# Thermal hydraulics of liquid metals by qdk21196

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```									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.
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
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
Sodium
8000                                             1000

6000                                             800

600
4000
400
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
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

650
0.4      0.2    0.0    0.2      0.4
z m
∆TPb= 100 K -> P/D = 1.62

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

40                                        k Na (T ) = 109.7 − 0.0645T +1.173 ×10−5 T 2
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

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

800
Pb                Due to larger hydraulic diameter
780
Na
(ratio between ﬂow area and wetted
740
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

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

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 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

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.)

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

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

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