First page

1. First page An Advanced Course For graduates

2. Refer to Table 0.2-1 On Page 4 We will study the topics by rows with shadowed chapters skipped

3. Basic Concepts Basic Concepts of Transport Phenomena

4. The First Question What does the terminology “Transport Phenomena” stand for ?

5. Answer - A Definition It stands for “The transporting processes, in time and space, of three basic physical entities which the substance possesses, momentum, energy and mass, due to the non-uniform distribution of physical quantities in the space.”

6. Point 1 On the ground of the definition, “The transport phenomena are dynamical processes or rate processes occurring in non-equilibrium state.”

7. Point 2 On the ground of the definition, “The direction, in which transport phenomena proceed, is determined in a rational and natural way: From a non-equilibrium state towards an equilibrium one, as they proceed spontaneously”

8. Explanation For Energy In transport phenomena , the energy being transported specializes the internal energy of substance, but the energy in the driving force for transporting processes, may involve many forms of energy, such as mechanical energy, electric energy, magnetic energy.

9. A Deduction “Momentum, Energy or mass should be transported spontaneously from a region of high density to a region of low density as the non-uniform distribution of the density of entity itself is the only driving-force.”

10. Momentum and Its Density • Momentum, M = mv, is a vector. • Velocity, v = M/ m, is equivalent to the momentum of unit mass, so it gives a representation of momentum density. • Velocity distribution in the space represents the distribution of momentum density.

11. Energy and Its Density • Internal energy of substances, U=mCVT, is a scalar. • Temperature is proportional to the internal energy of unit mass, so gives a representation of energy density. • Temperature distribution in the space represents the distribution of internal energy density.

12. Mass and Its Density • Mass in transport phenomena specializes the mass of chemical species in a mixture. • The mass density of a chemical component can be represented by its concentration, mass fraction or molecular fraction.

13. Ordinary Transport Phenomena “Ordinary Transport Phenomena” deals with “The transporting processes, in time and space, of momentum, internal energy or chemical species, due to the non-uniform distribution of velocity, temperature or composition in space.”

14. Three Levels • Transport Phenomena can be described in 3 levels: • Macroscopic level • Microscopic level • Molecular level

15. Macroscopic Level Consider the change of total quantity of momentum, energy and mass in a system via exchanging these entities with the surroundings • Being involved mainly in engineering applications.

16. Microscopic Level • Consider the distribution of momentum, energy and mass in time and space to give the profiles of related quantities within a system. • The field description is employed • The centric area in this course

17. Molecular Level Seek a fundamental understanding of the mechanisms of momentum, energy and mass transport phenomena in terms of material structures and intermolecular interactions. • The statistical physics method is employed • Being involved mainly in calculation of transport properties.

18. Three Mechanisms of Transport (1) • Convective transport The entity is carried by the flowing fluid into or out of a region, to cause the transport of the entity in space. • Molecular transport The transport of entities caused by the Brownian motion of molecules. These two mechanisms are classified as the short-range transport, in which the entity is transported point by point in space.

19. Three Mechanisms of Transport (2) • Long-range transport The entity to be transported is converted in a location into a kind of potential energy field such as gravity, electric, magnetic and electromagnetic fields. The energy is transmitted to another location, then is received by molecules of substance and re-converted into the entity again. It looks as the entity is transported point to point in the space.

20. The Flux of Physical Entity (1) Let be a physical entity which can be described mathematically as a tensor field of mth-order, and be the quantity of being transported in unit time and through unit area of a coordinate surface perpendicular to base vector i . The set composes a tensor field of (m+1)th-order, known as the flux of entity .

21. The Flux of Physical Entity (2) Proof The balance of in a given volume region, V, may be written as Let V be a tetrahedron, the production rate of in unit volume and the accumulation rate of in unit volume.

22. The Flux of Physical Entity (3) Then the balance equation may be expressed as (1) in which is the quantity of being transported in unit time and through unit area of a surface perpendicular to a vector n.

23. The Flux of Physical Entity (4) , , and are continuous, with the mean value theorem for integrals, Eq.(1) may be rewritten as (2) in which rirepresents a point on Si , rnrepresents a point on Sn , rp represents a point in V andrA represents a point in V .

24. The Flux of Physical Entity (5) From the geometric relations, we have and Inserting these into Eq.(2), we have (3) Taking the limit of Eq.(3) as xi0 , which means that ri , rp , rAand rn all approach to a common point, the origin O, we obtain an identity,

25. The Flux of Physical Entity (6) or Because ni stand for a vector and a tensor of mth-order, the quantities must be the components of a tensor of (m+1)th-order according to the tensor recognition theorem 1. This tensor of (m+1)th-order is named the flux tensor of , denoted as .

26. The Flux of Physical Entity (7) For example, 1) Let be the volume of fluid, a scalar, then is the volume flux tensor, a vector, which is called the velocity of fluid. 2) Let be the momentum of fluid, a vector, then is the momentum flux tensor, a Tensor of second-order.

27. Momentum and Force (1) The mansion of classical mechanics is based on the three Newton’s laws of motion. • The first law ( the law of inertia ) • Every body continues in its state of rest , or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed on it.

28. Momentum and Force (2) • The second law ( the law of acceleration) The acceleration of a body is equal to the resultant of all external forces exerted on it divided by its mass. • The third law ( the law of action and reaction ) • To every action there is always an equal opposed reaction.

29. Momentum and Force (3) What is the physical signification of force ? Is the force a property of materials ? Force is a phenomenological quantity which people use to describe the change of properties of materials. What is the physical entity behind force ?

30. Momentum and Force (4) In the case that the relativistic effect is negligible, the mass of a body does not change due to its motion. So Inserting it into the second law, we have It says that the force exerted on a body is equal to the rate of momentum transporting to it. It is a definition of force.

31. Momentum and Force (5) With the definition of force, we can give alternative explanations of other two Newton’s laws as momentum conservation. • The first law • The momentum of a body keeps constant unless the extra momentum is transported to it. • The third law • The momentum gain of a body is always equal to the momentum loss of another body in their interaction.

32. Momentum Flux and Stress The stress is defined as the force exerted on a unit area of surface, with two characteristic directions: the direction of normal vector of the surface and that of the force. xy represents the force in y direction exerted on a unit area of surface perpendicular to x . By the second Newton’s law, xy may be explained as the momentum inydirectiontransported in unit time and through a unit area of surface perpendicular to x , that is the xy component of momentum flux .

33. Newton’s Viscosity Law We have learnt the famous Newton’s viscosity law for molecular transport of momentum: It says that the momentum flux is a linear homogeneous function of the gradient of momentum density. This law is concluded from the observation of the molecular transport of momentum in 1-dimensional straight flow, so it cannot be extended simply to 3-dimensional or curving flow.

34. Simple Illustrations SHELL BALANCE Method For 1-dimensional transport

35. E gain by convective transport E gain by molecular transport E gain by local production + + = 0 Shell Balance Method (1) For the steady situation that the transport proceeds only in one direction, the shell balance method can be employed. • Considering a thin layer of space region (so called shell), of which the two main surfaces are perpendicular to the direction of transport, we can write out a balance equation of entity E as:

36. Shell Balance Method (2) • Let the thickness of the shell approach zero and make use of the definition of the first derivative to obtain an ordinary differential equation (ODE) for the entity flux. • Integrate the ODEto get the entity flux distribution. • Insert an expression for the entity flux by molecular transport to obtain an ODEfor the entity density.

37. Shell Balance Method (3) • Integrate the ODEto get the entity density distribution. • Use the entity densitydistribution to calculate the relative physical quantities. Consider the entity being momentum

38. 1-dimensional Momentum Transport For Isothermal Systems with Uniform Composition

39. Shell Balance Method Application to momentum (1) Procedure to modeling & solving a problem: • Choose a coordinate system that the fluid flows along one of the coordinate surface families. • Set up a shell of which main surfaces are two coordinate surfaces in the above family. • Write a momentum balance over the shell. • Obtain an ODEfor the momentum flux by letting the shell thickness approach zero.

40. Shell Balance Method Application to momentum (2) • Integrate the ODE with the boundary conditions (B.C.s) to get distribution of momentum flux. • Insert the Newton’s law of viscosity to obtain an ODE for velocity. • Integrate the ODE with B.C.s to get distribution of velocity. • Calculate the related quantities such as maximum & average velocity, flow rate, pressure change, force on solid surfaces.

41. §2.2Flow of A Falling Film (1) Consider the middle part in which the effect of end disturbances is negligible.

42. §2.2Flow of A Falling Film (2) It is appropriate to choose a Cartesian system of coordinates and set up a shell in the way shown in the figure.

43. §2.2Flow of A Falling Film (3) • The momentum gain by convective transport • The momentum gain by molecular transport • The momentum gain by long-range transport

44. §2.2Flow of A Falling Film (4) • The momentum balance equation reduces to • Taking the limit as x approaches zero, we get ODE (2.2-10) with (2.2-12) It is called a free surface B.C.

45. §2.2Flow of A Falling Film (5) • The solution for momentum flux distribution is (2.2-13) • Inserting The Newton’s law of viscosity, we have (2.2-15) (2.2-17) with It is called as no-slip B.C.

46. §2.2Flow of A Falling Film (6) • The solution for velocity distribution is (2.2-18) • The velocity distribution is the key point to solve a problem of momentum transport. With Eq.(2.2-18), all other related quantities are easy to calculate

47. §2.2Flow of A Falling Film (7) • The mass rate of flow is (2.2-22) • The shear force exerted on the solid surface by the fluid is (2.2-23)

48. §2.2Flow of A Falling Film (8) • Check the validity of the solution by comparing it with experimental observations: • Re < 20, • Laminar flow with negligible rippling. • 20 < Re < 1500 • Laminar flow with pronounced rippling. • Re > 1500 • Turbulent flow It is found that the above solution is valid only for the first regime because of the postulate that the upper surface of liquid is a free surface boundary and end disturbances are negligible.

49. §2.3Flow through a circular tube(1) A fluid of constant density and viscosity flows down a vertical circular tube. We postulate: laminar flow; steady-state; L>>R so that “end effects ” is negligible in the middle section of tube. Then the flow will be only in the direction of centric axis of tube.

50. §2.3Flow through a circular tube(2) For 1-dimensional flow along a cylindrical surface, the cylindrical coordinates and a cylindrical shell should be employed, as shown on the right side.