5.1 Heat removal by coolant flow

5. Convective Heat Transfer

Temperature at center of fuel pelletfcT

Temperature at outside surface of fuel pellet

Temperature at inside surface of fuel coolant

Temperature at outside surface of cladding tube

Coolant temperature

fT

cT

bT

mT

Fuel pellet

Bond layer

Cladding tube

Coolant flow

Heat is transferred from the surfaces of the fuel rods to the coolant.

1. Low neutron absorption cross section2. Low radiation-induced radioactivity 3. Low damage by irradiation4. High moderating power for thermal neutron

reactor, and low moderating power for fast neutron reactor

Requirement of nuclear properties

5.2 Properties of coolant

Requirement of thermal-hydraulic properties

1. High thermal conductivity and specific heat, and as a result high heat transfer coefficient

2. Low melting point in case of coolant which is solid at room temperature

3. Low viscosity (low friction pressure drop and strong turbulence)

4. No thermal decomposition5. Chemically inactive with air and water

-Low specific heat and thermal conductivity

-Required heat transfer is obtained by pressurization

Characteristics of gas coolant

Characteristics of water: light water (H2O) and heavy water (D2O)

-High pressure operation needed because of high saturation vapor pressure

-Good thermal properties-High moderating power

-Low saturation pressure (Low pressure operation is possible at high temerature)

-Low Prandtl number, high thermal conductivity, and low density

-Low corrosion rate under good control of impurity (Steel corrosion in Li is higher than in Na.)

-Chemically active with water and oxygen in case of alkali metals (Li reacts with nitrogen mildly)

Characteristics of liquid metal: alkali metals, sodium and lithium

-High specific heat (Good for heat transportation)

-High viscosity (Laminarization)-High Prandtl number (Not good for heat

transfer media)-Nickel-base alloy (Hastelloy) is compatible

with molten salt

Characteristics of molten salts:Flibe (LiF-BeF2), HTS (NaNO3-KNO3-NaNO2), Flinak(LiF-NaF-KF), Li2CO3-Na2CO3-K2CO3, etc.

0.19 26640.110.13 221.8Thermal diffusivity (10-6m2/s)

2.34.231.266.195.695.271.17Specific heat (kJ/kgK) 1490036025098 89 3934Viscosity (10-6Pa･s)

7.30.740.300.130.13161.2Kinematic viscosity (10-6m2/s)

380.0290.00461.20.980.73 0.67Pr

0.92952.767.40.5360.5360.280.059Thermal conductivity (W/mK)

20404838327707122.428Density (kg/m3)

14351342881101100－269 －56.6 at 0.53Pa

Boiling point (℃)

458 180.7 97.9 3.9 0 Melting pint (℃) FlibeLi NaD2O H2O He CO2

Table 5.1 Properties of various coolants

Note: Temperature and pressure dependency

5.3 Approach to convective heat transfer (Single-phase flow)

Solid

Velocity v

Temperature T

Coolant flow

Heat flux q”

How to calculate the heat transfer coefficient, or the temperature profile close to the solid surface, i. e. the fuel rod surfaces in coolant?

Continuity equation

TkDtDTCv

2∇=ρ

ρ μ ρDDt

pv v g= −∇ + ∇ +2

DDtρ ρ= − ∇ ⋅( )v

zv

yv

xv

tDtD

zyx ∂∂φ

∂∂φ

∂∂φ

∂∂φφ

+++≡

zv

yv

xv yyx

∂

∂+

∂

∂+

∂∂

=⋅∇ v2

2

2

2

2

22

zyx ∂∂

+∂∂

+∂∂

≡∇φφφφ

Momentum equation

Energy equationwhere

Navier-Stokes equations

Boundary conditions v = 0, given q” at solid surface

fluid ofty conductivi Thermal: fluid of viscosityDynamic :

kμ

Calculation of velocity and temperature field

∇⋅+∂∂

= vtDt

D

' ,' ,' wwwvvvuuu +=+=+=

yu

yuvu MT ∂

∂ρε∂∂μρτ ==−= ''

Eddy diffusivity of momentum

ylength Mixing

,2

κ∂∂ρμ

=

=

m

mT

lyul

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

∂∂

+∂∂

−= jii

j

j

i

ji

i vvxv

xv

xxP

DtvD ''1 ν

ρ

Reynolds equation (Time average)

Average velocity and fluctuation

Prandtl’s mixing length model

Reynolds stress (additional terms)

Mε

Approach to turbulent flow

Time t

u’

u

Turbulence model

41.0=κ

Nikuladse)(by flow pipein ,106.0108.014.042

⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −−=

Ry

Ry

Rlm

u y+ += 8 57 1 7. ( ) /7/1

max

1 ⎟⎠⎞

⎜⎝⎛ −=

Rr

uu

The 1/7th law of velocity profile

Calculation of velocity profile in turbulent flow

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

r/R

ΔΔ

px

uD

=0 3164 1

21

1 42.

Re / ρ

( )

ρτ

ν

τ

τ

τ

w

n

u

yu

uuu

yu

≡

≡≡

=

+

++

,y ,

,

+

Derived

100 101 102 1030

5

10

15

20

25

u+

y+

u+=5.5+2.5ln(y+)

u+=5.0+5.0ln(y+/5)

u+=y+

5 30

+

+

=

<<

yy

+u ,50

)5/ln(0.50.5u ,305

+ +

+

+=

<<

yy

)ln(5.25.5u ,30

+ +

+

+=

<

yy

wwu

yvuuuu

τρττ

ττ

stressshear Wall,/ elocity Friction v

,)/(y ,/ velocity ldimensiona-Non +

≡

≡≡+

von Karman’s universal velocity profile (Log law)Viscous sublayer

Transition region

Turbulence region

where

Derived from Prandtl’s mixing length model

LES：Large eddy simulationDNS: Direct numerical simulation

Empirical heat transfer correlations

Approximation of Boundary layer

Navier-Stokes equations

Mixing length model

・Two equation model・Reynolds stress model

・LES・DNS

Solid (Fuel rod)Heat conduction

Fluid (Coolant)

Experiment

Turbulence models

Theoretical heat transfer equations

Various approach

Heat from fuel rods is transferred to coolant flowing along the fuel rods.

q h T Tw b' ' ( )= −q”: Heat flux, W/m2

Tw: Surface temperatureTb: Bulk temperature of coolant

(Mixed average temperature）

Heat transfer coefficient W/m2K

5.4 Theoretical and empirical heat transfer coefficient

5.4.1 Measurement of heat transfer coefficient

DLqTTDuc inoutp ππρ '')(4

2=−

Heat balance

Smooth channel with uniform heat flux or uniform wall temperature

Hydro-dynamically and thermally fully developed flow

''qU niform heat flux

uFluid

δ

inT outT

L

D

Determination of heat transfer coefficient

2/)(''

outinw TTTqh+−

=

Heat flux q’’determined from input power,

radial temperature gradient in channel wall, andenthalpy increase from inlet to outlet: temperature

difference and flow rate Inner surface temperature Tw

determined from extrapolation of radial temperature gradient in channel wall

Fluid temperature (Tin+Tout)/2

Reynolds number：Ratio of inertia term to diffusion term in momentum equation

5.4.2 Non-dimensional correlations

Hydraulic diameter： Characteristic length of flow channel

D AFe ≡

4

Nu hDk

≡

Re ≡ =ρμ νuD uD

Pr ≡ =c

kpμ ν

α

Prandtl number：Ratio of velocity boundary layer thickness to temperature one

Nusselt number：Non-dimensional heat transfer coefficient

Empirical heat transfer correlations

Laminar flow -Heated plate-Constant heat flux

3/12/1 PrRe916.0=Nu

5.0Pr >

Turbulent flow (Dittus-Boelter’s equation)-Uniform wall surface temperature

nNu PrRe023.0 8.0=

3.0Cooling4.0Heating

==

nn

　

　

510Re > 60/ >DL100Pr7.0 <<

Liquid metal flow (Subbotin’s equation)

8.0025.05 PeNu +=

1.0Pr <510PrRe >≡Pe

Length of entrance region

[ ] 275.08.02.0 )//(Re11.1 / DLNuNum =∞

4/1Re623.0/ =DL

In case of undeveloped thermal boundary layer

Nu=0.916Re1/2Pr1/3Pr>0.5Constant q”

AvarageNu=0.664Re1/2Pr1/3Pr>0.6Constant TwPlateNu=3.65Constant Tw

Fully developed

Nu=4.36Constant q”Pipe

CommemtsCorrelation

kLhNu /≡

Table 5.2 Heat transfer correlation for forced convective laminar flow

Num/Nu0=1.11[Re0.2/(L/D)]0.275

Nu=5+0.025Pe0.8

Nu=0.023Rex0.8Prn

Nu=0.0288Rex0.8Pr1/3

Correlation

Pipe

Plate

Undeveloped, Entrance length L/D=0.623Re1/4

Developed, Subbotin’s equatin

0.1>PrPe=RePr>105

Constant q”

Developed, Dittus-Boelter’s equationn=0.4 for heatingn=0.3 for cooling

0.7<Pr<100Re>105

L/D>60

Constant Tw

Developed, LocalNux=hxx/k

CommentsConditions

Table 5.3 Heat transfer correlation for forced convective turbulent flow

102 103 104 105100

101

102

Reynolds number　Re

Nus

selt

num

ber　

Nu Nu=0.023Re0.8Pr0.4

（Turbulent flow,　Dittus-Boelter）

Nu=5+0.025Re0.8Pr0.8

(Turbulent flow,　Subbotin, Liquid metal)

Nu=4.36 （Laminar, q'':Const.）

Nu=3.65（Laminar flow, T:Const）

5.4.3 Check points for use of heat transfer correlations

- Experimental conditions in derivation of the correlation

- Applicability range in Re and Pr- Laminar flow or turbulent flow?- Fully developed or undeveloped?- Constant wall temperature or constant heat

flux?- Definition of characteristic length such as

hydraulic diameter

5.5 Calculation of temperature increasein flow direction

Heat balance condition in steady state conditionIncrease of cross-sectional average temperature

coolant along fuel rod

dzzqR

dzzRq

dzzqzdTmcp

)('''

)(''2

)(')(

2π

π

=

=

=&

∫+= −zH

pdzzq

mcTzT 2/1 )('1)(

&

z

dz

T1Fuel rod

q’

T

Prandtl’s boundary layer theory for laminar flow

2

2

yu

dxdpg

yuv

xuu c

∂∂ν

ρ∂∂

∂∂

+−=+

Momentum equation in x-direction

5.6 Calculation of Pressure drop in fuel bundle

0=+yv

xu

∂∂

∂∂dp

dy= 0

Boundary layer approximation

x

yu

Basic theory for friction shear stress

Continuity equation

∞V

sτ

L

Lff dxc

LC

Re328.11

0 =∫≡2/2∞∞

≡V

c sf

ρτ

2

2uc fwρτ =

Friction pressure loss coefficient in laminar flow

844

2

2

uxPD

DxPD

wρλ

π

π

τ =ΔΔ

=ΔΔ

=

Relation between cf and λ

4fc

=∴λ

wτ

wτDu

)10<Re<103( Re ,Re03955.0 532

4/1 ×≡=ν

ρτ duuw

Friction pressure loss coefficient in turbulent flow(Blasius’s equation)

: Fluid density p : Cross sectional average pressureu : Cross sectional average velocityDe: Hydraulic diameter

: Frictional pressure loss coefficientx : Distance in flow direction

eDu

zp 1

21 2ρλ=

ΔΔ

−

Friction pressure loss in fuel bundle

ρ

λ

z

Fuel rodp

zΔ

Re64

=λ

25.0Re3164.0

=λ2300Re >

2300Re <Laminar flow

Turbulent flow（Blasius’s equation）

Frictional pressure loss coefficient

102 103 104 105

10-2

10-1

Reynolds number　Re

Fric

tion

loss

coe

ffic

ient　λ

λ=0.3164/Re0.25

(Turbulent flow, Blasius)

λ=64/Re(Laminar, Theory)

Rough wall　λ=Constant

Pressure loss in various channel geometries

2

21 u

zp ρξ=

ΔΔ

For grid spacer, orifice, venturi, elbow, etc.

Problem 3: A single-phase liquid flow in an annular channelA coolant flows through an annular channel between a straight

circular heater rod and a straight circular tube at a mass flow rate of W, where the heater rod is installed axis-symmetrically at the center of the straight tube with two spacers. The length of the channel is L, the outer diameter of the heater rod is d, and the inner diameter of the tube is D.

1)Write expressions for the cross sectional area A, the hydraulic equivalent diameter De and the thermal equivalent diameter Dh of the annular channel.

2) Determine an expression for the mean velocity u, and then an expression for the pressure loss in the channel for a fully developed flow using the mean velocity u, the fluid density ρ, the pressure loss coefficient for one spacer ζ, and the friction factor λ.

Problem 4 : Heat removal in coreAPWR core consists of fuel rods with the arrangement given in Table 5.4. Heat is generated uniformly in the fuel pellets with the heat generation rate of 400MW/m3, and the outside surfaces of the cladding tubes are cooled by the coolant. The properties of the coolant are given in Table 5.5. Calculate the following variables to three significant digits:

1) Hyraulic diameter, Reynolds number and Prandtlnumber.2）Nusselt number and heat transfer coefficient at the cladding surface using the Dittus-Boelter’sequation.2）Temperature at the surfaces of fuel rods.

P = 13 mm

9 mmDiameter of fuel pellet10 mmDiameter of cladding tube, D

193 Number of fuel bundles in core 17x17 Arrangement of rod bundle 13 mm Pitch in the arrangement of fuel rods, P

1.9 kg/s Total coolant mass flow rate in core 300 ℃Average coolant temperature

Table 5.4 Cooling conditions

0.54 W/m･K Thermal conductivity, k9.6x10-5 Pa･s Dynamic viscosity, μ5.73 J/kg･K Specific heat cp

714 kg/m3Density, ρ

Table 5.5 Physical properties of coolant at 300 ℃