Skip to main content

Transport Phenomena - short answer questions from AMIE exams (Winter 2020)

Answer the following briefly (20 marks)

Dependence of boundary thickness with the distance x from x = 0 for a flat plate.

The thickness of the boundary layer increases as the square root of the distance from the leading edge of the plate.

\delta (x) = 1.72\sqrt {\left( {\frac{{vx}}{{{U_0}}}} \right)}

Express Nu as the ratio of two temperature gradients.

Nu = \frac{{hl}}{k} = \frac{{ - (\partial t/\partial y)}}{{({t_s} - {t_\infty })/l}}

The physical significance of St; Ra.

Stanton number indicates the degree of amount of heat delivered by the fluid when there is heat transfer between solid surface and fluid. The greater the Stanton number is, the more effectively heat is transferred. 

Rayleigh number is defined as the product of Grashof number and Prandtl number. The rayleigh number is denoted by the symbol Ra. Rayleigh number also to find the type of fluid flow,
If Ra < 10⁹, then the flow is laminar and If Ra > 10⁹ then the flow is turbulent.

Momentum equation (one-directional)

\sum {{F_x} = \frac{{dM}}{{dt}}}

Similarities of kinematic viscosity and mass diffusivity.

The kinematic viscosity is also referred to as the momentum diffusivity of the fluid, i.e. the ability of the fluid to transport momentum. 

Schmidt number (Sc) is a dimensionless number defined as the ratio of momentum diffusivity (kinematic viscosity) and mass diffusivity, and it is used to characterize fluid flows in which there are simultaneous momentum and mass diffusion convection processes. 

Sc = \upsilon /D

Fick’s second law of diffusion.

Fick's 2nd law of diffusion describes the rate of accumulation (or depletion) of concentration within the volume as proportional to the local curvature of the concentration gradient. 

\frac{{\partial C}}{{\partial t}} = D\frac{{{\partial ^2}C}}{{\partial {x^2}}}

Creeping flow.

Stokes flow also named creeping flow or creeping motion, is a type of fluid flow where advective inertial forces are small compared with viscous forces. The Reynolds number is low, i.e. Re << 1. This is a typical situation in flows where the fluid velocities are very slow, the viscosities are very large.

Prandtl’s mixing length.

According to Prandtl, the mixing length l, is that distance between two layers in the transverse direction such that lumps of fluid particles from one layer could reach the other layer and the particles are mixed in the other layer in such a way that the momentum of the particles in the direction of x is same.

The relation between shear stress and mixing length is given by

\tau  = \rho {l^2}{\left( {\frac{{du}}{{dy}}} \right)^2}

Express Grashof no as a function of three forces.

Grashof number, Gr, as the ratio between the buoyancy force and the viscous force:

Gr = \frac{{{l^3}{\rho ^2}\beta \Delta T}}{{{\mu ^2}}}
It can be written as
Gr = (\rho {l^3}\beta \Delta T)\left( {\frac{{\rho {v^2}{l^2}}}{{{{(\mu vl)}^2}}}} \right)
= bupyant force x (inertia force/visous force2)

Express Nu/Bi as a ratio of two conductivities.

Bi = hl/{k_s}
Nu = \frac{{h{k_f}}}{l}
From these
Nu/Bi = {k_f}/{k_s}

---

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...

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...

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...