# Engineering Hydrology Questions and Answers – Steady Flow into a Well – Set 2

This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Steady Flow into a Well – Set 2”.

1. Identify the Dupit’s equation for steady state discharge in an unconfined aquifer of permeability K and saturated thickness H. Let h be the depth of water in pumping well of radius r and R be the radius of influence.
a) $$\frac{πKH(H-h)}{ln⁡ \frac{R}{r}}$$
b) $$\frac{πK(H^2-h^2)}{ln⁡\frac{R}{r}}$$
c) $$\frac{πKH(H-h)}{ln⁡ \frac{R^2}{r^2}}$$
d) $$\frac{πK(H^2-h^2)}{ln⁡ \frac{R^2}{r^2}}$$

Explanation: The discharge is computed using the velocity of flow from Darcy’s law. This discharge is obtained by integrating the equation between the well and the edge of the radius of influence and their respective water table depths. The equilibrium equation thus obtained is the Dupit’s equation given as,
Q=$$\frac{πK(H^2-h^2)}{ln⁡\frac{R}{r}}$$.

2. The water table profile of an unconfined aquifer during steady state well pumping as per Dupit’s assumptions, will be least accurate at which location(s)?
a) Near the well
b) Near the edge of zone of influence
c) Near the well and near the edge of zone of influence
d) All points other than near the well and near the edge of zone of influence

Explanation: The steady state discharge for unconfined flow are obtained using Dupit’s assumptions. Due to this there is an inaccuracy and deviation of the actual water table profile from the true profile. This variation is most prominent near the circumference of the well and decreases as me move radially outward to the edge of the zone of influence.

3. A 64 cm diameter well is inserted into an unconfined aquifer of permeability 1 m/hr and saturated thickness 33 m. The steady state depth in the well is 27 m. If the radius of influence is 320 m, what is the discharge?
a) 82 m3/hr
b) 91 m3/hr
c) 164 m3/hr
d) 182 m3/hr

Explanation: Given r = 64/2 = 32 cm = 0.32 m, K = 1 m/hr, H = 33 m, h = 27 m, R = 320 m.
From Dupit’s equation for unconfined aquifer,
$$Q=\frac{πK(H^2-h^2)}{ln⁡ \frac{R}{r}} = \frac{π*1*(33^2-27^2)}{ln⁡ \frac{320}{0.32}} = \frac{360π}{ln⁡1000}$$ = 163.725 m3/hr≅164 m3/hr

4. A pumping out field test was conducted in a 50 m deep unconfined aquifer with a main well and two observation wells. Water was pumped at a steady discharge of 2200 L/min and the water table heights at the two observation wells were noted as 33.5 m and 37 m. If the observations wells are at a distance of 50 m and 100 m respectively in the same direction, what is the permeability (in m/day) of the aquifer?
a) 1.97
b) 2.83
c) 3.39
d) 6.77

Explanation: Given H = 50 m, Q = 2200 L/min = 2.2 m3/min, h1 = 33.5 m, h2 = 37 m, r1 = 50 m, r2 = 100 m.
From Thiem’s equation for steady pumping in an unconfined aquifer,
K=$$\frac{Q}{π(h_2^2-h_1^2)} ln⁡ (\frac{r_2}{r_1})=\frac{2.2}{π*(37^2-33.5^2)} ln⁡(\frac{100}{50})$$=1.967*10-3 m/min
=2.83 m/day

5. A well is used for long hours of pumping at 160 m3/hr from an unconfined aquifer of saturated depth 28 m. The drawdown due to this pumping at 40 m and 90 m from the main well was noted to be 6.2 m and 5.4 m, respectively. What is the transmissibility of the aquifer?
a) 0.54 m2/min
b) 1.16 m2/min
c) 69.6 m2/hr
d) 499 m2/day

Given: Q = 160 m3/hr, r1 = 40 m, r2 = 90 m, h1 = 28 – 6.2 = 21.8 m, h2 = 28 – 5.4 = 22.6 m.
From Thiem’s equation for steady pumping in unconfined aquifer,
K=$$\frac{Q}{π(h_2^2-h_1^2)} ln⁡ (\frac{r_2}{r_1})=\frac{160}{π*(22.6^2-21.8^2)} ln⁡(\frac{90}{40})$$=1.163 m/hr
⇒T=KH=1.163*28=32.6 m2/hr =782 m2/day=0.54 m2/min

6. The steady state discharge pumped from an unconfined aquifer using a 20 cm diameter well is 975 L/min. If the well is replaced by another well of radius 25 cm, find the new discharge. Assume that all other parameters remain same for both wells and radius of influence as 500 m.
a) 857 L/min
b) 992 L/min
c) 1093 L/min
d) 1115 L/min

Explanation: Let us assume that the permeability, water table profile and radius of influence remains same.
Then from Thiem’s equation $$Q∝\frac{1}{ln⁡ \frac{R}{r}}$$
Given R = 500 m, Q1 = 975 L/min, r1 = 20/2 = 10 cm = 0.1 m, r2 = 25 cm = 0.25 m.
$$\frac{Q_1}{Q_2} = \frac{ln⁡(\frac{R}{r_2}}{ln⁡(\frac{R}{r_1})} = \frac{ln⁡(\frac{500}{0.25})}{ln⁡(\frac{500}{0.1})}$$ =0.8924
∴Q2=$$\frac{Q_1}{0.8924}=\frac{975}{0.8924}$$≅1093 L/min

7. In a well penetrated in an unconfined aquifer of saturated depth 25 m, the steady state drawdown in the well is 3.5 m for a pumping discharge of 88 m3/hr. If the pumping rate is increased two fold, what is the new drawdown in the well? Assume all other parameters remain same.
a) 6.8 m
b) 7.3 m
c) 7.7 m
d) 8.2 m

Explanation: Let us assume that the permeability and radius of influence remains same.
Then from Thiem’s equation Q∝(H2-h2)
Given Q1 = 88 m3/hr, H = 25 m, h1 = 25 – 3.5 = 21.5 m, Q2 = 2 x 88 = 176 m3/hr.
$$\frac{Q_1}{Q_2} = \frac{H^2-h_1^2}{H^2-h_2^2}=\frac{25^2-21.5^2}{25^2-h_2^2}$$
⇒$$\frac{88}{176}=\frac{162.75}{625-h_2^2}$$
$$⇒625-h_2^2=\frac{162.75*176}{88}=325.5 ⇒h_2=\sqrt{625-325.5}=17.3 m$$
∴New drawdown=25-17.3=7.7 m

8. A well is completely inserted in an unconfined aquifer of saturated depth 38 m. After pumping, the steady state drawdown in the well is 7.3 m. If the steady state drawdown is to be doubled, what is the required increase in pumping rate? Assume all other parameters remain same.
a) 13%
b) 38%
c) 56%
d) 79%

Explanation: : Let us assume that the permeability and radius of influence remains same.
Given Q1 = 1600 m3/hr, H = 38 m, h1 = 38 – 7.3 = 30.7 m, h2 = 38 – (2 x 7.3) = 23.4 m.
Now,
$$\frac{Q_1}{Q_2} = \frac{H^2-h_1^2}{H^2-h_2^2}=\frac{38^2-30.7^2}{38^2-23.4^2}$$=0.5595
∴Increase in discharge=$$(\frac{Q_2-Q_1}{Q_1})*100=(\frac{Q_2}{Q_1} – 1)*100=(\frac{1}{0.5595}-1)*100$$≅79%

9. A 1 foot diameter tube well fully penetrates an unconfined aquifer. A steady pumping at 3000 L/min for a long time results in a drawdown of 4.2 m and 3.2 m at observation wells 48 m and 88 m, respectively from the main well. The bedrock lies at a depth of 60 m from the ground level and the water table lies 15 m below the ground level. What is the steady state drawdown in the main well?
a) 13.9 m
b) 14.8 m
c) 15.4 m
d) 16.8 m

Explanation: Given r = 0.5 ft = 0.15 m , H = 60 – 15 = 45 m, Q = 3 m3/min, h1 = 45 – 4.2 = 40.8 m, r1 = 48m, h2 = 45 – 3.2 = 41.8 m, r2 = 88 m.
From Thiem’s equation, considering the two observation wells,
K=$$\frac{Q}{π(h_2^2-h_1^2)} ln⁡(\frac{r_2}{r_1})=\frac{3}{π*(41.8^2-40.9^2)} ln⁡(\frac{88}{48})$$=7.0075*10-3m/min
Now considering the main well and first observation well,
K=$$\frac{Q}{π(h_1^2-h^2)}{ln⁡(\frac{r_1}{r})}$$
⇒$$h_1^2-h^2=\frac{Q}{πK} ln⁡(\frac{r_1}{r})=\frac{3}{π*7.0075*10^{-3})}*ln⁡(\frac{48}{0.15})$$=786.064
⇒h=$$\sqrt{40.8^2-786.064}$$=29.64 m
∴Drawdown in well=45-29.64=15.36 m≅15.4 m

10. The initial water table of an unconfined aquifer lies 6 m below the ground level. A discharge of 0.04 m3/s is pumped from the aquifer using a well of diameter 70 cm to achieve a steady state drawdown of 9 m in the well. What is the depth of the water table (from ground level) at a distance of 100 m from the well at steady state? Take the saturated thickness of aquifer as 36 m and transmissibility as 0.008 m2/s.
a) 3.5 m
b) 5.5 m
c) 7.5 m
d) 9.5 m

Explanation: Given Q = 0.04 m3/s, r = 70/2 = 35 cm = 0.35 m, H = 36 m, h = 36 – 9 = 27 m, r1 = 100 m, T = 0.004 m2/s.
So, K=$$\frac{T}{H}=\frac{0.004}{36}$$=2.22*10-4 m/s
Q=$$\frac{πK(h_1^2-h^2)}{ln \frac{r_1}{r}} ⇒h_1^2-h^2=\frac{Q}{πK}ln⁡(\frac{r_1}{r})$$
⇒$$h_1^2-h^2=\frac{0.04}{π*2.22*10^{-4}} ln⁡(\frac{100}{0.35})$$=324.33
⇒h1=$$\sqrt{324.33+27^2}$$=32.5 m
∴Depth of WT from ground=3.5+6=9.5 m

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