# Viscous, Turbulent Flow and Boundary Layer Theory

 Question 1
The figure shows two fluids held by a hinged gate. The atmospheric pressure is $P_a = 100 kPa$. The moment per unit width about the base of the hinge is ________ kNm/m. (Rounded off to one decimal place)
Take the acceleration due to gravity to be $g = 9.8 m/s^2$ . A 45.8 B 57.2 C 68.2 D 87.6
GATE ME 2023   Fluid Mechanics
Question 1 Explanation: Forces from Pressure diagram
$\mathrm{F}_{1}=$ Area of $\triangle \mathrm{ADE} \times$ width
$F_{1}=\frac{1}{2} \times A D \times D E \times$ width
$=\frac{1}{2} \times 1 \times 9810 \times 1$
$\mathrm{F}_{1}=4905 \mathrm{~N}$
$\mathrm{F}_{2}=$ Area of rectangle $\mathrm{DEBF} \times$ width
$F_{2}=2 \times 9810 \times 1$
$\mathrm{F}_{2}=19620 \mathrm{~N}$
$\mathrm{F}_{3}=$ Area of triangle $\Delta \mathrm{EFC}$
$=\frac{1}{2} \times \mathrm{EF} \times \mathrm{FC} \times$ width
$=\frac{1}{2} \times 2 \times 39240 \times 1$
$\mathrm{F}_{3}=39240 \mathrm{~N}$

Location of forces from top
\begin{aligned} & \mathrm{F}_{1} \rightarrow \frac{2}{3} \times 1=\frac{2}{3} \mathrm{~m} \\ & \mathrm{~F}_{2} \rightarrow\left(1 \mathrm{~m}+\frac{2 \mathrm{~m}}{2}\right)=2 \mathrm{~m} \\ & \mathrm{~F}_{3} \rightarrow 1 \mathrm{~m}+2 \mathrm{~m} \times \frac{2}{3}=\frac{7}{3} \mathrm{~m} \end{aligned}
Pressure at depth $1 \mathrm{~m}=\rho_{1} \mathrm{gh}_{1}$
$1000 \times 9.81 \times 1=9810 \frac{N}{m^{2}}$
$\mathrm{DE}=9810 \frac{\mathrm{N}}{\mathrm{m}^{2}}$
$BF=9810 \frac{N}{m^{2}}$

Pressure at depth $3 m$
\begin{aligned} & \rho_{1} g h_{1}+\rho_{2} g h_{2} \\ & 9810+2000 \times 9.81 \times 2 \\ & 9810+39240=49050 \frac{\mathrm{N}}{\mathrm{m}^{2}} \end{aligned}
$(B F)+(F C)=B C$

Location of forces from bottom hinge
\begin{aligned} & \mathrm{F}_{1} \rightarrow 3 \mathrm{~m}-\frac{2}{3} \mathrm{~m}=\frac{7}{3} \mathrm{~m} \\ & \mathrm{~F}_{2} \rightarrow 3 m-2 m=1 \mathrm{~m} \\ & \mathrm{~F}_{3} \rightarrow 3 m-\frac{7}{3} \mathrm{~m}=\frac{2}{3} \mathrm{~m} \end{aligned}

Moment about hinge $\mathrm{F}_{1} \times \frac{7}{3}+\mathrm{F}_{2} \times 1+\mathrm{F}_{3} \times \frac{2}{3}$
$4905 \times \frac{7}{3}+19620+39240 \times \frac{2}{3}$
$\Rightarrow \quad 57225 \mathrm{Nm}=57.22 \mathrm{kNm} / \mathrm{m}$
 Question 2
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 2 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 3
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
GATE ME 2022 SET-2   Fluid Mechanics
Question 3 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 4
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 $\frac{1}{12}$
GATE ME 2022 SET-1   Fluid Mechanics
Question 4 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 5
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.
GATE ME 2022 SET-1   Fluid Mechanics
Question 5 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.



There are 5 questions to complete.

### 5 thoughts on “Viscous, Turbulent Flow and Boundary Layer Theory”

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

2. I have also the same doubt , but everywhere the answer given is Reynolds forces ,

3. Yes bro u are ✅️ right

4. • 