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tiL12214_appe_643-698.indd Page 697 09/10/10 8:37 AM user-f463
tiL12214_appe_643-698.indd Page 697 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

d _
Calculate Q G . t = 12 h v =

t

P = 1,143 mm Q G = P − (AE + 53% (P−AE)) d = 14.7 km _
14.7 km

AE = 558 mm
Q G = 1,143 mm − (558 mm + 0.53 × v = ? = 12 h

Q G = ? (1,143 mm − 558 mm)) _
km

= 1.2
Q G = 1,143 mm − (558 mm + 310 mm) h

Q G = 275 mm 24.7. Groundwater flow velocity determined from tracer testing can be
calculated by using the velocity equation (see chapter 2).
Calculate Q S
d _
P = 1,143 mm Q S = 53% (P − AE) t = 429 days v =

AE = 558 mm = 0.53 (1,143 mm − 558 mm) d = 229 m t
229 m

Q S = ? = 0.53 (585) v = ? = _
429 days
= 310 mm −1 m _

= 5.34 × 10
The percentage is the ratio of each component to P times day

100 percent. Calculate the percentage of each component of
24.8. Determine the volume of the void space (V V ) by taking the
the watershed budget.

difference between the dry and saturated masses and dividing
_ Q S _
AE
%AE = × 100% %Q S = × 100% by the density of water.

P P
_ 310 mm m dry = 18.02 g _
558 mm
×
m wet − m dry
×

= 100% = _ 100% V V =

1,143 mm 1,143 mm m wet = 20.46 g ρ water
__
= 48.8% = 27.1% V V = ? = 20.46 g − 18.02 g

g
_
1.00

Q G _ 3

%Q G = × 100% cm
P _
2.44 g
=

g
_ _
275 mm
×

= 100% 1.00

1,143 mm cm 3
= 24.1% = 2.44 cm 3
24.4. Rearrange the watershed budget equation to derive an equation Th en divide V V by the total volume of the specimen to
for actual evapotranspiration (evaporation and transpiration) determine porosity.
and substitute the values provided. 3
V V = 2.44 cm ϕ =
V V _
V T = 9.24 cm 3
P = AE + Q S + Q G V T
2.44 cm
∴ AE = P − (Q S + Q G ) ϕ = ? = _ 3

AE = 254 mm − (30 mm + 21 mm) 9.24 cm 3
= 203 mm = 0.264
The percentage is the ratio of each component to P times 24.9. Determine the total volume (V T ) of the aquifer.

100 percent. Calculate the percentage of each component of A = 25 km 2 V T = Az
the watershed budget. z = 12.3 m Convert km to m : 2
2
_ V T = ? 6 2
AE
1 × 10 m

%AE = × 100% Q S _ 2 ( _

P %Q S = × 100% 25 km × 2 )

P 1 km
_
203 mm

= × 100% _ 7 2
30 mm

254 mm = × 100% 2.5 × 10 m

254 mm 7 2
= 79.9% V T = 2.5 × 10 m (12.3 m)
= 11.8% = 3.1 × 10 m 3
8
Q G _ Determine the volume of the void space (V V ) by multiplying

%Q G = × 100%
P the porosity by the volume of the aquifer.
_
21 mm
= × 100% ϕ = 0.21 V V _

254 mm 8 3 ϕ = ∴ V V = ϕV T

V T = 3.1 × 10 m V T
= 8.3% 8 3
V V = ? V V = 0.21 (3.1 × 10 m )
24.5. Q S = 737 mm Volume = Q G A = 6.5 × 10 m 3
7
A = 1,157 km 2 Convert mm to m:
Volume = ? 24.10. In order for the well to be artesian, the static water level in the
1 m
3 )
(

737 mm _ well must be above the upper contact of the confi ned aquifer.
1 × 10 mm
−1
7.37 × 10 m Determine the elevation of the upper contact of the aquifer by
subtracting the depth at which the aquifer was encountered
2
Convert km to m : 2 from the surface elevation.
2)
(
2 __
2
1 m

1,157 km surface elevation = 187.2 m

−6
1 × 10 km
confined aquifer depth = 42.5 m
9
1.157 × 10 m 2 minimum water elevation = ?
9
2
−1
Volume = 7.37 × 10 m (1.157 × 10 m ) minimum water
8
= 8.53 × 10 m 3 elevation = surface elevation − confined aquifer depth
24.6. Groundwater flow velocity determined from tracer testing can be = 187.2 m − 42.5 m

calculated by using the velocity equation (see chapter 2). = 144.7 m
E-55 APPENDIX E Solutions for Group A Parallel Exercises 697