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: 
Continuity equation for unsteady flow
\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.

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