Viscous, Turbulent Flow and Boundary Layer Theory

Question 1
Which of the following is responsible for eddy viscosity (or turbulent viscosity) in a turbulent boundary layer on a flat plate?
A
Nikuradse stresses
B
Reynolds stresses
C
Boussinesq stresses
D
Prandtl stresses
GATE ME 2021 SET-2   Fluid Mechanics
Question 1 Explanation: 
Reynolds stresses are responsible for eddy viscosity
\tau _{\text{Reynolds}}=\eta \frac{\partial \bar{u}}{\partial y}
where \eta = Eddy viscosity.
Question 2
Consider fully developed, steady state incompressible laminar flow of a viscous fluid between two large parallel horizontal plates. The bottom plate is fixed and the top plate moves with a constant velocity of U = 4 m/s. Separation between the plates is 5 mm. There is no pressure gradient in the direction of flow. The density of fluid is 800 kg/m^3, and the kinematic viscosity is 1.25 \times 10^{-4} \; m^2/s. The average shear stress in the fluid is ________Pa (round off to the nearest integer).
A
58
B
64
C
92
D
80
GATE ME 2021 SET-1   Fluid Mechanics
Question 2 Explanation: 


\begin{array}{l} V=4 \mathrm{~m} / \mathrm{s} \\ \rho=800 \mathrm{~kg} / \mathrm{m}^{3} \\ \mathrm{v}=1.25 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s} \\ h=5 \mathrm{~mm} \\ \tau=\mu \cdot \frac{d u}{d y} \\ \tau=\left[800 \times 1.25 \times 10^{-4}\right] \times \frac{4}{5 \times 10^{-3}} \\ \tau=80 \mathrm{~N} / \mathrm{m}^{2} \\ =80 \mathrm{~Pa} \end{array}
Question 3
Water (density 1000 kg/m^3) flows through an inclined pipe of uniform diameter. The velocity, pressure and elevation at section A are V_A = 3.2 m/s, p_A =186 kPa and z_A=24.5 m respectively, and those at section B are V_B = 3.2 m/s, p_B =260 kPa and z_B=9.1 m, respectively. If acceleration due to gravity is 10 m/s^2 then the head lost due to friction is _________ m (round off to one decimal place).
A
8
B
-8
C
-4
D
4
GATE ME 2020 SET-2   Fluid Mechanics
Question 3 Explanation: 


Energy at 'A' head =\frac{P_{A}}{\rho g}+\frac{V_{A}^{2}}{2 g}+Z_{A}=\frac{186 \times 10^{3}}{1000 \times 10}+\frac{3.2^{2}}{2 \times 10}+24.5
=18.6+0.512+24.5=43.612
Energy at 'b' head =\frac{P_{b}}{\rho g}+\frac{V_{b}^{2}}{2 g}+Z_{b}=\frac{286 \times 10^{3}}{1000 \times 10}+\frac{3.2^{2}}{2 \times 10}+9.1
=26+0.512+9.1=35.612
E_{A}>E_{B^{\prime}} so flow from 'A' to 'B'
Heat loss =E_{A}-E_{B}=43.612-35.612=8 \mathrm{m} of water head
Question 4
Consider steady, viscous, fully developed flow of a fluid through a circular pipe of internal diameter D. We know that the velocity profile forms a paraboloid about the pipe centre line, given by: V=-C\left (r^2-\frac{D^2}{4} \right ) m/s, where C is a constant. The rate of kinetic energy (in J/s) at the control surface A-B, as shown in the figure, is proportional to D^n. The value of n is________.
A
2
B
4
C
8
D
16
GATE ME 2020 SET-1   Fluid Mechanics
Question 4 Explanation: 
\begin{aligned} v &=-C\left[r^{2}-\frac{D^{2}}{4}\right]=-C\left[r^{2}-R^{2}\right] \\ v &=C\left[R^{2}-r^{2}\right]=C R^{2}\left[1-\frac{r^{2}}{R^{2}}\right] \\ \left[K\dot{E}\right]&=\frac{1}{2} \int_{0}^{R} \rho v^{3} d A=\rho \int_{0}^{R} C^{3} R^{6}\left[1-\frac{r^{2}}{R^{2}}\right]^{3} \cdot 2 \pi r \cdot d r \\ &=\rho C^{3} R^{6} 2 \pi \int_{0}^{R}\left[1-\frac{r^{2}}{R^{2}}\right]^{3} r \cdot d r \\ &=\rho C^{3} R^{6} 2 \pi \int_{0}^{R}\left[1-\frac{r^{6}}{R^{6}}-\frac{3 r^{2}}{R^{2}}\left[1-\frac{r^{2}}{R^{2}}\right]\right] r \cdot d r \\ &=\rho C^{3} R^{6} 2 \pi \int_{0}^{R}\left[1-\frac{r^{6}}{R^{6}}-\frac{3 r^{2}}{R^{2}}+\frac{3 r^{4}}{R^{4}}\right] r \cdot d r \\ &=\rho C^{3} 2 \pi R^{6}\left[\frac{R^{2}}{2}-\frac{R^{2}}{8}-\frac{3 R^{2}}{4}+\frac{R^{2}}{2}\right] \\ &=\rho C^{3} 2 \pi R^{6}\left[R^{2}-\frac{R^{2}}{8}-\frac{3 R^{2}}{4}\right]=\rho C^{3} 2 \pi R^{6}\left[\frac{R^{2}}{8}\right] \\ &=\frac{\rho C^{3} 2 \pi}{8} R^{8}=\frac{\rho C^{3} 2 \pi}{8 \times(2)^{8}} \times D^{8} \\ n &=8 \end{aligned}
Question 5
Match the following non-dimensional numbers with the corresponding definitions:
A
P-1, Q-3, R-2, S-4
B
P-3, Q-1, R-2, S-4
C
P-4, Q-3, R-1, S-2
D
P-3, Q-1, R-4, S-2
GATE ME 2020 SET-1   Fluid Mechanics
Question 6
Froude number is the ratio of
A
buoyancy forces to viscous forces
B
inertia forces to viscous forces
C
buoyancy forces to inertia forces
D
inertia forces to gravity forces
GATE ME 2020 SET-1   Fluid Mechanics
Question 7
The aerodynamic drag on a sports car depends on its shape. The car has a drag coefficient of 0.1 with the windows and the roof closed. With the windows and the roof open, the drag coefficient becomes 0.8. The car travels at 44 km/h with the windows and roof closed. For the same amount of power needed to overcome the aerodynamic drag, the speed of the car with the windows and roof open (round off to two decimal places), is ________ km/h (The density of air and the frontal area may be assumed to be constant).
A
12.60
B
32.80
C
22.00
D
18.90
GATE ME 2019 SET-2   Fluid Mechanics
Question 7 Explanation: 
Given data :
(i) When the windows and roof of the car are closed (CD)C = 0.1 and V_{c} = 44 km/h
(ii) When the windows and roof of the car are open, (CD)0 = 0.8 and V_{0} = ?
(iii) \rho c = \rho 0 , Ac = A0 and P_{c} = P_{0} where P is the power required to overcome the aerodynamic drag.
We know that
\mathrm{P}=\mathrm{F}_{\mathrm{D}} \times \mathrm{V}=\mathrm{C}_{\mathrm{D}} \times \frac{1}{2} \rho \mathrm{AV}^{2} \times \mathrm{V}=\frac{1}{2} \mathrm{C}_{\mathrm{D}} \rho \mathrm{AV}^{3}
For \rho and A to be constant
\mathrm{P} \propto \mathrm{C}_{\mathrm{D}} \mathrm{V}^{3}
Also, P is the same for both cases.
Hence, C_{D} V^{3}= constant
or \frac{V_{0}}{V_{c}}=\left(\frac{C_{D C}}{C_{D O}}\right)^{1 / 3}=\left(\frac{0.1}{0.8}\right)^{1 / 3}=\frac{1}{2}
\Rightarrow \mathrm{V}_{\mathrm{o}}=\frac{\mathrm{V}_{\mathrm{C}}}{2}=\frac{44}{2}=22.00 \mathrm{km} / \mathrm{h}
Question 8
Water enters a circular pipe of length L=5.0 m and diameter D= 0.20 m with Reynolds number R_{eD}=500. The velocity profile at the inlet of the pipe is uniform while it is parabolic at the exit. The Reynolds number at the exit of the pipe is______.
A
200
B
1000
C
500
D
250
GATE ME 2019 SET-2   Fluid Mechanics
Question 8 Explanation: 


Given data,
Circular pipe, L = 5 m,
D = 0.2 m,
Re = 500 at inlet
At inlet velocity is uniform while at exit velocity profile is parabola.
Since diameter of the pipe does not change, the average velocity at exit will be the same as that at inlet, for the same discharge. This will result in Re to be same as that at inlet.
Question 9
A two-dimensional incompressible frictionless flow field is given by \vec{u}=x\hat{i}-y\hat{j}. If \rho is the density of the fluid, the expression for pressure gradient vector at any point in the flow field is given as
A
\rho (x\hat{i}+y\hat{j})
B
-\rho (x\hat{i}+y\hat{j})
C
\rho (x\hat{i}-y\hat{j})
D
-\rho (x^2 \hat{i} + y^2 \hat{j})
GATE ME 2019 SET-2   Fluid Mechanics
Question 9 Explanation: 
Given, 2-D incompressible frictionless fluid flow.
\overrightarrow{\mathrm{u}}=x \hat{\mathrm{i}}-y \hat{\mathrm{j}}
Thus, velocity components in x and y directions are :
\mathrm{u}=\mathrm{x} \text { and } \mathrm{v}=-\mathrm{y}
Navier-Stokes equation for incompressible, frictionless fluid flow reduces to
\rho \frac{\mathrm{DV}}{\mathrm{Dt}}=-\nabla \overrightarrow{\mathrm{P}}+\overrightarrow{\rho \mathrm{g}}
There are no components of body force in x and y direction. Hence,
\rho \frac{\mathrm{D} \overrightarrow{\mathrm{V}}}{\mathrm{Dt}}=-\nabla \overrightarrow{\mathrm{P}}
where, \nabla \overrightarrow{\mathrm{P}} is the pressure gradient vector
Hence,
\begin{aligned} \bigtriangledown \vec{P}&=\rho \left [ \left ( u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right )\hat{i}+\left ( u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\right ) \hat{j}\right ]\\ &=-\rho \left [ \{ x(1)+(-y)(0)\}\hat{i}+\{ x(0)+(-y)(-1)\} \hat{j}\right ]\\ &=-\rho (x\hat{i}+y\hat{j}) \end{aligned}
Question 10
Two immiscible, incompressible, viscous fluids having same densities but different viscosities are contained between two infinite horizontal parallel plates, 2 m apart as shown below. The bottom plate is fixed and the upper plate moves to the right with a constant velocity of 3 m/s. With the assumptions of Newtonian fluid, steady, and fully developed laminar flow with zero pressure gradient in all directions, the momentum equations simplify to

\frac{d^2u}{dy^2}=0.

If the dynamic viscosity of the lower fluid, \mu _2, is twice that of the upper fluid, \mu _1, then the velocity at the interface (round offto two decimal places) is ________ m/s.
A
0.22
B
0.65
C
0.48
D
1.00
GATE ME 2019 SET-1   Fluid Mechanics
Question 10 Explanation: 


The shear stress must be same at the interface of fluids.
\tau_{1}=\tau_{2}
Let V be the velocity of the interface.
i.e. \mu_{1} \frac{\mathrm{U}-\mathrm{V}}{\mathrm{h}_{1}}=\mu_{2} \frac{\mathrm{V}-0}{\mathrm{h}_{2}}
\mu_{1} \frac{(3-\mathrm{V})}{1}=2 \mu_{1}\left(\frac{\mathrm{V}}{1}\right)
3-V=2 V
y=1 \mathrm{m} /\mathrm{s}
There are 10 questions to complete.

Leave a Comment

Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.