# Engineering Hydrology Questions and Answers – Aquifer Properties – Set 2

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

1. What is the correct dimension of intrinsic permeability?
a) [L]
b) [L2]
c) [LT-1]
d) [L2T-1]

Explanation: Intrinsic permeability is directly proportional to the mean particle size of a porous medium and this has dimension of [L2]. It is generally represented by units like cm2 or m2.

2. A Darcy is a common unit for which of the following?
a) Seepage velocity
c) Permeability
d) Intrinsic permeability

Explanation: The common unit for intrinsic permeability alongside cm2 and m2 is Darcy. Since the values of intrinsic permeability obtained with m2 are very small, it is converted to Darcy to represent the quantity is a more workable range. 1 Darcy=9.87*10-13 m2.

3. What is the definition of transmissibility?
a) Discharge through an aquifer one unit width and unit thickness
b) Discharge through an aquifer of unit width under unit hydraulic gradient
c) Discharge through an aquifer of unit thickness under unit hydraulic gradient
d) Discharge through an aquifer of unit width and unit thickness under unit hydraulic gradient

Explanation: The discharge through an aquifer of unit width and dome thickness (B) under a unit hydraulic gradient is called as transmissibility (T). It is equal to the permeability (K) for unit thickness of aquifer. It is given as T = KB, and has units of m2/s or m3/day/m.

4. In an unconfined aquifer of area 1.6 km2, water table elevation is 98.6 m. After a natural recharge of 1.25 Mm3 the water table rose to 101.3 m. What is the specific retention of the aquifer?
a) 12%
b) 17%
c) 23%
d) 29%

Explanation: Change in volume of aquifer, V=(101.3-98.6)*1.6*106=4.32*106 m3
Volume of water retained, Vr=1.25*106 m3
∴Specific yield,yr=$$\frac{V_r}{V}=\frac{1.25*10^6}{4.32*10^6}$$=0.2894≅0.29 or 29%

5. The water table of an aquifer, of specific yield 0.27 and area 7 km2, dropped by 2.5 m after water pumping for few hours. How many million litres of water was pumped out from the aquifer?
a) 3850
b) 4725
c) 6481
d) 7560

Explanation: Given ys = 0.27, A = 7 x 106 m2
Volume of water yielded, Vy = specific yield x change in aquifer volume
∴Vy=ys*V=0.27*(2.5*7*106)=4.725*106 m3
=4.725*109 litres
=4725 million litres

6. A 6 hour localized rainfall of uniform intensity 3.2 mm/hr occurred over an unconfined aquifer of 12 ha area. What will be the rise in the elevation of water table of the aquifer, if its specific retention is 12%?
a) 16 cm
b) 19 cm
c) 23 cm
d) 26 cm

Explanation: Given yr = 0.12, A = 12 x 104 m2
Volume of water added, Vr=3.2*10-3 m/hr*6 hr*12*104 m2=2304 m3
Now, yr=$$\frac{Volume \, of \, water \, added}{Chnage \, in \, aquifer \, volume}$$
So change in aquifer volume, V=$$\frac{V_r}{y_r} =\frac{2304}{0.12}$$=19200 m3
∴Change in elevation of water table=$$\frac{19200}{12*10^4}$$=0.16 m=16 cm

7. Three equally spaced observation wells A, B and C are inserted in an unconfined aquifer, in that order. The head difference between the first and last well, which are 1.5 km apart, is 8 m. The permeability and porosity of the aquifer is 12 m/day and 30%, respectively. How many days will a tracer injected at well A take to be observed at well B?
a) 875
b) 1750
c) 3500
d) 7000

Explanation: Given n = 0.3, K = 12 m/day, H = 8 m, L = 1.5 km = 1500 m
Hydraulic gradient, i=$$\frac{H}{L}=\frac{8}{1500}$$=5.33*10-3
Discharge velocity, V=Ki=12 m/day *5.33*10-3=0.06396 m/day
Seepage velocity or velocity of tracer, Vs=$$\frac{V}{n}=\frac{0.06396}{0.3}$$=0.2132 m/day
∴Time taken from A to B=$$\frac{Distance}{Velocity}=\frac{\frac{L}{2}}{V_s}=\frac{1500}{2}*\frac{1}{0.2132}$$
=3517.8≅3500 days

8. Data regarding the water table position of an aquifer of 10 km2 area during a week is given below.

 Day Mon Tue Wed Thurs Fri Sat Sun Mon Depth of W.T. from ground (in m) 25 22.5 23 22 20.5 23.5 25.5 25

If the porosity and specific yield of the aquifer is 35% and 20% respectively, which of the following is incorrect? Assume the measurements are taken at the beginning of the day.
a) The aquifer yielded least amount of water on Tuesday
b) The aquifer yielded 2.75 Mm3 more water than it gained during the week.
c) The aquifer gained 1.5 Mm3 of water on Wednesday
d) The aquifer yielded a total of 9.5 Mm3 of water during Friday and Saturday

Explanation: Given n = 0.35, ys = 0.2, yr = 0.35 – 0.2 = 0.15, A = 10 x 106 m3
Monday – change from 25 to 22.5 m = W.T. rises by 2.5 m
Tuesday – change from 22.5 to 23 m = W.T. drops by 0.5 m
⇒Water lost=ys*0.5*A=0.2*0.5*10*106=1 Mm3
Wednesday – change from 23 to 22 m = W.T. rises by 1 m
Thursday – change from 22 to 20.5 m = W.T. rises by 1.5 m
Friday – change from 20.5 to 23.5 m = W.T. drops by 3 m
⇒Water lost=ys*3*A=0.2*3*10*106=6 Mm3
Saturday – change from 23.5 to 25.5 m = W.T. drops by 2 m
⇒Water lost=ys*2*A=0.2*2*10*106=4 Mm3
Sunday – change from 25.5 to 25 m = W.T. rises by 0.5 m
Total amount of water gained/retained during the week = 3.75 + 1.5 + 2.25 + 0.75 = 8.25 Mm3
Total amount of water lost/yielded during the week = 1 + 6 + 4 = 11 Mm3
So, the aquifer yielded (11 – 8.25 =) 2.75 Mm3 water more than it gained during the week.

9. After a natural recharge of 2 Mm3 in an aquifer of area 5 km2, the water table rose by 2.2 m. After pumping, it was observed that the water table dropped by 2.6 m. What is the volume of water (in Mm3) that was pumped after the natural recharge?
a) 1.22
b) 1.69
c) 2
d) 2.37

Explanation: Given A = 5 x 106 m2
Specific retention, y=$$\frac{2*10^6}{2.2*5*10^6}$$=0.182
Since the water table dropped by 2.6 m after pumping the volume of water pumped is,
V=y*2.6*A=0.182*2.6*5*106=2.366*106 m3≅2.37 Mm3

10. Two observation wells 50 m apart are inserted in an aquifer of porosity 40%. The water level difference in the wells is 65 cm. The time of travel of a tracer between the wells is 8 hours. What is the intrinsic permeability of the aquifer? Take kinematic viscosity of water as 0.01 cm2/s.
a) 4307
b) 5545
c) 7172
d) 9323

Explanation: Given n = 0.4, L = 50 m, H = 65 cm = 0.65 m, t = 8 hours, v = 0.01 cm/s
Seepage velocity, Vs=$$\frac{L}{t}=\frac{50 m}{8 hr}=\frac{50*10^2 cm}{8*3600 sec}$$=0.1736 cm/sec
Discharge velocity, V=n*Vs=0.4*0.1736=0.06944 cm/sec
Hydraulic gradient, i=$$\frac{H}{L}=\frac{0.65}{50}$$=0.013
Permeability, K=$$\frac{V}{i}=\frac{0.06944}{0.013}$$=5.34 cm/sec
∴Intrinsic permeability,K0=$$\frac{Kv}{g}=\frac{5.34*0.01}{981}$$=5.44*10-5 cm2
Since 1 darcy = 9.87 x 10-9 cm2, K0=$$\frac{5.44*10^{-5} cm^2}{9.87*10^{-9} cm^2/darcy}$$≅5545 darcys

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