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
x^2y=1 | |
xy^2=1 | |
x^2y^2=1 | |
x/y^2=1 |
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
\frac{\partial (u^2)}{\partial x}+\frac{\partial (uv)}{\partial y}
can be simplified to
u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y} | |
2u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y} | |
2u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y} | |
u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y} |
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}
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)
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)
1522.12 | |
1.52 | |
321.32 | |
3.21 |
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?
\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?
-4xz+6xy | |
-4xy-4xz | |
4xz-6xy | |
4xy+4xz |
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)
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
2 | |
1 | |
2\sqrt{5} | |
0 |
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}
\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_{1}+\, b_{1}=0 | |
a_{1}+\, b_{2}=0 | |
a_{2}+\, b_{2}=0 | |
a_{2}+\, b_{1}=0 |
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
\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
(2)The flow is unsteady.
(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}}
(1)The flow is incompressible
(2)The flow is unsteady.
(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}}
(2) and (3) | |
(1) and (3) | |
(1) and (2) | |
(3) and (4) |
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}
\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
x^{2}y^{2} = constant | |
xy^{2} =constant | |
2xy-y^{2} =constant | |
xy =constant |
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}
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
|\Psi 1+\Psi 2| | |
\Psi 1\Psi 2 | |
\Psi 1/\Psi 2 | |
|\Psi 1-\Psi 2| |
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
v/u | |
-u/v | |
v^{2}/u^{2} | |
u/v |
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.
In question 18, the vorticity vector is to be found, but its written ‘velocity vector’. Please rectify that.
Thank you Spandan Samanta,
We have updated the answer.
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
Question no 26 wrong answer is given. Please corrrect it..