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

An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours. $\psi$index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.

Peak discharge of the resulting direct runoff hydrograph shall be

  1. 150 cumecs

  2. 225 cumecs

  3. 230 cumecs

  4. 360 cumecs
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 

Multiple choice

An average rainfall of 16 cm occurs over a catchment during a period of 12 hours with uniform intensity. The unit hydrograph (unit depth = 1 cm, duration = 6 hours) of the catchment rises linearly from 0 to 30 cumecs in six hours and then falls linearly from 30 to 0 cumecs in the next 12 hours. $\psi$index of the catchment is known to be 0.5 cm/hr. Base flow in the river is known to be 5 cumecs.

Area of the catchment in hectares is

  1. 97.20

  2. 270

  3. 9720

  4. 2700
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 

Multiple choice

Water is pumped from a well tapping an unconfined aquifer at a certain discharge rate and the steady state drawdown (X) in an observation well is monitored. Subsequently, the pumping discharge is doubled and the steady state drawdown in the same observation well is found to be more than double (i.e. more than 2X). This disproportionate drawdown is caused by

  1. well losses

  2. decrease in the saturated thickness of the aquifer

  3. nonlinear flow

  4. delayed gravity yield
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 

Multiple choice

Direct step method of computation for gradually varied flow is

  1. applicable to non-prismatic channels

  2. applicable to prismatic channels

  3. applicable to both prismatic and non-prismatic channels

  4. not applicable to both prismatic and non-prismatic channels
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Direct step method, employs the formula $dx = \frac{\triangle E}{S_0 - S_1}$ and only applicable when, there is no change in the cross-sectional area of channel, i.e., Prismatic channel.

Multiple choice

The depth of flow in an alluvial channel is 1.5 m. If critical velocity ratio is 1.1 and Manning’s n is 0.018, the critical velocity of the channel as per Kennedy’s method is

  1. 0.713 m/s

  2. 0.784 m/s

  3. 0.879 m/s

  4. 1.108 m/s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 $y = 1.5m, n = 0.018\\ \text{Critical velocity as per kennedy's equation}\\ \hspace{1.7cm}V = 0.55 my^{0.64}\\ \hspace{2cm} = 6.55 \times 1.1 \times (1.5)^{0.64}\\ \hspace{2cm} = 0.784 m/s$

Multiple choice

Direction: One hour triangular unit hydrograph of a watershed has the peak discharge of 60 m3/sec.cm at 10 hours and time base of 30 hours. The $\phi$ index is 0.4 cm per hour and base flow is 15 m3m/sec.

The catchment area of the watershed is

  1. 3.24 km2

  2. 32.4 km2

  3. 324 km2

  4. 3240 km2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Peak discharge of unit hydrograph = 60 m^3/s. Time base = 30 hours. Area under triangular hydrograph = 0.5 * base * height = 0.5 * 30 * 3600 * 60 = 3,240,000 m^3. Since it is a 1 cm unit hydrograph, 1 cm = 0.01 m. Area = Volume / Depth = 3,240,000 / 0.01 = 324,000,000 m^2 = 324 km^2.

Multiple choice

Direction: One hour triangular unit hydrograph of a watershed has the peak discharge of 60 m3/sec.cm at 10 hours and time base of 30 hours. The $\phi$ index is 0.4 cm per hour and base flow is 15 m3m/sec.

If there is rainfall of 5.4 cm in 1 hour, the ordinate of the flood hydrograph at 15th hour is

  1. 225 m3/sec

  2. 240 m3/sec

  3. 249 m3/sec

  4. 258 m3/sec
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Effective rainfall = 5.4 - (0.4 * 1) = 5.0 cm. The ordinate of the flood hydrograph = (Effective Rainfall * Unit Hydrograph Ordinate) + Base Flow. At 15 hours, the UH ordinate is calculated by linear interpolation: at 10h it is 60, at 30h it is 0. So at 15h, it is 60 * (30-15)/(30-10) = 60 * 15/20 = 45. Flood hydrograph = (5 * 45) + 15 = 225 + 15 = 240 m^3/s.

Multiple choice

Consider the turbulent flow of a fluid through a circular pipe of diameter, D. Identify the correct pair of statements.

I. The fluid is well-mixed. II. The fluid is unmixed. III. ReD < 2300 IV. ReD < 2300

  1. I, III

  2. II, IV

  3. II, III

  4. I, IV
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For flow through a circular pipe, the flow will be _____. (A) Leminar flow Reynolds no. (Re) < 2300 Fluid flow is unmixed. (B) Turbulent flow Reynolds no. (Re) > 2300 Fluid flow is mixed.

Multiple choice

Kaplan water turbine is commonly used when the flow through its runner is

  1. axial and the head available is more than 100 m

  2. axial and the head available is less than 10 m

  3. radial and the head available is more than 100 m

  4. mixed and the head available is about 50 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Kaplan water turbine is a kind of reaction turbine in which axial flow took place & it is used where head available is less than 10 m.