Turbulent Flow MCQ Quiz in मल्याळम - Objective Question with Answer for Turbulent Flow - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Formula:

Friction factor (F) for turbulent flow in rough pipes,

\(\frac{1}{\sqrt{\text{F}}}=2{{\log }_{10}}\left( \frac{\text{R}}{\text{K}} \right)+1.74\)

Where,

F = Friction factor, R = Radius of pipe, and K = Roughness height

Given:

d = 50 cm, R = 25 cm, and K = 0.25cm

Calculation:

\(\frac{1}{\sqrt{\text{F}}}=2{{\log }_{10}}\left( \frac{25}{0.25} \right)+1.74\)

= 2 log₁₀ 100 + 1.74 = 2 log₁₀ 10² + 1.74 = 2 × 2 + 1.74 = 5.74

\(\therefore \text{F}={{\left( \frac{1}{5.74} \right)}^{2}}=0.03035\)

Other Important formula:

Friction factor (F) for turbulent flow in smooth pipes,

\(\text{F}=\frac{0.316}{{{\left( \text{Re} \right)}^{1/4}}}\)

Where, R_e = Reynold’s Number and ‘F’ for laminar flow, \(\text{F}=\frac{64}{{{\text{R}}_{\text{e}}}}\)

Explanation:

Velocity distribution for turbulent flow in smooth pipe:

\(u = \frac{{{u_\infty }}}{k}\ln y + c\)

Velocity distribution for laminar flow:

\(u = - \frac{1}{{4\mu }}\frac{{\partial p}}{{\partial x}}\left[ {{R^2} - {r^2}} \right]\)

The velocity distribution through circular pipes for different types of flow is as follows:

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

Concept:

Head loss in the pipe due to friction is given as:

\(h_f = \frac {f lV^2}{2gD}=\frac {8}{\pi ^2g}\frac {flQ^2}{D^5} \)

\(Head~loss \propto \frac 1{D^5}\)

Calculation:

Given:

\(d_2 = \frac{d_1}{2}\)

From diameter proportional relation,

\(h_1 \propto \frac{1}{d_1^5}\)

\(h_2 \propto \frac{1}{d_2^5}\)

\(\Rightarrow h_2 \propto \frac{2^5}{(d_1)^5}\)

\(\Rightarrow \frac{h_2}{h_1} = 32\)

h₂ = 32h₁

The frictional drop will change by 32 times.

Explanation:

Turbulent flow – Turbulent flow occurs at relatively larger velocities and is characterized by chaotic behaviour, irregular motion, large mixing and eddies. For such flows, inertial effects are more pronounced than the viscous effects.

• Mathematically the velocity field of turbulent flow is represented as\(V = \bar V + V'\) or the velocity fluctuates at small time scales around a large time-averaged velocity.
• Similarly,  \(P = \bar P + P',T = \bar T + T'\)etc.
• Parameter Reynolds number \(\left( {\frac{{\rho vd}}{\mu }} \right)\) is used to characterize Laminar and Turbulent flow.
• If R_e < 2100 for pipe flow the flow is laminar and if R_e > 10⁴ the flow is turbulent.

For Turbulent flow, the velocity or pressure field may not be exactly (Analytically) represented. So we will solve it using dimensional analysis.

   \({\rm{\Delta }}P = {\rm{\Delta }}P\left( {D,L,\overline {V,} \rho ,\mu ,\varepsilon } \right)\)      where ε = Surface roughness

\(\frac{{{\bf{\Delta }}P}}{{\rho \overline {{V^2}} }} = f\left( {\frac{{\rho vd}}{\mu },\frac{L}{D},\frac{\varepsilon }{D}} \right) = f\left( {{R_e},\frac{L}{D},\frac{\varepsilon }{D}} \right)\),  the turbulent flow through the pipe turbulent pressure drop is a function of the square of velocity.

• The experimental observations suggest that pressure drop in the pipe flow under turbulent conditions depends upon Reynolds number and surface roughness.

Explanation:

Explanation:

Loss of pressure head in turbulent flow 

\(h_L= \frac{{fL{V^2}}}{{2gd}}\)

More exactly the loss of head ∝ Vn where n varies from 1.75 to 2.0

Additional Information

Other important dimensionless numbers are described in the table below:

Friction factor (f) for turbulent flow rough pipe is a function of relative roughness only.

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

for laminar flow Friction factor ⇒ \(f = \frac{{64}}{{Re}}\) 

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

Frictional resistance in a flow increases if the contact of the fluid increases with the boundary surface.

For turbulent flow, turbulent shear stress at the contact surface is directly proportional to the density of the fluid

\(\tau = \rho\)

Therefore, frictional resistance for turbulent flow is directly proportional to the density of the fluid.

Increase in the temperature decreases the viscosity of the fluid and hence decrease in frictional resistance.

The phenomenological Colebrook–White equation (or Colebrook equation) expresses the Darcy friction factor f as a function of Reynolds number R_e and pipe relative roughness k_s / D_h.

It's equation fitting the data of experimental studies of turbulent flow in smooth and rough pipes.

For a conduit flowing completely full of fluid at Reynolds numbers greater than 4000, its mathematically equivalent forms is expressed as:

\(\frac{1}{{\sqrt f }} = 1.14 - 2\log \left( {\frac{{{k_s}}}{D} + \frac{{9.35}}{{\sqrt f }}} \right)\)

Haaland equation: It is an approximation of the Colebrook equation.

\(\frac{1}{{\sqrt f }} = - 1.8\log \left[ {{{\left( {\frac{{\frac{{{k_s}}}{D}}}{{3.7}}} \right)}^{1.11}} + \frac{{6.9}}{{{R_e}}}} \right]\)

For a smooth or rough pipe, the difference of velocity at any point and average velocity is given by:

\(\frac{{u - \bar U}}{{{u_*}}} = 5.75\log_{10} \left( {\frac{y}{R}} \right) + 3.75\)

Local velocity = Average velocity ⇒ u = U̅

\(5.75{\log _{10}}\left( {\frac{y}{R}} \right) + 3.75 = 0 \Rightarrow {\log _{10}}\left( {\frac{y}{R}} \right) = - \frac{{3.75}}{{5.75}} = - 0.6522\)

Use log10 x = y ⇒ y = 10x

\(\frac{y}{R} = {10^{ - 0.6522}} = 0.2228\)

Y = 0.2228 R (Y is the distance from the pipe boundary)

Important Points Distance from center= r = (R – y) = 0.778 R

Distance from center = r = 0.778