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.
There are 5 questions to complete.