Fluid Dynamics MCQ Quiz - Objective Question with Answer for Fluid Dynamics - Download Free PDF

Explanation:

Pitot Tube is a device used for calculating the velocity of flow at any point in a pipe or a channel. 

The pitot tube is used to measure velocity at a point.

In the question velocity at the stagnation point is given which is zero. So here stagnation pressure will be the correct answer. Because this stagnation pressure head is used to calculate the velocity at a point.

V = \(\sqrt{2gh}\)

It is based on the principle that if the velocity of flow at a point becomes zero the pressure there is increased due to the conversion of the kinetic energy into pressure energy. 

Working:

• The liquid flows up the tube and when equilibrium is attained the liquid reaches a height above the free surface of the water stream
• Since the static pressure under this situation is equal to the hydrostatic pressure due to its depth below the free surface the difference in level between the liquid in the glass tube and the free surface becomes the measure of dynamic pressure \(p_0-p=\frac{\rho V^2}{2} = h\rho g\)  where p0 p and V are the stagnation pressure static pressure and velocity respectively at point A
• Such a tube is known as a Pitot tube and provides one of the most accurate means of measuring the fluid velocity
• For an open stream of liquid with a free surface this single tube is sufficient to determine the velocity but for a fluid flowing through a closed duct the Pitot tube measures only the stagnation pressure and so the static pressure must be measured separately.

Mistake PointsIn the option velocity at the stagnation point is mentioned at the stagnation point velocity is already zero there is no need to measure velocity at the stagnation point. the pitot tube is used to measure velocity at any point by measuring the stagnation pressure.

Explanation:

Bernoulli's theorem:

• It states that the total mechanical energy of the flowing fluid comprising the energy associated with fluid pressure the gravitational potential energy of elevation and the kinetic energy of fluid motion remains constant.
• It is based on conservation of energy.

Assumptions of Bernoulli's theorem i.e. Bernoulli's theorem is Valid for:

• Flow is ideal i.e inviscous.
• Flow is steady i.e. time variation is zero.
• Flow is incompressible i.e. ρ is constant.
• Flow is irrotaional i.e. ωx = ωy = ωz = 0.
• All the other external forces except gravity and pressure forces should be zero.
• The energy of the system is constant hence there should be no loss of energy.

Bernoulli's Equation:

Bernoulli's equation is obtained by integrating the Euler's equation of motion which is given by

Euler's Equation:

\(\frac{{dp}}{ρ } + gdz + vdv = 0\;\;\;\;\;(1)\)

In Euler's equation of motion the forces due to gravity and pressure are taken into consideration and which is derived considering the motion of a fluid element along a streamline.

Integrating the above equation (1):

\(\smallint \frac{{dp}}{ρ } + \smallint gdz + \smallint vdv = 0\)

\(\frac{p}{ρ } + gz + \frac{{{v^2}}}{2} = C\)

\(\frac {P}{ρ g}+ \frac {v^2}{2g}+ Z = Constant\)

This equation is called the Bernoulli's Equation.

where

 \(\frac{p}{ρg}\) = Pressure head or pressure energy per unit weight

 \(\frac{v^2}{2g}\) = Kinetic head or kinetic energy per unit weight

z = Potential head or potential energy by unit weight.

P = Pressure of fluid at given section V = Flow velocity at given section Z = Potential head at given section ρ = Density of fluid

Concept Used:

When a U-tube containing a liquid is rotated about a vertical axis passing through one of its arms, a pressure difference is created due to the centrifugal force.

The pressure at a point in the rotating fluid is given by:

P = P₀ + ρ ω² x² / 2

where:

ρ = density of liquid

ω = angular velocity of rotation

x = distance from the axis of rotation

For equilibrium, the pressure difference between the two arms is balanced by the hydrostatic pressure difference due to height variation.

Calculation:

The height difference H between the two liquid columns can be obtained by equating the pressure difference:

ρ g H = (ρ ω² L²) / 2

Solving for H:

H = (ω² L²) / (2 g)

∴ The difference in the levels of the liquid column is (ω² L²) / (2 g).

Explanation:

Bernoulli's Equation:

Bernoulli's equation is obtained by integrating the Euler's equation of motion which is given by-

\(\frac{{dp}}{ρ } + gdz + vdv = 0\)       ....(1)

In Euler's equation of motion, the forces due to gravity and pressure are taken into consideration and which is derived considering the motion of a fluid element along a stream-line.

Integrating the above equation(1):

\(\smallint \frac{{dp}}{ρ } + \smallint gdz + \smallint vdv = 0\)

\(\frac{p}{ρ } + gz + \frac{{{v^2}}}{2} = C\)

\(\frac{p}{{ρ g}} + \frac{{{v^2}}}{{2g}} + z = C\)        ....(2)

where p/ρg = pressure head or pressure energy per unit weight, v2/2g = kinetic head or kinetic energy per unit weight z = potential head or potential energy by unit weight.

Following assumptions are made in the derivation of Bernoulli's equation:

1. Flow is ideal.
2. Flow is steady i.e. time variation is zero.
3. Flow is incompressible i.e. ρ is constant.
4. Flow is irrotaional i.e. ωx = ωy = ωz = 0.

Concept:

Given that:

\(Q = 0.124 \, \text{m}^3/\text{min}, \quad P = 18 \, \text{kW}\)

Power: = \(\rho g H Q\)

Pressure = \(\frac{18 \times 10^3 \times 60}{0.124}\) = 8709 kPa

CONCEPT:

Bernoulli's principle: For a streamlined flow of an ideal liquid in a varying cross-section tube the total energy per unit volume remains constant throughout the fluid.

• This means that in steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline.

From Bernoulli's principle

\(\frac{{{{\rm{P}}_1}}}{{\rm{\rho }}} + {\rm{g}}{{\rm{h}}_1} + \frac{1}{2}{\rm{v}}_1^2 = \frac{{{{\rm{P}}_2}}}{{\rm{\rho }}} + {\rm{g}}{{\rm{h}}_2} + \frac{1}{2}{\rm{v}}_2^2\)

\(\frac{{\rm{P}}}{{\rm{\rho }}} + {\rm{gh}} + \frac{1}{2}{{\rm{v}}^2} = {\bf{constant}}.\)

EXPLANATION:

• From above it is clear that Bernoulli's equation states that the summation of pressure head, kinetic head, and datum/potential head is constant for steady, incompressible, rotational, and non-viscous flow.
• In other words, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy i.e. the total energy of a flowing system remains constant until an external force is applied.
• So Bernoulli’s equation refers to the conservation of energy.
• All of the above are the measuring devices like Venturimeter, Orifice meter, and Pitot tube meter works on the Bernoulli’s theorem. Therefore option 4 is correct.

Explanation:

Pitot Tube is a device used for calculating the velocity of flow at any point in a pipe or a channel. 

The pitot tube is used to measure velocity at a point.

In the question velocity at the stagnation point is given which is zero. So here stagnation pressure will be the correct answer. Because this stagnation pressure head is used to calculate the velocity at a point.

V = \(\sqrt{2gh}\)

It is based on the principle that if the velocity of flow at a point becomes zero, the pressure there is increased due to the conversion of the kinetic energy into pressure energy. 

Working:

• The liquid flows up the tube and when equilibrium is attained, the liquid reaches a height above the free surface of the water stream
• Since the static pressure, under this situation, is equal to the hydrostatic pressure due to its depth below the free surface, the difference in level between the liquid in the glass tube and the free surface becomes the measure of dynamic pressure \(p_0-p=\frac{\rho V^2}{2} = h\rho g\)  where p₀, p and V are the stagnation pressure, static pressure and velocity respectively at point A
• Such a tube is known as a Pitot tube and provides one of the most accurate means of measuring the fluid velocity
• For an open stream of liquid with a free surface, this single tube is sufficient to determine the velocity, but for a fluid flowing through a closed duct, the Pitot tube measures only the stagnation pressure and so the static pressure must be measured separately.

Mistake PointsIn the option velocity at the stagnation point is mentioned, at the stagnation point velocity is already zero there is no need to measure velocity at the stagnation point. the pitot tube is used to measure velocity at any point by measuring the stagnation pressure. Hence the best possible option out of the provided options is option B.

Concept:

Continuity equation:

A1V1 = A2V2

Bernoulli equation:

\(\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + z = {\rm{\;constant}}\)

Calculation:

Given:

P1 = 2.5 kPa, P2 = 1.5 kPa, V₁ = 2 m/s, V2 = ?

Bernoulli’s equation:

\(\frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {Z_2}\)

\(\therefore \frac{{{P_1}}}{{\rho g}} + \frac{{{V_1^2}}}{{2g}} = \frac{{{P_2}}}{{\rho g}} + \frac{{{V_2^2}}}{{2g}}\)

\(\therefore \frac{{{2.5 \times 10^3}}}{{1000 \times g}} + \frac{{{2^2}}}{{2g}} = \frac{{{1.5 \times 10^3}}}{{1000 \times g}} + \frac{{{V_2^2}}}{{2g}}\)

\(\therefore 2.5 + 2 = 1.5 + \frac{{{V_2^2}}}{{2}}\)

\(\therefore V_2^2=6\)

\(\therefore V_2=\sqrt6~m/s\)

Concept:

Bernoulli's Equation:

" In an ideal incompressible fluid when the flow is steady and continuous, the sum of pressure energy, kinetic energy, and potential(or datum) energy is constant along a streamline." 

\({P\over ρ g}+{v^2\over 2g}+z=Constant\)

where \({P\over ρ g}=Pressure\, head\)

\({v^2\over 2g}=Kinetic\, head\)

z = Head from datum line

ρ = Density of fluid

g = Acceleration due to gravity

v = velocity of the fluid

Calculation:

Given data:

Diameter of pipe (D) = 6 cm

Pressure (P) = 200 × 10³ N/m²

Mean velocity (v) = 2.0 m/s

Total head or total energy per unit weight (H) =?

\(Total\, head(H)={P\over ρ g}+{v^2\over 2g}+z\)

\(Total\, head(H)={200\times 10^3\over 1000\times 9.81}+{2^2\over 2\times 9.81}+6\)

\(Total\, head(H)=20.387+0.203+6=26.59\, m\)

\(Total\, head(H)=26.59\, m\)

Total head or total energy per unit weight (H) = 26.59 m

Concept:

Bernoulli’s equation is given by:   

\(\frac{P}{{\rho g}} + \frac{{{V^2}}}{{2g}} + Z = constant\)

where,

\(\frac{P}{{\rho g}} = \) Pressure head, \(\frac{{{V^2}}}{{2g}} = \) Kinetic head,   Z = Datum head

Calculation:

Given:

d = 5 cm, P = 20 N/cm²,  V = 2 m/s

Kinetic head = \(\frac{V^2}{2g}=\frac{2^2}{2\times9.81}=0.204\;m\)

Explanation:

Bernoulli’s equation

Bernoulli’s Equation states that in a steady, ideal flow of an incompressible fluid, the total energy at any point of the fluid is constant. The total energy consists of pressure energy, kinetic energy, and potential energy or datum energy.

\(\frac{p}{\rho } + \frac{{{v^2}}}{2} + gZ =Constant\)

• Bernoulli's equation was derived on the assumption that fluid is non-viscous and therefore frictionless.
• The Bernoulli Equation can be considered to be a statement of the conservation of energy principle appropriate for flowing fluids.
• It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.
• It is represented in head form (the total energy per unit weight).
• Bernoulli's equation can be obtained by integrating Euler's Equation of motion.
• But all real fluid is viscous and hence offers resistance flow.
• Thus there are always some losses in fluid flow and hence in the application of Bernoulli's equation, these losses are taken into consideration.

For a compressible fluid, the density of the fluid (ρ) is not constant.

General Bernoulli's equation is:

\(\int\frac{dp}{ρ}~+~\int Vdv~+~\int gdz~=~constant~(c)\\ \int Vdv~=~\frac{V^2}{2}\\ \int gdz~=~gz\)

For adiabatic conditions:

We know that,

 \(\frac{P}{ρ ^k}~=~C\\ ρ~=~(\frac{P}{C})^{\frac{1}{k}}\)

Now put the value of ρ,

\(\int \frac{dP}{\rho}~=~\int \frac{dP}{P^{\frac{1}{k}}}~C^{\frac{1}{k}}~=~\int P^{\frac{-1}{k}} ~C^{\frac{1}{k}}~dP~=~(\frac{P^{\frac{-1}{k}~+~1}}{\frac{-1}{k}~+~1})~C^{\frac{1}{k}}~=~(\frac{k}{k~-~1})~\frac{P}{\rho}\)

\(i.e.~\int \frac{dP}{\rho}~=~(\frac{k}{k~-~1})~\frac{P}{\rho}\)

∴ Bernoulli's equation is:

\((\frac{k}{k~-~1})~(\frac{P}{\rho })~+~\frac{V^2}{2}~+~gz~=~C\)

⇒ \((\frac{k}{k~-~1})~(\frac{P}{\rho g})~+~\frac{V^2}{2g}~+~z~=~C\)

Explanation:

Bernoulli's equation 

\(\frac{v^2}{2g}+\frac{p}{\rho g}+z=Constant\)

Where, 

 \(\frac{v^2}{2g}\) = velocity head

\(\frac{P}{\rho g}\) = pressure head

z = datum head

The Sum of datum head and pressure head is called the piezometric head.

As per Bernoulli’s equation sum of the piezometric head and velocity head is constant during the flow.

So, for a constant piezometric head between two points velocity will remain constant.

Explanation:

Bernoulli’s equation:

\(\frac{P}{\rho g} + \frac{{{v^2}}}{{2g}} + Z = Constant\)

• It can be derived from the principle of conservation of energy.
• It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.
• It represented in head form (the total energy per unit weight).

Following assumption are made in deriving the Bernoulli’s equation:

1. Fluid is inviscid i.e. zero viscosity.
2. Flow is in steady-state.
3. Flow is incompressible.
4. Flow is irrotational.
5. Flow is along the streamline.
6. Only gravity and pressure force acting on the liquid, no other external force.

Important Points

Even if the flow is rotational, we can still apply the Bernoulli’s equation but only along a streamline.

Explanation:

Bernoulli's equation is obtained by integrating Euler's equation of motion which is given by-

Euler's Equation:

\(\frac{{dp}}{ρ } + gdz + vdv = 0\;\;\;\;\;(1)\)

In Euler's equation of motion, the forces due to gravity and pressure are taken into consideration and which is derived considering the motion of a fluid element along a stream-line.

Integrating the above eq(1):

\(\smallint \frac{{dp}}{ρ } + \smallint gdz + \smallint vdv = 0\)

\(\frac{p}{ρ } + gz + \frac{{{v^2}}}{2} = C\)

\(\frac{p}{{ρ g}} + \frac{{{v^2}}}{{2g}} + z = C\;\;\;\;(2)\)

This equation is called Bernoulli's Equation.

where \(\frac{p}{ρg}\) = pressure head or pressure energy per unit weight, \(\frac{v^2}{2g}\) = kinetic head or kinetic energy per unit weight z = potential head or potential energy by unit weight.

Following assumptions are made in the derivation of Bernoulli's equation:

1. Flow is ideal i.e inviscous.
2. Flow is steady i.e. time variation is zero.
3. Flow is incompressible i.e. ρ is constant.
4. Flow is irrotaional i.e. ωx = ωy = ωz = 0.
5. Gravity and Pressure forces are taken only hence all other external forces should be zero.
6. The energy of the system is constant hence there should be no loss of energy.

Concept:

The total energy at a point A (in m) = \(\frac{Pa}{\rho g}+\frac{V^{2}}{2g}\)+z

Where, 

\(\frac{Pa}{\rho g}\)is pressure head

\(\frac{V^{2}}{2g}\) is velocity head.

Calculation:

Given, 

Diameter = 0.3 m

Specific gravity =0.8

Flow velocity = 1.5 m/s 

Pressure at point A = 20 kN/m2 or 20 kPa

Datum head = 3 m

So, The total energy at a point A (in m) = \(\frac{Pa}{\rho g}+\frac{V^{2}}{2g}\)+z 

= \(\frac{20*1000}{0.8*1000*9.81 g}+\frac{1.5^{2}}{2*9.81}+3\) 

= 2.54 +0.11 + 3 =5.65 m, which is near to option 3