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.
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.
1.14 | |
2.25 | |
3.65 | |
8.25 |
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}
\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)
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)
64 | |
26 | |
84 | |
98 |
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}
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?
\tau _{w}/2 | |
\sqrt{2}\tau _{w} | |
2\tau _{w} | |
\tau _{w}/ \sqrt{2} |
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}}
\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
10\hat{j} | |
20\hat{i} | |
-6\hat{i}-6\hat{j} | |
-4\hat{i}-4\hat{j} |
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}
\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
\vec{V}=4(xy)\hat{i}+2(x^2-y^2)\hat{j}
Which of the following statement(s) is/are correct?
MSQ
The velocity field is one-dimensional. | |
The flow is incompressible | |
The flow is irrotational | |
The acceleration experienced by a fluid particle is zero at (x = 0, y = 0). |
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}
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.
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..
please correct the ques. no. 29
In question no. 24, this is vorticity vector at (1,1,1) not velocity vector
We have updated the question.