Hydraulic Conductivity
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A sample of drill core collected from an unconfined sandstone aquifer was tested using a constant head permeameter with a radius of 3cm at 20°C and provided the following data:
Flow rate = 0.18 cm3/min
Distance between manometers = 5 cm
Difference in head between manometers = 2.8 cm
A well in the same aquifer was pumped to a steady state, and the following information measured in a series of observation wells: (refer to attachments)
- Calculate the hydraulic conductivity of the rock sample.
- Use the data from the well pumping test to calculate the hydraulic conductivity of the aquifer as a whole.
- Suggest possible reasons for the differences in the results of your two calculations, and discuss the relative advantages and disadvantages of field and laboratory permeability tests.
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Solution Summary
The distances between manometers for hydraulic conductivity is examined.
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Calculate the hydraulic conductivity of the rock sample.
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(1) Q=KiA
Q= flow rate, K= hydraulic conductivity, i= hydraulic gradient = change in hydraulic head/change in distance, A = cross sectional areas
(2) A= r^2pi
Thus,
A = (3 cm)^2 x pi = 28.27 cm^2
rearranging equation (1), isolate for K, hydraulic conductivity
(3) K = Q/iA
K = 0.18 cm3/min / (2.8cm/5cm)*28.27 cm^2
K = 0.0114 cm/min
Use the data from the well pumping test to calculate the hydraulic conductivity of the ...
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