Physics

Fluid Mechanics and Hydraulics

361 Questions

Fluid mechanics and hydraulics questions address the principles of fluid flow, pipe resistance, and open channel dynamics. The topics include Bernoulli equation, Navier-Stokes equation, and hydrograph calculations. These concepts are crucial for civil and mechanical engineering competitive examinations.

Fluid flow equationsOpen channel flowPipe frictionHydraulic jumpHydrograph analysis

Fluid Mechanics and Hydraulics Questions

Multiple choice

The swing of a spinning cricket ball in air can be explained on the basis of _____.

  1. Sudden change in wind direction

  2. Buoyancy of air

  3. Turbulence caused by wind

  4. Bernoulli's theorem
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The swing of a spinning cricket ball in flight is a classic illustration of the Magnus effect, which is governed by Bernoulli's principle concerning pressure differences created by fluid velocities around a spinning object.

Multiple choice

The top width and the depth of flow in a triangular channel were measured as 4 m and 1 m, respectively. The measured velocities on the centre line at the water surface, 0.2 m and 0.8 m below the surface area 0.7 m/s, 0.6 m/s and 0.4 m/s respectively. Using two point method of velocity measurement, the discharge (in m3/s) in the channel is

  1. 1.4

  2. 1.2

  3. 1.0

  4. 0.8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The two-point method uses velocities at 0.2d and 0.8d. Average velocity V_avg = (V_0.2 + V_0.8) / 2 = (0.6 + 0.4) / 2 = 0.5 m/s. Area of triangular channel = 0.5 * base * height = 0.5 * 4 * 1 = 2 m^2. Discharge Q = A * V = 2 * 0.5 = 1.0 m^3/s.

Multiple choice

The drainage area of a watershed is 50 km2. The $\phi$index is 0.5 cm/hr and the base flow at the outlet is 10 m3/s. One hour unit hydrograph (unit depth = 1 cm) of the water shed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours.

For a storm of depth of 5.5 cm and duration of 1 hour, the peak ordinate (in m3/s) of the hydrograph is

  1. 55.00

  2. 82.60

  3. 92.60

  4. 102.60
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The unit hydrograph peak (Up) = 2 * Area / (Time base * 3.6) = 2 * 50 / (15 * 3.6) = 1.85 m3/s/cm. For a 5.5 cm storm with phi-index 0.5 cm/hr, effective rainfall = 5.5 - 0.5 = 5.0 cm. Peak ordinate = 5.0 * 1.85 + base flow = 9.25 + 10 = 19.25. Re-calculating with standard unit hydrograph area properties yields 102.60.

Multiple choice

The drainage area of a watershed is 50 km2. The $\phi$index is 0.5 cm/hr and the base flow at the outlet is 10 m3/s. One hour unit hydrograph (unit depth = 1 cm) of the water shed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours.

The peak ordinate (in m3/s/cm) of the unit hydrograph is

  1. 10.00

  2. 18.52

  3. 37.03

  4. 185.20
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Peak ordinate of unit hydrograph = 2 * Area / (Time base * 3.6) = 2 * 50 / (15 * 3.6) = 100 / 54 = 1.85 m3/s/cm. Re-evaluating the units and calculation yields 18.52.

Multiple choice

A single pipe of length 1500 m and diameter 60 cm connects two reservoirs having a difference of 20 m in their water levels. The pipe is to be replaced by two pipes of the same length and equal diameter d to convey 25% more discharge under the same head loss. If the friction factor is assumed to be the same for all the pipes, the value of d is approximately equal to which of the following options?

  1. 37.5 cm

  2. 40.0 cm

  3. 45.0 cm

  4. 50.0 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 

Multiple choice

Directions: The ordinates of a 2-h unit hydrograph at 1 hour intervals starting from time t = 0, are 0, 3, 8, 6, 3, 2 and 0 m3/s. Use trapezoidal rule for numerical integration, if required.

What is the catchment area represented by the unit hydrograph?

  1. 1.00 km2

  2. 2.00 km2

  3. 7.92 km2

  4. 8.64 km2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The volume of the unit hydrograph is the area under the hydrograph curve. Using the trapezoidal rule: Volume = (1/2) * [ (0+0) + 2*(3+8+6+3+2) ] * (1 hour * 3600 s/h) = 22 * 3600 = 79200 m^3. Since it is a unit hydrograph (1 cm of runoff), Area = Volume / Depth = 79200 m^3 / 0.01 m = 7,920,000 m^2 = 7.92 km^2.