1. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
Find the height of the groundwater table, y1, when the pumping rate is 400 gallons per minute. [pause] In this problem, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

2. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
an unconfined sand aquifer is pumped at state-state. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

3. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
At a pumping rate of 200 gallons per minute, the observed groundwater levels are 48 feet and 50 feet, at distances of --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

4. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
50 feet and 500 feet from the center of the well, respectively. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

5. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
At an increased pumping rate of 400 gallons per minute, the groundwater level 500 feet away, is measured to at 49.2 feet, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

6. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
and the problem asks to find the groundwater level at a radius of 50 feet from the well, when the pump operates 400 gallons per minute. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

7. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
Since this diagram is radially symmetric, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

8. Find: y1 [ft] @ Q=400 [gpm] 44 Sand 45 46 47 y2 y1 r1 r2 unconfined Q
steady-state y2 y1
pump
we only need to view one half. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
r1=50 [ft]
y1=48 [ft]
y1=? [ft]
r2=500 [ft]
y2=50 [ft]
y2=49.2 [ft]

9. Find: y1 [ft] @ Q=400 [gpm] y2 y1 r1 r2 Q steady-state pump
r1=50 [ft]
r2=500 [ft] y2 y1
pump
[pause] The steady-state pumping rate for an unconfined aquifer, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

10. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) Q= ln (r2/r1) y2 y1 r1 r2
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft] y2 y1
pump
whose drawdown is small compared to the aquifer thickness, is shown here. [pause] Solving for y1, the equation becomes, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

11. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2 y1
pump
the square root of y2 squared, minus the flowrate times the natural log of the ratio of the radii, divided by PI times K. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

12. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K ? Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2 y1 ?
pump
From our given data, we know all the variables needed to solve for y1, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

13. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K ? Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2 y1 ?
pump
except, for the value of K, the hydraulic conductivity. [pause] To solve for K, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

14. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K ? Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2 y1 ?
pump
we’ll need to use the data from associated with the smaller flow rate. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

15. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) Q= ln (r2/r1) y2 y1 r1 r2
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft] y2 y1
pump
Revisiting the initial equation, --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

16. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) Q= ln (r2/r1) y2 y1 r1 r2
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft] y2 y1
pump
the hydraulic conductivity is calculated. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

17. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * (y2-y1) Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1) K=
π * (y2-y1) 2 2 y2 y1
pump
[pause] The known values are plugged into the equation --- r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

18. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * (y2-y1) Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1) K=
π * (y2-y1) 2 2 y2 y1
pump
for r1, r2, y1 and y2. r1 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

19. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * (y2-y1) Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1) K=
π * (y2-y1) 2 2 y2 y1
pump 1
ft3 1
min
and flowrate substituted in after being converted to cubic feet per second. [pause] The hydraulic conductivity equals --- x x r1
gal
7.48 s 60 r2
Q = 200 [gpm]
Q = 400 [gpm]
y1=48 [ft]
y1=? [ft]
y2=50 [ft]
y2=49.2 [ft]

20. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * (y2-y1) Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1) K=
π * (y2-y1) 2 2 y2 y1
pump 1
ft3 1
min
feet per second. [pause] Since the hydraulic conductivity is equal for both --- x x r1
gal
7.48 s 60 r2
Q = 200 [gpm]
Q = 400 [gpm] ft K=
y1=48 [ft]
y1=? [ft] s
y2=50 [ft]
y2=49.2 [ft]

21. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2 y1
pump
Q equals 200 gallons per minute, and Q equals 400 gallons per minute, r1 r2
Q = 200 [gpm]
Q = 400 [gpm] ft K=
y1=48 [ft]
y1=? [ft] s
y2=50 [ft]
y2=49.2 [ft]

22. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2 y1
pump
the same value of K can be used for the larger flowrate. r1 r2
Q = 200 [gpm]
Q = 400 [gpm] ft K=
y1=48 [ft]
y1=? [ft] s
y2=50 [ft]
y2=49.2 [ft]

23. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K Q= ln (r2/r1)
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2 y1
pump 1
ft3 1
min
The other values are also plugged in, and the groundwater level, y1, at a pumping rate of --- x x r1
gal
7.48 s 60 r2
Q = 200 [gpm]
Q = 400 [gpm] ft K=
y1=48 [ft]
y1=? [ft] s
y2=50 [ft]
y2=49.2 [ft]

24. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K y1=45.02 [ft] Q=
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2
π * K y2
y1=45.02 [ft] y1
pump 1
ft3 1
min
400 gallons per minute, equals feet. x x r1
gal
7.48 s 60 r2
Q = 200 [gpm]
Q = 400 [gpm] ft K=
y1=48 [ft]
y1=? [ft] s
y2=50 [ft]
y2=49.2 [ft]

25. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K y1=45.02 [ft] Q=
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2 44 45 46 47
π * K y2
y1=45.02 [ft] y1
pump
When reviewing the possible solutions, r1 r2
Q = 200 [gpm]
Q = 400 [gpm] ft K=
y1=48 [ft]
y1=? [ft] s
y2=50 [ft]
y2=49.2 [ft]

26. Find: y1 [ft] @ Q=400 [gpm] π * K * (y2-y1) π * K y1=45.02 [ft]
steady-state 2 2 Q=
ln (r2/r1)
r1=50 [ft]
r2=500 [ft]
Q * ln (r2/r1)
y1=
y2- 2 44 45 46 47
π * K y2
y1=45.02 [ft] y1
pump
AnswerB
the answer is B. r1 r2
Q = 200 [gpm]
Q = 400 [gpm] ft K=
y1=48 [ft]
y1=? [ft] s
y2=50 [ft]
y2=49.2 [ft]

27. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4