# Fluid Kinematics

 Question 1
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 Mechanics
Question 1 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 2
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 Mechanics
Question 2 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 3
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 Mechanics
Question 3 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, $\tau _ {w_2}=\frac{tau _{w_1}}{\sqrt{2}}$
 Question 4
The steady velocity field in an inviscid fluid of density 1.5 is given to be $\vec{V}=(y^2-x^2)\hat{i}+(2xy)\hat{j}$ Neglecting body forces, the pressure gradient at $(x = 1, y = 1)$ is
 A $10\hat{j}$ B $20\hat{i}$ C $-6\hat{i}-6\hat{j}$ D $-4\hat{i}-4\hat{j}$
GATE ME 2022 SET-2   Fluid Mechanics
Question 4 Explanation:
By Euler's equation of motion,
\begin{aligned} x:-\frac{\partial p}{\partial x}+\rho g_x&=\rho \frac{Du}{Dt}\\ y:-\frac{\partial p}{\partial y}+\rho g_y&=\rho \frac{Dv}{Dt}\\ \end{aligned}
Neglecting body forces (i.e. $g_x=g_y=0$)
\begin{aligned} \frac{\partial p}{\partial x}&=-\rho \frac{Du}{Dt}=-\rho \left [ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(-2x)+(2xy)(2y)]\\ &=-\rho[-2xy^2+2x^3+4xy^2]\\ &=-\rho(2xy^2+2x^3)\\ &=-1.5 \times (2 \times 1 \times 1^2 +2\times1^3 )\\ &=-6 Pa/m \end{aligned}
Similarly,
\begin{aligned} \frac{\partial p}{\partial y}&=-\rho \frac{Dv}{Dt}\\ &=-\rho \left [ u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(2y)+(2xy)(2x)]\\ &=-\rho[2y^3-2x^2y+4x^2y]\\ &=-\rho(2y^3+2x^2y)\\ &=-1.5 \times (2 \times 1^3 +2\times1^2 \times 1 )\\ &=-6 Pa/m \end{aligned}
The pressure gradient vector is given by
$\triangledown p=\frac{\partial p}{\partial x}\hat{i}+\frac{\partial p}{\partial y}\hat{j}=-6\hat{i}-6\hat{j}$
 Question 5
The velocity field in a fluid is given to be
$\vec{V}=4(xy)\hat{i}+2(x^2-y^2)\hat{j}$
Which of the following statement(s) is/are correct?

MSQ
 A The velocity field is one-dimensional. B The flow is incompressible C The flow is irrotational D The acceleration experienced by a fluid particle is zero at (x = 0, y = 0).
GATE ME 2022 SET-2   Fluid Mechanics
Question 5 Explanation:
For given flow,
$u=4xy, v = 2(x^2- y^2)$
As velocity field is function of two space variables, flow is two dimensional.
$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=4y-4y$
Therefore, flow is incompressible.
$\omega _z=\frac{1}{2}\left ( \frac{\partial v}{\partial x} -\frac{\partial u}{\partial y}\right )=\frac{1}{2}(4x-4x)=0$
Therefore, flow is irrotational.
\begin{aligned} a_x &=u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}\\ &= 4xy(4y)+2(x^2-y^2)(4x)\\ &= 16xy^2+8x^3-8xy^2 \\ &=16 \times 0 \times 0^2+8 \times 0^3-8 \times 0 \times 0^2 \\ &= 0 \end{aligned}
\begin{aligned} a_y &=u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\\ &= 4xy(4x)+2(x^2-y^2)(-4y)\\ &= 16x^2y-8x^2y+8y^3 \\ &=16 \times 0^2 \times 0-8 \times 0^2 \times 0+8 \times 0^3 \\ &= 0\\ |\vec{a}|&=\sqrt{a_x^2+a_y^2}=0 \end{aligned}

There are 5 questions to complete.

### 9 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,

2. In question 23, the question and the answer given are completely different, please check that.

• Thank you Spandan Samanta,
We have updated the answer as option A.

• still not solved..kindly read the question and explaination first

3. 4. please correct the ques. no. 29

5. • 