Lectures on Nuclear Power Safety Lecture No 6 Title: Introduction to Thermal-Hydraulic Analysis of Nuclear Reactor Cores Department of Energy Technology KTH Spring 2005 Slide No 1
Outline of the Lecture Thermal Issues in Nuclear Reactor Cores Power Generation in Nuclear Reactor Cores Principle of Fluid-Flow and Heat Transfer Slide No 2
Thermal Issues in Nuclear Reactors (1) Overview of Main Reactor Components in Light Water Reactors (LWR) Reactor pressure vessel: In Pressurized Water Reactors (PWR) In Boiling Water Reactors (BWR) Fuel Assemblies In BWR In PWR Specific thermal issues in LWRs Slide No 3
Schematic of Nuclear Power Plant (PWR) Slide No 4
Schematic of Nuclear Power Plant (BWR) Slide No 5
Nuclear Reactor Vessel (PWR) Slide No 6
Nuclear Reactor Vessel (BWR) Slide No 7
Fuel Assemblies (PWR) Slide No 8
Fuel Assemblies (BWR) Slide No 9
Energy Cycles in LWRs Fission-product energy in fuel Conduction through fuel Transfer across fuel-clad gas gap Forced convection in coolant in fuel assemblies Transfer from clad surface to coolant Conduction across clad Steam generation due to boiling Steam separation Slide No 10
Specific Thermal Issues in LWRs (1) There are several existing reactor concepts in which heat transfer solutions depend on particular design concepts and the choice of coolant Although each reactor design has its own specific thermal problems, the solution of these problems can be approached in standard engineering manners that involve the fluid flow and heat transfer analysis The effort to integrate the heat transfer and fluid mechanics principles to accomplish the desired rate of heat removal from the reactor fuel is termed as the thermal-hydraulic design of a nuclear reactor Slide No 11
Specific Thermal Issues in LWRs (2) Important difference between nuclear power plants and conventional power plants: in conventional power plants the temperature is limited to that resulting from combustion of coal, oil or gas temperature can increase continuously in a nuclear reactor when the rate of heat removal is less than the rate of heat generation Such situation could lead to the core damage Slide No 12
Specific Thermal Issues in LWRs (3) For a given reactor design, the maximum operating power is limited by some temperature in the system There are several possible factors that will set the limit temperature (and thus the reactor power): material changes of some construction material in the reactor allowable thermal stresses or influence of temperature on corrosion Thus the maximum temperature in a nuclear reactor core must be definitely established under normal reactor operation and this is the goal of the thermal-hydraulic design and analysis of nuclear reactor cores Slide No 13
Specific Thermal Issues in LWRs (4) In nuclear reactors the construction materials must be chosen not only on the basis of the thermo-mechanical performance, but also (and often exclusively) on the basis of the nuclear properties Beryllium metal, for example, is an excellent material for use as moderator and reflector, but it is relatively brittle Austenitic stainless steels are used as the cladding for fast reactor fuels, but they tend to swell as a result of exposure at high temperatures to fast neutrons Slide No 14
Specific Thermal Issues in LWRs (4) Another peculiarity (and factor that adds problem) of nuclear reactors is that the power densities (e.g. power generated per unit volume) are very high This is required by economical considerations That leads to power densities of approximately 100 MWt/m 3 in PWRs and 55 MWt/m 3 in BWRs A typical sodium-cooled commercial fast breeder reactor has a power density as high as 500 MWt/m 3 In conventional power plants is of the order of 10 MWt/m 3. Slide No 15
Power Generation in Nuclear Reactor Cores (1) Chain reaction is the primary source of energy in nuclear reactors Slide No 16
Power Generation in Nuclear Reactor Cores (2) The energy released in nuclear fission reaction is distributed among a variety of reaction products characterized by different range and time delays In thermal design, the energy deposition distributed over the coolant and structural materials is frequently reassigned to the fuel in order to simplify the thermal analysis of the core Slide No 17
Power Generation in Nuclear Reactor Cores (3) The volumetric fission heat source in the core can be found as, q ( r) = i () ( i) r deσ ( E)( φ r E) ( i) w f N i f, 0 w f (i) here is the recoverable energy released per fission event for i-th fissionable material Slide No 18
Power Generation in Nuclear Reactor Cores (4) The simplest model of fission heat distribution would correspond to a bare, homogeneous core One-group flux distribution for such geometry is given as, ν 0r πz φ( r, z) = J 0 ~ cos ~ R H here R ~ and H ~ are effective core dimensions that include extrapolation lengths as well as an adjustment to account for a reflected core Slide No 19
Power Generation in Nuclear Reactor Cores (5) For a fuel rod located at r = r f distance from the centerline of the core, the volumetric fission heat source becomes a function of the axial coordinate, z, only: q ( z) = w 2.405r ~ R π z cos H f f Σ f φ0 J 0 ~ Slide No 20
Power Generation in Nuclear Reactor Cores (6) There are numerous factors that perturb the power distribution of the reactor core, and the above equation will not be valid For example fuel is usually not loaded with uniform enrichment At the beginning of core life, higher enrichment fuel is loaded toward the edge of the core in order to flatten the power distribution Other factors include the influence of the control rods and variation of the coolant density Slide No 21
Power Generation in Nuclear Reactor Cores (7) All these power perturbations will cause a corresponding variation of temperature distribution in the core A usual technique to take care of these variations is to estimate the local working conditions (power level, coolant flow, etc) which are the closest to the thermal limitations Such part of the core is called hot channel and the working conditions are related with so-called hot channel factors Slide No 22
Power Generation in Nuclear Reactor Cores (8) One common approach to define hot channel is to choose the channel where the core heat flux and the coolant enthalpy rise is a maximum Working conditions in the hot channel are defined by several ratios of local conditions to core-averaged conditions These ratios, termed the hot channel factors or power peaking factors will be considered in more detail in coming lectures However, it can be mentioned already here that the basic initial plant thermal design relay on these factors Slide No 23
Principles of Fluid-Flow and Heat Transfer Heat Conduction Laminar and Turbulent Flows Solid-Fluid Heat Transfer for Single-Phase Flows Slide No 24
Heat Conduction (1) Heat conduction refers to the transfer of heat by means of the molecular interactions The central role in heat conduction plays the Fourier law: ( r, t ) = T ( r, t) q λ ( ) here q r,t is the heat flux vector at location r and time t, T(r,t) is the local temperature and λ is the thermal conductivity Slide No 25
Heat Conduction (2) A non-stationary heat balance in a finite volume dv is as follows: z q z +dq z q x q y q x +dq x q y +dq y y x q z q dydzdt + q dxdzdt + q dxdydt + q dxdydzdt x y z ( q + dq ) dydzdt ( q + dq ) dxdzdt ( q + dq ) dxdydt = ρc dtdxdydz x x y y z z p Slide No 26
Heat Conduction (3) Equation can be simplified as q dxdydzdt dq dydzdt x dq dxdzdt y dq dxdydt z = ρc p dtdxdydz and dividing by dxdydzdt q q x y q z q = ρc x y z or p T t ρc p T t = q q ρ c p T t = q + λ T Slide No 27
Heat Conduction (4) In general case, when solid properties are time-andspace dependent, the conduction equation can be written as, t [ ρ( r, t) c( r, t) T ( r, t) ] [ λ( r, t) T ( r, t) ] = q ( r, t) This equation can be solved if proper initial and boundary conditions are specified: Initial conditions specify the temperature distribution at initial time, typically at t = 0, e.g. T(r,0) = f(r) is a given function Boundary conditions describe the temperature behaviour at the boundary. They are of four different kinds, described below Slide No 28
Heat Conduction (5) Boundary conditions: Of the first kind (Dirichlet b.c.) when temperature value is given at the boundary, e.g., T ( r, t) = ϕ( r, t) r r= B Of the second kind (Neuman b.c.) when the heat flux at the boundary is given, e.g., q ( r t) ( r, t) T, λ = φ( r, t) r= rb n Of the third kind (Newton b.c.) with convective heat transfer T λ n ( r, t) r= r B = r= r B [ ( ) ( ) ] r, t T r, t * h T r= r B f r= r B Slide No 29
Heat Conduction (6) Boundary conditions (cont ed) Of the fourth kind when two solids are in contact, T Example: ( r t) = T ( r, t) T and λ ( r, t) T ( r, t) 1 2 1, = r 2 1 λ2 = rb r= rb n r= r B Steady-state heat conduction through an infinite cylindrical wall. Figure below shows the geometry and the boundary conditions of the case considered in the present example. n r= r B q T f r 1 r 2 r Slide No 30
Heat Conduction (7) Conduction equation becomes: d dr r dt dr = 0 T = C ln r + D And the boundary conditions give the constants C and D: dt dt λ = q, λ = h T r= r dr r= r dr r= r 1 2 ( Tf ) 2 C λ r 1 C λ r 2 = = q r1 q, C = λ h ( C ln r + D T ) 2 f, D C = λ hr 2 C ln r 2 + T f D = T f + q r1 λ λ hr2 + ln r 2 Slide No 31
Heat Conduction (8) The solution can be thus written as: q r q r q r q r r q 1 1 1 1 2 r1 T = ln r + ln r2 + + T f = ln + + T λ λ hr λ r hr 2 2 f Slide No 32
Laminar and Turbulent Flows (1) To describe the fluid motion, the basic laws for conservation of mass, momentum and energy are invoked For incompressible (e.g. with constant density) flows the equations are as follows: Mass Momentym Energy div( v) = 0 Dv ρ Dt DH ρ Dt = ρg + τ p Dp = + div T Dt ( λ ) + Φ Slide No 33
Laminar and Turbulent Flows (2) The majority of fluids obey the Newton s eqation that relates the stress tensor in terms of the strain tensor. Such fluids are termed as the Newtonian fluids τ = µ [ ( ) ] T v + v For a one-dimensional laminar flow (e.g. in pipes), the stress-strain relation becomes: dr dv z τ = µ Slide No 34
Laminar and Turbulent Flows (3) Assuming laminar, stationary, adiabatic flow of a Newtonian fluid in a vertical tube, the conservation equations reduce to: 0 = p + ρg + µ 2 v In cylindrical coordinates we have, 2 = 1 r d dr r d dr Slide No 35
Laminar and Turbulent Flows (4) Since the velocity in such a flow has only a one component, v z, the momentum conservation equation simplifies to: p ρ g µ which can be solved as: Π = 1 r d dr r dv dr z d dr r 2 2 dvz dvz Πr dvz Πr C Πr = Πr r = + C = + vz = + C ln r + dr dr 2 dr 2 r 4 D D = 2 2 ΠR ΠR r v z = 1 4 4 R 2 Slide No 36
Laminar and Turbulent Flows (5) The average velocity in the tube can be found as U 1 πr 2 0 R v z ( r) 2πrdr = Π R 8µ 2 And the wall shear stress can be calculated as τ w = dvz ΠR τ ( r) µ = = r= R dr 2 r= R 4µ U R Slide No 37
Laminar and Turbulent Flows (6) There are two definitions commonly used in the literature to define friction factors: Darcy friction factor Fanning friction factor λ 4τ 2 ρu w 2 C f τ w = ρ U 2 2 1 λ 4 Slide No 38
Laminar and Turbulent Flows (7) For laminar flow: λ 64 Re = C f = 16 Re Where Re is the Reynolds number Re = ρud µ Slide No 39
Laminar and Turbulent Flows (8) For turbulent flows in smooth tubes, Blasius proposed the following correlation: C f = 0.0791 0.25 Re For commercial tubes, which have wall roughness k, Colebrook devised the following formula 1 = 2.0log λ 10 k D 3.7 + 2.51 Re λ Slide No 40
Laminar and Turbulent Flows (9) Colebrook s formula is implicit: this annoyance can be avoided by instead using the Haaland formula: 1 = 1.8log λ 10 k D 3.7 1.11 + 6.9 Re Slide No 41
Solid-Fluid Heat Transfer for Single-Phase Flows (1) Heat transfer from the clad surface to the single-phase coolant is often described by the Newton s law of cooling: q = h ( ) T w T f Here h is the heat transfer coefficient. It can be found from the Dittus-Boelter correlation: Nu = 0.023Re 0.8 Pr 0.33 Nu = hd λ Pr = c p µ λ Slide No 42
Exercises (1) Exercise 13. A cylindrical core has the extrapolated height and extrapolated radius equal to 3.8 m and 3.2 m, respectively. Calculate the volumetric heat source ratio between the point located in the middle of the central fuel assembly and the point located in a fuel assembly with a distance r f = 1.5 m from the core center and at z = 2.5 m from the inlet of the assembly. Slide No 43
Exercises (2) Exercise 14. A cylindrical wall with the internal diameter equal to 8 mm and thickness 0.5 mm is heated on the internal surface with a heat flux equal to 1 MW m -2. Calculate the temperature difference between the inner and the outer surface of the wall, assuming that the thermal conductivity of the wall material is equal to 16 W m -1 K -1. What will be the heat flux on the outer surface of the wall? Slide No 44
Exercises (3) Exercise 15. In a horizontal pipe with an internal diameter equal to 10 mm flows water. Calculate the wall shear stress change when mass flux of water increases from 50 to 2000 kg m -2 s -1. Assume that the water density and the dynamic viscosity are constant and equal to 740 kg m -3 and 1.0x10-4 Pa s, respectively. Slide No 45
Exercises (4) Exercise 16. For geometry as in the picture and given Q = 2.5 x 10 4 W, wall conductivity coefficient λ 1 = 16 Wm -1 K -1, liquid temperature T L = 323 C, and mass flow rate G = 2000 kg m -2 s -1 : 1. Write shell balance for energy diffusion in the cylindrical solid, 2. state proper boundary conditions (boundary value problem), 3. calculate temperature distribution in the cylindrical solid, Obtain the heat transfer coefficient from the Dittus-Boelter correlation. Conductivity, heat capacity and dynamic viscosity of water are constant values: λ 2 = 0.5 Wm - 1 K -1, c p = 6 x 10 3 Jkg -1 K -1, µ = 8.5 x 10-5 r 3 =18 Q T L L = 1000 kgm -1 s -1. r 1 = 8 r 2 = 10 r Slide No 46
Exercises (5) Exercise 17. For a channel as shown in the figure calculate the frictional pressure drop Given:L = 1 m, D 1 = 8 mm, D 2 = 16 mm, G = 103 kgs -1 m -2, µ = 8.53 x 10-5 kgm -1 s -1, ρ = 660 kgm -3. To obtain the Fanning friction factor use the Blasius formula for smooth tubes L D 2 D 1 Slide No 47
Home Assignment #3 (1) Problem 1 (5 points): A pipe with an internal diameter 8 mm and wall thickness 1 mm is internally cooled with sub-cooled water and heated with an uniform heat flux 1MWm -2 at the outer wall. Mass flux of the water is 1200 kg m -2 s -1, density 600 kg m -3, dynamic viscosity 6.8 10-5 Pa s, specific heat 8950 J kg -1 K -1, heat conductivity 0.46 W m -1 K -1. Calculate the inner and the outer wall surface temperature in the pipe at the location where the bulk water temperature is equal 320 C. Assume the thermal conductivity of wall 16 Wm -1 K -1 Slide No 48
Home Assignment #3 (2) Problem 2 (5 points): Calculate the total pressure drop in a vertical tube with total length 3.6 m (see Figure on the next page). The lower inlet part of the tube has the diameter equal to 55 mm. The upper part of the tube has the diameter equal to 57.5 mm. The sudden area change is located at 2.7 m from the inlet. The upward flowing fluid is water with the inlet mass flux G = 1200 kg m -2 s -1. Neglect inlet and outlet local losses. Neglect phase-change effects and assume constant fluid properties along the channel. Use the Blasius formula for the friction coefficient. Slide No 49
Home Assignment #3 (3) Problem 2 (cont ed): Water properties: density 740 kg m -3, dynamic viscosity 9.12 10-5 Pa s 3.6 m 2.7 m G = 1200 kg m -2 s -1 Slide No 50