# Engineering Hydrology Questions and Answers – Groundwater – Open Wells – Set 2

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

1. What is the discharge form an open well corresponding to critical depression head known as?
a) Minimum yield
b) Maximum yield
c) Safe yield
d) Working yield

Explanation: The critical depression head is the maximum safe drawdown of water in the open well. The discharge obtained corresponding to this critical depression head is known as critical yield or maximum yield.

2. What is the general value of factor of safety adopted with regards to critical depression head in open wells?
a) 1
b) 1.5
c) 3
d) 6

Explanation: The critical depression head is divided by a factor of safety, normally taken as 2.5 to 3, and the obtained head is called the working head. The corresponding discharge obtained using this head is known as safe yield.

3. What is the general order of values of specific capacity per unit well area for an open well in hour-1?
a) 10-4
b) 10-2
c) 100
d) 102

Explanation: As per Marriot, the values of specific capacity per unit area for different types of soil are – 0.25 for clay, 0.5 for fine sand and 1 for coarse sand. These values may be used when recuperation test data is not available.

4. A recuperation test was conducted on an open well with a depression head of 2.8 m. After allowing 75 minutes for recuperation the water level in the well rose by 1.2 m. What is the specific capacity per unit area of the open well?
a) 0.011
b) 0.447
c) 0.678
d) 0.746

Explanation: Given h1 = 2.8 m, h2 = 2.8 – 1.2 = 1.6 m, T = 75 mins = 1.25 hours.
The specific capacity per unit area is given as,
$$K_s=\frac{1}{T} ln⁡(\frac{h_1}{h_2})=\frac{1}{1.25} ln⁡(\frac{2.8}{1.6})=0.447 h^{-1}$$

5. During a recuperation test, water level in a well was lowered by 3.2 m and then recuperated for 85 minutes until it reached a depression head of 1.5 m. What should be the diameter of the well in order to yield a discharge of 30 m3/h under a depression head of 2.5 m?
a) 4.72 m
b) 5.07 m
c) 5.34 m
d) 5.85 m

Explanation: Given T = 85 min = 1.417 hr, h1 = 3.2 m, h2 = 1.5 m, H = 2.5 m, Q = 30 m3/h.
Specific capacity per unit area is,
$$K_s=\frac{1}{T} ln⁡(\frac{h_1}{h_2})=\frac{1}{1.417} ln⁡(\frac{3.2}{1.5})=0.535 h^{-1}$$
Now, Q=Ks AH ⇒A=$$\frac{Q}{K_s H}=\frac{30}{0.535*2.5}$$=22.43 m2
⇒$$\frac{πd^2}{4}=22.43 ⇒d^2=\frac{22.43*4}{π}$$=28.56
∴Diameter=$$\sqrt{28.56}$$=5.34 m

6. A 3.5 m diameter open well recuperated 1.5 m of water in 2 hours from a depression head of 4.5 m. What will be the discharge (in m3/h) from the well for a depression head of 3.5 m?
a) 6.83
b) 12.11
c) 18.49
d) 27.29

Explanation: Given r = 3.5/2 = 1.75 m, T = 2 hr, h1 = 4.5 m, h2 = 4.5 – 1.5 = 3 m, H = 3.5 m.
The area of the well is, A=πr2=π*1.752=9.62 m2
Specific capacity per unit area is,
$$K_s=\frac{1}{T} ln⁡(\frac{h_1}{h_2})=\frac{1}{2} ln⁡(\frac{4.5}{3})=0.2027 h^{-1}$$

∴Discharge, Q=KsAH=0.2027*9.62*3.5=6.83 m3/h

7. A 3 m diameter open well with a depression head of 2.8 m recuperated 1.9 m in 100 minutes. What will be the safe yield from the well if the critical depression head is 4.5 m? Take factor of safety as 2.5.
a) 2.4 L/s
b) 2.96 m3/h
c) 6 L/s
d) 7.4 m3/h

Explanation: Given r = 3/2 = 1.5 m, T = 100 min = 1.67 hr, h1 = 2.8 m, h2 = 2.8 – 1.9 = 0.9 m, Hc = 4.5 m.
The area of the well is, A=πr2=π*1.52=7.07 m2
Specific capacity per unit area is,
$$K_s=\frac{1}{T} ln⁡(\frac{h_1}{h_2})=\frac{1}{1.67} ln⁡(\frac{2.8}{0.9})=0.68 h^{-1}$$
Working head is, H=$$\frac{Critical head}{Factor of safety}=\frac{H_c}{FOS}=\frac{4.5}{2.5}$$=1.8 m
∴Safe yield=Ks AH=0.68*7.07*1.8=8.65 m3/h=2.4 L/s

8. During a recuperation test, the water in a well was lowered by 2.4 m and it recuperated 60 cm in half an hour. What will be the side of a square well that will yield 7.5 L/sec of discharge under a depression head of 2 m?
a) 3.12 m
b) 4.11 m
c) 4.84 m
d) 5.47 m

Explanation: Given T = 30 min = 0.5 hr, h1 = 2.4 m, h2 = 2.4 – 0.6 = 1.8 m, H = 2 m, Q = 7.5 L/s = 27 m3/h.
Specific capacity per unit area is,
$$K_s=\frac{1}{T} ln⁡(\frac{h_1}{h_2})=\frac{1}{0.5} ln⁡(\frac{2.4}{1.8})=0.5754 h^{-1}$$
Now, Q=Ks AH ⇒A=$$\frac{Q}{K_s H}=\frac{27}{0.5754*2}$$=23.462 m2
∴Side of square=$$\sqrt{23.462}$$=4.84 m

9. Water was pumped from a 2 m diameter open well until it achieved a depression head of 2 m. After 1 hour of recuperation, the well was recharged with 3000 litres of water. What will be the per day yield from the well under a depression head of 1 m?
a) 44 m3
b) 49 m3
c) 52 m3
d) 56 m3

Explanation: r = 2/2 = 1m, T = 1hr, h1 = 2 m, H = 1 m, area of the well A=πr2=π*12=3.14 m2.
Depth recuperated =$$\frac{Volume recuperated}{Area of well}=\frac{3000 L}{3.14 m^2}=\frac{3 m^3}{3.14 m^2}$$=0.955 m
⇒Depression head after recuperation,h2=2-0.955=1.045 m
$$K_s=\frac{1}{T} ln⁡(\frac{h_1}{h_2})=\frac{1}{1} ln⁡(\frac{2}{1.045})=0.649 h^{-1}$$
⇒Discharge, Q=Ks AH=0.649*3.14*1=2.04 m3/h
∴Per day yield=2.04*24=48.96 m3≅49 m3

10. A recuperating open well elevated its depression head from 2.6 m to 2.1 m in 24 minutes. The diameter of the well is 2.5 m and its safe yield is 6 m3/h. If the critical depression head is 3.5 m, what is the factor of safety under which the well is operating?
a) 1.53
b) 2.01
c) 2.45
d) 3.14

Explanation: Given r = 2.5/2 = 1.25 m, T = 24 min = 0.4 hr, h1 = 2.6 m, h2 = 2.1 m, Hc = 3.5 m, Q = 6 m3/h.
The area of the well is, A=πr2=π*1.252=4.909 m2
Specific capacity per unit area is,
$$K_s=\frac{1}{T} ln⁡(\frac{h_1}{h_2})=\frac{1}{0.4} ln⁡(\frac{2.6}{2.1})=0.649 h^{-1}$$
Now, Q=Ks AH
⇒$$H=\frac{Q}{K_s A}=\frac{6}{0.534*4.909}$$=2.289 m
∴FOS=$$\frac{H_c}{H}=\frac{3.5}{2.289}$$=1.53

Sanfoundry Global Education & Learning Series – Engineering Hydrology.

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