fundamental momentum transfer

8/2/2019 Fundamental Momentum Transfer

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Transport Process(KNC2153)

Dr Khairuddin Sanaullah

Dr Ivy Tan Ai Wei

Dept. of Chemical Engineering & Energy SustainabilityFaculty of Engineering

Semester II, 2011-2012



Course Outline

Chapter 1: Fundamentals of Momentum Transfer & Transport Properties

(Wk 1): Modes of momentum transfer, Newton‘s Law of viscosity, Types of fluid flow &

Reynolds Number, Momentum Transfer in a Fluid, Continuity Equation.

Chapter 2: Momentum Equation (Navier Stoke‘s) & It‘s Applications (Wk 2 & 3):

Derivation of Navier Stoke‘s Equation, Shell momentum balance & velocity

profile in laminar flow, Design equations for laminar & turbulent flow in

pipes, Compressible flow of gases, Boundary-layer flow & Turbulence.

To be Continued



Course Outline

Books

Geankoplis, C.J., [2003],Transport Processes and Separation Process Principles ,

Prentice Hall

Joel Plawsky [2010], Transport Phenomena Fundamentals, CRC Press

Bird, R., [2002] Transport Phenomena, John Willey

Holland & Bragg [1995], Fluid Flow for Chemical Engineers (2nd Ed.), Butterworth.



Fundamental Principles of Momentum Transfer

Modes of Momentum Transfer

• Shear

• Pressure

• Convection

Shear & Normal Stresses (Internal Fluid Forces): Differences in shear and normal

stresses across the control volume result in a flow of momentum from the higher to thelower stress. Shear & Normal stresses normally arise from two sources:

1. Pressure in the fluid

2. Velocity Gradients

The pressure gives rise to a normal stress component whereas velocity gradients give riseto the shear stress components and can contribute to the total normal stress.

External Foreces: The most important foreces are due to the pressure gradient

& gravity (body force - ρg). Both are generation terms.




Modes of Momentum Transfer

Convection: Momentum may flow into and out of the control volume in a variety of

ways:

(i) By convection (i.e. by bulk fluid flow)

(ii) By molecular transfer (i.e. by velocity gradients)

(iii) By external forces acting on the body such as gravity or pressure (e.g. Pump).

The first mechanism mentioned above arises due to the fluid acceleration through the

control volume and the second arises from shear stress acting on the fluid.



Transport Property (Momentum)

Momentum Transport –Newton’s Law of Viscosity:

When a fluid is flowing through a pipe or between two flat plates, either of two types o

flow may occur, depending on the velocity of this fluid (i.e. Laminar or turbulent). Here

the discussion is limited to laminar flow.

An elastic solid deforms by an amount proportional to the applied stress. However, a fluid

when subjected to a similar applied stress will continue to deform, i.e. to flow at a

velocity that increases with increasing stress. A fluid exhibits to this stress, Viscosity is

that property of a fluid which gives rise to forces that resist the relative movement o

adjacent layers in the fluid.

The ideas can be clarified by a more quantitative discussion of viscosity: refer figure in

the next slide.




Types of Fluid Flow & Reynolds Number

Laminar flow ─ the flow is characterized

by smooth streamlines and

highly-ordered motion.

Turbulent flow ─ the flow is

characterized by velocity

fluctuations and

highly-disordered motion.

The transition from laminar

to turbulent flow does not

occur suddenly.




Types of Fluid Flow & Reynolds NumberThe velocity profile in turbulent flow is much fuller than that in laminar flow, with a

sharp drop near the surface.The turbulent boundary layer can be considered to consist of four regions:

Viscous sublayer

Buffer layer

Overlap layer

Turbulent layerThe intense mixing in turbulent flow enhances heat and momentum transfer, which

increases the friction force on the surface and the convection heat transfer rate.




Inertia forces

ReViscous forces

c cVL VL



Transport Property (Mass)




Example: A mixture of He and gas is contained in a pipe at 298 K and 1 atm total

pressure which is constant throughout. At one end of the pipe at point 1 the partial

pressure of He is 0.60 atm and at the other end 0.2 m (20 cm) = 0.20 . Calculate the flux of He at steady state if of the mixture is

0.687×10− /. Use SI units.

Solution

Since total pressure P is constant, then the concentration c is constant, where c is as

follows for a gas from the ideal gas law. = ≫

=

=

Where n is kg mol a + b, V is volume in , T is temperature in K, R is gas constant,

expressed as 8314.3.

.or 82.057 × 10− .

.and c is kg mol a + b/ .

For steady state the flux in equation (4) is constant, also for a gas is constant.Rearranging equation (4) and integrating,

=

=(−)

− …. (i)




Also, from the ideal gas law, = , and

=

=

…. (ii)

Substituting (ii) into (i) = (−)(−)

Now substitute values in this equation, = 0.6 atm = 0.2 atm. Here pressures in

atm are used with SI units.

=(.6×)(.6−.)

(.6×)()(.−) = . × −

.



Transport Property (Energy)



i



Transport Properties

T P i



Transport Properties

fundamental momentum transfer

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