Question 1 |
An explosion at time t=0 releases energy E at the origin in a space filled with
a gas of density \rho . Subsequently, a hemispherical blast wave propagates
radially outwards as shown in the figure.
Let R denote the radius of the front of the hemispherical blast wave. The radius R follows the relationship R =kt^aE^b\rho ^c, where k is a dimensionless constant.
The value of exponent a is _____. (Rounded off to one decimal place)
Let R denote the radius of the front of the hemispherical blast wave. The radius R follows the relationship R =kt^aE^b\rho ^c, where k is a dimensionless constant.
The value of exponent a is _____. (Rounded off to one decimal place)
0.1 | |
0.4 | |
0.6 | |
0.8 |
Question 1 Explanation:
Radius, R=f(t, E, \rho)
Where, \mathrm{t}= time, \mathrm{E}= Release energy \& \rho= density
Dimension, \mathrm{R}=[\mathrm{L}]
\begin{aligned} & t=[\mathrm{T}] \\ & E=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \\ & \rho=\left[\mathrm{ML}^{-3}\right] \end{aligned}
Fundamental dimensions are \mathrm{M}, \mathrm{L}, \mathrm{T} (basic)
Total no. of variables =4
No. of \pi. terms =4-3=1
So, R=t^{a} E^{b} \rho^{c}
[\mathrm{L}]=[\mathrm{T}]^{\mathrm{a}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]^{\mathrm{b}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{c}}
Comparing both side indices of M, L, T
a-2b = 0, 2b-3c = 1, b+c = 0 \Rightarrow b-c
a = 2 b, 2(-c) - 3c = 1
-5c = 1, c =(-1)/5 \Rightarrow b=1/5
So, a=2 \times \frac{1}{5} = \frac{2}{5}
Relation, R=kt^{2/5}E^{1/5}\rho ^{-1/5}=\frac{kt^{2/5}E^{1/5}}{\rho ^{1/5}}
Where, a=\frac{2}{5}=0.40
Question 2 |
The figure shows two fluids held by a hinged gate. The atmospheric pressure is P_a = 100 kPa. The moment per unit width about the base of the hinge is
________ kNm/m. (Rounded off to one decimal place)
Take the acceleration due to gravity to be g = 9.8 m/s^2 .
Take the acceleration due to gravity to be g = 9.8 m/s^2 .
45.8 | |
57.2 | |
68.2 | |
87.6 |
Question 2 Explanation:
Forces from Pressure diagram
\mathrm{F}_{1}= Area of \triangle \mathrm{ADE} \times width
F_{1}=\frac{1}{2} \times A D \times D E \times width
=\frac{1}{2} \times 1 \times 9810 \times 1
\mathrm{F}_{1}=4905 \mathrm{~N}
\mathrm{F}_{2}= Area of rectangle \mathrm{DEBF} \times width
F_{2}=2 \times 9810 \times 1
\mathrm{F}_{2}=19620 \mathrm{~N}
\mathrm{F}_{3}= Area of triangle \Delta \mathrm{EFC}
=\frac{1}{2} \times \mathrm{EF} \times \mathrm{FC} \times width
=\frac{1}{2} \times 2 \times 39240 \times 1
\mathrm{F}_{3}=39240 \mathrm{~N}
Location of forces from top
\begin{aligned} & \mathrm{F}_{1} \rightarrow \frac{2}{3} \times 1=\frac{2}{3} \mathrm{~m} \\ & \mathrm{~F}_{2} \rightarrow\left(1 \mathrm{~m}+\frac{2 \mathrm{~m}}{2}\right)=2 \mathrm{~m} \\ & \mathrm{~F}_{3} \rightarrow 1 \mathrm{~m}+2 \mathrm{~m} \times \frac{2}{3}=\frac{7}{3} \mathrm{~m} \end{aligned}
Pressure at depth 1 \mathrm{~m}=\rho_{1} \mathrm{gh}_{1}
1000 \times 9.81 \times 1=9810 \frac{N}{m^{2}}
\mathrm{DE}=9810 \frac{\mathrm{N}}{\mathrm{m}^{2}}
BF=9810 \frac{N}{m^{2}}
Pressure at depth 3 m
\begin{aligned} & \rho_{1} g h_{1}+\rho_{2} g h_{2} \\ & 9810+2000 \times 9.81 \times 2 \\ & 9810+39240=49050 \frac{\mathrm{N}}{\mathrm{m}^{2}} \end{aligned}
(B F)+(F C)=B C
Location of forces from bottom hinge
\begin{aligned} & \mathrm{F}_{1} \rightarrow 3 \mathrm{~m}-\frac{2}{3} \mathrm{~m}=\frac{7}{3} \mathrm{~m} \\ & \mathrm{~F}_{2} \rightarrow 3 m-2 m=1 \mathrm{~m} \\ & \mathrm{~F}_{3} \rightarrow 3 m-\frac{7}{3} \mathrm{~m}=\frac{2}{3} \mathrm{~m} \end{aligned}
Moment about hinge \mathrm{F}_{1} \times \frac{7}{3}+\mathrm{F}_{2} \times 1+\mathrm{F}_{3} \times \frac{2}{3}
4905 \times \frac{7}{3}+19620+39240 \times \frac{2}{3}
\Rightarrow \quad 57225 \mathrm{Nm}=57.22 \mathrm{kNm} / \mathrm{m}
Question 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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}}
There are 5 questions to complete.
q142 diagram missing
There is no diagram in the official question paper.
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