.9 mo 7

t 2 is 5.9 mo or 1.53 × 10  s 3

7

9

V tp2 = (320 ft /s× 1.53× 10 s) / 2.3026 = 2.13 × 10  ft

3

The amount of recharge between recessions is V tp2 -V t  9

3

9

3

8

3

V tp2 -V t = 2.13× 10  ft - 1.19× 10  ft V tp2 -V t  = 2.00 × 10 ft

3.

A constant-head permeameter has a cross-sectional area of 78.5 cm². The sample is 23 cm long. At a head of 3.4 cm, the permeameter discharges 50 cm³ in 38s. (A) What is the hydraulic conductivity in centimeters per second and feet peer day? (B) What is the intrinsic permeability if the hydraulic conductivity was measured at 15 ? (C) From the hydraulic conductivity value, name the type of soil.

Solution: (A) K  =

VL  Ath

Where V  = 50 cm 3  L

=

23 cm

 A = 78.5 cm 2 t  = 38 s h = 3.4 cm -3-

Homework 1 Solution

K

=

2006 Hydrogeology Due 3/14/2006

50cm3 23cm i

2

78.5cm 38 s 3.4 cm i

= 0.11 cm

s = 316.8 ft d

 

i

(B)  ρ 0.999099 g / cm3 and µ  = 0.011404 g / (cm ⋅ s )

 At 15



k=

K µ 

⇒ k =

 ρ g

0.11cm / s ×0.011404 g /( cm ⋅ s) 0.999099 g / cm

3

×980 cm /

s

2

= 1.3×10

−6

cm2

(C) K

4.

= 0.11cm / s,

soil is well sorted sand or gravel.

An aquifer has a specific yield of 0.19. During a drought period, the following average declines in the water table were noted: Area

Size (km²)

A

15

Decline (m) 2.34

B

7.5

1.22

C

18.3

0.76

D

22.5

3.44

E

9.44

F

1.89

22.7

0.35

What was the total volume of water represented by the decline in the water table? Solution: Vw = SA∆h

Where S  = 0.19 2

= 1.5 × 10

7

m2

 A A

= 15 km

 A B

=

 AC 

= 18.3 km

 A D

=

22.5 km 2

=

2.25 × 10 7 m 2

 A E 

=

9.44 km 2

=

9.44 × 10 6 m 2

 AF 

=

22.7 km 2

=

2.27 × 10 7 m 2

7.5 km 2

and

= 2

∆h A =

7.5 × 10 6 m 2

= 1.83 × 10

7

m2

2.34 m

∆h B = 1.22 m ∆hC  =

0.76 m

∆h D =

3.44 m

∆h E  = 1.89 m ∆hF  =

0.35 m

-4-

Homework 1 Solution

Area A : V W  A

2006 Hydrogeology Due 3/14/2006

0.19 ⋅ 1.5 × 10 7 m 2 ⋅ 2.34 m

=

6.67 × 10 6 m 3

V W  A

=

V W  B

= 1.74 × 10

V W C 

=

V W  D

= 1.47 × 10

V W  E 

=

V W F 

= 1.51 × 10

6

m3

2.64 × 10 6 m 3 7

m3

3.39 × 10 6 m 3 6

m3

Total volume of water the decline of head in the aquifer is 3.07×107 m3 5.

3 1 A confined aquifer has a specific storage of 7.239×10− m− and a porosity of

0.274. The compressibility of water is

4.6×10−10 m2 / N  . What is the

compressibility of the aquifer skeleton? Solution: Ss

= ρ g

(α + nβ ) = 7.239 ×10−3 m−1

= 1000

(

kg / m 3 ×9.81 m / s 2 × α  + 0.274 ×4.6 ×10 −10 m2 / N −7

⇒ α  = 7.380 ×10

6.

)

 

m 2 / N 

An aquifer has three different formations. Formation A has a thickness of 22 ft and a hydraulic conductivity of 17.0 ft/d. Formation B has a thickness of 3.5 ft and a conductivity of 99 ft/d. Formation C has a thickness of 26 ft and a conductivity of 22 ft/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.

Solution: Fm

K ( ft / d )

A

17.0

B

99

C

22

bm ( ft )

22 3.5 26

and total thickness = 51.5 ft

-5-

Homework 1 Solution

n

K haverage

=



2006 Hydrogeology Due 3/14/2006

khm bm b

m=1

=

K vaverage

=

17.0  ft d 22 ft i

51.5 ft

99 ft d 3.5 ft i

51.5 ft

+

22 ft d 26 ft  i

51.5 ft 

= 25  ft



b n

bm

∑ K  m=1

=

vm

22 ft 17.0  ft d

7.

+

+

51.5 ft  3.5 ft 99 ft d

+

26 ft 

=

21 ft d 

22 ft d 

Use the Hazen method to estimate the hydraulic conductivity of the sediments graphed in Figure 3.33.

Solution:

K  = C (d 10 ) 2 d10

= 0.16 mm ×

1cm 10mm

= 0.016 cm

C  100 

K

2

= 100 ×(0.016) = 2.6 ×10

−2

cm / s

-6-

Homework 1 Solution

8.

2006 Hydrogeology Due 3/14/2006

Given the following set of data representing the hydraulic conductivity of core samples from the same formation, perform the following: Sample #

K (m/s)

1

4.3×10−4

2

6.1×10−3

3

2.5×10−5

4

1.2×10−4

5

1.0×10−6

6

7.1×10−3

7

9.1×10−6

8

2.2×10−3

9

4.2×10−5

10

8.7×10−4

11

3.5×10−5

(A) Find the arithmetic mean of the data set. (B) Find the geometric mean of the data set. (C) Make a histogram of the data set. (D) Make a histogram of the log transformed data.

-7-

Homework 1 Solution

2006 Hydrogeology Due 3/14/2006

Solution: Sample #

K (m/s)

ln (K)

K bin

ln (K) bin

1

4.30E-04

− 7.75

1.00E-03



2

6.10E-03

− 5.10

5.00E-04

−5

3

2.50E-05

− 10.60

1.00E-04

−6

4

1.20E-04

− 9.03

5.00E-05

−7

5

1.00E-06

− 13.82

1.00E-05

−8

6

7.10E-03

− 4.95

5.00E-06

−9

7

9.10E-06

− 11.61

1.00E-06

− 10

8

2.20E-03

− 6.12

− 11

9

4.20E-05

− 10.08

− 12

10

8.70E-04

− 7.05

11

3.50E-05

− 10.26

1.69E-02

− 96.35

sum

4

Arithmetic mean = 1.54E - 03 m s Geometric mean = 1.57E - 04 m s histogram data for K

histogram data for ln (K)

 Bin

Frequency

Bin

Frequency

1.E-06

1

− 12

1

5.E-06

0

− 11

1

1.E-05

1

− 10

3

5.E-05

3

−9

1

1.E-04

0

−8

0

5.E-04

2

−7

2

1.E-03

1

−6

1

> 5.E - 3

3

−5

1

4

1



-8-

Homework 1 Solution

2006 Hydrogeology Due 3/14/2006

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