Turbulent Flow MCQ Quiz - Objective Question with Answer for Turbulent Flow - Download Free PDF

Last updated on Jun 11, 2025

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Latest Turbulent Flow MCQ Objective Questions

Turbulent Flow Question 1:

Rouse distance (x) required for establishment of a fully developed turbulent flow in a pipe of diameter D is given by

  1. \(\rm \frac{x}{D}=(50)^2\)
  2. \(\rm \frac{x}{D}= \sqrt{50}\)
  3. \(\rm x= \frac{50}{D}\)
  4. \(\rm \frac{x}{D}=50\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{x}{D}=50\)

Turbulent Flow Question 1 Detailed Solution

Explanation:

Rouse Distance (x) for Fully Developed Turbulent Flow:

  • The Rouse distance is the length of pipe required for the flow to transition from an initial state to a fully developed turbulent flow. When fluid flows through a pipe, it takes a certain distance for the velocity profile to stabilize and for turbulence to reach a fully developed state. This distance depends on the pipe diameter (D) and the Reynolds number.

xD=50" id="MathJax-Element-374-Frame" role="presentation" style="position: relative;" tabindex="0">Empirical Relation:

\( \frac{x}{D} = 50 \)

  • This relationship describes the Rouse distance for the establishment of fully developed turbulent flow in a pipe. The dimensionless ratio xD" id="MathJax-Element-375-Frame" role="presentation" style="position: relative;" tabindex="0">xD signifies the distance required for the flow to stabilize in terms of pipe diameters. This is consistent with empirical studies and engineering principles, which indicate that turbulent flow typically becomes fully developed after traveling a length of approximately 50 pipe diameters
  • When fluid enters a pipe, the flow profile is disturbed due to the boundary layer development near the pipe walls. The transition from laminar or disturbed flow to fully developed turbulent flow occurs over a distance proportional to the pipe diameter. The length required depends on flow conditions such as the Reynolds number and pipe roughness.

Importance in Engineering Applications:

  • Ensuring fully developed turbulent flow is critical for accurate calculations of pressure drop and flow rates.
  • Designing pipe lengths based on xD" id="MathJax-Element-378-Frame" role="presentation" style="position: relative;" tabindex="0">xD helps optimize fluid transport systems, ensuring efficient and predictable performance.
  • Understanding the Rouse distance aids in the placement of flow measurement devices, ensuring data is collected under stable flow conditions.

Turbulent Flow Question 2:

The intensity of turbulence flow refers to -

  1. The turbulent, eddy or Reynolds stresses
  2. Average kinetic energy of turbulence per unit mass
  3. A correlation between the fluctuating velocity components
  4. Root mean square of turbulent velocity fluctuations

Answer (Detailed Solution Below)

Option 4 : Root mean square of turbulent velocity fluctuations

Turbulent Flow Question 2 Detailed Solution

Explanation:

Intensity of turbulence flow:

In fluid dynamics, turbulence is characterized by chaotic changes in pressure and flow velocity. The intensity of turbulence is an important parameter that helps in quantifying the level of turbulence in the flow. The correct answer to the question is:

Root mean square of turbulent velocity fluctuations

  • The intensity of turbulence flow is measured by calculating the root mean square (RMS) of the velocity fluctuations from the mean flow velocity. This gives a measure of the average magnitude of the fluctuations, which is a direct indicator of the turbulence intensity.

The turbulence intensity, also often referred to as turbulence level, is defined as:
\(I = \frac{u'}{U}\)
where u' is the root-mean-square of the turbulent velocity fluctuations and U  is the mean velocity.
\(u' = \sqrt{\frac{1}{3} \left( u_x'^2 + u_y'^2 + u_z'^2 \right)} = \sqrt{\frac{2}{3}k} \)
\(U = \sqrt{U_x^2 + U_y^2 + U_z^2}\)

Turbulent Flow Question 3:

Identify the flow shown in the following figure.
qImage6735ea50d11fb14de6de2b4b

  1. Turbulent flow
  2. Laminar flow
  3. Recoil flow
  4. Transient flow

Answer (Detailed Solution Below)

Option 1 : Turbulent flow

Turbulent Flow Question 3 Detailed Solution

Explanation:
In the given figure:

  • The flow lines are irregular and wavy.
  • There are random vortices and swirling eddies shown throughout the flow.
  • The flow direction changes rapidly, and the streamlines are no longer smooth and parallel.
  • These are classic characteristics of turbulent flow, where:
    • Fluid particles move in a chaotic and disorderly manner.
    • There is cross mixing and rotational motion.
    • The Reynolds number is typically greater than 4000.

In contrast:

  • Laminar flow would have smooth, parallel streamlines.
  • Transient flow occurs during the shift from laminar to turbulent, and wouldn’t be this chaotic.
  • Recoil flow is not a standard term in fluid dynamics for describing this pattern.

qImage67f7c91d08fa0d93dbd5f76b

Turbulent Flow Question 4:

In a glass tube with water flowing, a liquid dye with the same specific weight is introduced. The resulting dye filament pattern is observed as shown in the figure. Identify the type of flow from the following options.
qImage6735ea32641ba37d353303ec

  1. Laminar flow
  2. Recoil flow
  3. Transient flow
  4. Turbulent flow

Answer (Detailed Solution Below)

Option 4 : Turbulent flow

Turbulent Flow Question 4 Detailed Solution

Explanation:

In the given image, the dye filament introduced into the flow becomes diffused and spreads irregularly throughout the fluid. This is a clear indication of turbulent flow, where fluid particles move in a chaotic manner and mixing occurs in all directions.

Characteristics of Turbulent Flow:

  • Irregular and chaotic fluid motion.
  • High Reynolds number (typically > 4000).
  • Dye spreads out diffusely, as seen in the image.

In contrast:

  • Laminar flow would show the dye filament moving smoothly in a straight line.
  • Recoil flow and Transient flow are not standard terms in this context, and do not match the observed behavior.

Turbulent Flow Question 5:

The frictional resistance for turbulent flow is

  1. proportional to the density of fluid
  2. dependent of pressure
  3. independent to the area of surface in contact
  4. independent to the nature of surface

Answer (Detailed Solution Below)

Option 1 : proportional to the density of fluid

Turbulent Flow Question 5 Detailed Solution

Explanation:

  • When the water is flowing in a pipe, it experiences some resistance to its motion, whose effect is to reduce the velocity and ultimately the head of water available.
  • Though there are many types of losses, the major loss is due to the frictional resistance of the pipe only.
  • The frictional resistance of a pipe depends upon the roughness of the inside surface of the pipe, the density, and the velocity of water.
  • It has been experimentally found that the more the roughness of the inside surface of the pipe, the greater the resistance.
  • This friction is known as fluid friction and resistance is known as frictional resistance.
  • The earlier experiments on fluid friction were conducted by ‘Froude who concluded that:
  1. The frictional resistance varies approximately with the square of the velocity of the liquid.
  2. The frictional resistance vanes with the nature of the surface.

Top Turbulent Flow MCQ Objective Questions

In the moody diagram is shown in the figure below, the friction factor for turbulent flow in the smooth pipe is given by the curve

F9 Tabrez 12-5-2021 Swati D1

  1. A
  2. B
  3. C
  4. D

Answer (Detailed Solution Below)

Option 4 : D

Turbulent Flow Question 6 Detailed Solution

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Explanation

Moody’s diagram is used to calculate the friction factor of commercial pipes.

It is drawn between friction factor and Reynolds number for various relative roughness.

It is derived for the circular pipe but applicable to other cross-sections as well provided that the diameter is replaced by 4 times the hydraulic radius.

Thus the moody’s chart is a graphical representation between the friction factor, relative roughness, and Reynolds number

F9 Tabrez 12-5-2021 Swati D4

From the above Moody diagram, we can  conclude that:

F9 Tabrez 12-5-2021 Swati D1

  • Curve A represents laminar flow.
  • Curve B and C represent Turbulent flow in the rough pipe.
  • Curve D and E represent turbulent flow in smooth pipe.

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, II
  2. II, IV
  3. II, III
  4. I, IV

Answer (Detailed Solution Below)

Option 4 : I, IV

Turbulent Flow Question 7 Detailed Solution

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Explanation:

  • The Reynold number for the flow through the pipe can be defined as,

\({\rm{Re}} = \frac{{{\rm{ρ VD}}}}{{\rm{μ }}}\)

  • where, ρ = density of the fluid, V = velocity of the fluid, D = characteristic length (Diameter of the circular pipe). μ = dynamic viscosity
  • For Turbulent flow through a pipe ⇒ ReD > 2300 
  • The turbulent flow is considered as a well-mixed flow because in it there is a huge order of intermixing of fluid particles and due to intermixing, they mixed with each other very strongly.

Additional Information

PLATE

Laminar Flow

Re < 5×105

Turbulent Flow

Re > 5×105

PIPE

Laminar Flow

Re < 2000

Turbulent Flow

Re > 4000

A fluid density 1200 kg/m3 and viscosity 0.8 poise is flowing at rate of 10 m3/min in circular pipe of a cross-section of 1 m2. Is the flow laminar or turbulent? Also find maximum velocity.

  1. Transition, 0.33
  2. Transition, 0.90
  3. Turbulent, 0.33
  4. Laminar, 0.80

Answer (Detailed Solution Below)

Option 1 : Transition, 0.33

Turbulent Flow Question 8 Detailed Solution

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Concept:

Laminar flow through a circular pipe follows parabolic velocity distribution.

\(Re=\frac{\rho VD}{\mu}\)

For the flow through a circular pipe,

Laminar flow – Reynold’s Number is less than 2000

Turbulent Flow – Reynold’s Number is greater than 4000

Transition Flow – Reynold’s Number is between 2000 to 4000

For a fully developed laminar viscous flow through a circular pipe, the maximum velocity is equal to twice the average velocity.

i.e. \({U_{max}} = 2 × {V_{avg}}\)

\({U_{max}} = \frac{1}{{4μ }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}\)

\(U_{avg}=V = \frac{1}{{8μ }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}\)

Hence \(\frac{{{U_{max}}}}{{V}} = 2\)

Calculation:

Given:

ρ =1200 kg/m3 , μ = 0.8 Poise = 0.08 kg/ms, Q = 10 m3/min = 1/6 m3/sec, A = 1 m2

Discharge, \(Q=A \times V=\frac{\pi}4{D^2} \times V\)

\(\frac16 =1 \times V\)

\(V = \frac 16~m/s\)

\(\frac 16=\frac{\pi}{4} \times D^2 \times \frac 16\)

\(D=1.128~m\)

\(Re=\frac{\rho VD}{\mu}=\frac{1200 × \frac16 ×1.128}{0.08}=2820\)

If Reynold’s Number is between 2000 to 4000, then the flow is Transition Flow.

\(\frac{{{U_{max}}}}{{V}} = 2 \Rightarrow U_{max}=2V\)

\(\Rightarrow U_{max}=2V=2 \times \frac16=\frac13=0.33\)

26 June 1

For the fully developed laminar flow through the parallel plates, the maximum velocity is equal to the (3/2) times of the average velocity.

i.e. \({U_{max}} = \frac{3}{2} × {U_{avg}}\)

In turbulent flow in a pipe, we know the 

  1. Reynolds number is greater than 10,000 
  2. fluid particles move in straight line
  3. head loss varies linearly with flow rate 
  4. shear stress varies linearly with radius

Answer (Detailed Solution Below)

Option 4 : shear stress varies linearly with radius

Turbulent Flow Question 9 Detailed Solution

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Explanation:

Turbulent Flow

  • Turbulent flow is the random, disordered and dis-organised flow which has bulk and or macroscopic mixing. It occurs at higher flow velocities compared to laminar flow. In turbulent flow, inertia forces are significant as compared to viscous forces.
  • For flow in the pipes if Reynold's number is less than 2000 the flow is called the laminar and if it is more than 4000, the flow is called turbulent flow. If the Reynolds number lies between 2000 and 4000 the flow may be laminar or turbulent  ( also known as transition period).
  • In Turbulent pipe flow shear stress is given by,

  • \(τ = {τ _o}\left( {\frac{r}{R}} \right) = {τ _o}\left( {\frac{{R - y}}{R}} \right) = {τ _o}\left( {1 - \frac{y}{R}} \right)\)
  • where τo = Turbulent shear stress at the pipe boundary i.e at y = 0 or r = R
  • Hence in the turbulent pipe flow shear stress varies linearly with the radius.

SSC JE MEchanical 2 10

F1 Ashik Madhu 13.10.20 D1

The Blasius equation for friction factor f in turbulent flow through pipes relates f to the Reynolds number Re as f = ______.

  1. 64/Re
  2. 0.316/Re1/4
  3. 1.328/Re1/2
  4. 0.316/Re1/5

Answer (Detailed Solution Below)

Option 2 : 0.316/Re1/4

Turbulent Flow Question 10 Detailed Solution

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Concept:

Blasius result for different flow conditions:

For laminar flow:

Friction factor (f) \(= \frac{{64}}{{{{\rm{R}}_{\rm{e}}}}}\)

For turbulent flow:

Case I: Smooth Pipe

Friction factor (f) \( = \frac{{0.316}}{{{{\left( {{{\rm{R}}_{\rm{e}}}} \right)}^{1/4}}}}\)

Case II: Rough Pipe

\(\frac{1}{{\sqrt {\rm{f}} }} = 2{\log _{10}}\left( {\frac{{\rm{r}}}{{\rm{k}}}} \right) + 1.74\)

Case III: In between smooth & rough (Transition)

Friction (f) = \(\rm{\phi (R_e, \frac{k}{D})}\)

It is the function of Reynold's No and roughness depth value. The exact value is obtained from Moody's Diagram.

Additional Information

Blasius results for Laminar and Turbulent flow in boundary layer:

Laminar Turbulent
 1. \(\frac{\delta }{x} = \frac{5}{{\sqrt {{{{\mathop{\rm Re}\nolimits} }_x}} }}\)  1. \(\frac{\delta }{x} = \frac{{0.376}}{{{{\left( {{{{\mathop{\rm Re}\nolimits} }_x}} \right)}^{1/5}}}}\)
 2. \({C_{fx}} = \frac{{0.664}}{{{{\left( {{R_{ex}}} \right)}^{1/2}}}}\)  2. \({C_{fx}} = \frac{{0.059}}{{{{\left( {{R_{ex}}} \right)}^{1/5}}}}\)
 3. \({C_D} = \frac{{1.328}}{{{{\left( {{R_{eL}}} \right)}^{1/2}}}}\)  3. \({C_{fx}} = \frac{{0.074}}{{{{\left( {{R_{eL}}} \right)}^{1/5}}}}\)

For fully developed turbulent flow though rough pipe at very large Reynolds numbers, friction factor is

  1. Dependent on Reynolds numbers only
  2. Dependent on relative roughness of pipe only
  3. Independent of both Reynolds numbers and relative roughness of pipe
  4. Dependent on density of fluid only

Answer (Detailed Solution Below)

Option 2 : Dependent on relative roughness of pipe only

Turbulent Flow Question 11 Detailed Solution

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Explanation:

The Colebrook–White equation, sometimes referred to simply as the Colebrook equation is a relationship between the friction factor and the Reynolds number, pipe roughness, and inside diameter of pipe.

For full flow (closed conduit)

\(\frac{1}{{\sqrt f }} =-2log[\frac{K}{3.7D} + \frac{2.51}{Re\sqrt{f}}]\)

Referring to the diagram,

F1 Tabrez Madhuri 04.03.2022 D1

  • For laminar flow friction factor depends only upon the Reynolds number of the flow and is independent of the contact surface. For laminar flow Friction factor,

⇒ \(f = \frac{{64}}{{Re}}\) 

  • Friction factor (f) for turbulent flow through rough pipe is a function of relative roughness only. For turbulent flow, It can be seen from the above graph that 'f' increases with relative surface roughness.
  • Friction factor (f) for turbulent flow rough pipe is a function of relative roughness only.

\(\frac{1}{{\sqrt f }} = 2\log \left( {\frac{R}{K}} \right) + 1.74\)

If k is the height of roughness projection and δ the thickness of laminar sub layer, for a pipe surface to be hydraulically smooth, condition is:

  1. k/δ < 0.25
  2. 1.0 < k/δ < 1.5
  3. 0.5 < k/δ < 1.0
  4. 0.25 < k/δ < 0.5

Answer (Detailed Solution Below)

Option 1 : k/δ < 0.25

Turbulent Flow Question 12 Detailed Solution

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Concept:

Hydro-dynamically smooth:

If the average height of irregularities (k) is much lesser than the thickness of the laminar sub-layer (δ), then the boundary is called hydro-dynamically smooth.

Hydro-dynamically rough:

If the average height of irregularities (k) is much greater than the thickness of the laminar sub-layer (δ), then the boundary is called hydro-dynamically rough.

According to NIKURDE's Experiment, the boundary is classified as:

Hydrodynamically smooth when

\(\frac{{\bf{k}}}{{\bf{\delta }}} < 0.25\)

Boundary transition condition, when

\(0.25 < \frac{{\bf{k}}}{{\bf{\delta }}} < 6\)

Hydrodynamically rough when

\(\frac{{\bf{k}}}{{\bf{\delta }}} > 6\)

In a fully developed rough–turbulent regime in pipe flow:

  1. rough and smooth pipes have the same friction factor
  2. the laminar sublayer is thicker than the roughness projections
  3. the friction factor is independent of the Reynolds number
  4. the friction factor is independent of the relative roughness

Answer (Detailed Solution Below)

Option 3 : the friction factor is independent of the Reynolds number

Turbulent Flow Question 13 Detailed Solution

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Explanation:

For rough pipe, the relation between friction factor and surface roughness of pipe is based on Nikursade’s experiment and it is given by:

\(\dfrac{1}{{\sqrt f }} = 2\ {\log _{10}}\left( {\dfrac{D}{{2K}}} \right) + \;1.74\) ;

Relative Roughness = K/D

It is clear from above that friction factor is the non-linear function of relative roughness (K/D).

For Smooth pipe, the relation between friction factor and Reynolds No. of pipe is based on Nikursade’s experiment and it is given by:

\(\dfrac{1}{{\sqrt f }} = 2\ {\log _{10}}\left( {Re\sqrt f } \right) - 0.8\) ;

It is clear from above that friction factor is the non-linear function of Reynolds Number for the smooth pipe.

Note:

(i) For rough turbulent regime, friction factor depends on the relative roughness whereas, for smooth turbulent regime, friction factor depends only on the Reynolds number.

In turbulent flow in a pipe

  1. Shear stress varies linearly with radius
  2. head loss varies linearly with a flow rate
  3. Fluid particles move in a straight line
  4. Reynolds number is less than 1000

Answer (Detailed Solution Below)

Option 1 : Shear stress varies linearly with radius

Turbulent Flow Question 14 Detailed Solution

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Explanation:

Shear stress in turbulent flow:

Perhaps the first thought that comes to mind is to determine the shear stress in an analogous manner to laminar flow from

\(τ =\mu\frac{du}{dr}\)

where u (r) is average velocity profile for turbulent flow.

But the experimental studies show that this is not the case, and the shear stress is much larger due to the turbulent fluctuations.

Therefore, it is convenient to think of the turbulent shear stress as consisting of two parts:

The laminar component, which accounts for the friction between layers in the flow direction 

\(τ_{laminar} =\mu\frac{du}{dr}\)

And the turbulent component, which accounts for the friction between the fluctuating fluid particles and the fluid body.

Total shear stress in a flowing fluid in a pipe is given by:

\(τ=τ_v\;+\;τ_t\)

where τv = shear stress due to viscosity and τt = shear stress due to turbulence.

F1 Ashik Madhu 13.10.20 D1

The shear stress can be treated as linear in both axes i.e zero at the centre and varying linearly up to the wall.

Additional Information

 Reynolds Expression for Turbulent Shear stress:

Reynold's through his experiment developed an expression between two layers of a fluid at a small distance apart, which is given as:

\(τ_t=\rho{u'}{v'}\)

where u', v' = fluctuating component of velocity in the direction of x and y respectively.

u' and v' are both varying and hence τ will also vary. To calculate τ, time average is done on both sides of the equation.

\(τ_t=\bar{\rho}\bar{u}'\bar{v}'\)

Due to difficulty in calculating time-varying velocities, Prandtl presented a mixing length hypothesis that can be used to express turbulent shear stress in terms of measurable quantities.

According to Prandtl, the mixing length l, is that distance between two layers in the transverse direction such that lumps of fluid particles from one layer could reach the other layer and the particles are mixed in the other layer in such a way that the momentum of the particles in the direction of x is same.

\(u'=l\frac{du}{dy}\;and\;v'=l\frac{du}{dy}\)

The above equation gives velocity fluctuation in x and y direction in terms of mixing length l.

\(\bar{u}\;\times\;\bar{v}=l\frac{du}{dy}\;\times\;l\frac{du}{dy}\Rightarrow\;l^2(\frac{du}{dy})^2\)

∴ turbulence shear stress is given by:

\(τ_t=\bar{\rho}\bar{u}'\bar{v}'\Rightarrow\rho{l^2}(\frac{du}{dy})^2\)

Total shear stress acting on the fluid is:

\(τ=τ_v\;+\;τ_t\)

\(τ=\mu\frac{du}{dy}\;+\;\rho{l^2}(\frac{du}{dy})^2\)

In the case of turbulent flow, viscous shear stress is negligible except near the boundary. Hence, it can be assumed that shear stress is purely dominated by \(τ=\rho{l^2}(\frac{du}{dy})^2.\)

∵ mixing length l is the distance measured in the transverse direction (perpendicular to flow) where lumps reach the other layer, and particles are mixed in the longitudinal direction (parallel to flow), the shear stress can be treated as linear in both axes i.e zero at centre and varying linearly up to the wall.

Important Points

Laminar flow

Turbulent flow

Laminar flow is the flow in which fluid particles flow in layers. Each layer moves smoothly past the adjacent layer with little or no intermixing. 

Turbulent flow is the flow in which fluid particles flow in a random fashion where intermixing/intermingling is present.

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The entire fluid layers move parallel to each other. They can’t cross each other.

Fluid layers are not parallel to each other. They can cross each other.

Reynolds’s number for a pipe is less than 2000 for laminar flow.

Reynolds’s number for a pipe is more than 4000 for turbulent flow.

The instantaneous stream-wise velocity of a turbulent flow is given as follows:

\(u\left( {x,\;y,\;z,\;t} \right) = \bar u\;\left( {x,\;y,\;z} \right) + u'\left( {x,\;y,\;z,\;t} \right)\)

The time-average of the fluctuating velocity u’ (x, y, z, t) is

  1. u’/2
  2. \(- \bar u/2\)
  3. zero
  4. \(\bar u/2\)

Answer (Detailed Solution Below)

Option 3 : zero

Turbulent Flow Question 15 Detailed Solution

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Concept:

For turbulent flow

\(u = \bar u + u'\)

\(\bar{u}=\frac{1}{T} \int_0^T u d t\)

where,

u = velocity in x-direction at any instant.

\(\bar u\) = Average velocity in x-direction.

u’ = Fluctuating component of velocity in x-direction.

Time average of fluctuating component of velocity,

\(\overline{u^{\prime}}=\frac{1}{T} \int_0^T u^{\prime} d t=\frac{1}{T} \int_0^T(u-\bar{u}) d t\)

\(\overline{u^{\prime}}=\frac{1}{T} \int_0^T u d t-\frac{1}{T} \int_0^T \bar{u} d t=\bar{u}-\left[\frac{1}{T} \times(\bar{u} T)\right]=0\)

Important Points

Interaction effect between two fluctuating components over a long period is non-zero.

\(\mathop \smallint \limits_{ - T}^T u'v'dt \ne 0\)

where v’ → fluctuating component of velocity in y-direction

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