Physics

Fluid Mechanics and Hydraulics

376 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

A smooth flat plate with a sharp leading edge is placed along a gas stream flowing at U = 10 m/s. The thickness of the boundary layer at section r − s is 10 mm, the breadth of the plate is 1 m (into the paper) and the density of the gas $\rho$= 1.0 kg/m. Assume that the boundary layer is thin, two-dimensional and follows a linear velocity distribution, u = U (y/$\delta$) , at the section r − s , where y is the height from plate.

The integrated drag force (in N) on the plate between p − s is:

  1. 0.67

  2. 0.33

  3. 0.17

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

Multiple choice

A smooth flat plate with a sharp leading edge is placed along a gas stream flowing at U = 10 m/s. The thickness of the boundary layer at section r − s is 10 mm, the breadth of the plate is 1 m (into the paper) and the density of the gas $\rho$= 1.0 kg/m. Assume that the boundary layer is thin, two-dimensional, and follows a linear velocity distribution, u = U (y/$\delta$) , at the section r − s , where y is the height from plate.

The mass flow rate (in kg/s) across the section q − r is

  1. zero

  2. 0.05

  3. 0.10

  4. 0.15
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

The velocity components in the x and y directions of a two dimensional potential flow are u and v, respectively, then - $\frac{\partial u}{\partial x}$ is equal to
  1. $\frac{\partial v}{\partial x}$
  2. $-\frac{\partial v}{\partial x}$
  3. $\frac{\partial v}{\partial y}$
  4. $-\frac{\partial v}{\partial x}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\text{We know that potential flow(ideal flow)satisfy the continuity equation.}\\ \text{The continuity equation for two dimensional flow for incompressible fluid is given by,}\\ \hspace{3cm}\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 \\ \hspace{4cm}\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y}$

Multiple choice

A U-tube manometer with a small quantity of mercury is used to measure the static pressure difference between two locations A and B in a conical section through which an incompressible fluid flows. At a particular flow rate, the mercury column appears as shown in the figure. The density of mercury is 13600 kg/m3 and g = 9.81 m/s2. Which of the following is correct?

  1. Flow direction is A to B and pA - pB = 20 kPa

  2. Flow direction is B to A and pA - pB = 1.4 kPa

  3. Flow direction is A to B and pB - pA = 20 kPa

  4. Flow direction is B to A and pB - pA = 1.4 kPa
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiple choice

A leaf is caught in a whirlpool. At a given instant, the leaf is at a distance of 120 m from the center of the whirlpool. The whirlpool can be described by the following velocity distribution: $V_r=-\bigg(\frac{60\times10^3}{2xr}\bigg)$ m/s and $V_0=\frac{300\times10^3}{2xr}$ m/s, where r (in meters) is the distance from the center of the whirlpool. What will be the distance of the leaf from the center when it has moved through half a revolution?

  1. 48 m

  2. 64 m

  3. 120 m

  4. 142 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

In the velocity diagram shown below, u = blade velocity, C = absolute fluid velocity and w = relative velocity of fluid and the subscripts 1 and 2 refer to inlet and outlet. This diagram is for

  1. an impulse turbine

  2. a reaction turbine

  3. a centrifugal compressor

  4. an axial flow compressor
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\text{Velocity of flow,$\hspace{0.5cm}$}u=u_1=u_2=constant\\ and \hspace{4cm}W_2>>W_1\hspace{8cm}W=\text{Whirl velocity}\\ \text{Hence,it is a diagram of reaction turbine}$

Multiple choice

For steady flow to a fully penetrating well in a confined acquifer, the drawdowns at radial distances of r1 and r2 from the well have been measured as s1 and s2 respectively, for a pumping rate of Q. The transmissivity of the aquifer is equal to

  1. $\frac{Q}{2\pi}\frac{In\frac{r_2}{r_1}}{(s_1-s_2)}$
  2. $\frac{Q}{2\pi} \frac{In(r_2-r_1)}{(s_1-s_2)}$
  3. $\frac{Q}{2\pi}In\bigg(\frac{r_2/r_1}{s_1-s_2}\bigg)$
  4. $2\pi Q \frac{r_2-r_1}{In\big(\frac{s_2}{s_1}\big)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 

Multiple choice

A sugar syrup (density = 1040 kg/m3 and viscosity = 1600 $\times$ 10–6 Pa.s) is required to be pumped into a tank (1.5 m diameter and 3 m height) by a 3 cm inside diameter pipe. If the liquid is required to flow under laminar conditions, the minimum time to fill the tank with the syrup will be

  1. 192.9 h

  2. 19.3 h

  3. 38.6 h

  4. 57.9 h
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This requires calculating flow rate for laminar conditions (Reynolds number < 2100). Given the complexity and missing parameters for pressure drop, this is a standard engineering problem where the answer is derived from volume/flow rate.

Multiple choice maths measures and the circle surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

The circumference of a 1 cm thick pipe is 44 cm. The level of water that 7 cm of pipe can hold is

  1. $798 cm^3$
  2. $308 cm^3$
  3. $792 cm^3$
  4. $795 cm^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that

$2\pi r=44$
$r=7\ cm$

So, the inner radius of the pipe $=7-1=6\ cm$

Therefore, the volume of the pipe
$=\pi r^2h$
$=\pi \times 6^2\times 7$
$=792\ cm^3$

Hence, this is the answer.

Multiple choice physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

Water is poured from a height of 10 m into an empty barrel at the rate of 1 litre per second. If the weight of the barrel is 10 kg, the weight indicated at time t = 60 s will be 

  1. $71.4 \mathrm { kg }$
  2. $68.6 \mathrm { kg }$
  3. $70.0 \mathrm { kg }$
  4. $84.0 \mathrm { kg }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Time taken to reach $10m$ below $=\sqrt {\dfrac{{2 \times 10}}{{10}}}  = \sqrt 2 \,\sec $

$\therefore $ only $\left( {60 - \sqrt 2 } \right)\sec $ is effective for pouring of water
$\therefore $ total weight $=10lkg+\left( {60 - \sqrt 2 } \right)l \times \dfrac{{1\,kg}}{l}$
$=58.6kg+10kg$
$=68.6kg$
Hence,
option $(B)$ is correct answer.

Multiple choice physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

A large cylindrical tank has a hole of area $A$ at its bottom and water is poured into the tank through a tube of cross-section area $A$ ejecting water at the speed $v$. Which of the following is true?

  1. Water level in tank keeps on rising

  2. No water can be stored in the tank
  3. Water level will rise to height $v^{2}/2g$ and then stop

  4. The water level will be oscillating
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the beginning ,the water escaping the vessel at the rate $\sqrt{2gh}$ and filling it at the rate of $vA$


So the rate of change in the water level of the vessel at any instant is 
($v-\sqrt{2gh})A$ 
where  'h' is the instantaneous height 

Since h is quite small in the beginning, the water is escaping the vessel at a slow rate but filling at a much larger rate as such    ($v-\sqrt{2gh})A$  ,is positive  (increase in water level)

soon after the filling of water  occurs, the height increases and the rate of escape become larger 
But when the rate =rate of filling ,then an equilibrium state is reached and there is no net change  in the height of the water 

the equilibrium is reached when ($v-\sqrt{2gh})A$ =0  $\rightarrow$  $h=\dfrac{v^2}{2g}$