1. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
Find the pore water pressure at point B in kilopascals. [pause] In this problem,
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

2. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
an aquifer sits below,
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

3. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
a 3 meter thick layer of silt, which is below
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

4. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
a 2 meter thick layer of sand, upon which
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

5. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
a body of water of unknown depth resides.
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

6. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
The hydraulic conductivities of the sand and silt layers are given.
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

7. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
Points A, B and C are positioned at the layer boundaries, ---
3 [m]
Ksilt=1*10-6 [m/s] V V
well A
Aquifer
pump

8. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
and point D is at the surface of the water.
3 [m]
Ksilt=1*10-6 [m/s] V V
well A
Aquifer
pump

9. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
A well pump, pumps water from the aquifer,
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

10. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
which holds the pore water pressure at point A to 24.5 kilopascals.
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

11. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x 15 63 245
V=0.2 [m3 / day * m2] D
Water x
uA=24.5 [kPa] V C
Sand
2 [m] 15 63
245
447
Ksand=1*10-5 [m/s] V B
Silt
This pumping causes water to flow down through the soil profile at a given seepage velocity, v. [pause]
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

12. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x V C
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
The pore water pressure at point B is equal to the density of water times acceleration from gravity times the pressure head at point B, ---
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

13. Find: u [kPa] at point B D C B A Water 1 Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x MG m3 V 1 C
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
where the density of water is 1 mega gram per meter cubed, ---
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

14. Find: u [kPa] at point B D C B A Water 1 Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x m
9.81 MG V 1 s2 C
Sand m3
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
the acceleration from gravity is 9.81 meters per second squared, ---
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

15. ? Find: u [kPa] at point B D C B A Water 1 Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x m
9.81 MG V 1 s2 C
Sand m3
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
and the pressure head at point B still needs to be computed. [pause] At this point, it may be convenient to ---
3 [m]
Ksilt=1*10-6 [m/s] V
well A
Aquifer
pump

16. ? Find: u [kPa] at point B D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] V C A B C D
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
tabulate the head values for points A through D.
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

17. ? Find: u [kPa] at point B D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] V C A B C D
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
This includes the total head, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

18. ? Find: u [kPa] at point B D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] V C A B C D
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
the elevation head, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

19. ? Find: u [kPa] at point B D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] V C A B C D
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
and the pressure head. [pause] Since we neglect velocity head, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

20. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
The total head equals the elevation head plus the pressure head.
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

21. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
We assume a vertical datum at point A to equal zero. So we assign point A to have ---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

22. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D B C D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
an elevation head of 0 meters.
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

23. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D B C D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Since the silt layer is 3 meters thick, ---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

24. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D B C D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
point B has an elevation head of 3 meters. ---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

25. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 C D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

26. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 C D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Similarly, since the sand layer is 2 meters thick,
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

27. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 C D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
point C has an elevation head of 5 meters. ---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

28. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5 D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

29. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5 D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Since the depth of the water is x meter deep, ---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

30. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5 D B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
The elevation head at point D equals, 5 plus x meters.
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

31. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

32. ? Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer
uB = ρw * g * hP,B D
Water x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Next, let’s look at the pressure head values, ---
3 [m]
Ksilt=1*10-6 [m/s]
hZ [m]=0 V A
Aquifer

33. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
and define atmospheric pressure, also known as gauge pressure, as zero.
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

34. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B C D
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
So the pressure head at point D is zero meters.
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

35. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B C
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

36. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B C
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Since the water is x meters deep,
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

37. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B C
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
The pressure head at point C is x meters.
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

38. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

39. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
We currently don’t know the pressure head at point B, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

40. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
so we’ll simply write in ----
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

41. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
h sub P, comma B, in the table, as a place holder. [pause]
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

42. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
We don’t know the pressure head at point A, either, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

43. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
however, we were given that the pore water pressure at point A is 24.5 kilopascals, ---
3 [m]
Ksilt=1*10-6 [m/s]
uA=24.5 [kPa] V A
Aquifer

44. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
and the pressure head can be computed by knowing this pore water pressure.
3 [m]
Ksilt=1*10-6 [m/s]
uA=24.5 [kPa] V
uA = ρw * g * hP,A A
Aquifer

45. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Plugging in our values for the pore water pressure, ---
3 [m]
Ksilt=1*10-6 [m/s]
uA=24.5 [kPa] V
uA = ρw * g * hP,A A
Aquifer

46. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
the density of water, ---
3 [m]
Ksilt=1*10-6 [m/s]
uA=24.5 [kPa] V
uA = ρw * g * hP,A A
Aquifer
1 [MG/m3]

47. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
and the gravitational acceleration, ---
3 [m]
Ksilt=1*10-6 [m/s]
uA=24.5 [kPa]
9.81 [m/s2] V
uA = ρw * g * hP,A A
Aquifer
1 [MG/m3]

48. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer hP [m]=0x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand
2.5 B C D 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
we compute a pressure head at point A to equal 2.5 meters. [pause]
3 [m]
Ksilt=1*10-6 [m/s]
uA=24.5 [kPa]
9.81 [m/s2] V
uA = ρw * g * hP,A A
Aquifer
1 [MG/m3]

49. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand B C D 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Remembering the total head equals the elevation head plus the pressure head, we can compute the total head values points, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

50. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand
2.5 B C D 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt A,
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

51. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand
2.5
3+hP,B C D 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt B,
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

52. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand
2.5
3+hP,B
5+x D 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt C,
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

53. Find: u [kPa] at point B = + D C B A Water Sand Silt Aquifer x
hT[m] hZ[m] hP[m] = + V C A B C D
Sand
2.5
3+hP,B
5+x 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
and D. [pause] Next we’ll set find the relationship between ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

54. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x
hT[m] hZ[m] hP[m] V C A B C D
Sand
2.5
3+hP,B
5+x 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
the total head at point B, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

55. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer x
hT[m] hZ[m] hP[m] V C A B C D
Sand
2.5
3+hP,B
5+x 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
and the depth of water, x, ---
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

56. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSiltx
hT[m] hZ[m] hP[m] V C A B C D
Sand
2.5
3+hP,B
5+x 3 5
5+x
2.5
hP,B x
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
based on the principle that the flowrate through the sand layer is equivalent to the flowrate through the silt layer.
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

57. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSiltx V C
Sand
2 [m]
Ksand=1*10-5 [m/s] V B
Silt
Let’s refer to the sand properties as CB, meaning between points C and B, and the silt properties will be BA.
3 [m]
Ksilt=1*10-6 [m/s] V A
Aquifer

58. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x V C
Sand
2 [m]
KBC=1*10-5 [m/s] V B
Silt
So we have Q sub BC equals Q sub AB. Where the flowrate for each layer is, ---
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

59. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
KBC=1*10-5 [m/s] V B
Silt
the hydraulic conductivity times the gradient times the area.
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

60. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s] V B
Silt
Since the cross-sectional area will be the same for both layers,
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

61. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s] V B
Silt
they cancel out of the equation.
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

62. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB V B
Silt
Now, we’re left with, ---
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

63. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB V B
Silt
the hydraulic conductivities, ---
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

64. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB V B
Silt
both of which we’ve been given.
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

65. Find: u [kPa] at point B D C B A Water Sand Silt Aquifer Qsand = QSilt
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB V B
Silt
and the gradients.
3 [m]
KAB=1*10-6 [m/s] V A
Aquifer

66. Find: u [kPa] at point B D C B = KAB* A Water Sand Silt Aquifer
Qsand = QSilt D
Water
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB B V
hT,B-hT,C
hT,A-hT,B
Silt
The gradient is the total headloss divided by the length over which the headloss occurs. [pause]
3 [m]
KBC*
= KAB*
LBC
LAB
KAB=1*10-6 [m/s] V A
Aquifer

67. Find: u [kPa] at point B D C B = KAB* A Water Sand Silt Aquifer
Qsand = QSilt D
Water
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB B V
hT,B-hT,C
hT,A-hT,B
Silt
These lengths correspond to the layer thicknesses, ---
3 [m]
KBC*
= KAB*
LBC
LAB
KAB=1*10-6 [m/s] V A
Aquifer

68. Find: u [kPa] at point B D C B = KAB* A Water Sand Silt Aquifer
Qsand = QSilt D
Water
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB B V
hT,B-hT,C
hT,A-hT,B
Silt
which are given.
3 [m]
KBC*
= KAB*
LBC
LAB
KAB=1*10-6 [m/s] V A
Aquifer

69. Find: u [kPa] at point B D C B = KAB* A Water Sand Silt Aquifer
Qsand = QSilt D
Water
QBC = QAB x
KBC*iBC*ABC = KAB*iAB*AAB V C
Sand
2 [m]
ABC = AAB
KBC=1*10-5 [m/s]
KBC*iBC = KAB*iAB B V
hT,B-hT,C
hT,A-hT,B
Silt
We also notice the total head values, which still need to be determined.
3 [m]
KBC*
= KAB*
LBC
LAB
KAB=1*10-6 [m/s] V A
Aquifer

70. Find: u [kPa] at point B = KAB* Qsand = QSilt QBC = QAB
KBC*iBC*ABC = KAB*iAB*AAB
ABC = AAB
KBC*iBC = KAB*iAB
hT,B-hT,C
hT,A-hT,B
If we solve the equation ---
KBC*
= KAB*
LBC
LAB
KBC*LAB*(hT,B-hT,C)=KAB*LBC *(hT,A-hT,B)

71. Find: u [kPa] at point B = KAB* Qsand = QSilt QBC = QAB
KBC*iBC*ABC = KAB*iAB*AAB
ABC = AAB
KBC*iBC = KAB*iAB
hT,B-hT,C
hT,A-hT,B
for the total head at point B ----
KBC*
= KAB*
LBC
LAB
KBC*LAB*(hT,B-hT,C)=KAB*LBC *(hT,A-hT,B)
KBC*LAB*(hT,B-hT,C)=KAB*LBC *(hT,A-hT,B)

72. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
hT[m] A B C D
2.5
3+hP,B
5+x
and plug our values in, ---
KBC*LAB*(hT,B-hT,C)=KAB*LBC *(hT,A-hT,B)
KBC*LAB*(hT,B-hT,C)=KAB*LBC *(hT,A-hT,B)

73. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
hT[m] A B C D
2.5
3+hP,B
5+x
remembering we already solved for the total head at at points A and C, from the table, ---
KBC*LAB*(hT,B-hT,C)=KAB*LBC *(hT,A-hT,B)
KBC*LAB*(hT,B-hT,C)=KAB*LBC *(hT,A-hT,B)

74. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
1*10-5 [m/s] * 3 [m] * (5 [m] + x)
+ 1*10-6 [m/s] * 2 [m] * 2.5 [m]
hT,B=
1*10-5 [m/s] * 3 [m] + 1*10-6 [m/s] * 2 [m]
we find the total head at point B equals ---

75. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
1*10-5 [m/s] * 3 [m] * (5 [m] + x)
+ 1*10-6 [m/s] * 2 [m] * 2.5 [m]
hT,B=
1*10-5 [m/s] * 3 [m] + 1*10-6 [m/s] * 2 [m]
hT,B=4.84[m]+0.937*x
4.84 meters plus times the water depth x, in meters. [pause] In the problem statement, we were also given that ---

76. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
1*10-5 [m/s] * 3 [m] * (5 [m] + x)
+ 1*10-6 [m/s] * 2 [m] * 2.5 [m]
hT,B=
1*10-5 [m/s] * 3 [m] + 1*10-6 [m/s] * 2 [m]
hT,B=4.84[m]+0.937*x
the measured seepage velocity, v, downward through the soil is 0.2 meters cubed per day, per meter squared. Or more simply, ---
V=0.2 [m3/day*m2]

77. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
1*10-5 [m/s] * 3 [m] * (5 [m] + x)
+ 1*10-6 [m/s] * 2 [m] * 2.5 [m]
hT,B=
1*10-5 [m/s] * 3 [m] + 1*10-6 [m/s] * 2 [m]
hT,B=4.84[m]+0.937*x
0.2 meters per day. This seepage velocity is equal to the ---
V=0.2 [m3/day*m2]
=0.2 [m/day ]

78. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
1*10-5 [m/s] * 3 [m] * (5 [m] + x)
+ 1*10-6 [m/s] * 2 [m] * 2.5 [m]
hT,B=
1*10-5 [m/s] * 3 [m] + 1*10-6 [m/s] * 2 [m]
hT,B=4.84[m]+0.937*x
hydraulic conductivity times the gradient, in the direction of flow. If we evaluate this expression ---
V=K * (-i)
V=0.2 [m3/day*m2]
=0.2 [m/day ]

79. Find: u [kPa] at point B KBC*LAB*hT,C+KAB*LBC *hT,A hT,B=
KBC*LAB+KAB LBC
1*10-5 [m/s] * 3 [m] * (5 [m] + x)
+ 1*10-6 [m/s] * 2 [m] * 2.5 [m]
hT,B=
1*10-5 [m/s] * 3 [m] + 1*10-6 [m/s] * 2 [m]
hT,B=4.84[m]+0.937*x
for silt layer, BC, ---
V=K * (-i)
V=0.2 [m3/day*m2]
hT,C-hT,B
VBC=KBC *
=0.2 [m/day ]
LBC

80. Find: u [kPa] at point B hT,C-hT,B LBC VBC=KBC *
we can make a couple substitutions,
hT,C-hT,B
VBC=KBC *
LBC

81. Find: u [kPa] at point B hT,C-hT,B LBC VBC=KBC *
and solve for the depth of water, x, explicitly.

82. Find: u [kPa] at point B hT,C-hT,B LBC s VBC=KBC * 86,400 day
0.2 [m/day]=1*10-5 [m/s] *
(5[m]+x)-(4.84[m]+0.937*x)
2[m]
After converting the units and little algebra, ---

83. Find: u [kPa] at point B hT,C-hT,B LBC s VBC=KBC * 86,400 day
0.2 [m/day]=1*10-5 [m/s] *
(5[m]+x)-(4.84[m]+0.937*x)
2[m]
* 2[m] = (5[m]+x)-(4.84[m]+0.937*x)
we find that the depth of water, x, ----

84. Find: u [kPa] at point B hT,C-hT,B LBC x=4.81[m] s VBC=KBC * 86,400
day
0.2 [m/day]=1*10-5 [m/s] *
(5[m]+x)-(4.84[m]+0.937*x)
2[m]
* 2[m] = (5[m]+x)-(4.84[m]+0.937*x)
is 4.81 meters. [pause] Plugging this value into our previously derived ---
x=4.81[m]

85. Find: u [kPa] at point B hT,C-hT,B LBC x=4.81[m] hT,B=4.84[m]+0.937*xs
VBC=KBC *
86,400
LBC
day
0.2 [m/day]=1*10-5 [m/s] *
(5[m]+x)-(4.84[m]+0.937*x)
2[m]
* 2[m] = (5[m]+x)-(4.84[m]+0.937*x)
equation for the total head at B, we find the total head at point B to be ---
x=4.81[m]
hT,B=4.84[m]+0.937*x

86. Find: u [kPa] at point B hT,C-hT,B LBC x=4.81[m] hT,B=4.84[m]+0.937*xs
VBC=KBC *
86,400
LBC
day
0.2 [m/day]=1*10-5 [m/s] *
(5[m]+x)-(4.84[m]+0.937*x)
2[m]
* 2[m] = (5[m]+x)-(4.84[m]+0.937*x)
9.35 meters. [pause] If we return to our table of pressures ---
x=4.81[m]
hT,B=4.84[m]+0.937*x
hT,B=9.35 [m]

87. Find: u [kPa] at point B hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 3 53 5
5+x
2.5
hP,B x
we identified the total head at point B as equivalent to ---
hT,B=9.35 [m]

88. Find: u [kPa] at point B hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 3 53 5
5+x
2.5
hP,B x
hT,B = 3 + hP,B
3 meters plus the pressure head at point B. If we plug in our value for the total head, ---
hT,B=9.35 [m]

89. Find: u [kPa] at point B hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 3 53 5
5+x
2.5
hP,B x
hT,B = 3 + hP,B
we find the pressure head at point B equals, ---
hT,B=9.35 [m]

90. Find: u [kPa] at point B hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 3 53 5
5+x
2.5
hP,B x
hT,B = 3 + hP,B
6.35 meters. [pause]
hP,B = 6.35 [m]
hT,B=9.35 [m]

91. Find: u [kPa] at point B hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 3 53 5
5+x
2.5
hP,B x
uB = ρw * g * hP,B
hT,B = 3 + hP,B
To find the pore water pressure at point B, ---
hP,B = 6.35 [m]
hT,B=9.35 [m]

92. Find: u [kPa] at point B hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 3 53 5
5+x
2.5
hP,B x
uB = ρw * g * hP,B
hT,B = 3 + hP,B
we multiply the pressure head at point B ---
hP,B = 6.35 [m]
hT,B=9.35 [m]

93. Find: u [kPa] at point B 1 hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 33 5
5+x
2.5
hP,B x MG 1 m3 m
9.81 s2
uB = ρw * g * hP,B
hT,B = 3 + hP,B
by the density of water and gravitational acceleration to find ---
hP,B = 6.35 [m]
hT,B=9.35 [m]

94. Find: u [kPa] at point B 1 hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 33 5
5+x
2.5
hP,B x MG 1 m3 m
9.81 s2
uB = ρw * g * hP,B
hT,B = 3 + hP,B
The pore water pressure at point B to equal 62.3 kilopascals.
hP,B = 6.35 [m]
uB = 62.3 [kPa]
hT,B=9.35 [m]

95. Find: u [kPa] at point B 1 hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 33 5
5+x
2.5
hP,B x MG 1 m3 m
9.81 s2 15 63
245
447
uB = ρw * g * hP,B
hT,B = 3 + hP,B
When compared with the possible solutions,
hT,B = 6.35 [m]
uB = 62.3 [kPa]
hT,B=9.35 [m]

96. Find: u [kPa] at point B 1 hT[m] hZ[m] hP[m] A B C D 2.5 3+hP,B 5+x 33 5
5+x
2.5
hP,B x MG 1 m3 m
9.81 s2 15 63
245
447
uB = ρw * g * hP,B
hT,B = 3 + hP,B
the answer is B.
hT,B = 6.35 [m]
uB = 62.3 [kPa]
AnswerB
hT,B=9.35 [m]

97. ? 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