J
Josh the Toad
In response to owillis28,
In response to your questions:
1. You need to think about this problem backwards in order to understand it. Basically, you are looking for the maximum depth of the landfill containment cell based on the water table elevation. The water table elevation is obtained from the data given. (FYI: The Land Surface Elevation information is a distractor !!!).
2. The parameter 'i' is called the hydraulic gradient. It represents the average slope of the water table across a specified distance. If you are familar with transportation problems (horizontal/vertical curve) - think of it as the rate of change over the linear length of water surface elevation.
So, distance between Monitoring Well A & B = SQRT(1500^2 + 300^2) = 1530 ft
Therefore, i = (229.75 ft - 222.25 ft)/1530 ft = 0.0049 ft/ft (unitless)
3. You are correct, Point C = Southeast Corner of Cell that corresponds with the point where the elevation of the groundwater table is at a maximum.
4. So, to complete the solution,
Distance between MW A, Point C = SQRT(100^2 + 500^2) = 510 ft
Groundwater Table elevation at Point C = GW Elevation MW A - slope (think vertical curve - sorta)
GW Table Elevation Point C = 229.75 ft - (0.0049 ft/ft)*510 ft = 227.25 ft
However, the problem statement indicated that the bottom of the containment cell was at least 5 ft above the grounwater level, so
Bottom of Containment Cell = 227.25 ft + 5 ft = 232.25 ft ---> Answer = B
My written solution is attached as well. Please feel free to ask any questions as a follow-up.
Toady.
In response to your questions:
1. You need to think about this problem backwards in order to understand it. Basically, you are looking for the maximum depth of the landfill containment cell based on the water table elevation. The water table elevation is obtained from the data given. (FYI: The Land Surface Elevation information is a distractor !!!).
2. The parameter 'i' is called the hydraulic gradient. It represents the average slope of the water table across a specified distance. If you are familar with transportation problems (horizontal/vertical curve) - think of it as the rate of change over the linear length of water surface elevation.
So, distance between Monitoring Well A & B = SQRT(1500^2 + 300^2) = 1530 ft
Therefore, i = (229.75 ft - 222.25 ft)/1530 ft = 0.0049 ft/ft (unitless)
3. You are correct, Point C = Southeast Corner of Cell that corresponds with the point where the elevation of the groundwater table is at a maximum.
4. So, to complete the solution,
Distance between MW A, Point C = SQRT(100^2 + 500^2) = 510 ft
Groundwater Table elevation at Point C = GW Elevation MW A - slope (think vertical curve - sorta)
GW Table Elevation Point C = 229.75 ft - (0.0049 ft/ft)*510 ft = 227.25 ft
However, the problem statement indicated that the bottom of the containment cell was at least 5 ft above the grounwater level, so
Bottom of Containment Cell = 227.25 ft + 5 ft = 232.25 ft ---> Answer = B
My written solution is attached as well. Please feel free to ask any questions as a follow-up.
Toady.
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