Thermal hydraulics of liquid metals by qdk21196

VIEWS: 7 PAGES: 20

									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

								
To top