Skip to main content

Mechanics of Fluid - short answer questions (AMIE Winter 2020)


Explain the following in brief (5 x 4)
 

Specific weight and specific volume of a fluid

Specific weight, sometimes referred to as unit weight, is simply the weight of fluid per unit volume. It is usually denoted by the Greek letter γ (gamma) and has dimensions of force per unit volume.
Its units in the SI system are N/m³ or kg/m²sec² 

Specific volume is the volume of a unit mass of a substance. In other words, it is the ratio of a substance’s volume to its mass. The unit of measurement is (m³/kg).

Buoyancy and centre of buoyancy

Buoyancy is an upward force exerted by a fluid on an immersed object in a gravity field. In fluids, pressure increases with depth. Hence, when an object is immersed in a fluid, the pressure exerted on its bottom surface is higher than the pressure exerted on its top surface.
This difference in pressure leads to a net upward force (buoyancy force).
 
 
Buoyancy force will act through the centre of gravity of the displaced fluid and that point i.e. centre of gravity of the displaced fluid will be termed as centre of buoyancy. Therefore, we can define the term centre of buoyancy as the point through which the force of buoyancy is supposed to act.
 

Centre of buoyancy = Centre of gravity of the displaced fluid = Centre of gravity of the portion of the body immersed in the liquid

Streamline and streak line

Streamline
  • It is an imaginary line showing the positions of various fluid particles.
  • Particles may change streamline depending on the type of flow
  • Streamlines cannot intersect with each other, they are always parallel.
  • No flow across streamline. 

Streak Line

  • It is a real line showing instantaneous positions of various particles.
  • May change from instant to instant.
  • Streak line changes with time. Two streak lines may intersect each other.
  • Flow across the streak line is possible.

Reynold’s number and Mach number

Reynolds Number
The Reynolds number is the ratio of inertial forces to viscous forces. The Reynolds number is a dimensionless number used to categorize the fluids in systems in which the effect of viscosity is important in controlling the velocities or the flow pattern of a fluid. Mathematically, the Reynolds number, Re, is defined as

Re = ρvd/μ
 
where
ρ = density
v = velocity
d  = diameter
μ = viscosity

Mach Number
Mach number is used to compare the speed of any object with the speed of sound. It is mainly used to know the proper idea of the motion of aeroplanes and rockets. Mach number is a dimensional less quantity expressing the ratio of the velocity of an object in a medium to the speed of sound. 

Classification of Mach Number 
 
According to the variation of Mach number, fluid flow is classified.
  • Subsonic: The Mach number remains below 0.8 in subsonic flow. Commercial aeroplanes flowing in this range have a round nose and leading edges.
  • Sonic: The Mach number is one in sonic flow. In these types of flow, the velocity of an object is equal to the velocity of sound in the local medium.
  • Supersonic: If an object is flowing with greater velocity than the sound, it is said to have supersonic velocity. It has a Mach number greater than one. Aeroplanes have proper designs to undergo supersonic speed.
  • Hypersonic: If the speed of an object is higher than the speed of sound, then the flow is called hypersonic. In this type of flow, the Mach number is five to ten.
Mach number is represented with the symbol ‘M’.

Mach Number= Speed of object / Speed of sound
---

The study material for AMIE/Junior Engineer exams is available at https://amiestudycircle.com

Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...

Design of Electrical Systems (Solved Numerical Problems)

Important note There is something wrong with this question paper. It seems that instead of "Design of Electrical Systems" the IEI has given problems from "Electrical Machines". You should raise a complaint to director_eea@ieindia.org in this regard. Numerical A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb?​ (AMIE Summer 2023, 8 marks) Solution The back EMF (E b ) of a DC motor can be calculated using the formula: E b = V - I a R a   Given: V = 120 V I a = 200 A R a = 0.02 ohms Substituting the values into the formula: E b = 120 − 200 × 0.02 = 120 − 4​ = 116 V Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ p ). The formula to calculate the speed of a DC motor is: N = 60×E b /(P×φ p ) Wh...