Question 1 |
Two discrete spherical particles (P and Q) of equal mass density are
independently released in water. Particle P and particle Q have diameters of
0.5 mm and 1.0 mm, respectively. Assume Stokes' law is valid.
The drag force on particle Q will be________ times the drag force on particle P. (round off to the nearest integer)
The drag force on particle Q will be________ times the drag force on particle P. (round off to the nearest integer)
4 | |
8 | |
10 | |
12 |
Question 1 Explanation:
In case of discrete particle settling and Stoke's
law valid, at terminal velocity, since there is no
change in velocity, the net force on the body
is zero. Hence,
F_D=\left ( \frac{\pi}{6}D^3\rho _sg-\frac{\pi}{6}D^3\rho _sg \right )=\frac{\pi}{6}gD^3(\rho _s-\rho _f)
For density of medium (\rho _f) and mass density of sphere (\rho _s) constant,
Drag force (F_D)\propto D^3
For particle P, diameter (D_P)=0.5 mm
For particle Q, diameter (D_Q)=1 mm
\Rightarrow \frac{(F_D)_Q}{(F_D)_P}=\frac{(D_Q)^3}{(D_P)^3}=\left ( \frac{1}{0.5} \right )^3=8
\Rightarrow (F_D)_Q=8 \times (F_D)P
F_D=\left ( \frac{\pi}{6}D^3\rho _sg-\frac{\pi}{6}D^3\rho _sg \right )=\frac{\pi}{6}gD^3(\rho _s-\rho _f)
For density of medium (\rho _f) and mass density of sphere (\rho _s) constant,
Drag force (F_D)\propto D^3
For particle P, diameter (D_P)=0.5 mm
For particle Q, diameter (D_Q)=1 mm
\Rightarrow \frac{(F_D)_Q}{(F_D)_P}=\frac{(D_Q)^3}{(D_P)^3}=\left ( \frac{1}{0.5} \right )^3=8
\Rightarrow (F_D)_Q=8 \times (F_D)P
Question 2 |
Velocity distribution in a boundary layer is given by \frac{u}{U_\infty }=\sin \left ( \frac{\pi}{2}\frac{y}{\delta } \right ), where u is the
velocity at vertical coordinate y, U_\infty is the free stream velocity and \delta is the boundary
layer thickness. The values of U_\infty and \delta are 0.3 m/s and 1.0 m, respectively. The velocity gradient \left ( \frac{\partial u}{\partial y} \right ) (in s^{-1}, round off to two decimal places) at y= 0, is ________.
0.24 | |
0.38 | |
0.47 | |
0.58 |
Question 2 Explanation:
Given:
\begin{aligned} \frac{u}{u_\infty }&=\sin \left ( \frac{\pi}{2} \frac{y}{\delta }\right )\\ u_\infty &=0.3 m/s\\ \delta &=1m\\ \frac{du}{dy}&=\frac{d}{dy}u_\infty \cdot \sin\left ( \frac{\pi}{2} \frac{y}{\delta } \right )\\ &=\frac{u_\infty \cdot \pi}{2\delta } \cos \left ( \frac{\pi}{2} \frac{y}{\delta } \right )\\ \text{At y=0 and} & \; \delta=1 \\ \left.\begin{matrix} \frac{du}{dy} \end{matrix}\right|_{y=0}&=\frac{0.3 \pi}{2(1)} \cos \left ( \frac{\pi}{2}\frac{0}{1} \right )\\ &=0.47 s^{-1} \end{aligned}
\begin{aligned} \frac{u}{u_\infty }&=\sin \left ( \frac{\pi}{2} \frac{y}{\delta }\right )\\ u_\infty &=0.3 m/s\\ \delta &=1m\\ \frac{du}{dy}&=\frac{d}{dy}u_\infty \cdot \sin\left ( \frac{\pi}{2} \frac{y}{\delta } \right )\\ &=\frac{u_\infty \cdot \pi}{2\delta } \cos \left ( \frac{\pi}{2} \frac{y}{\delta } \right )\\ \text{At y=0 and} & \; \delta=1 \\ \left.\begin{matrix} \frac{du}{dy} \end{matrix}\right|_{y=0}&=\frac{0.3 \pi}{2(1)} \cos \left ( \frac{\pi}{2}\frac{0}{1} \right )\\ &=0.47 s^{-1} \end{aligned}
Question 3 |
Consider a laminar flow in the x-direction between two infinite parallel plates (Couette flow). The lower plate is stationary and the upper plate is moving with a velocity of 1 cm/s in the x-direction. The distance between the plates is 5 mm and the dynamic viscosity of the fluid is 0.01 N-s/m^2. If the shear stress on the lower plate is zero, the pressure gradient, \frac{\partial p}{\partial x}, (in N/m^2 per m,round off to 1 decimal place)is _______
2.4 | |
8.0 | |
6.2 | |
9.8 |
Question 3 Explanation:
Given flow is a couette flow with pressure gradient
\mu =0.01N-s/m^2
h=5mm
if \tau _{y=0}=0, \; \text{find}\; \frac{dp}{dx}
The velocity distribution, for the couette flow with pressure gradient is given by
u=\frac{wy}{h}-\frac{1}{2\mu }\frac{dp}{dx}[hy-y^2]
where w is the velocity of the top plate and h is the gap between the plates
\begin{aligned} \frac{du}{dy}&=\frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}[h-2y] \\ \left.\begin{matrix} \frac{du}{dy} \end{matrix}\right|_{y=0} &=\frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}h \\ \tau _{y=0}&=\mu \left [ \frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}h \right ] \\ \tau _{y=0}&=\frac{w\mu }{h}-\frac{1}{2}\frac{dp}{dx}h \\ \text{when}\; \tau _{y=0}&=0, \; \text{we will have}\\ \frac{1}{2}\frac{dp}{dx}h&=\frac{w\mu }{h}\\ \frac{dp}{dx}&=\frac{2w\mu }{h^2} \\ &=\frac{2 \times 1 \times 10^{-2}\times 0.01}{(5 \times 10^{-3})^2} \\ &=8 N/m^2 \; \text{per} \; m \end{aligned}
\mu =0.01N-s/m^2
h=5mm
if \tau _{y=0}=0, \; \text{find}\; \frac{dp}{dx}
The velocity distribution, for the couette flow with pressure gradient is given by
u=\frac{wy}{h}-\frac{1}{2\mu }\frac{dp}{dx}[hy-y^2]
where w is the velocity of the top plate and h is the gap between the plates
\begin{aligned} \frac{du}{dy}&=\frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}[h-2y] \\ \left.\begin{matrix} \frac{du}{dy} \end{matrix}\right|_{y=0} &=\frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}h \\ \tau _{y=0}&=\mu \left [ \frac{w}{h}-\frac{1}{2\mu }\frac{dp}{dx}h \right ] \\ \tau _{y=0}&=\frac{w\mu }{h}-\frac{1}{2}\frac{dp}{dx}h \\ \text{when}\; \tau _{y=0}&=0, \; \text{we will have}\\ \frac{1}{2}\frac{dp}{dx}h&=\frac{w\mu }{h}\\ \frac{dp}{dx}&=\frac{2w\mu }{h^2} \\ &=\frac{2 \times 1 \times 10^{-2}\times 0.01}{(5 \times 10^{-3})^2} \\ &=8 N/m^2 \; \text{per} \; m \end{aligned}
Question 4 |
An automobile with projected area 2.6 m^{2} is running on a road with a speed of 120 km per hour. The mass density and the kinematic viscosity of air are 1.2 kg/m^{3} and 1.5 \times 10^{-5} m^{2}/s, respectively. The drag coefficient is 0.30.
The metric horse power required to overcome the drag force is
The metric horse power required to overcome the drag force is
33.23 | |
31.23 | |
23.23 | |
20.23 |
Question 4 Explanation:
Power required to overcome the drag = Drag force \times velocity
\begin{aligned} &=520 \times \frac{120 \times 1000}{60 \times 60}= 17333.33 \mathrm{watt} \\ &\quad (1 \mathrm{mhp}=735.5 \mathrm{watt})\\ &=\frac{17333.33}{735.5}\text{ metric horse power }\\ &=23.567 \text{ metric horse power} \end{aligned}
\begin{aligned} &=520 \times \frac{120 \times 1000}{60 \times 60}= 17333.33 \mathrm{watt} \\ &\quad (1 \mathrm{mhp}=735.5 \mathrm{watt})\\ &=\frac{17333.33}{735.5}\text{ metric horse power }\\ &=23.567 \text{ metric horse power} \end{aligned}
Question 5 |
An automobile with projected area 2.6 m^{2} is running on a road with a speed of 120 km per hour. The mass density and the kinematic viscosity of air are 1.2 kg/m^{3} and 1.5 \times 10^{-5} m^{2}/s, respectively. The drag coefficient is 0.30.
The drag force on the automobile is
The drag force on the automobile is
620N | |
600N | |
580N | |
520N |
Question 5 Explanation:
The drag force on the automobile may be given as
\begin{aligned} F_{D} &=C_{D} A \times \frac{\rho V^{2}}{2} \\ \text { Here, } C_{D} &=0.30, \\ A&=2.6 \mathrm{m}^{2}, \\ \rho&=1.2 \mathrm{kg} / \mathrm{m}^{3}, \\ V&=120 \mathrm{kmph} \\\text{So, }F_{D}&=0.30 \times 2.6 \times \frac{1.2 \times(120 \times 1000)^{2}}{2 \times(60 \times 60)^{2}}\\ &=520N \end{aligned}
\begin{aligned} F_{D} &=C_{D} A \times \frac{\rho V^{2}}{2} \\ \text { Here, } C_{D} &=0.30, \\ A&=2.6 \mathrm{m}^{2}, \\ \rho&=1.2 \mathrm{kg} / \mathrm{m}^{3}, \\ V&=120 \mathrm{kmph} \\\text{So, }F_{D}&=0.30 \times 2.6 \times \frac{1.2 \times(120 \times 1000)^{2}}{2 \times(60 \times 60)^{2}}\\ &=520N \end{aligned}
Question 6 |
The thickness of the laminar boundary layer on a flat plate at a point A is 2 cm
and at a point B, 1 m downstream of A, is 3 cm. What is the distance of A from the leading edge of the plate ?
0.5m | |
0.8m | |
1m | |
1.25m |
Question 6 Explanation:
Laminar boundary layer thickness,
\begin{aligned} \delta &=5 \sqrt{\frac{x v}{V}} \\ \Rightarrow \quad \frac{\delta_{A}}{\delta_{B}} &=\sqrt{\frac{x_{A}}{x_{B}}} \\ \Rightarrow \quad \frac{2}{3} &=\sqrt{\frac{x_{A}}{x_{A}+1}} \\ \Rightarrow \quad \frac{4}{9}&=\frac{x_{A}}{x_{A}+1} \\ \Rightarrow \quad 5x_{A}&=4 \\ \Rightarrow x_{A}&= 0.8 \mathrm{m} \end{aligned}
\begin{aligned} \delta &=5 \sqrt{\frac{x v}{V}} \\ \Rightarrow \quad \frac{\delta_{A}}{\delta_{B}} &=\sqrt{\frac{x_{A}}{x_{B}}} \\ \Rightarrow \quad \frac{2}{3} &=\sqrt{\frac{x_{A}}{x_{A}+1}} \\ \Rightarrow \quad \frac{4}{9}&=\frac{x_{A}}{x_{A}+1} \\ \Rightarrow \quad 5x_{A}&=4 \\ \Rightarrow x_{A}&= 0.8 \mathrm{m} \end{aligned}
Question 7 |
A thin flat plate 0.5m\times 0.7m in size settles in a large tank of water with a terminal velocity of 0.12 m/s. The coefficients of drag C_{D}=\frac{1.328}{\sqrt{R_{L}}} for a laminar boundary layer and C_{D}=\frac{0.072}{(R_{L})^{1/5}} for a turbulent boundary layer, where R_L is the plate Reynolds number.
Assume \mu=10^{-3} N-s/m^{2} and \rho =1000 kg/m^{3}
The submerged weight of the plate is
Assume \mu=10^{-3} N-s/m^{2} and \rho =1000 kg/m^{3}
The submerged weight of the plate is
0.0115N | |
0.0118N | |
0.0231N | |
0.0376N |
Question 7 Explanation:
The critical Reynolds number for flat plates is 5 \times 10^{5}
As R_{L} \lt critical Reynolds number thus laminar boundary layer will grow through out the length
\begin{aligned} \therefore C_{0}&=\frac{1.328}{\sqrt{R_{L}}}=\frac{1328}{\sqrt{84 \times 10^{3}}}=4.58 \times 10^{-3}\\ \therefore &\quad \text{Drag force.}\\ F_{0} &=2 C_{D} A\left(\frac{\rho V^{2}}{2}\right) \\ &=4.58 \times 10^{-3} \times 0.5 \times 0.7 \times 1000 \times(0.12)^{2} \\ &=0.0231 \mathrm{N} \end{aligned}
As R_{L} \lt critical Reynolds number thus laminar boundary layer will grow through out the length
\begin{aligned} \therefore C_{0}&=\frac{1.328}{\sqrt{R_{L}}}=\frac{1328}{\sqrt{84 \times 10^{3}}}=4.58 \times 10^{-3}\\ \therefore &\quad \text{Drag force.}\\ F_{0} &=2 C_{D} A\left(\frac{\rho V^{2}}{2}\right) \\ &=4.58 \times 10^{-3} \times 0.5 \times 0.7 \times 1000 \times(0.12)^{2} \\ &=0.0231 \mathrm{N} \end{aligned}
Question 8 |
An aircraft is flying in level flight at a speed of 200 km/hr through air (density,
\rho =1.2 kg/m^{3}, and viscosity \mu =1.6 \times 10^{-5} Ns/m^{2}). The lift coefficient at this speed is 0.4 and the drag coefficient is 0.0065. The mass of the aircraft is 800 kg. The effective lift area of the aircraft is
21.2m^{2} | |
10.6m^{2} | |
2.2m^{2} | |
1.1m^{2} |
Question 8 Explanation:
The lift force will be equal to self weight of the aircraft for equilibrium
\begin{aligned} F_{L} &=W \\ \Rightarrow C_{L} A_{L} \frac{\rho V^{2}}{2} &=m g\\ \Rightarrow \quad \frac{0.4 \times A_{L} \times 1.2 \times\left(\frac{200 \times 10^{3}}{60 \times 60}\right)^{2}}{2}&=800 \times 9.81\\ A_{L}&=10.6 \mathrm{m}^{2} \end{aligned}
\begin{aligned} F_{L} &=W \\ \Rightarrow C_{L} A_{L} \frac{\rho V^{2}}{2} &=m g\\ \Rightarrow \quad \frac{0.4 \times A_{L} \times 1.2 \times\left(\frac{200 \times 10^{3}}{60 \times 60}\right)^{2}}{2}&=800 \times 9.81\\ A_{L}&=10.6 \mathrm{m}^{2} \end{aligned}
Question 9 |
A solid sphere (diameter 6 mm) is rising through oil (mass density 900 kg/m^{3}, dynamic viscosity 0.7 kg/ms) at a constant velocity of 1 cm/s. What is the specific weight of the material form which the sphere is made ? (Take g = 9.81 m/s^{2})
4.3kN/m^{3} | |
5.3kN/m^{3} | |
8.7kN/m^{3} | |
12.3kN/m^{3} |
Question 9 Explanation:
The solid sphere will be acted upon by three forces viz. drag force \left(\mathrm{F}_{\mathrm{D}}\right), force of buoyancy \left(\mathrm{F}_{\mathrm{B}}\right) and self weight of sphere (W)
\begin{array}{l} W=\rho_{s} V g \\ F_{B}=\rho_{0} V g \\ F_{D}=C_{D} A \frac{\rho_{0} V^{2}}{2} \end{array}
Reynolds Number,
\begin{aligned} \mathrm{Re} &=\frac{\rho_{0} \mathrm{VD}}{\mu} \\ &=\frac{900 \times 1 \times 10^{-2} \times 6 \times 10^{-3}}{0.7} \\ &=0.07714 \lt 0.2(\mathrm{ok}) \\\; \therefore \quad \mathrm{C}_{\mathrm{D}} &=\frac{24}{\mathrm{Re}}=\frac{24}{0.07714}=311.12 \end{aligned}
Now, from the free body diagram, we have W+F_{D}=F_{B}
\begin{aligned} \quad W&=F_{B}-F_{D} \\ \Rightarrow \quad \rho_{s} V g&=\rho_{0} V g-C_{D} A \frac{\rho_{0} V^{2}}{2} \\ \Rightarrow \quad w_{\mathrm{s}} V&=\rho_{0} V g-C_{D} \frac{A \rho_{0} V^{2}}{2} \\ \Rightarrow \quad w_{\mathrm{s}}&=\rho_{0} g-\frac{C_{\mathrm{D}} A \rho_{0} v^{2}}{2 V} \\ \Rightarrow \quad w_{\mathrm{s}}&=900 \times 9.81 \\ & -\frac{\left[\begin{array}{l}6 \times 311.12 \times \pi \times\left(6 \times 10^{-3}\right)^{2} \\ \times 900 \times\left(1 \times 10^{-2}\right)^{2}\end{array}\right]}{4 \times \pi \times\left(6 \times 10^{-3}\right)^{3}} \\ \Rightarrow \quad w_{\mathrm{s}}&=5328.9 \mathrm{N} / \mathrm{m}^{3}=5.3 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}
\begin{array}{l} W=\rho_{s} V g \\ F_{B}=\rho_{0} V g \\ F_{D}=C_{D} A \frac{\rho_{0} V^{2}}{2} \end{array}
Reynolds Number,
\begin{aligned} \mathrm{Re} &=\frac{\rho_{0} \mathrm{VD}}{\mu} \\ &=\frac{900 \times 1 \times 10^{-2} \times 6 \times 10^{-3}}{0.7} \\ &=0.07714 \lt 0.2(\mathrm{ok}) \\\; \therefore \quad \mathrm{C}_{\mathrm{D}} &=\frac{24}{\mathrm{Re}}=\frac{24}{0.07714}=311.12 \end{aligned}
Now, from the free body diagram, we have W+F_{D}=F_{B}
\begin{aligned} \quad W&=F_{B}-F_{D} \\ \Rightarrow \quad \rho_{s} V g&=\rho_{0} V g-C_{D} A \frac{\rho_{0} V^{2}}{2} \\ \Rightarrow \quad w_{\mathrm{s}} V&=\rho_{0} V g-C_{D} \frac{A \rho_{0} V^{2}}{2} \\ \Rightarrow \quad w_{\mathrm{s}}&=\rho_{0} g-\frac{C_{\mathrm{D}} A \rho_{0} v^{2}}{2 V} \\ \Rightarrow \quad w_{\mathrm{s}}&=900 \times 9.81 \\ & -\frac{\left[\begin{array}{l}6 \times 311.12 \times \pi \times\left(6 \times 10^{-3}\right)^{2} \\ \times 900 \times\left(1 \times 10^{-2}\right)^{2}\end{array}\right]}{4 \times \pi \times\left(6 \times 10^{-3}\right)^{3}} \\ \Rightarrow \quad w_{\mathrm{s}}&=5328.9 \mathrm{N} / \mathrm{m}^{3}=5.3 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}
Question 10 |
A flat plate is kept in an infinite fluid medium. The fluid has a uniform free-stream
velocity parallel to the plate. For the laminar boundary layer formed on the plate, pick the correct option matching Columns I and II.
P-1 Q-2 R-3 | |
P-2 Q-2 R-2 | |
P-1 Q-1 R-2 | |
P-2 Q-1 R-3 |
There are 10 questions to complete.
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