GATE CE · Environmental Engineering

Source Of Water Supply Distribution System And Well Hydraulics – GATE CE Environmental EngineeringPractice Questions & PYQs

Generate GATE-level questions on Source Of Water Supply Distribution System And Well Hydraulics in Environmental Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

17 questions · 17 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025NAT

A 60 cm diameter well completely penetrates a confined aquifer of permeability $5 \times 10^{-4} , \text{m/s}$ . The length of the strainer (spanning the entire thickness of the aquifer) is 10 m. The drawdown at the well under steady state pumping is 1.0 m. Assume that the radius of influence for this pumping is 300 m. The discharge from the well (in litres per minute) is ______ (round off to the nearest integer).

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📖 Explanation

Use Thiem's equation for a confined aquifer: $Q = \frac{2 \pi k b (H - h)}{\ln(R / r)}$ Where: $k = 5 \times 10^{-4} , \text{m/s}$ (permeability) $b = 10 , \text{m}$ (thickness of aquifer) $H - h = 1.0 , \text{m}$ (drawdown) $R = 300 , \text{m}$ (radius of influence) $r = \frac{60}{2 \times 100} = 0.30 , \text{m}$ (radius of well) Substituting the values: $Q = \frac{2 \pi (5 \times 10^{-4}) (10)(1)}{\ln(300 / 0.30)} = \frac{2 \pi (5 \times 10^{-3})}{\ln(1000)}$ $Q = \frac{0.0314}{6.9078} \approx 0.00455 , \text{m}^3/\text{s}$ Convert to litres per minute: $Q = 0.00455 \times 1000 \times 60 = 273 , \text{litres/minute}$

Q2GATE 2024NAT

A $10 \mathrm{~m}$ diameter well fully penetrates an unconfined aquifer of saturated thickness 20 $\mathrm{m}$ with hydraulic conductivity of $10 \mathrm{~m}$ day. Under the steady pumping rate for a long time, the drawdowns in two observation wells located at $10 \mathrm{~m}$ and $100 \mathrm{~m}$ from the pumping well are $5 \mathrm{~m}$ and $1 \mathrm{~m}$ . respectively. The corresponding pumping rate (in $\mathrm{m}^{3}$ day) from the well is (rounded off to 2 decimal places).

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📖 Explanation

Thickness of aquifer, $H=20 \mathrm{~m}$

\begin{aligned} & r_{1}=10 \mathrm{~m} \\ & r_{2}=100 \mathrm{~m} \\ & h_{2}=H-S_{w 1}=20-5=15 \mathrm{~m} \\ & h_{1}=H-S_{w 2}=20-1=19 \mathrm{~m} \end{aligned}

Now discharge for unconfined aquifer is given as

\begin{aligned} & Q=\frac{\pi \cdot K\left[h_{1}^{2}-h_{2}^{2}\right]\$}{\ln \left[\frac{r_{2}}{r_{1}}\right]}=\frac{\pi \times 10 \times \left[19^{2}-15^{2}\right]\$}{\ln \left[\frac{100}{10}\right]} \\ & Q=1855.55 \mathrm{~m}^{3} / \text { day } \end{aligned}

Q3GATE 2022NAT

The concentration $s(x,t)$ of pollutants in a one-dimensional reservoir at position $x$ and time $t$ satisfies the diffusion equation $\frac{\partial s(x,t)}{\partial t}=D\frac{\partial^2 s(x,t)}{\partial x^2}$ on the domain $0 \leq x \leq L$ , where $D$ is the diffusion coefficient of the pollutants. The initial condition $s(x, 0)$ is defined by the step-function shown in the figure. The boundary conditions of the problem are given by $\frac{\partial s(x,t)}{\partial x}=0$ at the boundary points $x = 0$ and $x = L$ at all times. Consider $D = 0. 1 m^2/s, s_0 = 5 \mu mol/m$ , and $L = 10 m$ . The steady-state concentration $\tilde{S}\left ( \frac{L}{2} \right )=S\left ( \frac{L}{2},\infty \right )$ , at the center $x=\frac{L}{2}$ of the reservoir (in $\mu$ mol/m) is ___________. (in integer)

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📖 Explanation

From figure s(x,t) at $x=0\Rightarrow s(0,t)=s_0=5$ at $x=0.4L\Rightarrow s(0.4L,t)=s_0=5$ at $x=L\Rightarrow s(L,t)=0$ From $x = 0\; to \;x = 0.4 L$ Concentration of pollutant $5\mu \; mol/m \times (0.4 \times 10m) =20 \mu \; mol$ $x = 0.4L\; to \;x = L$ Concentration of pollutant = 0 Total concentration of pollutant in 10 m $=20 \mu \; mol$ In infinite time this concentration will be diluted so concentration of pollutant per m $=\frac{20}{10} \mu \; mol/m=2 \mu \; mol/m$ Under steady state condition, concentration of pollutant will be uniformly distributed. Steady state concentration at $x=\frac{L}{2} =2\mu \; mol/m$

Q4GATE 2021NAT

A lake has a maximum depth of 60 m. If the mean atmospheric pressure in the lake region is 91 kPa and the unit weight of the lake water is $9790 \mathrm{~N} / \mathrm{m}^{3}$ , the absolute pressure (in kPa,round off to two decimal places) at the maximum depth of the lake is ___________

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📖 Explanation

Absolute pressure at maximum depth of the lake $=P_{\text {atm }}+\rho g h$ $=91+\frac{9790(60)}{1000}=678.4 \mathrm{kPa}$

Q5GATE 2018MCQ

Dupuit's assumptions are valid for

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Q6GATE 2016NAT

Water table of an aquifer drops by 100 cm over an area of 1000 $km^{2}$ . The porosity and specific retention of the aquifer material are 25% and 5%, respectively. The amount of water (expressed in $km^{3}$ ) drained out from the area is ______ .

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📖 Explanation

\begin{aligned} S_{\mathrm{r}}+S_{\mathrm{y}} &=n \\ \frac{5}{100}+\frac{V_{\mathrm{w}}}{10^{3} \times 10^{6} \times 1} &=\frac{25}{100} \\ V_{\mathrm{w}} &=0.2 \times 10^{9} \mathrm{m}^{3}=0.2 \mathrm{km}^{3} \end{aligned}

Q7GATE 2016MCQ

A tracer takes 100 days to travel from Well-1 to Well-2 which are 100 m apart. The elevation of water surface in Well-2 is 3m below that in Well-1. Assuming porosity equal to 15%, the coefficient of permeability (expressed in m/day) is

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📖 Explanation

Seepage velocity,

\begin{aligned} V_{s}&=\frac{v}{n}=\frac{\text { distance }}{\text { time }}\\ \text{as per Darcy},\\ v &=k i \\ \frac{k i}{n} &=\frac{100 \mathrm{m}}{100 \mathrm{days}} \\ i &=\frac{\text { head difference }}{\text { length }}=\frac{3}{100} \\ \frac{k \times 3}{0.15 \times 100} &=\frac{100}{100} \mathrm{m} / \mathrm{day} \\ k &=5 \mathrm{m} / \mathrm{day} \end{aligned}

Q8GATE 2015MCQ

The relationship between porosity ( $\eta$ ), specific yield ( $S_{y}$ ) and specific retention ( $S_{r}$ ) of an unconfined aquifer is:

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Q9GATE 2011MCQ

In an aquifer extending over 150 hectare, the water table was 20m below ground level. Over a period of time the water table dropped to 23m below the ground level. If the porosity of aquifer is 0.40 and the specific retention is 0.15, what is the change in ground water storage of the aquifer?

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📖 Explanation

We know that,

\begin{aligned} n&=S_{y}+S_{r} \\ \Rightarrow\quad S_{y}&=S_{t}-n=0.15-0.40=-0.25 \end{aligned}

The negative sign signifies the decrease in storage. Change in volume of aquifer $=150 \times (23-20)=450 \mathrm{ha}-\mathrm{m}$ Specific yield is nothing but the actual volume of water that can be extracted by force of gravity from a unit volume of aquifer. $\therefore$ Change in water storage $=450 \times 0.25=112.5 \mathrm{ha}-\mathrm{m}$

Q10GATE 2010MCQ

A well of diameter 20 cm fully penetrates a confined aquifer. After a long period of pumping at rate of 2720 litres per minute, the observations of drawdown taken at 10 m and 100 m distances from the centre of the well are found to be 3m and 0.5m respectively. The transmissivity of the aquifer is

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📖 Explanation

\begin{aligned} Q &=\frac{2 \pi T\left(s_{1}-s_{2}\right)}{\ln \left(\frac{r_{2}}{r_{1}}\right)} \\ T &=\frac{2720 \times 10^{-3} \times 24 \times 60 \times \ln \left(\frac{100}{10}\right)}{2 \times \pi \times (3-0.5)} \\ &=574.26 \mathrm{m}^{2} / \mathrm{day} \simeq 576 \mathrm{m}^{2} / \mathrm{day} \end{aligned}

Q11GATE 2009MCQ

The relationship among specific yield ( $S_{y}$ ), specific retention ( $S_{r}$ ) and porosity ( $\eta$ ) of an aquifer is

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Q12GATE 2008MCQ

A volume of $3.0 \times 10^{6} m^{3}$ of groundwater was pumped form an unconfined aquifer uniformly from an area of 5 $km^{2}$ . The pumping lowered the water table form initial level of 102 m to 99 m. The specific yield of the aquifer is

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📖 Explanation

Specific yield

\begin{aligned} &=\frac{\begin{array}{c} \text{Volume of water pumped }\\\text { out of the aquifer }\end{array}}{\text { Total volume of aquifer }} \\ &=\frac{3.0 \times 10^{6}}{5 \times 10^{6} \times (102-99)}=0.20 \end{aligned}

Q13GATE 2006MCQ

For steady flow to a fully penetrating well in a confined aquifer, the drawdowns at radial distances of $r_{1}$ and $r_{2}$ from the well have been measured as $s_{1}$ and $s_{2}$ respectively, for a pumping rate of Q. The transmissivity of the aquifer is equal to

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Q14GATE 2005MCQ

Match the following

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Q15GATE 2005MCQ

Two observation wells penetrated into a confined aquifer and located 1.5 km apart in the direction of flow, indicate head of 45 m and 20 m. If the coefficient of permeability of the aquifer is 30 m/day and porosity is 0.25, the time of travel of an inert tracer from one well to another is

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📖 Explanation

The average flow velocity or discharge velocity, $v=k i=30 \times \left(\frac{45-20}{1500}\right)=0.5 \mathrm{m} / \mathrm{da}$ Seepage velocity, $v_{s}=\frac{v}{n}=\frac{0.5}{0.25}=2 \mathrm{m} / \mathrm{day}$ $\therefore$ Travel time $=\frac{1500}{2}=750 \text{ days}$

Q16GATE 2003MCQ

Water is pumped from a well tapping an unconfined aquifer at a certain discharge rate and the steady state draw down (X) in an observation well is monitored. Subsequently, the pumping discharge is doubled and the steady state draw down in the same observation well is found to be more than double (i.e., more than 2X). This disproportionate draw down is caused by

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📖 Explanation

By Theim's formula, the discharge rate for an unconfined aquifer is given as, $Q=\frac{\pi \mathrm{k}\left(\mathrm{h}_{2}^{2}-h_{1}^{2}\right)}{2.3 \log _{10}\left(\frac{r_{2}}{r_{1}}\right)}$ From the above formula it is obvious that pumping rate in a well in an unconfined aquifer follows nonlinear relationship with drawdown.

Q17GATE 2002MCQ

When there is an increase in the atmospheric pressure, the water level in a well penetrating in a confined aquifer

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