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 ⋅

=ν

∂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

rλ

dr

2

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

74

dT

dr

rλ

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

rλ

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& ′

4π

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

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 ⋅

=ν

∂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

rλ

dr

2

Thermal hydraulics

Prof. Dr. Attila Aszódi, BME NTI

74

dT

dr

rλ

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

rλ

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& ′

4π

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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