An automobile with projected area 2.6 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 and , respectively. The drag coefficient is 0.30. The metric horse power required to overcome the drag force is

Power required to overcome the drag = Drag force velocity

A thin flat plate in size settles in a large tank of water with a terminal velocity of 0.12 m/s. The coefficients of drag for a laminar boundary layer and for a turbulent boundary layer, where is the plate Reynolds number. Assume and The submerged weight of the plate is

The critical Reynolds number for flat plates is As critical Reynolds number thus laminar boundary layer will grow through out the length

GATE CE · Fluid Mechanics And Hydraulics

Boundary Layer Theory Drag And Lift – GATE CE Fluid Mechanics And HydraulicsPractice Questions & PYQs

Generate GATE-level questions on Boundary Layer Theory Drag And Lift in Fluid Mechanics And Hydraulics. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

11 questions · 11 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2023MSQ

Which of the following statements is/are TRUE?

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📖 Explanation

$P_{H}=$ Total pressure on the projected are of the curved surface on the vertical plane. $P_{H}$ is equal to the total pressure exerted by the liquid on an imaginary vertically immersed plane surface which is the vertical projection of the curved surface, it will act at the center of pressure of the lane surface. $P_{v}=$ The weight $o$ the liquid contained in the portion extending vertically above the curved surface upto the free surface of the liquid. $P_{V}$ will act through the centre of gravity of the volume of liquid contained in the portion extending above the curved surface upto the free surface of the liquid (represented by the profile ABCDEFA in the present case). Momentum correction factor $(\beta)=\frac{1}{A V^{2}} \int \mathrm{v}^{2} d A$ {Sink Flow :} The sink flow is the flow in which fluid moves radially inwards towards a point where it disappears at a constant rate. This flow is just opposite to the source flow. Figure shows a sink flow in which the fluid moves radially inwards towards point $\mathrm{O}$ , where it disappears tat a constant rate. The pattern of stream lines and equipotential lines of a sink flow is the same as that of source flow. All the equations derived for a source flow shall hold to good for sink flow also except that in sink flow equations, $q$ is to be replaced by $(-q)$ Boundary layer thickness (Normal Boundary layer thickness disturbance thickness. Distance from the boundary surface in which the velocity reaches $99 \%$ of the stream velocity is the boundary layer thickness. Turbulent boundary layer on smooth plate. $\frac{\delta}{x}=\frac{0.376}{\left(R_{e x}\right)^{1 / 5}}, \quad R_{e x}=\frac{v_{0} x}{v}$ $[Valid for $R_{e x} \gt 5 × 10^{5}$ and $R_{e x} \lt 10^{7}$ ]$ $=\frac{0.22}{\left(R_{e x}\right)^{1 / 6}}, 10^{7} \lt R_{e x} \lt 10^{9}$

Q2GATE 2022NAT

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)

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📖 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 $(D_Q)=1 mm$ $\Rightarrow \frac{(F_D)_Q}{(F_D)_P}=\frac{(D_Q)^3}{(D_P)^3}=\left ( \frac{1}{0.5} \right )^3=8$ $\Rightarrow (F_D)_Q=8 \times (F_D)P$

Q3GATE 2020NAT

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 ________.

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📖 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}

Q4GATE 2019NAT

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 _______

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📖 Explanation

Given fl\ow 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}

Q5GATE 2008MCQ

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

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📖 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}

Q6GATE 2006MCQ

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 ?

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📖 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}

Q7GATE 2004MCQ

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

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📖 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}

Q8GATE 2004MCQ

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

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📖 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}

Q9GATE 2003MCQ

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 $m/s^{2}$ )

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📖 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}

Q10GATE 2003MCQ

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.

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Q11GATE 2002MCQ

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

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📖 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}

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Q1GATE 2023MSQ

Which of the following statements is/are TRUE?

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📖 Explanation