# Fluid Mechanics

 Question 1
An explosion at time $t=0$ releases energy $E$ at the origin in a space filled with a gas of density $\rho$. Subsequently, a hemispherical blast wave propagates radially outwards as shown in the figure.
Let $R$ denote the radius of the front of the hemispherical blast wave. The radius $R$ follows the relationship $R =kt^aE^b\rho ^c$, where $k$ is a dimensionless constant.
The value of exponent $a$ is _____. (Rounded off to one decimal place)

 A 0.1 B 0.4 C 0.6 D 0.8
GATE ME 2023      Fluid Properties
Question 1 Explanation:

Radius, $R=f(t, E, \rho)$
Where, $\mathrm{t}=$ time, $\mathrm{E}=$ Release energy $\& \rho=$ density
Dimension, $\mathrm{R}=[\mathrm{L}]$
\begin{aligned} & t=[\mathrm{T}] \\ & E=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \\ & \rho=\left[\mathrm{ML}^{-3}\right] \end{aligned}
Fundamental dimensions are $\mathrm{M}, \mathrm{L}, \mathrm{T}$ (basic)
Total no. of variables $=4$
No. of $\pi$. terms $=4-3=1$
So, $R=t^{a} E^{b} \rho^{c}$
$[\mathrm{L}]=[\mathrm{T}]^{\mathrm{a}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]^{\mathrm{b}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{c}}$
Comparing both side indices of $M, L, T$
$a-2b = 0, 2b-3c = 1, b+c = 0 \Rightarrow b-c$
$a = 2 b, 2(-c) - 3c = 1$
$-5c = 1, c =(-1)/5 \Rightarrow b=1/5$
So, $a=2 \times \frac{1}{5} = \frac{2}{5}$
Relation, $R=kt^{2/5}E^{1/5}\rho ^{-1/5}=\frac{kt^{2/5}E^{1/5}}{\rho ^{1/5}}$

Where, $a=\frac{2}{5}=0.40$
 Question 2
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      Viscous, Turbulent Flow and Boundary Layer Theory
Question 2 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 3
Consider a unidirectional fluid flow with the velocity field given by
$V(x,y,z,t)=u(x,t)\hat{i}$
where $u(0,t)=1$. If the spatially homogeneous density field varies with time $t$ as
$\rho (t)=1+0.2e^{-t}$
the value of $u(2,1)$ is _______. (Rounded off to two decimal places)

Assume all quantities to be dimensionless.
 A 1.14 B 2.25 C 3.65 D 8.25
GATE ME 2023      Fluid Kinematics
Question 3 Explanation:
$\rho \frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\rho \frac{\partial \mathrm{v}}{\partial \mathrm{y}}+\rho \frac{\partial \omega}{\partial \mathrm{z}}+\frac{\partial \rho}{\partial \mathrm{t}}=0$
Here $V(x, y, z, t)=u(x, t) \hat{i}$
So $v=0$
$\omega=0$
$\rho \frac{\partial u}{\partial x}+\frac{\partial P}{\partial t}=0$

Given $\rho(t)=1+0.2 e^{-t}$
So, $\left[1+0.2 e^{-t}\right] \frac{\partial u}{\partial x}+\frac{\partial}{\partial t}\left[1+0.2 e^{-t}\right]=0$
\begin{aligned} &\left(1+0.2 e^{-t}\right) \frac{\partial u}{\partial x}-0.2 e^{-t}=0 \\ &\left(1+0.2 e^{-t}\right) \frac{\partial u}{\partial x}=0.2 e^{-t} \\ & \frac{\partial u}{\partial x}=\left[\frac{0.2 e^{-t}}{1+0.2 e^{-t}}\right] \\ & u=\left[\frac{0.2 e^{-t}}{1+0.2 e^{-t}}\right] x+C \end{aligned}

Since
\begin{aligned} u(0, t) & =1 \text { so } C=1 \\ u & =\left[\frac{0.2 e^{-t}}{1+0.2 e^{-t}}\right] x+1 \\ u(2,1)(x, t) & =\left[\frac{0.2 e^{-1}}{1+0.2 e^{-1}}\right] \times 2+1 \\ u(2,1) & =1.137 \mathrm{~m} / \mathrm{s} \approx 1.14 \mathrm{~m} / \mathrm{sec} \end{aligned}
 Question 4
The velocity field of a certain two-dimensional flow is given by
$V(x,y)=k(x\hat{i}-y\hat{j})$
where $k=2s^{-1}$. The coordinates $x$ and $y$ are in meters. Assume gravitational effects to be negligible.
If the density of the fluid is $1000kg/m^3$ and the pressure at the origin is 100 kPa, the pressure at the location (2 m, 2 m) is _____________ kPa. (Answer in integer)
 A 64 B 26 C 84 D 98
GATE ME 2023      Fluid Kinematics
Question 4 Explanation:
To find the pressure at location $(2 m, 2 m)$ we apply Bernouli's equation

$P_{1}+\frac{1}{2} \rho V_{1}^{2}=P_{2}+\frac{1}{2} P V_{2}^{2}$

We will apply this equation between two points Origin $(0,0)$ and location $(2 \mathrm{~m}, 2 \mathrm{~m})$

At Origin $(0,0)$
\begin{aligned} & V=k(x \hat{i}-y \hat{j}) \\ & V_{1}=2(0-0)=0 \\ & P_{1}=100 \mathrm{kPa} \end{aligned}

At Iocation $(2,2)$
\begin{aligned} & V=2(2 \hat{i}-2 \hat{j}) \\ & \vec{V}=4 \hat{i}-4 \hat{j} \end{aligned}

magnitude of velocity
\begin{aligned} & V_{2}=\sqrt{4^{2}+4^{2}}=\sqrt{32} \\ & V_{2}=\sqrt{16 \times 2}=4 \sqrt{2} \end{aligned}

Applying Bernouli's theorem
$100,000+\frac{1}{2} \times 1000 \times 0=P_{2}+\frac{1}{2} \times 1000 \times 32$
So $P_{2}+16,000=100,000$
$P_{2}=100,000-16,000$
$P_{2}=84,000 \mathrm{~Pa}=84 \mathrm{kPa}$
 Question 5
Air (density = $=1.2kg/m^3$, kinematic viscosity $=1.5 \times 10^{-5} m^2/s$) flows over a flat plate with a free-stream velocity of $2 m/s$. The wall shear stress at a location $15 mm$ from the leading edge is $\tau _w$. What is the wall shear stress at a location $30 mm$ from the leading edge?
 A $\tau _{w}/2$ B $\sqrt{2}\tau _{w}$ C $2\tau _{w}$ D $\tau _{w}/ \sqrt{2}$
GATE ME 2023      Fluid Kinematics
Question 5 Explanation:
Step-1: First check type of flow by Reynold No
\begin{aligned} Re&\propto \frac{u_\infty L}{v} \\ Re &= \frac{2 \times 0.0.3}{1.5 \times 10^{-5}}=4000 \end{aligned}
As Reynold no. is less than $5 \times 10^5$ , it is laminar flow

Step-2: Wall shear stress in laminar flow -
$\tau _w\propto \left ( \frac{1}{\sqrt{x}} \right )$

$\frac{\tau _ {w_2}}{\tau _{w_1}}=\sqrt{\frac{x_1}{x_2}}=\sqrt{\frac{15}{30}}=\sqrt{\frac{1}{2}}$
$\frac{\tau _ {w_2}}{\tau _{w_1}}=\sqrt{\frac{1}{2}}$
So, $\tau _ {w_2}=\frac{tau _{w_1}}{\sqrt{2}}$

There are 5 questions to complete.

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1. q142 diagram missing