VIEWS: 7 PAGES: 20 CATEGORY: Technology POSTED ON: 6/15/2010 Public Domain
Thermal hydraulics of liquid metals Janne Wallenius Reactor Physics, KTH Objectives After this meeting and home assignment you will be able to Calculate coolant and clad temperatures in liquid metal cooled reactors Assess the impact of coolant velocity and pin pitch on pressure drop. Estimate natural convection ﬂow velocity in liquid metal cooled reactors. Fast neutron reactor geometry Hexagonal geometry (triangular unit cell) standard choice for fast Diameter neutron reactors Close packing, minimises leakage Pitch In hexagonal geometry, one Area coolant channel transports heat from half a pin. Square lattice suggested for lead cooled reactors (BREST, ELSY) Methodology Calculate mass ﬂow corresponding to the desired temperature increase in the coolant. Applying maximum permitted coolant velocity, ﬁnd minimum P/D satisfying temperature limits for the cladding: Calculate axial temperature proﬁle in bulk coolant Calculate axial dependence of temperature drop in ﬁlm between clad and bulk coolant Axial power proﬁle Linear power [kW/m] 35 Axial power distribution in fast 30 reactors well described by cosine function (far away from control rods). 25 20 Extrapolation length in sodium: 135% 15 of fuel column height. 10 Axial peaking factor: ~ 1.26 5 z [m] Extrapolation length in lead: 0 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 150% of fuel column height Axial peaking factor: ~ 1.20 Coolant temperature (hexagonal geometry) Coolant temperature K Half the linear power is removed by one coolant channel: 800 z 1 χ (z) T (z) = Tin + ∫ 2C dz m −L /2 p 750 Mass ﬂow is a constant, determined e.g. by density and velocity at the inlet of the coolant channel. 700 To perform analytical integration, we may approximate Cp with its value at 650 the inlet 0.4 0.2 0.0 0.2 0.4 πz z m sin L χ ave 1+ L0 T (z) = Tin + 4 ρinVin A C p π L sin 2L0 Physical properties The density of liquid lead is one order of magnitude higher than the density of liquid sodium. The opposite holds for the respective heat capacities The product of density and heat capacity is ~ 40% higher for lead. Density [kg/m3] Heat capacity [J/(kg K)] 12 000 1400 10 000 Lead 1200 Sodium 8000 1000 6000 800 600 4000 400 2000 Sodium Lead 200 0 0 600 700 800 900 1000 1100 600 700 800 900 1000 1100 Temperature [K] Temperature [K] Density Sodium density as function of temperature: ρ Na (T ) = 1012 − 0.2205T −1.923 ×10−5 T 2 + 5.637 ×10−9 T 3 Lead density as function of temperature: ρ Pb (T ) = 11367 −1.1944T Relative expansion of sodium larger than for lead (2.2% for ∆T = 100 K) Absolute expansion of lead larger than for sodium (120 kg/m3 for ∆T = 100 K) Velocity limitation Upper limit for sodium velocity: 7–9 m/s Limiting phenomenon: mechanical vibration of fuel assembly Upper limit for lead velocity: 1.8 – 1.9 m/s Erosion of protective iron oxide ﬁlm on steel cladding surface Q Q ΔTcool = = mC p ρ AvC p Coolant ﬂow area in lead must be ~ 3 times larger than in sodium to achieve the same heat removal rate! Coolant channel dimensions Refence design parameters Diameter Pitch Cladding diameter: 8.6 mm Area Average linear power: 30 kW/m Fuel column height: 1.0 m Coolant temperature K Sodium velocity: 8.0 m/s 800 Hexagonal geometry 750 ∆TNa = 100 K -> P/D = 1.20 700 Lead velocity: 1.75 m/s 650 0.4 0.2 0.0 0.2 0.4 z m ∆TPb= 100 K -> P/D = 1.62 Clad temperature Heat transfer from fuel clad to bulk coolant leads to temperature difference: χ (z) Nu × k h (Tsurf − Tcool ) = h= πD Dh The heat transfer coefﬁcient depends on thermal conductivity of coolant, on the dimensionless Nusselt number and on the so called hydraulic diameter: 4A Dh = Pwet Dh 1 Pwet = πD 2 Physical properties Thermal conductivity [W/(m K)] 80 Thermal conductivity of sodium larger 60 Sodium than for lead 40 k Na (T ) = 109.7 − 0.0645T +1.173 ×10−5 T 2 Lead 20 kPb (T ) = 9.2 + 0.011T T[K] 0 700 800 900 1000 1100 1200 Nusselt number depends on density, heat capacity, velocity, geometry and thermal Nusselt number 40 conductivity. 30 Lead Nu = 0.047(1 − e−3.8(P / D−1) )(Pe0.77 + 250) ρC p vDh 20 Pe = k Sodium 10 Reference conﬁguration: Nu[Pb] > Nu[Na] T[K] 0 700 800 900 1000 1100 1200 Temperature difference Tsurf - Tcool [K] 4A 40 χ (z) Dh Dh = ∆ T (z) = π D Nu × k Pwet 30 Lead Nu x k ~ same for lead and sodium 20 Sodium Dh ~ 4 times larger for lead 10 T [K] For same bulk coolant temperature, 0 700 800 900 1000 1100 1200 temperature difference between clad and coolant several times larger in lead Cladding temperatures Clad surface temperature [K] 800 Pb Due to larger hydraulic diameter 780 Na (ratio between ﬂow area and wetted 760 perimeter), cladding temperatures 740 are signiﬁcantly higher with lead coolant, even if bulk coolant 720 temperature is the same. 700 680 z [m] Maximum cladding temperature is -0.4 -0.2 0.0 0.2 0.4 located below the top of the fuel column. Natural circulation Hot coolant above core has lower density than cold coolant at the outlet of the heat exchanger Buoyancy pressure head Pb = g H xc (ρin − ρ out ) For fully established natural convection, buoyancy pressure equals pressure losses in core and heat exchanger. Pressure drop proportional to coolant velocity square, e.g. is the channel friction pressure drop: H ch ρ v2 H ch m 2 ΔPch = f =f 2D h 2D h ρ A 2 Natural circulation & lead Pb = g H xc (ρin − ρ out ) H ch ρ v2 H ch m 2 ΔPch = f =f 2D h 2D h ρ A 2 Larger volumetric expansion of lead Larger hydraulic diameter of lead core Potential for heat removal by natural convection considerably larger! During normal operation, natural convection contributes with only a few percent of ﬂow velocity Heat removal during transient If pumping power is lost without scram of reactor (Unprotected Loss of Flow, ULOF), coolant temperatures will increase. Buoyancy forces increase, leading to increase in natural convection ﬂow velocity. Finally, equilibrium is obtained and temperatures stabilise! For lead cooled reference core: vPb ~ 0.4 m/s, ∆TPb ~ 500 K Short term (30 minutes) permitted clad temperatures: 1000 K for ferritic martensitic steels (may not survive) 1200 K for austenitic steels (would probably survive!) Assignment, part 1 Calculate the pin pitch (in units of D) required to stay below the long term clad failure limit for lead and sodium, assuming a square pin lattice and D = 8.6 mm, H = 100 cm. Velocity[Pb] = 1.75 m/s, Velocity[Na] = 8.0 m/s Pin average power density = 30 kW/m Inlet temperature of 670 Kelvin. The long term limit for clad temperature (see chapter on clad materials) is 820 K for lead 870 K for sodium Assignment, part 2. Calculate the vertical elevation of the heat exchanger required for the cladding to survive an unprotected loss of ﬂow accident (ULOF) with lead and sodium coolants (reference conﬁguration as presented in the lecture notes, same coolant channel height.) Adopt austenitic steel cladding, for which Tclad should be kept below 1200 K during the 30 minute transient. Assume that the lead velocity of 1.75 m/s during normal operation leads to a total pressure drop in core and heat exchanger of 350 kPa. Report Individually written report (3–4 pages) with graphs showing axial dependence of coolant temperature, ∆T between coolant and clad and resulting clad temperature. Introduction Explanation of methodology Input data (physical properties, correlations) Results Interpretation of results (explain differences between results for lead and sodium). Report will be checked by other student, to be selected at next meeting. My solution will be provided as example