Laminar Flow MCQ Quiz - Objective Question with Answer for Laminar Flow - Download Free PDF
Last updated on Apr 29, 2025
Latest Laminar Flow MCQ Objective Questions
Laminar Flow Question 1:
In a laminar flow between two fixed plates held parallel to each other at a distance d, the shear stress is:
1) Maximum at plane \(\frac{d}{2}\) away from each plate and zero at the plate boundaries.
2) Zero throughout the passage.
3) Maximum at the plate boundaries and zero at plane \(\frac{d}{2}\) away from each plate.
Which of the above statements is/are correct?Answer (Detailed Solution Below)
Laminar Flow Question 1 Detailed Solution
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.
∴ Shear stress is maximum at the plate boundaries and zero at plane \(\frac{d}{2}\) away from each plateLaminar Flow Question 2:
In a laminar flow between two parallel plates, where one plate is stationary and the other moves at a constant velocity, how does Newton's Law of Viscosity describe the velocity gradient in the fluid?
Answer (Detailed Solution Below)
Laminar Flow Question 2 Detailed Solution
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.
Laminar Flow Question 3:
Which of the following hypothesis satisfies the attraction between two closely parallel moving boats?
Answer (Detailed Solution Below)
Laminar Flow Question 3 Detailed Solution
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.
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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.
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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.
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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
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Macaulay’s equation (Incorrect)
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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.
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Euler’s equation (Incorrect)
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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.
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Bernoulli’s equation (Correct)
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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.
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Momentum equation (Incorrect)
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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.
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Laminar Flow Question 4:
In the pipe network shown in the figure, all pipes have the same cross-section and can be assumed to have the same friction factor. The pipes, connecting points W, N, and S with point J have an equal length L. The pipe connecting points J and E has a length 10L. The pressure at the ends N, E, and S are equal. The flow rate in the pipe connecting W and J is Q. Assume that the fluid flow is steady, incompressible, and the pressure losses at the pipe entrance and junction are negligible. Consider the following statements:
I. The flow rate in pipe connecting J and E is Q/21.
II. The pressure difference between J and N is equal to the pressure difference between J and E.
Which one of the following option is CORRECT?
Answer (Detailed Solution Below)
Laminar Flow Question 4 Detailed Solution
Explanation:
PN - PJ = PE - PJ
⇒ \(\frac{P_{N}-P_{J}}{\rho g}=\frac{P_{E}-P_{J}}{\rho g}\)
⇒ \(\left(h_{f}\right)_{N J}=\left(h_{f}\right)_{E J}\)
⇒ \(\frac{32 \mu v_{1} L}{\rho g d}=\frac{32 \mu v_{2}(10 L)}{\rho g d}\)
⇒ 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
⇒ PN - PE
⇒ PJ - PN = PJ - PE
Hence, the pressure difference between J and N is equal to the pressure difference between J and E.
So, statement II is true.
Laminar Flow Question 5:
Which of the following types of flow is also known as viscous flow?
Answer (Detailed Solution Below)
Laminar Flow Question 5 Detailed Solution
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:
Compressible flow |
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Incompressible flow |
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Uniform flow |
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Non - uniform flow |
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Steady flow and |
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Unsteady flow |
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Rotational flow |
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Irrotational flow |
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Laminar flow |
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Turbulent flow |
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1D flow |
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2D flow |
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3D flow |
Top Laminar Flow MCQ Objective Questions
In which of the following case flow net cannot be drawn?
Answer (Detailed Solution Below)
Laminar Flow Question 6 Detailed Solution
Download Solution PDFA 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.
∴ Thus flow net cannot be drawn when the flow is governed by gravity.For laminar flow, Kinetic energy correction factor is:
Answer (Detailed Solution Below)
Laminar Flow Question 7 Detailed Solution
Download Solution PDFExplanation:
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
The discharge in m3/s for laminar flow through a pipe of diameter 0.04 m having a centre line velocity of 1.5 m/s.
Answer (Detailed Solution Below)
Laminar Flow Question 8 Detailed Solution
Download Solution PDFConcept:
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.
The Reynold's number for the flow of a fluid in a horizontal circular tube of constant diameter is 1200. If the diameter of the tube and the kinematic viscosity of the fluid are doubled and that discharged at the pipe exit is unchanged, then the new Reynold's number for the flow in the tube will be
Answer (Detailed Solution Below)
Laminar Flow Question 9 Detailed Solution
Download Solution PDFConcept:
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.
The shear stress between two fixed parallel plates with a laminar flow between them
Answer (Detailed Solution Below)
Laminar Flow Question 10 Detailed Solution
Download Solution PDFExplanation:
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.
A plate at a distance of 0.03 mm from a fixed plate moves at 0.8 m/s and requires a force of 1.50 N/m2 area of plate. Determine the dynamic viscosity of liquid between the plates.
Answer (Detailed Solution Below)
Laminar Flow Question 11 Detailed Solution
Download Solution PDFConcept:
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) =?
\(Shear\, stress(\tau)=\mu× ({v\over y})\)
\(1.5=\mu× ({0.8\over 3\times 10^{-5}})\)
\(\mu={1.5\times 3\times 10^{-5}\over 0.8}\)
\(\mu=5.625\times 10^{-5}\, N-s/m^2\)
\(\mu=56.25\times 10^{-6}\, N.s/m^2\)
At the centre line of a pipe flowing under pressure where the velocity gradient is zero, the shear stress will be ________.
Answer (Detailed Solution Below)
Laminar Flow Question 12 Detailed Solution
Download Solution PDFExplanation:
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
For a laminar flow through a circular pipe of diameter 400 mm, the maximum velocity is 2 m/s. What will be the velocity at 8 cm from the wall of the pipe?
Answer (Detailed Solution Below)
Laminar Flow Question 13 Detailed Solution
Download Solution PDFConcept:
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), umax = 2 m/s
\(u\left( r \right) = 2( {1 - \frac{{{120^2}}}{{{200^2}}}} ) = 1.28 \ m/s\)
Laminar developed flow at an average velocity of 5 m/s occurs in a pipe of 10 cm radius. The velocity at 5 cm radius is
Answer (Detailed Solution Below)
Laminar Flow Question 14 Detailed Solution
Download Solution PDFConcept:
Distribution of velocity profile:
In a fully developed laminar pipe flow, the average velocity is one half of the maximum velocity i.e. Umax = 2Uavg
\(\begin{array}{l} U (r) = U_{max}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}\)
\(\begin{array}{l} U (r) = 2{U_{av}}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}\)
Calculation:
Given:
Uavg = 5 m/s, R = 10 cm, r = 5 cm
\(U = 2 \times 5\left[ {1 - {{\left( {\frac{5}{{10}}} \right)}^2}} \right] = 7.5\;m/s\)
Glycerine (μ = 1.50 Pa.s: ρ = 1260 kg/m3) flows at a velocity of 6.0 m/s in 10 cm diameter pipe. Head loss in a length of 7 m pipe will be (g = 10 m/s2).
Answer (Detailed Solution Below)
Laminar Flow Question 15 Detailed Solution
Download Solution PDFConcept:
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 \(= \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32 \times 6\times1.50\times 7 }}{{1260\times 10\times 0.01{}}} = 16 ~m\)