Laminar Flow MCQ Quiz in বাংলা - Objective Question with Answer for Laminar Flow - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

Shear stress at any distance ‘r’ from the center of the pipe is given by.

\(τ = \frac{{ - r}}{2}\left( {\frac{{\partial p}}{{\partial x}}} \right)\)

At r = R, i.e. at the, pipe wall, shear stress is maximum and is given by

\({τ _{max}} = \frac{{ - R}}{2}\left( {\frac{{\partial p}}{{\partial x}}} \right)\)

Where, R = Radius of the pipe

\(\frac{{\partial p}}{{\partial x}}\) = Pressure gradient over the length of the pipe.

So, from the above, τ ∝ R

Calculation:

Given:

τ = 20 Pa, r = 5 cm, R = 10 cm

\(\frac{τ _{max}}{τ} = \frac Rr\)

\(\frac{τ _{max}}{20} = \frac {10}{5}\)

τ_max = 40 Pa 

Explanation:

Laminar 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. There is no intermingling of fluid particles across the cross-section. So laminar flow resembles streamline flow.

Laminar flow through a circular pipe follows parabolic velocity distribution.

From the above diagram, we can see that the velocity distribution in laminar flow is parabolic and the velocity distribution in Turbulent flow is Logarithmic. As in the question figure the velocity distribution parabolic, Hence the given flow is laminar.

Additional Information

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

Explanation:

Reynolds Number

Reynold number is a dimensionless number that helps to predict flow patterns in different fluid flow situations.

\({\rm{Re}} = \frac{{{\rm{Inertia\;force}}}}{{{\rm{Viscous\;force}}}} = {\rm{\;}}\frac{{{\rm{\rho V}}{{\rm{L}}_{\rm{c}}}}}{{\rm{\mu }}} = \frac{{{\rm{V}}{{\rm{L}}_{\rm{c}}}}}{{\rm{\nu }}}\)

Where,

Re = Reynolds number, ρ = density, V = velocity of flow

μ = dynamic viscosity ν = kinematic viscosity, LC = characteristic linear dimension

For pipe flow:

LC = diameter of pipe = D

Reynold number = \(Re = \frac{{{\rm{\rho VD}}}}{\mu } = \frac{{{\rm{VD}}}}{\nu }\)

• Laminar flow  Re ≤ 2000
• Transition flow 2000 ≤ Re ≤ 4000
• Turbulent flow Re ≥ 4000

Additional Information

The Reynolds number value below which flow can be definitely considered to be Laminar flow is called  Critical Reynolds number Rc.

The values of the critical Reynolds number are listed below:

• For Pipe flow Rc = 2300
• For Flow between plate Rc = 1000
• For open channel flow Rc = 500
• For flow over plate Rc = 5 × 105
• For flow over sphere Rc = 1

Different dimensionless numbers:

• \(Weber\;Number = \frac{{Inertia\;force}}{{Surface\;tension\;force}}\)
• \(Reynold's\;number = \frac{{Inertia\;force}}{{Viscous\;force}}\)
• \(Mach\;number = \frac{{Inertia\;force}}{{Elastic\;force}}\)
• \(Euler's\;number = \frac{{Inertia\;force}}{{Pressure\;force}}\)
• \(Froude\;number = \frac{{Inertia\;force}}{{Gravity\;force}}\)
• Biot number  → Ratio of internal thermal resistance to boundary layer thermal resistance
• Grashof number  → Ratio of buoyancy to viscous force
• Prandtl number  → Ratio of momentum diffusivity (ν) and thermal diffusivity (α). Pr =ν/α 
• Reynolds number  → Ratio of inertia force to viscous force
• Nusselt number → Ratio of Convective heat transfer to Conduction heat transfer
• \(Nu = \frac{{h{L_c}}}{k}\)

Explanation:

Concept:

Head loss due to friction in a pipe:

\({h_f} = \frac{{f'L{V^2}}}{{2gd}}\)

f’ = Friction factor, L = Length of the pipe, V = average velocity of the fluid in the pipe, g = Acceleration due to gravity, d = Diameter of the pipe

Calculations:

Given:

f’ = 4f = 0.06 (f is friction coefficient), L = 9.81 m, V = 2 m/s, g = 9.81 m/s², d = 1 m

\({h_f} = \frac{{f'L{V^2}}}{{2gd}}=\frac{0.06~\times~9.81~\times~4}{2~\times~9.81~\times~1}\)

Concept:

Reynolds Number R_e is defined as the ratio of the Inertia force to viscous force.

Inertia force (F_i) = mass × acceleration

\({{\rm{F}}_{\rm{i}}} = {\rm{ρ }} \times {\rm{volume}} \times \frac{{{\rm{velocity}}}}{{{\rm{time}}}} = {\rm{ρ }} \times {\rm{velocity}} \times \frac{{{\rm{volume}}}}{{{\rm{time}}}} = {\rm{ρ }} \times {\rm{V}} \times {\rm{AV}}\)

Viscous force (F_v) = shear stress × area \(= {\rm{\;}}\tau \times {\rm{A}} = \mu \times \frac{{\rm{V}}}{{\rm{L}}} \times {\rm{A}}\)

Reynolds Number (R_e) = \(\frac{{ρ \times {\bf{V}} \times {\rm{L}}}}{\mu }\)

Calculation:

\({{\rm{R}}_{\rm{e}}} = \frac{{{\rm{ρ }} \times {\rm{V}} \times {\rm{d}}}}{{\rm{\mu }}}\)

ρ = 0.9 g/cm³ = 900 kg/m³, V = 4 m/s, d = 30 cm = 0.3 m, μ = 0.8 poise = 0.08 Ns/m²

\({{\rm{R}}_{\rm{e}}} = \frac{{900 \times 4 \times 0.3}}{{0.08}} = 13500\)

Explanation:

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

Bernoulli’s equation:

• 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 Bernoulli’s equation:

Bernoulli’s equation assumptions:

1. The fluid is ideal, i.e. fluid has zero viscosity
2. Flow is steady
3. Flow is continuous
4. Flow is incompressible
5. Flow is irrotational
6. Flow is along a streamline.

Within a boundary layer for a steady incompressible flow, viscosity is present and the viscous forces dominate over inertia forces.

Thus, the Bernoulli equation does not hold within a boundary layer for a steady incompressible flow.

Explanation:

Reynold's number: 

• It is a dimensionless number that determines the nature of the flow of liquid through a pipe or flat plate. 
• It is defined as the ratio of the inertial force to the viscous force for a flowing fluid.

Reynold's number is written as Re.

\({Re} = \frac{{{\rm{Inertial\;force}}}}{{{\rm{Viscous\;force}}}}\Rightarrow\frac{ρ Vx}{μ}=\frac{Vx}{ν}\)

where ρ = density of fluid, V = velocity of flowing fluid, μ = dynamic viscosity of fluid, ν = kinematic viscosity of fluid and x = characteristic diameter.

where \(x=\frac{4A}{P}\)

and A = area of cross-section and P = perimeter.

Explanation:

Data given,

Velocity of flow, v = 0.15 m/sec

viscosity of water µ = 0.001 Ns/m2

Length of plate, l  = 1 m

Reynold's number = ?

Solution:

We know that,

R_e = \(ρ.v.l \over µ\) = \(1000 \times 0.15 \times 1 \over 0.001\) = 150,000