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

The velocity distribution through laminar flow in fixed parallel plates is given by:

\(u = \frac{1}{{2\mu }}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)(dy - {y^2})\)

Now

\(\frac{{du}}{{dy}} = \frac{1}{{2\mu }}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( {d - 2y} \right)\)

Shear stress distribution, τ is given by

\(\tau = \frac{{\mu \;du}}{{dy}}\)

\(\tau = \frac{1}{2}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( {d - 2y} \right)\)

From above, following can be concluded:

1. Shear Stress distribution, ‘τ’ is linear.

2. At y = d/2 i.e. mid point; τ = 0 i.e. At center shear stress = 0.

3. At y = 0 i.e. at boundary; \(\tau = {\tau _{max}} = \frac{1}{2}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( d \right)\) i.e. shear stress is maximum at boundary.

Explanation:

Viscosity:

Viscosity is the physical property that characterizes the flow resistance of simple fluids.

Newton’s law of viscosity defines, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers.

The ratio of shear stress to shear rate is a constant, for a given temperature and pressure, and is defined as the viscosity or coefficient of viscosity.

Shear stress between two layers of fluid is directly proportional to the rate of change of velocity with respect to perpendicular distance from the fixed point (Velocity Gradient)

\({\rm{\tau \;\propto \;}}\frac{{{\rm{du}}}}{{{\rm{dy}}}}\)

\({\rm{\tau }} = {\rm{\mu \;}}\frac{{{\rm{du}}}}{{{\rm{dy}}}}\)

\({\rm{\mu }} = \frac{{\rm{\tau }}}{{\frac{{{\rm{du}}}}{{{\rm{dy}}}}}}{\rm{\;}}\)

where, τ = Shear stress and \(\frac{{{\rm{du}}}}{{{\rm{dy}}}}\) = Velocity gradient.

Explanation:

Attraction between Two Closely Parallel Moving Boats

The phenomenon where two closely parallel moving boats experience an attraction towards each other can be explained using Bernoulli's equation. This principle is rooted in fluid dynamics and describes the behavior of fluid flow.

• Bernoulli's Principle: It states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

• Application to Boats: When two boats move parallel to each other, the water flow between the boats speeds up, leading to a decrease in pressure between them.

• Resulting Attraction: The lower pressure between the boats compared to the higher pressure on their outer sides creates a net force that pulls the boats towards each other.

Analyzing the Given Options

1. Macaulay’s equation  (Incorrect)

   • Macaulay’s equation is used in structural analysis for determining deflections and moments in beams, which is unrelated to the fluid dynamics of moving boats.

2. Euler’s equation  (Incorrect)

   • Euler’s equation describes the motion of a fluid, but it does not specifically explain the pressure changes due to fluid speed in the context of two parallel moving boats.

3. Bernoulli’s equation  (Correct)

   • Bernoulli’s equation directly relates the speed of the fluid to the pressure, explaining the attraction due to the faster water flow between the boats creating lower pressure.

4. Momentum equation  (Incorrect)

   • The momentum equation deals with the conservation of momentum in fluid flow but does not specifically describe the pressure variations causing the boats to attract each other.

Explanation:

P_N - P_J = P_E - P_J

⇒ \(\frac{P_{N}-P_{J}}{\rho g}=\frac{P_{E}-P_{J}}{\rho g}\)

⇒ \(\left(h_{f}\right)_{N J}=\left(h_{f}\right)_{E J}\)

⇒ \(\frac{32 \mu v_{1} L}{\rho g d}=\frac{32 \mu v_{2}(10 L)}{\rho g d}\)

⇒ v₁ = 10v₂

Using continuity equation,

Q = 2Q₁ + Q₂

⇒ \(\frac{\pi}{4} d^{2} v=\frac{\pi}{4} d^{2}\left[2 v_{1}+v_{2}\right]\) 

⇒ v = 2v₁ + v₂

Using equation (i) and (ii)

\(v_{2}=\frac{1}{21} v \text { and } v_{1}=10 v_{2}=\frac{10}{21} v\) 

Discharge, \(Q_{2}=\frac{\pi}{4} d^{2} v_{2}=\frac{\pi}{4} d^{2}\left(\frac{1}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right) \frac{1}{21}=\frac{Q}{21}\)

Similarly, \(Q_{1}=\frac{\pi}{4} d^{2} v_{1}=\frac{\pi}{4} d^{2}\times\left(\frac{10}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right)\times \frac{10}{21}=\frac{10 Q}{21}\)

Now, clearly \(Q_{2}=\frac{Q}{21}\)

Hence, statement I is correct.

and (Pressure)_N = (Pressure)_E

⇒ P_N - P_E

⇒ P_J - P_N = P_J - P_E

Hence, the pressure difference between J and N is equal to the pressure difference between J and E.

So, statement II is true.

Explanation:

Laminar flow is also known as viscous flow because in this type of flow:

• The fluid particles move in parallel layers without mixing.
• There is no turbulence, and the flow is smooth and orderly.
• Viscous forces dominate over inertial forces, meaning that fluid friction (viscosity) plays a major role in determining the motion of the fluid.

There are six types of fluid flow:

A flow net is a grid obtained by drawing a series of equipotential lines and streamlines. Flow net is very much useful in analyzing the two dimensional, irrotational flow problems.

The flow nets can be constructed only in the following situations -

(i) The flow should be steady. This is so because streamline pattern for unsteady flow does not remain constant, it changes from instant to instant.

(ii) The flow should be irrotational, which is possible only when flowing fluid is an ideal fluid (having no viscosity) or it has negligible viscosity.

(iii) The flow should not be governed by the force of gravity, because under the action of gravity, the shape of the free surface changes constantly and hence no fixed flow net pattern can be obtained.

Explanation:

Kinetic energy correction factor(α): 

• It is defined as the ratio of kinetic energy/second based on actual velocity to the kinetic energy/second based on average velocity.
• \(\alpha = \frac{1}{A}\mathop \smallint \limits_A^{} {\left( {\frac{u}{V}} \right)^3}dA\)
• where A = area, V= average velocity, u= local velocity at distance r.
• In circular pipe laminar flow α = 2

Concept:

For laminar flow, Maximum velocity occurs at centre.

∴ Umax = Centre–line velocity 

For laminar flow, \({U_{avg}} = \frac{{{U_{max}}}}{2} \)

Discharge (Q) = Area × Average velocity

Calculation:

Given:

Pipe diameter, d = 0.04 m, Umax = Centre–line velocity = 1.5 m/s

\({U_{avg}} = \frac{{{U_{max}}}}{2} =\frac{{1.5}}{{2}}\Rightarrow 0.75\) m/s

Discharge (Q) = Area × Average velocity = \( \frac{\pi }{4} \times {\left( {0.04} \right)^2} \times 0.75 =\frac{{3\pi}}{{10000}}\)  m³/s.

Concept:

Reynolds Number:

• Reynold's number is defined as the ratio of inertia force to viscous force.

\(Re~=~\frac{F_i}{F_v}\) = \(\frac{ρ× V× D}{μ}\) ; where ρ = density, V = velocity, D = diameter of pipe, μ = Dynamic viscosity of fluid

∵ Kinematic viscosity, = \(\frac{Dynamic~viscosity}{density}~= \frac{\mu}{\rho}\)

∴ \(Re~=~\frac{VD}{ν}\) ; Clearly, Reynold's number is directly proportional to the velocity and diameter, and inversely proportional to the kinematic viscosity.

Calculation:

Given:

Initially Reynold's number, Re₁ = 1200, D₁ = D, D₂ = 2 × D, (kinematic viscosity)₂ = 2 × (kinematic viscosity)₁ ⇒ \(\nu_2~=~2~\times~\nu_1\)

Also, discharge at the pipe exit is unchanged.

So, \(Q_1~=~Q_2\\A_1~\times~V_1~=~A_2~\times~V_2\\{D_1}^2~\times~V_1~=~{D_2}^2~\times~V_2\)

putting the value of D₁ and D₂ in the above equation, we get

\(V_2~=~\frac{V_1}{4}\)

∵ \(Re~=~\frac{VD}{ν}\)

\(\frac{Re_1}{Re_2}~=~\frac{D_1~\times~V_1}{\nu_1}~\times~\frac{\nu_2}{D_2~\times~V_2}~=~\frac{D~\times~V_1}{\nu_1}~\times~\frac{2~\times~\nu_1}{2~\times~D~~\times~V_1\over 4~}~\)

\(Re_2~=~\frac{Re_1}{4}~=~300\)

So, New Reynold's number for the flow in the tube will be 300.

Explanation:

For the flow between two parallel spaced plates (as shown in the figure below): 

The velocity distribution is given by:

\(\rm{v = \frac{1}{{2\mu }}\left( {\frac{{ - \partial p}}{{\partial x}}} \right)\left( {By - {y^2}} \right)}\)

Shear stress is given by:

\(\rm{\begin{array}{l} \rm{\tau = \mu \frac{{du}}{{dy}}}\\ \rm{\tau= \frac{1}{2} \times \left( { - \frac{{\partial p}}{{\partial x}}} \right)\left( {B - 2y} \right)}\\ y = \frac{B}{2},\;\tau = 0 \end{array}}\)

At y = 0, τ = τmax

Hence, shear stress varies directly as the distance from the midplane.

Thus, the stress distribution across a section is depicted below:

∴ The shear stress is maximum at the boundaries and zero at the centre.

Concept:

According to newton's law of viscosity-

This law states that the shear stress (\(\tau\)) on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the coefficient of viscosity.

\(\tau=\mu\, .({du\over dy})\)

Where \(\tau=\) shear stress \(={Force\over Area}\)

\(\mu=\) dynamic viscosity

\({du\over dy}=\) velocity gradient

Calculation:

Given data:

Distance between plates (y) = 0.03 mm or 3 × 10^-5 m

Velocity of moving plate (\(v\)) = 0.8 m/s

Force per unit area on plate (F/A) = 1.5 N/m²

Dynamic viscosity (μ) =?

\(Shear\, stress(\tau)=\mu× ({v\over y})\)

\(1.5=\mu× ({0.8\over 3\times 10^{-5}})\)

\(\mu={1.5\times 3\times 10^{-5}\over 0.8}\)

\(\mu=5.625\times 10^{-5}\, N-s/m^2\)

\(\mu=56.25\times 10^{-6}\, N.s/m^2\)

Explanation:

Shear stress equation for a laminar flow through the pipe is given by:

\({\rm{\tau }} = - \frac{{\rm{r}}}{2} \times \frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}}\)

Where,

τ = Shear stress at any distance “r” from the center of the pipe

∂P/∂x = Pressure gradient

r = distance from the center of the pipe,

At r = 0, τ = 0,

Hence, at the center line of a pipe flowing under pressure where the velocity gradient is zero, the shear stress will be zero.

The velocity is zero at the wall of the pipe increasing to a maximum at the center, then symmetrically to the other wall, and the velocity distribution is parabolic.

Shear stress maximum at the wall of the pipe decreases to a minimum (zero) at the center, then symmetrically to the other wall, shear stress increase and shear distribution is linear

Concept:

When a flow is fully developed steady laminar flow of an incompressible fluid in a pipe (circular).

Then velocity profile at any radius r:

\(U = - \frac{1}{{4\mu }}\left( {\frac{{dP}}{{dx}}} \right)\left( {{R^2} - {r^2}} \right)\)

For r = 0

\(\)\(u\left( r \right) = u_{max}( {1 - \frac{{{r^2}}}{{{R^2}}}} )\)

Calculation:

Given :

r = R/2; u = ? at r = 200 - 80 = 120 mm (radial distance from the central line), u_max = 2 m/s

\(u\left( r \right) = 2( {1 - \frac{{{120^2}}}{{{200^2}}}} ) = 1.28 \ m/s\)

Concept:

Distribution of velocity profile:

In a fully developed laminar pipe flow, the average velocity is one half of the maximum velocity i.e. U_max = 2U_avg

\(\begin{array}{l} U (r) = U_{max}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}\)

\(\begin{array}{l} U (r) = 2{U_{av}}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}\)

Calculation:

Given:

Uavg = 5 m/s, R = 10 cm, r = 5 cm

\(U = 2 \times 5\left[ {1 - {{\left( {\frac{5}{{10}}} \right)}^2}} \right] = 7.5\;m/s\)

Concept:

Hagen Poiseuille equation

Loss of pressure head  \(= \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32μ VL}}{{ρ g{D^2}}} = {h_f}\)

Calculation:

Given:

Viscosity μ = 1.50 Pa.s, diameter d = 0.1 m, V = 6 m/s, L = 7 m, ρ = 1260 kg/m³

Loss of pressure head  \(= \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32 \times 6\times1.50\times 7 }}{{1260\times 10\times 0.01{}}} = 16 ~m\)