# Fluid Kinematics

 Question 1
A two dimensional flow has velocities in $x$ and $y$ directions given by $u=2xyt$ and $v=-y^2 t$, where $t$ denotes time. The equation for streamline passing through $x=1, \; y=1$ is
 A $x^2y=1$ B $xy^2=1$ C $x^2y^2=1$ D $x/y^2=1$
GATE ME 2021 SET-2   Fluid Mechanics
Question 1 Explanation:
\begin{aligned} u &=2 x y t \\ v &=-y^{2} t \\ \frac{d x}{u} &=\frac{d y}{v}=\frac{d z}{w} \\ \frac{d x}{2 x y t} &=\frac{d y}{-y^{2} t} \\ -y d x &=2 x d y \\ \ln x y^{2} &=c \\ x y^{2} &=1 \end{aligned}
 Question 2
For a two-dimensional, incompressible flow having velocity components $u$ and $v$ in the $x$ and $y$ directions, respectively, the expression

$\frac{\partial (u^2)}{\partial x}+\frac{\partial (uv)}{\partial y}$
can be simplified to
 A $u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y}$ B $2u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y}$ C $2u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y}$ D $u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y}$
GATE ME 2021 SET-2   Fluid Mechanics
Question 2 Explanation:
$\frac{\partial\left(u^{2}\right)}{\partial x}+\frac{\partial(u v)}{\partial y}$
By differentiating:
$\Rightarrow 2 u\left[\frac{\partial u}{\partial x}\right]+u \frac{\partial v}{\partial y}+v \frac{\partial u}{\partial y}$
$\Rightarrow u \frac{\partial u}{\partial x}+\left[u \frac{\partial u}{\partial x}+u \frac{\partial v}{\partial y}\right]+v \frac{\partial u}{\partial y}$
According to continity eq. : $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$
So, $u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}$
 Question 3
Consider a flow through a nozzle, as shown in the figure below:

The air flow is steady, incompressible and inviscid. The density of air is $1.23 kg/m^3$. The pressure difference ($p_1 - p_{atm}$) is _______ kPa (round off to the nearest integer)
 A 1522.12 B 1.52 C 321.32 D 3.21
GATE ME 2020 SET-2   Fluid Mechanics
Question 3 Explanation:

\begin{aligned} A_{1} V_{1} &=A_{2} V_{2} \\ 0.2 \times V_{1} &=0.02 \times 50 \\ V_{1} &=\frac{1}{10} \times 50=5 \mathrm{m} / \mathrm{s} \end{aligned}
Applying BE
\begin{aligned} \frac{P_{1}}{w}+\frac{V_{1}^{2}}{2 g}+z_{1} &=\frac{P_{2}}{w}+\frac{V_{2}^{2}}{2 g}+z_{2} \; \; \; (\because z_{1}=z_{2})\\ \frac{P_{1}-P_{2}}{\rho_{\text {air }} g} &=\frac{V_{2}^{2}-V_{1}^{2}}{2 g} \\ P_{1}-P_{2} &=\left(\frac{50^{2}-5^{2}}{2}\right) \times 1.23\\ &=1522.125 \mathrm{Pa}=1.52 \mathrm{kPa} \end{aligned}
 Question 4
The velocity field of an incompressible flow in a Cartesian system is represented by

$\vec{V}=2(x^2-y^2)\hat{i}+v\hat{j}+3\hat{k}$

Which one of the following expressions for v is valid?
 A $-4xz+6xy$ B $-4xy-4xz$ C $4xz-6xy$ D $4xy+4xz$
GATE ME 2020 SET-1   Fluid Mechanics
Question 4 Explanation:
$\vec{V}=\left(2 x^{2}-2 y^{2}\right) \hat{i}+v \hat{j}+3 \hat{k}$
For Incompressible flow
\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial V}{\partial y}+\frac{\partial W}{\partial z} &=0 \\ 4 x+\frac{\partial V}{\partial y} &=0 \\ \frac{\partial V}{\partial y} &=-4 x \\ v &=-4 x y+f(x, z) \end{aligned}
f(x,z) is an arbitary function of x and z
Hence the most suitable answer is option (B)
 Question 5
For a steady flow, the velocity field is $\overrightarrow{\textrm{v}}=(-x^{2}+3y)\hat{i}+(2xy)\hat{j}$ , The magnitude of the acceleration of a particle at (1, -1) is
 A 2 B 1 C $2\sqrt{5}$ D 0
GATE ME 2017 SET-1   Fluid Mechanics
Question 5 Explanation:
Given velocity field,
$\vec{V}=\left(-x^{2}+3 y\right) \hat{i}+(2 x y) \hat{j}$
$where \quad u=-x^{2}+3 y \text { and } v=2 x y$
The acceleration components along x and y -axis.
\begin{aligned} a_{x}&=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}\\ \text{and}\quad a_{y} &=u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y} \\ &=\left(-x^{2}+3 y\right) \times(-2 x)+2 x y \times 3 \\ &=2 x^{3}-6 x y+6 x y=2 x^{3}\\ \text{and}\quad a_{y} &=\left(-x^{2}+3 y\right) \times 2 y+2 x y \times 2 x \\ &=-2 y x^{2}+6 y^{2}+4 x^{2} y=2 y x^{2}+6 y^{2} \end{aligned}
At point (1,-1)
\begin{aligned} a_{x}&=2\\ \text{and}\quad a_{y} &=2 \times(-1) \times 1+6 \times(-1)^{2} \\ &=-2+6=4 \\ \vec{a} &=a_{x} \hat{i}+a_{y} \hat{j}=2 i+4 \hat{j} \end{aligned}
Resultant acceleration:
\begin{aligned} a &=\sqrt{4+16}=\sqrt{20} \\ &=\sqrt{4 \times 5}=2 \sqrt{5} \end{aligned}
 Question 6
Consider the two-dimensional velocity field given by $\overrightarrow{\mathrm{v}}=(5+a_{1}x+b_{1}y)\hat{i}\: + \: (4+a_{2}x+b_{2}y)\hat{j}$ ,where $a_1,b_1,a_2 text{ and }b_2$ are constants. Which one of the following conditions needs to be satisfied for the flow to be incompressible?
 A $a_{1}+\, b_{1}=0$ B $a_{1}+\, b_{2}=0$ C $a_{2}+\, b_{2}=0$ D $a_{2}+\, b_{1}=0$
GATE ME 2017 SET-1   Fluid Mechanics
Question 6 Explanation:
Given that the flow is incompressible.
$\therefore \quad \operatorname{div} \bar{V}=0$
$\frac{\partial}{\partial x}\left(5+a_{1} x+b_{1} y\right)+\frac{\partial}{\partial y}\left(8+a_{2} x+b_{2} y\right)=0$
$a_{1}+b_{2}=0$
 Question 7
For a two-dimensional flow, the velocity field is $\vec{u}=\frac{x}{x^{2}+y^{2}}\hat{i}+\frac{y}{x^{2}+y^{2}}\hat{j}$, where $\hat{i}$ and $\hat{j}$ are the basis vectors in the x-y Cartesian coordinate system. Identify the CORRECT statements from below.
(1)The flow is incompressible
(3) y-component of acceleration, $a_{y}=\frac{-y}{(x^{2}+y^{2})^{2}}$
(4) x-component of acceleration,$a_{x}=\frac{-(x+y)}{(x^{2}+y^{2})^{2}}$
 A (2) and (3) B (1) and (3) C (1) and (2) D (3) and (4)
GATE ME 2016 SET-3   Fluid Mechanics
Question 7 Explanation:
For $2 \mathrm{D}$ -flow velocity field is:
\begin{aligned} \vec{V}&=\frac{x}{x^{2}+y^{2}} \hat{i}+\frac{y}{x^{2}+y^{2}} \hat{j}\\ so \quad u&=\frac{x}{x^{2}+y^{2}} and v=\frac{y}{x^{2}+y^{2}} \\ a_{x} &=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y} \\ &=\frac{x}{x^{2}+y^{2}} \frac{\partial}{\partial x}\left[\frac{x}{x^{2}+y^{2}}\right] \\ &+\frac{y}{x^{2}+y^{2}} \frac{\partial}{\partial y}\left[\frac{x}{x^{2}+y^{2}}\right] \\ &= \frac{x}{x^{2}+y^{2}}\left[\frac{x(-2 x)}{\left(x^{2}+y^{2}\right)^{2}}+\frac{1}{x^{2}+y^{2}}\right]\\ &+\frac{y}{x^{2}+y^{2}} \left[\frac{x(2 y)}{\left(x^{2}+y^{2}\right)^{2}}\right]\\ &=\frac{-2 x^{3}}{\left(x^{2}+y^{2}\right)^{3}}+\frac{x}{\left(x^{2}+y^{2}\right)^{2}}\\ &+ \frac{2xy^{2}}{\left(x^{2}+y^{2}\right)^{3}}\\ &=\frac{-2 x^{3}+x\left(x^{2}+y^{2}\right)-2 x y^{2}}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-2 x^{3}+x^{3}+x y^{2}-2 x y^{2}}{\left(x^{2}+y^{2}\right)^{3}}\\ &=\frac{-x^{3}-x y^{2}}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-x\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \end{aligned}
\begin{aligned} a_{x} &=\frac{x}{\left(x^{2}+y^{2}\right)^{2}} \\ a_{y} &=u \frac{\partial v}{\partial x}+u \frac{\partial v}{\partial y} \\ &=\frac{x}{x^{2}+y^{2}} \frac{\partial}{\partial x}\left[\frac{y}{x^{2}+y^{2}}\right] \\& +\frac{y}{x^{2}+y^{2}} \frac{\partial}{\partial y}\left[\frac{y}{x^{2}+y^{2}}\right] \\ &=\frac{x}{x^{2}+y^{2}} \frac{(-2 x y)}{\left(x^{2}+y^{2}\right)^{2}}\\& +\frac{y}{x^{2}+y^{2}} \times\left[\frac{-2 y^{2}}{\left(x^{2}+y^{2}\right)^{2}}+\frac{1}{x^{2}+y^{2}}\right]\\ &=\frac{-2 x^{2} y}{\left(x^{2}+y^{2}\right)^{3}}+\frac{2 y^{3}}{\left(x^{2}+y^{2}\right)^{3}}\\&+\frac{y}{\left(x^{2}+y^{2}\right)^{2}} \\ &=\frac{-2 x^{2} y-2 y^{3}+y\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-x^{2} y-y^{3}}{\left(x^{2}+y^{2}\right)^{3}}=\frac{y\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \\ a_{y}&=\frac{-y}{\left(x^{2}+y^{2}\right)^{2}} \\ \end{aligned}
 Question 8
For a certain two-dimensional incompressible flow, velocity field is given by $2xy\hat{i}-y^{2}\hat{j}$. The streamlines for this flow are given by the family of curves
 A $x^{2}y^{2}$ = constant B $xy^{2}$ =constant C $2xy-y^{2}$ =constant D $xy$ =constant
GATE ME 2016 SET-3   Fluid Mechanics
Question 8 Explanation:
$\vec{v}=2 x y \hat{i}-y^{2} \hat{j}$
where $u=2 x y$ and $v=-y^{2}$
Equation of stream line,
\begin{aligned} \frac{d x}{u} &=\frac{d y}{v} \\ \frac{d x}{2 x y} &=\frac{d y}{-y^{2}} \\ \text{or}\quad \frac{d x}{2 x} &=-\frac{d y}{y} \end{aligned}
On integrating, we get
\begin{aligned} \frac{1}{2} \log _{e} x&=-\log _{e} y+\log _{e} c \\ \log _{e} x^{1 / 2}+\log _{e} y&=\log _{e} c \\ \log _{e} x^{1 / 2} y&=\log _{e} c \\ \text { or } \quad x^{1 / 2} y&=c \end{aligned}
 Question 9
The volumetric flow rate (per unit depth) between two streamlines having stream functions $\Psi_{1}$ and $\Psi_{2}$ is
 A $|\Psi 1+\Psi 2|$ B $\Psi 1\Psi 2$ C $\Psi 1/\Psi 2$ D $|\Psi 1-\Psi 2|$
GATE ME 2016 SET-2   Fluid Mechanics
 Question 10
If the fluid velocity for a potential flow is given by $V(x,y)=u(x,y)i+v(x,y)j$ with usual notations, then the slope of the potential line at $(x,y)$ is
 A $v/u$ B $-u/v$ C $v^{2}/u^{2}$ D $u/v$
GATE ME 2015 SET-2   Fluid Mechanics
Question 10 Explanation:
\begin{aligned} d \phi&=\frac{\partial \phi}{\partial x} d x+\frac{\partial \phi}{\partial y} d y\\ \therefore \qquad \frac{\partial \phi}{\partial x}&=-u\\ \frac{\partial \phi}{\partial y}&=-v \\ \therefore \qquad d \phi&=-u d x-v d y \\ \text { If } \phi&=\text { constant, } d \phi=0 \\ \therefore \quad\left(\frac{d y}{d x}\right)_{\phi}&=-\frac{u}{v} \end{aligned}
There are 10 questions to complete.

### 5 thoughts on “Fluid Kinematics”

1. In question 18, the vorticity vector is to be found, but its written ‘velocity vector’. Please rectify that.

• Thank you Spandan Samanta,