Question 1 |
The velocity components in the x and y directions for an incompressible flow are given
as u=(-5+6x) and v=-(9+6y), respectively. The equation of the streamline is
(-5+6x)-(9+6y)= constant | |
\frac{-5+6x}{9+6y}=constant | |
(-5+6x)(9+6y)= constant | |
\frac{9+6y}{-5+6x}=constant |
Question 1 Explanation:
\begin{aligned} \text{Given } \; \; \; & u=-5+6x \\ & v=-(9+6y) \end{aligned}
Equation of streamline
\begin{aligned} \frac{dx}{u}&=\frac{dy}{u} \\ \frac{dx}{-5+6x}&=\frac{dy}{-(9+6y)} \\ \text{Integrating it,} \\ \ln(-5+6x)^{1/6}&= -\ln(9+6y)^{1/6}+ \ln C^{1/6}\\ \frac{1}{6} \ln (-5+6x)(9+6y)&=\frac{1}{6} \ln C \\ \text{Take antilog,} \\ (-5 + 6x)(9 + 6y) &= \text{constant} \\ \text{u.v} &= \text{constant} \end{aligned}
Equation of streamline
\begin{aligned} \frac{dx}{u}&=\frac{dy}{u} \\ \frac{dx}{-5+6x}&=\frac{dy}{-(9+6y)} \\ \text{Integrating it,} \\ \ln(-5+6x)^{1/6}&= -\ln(9+6y)^{1/6}+ \ln C^{1/6}\\ \frac{1}{6} \ln (-5+6x)(9+6y)&=\frac{1}{6} \ln C \\ \text{Take antilog,} \\ (-5 + 6x)(9 + 6y) &= \text{constant} \\ \text{u.v} &= \text{constant} \end{aligned}
Question 2 |
Uniform flow with velocity U makes an angle \theta with the y-axis, as shown in the figure
The velocity potential (\Phi), is
The velocity potential (\Phi), is
\pm U(x \sin \theta +y \cos \theta ) | |
\pm U(y \sin \theta -x \cos \theta ) | |
\pm U(x \sin \theta -y \cos \theta ) | |
\pm U(y \sin \theta +x \cos \theta ) |
Question 2 Explanation:
Velocity in x-depth, u_x=u\sin \theta
Velocity in y-depth, u_y=u\cos \theta
\begin{aligned} -\frac{\partial \phi }{\partial x}&=u_x \\ &{Integrating it} \\ \phi &= -u_xx+f(y)+c\\ &= -(u \sin \theta )x+f(y)+c...(i)\\ -\frac{\partial \phi }{\partial y}&=u_y \\ &\text{Integrating it} \\ \phi &= -u_y y+f(x)+c\\ &= -(u \cos \theta )y+f(x)+c...(ii)\\ & \text{By equation (i) and (ii),}\\ \phi &= -u(x \sin \theta+y \cos \theta )\\ &\text{If we take,} \\ \frac{\partial \phi }{\partial x}&= u_x \; and \; \frac{\partial \phi }{\partial y}= u_y\\ \text{then}, \; \phi &= u(x \sin \theta +y \cos \theta )\\ \text{So}, \; \phi &= \pm u(x \sin \theta +y \cos \theta ) \end{aligned}
Velocity in y-depth, u_y=u\cos \theta
\begin{aligned} -\frac{\partial \phi }{\partial x}&=u_x \\ &{Integrating it} \\ \phi &= -u_xx+f(y)+c\\ &= -(u \sin \theta )x+f(y)+c...(i)\\ -\frac{\partial \phi }{\partial y}&=u_y \\ &\text{Integrating it} \\ \phi &= -u_y y+f(x)+c\\ &= -(u \cos \theta )y+f(x)+c...(ii)\\ & \text{By equation (i) and (ii),}\\ \phi &= -u(x \sin \theta+y \cos \theta )\\ &\text{If we take,} \\ \frac{\partial \phi }{\partial x}&= u_x \; and \; \frac{\partial \phi }{\partial y}= u_y\\ \text{then}, \; \phi &= u(x \sin \theta +y \cos \theta )\\ \text{So}, \; \phi &= \pm u(x \sin \theta +y \cos \theta ) \end{aligned}
Question 3 |
The velocity field in a flow system is given by v=2i+(x+y)j+(xyz)k. The acceleration of the fluid at (1,1,2) is
2i + 10k | |
4i + 12k | |
l + k | |
4j + 10k |
Question 3 Explanation:
\begin{aligned}
\vec{v} &=2\hat{i}(x+y)\hat{j}+xyz\hat{k} \\
u&=2 \\
v&=x+y \\
w&=xyz \\
a_x&=u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}+\frac{\partial u}{\partial t}=0\\
a_y&=x+y+2 \\
a_z&= u\frac{\partial w}{\partial x} +v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z}+\frac{\partial w}{\partial t}\\
&= 2(yz)+(x+y)(xz)+xyz(xy)\\
a_z&=2yz+x^2z+xyz+x^2y^2z \\
&At\;(1,1,2) \\
a_y&=11+2=4\\
a_z&=2(1)(2)+(1)^2(2)+(1)(1)(2)+(1)^2(1)^2(2)\\
&=4+2+2+2=10\\
\vec{a}&=4\hat{j}+10\hat{k}
\end{aligned}
Question 4 |
A solid sphere of radius, r, and made of material with density, \rho _s, is moving through the atmosphere (constant pressure, p)with a velocity, v. The net force ONLY due to atmospheric pressure (F_p) acting on the sphere at any time, t, is
\pi r^2p | |
4 \pi r^2p | |
\frac{4}{3}\pi r^3 \rho _s\frac{dv}{dt} | |
zero |
Question 4 Explanation:
When object of any shape ( not only sphere) is subjected to uniform pressure over entire surface, the net force is zero because pressure force at any point is balanced by an equal and opposite force on opposite point.
Question 5 |
A flow field is given by u=y^{2} ,v=-xy,w=0. Value of the component of the angular velocity(in radians per unit time,up to two decimal places)at the point (0,-1,1) is_____
0.5 | |
1 | |
1.5 | |
2 |
Question 5 Explanation:
\begin{aligned} \omega_{z} &=\frac{1}{2}\left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right] \\ &=\frac{1}{2}\left[\frac{\partial}{\partial x}(-x y)-\frac{\partial}{\partial y}(y)^{2}\right] \\ &=\frac{1}{2}[-y-2 y] \\ &=-\frac{3 y}{2} \\ \text { At point } &(0,-1,1) \\ \omega_{z} &=-\frac{3}{2} \times-1=1.50 \mathrm{rad} / \mathrm{s} \end{aligned}
Note: Since the density variation is not given continuity equation of incompressible flow can not be applied directly to check the possibility of flow.
Note: Since the density variation is not given continuity equation of incompressible flow can not be applied directly to check the possibility of flow.
Question 6 |
The velocity components of a two dimensional plane motion of a fluid are: u=\frac{y^{3}}{3}+2x-x^{2}y and v=xy^{2}-2y-\frac{x^{3}}{3}.
The correct statement is:
The correct statement is:
Fluid is incompressible and flow is irrotational | |
Fluid is incompressible and flow is rotational | |
Fluid is compressible and flow is irrotational | |
Fluid is compressible and flow is rotational |
Question 6 Explanation:
Continuity equation
\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y} &=0 \\ \Rightarrow \quad(2-2 x y)+(2 x y-2) &=0 \\ 0 &=0 \end{aligned}
Hence, incompressible fluid.
\begin{aligned} \omega_{z} &=\frac{1}{2}\left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right] \\ &=\frac{1}{2}\left[y^{2}-\frac{3 x^{2}}{3}-\left(\frac{3 y^{2}}{3}-x^{2}\right)\right] \\ &=\frac{1}{2}\left[y^{2}-x^{2}-y^{2}+x^{2}\right] \\ &=0 \end{aligned}
Hence, flow is irrotational.
\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y} &=0 \\ \Rightarrow \quad(2-2 x y)+(2 x y-2) &=0 \\ 0 &=0 \end{aligned}
Hence, incompressible fluid.
\begin{aligned} \omega_{z} &=\frac{1}{2}\left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right] \\ &=\frac{1}{2}\left[y^{2}-\frac{3 x^{2}}{3}-\left(\frac{3 y^{2}}{3}-x^{2}\right)\right] \\ &=\frac{1}{2}\left[y^{2}-x^{2}-y^{2}+x^{2}\right] \\ &=0 \end{aligned}
Hence, flow is irrotational.
Question 7 |
A nozzle is so shaped that the average flow velocity changes linearly from 1.5 m/s at the beginning to 15 m/s at its end in a distance of 0.375 m. The magnitude of the convective acceleration (in m/s^{2}) at the end of the nozzle is _________.
74 | |
54 | |
540 | |
740 |
Question 7 Explanation:
Convective acceleration =\frac{u \partial u}{\partial x}
=15 \times \frac{15-1.5}{0.375}=540 \mathrm{m} / \mathrm{s}^{2}
=15 \times \frac{15-1.5}{0.375}=540 \mathrm{m} / \mathrm{s}^{2}
Question 8 |
In a two-dimensional steady flow field, in a certain region of the x-y plane, the velocity component in the x-direction is given by v_{x}=x^{2} and the density varies as \rho =\frac{1}{x}. Which of the following is a valid expression for the velocity component in the y- direction, v_{y} ?
v_{y}=-x/y | |
v_{y}=x/y | |
v_{y}=-xy | |
v_{y}=xy |
Question 8 Explanation:
Continuity equation in steady flow.
\begin{aligned} \frac{\partial(\rho u)}{\partial x}+\frac{\partial(\rho v)}{\partial y} &=0 \\ \frac{\partial}{\partial x}\left[\frac{1}{x} \cdot x^{2}\right]+\frac{\partial}{\partial y}\left(\frac{1}{x} \cdot v_{y}\right) &=0 \\ \frac{\partial}{\partial x}(x)+\frac{1}{x} \frac{\partial v_{y}}{x \partial y} &=0 \\ 1+\frac{1}{x} \frac{\partial v_{y}}{\partial y} &=0 \\ -x &=\frac{\partial v_{y}}{\partial y} \end{aligned}
On integrating both sides
v_{y}=-x y
\begin{aligned} \frac{\partial(\rho u)}{\partial x}+\frac{\partial(\rho v)}{\partial y} &=0 \\ \frac{\partial}{\partial x}\left[\frac{1}{x} \cdot x^{2}\right]+\frac{\partial}{\partial y}\left(\frac{1}{x} \cdot v_{y}\right) &=0 \\ \frac{\partial}{\partial x}(x)+\frac{1}{x} \frac{\partial v_{y}}{x \partial y} &=0 \\ 1+\frac{1}{x} \frac{\partial v_{y}}{\partial y} &=0 \\ -x &=\frac{\partial v_{y}}{\partial y} \end{aligned}
On integrating both sides
v_{y}=-x y
Question 9 |
A plane flow has velocity components u=\frac{x}{T_{1}}, \; v=-\frac{y}{T_{2}} and w=0 along x, y and z directions respectively, whereT_{1}(\neq 0) and T_{2}(\neq 0) are constants having the dimensions of time. The given flow is incompressible if
T_{1}=-T_{2} | |
T_{1}=-\frac{T_{2}}{2} | |
T_{1}=\frac{T_{2}}{2} | |
T_{1}=T_{2} |
Question 9 Explanation:
For a flow to exist
\begin{aligned} \Rightarrow \quad \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}&=0 \\ \Rightarrow \quad \frac{1}{T_{1}}-\frac{1}{T_{2}}&=0 \\ \Rightarrow \quad T_{1}&=T_{2} \end{aligned}
\begin{aligned} \Rightarrow \quad \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}&=0 \\ \Rightarrow \quad \frac{1}{T_{1}}-\frac{1}{T_{2}}&=0 \\ \Rightarrow \quad T_{1}&=T_{2} \end{aligned}
Question 10 |
A particle moves along a curve whose parametric equations are: x=t^{3}+2t, y=-3e^{-2t} and z=2 \sin (5t) , where x, y and z show variations of the distance covered by the particle (in cm)with time t (in s). The magnitude of the acceleration of the particle (in cm/s^{2}) at t = 0 is ________
8 | |
10 | |
12 | |
16 |
Question 10 Explanation:
\begin{aligned} x &=t^{3}+2 t \\ y &=-3 e^{-2 t} \\ z &=2 \sin (5 t) \\ \frac{d x}{d t} &=3 e^{2}+2 \\ \Rightarrow\quad a_{x} &=\frac{d^{2} x}{d t^{2}}=6 t \\ \frac{d y}{d t} &=-3 e^{-2} \times(-2)=6 e^{-2 t} \\ \Rightarrow\quad a_{y} &=\frac{d^{2} y}{d t^{3}}=-12 e^{-2 t}\\ \Rightarrow\quad \frac{d z}{d t} &=2 \times 5 \cos (5 t) \\ &=10 \cos (5 t) \\ \Rightarrow\quad a_{z} &=\frac{d^{2} z}{d t^{2}}=-50 \sin 5 t \\ \vec{a} &=a_{x} \hat{i}+a_{y} \hat{j}+a_{z} \hat{k} \\ \vec{a} \text { at } t=0 &=0 \hat{i}-12 \hat{j}+0 \hat{k} \\ \vec{a} &=-12 \hat{j} \end{aligned}
\Rightarrow\quad Magnitude of acceleration at t=0
=12 \mathrm{cm} / \mathrm{s}^{2}
\Rightarrow\quad Magnitude of acceleration at t=0
=12 \mathrm{cm} / \mathrm{s}^{2}
There are 10 questions to complete.