# Engineering Hydrology Questions and Answers – Unsteady Flow in a Confined Aquifer

This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Unsteady Flow in a Confined Aquifer”.

1. When water is pumped from a well in a confined aquifer, which of the following is true until equilibrium is attained?
a) Steady flow, increasing cone of depression
b) Steady flow, decreasing cone of depression
c) Unsteady flow, increasing cone of depression
d) Unsteady flow, decreasing cone of depression

Explanation: When water is pumped from a well, the piezometric surface of a confined aquifer forms a cone of depression which gradually increases until equilibrium is attained. The flow during this stage is unsteady.

2. Which of the following is the correct integral form of the well function in terms of u=$$\frac{r^2 S}{4Tt}$$, where r = distance from well, S = storativity, T = transmissibility and t = time from pumping?
a) $$\int_{u}^{\infty} u e^u \, du$$
b) $$\int_{u}^{\infty} u e^{-u} \, du$$
c) $$\int_{u}^{\infty} \frac{e^u}{u} \, du$$
d) $$\int_{u}^{\infty} \frac{e^{-u}}{u} \, du$$

Explanation: For unsteady flow in a confined aquifer, an important term in estimating the drawdown at a point is the well function. It is an integral function in terms of a parameter u. It is given as,
W(u)=$$\int_{u}^{\infty} \frac{e^{-u}}{u} \, du$$.

3. The drawdown at point for a pumping well in a confined aquifer is inversely proportional to the well function.
a) True
b) False

Explanation: The drawdown at a point due to pumping from a well in a confined aquifer is given as,
s=$$\frac{Q.W(u)}{4πT}$$, where Q = discharge rate, T = transmissibility, W(u) = well function. ⇒s∝W(u).

4. For values of parameter u less than 0.01, what will be the approximate expression for the well function?
a) 0.577-ln⁡u
b) -0.577-ln⁡u
c) -0.577-ln⁡u+u
d) -0.577-ln⁡u-u

Explanation: The well function in the form of a series is given as,
W(u)=$$-0.577216-ln⁡u+u-\frac{u^2}{2.2!}+\frac{u^3}{3.3!}-…$$
For small values of u ≤ 0.01, only the first two terms of the series can be considered.

5. What is the correct equation for drawdown s at a distance r from a well pumping at rate Q in a confined aquifer of transmissibility T and storativity S? Assume the parameter u=$$\frac{r^2 S}{4Tt}$$ to be very small. Take t as the time from start of pumping.
a) $$s=\frac{Q}{4πT} ln⁡(\frac{2.25Tt}{r^2 S})$$
b) $$s=\frac{Q}{4πT} ln⁡(\frac{7.12Tt}{r^2 S})$$
c) $$s=\frac{Q}{4πT} ln⁡(\frac{0.14r^2 S}{Tt})$$
d) $$s=\frac{Q}{4πT} ln⁡(\frac{0.45r^2 S}{Tt})$$

Explanation: The drawdown is given as,
s=$$\frac{Q.W(u)}{4πT}$$, where W(u)=-0.577216-ln⁡u for very small values of u
⇒ s=$$\frac{Q}{4πT}$$ [-0.577216-ln⁡u ]
⇒ s =$$\frac{Q}{4\pi T} \left[\ln(e^{-0.577216}) – \ln\left(\frac{r^2 S}{4Tt}\right)\right] ⇒ s = \frac{Q}{4\pi T} \left[\ln(0.5615) – \ln\left(\frac{r^2 S}{4Tt}\right)\right]$$
⇒ $$s = \frac{Q}{4\pi T} \ln\left(\frac{0.5615}{\frac{r^2 S}{4Tt}}\right)s = \frac{Q}{4\pi T} \ln\left(\frac{2.25Tt}{r^2 S}\right)$$

6. A well penetrating a confined aquifer is pumping at a rate of 1650 L/min. The drawdown at a point some distance away from the well is observed to be 2.2 m and 2.8 m after 1 hour and 2 hours of pumping, respectively. What is the transmissibility of the aquifer?
a) 2.53 x 10-4 m2/s
b) 5.06 x 10-3 m2/s
c) 0.32 m2/min
d) 9.1 m2/hr

Explanation: Given Q = 1650 L/min = 0.0275 m3/s, s1 = 2.2 m, s2 = 2.8 m, t1 = 60 mins, t2 = 120 mins.
Since no data is given let us assume that u is very small. So the difference in drawdown at a point is,
$$s_2 – s_1 = \frac{Q}{4\pi T} \ln\left(\frac{t_2}{t_1}\right)$$
(2.8 – 2.2) = $$\frac{0.0275}{4\pi T} \ln\left(\frac{120}{60}\right) ⇒T = \frac{0.0275}{4\pi \times 0.6} \ln(2)$$

∴T=2.528*10-3 m2/s=0.15 m2/min=9.1 m2/hr

7. A well is penetrated in a 15 m thick confined aquifer of permeability 14 m/day and storativity 0.002. If the pumping rate is 108 m3/hr, find the drawdown at 25 m from the well after 5 hours of pumping?
a) 2.28 m
b) 3.28 m
c) 4.28 m
d) 5.28 m

Explanation: Given B = 15 m, K = 14 m/day, S = 0.002, Q = 108 m3/hr = 0.03 m3/s, r = 25 m, t = 5 hrs.
Transmissibility, $$T = \frac{KB}{24 \times 60 \times 60} \times 15 = 2.43 \times 10^{-3} \, \text{m}^2/\text{s}$$
$$u = \frac{r^2 S}{4Tt} = \frac{25^2 \times 0.002}{4 \times (2.43 \times 10^{-3}) \times (5 \times 60 \times 60)} = 0.00714$$ < 0.01
W(u)=-0.5772-ln⁡u=-0.5772-ln⁡(0.00714)=4.365
$$s_{25} = \frac{Q \cdot W(u)}{4\pi T} = \frac{0.03 \times 4.365}{4\pi \times 2.43 \times 10^{-3}} = 4.28 \, \text{m}$$

8. A fully penetrating well in a confined aquifer of storativity 0.0007 and transmissibility of 20 m2/hr, is pumped at 72 m3/hr. What is the drawdown at a distance of 120 m away from the well after 10 hours of pumping?
a) 90 cm
b) 115 cm
c) 125 cm
d) 140 cm

Explanation: Given S = 0.0007, T = 20 m2/h = 5.56 x 10-3 m2/s, Q = 72 m3/h = 0.02 m3/s, r = 120 m, t = 10 h
u=$$\frac{r^2 S}{4Tt}=\frac{120^2*0.0007}{4*(5.56*10^{-3})*(10*60*60)}$$=0.004<0.01
⇒W(u)=-0.5772-ln⁡u=4.944
∴s120=$$\frac{Q.W(u)}{4πT}=\frac{0.02*4.944}{4π*5.56*10^{-3}}$$≅1.4 m=140 cm

9. What is the slope of the semi-log graph between residual drawdown in a pumping well (in a confined aquifer) and the ratio t/t’? t = time since start of pumping and t’ = time since stop of pumping. Take Q as pumping rate and T as transmissibility.
a) $$\frac{0.183Q}{T}$$
b) $$\frac{0.08Q}{T}$$
c) $$\frac{0.733Q}{T}$$
d) $$\frac{0.575Q}{T}$$

Explanation: The approximate expression for residual drawdown in a well is given as,
$$s’ = \frac{Q}{4\pi T} \ln\left(\frac{t}{t’}\right) = \frac{Q}{4\pi T} \times 2.303 \log\left(\frac{t}{t’}\right) = \frac{2.303Q}{4\pi T} \log\left(\frac{t}{t’}\right)$$
⇒Slope=$$\frac{2.303Q}{4πT}=\frac{0.183Q}{T}$$

10. A well in a confined aquifer was pumped at 1250 m3/day for 4 hours. Find the residual drawdown at 75 m from the well after 6 hours from start of pumping. Take transmissibility as 500 m2/day and storage coefficient as 1 x 10-4.
a) 11.3 cm
b) 21.9 cm
c) 29.5 cm
d) 34.2 cm

Explanation: Given Q = 1250 m3/day, t = 6 hr, t’ = 6 – 4 = 2 hr, T = 500 m2/day = 5.78 x 10-3 m2/s, S = 10-4
u=$$\frac{r^2 S}{4Tt}=\frac{75^2*0.0001}{4*(5.78*10^{-3})*(6*60*60)}$$=0.001126<0.01
W(u)=-0.5772-ln⁡u=6.21
u’=$$\frac{r^2 S}{4Tt’}=\frac{75^2*0.0001}{4*(5.78*10^{-3})*(2*60*60)}$$=0.00338<0.01
W(u’)=-0.5772-ln⁡ u’ =5.11
The residual drawdown at 75 m is,
$$s_{75}’ = \frac{Q}{4\pi T} [W(u) – W(u’)] = \frac{\frac{1250}{24 \times 60 \times 60}}{4\pi \times 5.78 \times 10^{-3}} [6.21 – 5.11]$$=0.219 m=21.9 cm

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