All questions of Dimensional Analysis for Civil Engineering (CE) Exam

Given:
Time taken by model to empty = 10 min
Model scale = 1:25

To find:
Time taken by prototype to empty

Solution:
Let's assume the volume of the reservoir in the model as Vm and the volume of the reservoir in the prototype as Vp.

We know that the volume of the model is 1/25 times the volume of the prototype.

Vm/Vp = 1/25

=> Vp = 25Vm

Now, let's assume the rate at which the model empties as Rm and the rate at which the prototype empties as Rp.

We know that the rate of the model is same as the rate of the prototype.

Rm = Rp

We also know that the time taken by the model to empty is 10 minutes.

Therefore, the volume of the model emptied in 1 minute is Vm/10.

Volume emptied in 1 minute by prototype = Vp/?

=> Vp/? = Vm/10

=> 25Vm/? = Vm/10

=> ? = 250 minutes

Therefore, the time taken by the prototype to empty itself is 250 minutes or 50 minutes (since 1 minute in the model corresponds to 25 minutes in the prototype).

Answer: Option B) 50 min

Given data:
Scale of the model: 1:30
Velocity of the submarine in water: 15 km/h
Kinematic viscosity of air: 1.51 x 10-5 m2/s
Kinematic viscosity of water: 1.02 x 10-6 m2/s

To achieve kinematic similarity between the model and the actual submarine, the velocities in the wind tunnel and in the ocean should be related by the equation:

(V_air_tunnel / V_water) = √(ν_air_tunnel / ν_water)

Where:
V_air_tunnel is the velocity of air in the wind tunnel
V_water is the velocity of water in the ocean
ν_air_tunnel is the kinematic viscosity of air in the wind tunnel
ν_water is the kinematic viscosity of water in the ocean

Let's calculate the values:

Convert the velocity of the submarine in water:
15 km/h = (15 * 1000) / (60 * 60) m/s = 4.17 m/s

Substitute the values into the equation:
(V_air_tunnel / 4.17) = √(1.51 x 10-5 / 1.02 x 10-6)

Simplify the equation:
V_air_tunnel = 4.17 * √(1.51 x 10-5 / 1.02 x 10-6)

Calculate the result:
V_air_tunnel = 4.17 * √14.8

V_air_tunnel ≈ 4.17 * 3.847

V_air_tunnel ≈ 16.08 m/s

Therefore, the velocity of air in the wind tunnel that should be maintained for kinematic similarity is approximately 16.08 m/s, which is closest to option C: 1850.5 m/s.

The most important force in the motion of submarines underwater is viscous force. This force is caused by the friction between the water and the surface of the submarine as it moves through the water. The following points explain why viscous force is the most important force in the motion of submarines underwater:

• Resistance to motion: Viscous force is the force that opposes the motion of an object through a fluid. In the case of submarines, the friction between the water and the surface of the submarine creates a resistance to motion. This resistance is known as hydrodynamic drag. Viscous force is the dominant force that causes this drag.

• Effect on speed: The drag caused by viscous force reduces the speed of the submarine. This is why submarines are designed to be streamlined to reduce the drag caused by viscous force. The more streamlined a submarine is, the less viscous force it experiences and the faster it can move through the water.

• Effect on maneuverability: Viscous force can also affect the maneuverability of submarines. As the submarine changes direction, it experiences different amounts of drag on different parts of its surface. This can cause the submarine to rotate or pitch, which can affect its stability. Submarines are designed to have a balance between stability and maneuverability to ensure they can perform their missions effectively.

In conclusion, viscous force is the most important force in the motion of submarines underwater because it creates the resistance to motion that reduces speed and affects maneuverability. Understanding the effects of viscous force is essential for designing submarines that can perform effectively in their underwater environment.

Ω, and the density of the fluid ρ, we can write the equation for thrust as:

T = f(D, V, ω, ρ)

To determine the relationship between thrust and each of the variables, we need to consider the physical principles involved.

1. Diameter (D): The thrust generated by a propeller is directly proportional to the square of its diameter. This can be expressed as:

T ∝ D^2

2. Speed of advance (V): The thrust generated by a propeller is directly proportional to the cube of the speed of advance. This can be expressed as:

T ∝ V^3

3. Angular velocity (ω): The thrust generated by a propeller is directly proportional to the square of the angular velocity. This can be expressed as:

T ∝ ω^2

4. Density of the fluid (ρ): The thrust generated by a propeller is directly proportional to the density of the fluid. This can be expressed as:

T ∝ ρ

Combining these relationships, we can write the equation for thrust as:

T = k * D^2 * V^3 * ω^2 * ρ

where k is a constant that takes into account other factors such as efficiency, blade shape, and propeller design.

Euler number: Eu = ρvL/μ

where ρ is the density of the fluid, v is the velocity of the fluid, L is a characteristic length, and μ is the dynamic viscosity of the fluid.

b) Pressure coefficient: Cp = (p - p_∞)/(1/2ρv^2)

where p is the pressure at a point on the surface, p_∞ is the freestream pressure, ρ is the density of the fluid, and v is the velocity of the fluid.