# Viscous, Turbulent Flow and Boundary Layer Theory

 Question 1
A uniform wooden rod (specific gravity = 0.6, diameter = 4 cm and length = 8 m) is immersed in the water and is hinged without friction at point A on the waterline as shown in the figure. A solid spherical ball made of lead (specific gravity = 11.4) is attached to the free end of the rod to keep the assembly in static equilibrium inside the water. For simplicity, assume that the radius of the ball is much smaller than the length of the rod.
Assume density of water $= 10^3 kg/m^3$ and $\pi = 3.14$.
Radius of the ball is _______ cm (round off to 2 decimal places). A 2.42 B 3.62 C 8.26 D 3.59
GATE ME 2022 SET-2   Fluid Mechanics
Question 1 Explanation: As the system is in equilibrium,
$\Sigma M_o=0$
$(F_{B1}-W_1) \times \frac{L}{2} \cos \theta +F_{B2}-W_2) \times L \times \cos \theta =0$
$(\rho gV_1-\rho _1gV_1) \times \frac{1}{2}+(\rho gV_2-\rho _2gV_2) =0$
$(\rho -\rho _1) \times g \times \frac{\pi d^2L}{4} \times \frac{1}{2}+(\rho -\rho _2) \times \frac{4}{3} \times \\pi R^3 \times g=0$
$(100-600) \times \frac{0.04^2 \times 8}{8} +(1000-11400) \times \frac{4}{3} \times R^3=0$
$\Rightarrow R=0.03587m=3.59cm$
 Question 2
Consider a steady flow through a horizontal divergent channel, as shown in the figure, with supersonic flow at the inlet. The direction of flow is from left to right. Pressure at location B is observed to be higher than that at an upstream location A. Which among the following options can be the reason?

MSQ
 A Since volume flow rate is constant, velocity at B is lower than velocity at A B Normal shock C Viscous effect D Boundary layer separation
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Question 2 Explanation:
If the supersonic flow enters to the diverging duct, it will act as nozzle the pressure of the flow is decreases in the nozzle. But it is given the pressure of the flow at B is more than that A It will happen only with the development of normal shoot in the diverging flow (the normal shock wave is the characteristic of only supersonic flow) Because of normal shock wave in the diverging nozzle pressure increases and velocity decreases. But mass flow rate remains unchanged.
 Question 3
A solid spherical bead of lead (uniform density $=11000\; kg/m^3$) of diameter $d = 0.1 mm$ sinks with a constant velocity $V$ in a large stagnant pool of a liquid (dynamic viscosity $=1.1 \times 10^{-3} kg.m^{-1}.s^{-1}$). The coefficient of drag is given by $C_D=\frac{24}{Re}$, where the Reynolds number ($Re$) is defined on the basis of the diameter of the bead. The drag force acting on the bead is expressed as $D=(C_D)(0.5 \rho V^2)\left ( \frac{\pi d^2}{4} \right )$, where $\rho$ is the density of the liquid. Neglect the buoyancy force. Using $g = 10 m/s^2$, the velocity $V$ is __________ m/s.
 A $\frac{1}{24}$ B $\frac{1}{6}$ C $\frac{1}{18}$ D $\frac{1}{12}$
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Question 3 Explanation:
As buoyancy force is neglected we can neglect density of fluid () in the following equation.
\begin{aligned} V&=\frac{1}{18\mu _f}(\rho _b-\rho _f)gd^2\\ &=\frac{11000 \times 10 \times (10^{-4})^2}{18 \times 1.1 \times 10^{-3}}\\ &=\frac{1}{18}m/s \end{aligned}
 Question 4
The figure shows a purely convergent nozzle with a steady, inviscid compressible flow of an ideal gas with constant thermophysical properties operating under choked condition. The exit plane shown in the figure is located within the nozzle. If the inlet pressure $(P_0)$ is increased while keeping the back pressure $(P_{back})$ unchanged, which of the following statements is/are true? MSQ
 A Mass flow rate through the nozzle will remain unchanged. B Mach number at the exit plane of the nozzle will remain unchanged at unity C Mass flow rate through the nozzle will increase. D Mach number at the exit plane of the nozzle will become more than unity.
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Question 4 Explanation:
\begin{aligned} \frac{\dot{m}}{A}&=\sqrt{\frac{\gamma }{R}}\frac{P_0}{\sqrt{T_0}}\frac{1}{\left ( \frac{\gamma +1}{2} \right )^{\frac{\gamma +1}{2(\gamma -1)}}}\\ \frac{\dot{m}}{A}&=0.685 \times \sqrt{P_0\rho _0} \;\;\;...\text{ for Air}\\ \frac{T_c}{T_0}&=\frac{T^*}{T_0}=\frac{2}{\gamma +1}=0.8333\\ &(\text{critical temp ratio})\\ \frac{P_c}{P_0}&=\frac{P^*}{P_0}=\left ( \frac{2}{\gamma +1} \right )^{\frac{\gamma }{\gamma -1}}=0.5282\\ &(\text{critical pressure ratio}) \end{aligned}
where $\gamma=$ Adiabatic index = 1.4
If $P_0$ increases , mass flow rate increases. Mach number at the exit plane of the nozzle will remain unchanged at unity.
Since the maximum expected velocity from the converging nozzle is sound velocity.


 Question 5
In the following two-dimensional momentum equation for natural convection over a surface immersed in a quiescent fluid at temperature $T_\infty ( g$ is the gravitational acceleration, $\beta$ is the volumetric thermal expansion coefficient, $v$ is the kinematic viscosity, $u$ and $v$ are the velocities in $x$ and $y$ directions, respectively, and $T$ is the temperature)
$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=g\beta (T-T_\infty )+v\frac{\partial u^2}{\partial y^2}$
the term $g\beta (T-T_\infty )$ represents
 A Ratio of inertial force to viscous force. B Ratio of buoyancy force to viscous force. C Viscous force per unit mass. D Buoyancy force per unit mass.
GATE ME 2022 SET-1   Fluid Mechanics
Question 5 Explanation:
$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=g\beta (T-T_\infty )+v\frac{\partial^2 u}{\partial y^2}$
This is the equation that governs the fluid motion in the boundary layer due to effect of Buoyancy
$g\beta (T-T_\infty )=\frac{m}{s^2} \times \frac{1}{K} \times K=m/s^2$
Unit of $g\beta (T-T_\infty )= m/s^2$
Buoyancy force is due to density difference and gravitational effect
$g\beta (T-T_\infty )$ represents Buoyancy force per unit mass.
 Question 6
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
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Question 6 Explanation:
Reynolds stresses are responsible for eddy viscosity
$\tau _{\text{Reynolds}}=\eta \frac{\partial \bar{u}}{\partial y}$
where $\eta =$ Eddy viscosity.
 Question 7
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
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Question 7 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 8
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
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Question 8 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 9
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
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Question 9 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 10
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
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### 3 thoughts on “Viscous, Turbulent Flow and Boundary Layer Theory”

1. IN Q.1. ITS BOUSINESSQ THEORY NOT REYNOLD’S…PLS CONFIRM AND RECTIFY!

2. 3. 