This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Groundwater – Open Wells”.
1. Open wells are most suitable for which of the following?
a) Shallow and low yield aquifers
b) Shallow and high yield aquifers
c) Deep and low yield aquifers
d) Deep and high yield aquifers
View Answer
Explanation: Open wells are commonly used in rural areas for water supply and in small scale farming works. As they are made by digging, it is most feasible for shallow aquifers and those that are low yielding.
2. The flow into open wells usually takes place from which part of the well?
a) Bottom
b) Upper portion
c) Bottom and sides
d) Side walls
View Answer
Explanation: Open wells are inserted upto depths of 10 m below the ground and the inflow into them is usually through joints, fractures and other openings present at the bottom and lower portions of the well.
3. What is the most common form of open wells used for alluvial soils?
a) Circular and lined
b) Circular and unlined
c) Rectangular and lined
d) Rectangular and unlined
View Answer
Explanation: In unconsolidated formations like alluvial soils, circular shaped and lined open wells are used which are dug to a depth of about 20 m below the water table. These wells tap water in unconfined aquifers.
4. What is the difference in water table elevation and water level inside the well known as?
a) Critical head
b) Specific head
c) Depression head
d) Maximum head
View Answer
Explanation: The water level inside an open wells is lowered as the water is pumped out. The difference between the new water level inside the open well and the elevation of the water table is known as depression head.
5. What is the relationship between flow into the well (Q) and depression head (H) for an open well?
a) Q∝H
b) Q∝\(\frac{1}{H}\)
c) Q∝Hn; n = constant > 1
d) Q∝(aH+b); a, b = constants
View Answer
Explanation: For an open well, the discharge into the well follows a directly proportional relationship with the depression head of the well, with a constant of proportionality.
6. The proportionality constant used for relating the inflow and depression head of open wells depends on which of the following?
a) Aquifer properties
b) Well area and depth
c) Aquifer properties and well area
d) Aquifer properties and well depth
View Answer
Explanation: The relation between inflow and depression head for open wells is given as Q=KH, where K is the constant of proportionality. K is known as specific capacity and depends on the aquifer characteristics and the area of the well.
7. Critical depression head is the minimum head below which any value would lead to erosion of surrounding soil.
a) True
b) False
View Answer
Explanation: Critical depression head is the maximum depression for an open well beyond which any lowering of water would cause the dislodging and displacement of surrounding soil particles due to high flow velocity.
8. The specific capacity of an open well is determined using which of the following tests?
a) Pumping out test
b) Open end test
c) Packer’s test
d) Recuperation test
View Answer
Explanation: The proportionality constant known as specific capacity is fixed for a given open well and is found using a recuperation test. It involves pumping water form the well and measuring depth of water at different time intervals after allowing it to recuperate.
9. Specific capacity per unit area of an open well has units similar to which of the following quantities?
a) Time
b) Kinematic viscosity
c) Permeability
d) Frequency
View Answer
Explanation: Specific capacity per unit area (Ks) for an open well is defined as the flow into the well per unit area under a unit depression head. If K is the specific capacity and A is the well area, then,
\(\)K_s=\frac{K}{A}=\frac{(Q/H)}{A}=\frac{Q}{AH}=\frac{m^3/s}{m^2*m}=s^{-1}=unit of frequency
10. What is the equation for specific capacity (K) of an open well as per recuperation test? T is the recuperation time, H is the initial drawdown, h is the drawdown after recharge and A is the well area.
a) K=\(\frac{1}{T} ln(\frac{h}{H})\)
b) K=\(\frac{1}{T} ln(\frac{H}{h})\)
c) K=\(\frac{A}{T} ln(\frac{h}{H})\)
d) K=\(\frac{A}{T} ln(\frac{H}{h})\)
View Answer
Explanation: By integrating the drawdown over the recuperation time T during which the water elevation rises from H to the final level h, the equation obtained is
\(\frac{K}{A}=\frac{1}{T} ln(\frac{H}{h})\)
This represents the specific capacity per unit area and K is the specific capacity (dimensions of L2T-1).
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]
