Question 1 |

Which of the following statements is/are TRUE?

The thickness of a turbulent boundary layer on a flat plate kept parallel to the flow direction is proportional to the square root of the distance from the leading edge | |

If the streamlines and equipotential lines of a source are interchanged with each other, the resulting flow will be a sink. | |

For a curved surface immersed in a stationary liquid, the vertical component of the force on the curved surface is equal to the weight of the liquid above it | |

For flow through circular pipes, the momentum correction factor for laminar flow is larger than that for turbulent flow |

Question 1 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 \times 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}

Question 2 |

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 2 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 3 |

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 3 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 4 |

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 4 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 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 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 5 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}

There are 5 questions to complete.

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