1、1,Chapter 5,Principles of Convection,College of Nuclear Science and Technology,2,Chapter 5,College of Nuclear Science and Technology,5-1 Introduction,In the preceding chapters, weve learned about the Mechanism & Calculation of conduction heart transfer.,Convection was considered as it related to the
2、 Boundary conditions,3,Chapter 5,College of Nuclear Science and Technology,Definition,Convection Heat-transfer: The process of heat transfer between the fluid and the surface of an object when the fluid flows over it.,What influences the convection heat-transfer process,The physical properties of th
3、e fluid The shape, size and arrangement of the surface The velocity of the flow The cause of the flow Whether there is a phase change,4,Chapter 5,College of Nuclear Science and Technology,How to get the convection heat transfer coefficient h?,What is the Newtons law of cooling?,Question,What are we
4、going to learn in the next following chapters is mainly about,5,Chapter 5,Consider some of the simple relations of fluid dynamics and boundary-layer analysis. Impose an energy balance on the flow system and determine the influence of the flow on temperature gradients in the fluid. Obtained a knowled
5、ge of temperature distribution.,College of Nuclear Science and Technology,Our Discussion in Chapter 5,6,Chapter 5,College of Nuclear Science and Technology,5-2 Viscous Flow,7,Chapter 5,College of Nuclear Science and Technology,8,Chapter 5,College of Nuclear Science and Technology,The influence of vi
6、scous force is felt at the leading edge of the plate.,Consider the flow over a plate.,These viscous forces are described in terms of a shear stress.,If this stress is assumed to be proportional to the normal velocity gradient.,9,Chapter 5,College of Nuclear Science and Technology,We have the definin
7、g equation for the viscosity.,Definition, - dynamic viscosity,10,Chapter 5,College of Nuclear Science and Technology,Boundary Layer: The region of flow that develops from the leading edge of the plate in which effects of viscosity are observed.,Definition,The y position where boundary layer ends: y
8、coordinate where velocity becomes 99% of the free-stream value,11,Chapter 5,Small disturbances in the flow begin to become amplified A transition process takes place,College of Nuclear Science and Technology,Laminar,Turbulent,The boundary-layer development,12,The turbulent-flow region may be picture
9、d as a random churning action with chunks of fluid moving to and fro in all direction.,Chapter 5,The transition from laminar to turbulent flow occurs when,College of Nuclear Science and Technology,13,The physical mechanism of viscosity is one of momentum exchange.,Chapter 5,There is a net transfer f
10、rom regions of high velocity to regions of low velocity.,The rate at which the momentum transfer takes place is independent on the rate at which the molecules move across the fluid layer,College of Nuclear Science and Technology,14,Chapter 5,Distinct fluid layers are no longer observed in turbulent-
11、flow region.,We can obtain a qualitative picture of the turbulent-flow by imaging macroscopic chunks of fluid transporting energy and momentum.,College of Nuclear Science and Technology,15,Chapter 5,Our Exception,Larger mass of macroscopic elements of fluid transport more energy and momentum than th
12、e individual molecules.,A larger viscous-shear force in turbulent flow than in laminar flow.,This exception is verified by experiment,College of Nuclear Science and Technology,16,Chapter 5,College of Nuclear Science and Technology,17,Chapter 5,Figure 5-3a: If the flow is laminar, a parabolic velocit
13、y profile is experienced.,Figure 5-3b: If the flow is turbulent, a somewhat blunter profile is observed.,College of Nuclear Science and Technology,18,Chapter 5,A criterion for laminar and turbulent flow is also Reynolds number. For,The flow is considered to be turbulent.,College of Nuclear Science a
14、nd Technology,19,Chapter 5,The generally accepted range of Reynolds numbers for transition is,College of Nuclear Science and Technology,20,Chapter 5,The continuity relation for one-dimensional flow in a tube is,College of Nuclear Science and Technology,21,Chapter 5,The continuity relation for one-di
15、mensional flow in a tube is,College of Nuclear Science and Technology,22,Chapter 5,We define the mass velocity as,Mass velocity =,So the Reynolds number may also be written as,College of Nuclear Science and Technology,23,Chapter 5,5-3 Inviscid Flow,Although no real fluid is inviscid, in some instanc
16、es the fluid may be treated as such, and its worthwhile to present some of the equations that apply in these circumstances.,College of Nuclear Science and Technology,24,Chapter 5,IF a balance of forces is made on elements of incompressible fluid and these forces are set equal to the change in moment
17、um of fluid element of the fluid element.,The Bernoulli equation for flow along a streamline,or,College of Nuclear Science and Technology,25,Chapter 5,Where = fluid density, kg/m3 p= pressure at particular point in the flow, Pa V= velocity of flow at that point.,College of Nuclear Science and Techno
18、logy,26,Chapter 5,For a one dimensional flow system, when the fluid is compressible, the steady-flow energy equation for a control volume is,Where i is the enthalpy defined by,College of Nuclear Science and Technology,27,Chapter 5,For ideal gas,College of Nuclear Science and Technology,28,Chapter 5,
19、Example 5-1,Water at 20 flows at 8 kg/s through the diffuser arrangement shown in Figure Example 5-1, The diameter at section 1 is 3.0cm, and the diameter at section 2 is 7.0cm. Determine the increase in static pressure between sections 1 and 2. Assume frictionless flow.,College of Nuclear Science a
20、nd Technology,29,Chapter 5,College of Nuclear Science and Technology,30,Chapter 5,Solution,The flow cross sectional areas are:,College of Nuclear Science and Technology,31,Chapter 5,The density of water at 20 is 1000kg/m, and we may calculate the velocities from the mass continuity relation,College
21、of Nuclear Science and Technology,32,Chapter 5,The pressure difference is obtained from Bernoulli Equation,College of Nuclear Science and Technology,33,Chapter 5,5-4 Laminar Boundary Layer on Flat Plate,College of Nuclear Science and Technology,34,Chapter 5,Consider the elemental control volume show
22、n in Figure 5-4, we derive the equation of motion for the boundary layer by making a force-and-momentum balance on this element.,ASSUMPTION: 1. The fluid is incompressible and the flow is steady. 2. There is no pressure variations in the direction perpendicular to the plate. 3. The viscosity is cons
23、tant. 4. Viscous-shear forces in the y direction are negligible.,College of Nuclear Science and Technology,35,Chapter 5,We apply Newtons 2nd Law of motion,For the system the force balance is written,=increase in momentum flux in x direction,The mass entering the left face of the element per unit tim
24、e is,College of Nuclear Science and Technology,36,Chapter 5,We assume unit depth in the z direction. The momentum flux entering the left face per unit time is,College of Nuclear Science and Technology,37,Chapter 5,The mass flow leaving the right face:,The momentum flux leaving the right face:,The ma
25、ss flow entering the bottom face is:,The mass flow leaving the top face is:,College of Nuclear Science and Technology,38,Chapter 5,A mass balance on the element yields,The mass continuity equation for the boundary layer,College of Nuclear Science and Technology,39,Chapter 5,The momentum flux in the
26、x direction that enters the bottom face:,The momentum in the x direction that leaves the top face :,College of Nuclear Science and Technology,40,Chapter 5,The pressure force on the left face is,The pressure force on the right face is,The net pressure force in the direction of motion is,College of Nu
27、clear Science and Technology,41,Chapter 5,The viscous-shear force on the bottom face is,Shear force on the top is,Net viscous-shear force in the direction of motion is,College of Nuclear Science and Technology,42,Chapter 5,Equating the sum of viscous-shear force and the Pressure forces to the net mo
28、mentum transfer In the x direction, we have,College of Nuclear Science and Technology,43,Chapter 5,Clearing terms, making use of the continuity relation and neglect second-order differentials, gives,The momentum equation of the laminar Boundary layer with constant properties.,College of Nuclear Scie
29、nce and Technology,44,Chapter 5,College of Nuclear Science and Technology,45,Chapter 5,We wish to make a momentum-and-force balance on the control system bounded by the planes 1,2,A-A, and the solid wall.,The free-stream velocity outside the boundary layer is,The boundary-layer thickness is ,Assumpt
30、ion： 1、The velocity components normal to the wall are neglected, and only those in x direction are considered. 2、H ,College of Nuclear Science and Technology,46,Chapter 5,The mass flow through the plane1 is,The momentum flow through plane1 is,The momentum flow through plane2 is,The mass flow through
31、 plane2 is,a,b,c,d,College of Nuclear Science and Technology,47,Chapter 5,Consider the conservation of mass and the fact that no mass can enter the control volume through solid wall, the additional mass flow in the expression(d) over that in a must enter through plane A-A. This mass flow carries wit
32、h it a momentum in the x direction equal to,College of Nuclear Science and Technology,48,Chapter 5,The net momentum flow out of the control volume is,Put it into a more useful form,College of Nuclear Science and Technology,49,Chapter 5,In the momentum expression given above, the integral,Is the func
33、tion and is the function. THUS,College of Nuclear Science and Technology,50,Chapter 5,The force on plane1 is the pressure force pH and that on plane 2 is p+(dp/dx)H. The shear force at the wall is,There is no shear force on plane AA since the velocity gradient is 0 outside the boundary layer. Settin
34、g the forces on the element equal to the net increase in the momentum and collecting terms gives,College of Nuclear Science and Technology,51,Chapter 5,The integral momentum equation for the boundary layer,For the constant-pressure condition the integral boundary-layer equation becomes,5-17,College
35、of Nuclear Science and Technology,52,Chapter 5,If the velocity profile were known, the appropriate function could be inserted in the equation. THE CONDITIONS that the velocity function must satisfy:,College of Nuclear Science and Technology,53,Chapter 5,For a constant-pressure condition,Since the ve
36、locities u and v are 0 at y=0. We assume that the velocity profiles at various x positions are similar; that is, they have the same functional dependence on the y coordinate. There are four conditions to satisfy. The simplest function we can choose to satisfy these conditions is a polynomial with fo
37、ur arbitrary constants.,College of Nuclear Science and Technology,54,Chapter 5,Applying the four conditions,Inserting the equation 5-17,College of Nuclear Science and Technology,55,Chapter 5,Carrying out the integration leads to,Since and u are constants , the variables may be separated to give,Coll
38、ege of Nuclear Science and Technology,56,Chapter 5,At x=0, =0,It may be written in terms of Reynolds number as,The exact solution of boundary-layer equations as given in Appendix B yields,College of Nuclear Science and Technology,57,Chapter 5,5-5 Energy Equation of the Boundary Layer,The foregoing a
39、nalysis considered the fluid dynamics of a laminar-boundary-layer flow system. We shall now develop the energy equation for this system and then proceed to an integral method of solution.,College of Nuclear Science and Technology,58,Chapter 5,College of Nuclear Science and Technology,59,Chapter 5,Co
40、nsider the elemental control volume shown in Figure 5-6. To simplify the analysis we assume:,ASSUMPTION: 1. Incompressible steady flow. 2. Constant viscosity thermal conductivity, and specific heat. 3. Negligible heat conduction in the direction of flow (x direction),College of Nuclear Science and T
41、echnology,60,Chapter 5,Then, for the element shown, the energy balance may be written: ENERGY convected in the left face + Energy convected in the bottom face + heat conducted in the bottom face + net viscous work done on element = energy convected out right face + energy convected out top face +ene
42、rgy conducted out top face,College of Nuclear Science and Technology,61,Chapter 5,The net viscous energy delivered to the element is,Writing the energy balance corresponding to the quantities shown in Figure 5-6, assuming unit depth in the z direction, and neglecting 2nd-order differentials yields,C
43、ollege of Nuclear Science and Technology,62,Chapter 5,Using the continuity relation and dividing by gives,The energy equation of the laminar boundary layer,The net transport energy into the control volume,The sum of the net heat conducted out of it and the net viscous work done on it,College of Nucl
44、ear Science and Technology,63,Chapter 5,Using an order-of-magnitude analysis of the two terms on the right side of the equation,So,College of Nuclear Science and Technology,64,Chapter 5,If,Then the viscous dissipation is small in comparison with the conduction term.,Introducing Prandtl number,Colleg
45、e of Nuclear Science and Technology,65,Chapter 5,The significance of Prandtl number,For laminar flow over a plate.,Momentum Equation,Energy Equation,It indicates that the laws of momentum transfer and energy transfer are similar to each other.,College of Nuclear Science and Technology,66,Chapter 5,T
46、he significance of Prandtl number,For fluid whose = a , its Pr=1, and its velocity field and the non-dimensional temperature field will completely similar. So, the Prandtl number represents the comparison of the momentum diffusion capacity and the thermal diffusion capacity OR the relative size of t
47、he flow boundary layer and the temperature boundary layer.,College of Nuclear Science and Technology,67,Chapter 5,The significance of Prandtl number,Liquid medal,Air Water vapor,water,High-viscosity oil,College of Nuclear Science and Technology,68,Chapter 5,Back to the analysis,Thus, for low-velocit
48、y incompressible flow, we have,5-25,There is great similarity between Equation 5-25 and the momentum equation for constant pressure,5-26,College of Nuclear Science and Technology,69,Chapter 5,The solution to the two equations will have exactly the same form when a=. Thus we should expect that the re
49、lative magnitudes of the thermal diffusivity and kinematic viscosity would have an important influence on convection heat transfer since these magnitudes relate the velocity distribution to the temperature distribution. This is exactly the case, and we shall see the role that these parameters play in the subsequent discussion.,College of Nuclear Science and Technology,70,Chapter 5,5-6 The Thermal Boundary
