This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Steady Flow into a Well”.
1. Identify the Thiem’s equation for steady state discharge in a confined aquifer. K = permeability of aquifer, T = transmissibility, S = drawdown difference in observation wells and r1, r2 = radial distance of observation wells from main well.
a) \(\frac{πKS}{ln\frac{r_2}{r_1}}\)
b) \(\frac{2πKS}{ln\frac{r_2}{r_1}}\)
c) \(\frac{πTS}{ln\frac{r_2}{r_1}}\)
d) \(\frac{2πTS}{ln\frac{r_2}{r_1}}\)
View Answer
Explanation: The discharge is computed using the velocity of flow from Darcy’s law. This discharge is obtained by integrating the equation between the piezometric head and radial distance of two observation wells. The equilibrium equation thus obtained is Thiem’s equation and is given as,
Q=\(\frac{2πKB(h_2-h_1 )}{ln\frac{r_2}{r_1}} =\frac{2πTS}{ln\frac{r_2}{r_1}}\)
2. Which of the following assumptions should be satisfied for Thiem’s equation to be applicable for finding well discharge in a confined aquifer?
a) Well should completely penetrate the aquifer and flow should be steady
b) Well should completely penetrate the aquifer and flow should be unsteady
c) Well should not completely penetrate the aquifer and flow should be steady
d) Well should not completely penetrate the aquifer and flow should be unsteady
View Answer
Explanation: For the applicability of the equilibrium equation (Thiem’s equation) and any of its forms, the well should be completely penetrating the confined aquifer and also the state of flow should be steady.
3. A 45 cm diameter well is inserted into a confined aquifer of permeability 25 m/day and thickness 20 m. The steady state drawdown in the well is 4 m. If the radius of influence is 180 m, what is the discharge?
a) 940 m3/day
b) 1049 m3/day
c) 1880 m3/day
d) 2097 m3/day
View Answer
Explanation: Given r = 45/2 = 22.5 cm = 0.225 m, K = 25 m/day, B = 20 m, S = 4 m, R = 180 m.
From Dupit’s equation for confined aquifer,
Q=\(\frac{2πKBS}{ln \frac{R}{r}} = \frac{2π*25*20*4}{ln \frac{180}{0.225}} = \frac{4000π}{ln800}\)=1879.9 m3/day≅1880 m3/day
4. A pumping out field test was conducted in a 15 m thick confined aquifer with a main well and two observation wells. Water was pumped at a steady discharge of 1050 L/min and the piezometric heads at the two observation wells were noted as 37.5 m and 41 m. If the observations wells are at a distance of 12 m and 40 m respectively in the same direction, what is the permeability (in m/day) of the aquifer?
a) 3.83
b) 5.52
c) 8.81
d) 9.21
View Answer
Explanation: Given B = 15 m, Q = 1050 L/min = 1.05 m3/min, h1 = 37.5 m, h2 = 41 m, r1 = 12 m, r2 = 28 m.
From Thiem’s equation for steady pumping in confined aquifer,
K=\(\frac{Q}{2πB(h_2-h_1)} ln(\frac{r_2}{r_1})=\frac{1.05}{2π*15*(41-37.5)} ln(\frac{40}{12})\)=3.83*10-3 m/min =5.52 m/day
5. A well is used for long hours of pumping at 2000 L/min from a confined aquifer. The drawdown due to this pumping at 20 m and 60 m from the main well was noted to be 4.2 m and 2.7 m, respectively. What is the transmissibility of the aquifer?
a) 0.47 m2/min
b) 13.9 m2/hr
c) 281 m2/day
d) 677 m2/day
View Answer
Given: Q = 2000 L/min = 2 m3/min, r1 = 20 m, r2 = 60 m, s1 = 4.2 m, s2 = 2.7 m.
From Thiem’s equation for steady pumping in confined aquifer,
Q=\(\frac{2πT(s_1-s_2 )}{ln \frac{r_2}{r_1}} ⇒T=\frac{Q}{2π(s_1-s_2)} ln(\frac{r_1}{r_2})\)
⇒T=\(\frac{2}{2π*(4.2-2.7)} ln(\frac{60}{20})\)=0.233 m2/min=13.98 m2/hr=335.5 m2/day
6. The steady state discharge pumped from a confined aquifer using a 50 cm diameter well is 1750 L/min. If the well is replaced by another well of radius 60 cm, find the new discharge. Assume that all other parameters remain same for both wells and radius of influence as 400 m.
a) 1625 L/min
b) 1795 L/min
c) 1986 L/min
d) 2558 L/min
View Answer
Explanation: Let us assume that the transmissibility, drawdown and radius of influence remains same.
Then from Thiem’s equation Q∝\(\frac{1}{ln\frac{R}{r}}\)
Given R = 400 m, Q1 = 1750 L/min, r1 = 50/2 = 25 cm = 0.25 m, r2 = 60 cm = 0.6 m.
\(\frac{Q_1}{Q_2} =\frac{ln(\frac{R}{r_2})}{ln(\frac{R}{r_1})} =\frac{ln(\frac{400}{0.6})}{ln(\frac{400}{0.25})}\) =0.88134
∴\(Q_2=\frac{Q_1}{0.88134}=\frac{1750}{0.88134}\)≅1986 L/min
7. In a well penetrated in a confined aquifer, the steady state drawdown is 2.9 m for a pumping discharge of 112 m3/hr. If the pumping rate is increased by 50%, what is the change in drawdown of the well? Assume all other parameters remain same.
a) Increases by 50%
b) Decreases by 50%
c) Remains same
d) Doubles
View Answer
Explanation: Let us assume that the transmissibility and radius of influence remains same.
Then from Thiem’s equation Q∝s
Given Q1 = 112 m3/hr, s1 = 2.9 m, Q2 = 1.5 x 112 =168 m3/hr.
\(\frac{Q_2}{Q_1} = \frac{s_2}{s_1} \)
⇒s2=\(\frac{Q_2}{Q_1}*s_1=\frac{168}{112}*2.9\)=4.35 m
∴Change in drawdown=\(\frac{s_2-s_1}{s_1}*100=\frac{4.35-2.9}{2.9}*100\)=50% increase
8. A well is used for steady state pumping from a confined aquifer. It has a radius of influence 10 times its diameter. If the well is replaced by another well double its size, what will be the increase in discharge if all other factors remain same?
a) 21%
b) 30%
c) 43%
d) 77%
View Answer
Explanation: Let us assume that the transmissibility, drawdown and radius of influence remains same.
Let radius of first well be r. Then r2 = 2r, R = 10 x (2r) = 20r.
\(\frac{Q_1}{Q_2} =\frac{ln(\frac{R}{r_2})}{ln(\frac{R}{r_1})} =\frac{ln(\frac{20r}{2r})}{ln(\frac{20r}{r})} =\frac{ln10}{ln20} =0.7686\)
∴Increase in discharge=(\(\frac{Q_2-Q_1}{Q_1})*100=(\frac{Q_2}{Q_1} -1)*100=(\frac{1}{0.7686}-1)*100\)≅30%
9. All other factors remaining constant, if the steady state drawdown of a well in a confined aquifer is doubled, then the discharge also doubles.
a) True
b) False
View Answer
Explanation: From Dupit’s equation for confined aquifer,
Q=\(\frac{2πKBS}{ln\frac{R}{r}}\)⇒Q∝S
This implies that the discharge increases linearly with the drawdown, given other factors are constant.
10. A 25 cm diameter tube well fully penetrates a confined aquifer of thickness 30 m. A steady pumping at 36 L/s for a long time results in a drawdown of 3.2 m and 2.2 m at observation wells 15 m and 35 m, respectively from the main well. What is the steady state drawdown in the main well?
a) 5.65 m
b) 7.85 m
c) 8.85 m
d) 9.65 m
View Answer
Explanation: Given r = 12.5 cm, B = 30 m, Q = 0.036 m3/s, s1 = 3.2 m, s2 = 2.2 m, r1 = 15 m, r2 = 35 m.
From Thiem’s equation, considering the two observation wells,
K=\(\frac{Q}{2πB(s_1-s_2)} ln(\frac{r_2}{r_1})=\frac{0.036}{2π*30*(3.2-2.2)} ln(\frac{35}{15})\)=1.618*10-4 m3/s
Now considering the main well and first observation well,
K=\(\frac{Q}{2πB(S-s_1)}ln(\frac{r_1}{r})\)
⇒S-s1=\(\frac{Q}{2πKB} ln (\frac{r_1}{r})=\frac{0.036}{2π*1.618*10^(-4)*30}*ln(\frac{15}{0.125})\)=5.65
∴S=5.65+3.2=8.85 m
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.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
