# Boundary Layer Theory, Drag and Lift

 Question 1
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 ________.
 A 0.24 B 0.38 C 0.47 D 0.58
GATE CE 2020 SET-2   Fluid Mechanics and Hydraulics
Question 1 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}
 Question 2
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 _______
 A 2.4 B 8 C 6.2 D 9.8
GATE CE 2019 SET-1   Fluid Mechanics and Hydraulics
Question 2 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}
 Question 3
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
 A 33.23 B 31.23 C 23.23 D 20.23
GATE CE 2008   Fluid Mechanics and Hydraulics
Question 3 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}
 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 drag force on the automobile is
 A 620N B 600N C 580N D 520N
GATE CE 2008   Fluid Mechanics and Hydraulics
Question 4 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}
 Question 5
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 ?
 A 0.5m B 0.8m C 1m D 1.25m
GATE CE 2006   Fluid Mechanics and Hydraulics
Question 5 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}
 Question 6
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
 A 0.0115N B 0.0118N C 0.0231N D 0.0376N
GATE CE 2004   Fluid Mechanics and Hydraulics
Question 6 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}
 Question 7
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
 A 21.2$m^{2}$ B 10.6$m^{2}$ C 2.2$m^{2}$ D 1.1$m^{2}$
GATE CE 2004   Fluid Mechanics and Hydraulics
Question 7 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}
 Question 8
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 $m/s^{2}$)
 A 4.3$kN/m^{3}$ B 5.3$kN/m^{3}$ C 8.7$kN/m^{3}$ D 12.3$kN/m^{3}$
GATE CE 2003   Fluid Mechanics and Hydraulics
Question 8 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}
 Question 9
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.
 A P-1 Q-2 R-3 B P-2 Q-2 R-2 C P-1 Q-1 R-2 D P-2 Q-1 R-3
GATE CE 2003   Fluid Mechanics and Hydraulics
 Question 10
Velocity distribution in a boundary layer flow over a plate is given by
$(u/u_{m})=1.5 \eta$
where, $\eta =y/\delta$; y is the distance measured normal to the plate; $\delta$ is the boundary layer thickness; and $u_m$ is the maximum velocity at $y =\delta$. If the shear stress $\tau$, acting on the plate is given by
$\tau =K(\mu u_{m})\delta$
where, $\mu$ is the dynamic viscosity of the fluid, K takes the value of
 A 0 B 1 C 1.5 D none of the above
GATE CE 2002   Fluid Mechanics and Hydraulics
Question 10 Explanation:
\begin{aligned} \text { Given, } \frac{u}{u_{m}}=1.5 \eta ; \; \; &\;\eta=\frac{y}{\delta} \\ \therefore \quad u &=1.5 u_{m} \times \frac{y}{\delta} \\ \frac{d u}{d y} &=1.5 \frac{u_{m}}{\delta} \\ \text { Now, }\quad \tau&=\mu \frac{d u}{d y}=1.5 \mu \frac{u_{m}}{\delta} \\ \therefore \quad k &=1.5 \end{aligned}
There are 10 questions to complete.

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