This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Groundwater – Equation of Motion”.

1. Which of the following differential equations represent unidirectional unsteady flow in a homogeneous confined aquifer? S_{s} = specific storage, K= permeability, S = storativity, T = transmissibility, h = piezometric head, t = time, (x, y, z) = directions.

a) \(\frac{∂^2 h}{∂x^2}=\frac{Ss}{S} \frac{∂h}{∂t}\)

b) \(\frac{∂^2 h}{∂z^2}=\frac{Ss}{K} \frac{∂h}{∂t}\)

c) \(\frac{∂^2 h}{∂x^2}+\frac{∂^2 h}{∂y^2}=\frac{S}{T} \frac{∂h}{∂t}\)

d) \(\frac{∂^2 h}{∂x^2}+\frac{∂^2 h}{∂z^2}=\frac{K}{T} \frac{∂h}{∂t}\)

View Answer

Explanation: The basic differential equation concerning unsteady flow in a confined aquifer is given as,

\(\frac{∂^2 h}{∂x^2}+\frac{∂^2 h}{∂y^2}+\frac{∂^2 h}{∂z^2}=\frac{Ss}{K} \frac{∂h}{∂t}=\frac{S}{T} \frac{∂h}{∂t}\)

Since the equation is asked for unidirectional flow, only one suitable direction is considered.

2. What is the name of the following equation concerning groundwater flow in confined aquifers.

∇^{2} h=\(\frac{S}{T} \frac{∂h}{∂t}\), where S = storativity, T = transmissibility, h = piezometric head, t = time.

a) Diffusion equation

b) Laplace equation

c) Dupit’s equation

d) Thiems’ equation

View Answer

Explanation: The given equation is the basic differential equation that represents unsteady flow of groundwater in an isotropic homogeneous confined aquifer. This form of the equation is also known as the diffusion equation.

3. Laplace equation is derived from the diffusion equation for the case of steady flow.

a) True

b) False

View Answer

Explanation: When the flow of groundwater in a confined aquifer is steady, the \(\)\frac{∂h}{∂t} term in the diffusion equation becomes zero, hence the new equation is ∇

^{2}h=0. This is known as the Laplace equation and is applicable to both confined and unconfined aquifers.

4. Which of the following is not an assumption made by Dupit for steady flow in unconfined aquifers?

a) Hydraulic grade line is equal to free surface slope

b) Free surface curvature is very small

c) Streamlines at all sections are horizontal

d) Hydraulic grade line varies with depth

View Answer

Explanation: Dupit suggested a simple method for steady state unconfined aquifer problems with two main assumptions – The free surface curvature is very small such that streamlines can be assumed as horizontal, and the hydraulic grade line equal to the slope of free surface and does not vary with depth.

5. Which of the following is the correct equation for steady 2D flow in an unconfined aquifer as per Dupit’s assumptions? h is the piezometric head and x, y are the two directions of flow.

a) \(\frac{∂h}{∂x}+\frac{∂h}{∂y}\)=0

b) \(\frac{∂^2 h}{∂x^2}+\frac{∂^2 h}{∂y^2}\)=0

c) \(\frac{∂^2 h^2}{∂x^2}+\frac{∂^2 h^2}{∂y^2}\)=0

d) \(\frac{∂^3 h}{∂x^3}+\frac{∂^3 h}{∂y^3}\)=0

View Answer

Explanation: Due to the difficulty of solving the steady flow problems in unconfined aquifer using Laplace equation in h, a modified equation was derived using Dupit’s assumptions and is given as,

\(\frac{∂^2 h^2}{∂x^2}+\frac{∂^2 h^2}{∂y^2}\)=0 ⇒∇

^{2}h

^{2}=0 . This is the Laplace equation is h

^{2}.

6. The differential equation for steady flow in an unconfined aquifer of permeability K undergoing infiltration at a rate R, under Dupit’s assumptions is given as ∇^{2} h^{2}=A. What is A?

a) 0

b) \(\frac{R}{K}\)

c) \(\frac{2R}{K}\)

d) \(\frac{R^2}{K}\)

View Answer

Explanation: For a steady flow problem in an unconfined aquifer, if there is an additional recharge from top at the rate R, the equation of motion would be given as,

\(\frac{∂^2 h^2}{∂x^2}+\frac{∂^2 h^2}{∂y^2}=\frac{2R}{K}\)

7. An unconfined aquifer is located on an impervious stratum between two rivers with a difference in their surface elevation. There is constant recharge of the water table from the top. What is the shape of the water table of the aquifer?

a) Circular arc

b) Elliptical arc

c) Parabolic arc

d) Linear

View Answer

Explanation: After integrating the Laplace equation based on Dupit’s assumptions and applying the suitable boundary conditions, the equation for the piezometric head is obtained as,

h

^{2}=\(-\frac{Rx^2}{K}-\frac{(h_0^2-h_1^2-\frac{RL^2}{K})}{L} x+h_0^2\)

Where R is the recharge rate, K is the permeability, h

_{0}and h

_{1}the heads of the upstream and downstream rivers respectively, L is the length of the aquifer and x is the distance from the upstream end. This equation represents an ellipse and hence the water table follows an upward elliptical arc from the upstream body to the downstream body.

8. In a case of steady flow in a recharging unconfined aquifer between two water bodies at different levels, what is a water divide?

a) Point of maximum piezometric head in the aquifer

b) Point of minimum piezometric head in the aquifer

c) Mid-point of the aquifer between the two bodies

d) Point on the water table that coincides with the average elevation of the two water bodies

View Answer

Explanation: The water table in the given situation follows the shape of an ellipse, with the head rising from the upstream end to a maximum point and then dropping to the downstream end. The location between the bodies when the water table reaches its maximum height is called the water divide.

9. Two rivers of depths 15 m and 10 m are separated by an unconfined aquifer of permeability 10 m/day lying atop a hard stratum. It is subjected to a uniform recharge of 0.01 m^{3}/day per m^{2} of its area. What is the water table profile if the distance between the rivers is 1.6 km? Take x as horizontal direction and y as vertical direction.

a) y^{2}=-0.001x^{2}-1.52x+100

b) y^{2}=-0.001x^{2}-1.68x+225

c) y^{2}=-0.001x^{2}+1.52x+225

d) y^{2}=-0.001x^{2}+1.68x+100

View Answer

Explanation: Given h

_{0}= 15 m, h

_{1}= 10 m, L = 1600m, K = 10 m/day, R = 0.01 m

^{3}/day/m

^{2}

For the given system, the water table equation is given as,

h

^{2}=\(-\frac{Rx^2}{K}-\frac{(h_0^2-h_1^2-\frac{RL^2}{K})}{L} x+h_0^2=-(\frac{0.01}{10}) x^2-\frac{1}{1600} (15^2-10^2-\frac{0.01*1600^2}{10})x+15^2\)

⇒h

^{2}=-0.001x

^{2}–\(\frac{1}{1600} (15^2-10^2-2560)x\)+225

∴h

^{2}=-0.001x

^{2}+1.52x+225

10. An unconfined aquifer present between two water bodies 900 m apart has a parabolic water table. One of the water bodies is 8 m deep and the other is 2.5 m shallower. If the permeability of the aquifer is 0.83 m/hr, what is the seepage discharge from one water body to the other per meter width of the aquifer?

a) 0.374 m^{3}/day

b) 0.639 m^{3}/day

c) 0.748 m^{3}/day

d) 1.278 m^{3}/day

View Answer

Explanation: Given h

_{0}= 8 m, h

_{1}= 8 – 2.5 = 5.5 m, L = 900m, K = 0.83 m/hr = 19.92 m/day

Since the water table profile is parabolic, it is an aquifer with no recharge.

The seepage discharge is given as,

q=\(\frac{(h_0^2-h_1^2 )}{2L} K=\frac{(8^2-5.5^2)}{2*900}*19.92=\frac{33.75}{1800}*19.92\)

=0.3735 m

^{3}/day per m≅0.374 m

^{3}/day per m

**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__.