Engineering Hydrology Questions and Answers – Compressibility of Aquifers – Set 2

This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Compressibility of Aquifers – Set 2”.

1. The storativity of an unconfined aquifer is approximately equal to which of the following?
a) Specific yield
b) Specific storage
c) Specific retention
d) Specific capacity
View Answer

Answer: a
Explanation: For an unconfined aquifer, the storage coefficient is given as the sum of the storativity of the confined aquifer and the specific yield of the aquifer. Since the specific yield term is significantly greater than the other term, the storativity is assumed to be equal to the specific yield.

2. Aquifers P and Q are unconfined and confined aquifers, respectively. Both are made of the same porous material and exist in similar surrounding conditions. If the saturated thickness of P and the thickness of Q are equal, what is the difference between the storativity of the two aquifers?
a) Specific storage of P
b) Specific storage of Q
c) Specific yield of P
d) Specific yield of Q
View Answer

Answer: c
Explanation: Given the information, it can be assumed that the porosity (n), compressibility (α), water compressibility (β) and depth (B) of both aquifers are same.
Storativity of P, SP=SyP+γ(nP βPP) BP, where SyP is the specific yield of aquifer P.
Storativity of Q, SQ=γ(nQ βQQ) BQ
∴Difference=SP-SQ=SyP+γ(nP βPP) BP-γ(nQ βQQ) BQ=SyP= specific yield of P.

3. The porosity of an unconfined aquifer is 44% and its specific retention is 13%. If the saturated depth of the aquifer is 6 m, what is the storage coefficient of the aquifer?
a) 0.13
b) 0.31
c) 0.44
d) 0.57
View Answer

Answer: b
Explanation: Since no data regarding the compressibility paraments is given, it is assumed that the storage coefficient is equal to the specific yield of the aquifer. Given n = 44% and Sr = 13%.
Specific yield, Sy=n-Sr=44-13=31%=0.31
Therefore, the storage coefficient is 0.31.
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4. Barometric efficiency of a confined aquifer is the ratio between which of the following?
a) Change in storativity to change in pressure head
b) Change in pressure head to change in water level
c) Change in pressure head to change in storativity
d) Change in water level to change in pressure head
View Answer

Answer: d
Explanation: Wells driven in confined aquifers show variation in water level in response to the changes in the atmospheric pressure. This relation is expressed in the form of barometric efficiency, which is the ratio of the change in water level to the change in pressure head.

5. The water level of a well drilled in a confined aquifer will drop if there is an increase in the atmospheric pressure.
a) True
b) False
View Answer

Answer: a
Explanation: As the atmospheric pressure increases, the loading of the aquifer increases. This implies that an increase in pressure is accompanied by a decrease in the well water level and vice-versa.

6. If n is the porosity, α is the pore compressibility and β is the water compressibility of an unconfined aquifer, what is the correct expression of barometric efficiency of the aquifer?
a) \(\frac{α}{α+nβ}\)
b) \(\frac{β}{α+nβ}\)
c) \(\frac{nα}{α+nβ}\)
d) \(\frac{nβ}{α+nβ}\)
View Answer

Answer: d
Explanation: The barometric efficiency of an aquifer, which is the ratio of the water level change to pressure change, can be expressed in terms of the compressibility parameters of the aquifer as,
Barometric efficiency=\(\frac{nβ}{α+nβ}\)

7. A 15 m thick aquifer made of dense sand has a porosity of 45% and an elastic bulk modulus of 6000 N/cm2. If the compressibility of water is 4.75×10-10 m2/N, what is the storage coefficient of the aquifer?
a) 0.0016
b) 0.0025
c) 0.0033
d) 0.0039
View Answer

Answer: b
Explanation: Given n = 0.45, Es = 6000 N/cm2 = 6×107 N/m2, β = 4.75×10-10 m2/N, B = 15 m.
⇒Compressibility of medium=α=\(\frac{1}{E_s} =\frac{1}{6*10^7}\)=1.667*10-8 m2/N
Storage coefficient= γ(nβ+α)B=9810*((0.45*4.75*10-10)+(1.667*10-8))*15
=9810*((2.1375*10-10)+(1.667*10-8))*15=9810*1.688375*10-8*15
=2.484*10-3≅0.0025
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8. Find the storativity of an unconfined aquifer with porosity 60%, specific yield 42%, elastic bulk modulus 250 N/cm2, water compressibility 4.8×10-6 cm2/N and saturated thickness 30 m. Take unit weight of water as 10 kN/m3.
a) 0.42
b) 0.47
c) 0.51
d) 0.54
View Answer

Answer: d
Explanation: Given n = 0.6, Es = 250 N/cm2 = 2.5×106 N/m2, β = 4.8×10-10 m2/N, B = 30 m, Sy = 0.42.
⇒Compressibility of medium=α=\(\frac{1}{E_s}=\frac{1}{2.5*10^6}\)=4*10-7 m2/N
Storativity=Sy+γ(nβ+α)B=0.42+[10000*((0.6*4.8*10-10)+(4*10-7))*30] =0.42+[10000*((2.88*10-10)+(4*10-7))*30] =0.42+0.1200864
=0.5400864≅0.54

9. The barometric efficiency (BE) of a rocky aquifer with a porosity of 38%, bulk modulus of elasticity 75000 N/cm2 and water compressibility 4.4×10-6 cm2/N, lies within which of the following ranges?
a) BE < 10%
b) 10% < BE < 15%
c) 15% < BE < 20%
d) BE > 20%
View Answer

Answer: b
Explanation: Given n = 0.38, Es = 75000 N/cm2, β = 4.4×10-6 m2/N.
⇒Compressibility of medium=α=\(\frac{1}{E_s} =\frac{1}{75000}\)=1.33*10-5 cm2/N
∴Barometric efficiency=\(\frac{nβ}{α+nβ}*100=\frac{0.38*4.4*10^{-6}}{(1.33*10^{-5})+(0.38*4.4*10^{-6})}*100\)
=\(\frac{0.38*4.4*10^{-6}}{1.4972*10^{-5}}*100\)=11.1675%≅11.2%
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10. The data of two confined aquifers A and B is given below.

A B
Porosity (%) 24 41
Thickness (m) 20 26
Pore compressibility (m2/N) 1.25×10-8 8×10-8
Water compressibility (m2/N) 4.68×10-10 4.42×10-10
Specific yield (%) 16 27

Which of the following statements with regards to the properties of A and B is correct? Take γ=10kN/m3.
a) Specific storage of A is greater than specific storage of B
b) Storativity of A if greater than storativity of B
c) Barometric efficiency of A is greater than barometric efficiency of B
d) Specific retention of A is greater than specific retention of B
View Answer

Answer: c
Explanation: Let Ss = specific storage, S = storativity, BE =barometric efficiency and Sr = specific retention.
For aquifer A, given nA = 0.24, αA = 1.25×10-8 m2/N, βA = 4.68×10-10 m2/N, BA = 20 m, SyA = 0.16.
SSA= γ(nA βAA)=10000*((0.24*4.68*10-10))+(1.25*10-8))=1.26*10-4 m-1
SA=γ(nA βAA) BA=1.26*10-4*20=2.52*10-3
BEA=\(\frac{n_A β_A}{α_A+n_A β_A}*100=\frac{0.24*4.68*10^{-10}}{1.25*10^{-8}+(0.24*4.68*10^{-10})}*100\)=0.89%
SrA=nA-SyA=24-16=8%
Now for aquifer B, given nB = 0.41, αB = 8×10-8 m2/N, βB = 4.42×10-10 m2/N, BB = 26 m, SyB = 0.27.
SSB = γ(nB βBB)=10000*((0.41*4.42*10-10)+(8*10-8))=8.02*10-4 m-1
SB=γ(nB βBB) BB=8.02*10-4*26=2.08*10-2
BEB=\(\frac{n_B β_B}{α_B+n_B β_B}*100=\frac{0.41*4.42*10^{-10}}{(8*10^{-8})+(0.41*4.42*10^{-10})}*100\)=0.23%
SrB=nB-SyB=41-27=14%
It can be observed from the above values that the specific storage, storativity and specific retention of aquifer B is more than that of aquifer A. However, the barometric efficiency of aquifer A is greater than that of aquifer B.

Sanfoundry Global Education & Learning Series – Engineering Hydrology.

To practice all areas of Engineering Hydrology, here is complete set of 1000+ Multiple Choice Questions and Answers.

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