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HUVINETT 2012/1

Thermal hydraulics
of nuclear reactors

General considerations

Prof. Dr. Attila ASZÓDI
Director
Budapest University of Technology and Economics
Institute of Nuclear Techniques (BME NTI)

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

1

Basic considerations of nuclear safety

– Safety objective is ranked higher than electricity production!


• Specialties of nuclear fuel and NPPs:





No real environmental risk of fresh (non-irradiated) fuel
High risk of irradiated fuel
High thermal density
Reactor power possible in a very wide range (even over nominal power, in a
short pulse hundreds of nominal power – see Chernobyl)
– Cooling of irradiated fuel needed after shut-down (removal of remanent (decay)
heat in the first ~5 years in water feasible, later in gas atmosphere over hundreds
of years possible)
Prof. Dr. Attila Aszódi, BME NTI

Prof. Dr. Attila Aszódi, BME NTI

2

Design of NPPs

• Nuclear Power Plant in normal operation: very low
emission, practically only thermal load to the environment.
• But high risk: Large amount of highly dangerous
radioactive material generated and accumulated in the
reactor core.
• Safety objective: protect the environment from this highly
dangerous radioactive material

Thermal hydraulics

Thermal hydraulics

3

Design necessary not only for normal
operation, but also for anticipated operational
transients and for wide range of so called


Design Basis Accidents.
– Beside operational systems separated
safety systems required
– Due to single failure criteria multiple
independent safety systems needed for
the same function
Construction of an NPP is much
more expensive than of a fossil power
plant, where safety is not that
critical.
Electricity generated in NPPs will be
competitive only if
high annual load factor is ensured
fuel cost has to be much lower than by
fossil power plants
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

4


Example: Bulgaria, Kozloduy NPP, VVER-1000

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

Example: Bulgaria, Kozloduy NPP, VVER-1000

5

Example: Bulgaria, Kozloduy NPP, VVER-1000

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

6

Example: Bulgaria, Kozloduy NPP, VVER-1000

7

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

8


Example: Bulgaria, Kozloduy NPP, VVER-1000

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

Example: Bulgaria, Kozloduy NPP, VVER-1000

9

Example: Bulgaria, Kozloduy NPP, VVER-1000

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

10

Basic safety functions
– Efficient control of the chain reaction and hence the
power produced.
– Fuel cooling assured under thermal hydraulic conditions
designed to maintain fuel clad integrity, thus constituting
an initial containment system.
– Containment of radioactive products in the fuel but
also in the primary coolant, in the reactor building
constituting the containment or in other parts of the plant
unit.

Independent and redundant water source to avoid loss of ultimate heat sink
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

11

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

12


Fuel design

Fuel design
• Possible objectives of thermal hydraulical design
of fuel

• Limits in design: thermal hydraulics
(cooling) is more limiting than reactor
physics

– Maximizing power density to decrease Reactor
Pressure Vessel (RPV) size
– Maximizing power density to decrease number of
fuel assemblies in the core
– Maximizing power density to increase reactor
power (by given RPV size)
– Maximizing coolant outlet temperature to increase
NPP efficiency
– Maximizing fuel burnup to enhance fuel economy

– Choice of coolant material (H2O, CO2, He,
Na, Pb), pressure, temperature
– Choice of flow velocity and other flow
parameters (turbulence, mixing, boiling)

• Many parameters have to be optimized:








Enrichment
Burn-up
Power, power density
Fuel and cladding temperature
Cladding integrity and durability
Coolant parameters
Etc.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

• Limits in thermal hydraulics design:
– Maximal fuel temperature (to avoid fuel melting),
– Maximal cladding temperature (to avoid cladding
oxidation, decrease in strength)
– Maximal coolant temperature (to avoid boiling or
boiling crisis)
– Other thermal limits (to keep reactivity feedbacks
between reactor physical limits)

13

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

14

Fundamentals of heat transport
• Heat: energy transport due to temperature difference.
• Conduction: In heat transfer, conduction (or heat conduction) is the transfer of
thermal energy between neighboring molecules in a substance due to a temperature
gradient. It always takes place from a region of higher temperature to a region of
lower temperature, and acts to equalize temperature differences. Conduction needs
matter and does not require any bulk motion of matter.
• Convection in the most general terms refers to the movement of molecules within
fluids (i.e. liquids, gases). Convection is one of the major modes of heat transfer and
mass transfer. In fluids, convective heat and mass transfer take place through both
diffusion – the random Brownian motion of individual particles in the fluid – and by
advection, in which matter or heat is transported by the larger-scale motion of
currents in the fluid. In the context of heat and mass transfer, the term "convection"
is used to refer to the sum of advective and diffusive transfer.
• Thermal radiation is electromagnetic radiation emitted from the surface of an
object which is due to the object's temperature. Infrared radiation from a common
household radiator or electric heater is an example of thermal radiation, as is the
light emitted by a glowing incandescent light bulb. Thermal radiation is generated
when heat from the movement of charged particles within atoms is converted to
electromagnetic radiation.

Fundamentals of heat
transport

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

15

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

16


The definition of heat transfer

Energy production and heat transfer

• Heat transfer: across the surfaces of two different
materials it is a complex physical process which combines
the three fundamental heat transport methods (conduction,
convection, radiation).
• In the technical practise: investigation of the heat transfer
process between solid wall and liquid, solid wall and
steam/gas, for example:
– Cooling the fuel rods;
– Heat transfer through a surface between the primary and secondary
side of a steam generator








Volumetric heat power rate:
Heat flux:
Linear heat power rate :
Pin power:
Core power:
Core volumetric power density:

r
q& ′′′(r )
r
q& ′′( A)
q& ′(z )
q&
Q&
Q& / V ≡ Q′′′

• Thermal hydraulics: coupled thermal- and hydrodynamics
analysis of the reactors
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

17

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

Energy production in fuel

Energy production and heat transfer

• Energy release in fission: 200 MeV/fission

• Thermal hydraulical design of the fuel:

– 92-94% discharged in fuel
– 2,5% discharged in moderator

r

r

A

V

• Metallurgical design:
r

∫ q&′( z )dz = ∫∫∫ q&′′′(r )dV
L

i

r
Φ th ( r ) Thermal neutron flux [1/cm2s]
Type of fissionable material (e.g. 235U, 239Pu, 241Pu) [-]
i
Energy release in one fission of an „i” type atom
[J/fission]
ci
r
r
r
[atoms/cm3]
N i ( r ) Number of „i” type atoms in 1 cm3 around
σ fi
Fission cross section of „i” type fissionable material [cm2]
r
Space coordinate
r
Prof. Dr. Attila Aszódi, BME NTI

r

∫∫ q& ′′( A) ⋅ ndA = ∫∫∫ q&′′′(r )dV

• Calculation of volumetric heat power generation
from the thermal neutron flux
r
r
r
q& ′′′ ( r ) = Φ th ( r )∑ c i ⋅ N i ( r ) ⋅ σ fi

Thermal hydraulics

18

19

V

• Pin power:
r
q& = ∫∫∫ q& ′′′(r )dV
V

• Core power:
N

Q& = ∑ q&n

where N: number of fuel pins inside the core

n =1

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

20


Newton's law of cooling

Newton's law of cooling 2.

• A related principle, Newton's law of cooling, states that the rate of heat loss of a
body is proportional to the temperature difference between the body and its

• In the technical practise the average heat transfer coefficient is often
used. In this case (Tw − T ∞ ) temperature difference is an averaged value
along the F surface.
• In some cases the heat transfer coefficient strongly alters along the
surface due to the changing influential properties. This is the local
heat transfer coefficient: at the x place on the dF surface the thermal

surroundings, or environment. The law is

Q& = α ⋅ F ⋅ (Tw − T ∞ )

where:

Q&

the thermal power from the solid surface to the fluid, W;

F

Surface area of the heat being transferred, m2;

Tw

Temperature of the object's surface, °C or K;

T∞

average Temperature of the environment , °C or K;

α

average Heat Transfer coefficient, W/(m2·K)

power is the following:
dQ& x = α x ⋅ (Tw ,x − T ∞ ) ⋅ dF where:
Tw,x
T∞

αx

(instead of α „h” is also used).

Thermal hydraulics

The aim of investigation on thermal processes is to describe the heat
transfer coefficient.

Prof. Dr. Attila Aszódi, BME NTI

21

50,000
2

α CFX =

Prof. Dr. Attila Aszódi, BME NTI

q& w ' '
Tw − Tnw

40,000

Spacer Grid 1
35,000

∫ w ⋅ ρ ⋅ h(T ) ⋅ dF = m& ⋅ h(T

30,000

α GEN

25,000

22

• The T∞ temperature is equal to that temperature which can
be measured far from the heated surface in case of flow in
half unlimited space.
• In closed channel flow, if the mass flow rate is m& , the
density is ρ, the velocity is w and the enthalpy is h = h(T) :

The distribution of the average heat
transfer coefficient
45,000

Thermal hydraulics

Newton's law of cooling 3.

Local heat transfer coefficient – example:

Avera
erage Heat Transfer Coefficient [W/(m
[W
K)]

local Temperature of the object's surface, °C or K;
average Temperature of the environment , °C or K;
local Heat Transfer coefficient, W/(m2·K).

q& w ' '
=
Tw − Tave



)

F

20,000

If the specific enthalpy is h = cp·T and cp= constant, than

SST Model - CFX Definition
SST Model - General Definition
Dittus-Boelter Correlation
Dittus-Boelter Correlation + 25%
Dittus-Boelter Correlation - 25%

15,000
10,000
5,000

∫w ⋅ ρ ⋅c

⋅ T ⋅ dF = m
& ⋅ c p ⋅ T∞

where:

F

0
0

0.5

1

1.5

2

2.5

Height [m]

• The average heat transfer coefficient calculated according to the general definition
agrees well with the value calculated by Dittus-Boelter correlation
Stockholm, 10.10.2006.

Thermal hydraulics

p

S. Tóth, A. Aszódi, BME NTI

Prof. Dr. Attila Aszódi, BME NTI

T∞ =
18

23

Thermal hydraulics

∫ w ⋅ ρ ⋅ T ⋅ dF
F

m
&
Prof. Dr. Attila Aszódi, BME NTI

24


The equilibrium equations - Boussinesq-approach
• The flow velocity field: w = w(r ) , in Descartes coordinate
system:
w x = w x (x , y, z ) w y = w y (x , y, z ) w z = w z (x , y, z )
• The temperature field: T = T (r ) = T (x , y, z )
• The pressure field:
p = p(r ) = p(x , y, z )
• Continuity (if the density is constant):
div(w ) = 0
• The general differential equation of heat
conduction (no heat source, the fluid heat conductivity
(λ), isobaric heat capacity (cp), density (ρ) are constant,
steady-state condition):
w∇T = a ⋅ ∇ 2T

Thermal hydraulics

, where: a =

λ
ρ ⋅ cp

(thermal diffusivity)

Prof. Dr. Attila Aszódi, BME NTI

25

The hydraulic boundary layer

The equilibrium equations - Boussinesq-approach
Navier-Stokes equations (steady-state, kinematic viscosity is const.):

(w ⋅ ∇ ) ⋅ w = ν ⋅ ∇ 2 w − ∇p + g − β ⋅ ∆T ⋅ g
ρ

Conditions for the above equations:
– The gravity has only a component in z direction, whereas

g = −g ⋅ k

, where: k is the unit vector in z direction;

– The density dependency on the temperature is only considered at the gravity term of
the above equation to consider the temperature difference (∆T) which causes a
buoyancy force due to the density difference (β is the volumetric heat expansion or
compressibility)
– ∆T is the temperature difference, T is the temperature and T∞ is the average
temperature difference.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

26

The hydraulic boundary layer in the flow
near a solid plane

• Ludwig Prandtl’s recognition (1904) – the flow can
be divided into two regions at the vicinity of solid wall:
– first: inside the boundary layer, where viscosity is
dominant and the majority of the drag experienced by a
body immersed in a fluid is created,
– second: outside the boundary layer where viscosity can
be neglected without significant effects on the solution.

Experiment:
This experiment has
been performed in
water where very
small hydrogen
bubbles have been
generated by
hydrolysis.

• The thickness of the hydraulic boundary layer
(δ): if the velocity deviate from the main flow velocity

The bubbles have
very slow lifting
movement so they are
suitable to visualize
the flow.

(w∞) more than 1% then it is inside the hydraulic boundary
layer.

Reference: Lajos Tamás: Az áramlástan alapjai (Fundamentals of fluid
mechanics – in Hungarian, CD), Műegyetemi Kiadó, Budapest, 2004.
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

27

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

28


The hydraulic boundary layer in the flow
near a solid plane

The hydraulic boundary layer in pipe flow
Experiment:
This experiment has
been performed in
water where very
small hydrogen
bubbles have been
generated by
hydrolysis.

Physical
model:

δx = 5⋅

ν ⋅x

The bubbles have
very slow lifting
movement so they are
suitable to visualize
the flow.

w∞

( x < x kr )

x kr = 3,2 ⋅ 10 5 ⋅

ν

Reference: Lajos Tamás: Az áramlástan alapjai (Fundamentals of fluid
mechanics – in Hungarian, CD), Műegyetemi Kiadó, Budapest, 2004.

w∞

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

29

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

The hydraulic boundary layer in pipe flow

The hydraulic boundary layer in pipe flow

Physical model:

Physical model:

• The boundary layer on the pipe wall abuts after a certain
distance from the entrance of the pipe.
• The constant velocity field before the entrance modify and a
characteristic velocity profile builds up after the suitable
distance from the entrance. After this certain length this is
the so called fully developed flow.
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

31

30

& , the density is ρ.
The flow area is A, the mass flow rate is m
m
w ⋅D
&
< Re kr = 2300 , where w =
The flow is laminar, if Re =
.
ρ⋅A
ν
The flow can be laminar along the whole pipe length, if the diameter:
.

D < 2300 ⋅
Thermal hydraulics

ν

w

Prof. Dr. Attila Aszódi, BME NTI

32


The thermal boundary layer

The heat transfer NUSSELT’s equation:

• The thermal boundary layer: this is analogue with
the hydraulic boundary layer. If the wall temperature (Tw) is
not equal to the bulk temperature (T∞), than:
– the fluid temperature is equal to the wall temperature on the solid
wall;
– approaching to the bulk from the wall the temperature approximate
the bulk temperature.

• The mechanism of heat transfer: if the heat radiation is negligible,
then the energy from the wall (which has Tw temperature) to the fluid
(which has T∞ temperature) transported by heat conduction through
the boundary layer at x position where the thickness of boundary layer
is δx. The δx alters along the heated length but the viscose sub layer
exists everywhere near the solid wall.
• At a certain place (x) the surface heat flux can be defined by two
ways, if the local heat transfer coefficient is αx:
q& W′′ (x ) = −λ ⋅ gradT (x ) W


′′ (x ) = α x ⋅ (TW − T∞ )
q& W

The heat transfer NUSSELT’s equation:

− λ ⋅ gradT (x ) W = α x ⋅ (TW − T∞ )

• The thermal boundary layer (δt): if the temperature deviate
from the main flow temperature (T∞ ) more than 1% then it is inside



λ is the heat conductivity of the fluid,

The heat transfer NUSSELT’s equation is not equal to the 3rd kind boundary condition.

the thermal boundary layer.
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

33

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

Methods to calculate the heat transfer coef.

Similarity of heat transfer processes 1.

• Solving the steady state equations by using certain
boundary conditions the velocity and temperature
field can be determined. Using the temperature field
and the NUSSELT equation the heat transfer
coefficient can be calculated.

• Two physical processes are similar, if

• Measuring the heat transfer coefficient and
the generalization of the experimental data,
using them for similar flows (Similarity
theory) is possible.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

35

34

– The differential equations are the same;
– The geometries are similar;
– The initial and boundary conditions of the differential
equations can be transformed in the same values using
proper ratios.
• The similarity theory enables to investigate an unmeasured
heat transfer process using a geometrically similar and
previously measured example.
• To characterize the similarity many similarity number are
used in practice.
• In the followings the practically important similarity
numbers are defined.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

36


Similarity of heat transfer processes 2.

Similarity of heat transfer processes 3.

• To define the similarity numbers let’s use the Boussinesqapproach of the steady state equations:
• The z component of Navier-Stokes equation:

• Let’s transform the steady state equations into
dimensionless form using normalization. The base value of
the normalization must be well measurable. L is the
characteristic geometrical parameter, w∞ is the characteristic
velocity, Tw is the surface temperature, T∞ is the flow
characteristic temperature!
• The dimensionless parameters:

wx ⋅

∂wz
∂wz
∂w z
+ wy ⋅
+ wz ⋅

∂x
∂y
∂z

 ∂ 2wz ∂ 2wz ∂ 2wz
⋅ 
+
+
2
∂y 2
∂z 2
 ∂x

 1 ∂p
 − ⋅
− g − g ⋅ β ⋅ ∆T
 ρ ∂z

• The energy equation:
wx ⋅

 ∂ 2T ∂ 2T ∂ 2T
∂T
∂T
∂T
= a ⋅  2 + 2 + 2
+ wz ⋅
+ wy ⋅
∂z
∂y
∂z
∂y
∂x
 ∂x





– Velocity:

• The continuity equation:
∂w x ∂w y ∂w z
+
+
=0
∂x
∂y
∂z

– Pressure:

• The Nusselt-equation:
− λ ⋅ gradT (x ) W = α x ⋅ (TW − T∞ )

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

37

ωx =

wx
w∞

ωy =

ωz =

w∞

wz
w∞

p
π =
ρ ⋅ w ∞2

– Temperature:

ϑ=

T − T∞
TW − T∞

– Space:

ξ =

x
L

Thermal hydraulics

wy

η=

y
L

ζ =

z
L

Prof. Dr. Attila Aszódi, BME NTI

38

Similarity of heat transfer processes 4.

Similarity of heat transfer processes 5.

• Replacing the dimensionless parameters into the Navier-Stokes
equations and reassembling:

• The dimensionless similarity parameters in the above mentioned
equations are the following:
– The Péclet number (this number shows a relationship between the
w ⋅L
velocity field and temperature field):
Pe = ∞

ωx ⋅

∂ω z
∂ω
∂ω
ν
+ ω y ⋅ z + wz ⋅ z =
∂ξ
L ⋅ w∞
∂η
∂ζ

 ∂ 2ω z ∂ 2ωz ∂ 2ωz
+
+
⋅ 
2
∂ζ 2
∂η 2
 ∂ξ

 ∂π L ⋅ g L ⋅ g
 −
− 2 − 2 ⋅ β ⋅ (TW − T∞ ) ⋅ ϑ
w∞
w∞
 ∂ζ

• Replacing the dimensionless parameters into the energy equation
and reassembling :
∂ϑ   ∂ 2ϑ ∂ 2ϑ ∂ 2ϑ 
∂ϑ
∂ϑ
w∞ ⋅ L 

+ ωz ⋅
+
+ ωy ⋅
+
⋅  ωx ⋅
=
∂ζ   ∂ξ 2 ∂η 2 ∂ζ 2 
∂η
∂ξ
a


– The Nusselt number (the criteria of the similarity of the heat
α ⋅L
transfer):
Nu =

∂ωx ∂ω y ∂ωz
+
+
=0
∂ξ
∂η
∂ζ

λ

– The Froude number (this number shows the relationship between
w∞
Fr =
the gravity and inertia forces):

• The Nusselt-equation:

Thermal hydraulics

– The Reynolds number (this number shows the relationship between
w ⋅L
the inertia and drag forces):
Re = ∞
ν

• The continuity equation:

gradϑ W = −

a

L⋅g

α ⋅L
⋅ϑ W
λ

– The Archimedes number (this number shows the relationship
between the buoyancy and inertia force): Ar = L ⋅ g ⋅ β ⋅ (TW − T∞ )
Prof. Dr. Attila Aszódi, BME NTI

39

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

w∞2

40


Similarity of heat transfer processes 6.

Similarity of heat transfer processes 8.

• Further practically used similarity numbers

• The Nusselt number (which consist of the α) depends on the
Reynolds-, Grasshoff- and Prandtl number, furthermore the geometry
and boundary conditions:

– The Prandtl number defines the ratio of the thickness of the
hydraulic and thermal boundary layer:
If Pr = 1, then δt / δh = 1

Pr =

Pe ν
=
Re a

– The Grasshoff number:

Gr = Re 2 ⋅ Ar =

– The Rayleigh number:

Ra = Gr ⋅ Pr

– The Stanton number:

St =

L3 ⋅ g ⋅ β ⋅ (TW − T∞ )

ν2

Nu
α
=
Re ⋅ Pr ρ ⋅ c p ⋅ w∞

• Hereafter the fluid fully fill the whole domain for all of the discussed
heat transfer cases (there is no free surface) or the solid body is fully
bounded by the fluid. Therefore the Froude number do not has an
important roll.
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

41

Nu = f (Re, Gr, Pr, Geometry, Boundary
conditions)
• But: the similarity numbers are defined using the Boussinesqapproach which has the constant material properties (ρ, λ, ν, cp) versus
the temperature. The temperature dependency of the material
properties influences the velocity and temperature fields, so finally the
heat transfer coefficient too.
• This effect can be considered by a ΦT correction factor:
Nu = f (Re, Gr, Pr, Geometry, Boundary conditions)· ΦT

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

Practical determination of α 1.

Practical determination of α 2.

• The similarity theory enable to generalize the experimental
results of heat transfer. If the similarity numbers are used
for the experimental results then the Nusselt number can be
expressed by the similarity numbers and their powers:

• The Nusselt number correlations are generally semi-empirical
equations using experimentally determined coefficient sets.

Nu = Rem ⋅ Gr n ⋅ Pr p ⋅ ΦT

• Forced convection (the flow created by external force)
often the buoyancy effect is negligible, therefore the Nusselt
number is independent from the Grasshoff number.
Nu = f (Re, Pr, Geometry, Boundary conditions)· ΦT

• Natural convection (due to inhomogeneous temperature
field the density field is inhomogeneous, which induces the
flow) here the Nusselt number is independent from the
Reynolds number

42

• The correlations are valid in their definition range.
• The error range of calculated α is typically ± 20…40%, at
complex phenomenon that could be higher (for example boiling,
condensation)!
• The values of the similarity numbers must be calculated at
reference temperature.
• All parameters must be used with correct units.
• Near the critical point of the fluid the following correlations and
equations are not valid.

Nu = f (Gr, Pr, Geometry, Boundary conditions)· ΦT
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

43

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

44


Heat transfer coefficient at forced convection

Heat transfer coefficient at forced convection

• Single phase heat transfer in horizontal tubes. The

• Single phase heat transfer in horizontal tubes.
• The α for laminar, hydraulically developed flow [1]:

characteristic of α strongly depends on the flow mode (laminar or
turbulent).
– Re < 2300 the flow is laminar, Re > 2300 the flow is turbulent.
– The averaged velocity in the Reynolds number is a value calculated
& ), the flow area (A) and the fluid
by the fluid mass flow rate (m
m
&
density (ρ): w =
ρ⋅A
– The characteristic length is the diameter (D), standard temperature is
the flow characteristic temperature (T∞), a needed additional size is
pipe length (L).

Nu = 3 3,663 + 1,613 ⋅ Re ⋅ Pr ⋅

D
⋅ ΦT
L

(constant wall temperature)

Valid, if Re < 2300 and 0,1< Re·Pr·D/L <. 104
Nu = 4,364

.
(constant
heat flux)
Valid, if Re < 2300 and 0,5 < Pr < 2000

• The α for turbulent, hydraulically developed flow [1]:
ξ 
  ⋅ (Re − 1000 ) ⋅ Pr
2/3
D
8

Nu =  
⋅ 1 +  ⋅ ΦT
L
ξ 
1 + 12,7 ⋅ Pr 2 / 3 − 1 ⋅
8

(

)

where

ξ=

1
(1,82 ⋅ log10 Re − 1,64 )2

Valid, if 3000 < Re < 5·106 and 0,1 < D/L and 0,5. < Pr < 2000

• Correlation for temperature:
For fluid:
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

45

Heat transfer coefficient at forced convection
• Single phase heat transfer in horizontal tubes.
• The α can be calculated for turbulent flows by Dittus-Boelter formula
[3]:
0,4 ha TW > T∞
Nu = 0,023 ⋅ Re0,8 ⋅ Pr n ⋅ ΦT , where n = 
0,3

Valid, if Re > 105 and 0,6 < Pr < 160 and L/D > 10.

ha TW < T∞

• If the tube cross section is not circular then the so called equivalent
diameter (De) must be used:
De = 4 ⋅

F
K

,
where F is the flow area, K is the wetted perimeter.
• For curved pipe the heat transfer coefficient (αR) is higher compared
to straight tube [2]:
D

α R = 1 + 1,77 ⋅  ⋅ α egyenes
R


,

where R is the curvature radius, D is the pipe diameter.
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

47

 Pr
ΦT = 
 PrW





0 ,14

, For gas:

T
ΦT =  ∞
 TW





0 ,12

,

where PrW is the Prandtl number at TW wall temperature.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

46

1st example – Heat transfer at the primary side of a steam generator
The diameter of heat exchanger pipe in a PWR steam generator is
10 mm. The pressure of the primary side is 124 bar and the velocity of
the primary water is 2 m/s. The characteristic temperature of the
primary water is 282 °C and the wall temperature at the primary side
is 260 °C. The length of the pipe is far higher than the diameter of the
pipe.
Define the heat transfer coefficient between the primary water and
wall!
The material properties of the water at 124 bar:
t, °C

ν, m2/s

λ, W/(m·K)

Pr

260

1,31·10-7

0,6198

0,8159

280

1,26·10-7

0,5915

0,8295

290

1,23·10-7

0,5747

0,8476

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

48


1st example – Heat transfer at the primary side of a steam generator
Solution.
Heat transfer in straight tube:
Characteristic temperature: t m = 282 °C .
Characteristic dimension: the inner diameter of the tube, D = 10 mm = 0,01
. m
t
=
282
°
C
Material properties at m
temperature with linear interpolation:

1st example – Heat transfer at the primary side of a steam generator
Solution (continuation):
For turbulent flow the following formulas must be used:
ξ=

1
1
=
= 0,01632
(1,82 ⋅ log10 Re − 1,64 )2 (1,82 ⋅ log10 1,595 ⋅ 105 − 1,64 )2

(

)

The Nusselt number:
t, °C

ν, m2/s

λ, W/(m·K)

Pr

282

1,254·10-7

0,5881

0,8331

The Prandtl number at the wall temperature:

The temperature correlation:

 Pr 

ΦT = 
 PrW 

0 ,14

ξ 
  ⋅ (Re − 1000 ) ⋅ Pr
2/3
D
8

Nu =  
⋅ 1 +  ⋅ ΦT ,
L
ξ 
1 + 12,7 ⋅ Pr 2 / 3 − 1 ⋅
8
ξ 
 0,01632 

(

0 ,14

= 1,003

Prof. Dr. Attila Aszódi, BME NTI

(

so

)

The heat transfer coefficient:

The Reynolds number: Re = w ⋅ D = 2 ⋅ 0,01 = 1,595 ⋅ 10 5 > 2300 ,
ν
1,254 ⋅ 10 −7
so the flow is turbulent.
Thermal hydraulics

)

D
≈0,
L

5
  ⋅ (Re − 1000 ) ⋅ Pr

 ⋅ 1,595 ⋅ 10 − 1000 ⋅ 0,8331
8
8


⋅ 1,003 = 289,1
⋅ ΦT =
Nu =
2
 23
 ξ

 0,01632
3




1 + 12,7 ⋅ Pr − 1 ⋅
1 + 12,7 ⋅ 0,8331 − 1 ⋅

 8


8





PrW = 0,8159

 0,8331 
=

 0,8159 

but: D<
α=
49

Nu ⋅ λ 289,1 ⋅ 0,5881
W
=
= 17001,0 2
0,01
D
m ⋅K

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

50

2nd example – Cooling a fuel rod

2nd example – Cooling a fuel rod
On the below figure a simplified sub-channel of a fuel rod
can be seen. The mass flow rate of the coolant (water) in the
core is 8800 kg/s at 124 bar. The number of fuel assemblies
is 349. Each fuel assembly has 126 fuel rods.
The inlet water temperature is 267 °C.
The outlet temperature is 297 °C.

rod

Calculate the heat transfer coefficient between the water and
wall outer surface!
(Neglect the correction factor coming from the difference
between the wall and water temperature!)
The material properties of the water at 124 bar:
t, °C

ρ, kg/m3

ν, m2/s

λ, W/(m·K)

Pr

260

793,18

1,31·10-7

0,6198

0,8159

280

759,75

1,26·10-7

0,5915

0,8295

740,98

1,23·10-7

0,5747

0,8476

290
hexagon

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

51

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

52


2nd example – Cooling a fuel rod

2nd example – Cooling a fuel rod

Solution.
12,2
Using these signs:
D = 9 mm m =
= 6,1 mm.
2
the flow area (F) and the equivalent diameter (De):

Solution (continuation):
The flow velocity:

co

D2 ⋅π
m
6,61 2 9 2 ⋅ π
F = 6 ⋅ T ∆ − TΟ = 6 ⋅
− co
= 6⋅

= 65,282 mm 2 = 6,5282 ⋅ 10 −5 m 2
4
4
3
3
4⋅F
4⋅F
4 ⋅ 65,282
De =
=
=
= 9,235 mm = 0,009235 m
K
Dco ⋅ π
9 ⋅π
2

The characteristic temperature

Tm =

T zóna
zóna ,ki
in ,be + Tout
2

=

m
8800
m
&
=
= 4,055
n p ⋅ n kaz ⋅ ρ ⋅ F 126 ⋅ 349 ⋅ 755,99 ⋅ 6,5282 ⋅ 10 −5
s

The Reynolds number:
Re =

w ⋅ De

ν

=

4,055 ⋅ 0,009235
= 2,986 ⋅ 105 > 2300 , so the flow is turbulent.
1,254 ⋅ 10 −7

The Nusselt number for turbulent flow by the Dittus-Boelter formula,
if ΦT = 1:

267 + 297
= 282 °C
2

(

The material properties at 124 bar, Tm = 282 °C with linear
interpolation:
t, °C

ρ, kg/m3

ν, m2/s

λ, W/(m·K)

Pr

282

755,99

1,254·10-7

0,5881

0.8331

Thermal hydraulics

w=

Prof. Dr. Attila Aszódi, BME NTI

Nu = 0,023 ⋅ Re 0.8 ⋅ Pr 0.4 = 0,023 ⋅ 2,986 ⋅ 10 5

) ⋅ (0,8331)
0.8

0.4

= 513,0

The heat transfer coefficient:
α =

53

λ
De

⋅ Nu =

Thermal hydraulics

0,5881
W
⋅ 513,0 = 32669,0
0,009235
m2 ⋅ K

Prof. Dr. Attila Aszódi, BME NTI

54

The thermal conductivity (λ) of UO2
Influencing effects on λ:
• temperature,
• porosity,
• O/U ratio,
• PuO2 ratio,
• Pellet crackings,
• Burn up.

The thermal properties
of UO2

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

55

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

56


Temperature dependency of λ Ι.

Temperature dependency of λ II.

Up to 1750
λ decreases if temperature increases, but above 1750
λ slightly increases. The uranium-dioxide has a lower heat conductivity
compared to the metals.
The Lyon formula:
3824
(T [°C])
λ (T ) =
+ 6.1256 ⋅10−11 (T + 273)3
oC

oC

The integral of Lyon formula:
T

K (T ) = ∫ λ (T )dT = 3824 ⋅ ln(402.4 + T ) + 1.5314 ⋅ 10 −11 (T + 273) 4 − 22934
0

402.4 + T

The
λ of thetényezõje
UO2 using
the Lyon formula
UO hõvezetési
a Lyon-összefüggés
alapján

összefüggés integrálja
The integral Lyon
of the
Lyon formula

8000

2

10

7000

9

6000

8
7

K [W/m]

5000

5

4000
3000

4
3

T [oC]

Prof. Dr. Attila Aszódi, BME NTI
T
[°C]

Temperature
dependency of λ
and its integral in
chart
T

Lambda

Integral

T

Lambda

[°C]

[W/mK]

[W/m]

[°C]

[W/mK]

670

3,6172

[W/m]

0

9,5042

10

9,2739

93,8814

680

3,5859

3796,3543

20

9,0546

185,5152

690

3,5553

3832,0596

30

8,8454

275,0067

700

3,5252

3867,4614

40

8,6456

362,4541

710

3,4958

3902,5660

50

8,4548

447,9490

720

3,4670

3937,3794

60

8,2722

531,5769

730

3,4387

3971,9072

70

8,0973

613,4180

740

3,4110

4006,1554

80

7,9297

693,5473

750

3,3839

4040,1294

90

7,7690

772,0353

760

3,3573

4073,8346

100

7,6146

848,9482

770

3,3312

4107,2765

Thermal hydraulics

0,0000

Integral

3760,3393

Prof. Dr. Attila Aszódi, BME

0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
420
430
440
450
460
470
480
490
500
510
520
530
540
550
560
570
580
590
600
610
620
630
640
650
NTI
660

57
Lambda
[W/mK]
9,5042
9,2739
9,0546
8,8454
8,6456
8,4548
8,2722
8,0973
7,9297
7,7690
7,6146
7,4664
7,3238
7,1866
7,0545
6,9272
6,8044
6,6860
6,5716
6,4612
6,3544
6,2512
6,1513
6,0546
5,9609
5,8702
5,7822
5,6969
5,6141
5,5338
5,4557
5,3799
5,3062
5,2346
5,1650
5,0972
5,0313
4,9671
4,9046
4,8437
4,7844
4,7266
4,6702
4,6152
4,5616
4,5093
4,4583
4,4084
4,3598
4,3123
4,2659
4,2206
4,1763
4,1330
4,0906
4,0493
4,0088
3,9692
3,9305
3,8927
3,8556
3,8193
3,7838
3,7491
3,7151
3,6818
3,6491

Integral
[W/m]
0,0000
93,8814
185,5152
275,0067
362,4541
447,9490
531,5769
613,4180
693,5473
772,0353
848,9482
924,3483
998,2944
1070,8418
1142,0429
1211,9471
1280,6012
1348,0496
1414,3342
1479,4951
1543,5701
1606,5953
1668,6050
1729,6319
1789,7072
1848,8605
1907,1204
1964,5138
2021,0668
2076,8041
2131,7495
2185,9259
2239,3549
2292,0575
2344,0538
2395,3632
2446,0042
2495,9946
2545,3516
2594,0916
2642,2307
2689,7842
2736,7667
2783,1927
2829,0758
2874,4293
2919,2661
2963,5986
3007,4388
3050,7982
3093,6881
3136,1194
3178,1026
3219,6478
3260,7651
3301,4638
3341,7534
3381,6429
3421,1410
3460,2562
3498,9969
3537,3709
3575,3861
3613,0502
3650,3704
3687,3541
3724,0081

T
[°C]
670
680
690
700
710
720
730
740
750
760
770
780
790
800
810
820
830
840
850
860
870
880
890
900
910
920
930
940
950
960
970
980
990
1000
1010
1020
1030
1040
1050
1060
1070
1080
1090
1100
1110
1120
1130
1140
1150
1160
1170
1180
1190
1200
1210
1220
1230
1240
1250
1260
1270
1280
1290
1300
1310
1320
1330

Lambda
[W/mK]
3,6172
3,5859
3,5553
3,5252
3,4958
3,4670
3,4387
3,4110
3,3839
3,3573
3,3312
3,3056
3,2806
3,2560
3,2319
3,2083
3,1851
3,1624
3,1401
3,1182
3,0968
3,0758
3,0552
3,0350
3,0152
2,9957
2,9767
2,9580
2,9396
2,9216
2,9040
2,8867
2,8697
2,8531
2,8368
2,8208
2,8052
2,7898
2,7747
2,7600
2,7455
2,7313
2,7174
2,7038
2,6905
2,6774
2,6646
2,6521
2,6398
2,6278
2,6160
2,6045
2,5932
2,5822
2,5714
2,5609
2,5505
2,5405
2,5306
2,5210
2,5116
2,5024
2,4934
2,4847
2,4761
2,4678
2,4597

Integral
[W/m]
3760,3393
3796,3543
3832,0596
3867,4614
3902,5660
3937,3794
3971,9072
4006,1554
4040,1294
4073,8346
4107,2765
4140,4601
4173,3906
4206,0729
4238,5118
4270,7121
4302,6785
4334,4154
4365,9273
4397,2186
4428,2936
4459,1563
4489,8110
4520,2616
4550,5120
4580,5660
4610,4276
4640,1004
4669,5880
4698,8940
4728,0219
4756,9751
4785,7571
4814,3712
4842,8206
4871,1086
4899,2383
4927,2128
4955,0352
4982,7085
5010,2356
5037,6195
5064,8629
5091,9689
5118,9400
5145,7792
5172,4890
5199,0721
5225,5311
5251,8687
5278,0874
5304,1897
5330,1780
5356,0549
5381,8227
5407,4839
5433,0407
5458,4956
5483,8507
5509,1084
5534,2709
5559,3405
5584,3193
5609,2094
5634,0131
5658,7324
5683,3695

T
[°C]
1340
1350
1360
1370
1380
1390
1400
1410
1420
1430
1440
1450
1460
1470
1480
1490
1500
1510
1520
1530
1540
1550
1560
1570
1580
1590
1600
1610
1620
1630
1640
1650
1660
1670
1680
1690
1700
1710
1720
1730
1740
1750
1760
1770
1780
1790
1800
1810
1820
1830
1840
1850
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000

Lambda
[W/mK]
2,4517
2,4440
2,4365
2,4292
2,4221
2,4152
2,4085
2,4019
2,3956
2,3894
2,3835
2,3777
2,3721
2,3667
2,3614
2,3564
2,3515
2,3468
2,3423
2,3379
2,3337
2,3297
2,3259
2,3222
2,3187
2,3154
2,3122
2,3092
2,3064
2,3037
2,3011
2,2988
2,2966
2,2945
2,2926
2,2909
2,2893
2,2879
2,2867
2,2855
2,2846
2,2838
2,2831
2,2826
2,2822
2,2820
2,2820
2,2821
2,2823
2,2827
2,2832
2,2839
2,2847
2,2857
2,2868
2,2880
2,2894
2,2909
2,2926
2,2944
2,2964
2,2985
2,3007
2,3031
2,3056
2,3083
2,3111

Integral
[W/m]
5707,9263
5732,4050
5756,8076
5781,1361
5805,3925
5829,5787
5853,6967
5877,7484
5901,7357
5925,6606
5949,5249
5973,3305
5997,0791
6020,7727
6044,4131
6068,0020
6091,5413
6115,0326
6138,4779
6161,8787
6185,2368
6208,5541
6231,8320
6255,0724
6278,2770
6301,4473
6324,5850
6347,6919
6370,7695
6393,8195
6416,8434
6439,8429
6462,8196
6485,7750
6508,7108
6531,6285
6554,5297
6577,4159
6600,2886
6623,1495
6645,9999
6668,8416
6691,6759
6714,5043
6737,3285
6760,1498
6782,9698
6805,7899
6828,6116
6851,4363
6874,2657
6897,1010
6919,9438
6942,7955
6965,6575
6988,5313
7011,4183
7034,3199
7057,2376
7080,1728
7103,1268
7126,1012
7149,0973
7172,1165
7195,1602
7218,2298
59
7241,3267

2000

1900

1800

1700

1600

1500

1400

1300

1200

1100

900

1000

800

700

600

500

400

300

0
200

2000

1900

1800

1700

1600

1500

1400

1300

1200

1100

900

1000

800

700

600

500

400

300

200

0

100

0

1000

0

1

Thermal hydraulics

2000

Reference:
N. E. Todreas, M. S. Kazimi:
Nuclear Systems I; Thermal
hydraulic fundamentals, 1990,
Hemisphere Publishing
Coproration, New York, ISBN
1-56032-051-6

2

100

[[W/mK]

6

T [oC]

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

58

The influence of Porosity (density) I.
• The fuel pellet is often made from powder of UO2 or MOX by high temperature
sintering.
• 90% or more can be reached compared to the theoretical density.
• If the dimension of the pores is smaller then theoretically, the heat conductivity
could be higher.
• But the gas phase fission products deforms the fuel pellet and causes over
pressure in the fuel pellet. The generated gas fills the pores, that is why there is
an optimal porosity for the manufacturing.
The porosity:

P=

the volume of the pores (V p )
the volume of the pores (V p ) and the volume of the solid (Vs )

P = 1−

=

Vp
V

=

V − Vs
V

ρ
ρ TD

where ρTD = theoretical maximal density = theoretical density
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

60


The influence of Porosity (density) II.

The influence of oxygen/metal atomic ratio

For sphere pores: α2 = 1,5:

Biancharia equation:



1− P
λ=
λTD
1 + 0,5 ⋅ P

1− P
λ=
λTD
1 + (α 2 − 1) P

Both, the hyper- and hipostoichiometric ratio
decreases the heat conductivity.

The λ of UO0.8Pu0.2O2±X versus O/(U+Pu) ratio
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

61

Thermal hydraulics

The influence of plutonium content







The heat
conductivity of (U,
Pu)O2 can be seen in
the function of the
content of PuO2 on
the right hand side.

Prof. Dr. Attila Aszódi, BME NTI

62

The influence of Pellet cracking

If the Pu content
increases then the
heat conductivity
decreases.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI



63

Thermal effects leads to thermal
stresses
Burn up → gas generation

Enhancement of pellet cracking
The heat conductivity of the pellet
decreases due to the increasing
dimensions of cracks.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

64


The influence of burn up (power reactors)





Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI






65

Certain fission products in gas phase can leave the UO2
fuel material at low temperature conditions.
At high temperature when structural modification can
be occurred, significant amount of fission products in
gas phase could leave the fuel material and mix with
the filling gas of the fuel rod.
The generated gas phase fission products causes
increased pressure in the fuel rod which effect have to
be considered during the design of the fuel rods.

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

The specific heat

The melting point
The melting point of UO2 is 2840 °C.

The heat capacity of the fuel play an important role during the transient calculations.

The melting point for the oxide of uranium-plutonium
mixture in the function of plutonium content
Thermal hydraulics

66

Heat capacity
He



The irradiation of the fuel causes: modification of the fuel microscopic
structure; in the fuel porosity; material content of the fuel and in the
stoichiometric ratio of the fuel.
Only moderate modification occurs in LWRs (the degree of burn up is
maximum 3%), for fast reactors this effect is higher (the 10% of
original U and Pu atoms could be burned in the nuclear fission
process).
The occurrence of fission products in the fuel slightly decreases the
heat conductivity.
The crackings of fuel pellets due to thermal fatigue decreases the
effective heat conductivity of fuel pellets.
If the nominal temperature exceeds a certain temperature (1400°C)
then the material structure of the oxide fuel modify which process
leads an increased density. This increased density – which occurs in
the inner region of the fuel pellet – has an influence on the thermal
conductivity and the temperature field too.

Temperature



Generation of gases (power reactors)

Prof. Dr. Attila Aszódi, BME NTI

Temperature
67

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

68


Thermal properties of different fuels
U

UO2

UC

UN

Theoretical maximal
density at room
temperature [kg/m3]

19040

10970

13630

14320

Metal density in the
crystal [kg/m3]

19040

9670

12970

13600

[oC]

1133

2800

2390

2800

Stability

Up to 665 oC

Up to melting
point

Up to melting
point

Up to melting
point

Average heat
conductivity between
200-1000 oC [W/moC]

32

3,6

23

21

Melting point

Heat capacity at 100 oC
[J/kgoC]

Density [kg/m3]

116

247

Linear thermal expansion
coefficient [1/oC]

Structure of crystals

Thermal properties of different cladding
materials (power reactors)

Below 655 oC:
a, orthorombic
Above 770 oC: g, body-

Melting point

[oC]

Heat conductivity at 400

146

0,0000101
(400-1400 oC)

0,0000111
(20-1600 oC)

0,0000094
(1000 oC)

cubic facecentered

cubic facecentered

cubic facecentered

110

62

Heat capacity at 400

oC

oC

[W/moC]

[J/kgoC]

Linear thermal expansion

coefficient[1/oC]

Zircaloy 2

SS 316

Aluminium*

6500

7800

2700

1850

1400

660

13

23

237 (25°C)

330

580

910

5.9E-06

1.8E-05

2.31E-05

*the values can differ for different alloys

centered cubic
Tensile strength [Mpa]
Thermal hydraulics

344-1380

Prof. Dr. Attila Aszódi, BME NTI

69

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

70

Subchannels

The temperature
distribution in the core

Hexagonal (for example VVER-440) and rectangular (for
example western PWRs) subchannels
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

71

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

72


The temperature distribution into the cross
section of the fuel rod

The temperature distribution of the fuel rod
General differential equation of transient heat conduction:

Cladding

Fuel pellet
Coolant

Fuel pellet

Cladding

r
r
r
∂T( r , t )
= ∇(λ(T ) ⋅ ∇T( r , t )) + q& ′′′( r , t )
ρ ⋅ cp
∂t
Tfo
Tci
Tco
Tm

Gap

Central bore

Radial temperature distribution in the fuel rod
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

73

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

The temperature distribution of the fuel rod

The temperature distribution of the fuel rod

In case of compact pellet (no central bore)
Rv=0

If L/D>10 then the axial temperature distribution can be considered as
uniform at the centre line of the fuel rod 1D:

r = 0 ⇒ q& ′′ = − λ

dT
1 d
(λ r
) + q& ′′′ = 0
r dr
dr
dT
r2
λr
+ q& ′′′ + c1 = 0
dr
2
r2
dT



+ q&
+ c1 = 0

dr
2
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

74

dT
dr



r2
dT
+ q& ′′′ + c1 = 0
2
dr

=0
r =0

c1=0

λ

75

Thermal hydraulics

dT
r
+ q& ′′′ = 0
dr
2

Prof. Dr. Attila Aszódi, BME NTI

76


The temperature distribution of the fuel rod

The temperature distribution of the fuel rod
Pellet with bore: (Rv ≠0 is the radius of the bore in the middle of the fuel)
r2
dT
Base equation:
+ q& ′′′ + c1 = 0

dr
2
dT
q& ′′ r = Rv = −λ
Rv≠0, but
r = Rv = 0
dr
2
q& ′′′Rv
c1 = −
2

In case of compact pellet (no central bore)
Tmax

∫ λdT =

T
Tmax

∫ λdT =

q& ′′′r 2
4
2
q& ′′′R fo

4

T fo

2
q& ′ = πR fo q& ′′′

q& ′

Tmax

r c
dT
+ q& ′′′ + 1 = 0
dr
2 r
T
 r 
q& ′′′ 2
2
r − Rv + c1 ln 
− ∫ λdT =
4
 Rv 
Tmax

∫ λdT = 4π

λ

T fo

λ (Tmax − T fo ) =

Using average heat transfer coefficient:
Thermal hydraulics

[

q& ′


Prof. Dr. Attila Aszódi, BME NTI

77

The temperature distribution of the fuel rod
Pellet with bore:
T

∫ λdT =



Tmax

[

]

  R  2   R  2  r 
v
v
1 −    − 2  ln 
  r    r   Rv 

2
q& ′′′R fo
λdT
=

4
Tfo

Writing in (*):

(*)

2

 R fo  

ln
 
Tmax
 Rv  
q& ′ 
∫T λdT = 4π 1 −  R  2 
fo
  fo  − 1

  Rv 



  R  2   R  2  R  2 
v
  −  v  ln fo  
1 − 
R
R
 Rv  

fo  

  fo 


(

78

Pellet with bore:

Let’s use r=Rfo then T=Tfo
Tmax

Prof. Dr. Attila Aszódi, BME NTI

The temperature distribution of the fuel rod

2
q& ′′′ 2
q& ′′′Rv
r
2
ln
r − Rv −
4
2
Rv

q& ′′′r 2
λ
dT
=

4
T

Tmax

Thermal hydraulics

]

)

2
2
q& ′ = π R fo − R v q& ′′′

⇒ q& ′′′R fo =
2

Thermal hydraulics

q& ′
  R 2
π 1 −  v  
  R fo  

Prof. Dr. Attila Aszódi, BME NTI

79

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

80


The temperature distribution of the fuel rod

Lessons

The void coefficient:
Fv(α,β)=1-ln(α2)/(β2(α2-1))
where α=(Rfo/Rv)
β= is the volumetric power density irregularity coefficient
β=1 if regular, constant
β>1 if the power density is higher in the inner zone
(reformed fuel pellet)

1. In case of equal Tmax limit: q& ′bore ⋅ Fv = q& ′compact if,
the Tfo and λ are the same. So : q& ′bore > q& ′compact (Fv < 1)
If Tmax, Tfo and λ are same then the fuel pellet
with bore can produce more power compared to
the pellet without bore (the higher grad T –
indicated with blue colour next to the red curve –
causes higher heat flux).

Üregtényező
Void
coefficient
1

q& ′

 R fo

∫ λdT = 4π ⋅ F  R
v

v

T fo


, 1


Fv ( , )

0.8

Tmax

so identical material and identical λ(T), Tfo:

0.6

β=10

0.4

β=2

Tmax

β=1.5
β=1.1

0.2

β=1

1

0.96

0.92

0.88

0.8

0.84

0.76

0.72

0.68

0.6

0.64

0.56

0.52

0.48

0.4

0.44

0.36

0.32

0.28

0.2

0.24

0.16

0.12

0.08

0.04

0

0

1/α
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

81

Axial temperature distribution in
the fuel rod

compact

Lower maximal nominal temperature should be
considered (see the figure on the right hand side T’max)!
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

82

– q& ′(z ) is a cosines-function:
πz

q& ′( z ) = q&o cos( )
Le

– where q&0′ is the maximum of the linear power density,
Le is that length where the heat generation higher than
zero (this could be longer than the active length, L),
– The parameters (heat conductivity, heat capacity, etc.)
of the coolant, the fuel and cladding are constant,
– Single phase flow.

d
hm = q& ′(z )
dz

Prof. Dr. Attila Aszódi, BME NTI

< Tmax

• Simplifications:

– If the mass flow rate is given then the enthalpy
rise depends only on q& ′(z ) .
– In nuclear reactors q& ′(z ) depends on the
neutron flux and the fuel distribution in the
zone.
Thermal hydraulics

bore

Axial temperature distribution in
the fuel rod

• The energy equation is the following for
steady state and single phase flow (the
pressure gradient and the drag are neglected):
m&


  Tmax
 Tmax

 λ dT  =  λ
bore
compact dT Fv



 T
T

  fo
 fo

2. In case of equal q& ′ :

83

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

84


Axial temperature distribution in
the fuel rod

Axial temperature distribution in
the fuel rod

d

• Replace the q& ′(z ) term in m& hm = q& ′(z ) and
dz
integrate that along the active length to express
the coolant temperature:
hm ( z )

m&

∫ dhm = q&o′

hin
Tm ( z )

m& c p

∫ dT = q&o′

Tin

Tm ( z ) − Tin =

• The outlet coolant temperature:
 q& ′  2 L 
πL
Tout = T ( L / 2) = Tin +  o  e  sin
 m& c  π 
2 Le
 p

 πz 
cos dz
 Le 
−L / 2
z



• If the heat generation length (Le) can be
modeled with the active length (L):

 πz 
cos dz
 Le 
−L / 2
z



Tout = Tin +

q& o′ Le
πz
πL
(sin + sin
)
m& c p π
Le
2 Le

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

85

2q&o′ L
πm& c p

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

86

Axial temperature distribution in
the fuel rod

Axial temperature distribution in
the fuel rod

• The cladding temperature:

• For the maximal cladding temperature:

α [Tco ( z ) − Tm ( z )] = q& ′′( z ) =
Tco ( z ) = Tm ( z ) +

q& ′( z )
2πRco

dTco
=0
dz

Thermal hydraulics

 πz  2πRco Leα
tan c  =
πm& c p
 Le 

zc =

 2πRco Leα 
tan −1 

π
 πm& c p 

Le

• Because all terms in the argument are positive
the zc is also positive, so the maximal cladding
temperature can occur in 0
 πz
πL 
1
πz 
 +
 sin + sin
cos 
Le
2 Le  2πRcoα
Le 

Prof. Dr. Attila Aszódi, BME NTI

d 2Tco
<0
dz 2

• From the above expression replacing Tco:

 πz 
q&o′
cos 
2πRcoα
 Le 

• Replace Tm with the above expression:
 L
Tco ( z ) = Tin + q&o′  e
πm& c p

and

87

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

88


Axial temperature distribution in
the fuel rod

Axial temperature distribution in
the fuel rod

• The centreline temperature in the fuel:

• The location of the maximal temperature at the
centreline:

– From the previously mentioned cladding temperature
and the radial temperature distribution:
 L  πz
πL   1
1
1  πz 
1  Rco 
+
TCL(z) = Tin + q&0′  e sin + sin  + 
+
ln  +
 cos 
2Le   2πRcoα 2πλc  Rci  2πRgαg 4πλf  Le 
& cp  Le
πm

where λ f and λc are heat conductivity of the fuel and
cladding materials, αg is heat transfer coefficient of the
gap filling gas.
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

89

Axial temperature distribution in
the fuel rod

Prof. Dr. Attila Aszódi, BME NTI

90

Figure: The axial temperature distributions in subchannel at the beginning and end of the burn up

Figure: Temperatures along the length of fuel rod
Prof. Dr. Attila Aszódi, BME NTI

Thermal hydraulics

Axial temperature distribution in
the fuel rod

Tm

Thermal hydraulics





Le −1 
Le

z f = tan 

π
πmc  1 + 1 ln Rco  + 1 + 1  
 & p  2πR α 2πλ  R  2πR α 4πλ  
c
g g
f 
 ci 
 co


91

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

92


Axial temperature distribution in
the fuel rod

Typical fuel geometries

Figure: The effect of partially inserted control rods on the power density and temperatures in axial
direction
Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

93

The types of nuclear reactors’ fuel

Prof. Dr. Attila Aszódi, BME NTI

94

Research reactor fuel - examples
OPAL reactor, Australia

The nuclear fuel (TRISO)

• TRISO fuel
• CERMET fuel
• Plate type fuel

Thermal hydraulics

Thermal hydraulics

Fuel assemblies each hold 21 aluminiumlaminated plates of uranium. The reactor
core contains 16 of these assembles
which are 8cm square and just under a
metre long. The 16 assembles sit in a
square array of four by four. The reactor
core sits in the middle of the reflector
vessel.

Prof. Dr. Attila Aszódi, BME NTI

95

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

96


BME training reactor

BME Training reactor
core design

• Location: Campus of the Budapest
University of Technology and
Economics (BME)
• Reactor type: pool-type reactor
(Hungarian design)
• First criticality: 1971.
• Nominal Power: 100 kW
• Moderator and coolant:
light water
• Horizontal and vertical irradiation
channels
• Pneumatic dispatch system for the
transfer of the samples into the core

• Assemblies: 24 EK-10 (Soviet type)
• 369 fuel pins
• D=10 mm, Lactive=500 mm
• Fuel: 10%-enriched UO2 in Mg matrix
• Cladding material: Al
• Heated length: 500 mm
• Total mass of uranium in the core: 29.5 kg
• Horizontal reflector material: graphite
• Vertical reflector material: water
• Highest thermal neutron flux
2.7*1012 n/cm2/s (measured in
one of the vertical irradiation channels)

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

97

Thermal hydraulics

The types of power reactors’ fuel

98

The structure of the fuel assembly

• PWR fuel
• BWR fuel
• CANDU fuel

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

• UO2 / MOX fuel pellet
• Zirconium rods (pellets
+ filling gas)
• Fuel bundles (rods,
spacers, claddings etc.)

Prof. Dr. Attila Aszódi, BME NTI

99

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

100


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