Download - Lecture 2 - Thermal Design Principles
MASTER NUCLEAR ENERGY
NUCLEAR THERMOHYDRAULICSLecture 2 - Thermal Design Principles
Nicola Pedroni
17/09/2012
Outline
� Overall plant characteristics influenced by thermal-hydraulic considerations
� Coolant selection
� Plant thermal performance
� Energy production and transfer parameters
� Volumetric energy (or heat) generation rate
� Surface heat flux
� Linear heat generation rate or power rating
� Rate of energy generation
22
� Rate of energy generation
� Core power
� Thermal design limits*
� Fuel Pins With Metallic Cladding
� Graphite-Coated Fuel Pellets
� Figures of merit of core thermal performance
� Core power density
� Specific power
� Examples
2
Overall Plant Characteristics
333
Overall Plant Characteristics Influenced by Thermal-Hydraulic Considerations
Overall Plant Characteristics Influenced by Thermal-Hydraulic Considerations
Primary system temperature Tp and pressure Pp
Coolant selection
Plant thermal performance
Plant thermal performance depends on:� maximum allowable primary coolant outlet temperature, Tp = Thot (heat source)� minimum achievable condenser coolant inlet temperature, Tcond (heat sink)
= Tp
Hot leg to the SG
q = UA∆Tm
444
Tcond fixed (atmosphere)
= Tp
Turbine
Condensate
Feedwater train
piping
Boiling
Pumping+pipingTcond
~ Power prod.
Heat tranfered ~
↑ thermal performance = ↑ Thot
Ts
Maximum allowable primary coolant outlet temperature, Tp = Thot
IT DEPENDS ON THE COOLANT TYPE
LIQUID COOLANTS
Key aspect: relation between saturation temperature and pressure (Ts-Ps)(which does not exist for gas)
“Ideal” condition: primary coolant liquid at the higher T possible!
GAS COOLANTS
Pressure and temperature “decoupled”
555
Pressure and temperature “decoupled”
Maximum allowable primary coolant outlet temperature, Tp = Thot
Liquid coolants
The Tsat-Psat relation
666
Water:� (Ideal) plant efficiency (Carnot) limited by critical point (Tmax ≈ 374 °C)
� Tmax achieved at very high Pp (≈22 Mpa)� For Light Water Reactors (LWRs), high Thot requires high Pp (≈7-15 MPa)
� high stored energy in primary coolant (dangerous in accidents)� increased structural piping and component wall thickness
Metals:� Psat less than atmospheric pressure at Tp of interest (≈500-550 °C)
� For Liquid Metal Fast Breeder Reactors (LMFBRs) Tp not limited by boilingpoint of coolant, but by creep lifetime characteristics of the stainless steelprimary system material
Maximum allowable primary coolant outlet temperature, Tp = Thot
Gas coolants
� Single-phase coolants offer the potential for high outlet temperatures Tp = Thot without such inherently coupled high primary pressure Pp
� System pressure Pp is dictated by the desired core heat transfer capabilities which strongly depend on pressure
� Tp = Thot = 635-750 °C and Pp = 4-5 MPa
777
No saturation line
General Considerations: PWR vs BWR Thermal Efficiencies
� Plant thermal efficiency depends on the maximum temperature in the secondary or power generation system (saturation temperature Ts at which vapor is produced)
� PWR: Ts ≠ Tp = Thot, due to the temperature diffeence needed to transfer heat between the primary and secondary systems in the steam generator
� BWR: Ts ≈ Tp = Thot (neglecting losses), limited to saturation condition (vapor directly in turbine: average quality x ≈ 15%)
Turbine steam conditions in PWR and BWR are similar(although T and P are very different)
888
PWR and BWR have similar thermal efficiencies
(although Tp and Pp are very different)
Power Spent per Power Produced
Tcold
Thot The fluid through the core suffers pressure drops and require pumping power:
vA∆p
W
flow ⋅⋅== timeunitperdistancethroughforce&
∆pwhere ∆p = pressure drop through the circuit
Aflow = cross-sectional area of the coolant passagev = average coolant velocity
W 2.0µ& LIQUID BETTER THAN GAS:
2
2v
D
Lfp
ρ=∆
999
It can be seen that:
pcQ
W
⋅∝
2.0
2.0
ρµ
&
& LIQUID BETTER THAN GAS:cpl >> cpg
ρl >> ρg
Coolant hi [W/m2] W/Q (normalized towater)
Water 2000-10000 1
Organic liquid 1000-4000 4-10
Liquid metals 3000-20000 3-7
Gas 50-500 100
Remember:
µρvD=Re2.0Re−⋅= Cf (smooth tubes)
Energy Production and Transfer Parameters
101010
Energy Production and Transfer Parameters
Energy Production and Transfer Parameters
� Energy production in a nuclear reactor is expressed by a variety of terms reflecting the multidisciplinary nature of the design process:
� Volumetric energy (or heat) generation rate
� Surface heat flux
( )rqr
'''
( )Sq ''r
[W/m3]
[W/m2]
1111
� Surface heat flux
� Linear heat generation rate or power rating
� Rate of energy generation (in the n-th fuel pin � )
� Core power
11
( )Sq ''
( )zq'
q& nq&
Q&
[W/m ]
[W/m]
[W]
[W]
Energy Production and Transfer ParametersVolumetric Energy (or heat) generation rate
� Volumetric energy (or heat) generation rate
� energy-generation rate per unit volume at position of the fuel material
� computed by the reactor physicists using fission reaction rates and flux
� Assumption: energy released by a fission reaction is recovered at the position of
( )rqr
''' [W/m3]
rr
1212
� Assumption: energy released by a fission reaction is recovered at the position of the fission event (except for the fraction carried away by neutrinos and the fraction deposited in nonfuel materials)
12
( ) ( )rRRrq f
rr ∝'''
( ) ( )rRRrdepositionEnergy f
rr ∝(Fission fragments, γ’s, β’s) Total fission reaction rate
rrV
Energy Production and Transfer ParametersVolumetric Energy (or heat) generation rate
� The total fission reaction rate is obtained by summing the fission reaction rates of all the isotopes j considered (e.g., U-233, U-235, Pu-239, …)
� Let
( ) ( )∑=j
jff rRRrRRrr
( )Ejaσ
= microscopic absorption cross section for isotope j [barn] = [10-24 cm2]= equivalent projected area of an atom for an absorbtion reaction= proportional to the probability that one atom of type j absorb an incident neutron
of energy E
131313
� Macroscopic fission cross section = sum of all microscopic cross section for a fission reaction due to all atoms of type j within a unit volume interacting with an incident neutron per unit time
of energy E
( ) ( )EE jc
jf σσ += (fissions + captures)
( ) ( ) ( )ErNEr jf
jjf σrr ≡Σ ,
jM
AN
j
jvj isotopeofdensityatomic==ρ
jj isotopeofdensitymass=ρ
( )molmoleculesAv /106.022numberAvogadro 23⋅=
jM j isotopeofmassmolecular=
Energy Production and Transfer ParametersVolumetric Energy (or heat) generation rate
� The fission rate of isotope j at position due to the netron flux within the interval of neutron energy of E to E + dE is obtained from
� The fission reaction rate for isotope j due to neutrons of all energies is
rr ( )Er ,
rφ
( ) ( ) ( )dEErErdEErRR jf
jf ,,,
rrr φΣ=
( ) ( )∫∞
=0
, dEErRRrRR jf
jf
rr
1414
� Defining the energy per fission reaction of isotope j as , the heat-generation rate per unit volume at due to isotope j is
� Summing over all the isotopes yields the total volumetric heat-generation rate:
14
0
rr
jfχ
( ) ( )∫∞
=0
,''' dEErRRrq jf
jfj
rr χ
( ) ( ) ( ) ( )∑∫∑∫∞∞
Σ==j
jf
jf
j
jf
jf dEErErdEErRRrq
00
,,,'''rrrr φχχ
Energy Production and Transfer ParametersVolumetric Energy (or heat) generation rate
� In practice, the energy is subdivided into a few intervals or groups; a multi-energy group model is then used to calculate neutron fluxes. Thus,
( ) ( ) ( )∑∑=
Σ=j
K
kk
jfk
jf rrrq
1
'''rrr φχ
where K = number of energy groups
= equivalent macroscopic fission cross section and neutron fluxrespectively for the energy group k
( ) ( )rr kjfk
rr φ,Σ
1515
� If we use a one energy group approximation (OK for homogeneous thermal reactors at locations far from the reactor core boundaries), we have:
� If we also assume uniform fuel material composition, then:
15
( ) ( ) ( )∑ Σ=j
jf
jf rrrq
rrr11''' φχ
( ) ( )∑ Σ=j
jf
jf rrq
rr11''' φχ
Energy Production and Transfer ParametersVolumetric Energy (or heat) generation rate
� Assuming that for all fissile material, the volumetric core heat-generation rate is given by:
fjf χχ =
( ) ( )rrq ff
rr11''' φχ Σ= ∑=
j
jff ΣΣ 11where
161616
j
NOTICE: the thermal neutron flux in a typical LWR is only 15% of total neutron flux!BUT
Fission cross sections of U-235 and Pu-239 are so large at thermal energiesthat thermal fissions are 85-90% of total fissions
Energy Production and Transfer ParametersSurface Heat Flux
� Fuel element surface heat flux
� Important to the thermal designer
� Related to the volumetric energy generation rate as
( )Sq ''r
[W/m2]
( )rqr
'''
( ) ( )∫∫∫∫∫ =⋅VS
dVrqdSnSqrrr
'''''
171717
Where: S is the surface area that bounds the volume V where the heat generation occurs
VS
is the outward unity vector normal to the surface S
rrV
nr
Snr
nr
Energy Production and Transfer ParametersLinear heat generation rate or power rating
� Linear heat generation rate or power rating
� Important to both the thermal and metallurgic designers
� Related to the volumetric energy generation rate as
( )zq' [W/m]
( )rqr
'''
( ) ( )∫∫∫∫ =VL
dVrqdzzqr
''''
181818
VL
Where: L is the length of the volume V bounded by the surface S where the heat generation occurs
rrV
Sz
L
Energy Production and Transfer ParametersRate of energy generation
� Rate of energy generation
( )∫∫∫=V
dVrqqr
& ''' Heat-generation rate in the heat-generating volume V
q& [W]
� If the volume V is taken as the entire heat-generating volume of the n-th fuel pin Vfn, the quantity is the heat-generation rate in the n-th pin
q&nq&
191919
( ) ( ) ( )∫∫∫∫∫∫ =⋅===LSV
n dzzqdSnSqdVrqqqnfn
''''''rrr
&&
Where: Vfn is the volume of the energy-generating region of the n-th fuel pin
is the outward unity vector normal to the surface Sn surrounding Vfnnr
L is the length of the active fuel element (pin)
These general expressions take different forms depending on the size and shapeof the region generating the heat (e.g., cylindrical pin, …)
Energy Production and Transfer ParametersThermal Parameters Averaged Over A Pin
� Define the following (mean) quantities:
( )fn
n
Vfnn V
qdVrq
Vq
fn
&r == ∫∫∫ '''1
'''
( )n
n
Snn S
qdSnSq
Sq
n
&rr =⋅= ∫∫ ''1
''
( )L
qdzzq
Lq n
Ln
&== ∫ '
1'
Mean volumetric generation rate in the n-th pin
Mean heat flux through surface Sn of the n-th pin
Mean linear power rating of the n-th pin
202020
For any cylindrical fuel rod
nfoncoconn qRLqRLqLq '''''2' 2ππ ==⋅=&
Energy Production and Transfer ParametersCore Power
� Core power
� It is obtained by summing the heat generated per pin over all the N pins in the core and the heat generated in the nonfueled regions (moderator and structural materials)
Q& [W]
N
∑
2121
� Defining γ as the fraction of power generated in the fuel:
21
nonfuel
N
nn QqQ &&& +=∑
=1
∑=
=N
nnqQ
1
1&&
γ ( )typereactortheondependsbut96.093.0 −≅γ
Energy Production and Transfer ParametersCore-Wide Thermal Parameters For An “Average” Pin
� Rimembering that nfoncoconn qRLqRLqLq '''''2' 2ππ ==⋅=&
∑∑∑∑====
==⋅==N
nnfo
N
nncoco
N
nn
N
nn qRLqRLqLqQ
1
2
111
'''1
''21
'11 π
γπ
γγγ&&
� Define the core-wide thermal parameters for an “average” pin
qqN
N
n && ≡∑1
(heat generation rate in the “average” pin)
222222
'''''2'1 2 qR
LqR
Lq
Lq
N
Qfococo π
γπ
γγγ==== &
&
N nn∑
=1
� NOTICE:
fuelfn V
Q
NV
&& γγ =='''- Core-average volumetric heat generation rate in the fuel:
- Core power density:
coreV
&='''
≠
(Vcore = Vfuel + Vmoderator + Vstructures)
Thermal Design Limits
232323
Thermal Design Limits
Thermal Design Limits
� All reactors (except High Temperature Gas Reactor-HTGR)
� Fuel pins with metallic cladding
2424
� HTGR
� Graphite-coated fuel pellets
24
Thermal Design Limits Fuel Pins With Metallic Cladding
� Thermal design limits � integrity of the clad
� In principle: limits expressed in terms of structural design parameters (e.g., strain, fatigue limits)
� In practice: complex behavior of materials in radiation and thermal environment � limits expressed on temperatures and heat fluxes
Cylindrical, metallic clad, oxide fuel pellets
252525
Thermal Design Limits Fuel Pins With Metallic Cladding - LWRs
� Clad average temperature, Tclad
� Steady-state: inherent characteristics of LWRs limit Tclad to a narrow band above the coolant saturation temperature � not necessary to set a
steady-state limit on Tclad
� Transient (in particular, Loss of Coolant Accident-LOCA): most important criterion is T < 2200°F to prevent extensive metal (Zirconium)-water
2626
criterion is Tclad < 2200°F to prevent extensive metal (Zirconium)-water reaction from occurring
26
Thermal Design Limits Fuel Pins With Metallic Cladding - LWRs
� Fuel centerline temperature
� It is maintained well within its design limit due to restrictions imposed by the critical heat flux limit
272727
Thermal Design Limits Fuel Pins With Metallic Cladding - LWRs
� Critical Heat Flux (CHF)
� It results from a sudden reduction of the heat transfer capabilities of the two-phase coolant (i.e., reduction of heat transfer coefficient h)
� For fuel rods:
fobco hD
Rq
h
qTT
2''''' ==− TbTb
282828
cobco hDh
TT ==−
Tco
TbTb
q’’ = “independent” parameter + Tb = nominally fixed (saturation)
↓ h � ↑ Tco
Thermal design limit on q’’
Thermal Design Limits Fuel Pins With Metallic Cladding - LWRs
� Critical Heat Flux (CHF)
� PWR: low void fractions � heated surface cooled by nucleate boiling
Departure from Nucleate Boiling (DNB)
(vapor blanket)
292929
DNB depends on local conditions� correlations and limits in terms of heat flux
ratios between limiting and operating
Thermal Design Limits Fuel Pins With Metallic Cladding - LWRs
� Critical Heat Flux (CHF)
� PWR: low void fractions � heated surface cooled by nucleate boiling
= Departure from Nucleate Boiling Ratio (DNBR)
Minimum Departure from Nucleate Boiling Ratio (MDNBR)
303030
Design limit: MDNBR ≥ 1.3 at 112% power (to account for overpower margin)
Thermal Design Limits Fuel Pins With Metallic Cladding - LWRs
� Critical Heat Flux (CHF)
� BWR: high void fractions � heated surface cooled by a liquid film
Dryout depends on the channel thermal hydraulicconditions upstream of the dryout location, ratherthan on the local condition of the dryout:
� correlations and limits in terms of power
313131
� correlations and limits in terms of power
ratios between limiting and operating
Design limit: MCPR ≥ 1.2 at 100% power(Minimum Critical Power Ratio)
(see second part of the course)
Thermal Design Limits Fuel Pins With Metallic Cladding - LMFBRs
� Present practice is to require a level of soobcoling such that the sodium temperature does not exceed its boiling temperature for transient conditions
� Additionally, considerable effort is applied to ensure that coolant voiding in accident situations can be satisfactoritly accommodated
LMFBR limits placed on fuel and clad temperatures
323232
Thermal Design LimitsGraphite-Coated Fuel Pellets
� Coated particles deposited in holes symmetrically drilled in graphite matrix blocks to provide passages for helium coolant
� Inner layer and fuel kernel designated to accommodate expansion of the particle and trap gaseous fission products
� Fission gas release due to:
333333
� Diffusion
� Coating rupture
� Limits imposed to control gas release rates from coatings
� Limits on fuel particle center temperature:
� 100% power � 1300°C (minimize steady-state diffusion)
� Peak transient � 1600°C (minimize cracking of particle coatings)
Figures of Merit for Core Thermal Performance
343434
Figures of Merit for Core Thermal Performance
Figures of Merit for Core Thermal Performance
�Core power density
3535
�Specific power
35
Figures of Merit for Core Thermal PerformanceCore Power Density
� Core power density
� It is a measure of the energy generated per unit volume of the core
� Since the size of the reactor vessel and hence the capital costs are related to the core size, the power density is a measure of the capital cost of a design concept
coreV
&=''' (Vcore = Vfuel + Vmoderator + Vstructures)
core
fuel
V
Vq
γ'''
=
363636
( ) ( )22
2 '1'''4141'''
P
q
dzP
dzqRQ fo
square γπ
γ=
⋅=−∞
( ) ( )2
2
23
'1
23
2
'''6131'''
P
q
dzPP
dzqRQ fo
triang γπ
γ=
⋅=−∞
( ) ( ) squaretriang QQ −∞−∞ ⋅= '''155.1'''(LMFBRs) (LWRs: loosely-packed
� better moderation)
Figures of Merit for Core Thermal PerformanceSpecific Power
� Specific power
� It is a measure of the energy generated per unit mass of fuel material
� It is usually expressed as watts per grams of heavy fuel atoms (U, Pu, Th, but not O, C, etc.)
� While power density is a measure of the capital cost, specific power has direct implications on the fuel cycle cost and core inventory requirements
atomsheavyofMass
QSP
&=
373737
fLR
Lq
QSP
pelletfo ρπγ 2
'1
atomsheavyofMass
=
=&
fuelofgrams
atomsheavyfuelofgrams
fueltheinatomsheavyoffractionmass
=
=f*
* See later
Figures of Merit for Core Thermal PerformanceSpecific Power
� Introduce the “smeared density” ρsmeared:
�
� ρsmeared includes the void that is present as the gap between the fuel pellet and the clad inside diameter
( ) fLR
LqSP
smearedgfo ρδπγ 2
'1
+=
3838
pellet and the clad inside diameter
38
( )22
gfo
pelletfosmeared
R
R
δπ
ρπρ
+=
Figures of Merit for Core Thermal PerformanceSpecific Power
�
� Heavy atoms include all the U, Pu, Th isotopes and are therefore composed of fissile atoms (Mff) and nonfissile atoms (Mnf), where M is the molecular weight
� Fuel is the entire fuel-bearing material, i.e., UO2 but not the clad
� For oxid fuel:
fuelofgrams
atomsheavyfuelofgramsfueltheinatomsheavyoffractionmass ==f
+
3939
� The enrichment (r) is the mass ratio of fissile atoms to total heavy atoms
39
22 OOnfnfffff
nfnfffff
MNMNMN
MNMNf
+++
= (N = atomic density [atoms/cm3])
nfnfffff
ffff
MNMN
MNr
+=
nfnfffff
nfnf
MNMN
MNr
+=−1and
Figures of Merit for Core Thermal PerformanceSpecific Power
� It can be observed that for UO2:
� Using the expressions
� Divide each term of by
nfffO NNN +=2
nfnfffff
ffff
MNMN
MNr
+=
nfnfffff
nfnf
MNMN
MNr
+=−1and
r
N
M
MrN nf
ff
nfff −
=
1we obtain ( )
r
N
M
MrN ff
nf
ffnf
−= 1
nfnfffff MNMNf
+= NNN +=
4040
� Divide each term of by
� Substitute the expressions for Nff, Nnf and (Nff + Nnf) into the expression for f to obtain
40
22 OOnfnfffff
nfnfffff
MNMNMN
MNMNf
+++
=
( )( ) ( ) ( )
( )( ) ( ) ( ) 2
2
1/1
/1
1/1
/1
Onfffnf
ffnfff
nfffnf
ffnfff
UO
MMrMMr
rM
MMrr
r
MrMMr
rM
MMrr
r
f+
−+−+
−+
−+−+
−+=
nfffO NNN +=2
Figures of Merit for Core Thermal PerformanceSpecific Power
� For the case where Mff ≈ Mnf
( )( ) ( ) ( )
( )( ) ( ) ( ) 2
2
1/1
/1
1/1
/1
Onfff
nfffnf
ffnfff
UO
MMrMMr
rM
MMrr
r
MrMMr
rM
MMrr
r
f+
−+−+
−+
−+−+
−+=
414141
( )( ) ( ) ( ) 21//1 Onfffnf
ffnfff
MMrMMr
MMMrr
+−+
+−+
( )( )
2
2 1
1
Onfff
nfffUO MMrrM
MrrMf
+−+−+
=
Examples
424242
Examples
Determination of the neutron flux at a given power in a thermal reactor – Example 3.1 Nuclear Systems I
� A large PWR designed to produce heat at a rate of 3083 MW has 193 fuel assemblies each loaded with 517.4 kg of UO2. If the average isotopic content of the fuel is 2.78 weight percent 235U, what is the average thermal neutron flux in the reactor?
4343
reactor?
Assume uniform fuel composition, and Xf = 190 MeV/fission (3.04 x 10-11 J/fission). Also assume that γ = 95% of the reactor energy, i.e., of the recoverable fission energy is from the heat generated in the fuel. The effective thermal fission cross section of 235U (σf
235) for this reactor is 350 barns.
43
Determination of the neutron flux at a given power in a thermal reactor – Solution
� Use to find
� Since in this reactor, then
� The core-average value of the neutron flux in a reactor with uniform density of 235U is obtained from the average heat-generation rate by
( ) ( )rrq ff
rr1''' φχ Σ= ( ) ( )
ff
rqr
Σ=
χφ
rr '''
1
235
ff Σ=Σ 235235235
ff N σ=Σ
'''q
4444
� Multiplying both the numerator and denominator by the UO2 volume (VUO2) we get:
44
2352351
'''
ff N
q
σχφ =
22
2
2352352352351
'''
UOff
fuel
UOff
UO
VN
Q
VN
Vq
σχσχφ
&
==
Determination of the neutron flux at a given power in a thermal reactor – Solution
� To use the above equation only N235 needs to be calculated. All other values are given. The value of N235 can be obtained from the uranium atomic density if the 235U atomic fraction (a) is known:
� The uranium atomic density is equal to the molecular density of UO2, as each molecule contains one uranium atom (NU = NUO2).
UaNN =235
4545
UO2, as each molecule contains one uranium atom (NU = NUO2). Thus:
� Multiplying each side by VUO2 we get
45
2
22235
UO
UOvUOU
M
AaaNaNN
ρ===
2
2
2
2
2
235
UO
UOv
UO
UO
UO M
mAaVaNVN ==
Determination of the neutron flux at a given power in a thermal reactor – Solution
� Now
� The molecular mass of UO2 is calculated from
g109858.9kg
g1000
assembly
kg4.517]assemblies[193 72
2⋅=
⋅
⋅= UO
mUO
]g/mole[0439.235235235 =MU
]g/mole[0508.238238
238 =MU]g/mole[9994.15Oxygen =OM
MMM 2+=
4646
� To obtain the value of a we use the known 235U weigth fraction, or enrichment (r):
46
OUUO MMM 22
+=( ) 238235 1 MaaMM U −+=( ) OUO MMaaMM 21 2382352
+−+=
( ) 238235
235235
1 MaaM
aM
M
Mar
U −+==
Determination of the neutron flux at a given power in a thermal reactor – Solution
� Expression can be solved for a:
� Then,
( ) 238235
235235
1 MaaM
aM
M
Mar
U −+==
( ) ( )028146.0
0278.010508.2380439.235
0278.0
0278.0
1238
235
=−+
=−+
=r
M
Mr
ra
( ) ]g/mole[9645.26921 2382352=+−+= OUO MMaaMM
g109858.9moleatoms
100225.6 723 ⋅⋅⋅mA
4747
� Now,
� Finally, the value for the average flux is obtained as:
47
Uatoms1027.6g/mole269.9645
g109858.9mole
100225.6028146.0 23527235
2
2
2⋅=
⋅⋅⋅==
UO
UOv
UO M
mAaVN
( )smneutrons/c1038.4
atoms1027.6neutronatom
cm10350
fissionJ
103.04
MWW
10MW308395.0
213
272
2411-
6
2352352352351
22
⋅⋅=
⋅
⋅⋅
⋅
⋅===
−UOff
reactor
UOff
fuel
VN
Q
VN
Q
σχγ
σχφ
&&
Heat transfer parameters in various power reactors – Example 3.2 Nuclear Systems I
� For the set of reactor parameters given below, calculate for each reactor type:
� Equivalent core diameter and core length
� Average core power density Q'" (MW/m3)
� Core-wide average linear heat-generation rate of a fuel rod , <q’> (kW/m)
� Core-wide average heat flux at the interface between the rod and the coolant, <q’’co> (MW/m2)
484848
Heat transfer parameters in various power reactors – Solution
� Only the PWR case is considered in detail here; the results for the other reactors are summarized
� Equivalent core diameter and length calculation
( ) 222 043.0207.0)spacinglateralassembly(areaassemblyFuel m===22 36.10assemblies241043.0areaassemblyFuelareaCore mmN assemblies =⋅=⋅=
mDmD
64.336.10areacore4
:diametercircularEquivalent 22
=⇒==π
mL 81.3lengthrodFuellengthCore ===
4949
� Average core power density
49
mL 81.3lengthrodFuellengthCore ===32 65.3981.336.10lengthcoreareacorevolumecoreTotal mmm =⋅=⋅=
3
3/85.95
65.39
3800
volumeCore
levelpowerCore''' mMW
m
MW
V
core
====&
Heat transfer parameters in various power reactors – Solution
� Average linear heat generation rate in a fuel rod:
� Average heat flux at the interface between rod and coolant
( ) mkWMW
NL
Qq /8.16
m/rod81.3assemblies241blyrods/assem236
380096.0' =
⋅⋅⋅==
&γ
5050
� Average heat flux at the interface between rod and coolant
50
( ) 23
/552.00097.0
/10/8.16'
2'' mMW
m
kWMWmkW
D
q
DLN
Q
RLN
cococo
co =⋅
====−
πππγ
πγ &&
Power density and specific power for a PWR – Example 2.1 Nuclear Systems I
� Confirm the core power density listed in Table 1 (Table 2.3 Nuclear Systems I) for the PWR case and comment the results. Calculate the specific power of the PWR with characteristics given in Tables 1 and then Table 2 (Tables 2.3 and 1.3, respectively, in Nuclear Systems
5151
then Table 2 (Tables 2.3 and 1.3, respectively, in Nuclear Systems I). Assume a pellet density of 95% of the UO2 theoretical density.
51
(Notice: consider γ = 1 in this case)
Power density and specific power for a PWR
525252
Table 1 (Table 2.3 Todreas & Kazimi)
Power density and specific power for a PWR
535353
Table 2 (Table 1.3 Todreas & Kazimi)
Power density and specific power for a PWR – Solution
� The core power density is given as
� From Table 1 (Table 2.3),
� From Table 2 (Table 1.3), P = 12.6 mm
� Then,
( )2
'1'''
P
qQ square γ
=−∞
mkWq /8.17' =γ
( ) ( )kW
orMWkWq
Q 6 11210112.08.17'1
''' =⋅===
5454
� Notice that Table 1 (Table 2.3) lists the average power density as Q’’’PWR = 105 kW/L. The difference arises because our calculation is based on the core as an infinite square array, whereas in practice a finite number of pins form each array so that edge effects within assemblies must be considered.
54
( ) ( ) Lor
mmPQ square 33
6
23211210112.0
106.12''' =⋅=
⋅==
−−∞ γ
Power density and specific power for a PWR – Solution
� The specific power is given by
� Evaluating f using for the enrichment r of 2.6% listed in Table 2 (Table 1.3)
fLR
LqQSP
pelletfo ρπγ 2
'1
atomsheavyofMass==
&
( )( )
21
1
Onfff
nfff
MMrrM
MrrMf
+−+−+
=
( )( ) 8815.0
0508.238)026.01(0439.235026.01=−+⋅=
−+= nfff MrrM
f
5555
� For a pellet density of 95% of the UO2 theoretical density of 10.97 g/cm3 and a fuel pellet diameter of 8.2 mm (Table 2), the specific power is
55
( )( ) 8815.0
9994.1520508.238)026.01(0439.235026.0
0508.238)026.01(0439.235026.0
12
=⋅+−+⋅
−+⋅=+−+
=Onfff
nfff
MMrrMf
atomsheavyfuelg
W7.36
8815.010
97.1095.02102.8
/1078.1'1
atomsheavyofMass3
26
23
4
2=
⋅
⋅
⋅===
−
−
m
gm
mW
fR
qQSP
pelletfo πρπγ
&
Power density and specific power for a PWR – Solution
� Alternatively, Table 1 (Table 2.3) lists the total core loading of fuel material as Mfm = 101∙103 kg of UO2. In this case,
� If Q is evaluated from the PWR dimensions as in Tables 1 and 2 (Tables 2.3 and 1.3, respectively)
fmMf
QQSP
⋅===
&&
atomsheavyfuelofloading
powercore
atomsheavyofMass
( ) tMWmmMWLNqQ 331926419366.3/0178.0'1 =⋅⋅⋅==γ
&
5656
� Then,
� Unlike the case of power density, specific power is closely estimated, as the fuel mass, not the core volume, is utilized.
56
γ
atomsheavyfuelofg
W28.37
101018815.0
3319
atomsheavyofMass 3=
⋅⋅=
⋅==
kg
MW
Mf
QQSP t
fm
&&