Heat Generation

8/10/2019 Heat Generation

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REACTOR HEAT GENERATION

The energy released from fission appears as kinetic energy of fission fragments, fission neutrons and gamma and

beta radiation. Approximately 193 Mev is released directly as a result of fission, and consists of that energy

produced promptly at the time of fission, and a later delayed component resulting from delayed neutrons and decay

of radioactive fission products. The prompt component is associated with the kinetic energy of fission products,

prompt fission neutrons and prompt fission gammas. An additional 7 Mev per fission is produced by the capture ofexcess fission neutrons and the subsequent gamma and beta decay of activation products. Thermal energy (heat) is

produced as these particles interact with, and transfer their kinetic energy to, the lattice atoms of the fuel and other

reactor materials. Of the 200 Mev available from a fission reaction, approximately 10 Mev is due to the kinetic

energy of neutrinos associated with beta decay and is unrecoverable.

Fission fragments and beta particles have very short ranges in reactor materials and their kinetic energy can be

considered absorbed at their point of origin. Approximately 160 Mev is carried as kinetic energy of the fission

fragments alone, and as such all of this energy would be deposited locally in the fuel. Gamma radiation on the other

hand has a relatively long range as compared to the dimensions of the reactor fuel and therefore deposits its energy

both in the fuel, the moderator and reactor structure. Due to the large mass of fuel in the core however, the majority

of the gamma energy is deposited in the fuel, though the point of interaction may be far from the point of origin. In

thermal reactors, the energy released by the fission neutrons is deposited primarily in the moderator as a result of the

thermalization process. In Light Water Reactors where the moderator and the coolant are the same, this results inapproximately 5 Mev per fission deposited directly into the coolant by neutron thermalization alone. The amount of

heat produced in various reactor components is a function of the reactor materials and configuration. This is

particularly true of the activation component. The following values however provide guidelines when more precise

information is unavailable.

ENERGY DISTRIBUTION

(a) Fuel 180 Mev/fission

(b) Moderator 8 Mev/fission

(c) Reactor structure 2 Mev/fission

(d) Neutrinos 10 Mev/fission

It is of interest to note, that of the approximately 190 Mev per fission that is recoverable, only 180 Mev or

approximately 95 % is produced in the fuel itself, with the remainder coming from outside the fuel. Of additional

interest is the heat produced by the decay of the radioactive fission fragments, their daughters and activation

products produced by excess fission neutrons. At the beginning of core life this energy is unavailable. However,

these radioactive products reach equilibrium after a short period of time, and at equilibrium constitute

approximately 7% of the reactor power.

Example:

Determine the energy released in the fission reaction

92 0 56U n Ba + Kr +2 n235 1 137

36

97

0

1

+

A mass balance on the reactants gives

235 0439 100867 136 9061 96 9212 2 100867

236 0526 235 8446

. . . . .

. .

+ + +


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m

amu

=

=

2360526 2358446

0 2080

. .

.

The energy associated with the mass deficit is then

0 2080 931 193 6. .amu

Mev

amu

Mev

fission =

As the overwhelming majority of the energy deposited in the fuel is due to the short ranged fission fragments and

beta particles, we approximate the heat generation rate in the reactor fuel as a constant times the fission rate

=

q r G r Mev

cmf( ) ( )

sec

v v

3

where:

q r( )v

= Volumetric heat generation rate (Energy/vol-time)

G = Fission energy absorbed by the fuel per fission (180 Mev/fission)

Both the cross section and the flux are in general functions of position, energy and time, i.e.

=q r E t G r E t r E t f( , , ) ( , , ) ( , , )v v v

(1)

where the flux and volumetric heat generation rate are now differential quantities with respect to energy. If we

assume steady-state operation, the heat generation rate is independent of time and the volumetric heat generation

rate as a function of position is

=

q r G r E r E dE f( ) ( , ) ( , )v v v 0

(2)

The energy dependence of the flux is usually treated by rewriting the integral as the sum of integrals over energy

bands or groups

=

=G

g

g

N

g

E

Ef dEErErGrq

1

1

),(),()( vvv

(3)

and defining a group averaged flux and cross section such that the fission rate over the group is preserved, i.e.

fE

E

f gr E r E dE r r

g

g

g( , ) ( , ) ( ) ( )

v v v v

1

(4)

where

1

),()(g

g

E

Eg dEErr

vv (5)

Note, this implies the appropriate group averaged cross section is given by


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f

f

E

E

E

E

f

E

E

gg

g

g

g

g

g

g

r

r E r E dE

r E dE

r E r E dE

r( )

( , ) ( , )

( , )

( , ) ( , )

( )

v

v v

v

v v

v =

1

1

1

(6)

The volumetric heat generation rate may then be written in terms of the group fluxes and group averaged cross

sections as

)()()(

1

rrGrq g

N

g

f

G

g

vvv

=

= (7)

From Equation 7, we can infer that to a very good approximation the spatial distribution of the heat generated in a

nuclear reactor is proportional to the spatial distribution of the fission rate. In thermal reactors, the fission rate is

dominated by neutrons in the thermal energy range, primarily due to the large macroscopic cross section associated

with thermal fission relative to the fission cross section at higher energies. The volumetric heat generation rate can

then be approximated as

)()()()()(

1

rrGrrGrq thfg

N

g

f th

G

g

vvvvv =

=

(8)

which implies the heat generation rate is proportional to the thermal neutron flux distribution in the fuel.


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HEAT GENERATED IN A FUEL ROD

Consider a long, thin fuel rod oriented vertically at some arbitrary point within a cylindrical core. As the neutron

flux is in general a function of space, the flux and therefore heat generation rate in any particular rod will be a

function of space.

Figure 1: Core Flux Distribution in the Vicinity of a Fuel Rod

Due to the relatively small cross sectional area of a fuel rod compared to that of the core, the flux in the vicinity of

the fuel rod may be considered constant in terms of the overall core radial behavior with the magnitude of the flux

governed by the rod position. The axial distribution in the rod however follows the axial distribution in the core.

At any location within the core, the thermal flux within an individual fuel rod is depressed radially due to the strong

neutron absorption in the fuel.

Thermal Flux

Fast Flux

Moderator

Fuel

Figure 2: Fast and Thermal Flux Distributions in a Fuel Rod

If we assume the flux within the fuel rod to be separable in the radial and axial directions then

)()(),( 0 zrzr = (1)


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with the amplitude 0 a function of core position. This implies the volumetric heat generation rate is also separable

in randzsuch that

)()(),( 0 zrqzrq = (2)

The total heat generated in the rod is the integral of the volumetric heat generation rate over the rod volume

q q r z dV

V

= ( , ) (3)

or for a cylindrical fuel element

=R H

dzzrdrrqq0 0

0 )(2)( (4)

Assume for sake of illustration that the flux in the fuel rod is uniform radially and cosine shaped axially with z = 0

the core mid plane, i.e.

= q r z q z

He( , ) cos0 (5)

where He is the extrapolated core height. The total heat generated in the fuel rod would be

q q Rz

H dz

eH

H

=

0 2

2

2

cos (6)

or

q q R H H

H

e

e

=

0

2 2

2

sin (7)

whereRis the fuel (pellet) radius. If the extrapolation distances are small relative to the dimensions of the reactor,

then

q q R H 022 (8)

Relationships which are useful in describing the heat generated in reactor fuel elements include:

Linear Heat Rate: The linear heat rate q is defined to be the heat generated per unit length in a fuel element.Local linear heat rate can be related to the heat produced at a specific location within a fuel

element and the volumetric heat generation rate at that location through the following

relationships. The total heat produced within a differential length dzaboutzis

=xA

dAdzzrqdzzq ),()( v

(9)

such that the local linear heat rate is

=xA

dAzrqzq ),()( v

(10)


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The total heat generated within the fuel element is then

q q z dz

H

= ( )0

. (11)

The average linear heat rate in a particular element is obtained by averaging the local linear heat

rate over the height.

= =q H q z dz qHH

1

0

( ) . (12)

The core averaged linear heat rate can be obtained from the core thermal output, and the number

of fuel elements in the core by recognizing that the total heat generated in the reactor fuel is

simply the sum of the heat generated in the individual fuel elements. If qi is the heat generated

in an arbitrary fuel element, then

HqqQ

n

i

i

n

i

if ==

==11

& (13)

where : Q& = Core thermal output from all sources

f = Fraction of core thermal energy generated in the fuel

n = Total number of fuel elements

If we define the core averaged linear heat rate as

= n

i

ic qnq1

1

(14)

then nHqQ cf =& (15)

or

=qQ

nHc

f&

(16)

It will be shown later, that linear heat rate can be related to fuel temperature. As a

result, maximum linear heat rate is usually set by fuel melt or other maximumtemperature considerations. For a given fuel height, the maximum linear heat rate

dictates the number of fuel elements in a core.

Heat Flux: The heat flux is the heat transfer rate per unit surface area. While the heat flux can be

referenced to any surface, in reactors the heat flux is most often referenced to the outer clad

surface, i.e. the clad/coolant interface. Assuming all heat transfer in a fuel element is in the


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radial direction, at steady-state the local heat flux can be related to local linear heat rate and

volumetric heat generation rate by the simple energy balance

dzzqdAdzzrqdzPzq

xAw )(),()( == (17)

or

)(),()( zqdAzrqPzq

xAw

== (18)

where wP is the heated peremiter.

We can then relate the heat flux and total heat generated in a fuel element by

=H

wdzPzqq0

)( . (19)

The average heat flux in a particular fuel element is then

sw

H

A

q

HP

qdzzq

Hq === 0 )(

1. (20)

As with the core averaged linear heat rate, the core averaged heat flux can be obtained from the

core thermal output, and the number of fuel elements in the core by

HnP

Q

nA

Qq

w

f

s

fc

&&== (21)

As will be shown later, maximum heat flux is usually set by critical heat flux (DNB,

Dryout) considerations. For a given fuel height and number of fuel elements, the

maximum heat flux dictates the cross sectional dimensions (radius, thickness, etc.) of

the fuel.


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HEAT GENERATED IN A REACTOR CORE

We have seen, that the total heat generated in the reactor fuel is simply the sum of the heat generated in the

individual fuel elements

f i

i

n

Q q& ==1

(1)

where again the magnitude of qi is a function of the fuel elements location in the core. This is equivalent to

integrating the fission rate over the entire core, i.e.

f f

V

Q G r r dV

core

& ( ) ( )= v v (2)

Due to the complex spatial distribution of the flux and the cross section, this integral can only be evaluated under

special conditions. Lets consider one such special case, where we assume the fission cross section is a constant in

the fuel, and zero if outside the fuel.

f

fo f

f

r

r r

r r

( )v

v v

v v=

0

(3)

wherevrf denotes locations within the fuel. We define an equivalent homogeneous cross section for the entire core

( = f fN ) such that the total number of fuel atoms is conserved. The total heat generated in the fuel can then

be written as

f f

V

Q G r dV

core

& ( )=

v (4)

If No is the fuel number density, then the total number of fuel atoms is

N V N V

N NV

V

o fuel core

o

fuel

core

=

= (5)

The equivalent homogeneous macroscopic cross section can then be written in terms of our original fuel

macroscopic cross section as

= = = f f f ofuel

corefo

fuel

core

N NV

V

V

V (6)

such that the total heat generated in the fuel is


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f fofuel

core

V

Q GV

Vr dV

core

& ( )=

v (7)

If we further assume, that the heat generation rate in the moderator and structural materials is proportional to the

core wide fission rate, then

& ( )Q GV

Vr dVfo

fuel

core

Vcore

=

v (8)

where G contains contributions to the heat generation rate from all sources. Evaluation of Equation 8 still requiresspecification of the core wide flux distribution. If we assume the local variations in the flux due to fuel elements is

small compared to the total flux, then we can treat the core as approximately homogeneous and use Diffusion

Theory or some other suitable neutron flux model to generate ( ).vr One group Diffusion Theory gives the

following simple flux shapes for idealized reactor geometries.

Flux Distributions in Ideal Geometries

Infinite Slab

0 cos x

ae

Parallelepiped

0 cos cos cosx

a

y

b

z

ce e e

Sphere

0

r R

r

Re esin

Finite Cylinder

0 0

2 405J

.cos

r

R

z

He e

Table 1 (All dimensions are extrapolated)

Flux shapes in actual reactor systems almost never follow the simple functional forms given in Table 1. To account

for power variations due to non ideal geometries and/or uncertainties due to manufacturing tolerances and physical

changes during operation, the concept of a Hot Spot or Power Peaking Factor is introduced. The Power Peaking

Factor is defined such that

Fq=Maximum Core Heat Flux

Core Averaged Heat Flux. (9)

This implies that the maximum local heat flux at any point in the core is

= q F qq cmax (10)

and since the core averaged heat flux is proportional to the linear heat rate

= q F qq cmax . (11)


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The simple flux shapes given in Table 1 can be used to estimate the power peaking factor as illustrated in the

following example.

Example:

Compute the power peaking factor for a cylindrical reactor having a neutron flux distribution

0 0

2405J

.cos

r

R

z

He e

wherez= 0 is at the core midplane.

SOLUTION

The local heat flux in a power reactor is proportional to the local fission rate and therefore the local flux, i.e.

=

q r zr

R

z

H

q r z C r

R

z

H

e e

e e

( , ).

cos

( , ).

cos

0 0

1 0

2405

2 405

J

J

For this flux distribution, the maximum heat flux occurs at r= 0, andz= 0, such that =q Cmax 1 . The core averagedheat flux is obtained by averaging the heat flux over the core volume

=

q V C rR zH dVcore e e

Vcore

1 2 4051 0J

.cos

=

qR H

Cr

R

z

H rdrdz

core e e

R

H

H core1 2 405

22 1 002

2

J

.cos

/

/

=

q

R HC

R R H R

R

H

Hcore

core e e core

e e

1 4

2 405

2405

221

.

.sinJ1

If the reactor dimensions are large compared to the extrapolation distances R Re core , H He and the abovereduces to

( ) =

q C1 4

240524051 ..J1

The power peaking factor is then

( ) ( )

Fq

q

C

C

q =

=

= =max

..

( . )

..

1

1

1 4

2 4052 405

2 405

4 2 405364

JJ

11


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The fuel in power reactors is normally loaded in such a way as to reduce the power peaking factor, with typical

values around 2.3.

Example:

A nuclear reactor is to be constructed to produce 3411 Mwt, 97.4 % of which is produced in the fuel. The fuel

elements are in the form of cylindrical rods. The power distribution predicted by a neutronics analysis indicates a

core power peak-to-average ratio of 2.5. Accident analyses place the maximum allowable linear heat rate at 13.58

kW/ft at any point in the core and requires the maximum core heat flux not exceed 474,500 Btu/hr-ft2at any point.

a) If the fuel rods are to be 12 feet long, how many fuel rods are required?

b) What diameter are the rods?

c) In light water reactors, the fuel rods are typically arranged in a square lattice. Assuming a rod pitch (center-to-

center spacing) of 0.496 inches, what is the effective diameter of the core?

SOLUTION

a) The number of rods in the core are related to the total core power and the linear heat rate through the relationship

f cq

Q nq H nq

FH& max= =

The number of rods in the core is then

nF Q

Hq

q f=

=

=

& ( . )( . )( )

( )( . ),

max

2 5 0 974 3411 10

12 13 5850 968

3

b) The rod diameter is dictated by the surface heat flux. The relationship between the surface heat flux and the

linear heat rate isDqPqq w maxmaxmax ==

such that the rod diameter is

D q

qft inches=

= = =maxmax

( . )( )

( , ). .

1358 3413

474 5000 03109 0 373

c) For a rod pitch of S= 0.496 inches, the total core area is

A nS ftcore= = =2 2 2

50 968 0 496 12 87 076( , )( . / ) .

The effective core diameter is then

Dft D fte e

22

487 076

4 87 07610 53= = =.

( )( . ).


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HEAT GENERATION DURING SHUTDOWN

In a reactor shutdown, the reactor power does not immediately go to zero, but falls off rapidly with a rate

governed by the longest lived delayed neutron precursor. This may be easily shown if we assume that reactor

power can be described by the point kinetics equations

+= i iiCPk

dtdP

l)( (1)

iiii CPk

dt

dC

=

l (2)

or taking advantage of the definition of reactivity

+=i

iiCPkk

dt

dP

l

)1( (3)

ii

ii

C

Pk

dt

dC

= l (4)

If we further assume that following control rod insertion, neutron multiplication is very small ( 0k ) then thesolution for the power may be approximated by

=+i

iiCP

dt

dP

l (5)

iii C

dt

dC= (6)

which has solution

[ ] +=i

tt

i

iit eeC

ePtP i l

l

l /

1

/

)(

)0()0()(

(7)

Note, l is the prompt neutron lifetime and is extremely short (4

10 seconds) such that exponential terms

containing l die out quickly leaving only those asssociated with the delayed neutron precursors. As time

progresses, the short lived delayed neutron precursors also die out such that eventually only terms associated with

the longest lived delayed neutron precursor remain and control the rate at which the reactor power decays. In

addition to this residual fission energy, the reactor also continues to generate heat due to the decay of fission

fragments and activation products built up during operation. The magnitude of this heat source and the rate at

which it decays depends partly on the operating history of the core. In particular, shutdown power (Ps) depends on

the operating power level (Po), the operating time (to) at which the reactor operated at power level Po, and the

shutdown time (ts). In reality, the decay heat source is the many beta and gamma transitions of the excited nuclei

formed as fission fragments or neutron capture products. To account for all, or even most of these decay chains is

impractical at best for routine estimates of the decay heat rate. As a result, empirical fits have been developed

which relate the ratio of the decay power or shutdown power of the reactor to the operating power in terms of the

operating and shutdown times. The spatial distribution of decay heat can be assumed to follow the operating power

distribution. To obtain an idea of the rate at which decay heat is built up in a reactor core the following table is

provided.

Ps/PoVersus Operating Time


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Example:

An empirical relationship for determining decay heat rates is given by

P

P

t t t t t t s o s o s s0

0 2 0 2 7 0 2 7 0 20 1 10 10 0 87 2 10 0 87 2 10= + + + + + + + . ( ) ( ) . ( ) . ( ). . . .

where tois the time (seconds) the reactor operated at power P0and tsis the time (seconds) since reactor shutdown.

a) Compute and plot the decay heat rate as a function of operating and shutdown times for operating times of

one day, one week, one month and one year.

b) Determine the operating time required for the decay heat rate to reach 95% of its equilibrium value.

c) A reactor operates for the first 6 months of a one year cycle at 50% power, and the remaining six months at

100 % power. Compare the decay heat rate at the end of the year to that which would be obtained had the

reactor operated for the entire year at 100 % power.

SOLUTION

a) The decay heat rate as a function of operating and shutdown time is given in the following graphs

0 5 10 15 20 25 30 35 40 45 500

0.01

0.02

0.03

0.04

0.05

0.06

Time (hours)

DecayHeatRate

Ptotk

tk

Decay heat rate for operation and shut down times of one day


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0 2 4 6 8 10 12 14 160

0.01

0.02

0.03

0.04

0.05

0.06

Time (days)

DecayHeatRate

Ptotk

tk

Decay heat rate for operation and shut down times of one week

0 10 20 30 40 50 60 700

0.01

0.02

0.03

0.04

0.05

0.06

Time (days)

DecayHeatRate

Ptotk

tk

Decay heat rate for operation and shut down times of one month


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0 100 200 300 400 500 600 700 8000

0.01

0.02

0.03

0.04

0.05

0.06

Time (days)

DecayHeatRate

Ptotk

tk

Decay heat rate for operation and shut down times of one year

b) The equilibrium decay heat rate (P) is 6 % of the operating power level for this correlation. The time to reach

95 % of this value (P = 5.7 %) is found by iteratively solving the above decay heat equation for to, with ts= 0.

The resulting operating time is 15.12 days.

c) For the first six months, the decay heat rate is due only to continuous operation at 50 % power. For the second

six months, the decay heat rate is the sum of the decay heat rates from operation at 50 % power for the total

operating time and the decay heat rate from operating at 50 % power for the second six month period. For the

decay heat rate written as

[ ]P P t t

t

P t t t t t t t

o o s

o

o s o s o s s o

( , , )

. ( ) ( ) . ( ) . ( ). . . .=

0 0

01 10 10 0 87 2 10 087 2 10 00 2 0 2 7 0 2 7 0 2

this may be expressed as

P t P t P t monthsop o o o( ) (. , , ) (. , , )= + 5 0 5 6 0

The reference decay heat rate is

P t P t ref o o( ) ( , , )= 1 0 .

The resulting decay heat rates for one year of operation are illustrated below.


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0 50 100 150 200 250 300 350 4000

0.01

0.02

0.03

0.04

0.05

0.06

Time (days)

DecayHeatRate

Popi

Prefi

ti

The initial steep rise in the decay heat rate upon initiation of reactor operation, and corresponding steep decline

in the decay heat rate upon reactor shutdown is due to the buildup and decay of the short lived fission products.

It is obvious from these graphs, that the short time behavior of the decay heat is dominated by the most recent

operating conditions.

Example:

A power reactor operates at 3400 Mwt for one year. Determine the decay heat rate 5 minutes following reactor

shutdown.

SOLUTION

We again use the empirical equation for decay heat

P

Pt t t t t t s o s o s s

0

0 2 0 2 7 0 2 7 0 20 1 10 10 0 87 2 10 0 87 2 10= + + + + + + + . ( ) ( ) . ( ) . ( ). . . .

to = = 1 316 107

year seconds.ts = =5 300minutes seconds

From the given equation for decay heat

P

PP Mwt s

os= = =0 028 3400 0 028 95. ( )( . )


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It should be obvious from this level of decay heat, that core cooling is necessary even when the reactor is shutdown.

In power reactors, the lack of sustained cooling can result in severe structural damage, including core melt.

Alternate Approach

An alternate approach to the purely empirical correlations given above, is to assume the numerous components

of the decay heat source can be lumped into a relatively small number of groups, similar to the approach takenwith delayed neutrons. If jq is the concentration of decay heat groupj, then jq is assumed to satisfy the simple

balance equation

43421

4434421

43421rateloss

rateproduction

ratefission

)( fjjffjfj

VqVtEdt

Vdq = (8)

where jE is the yield fraction and j the decay constant for decay heat groupj. As was shown previously, the heat

production rate is proportional to the fission rate, such that

32143421jo

fjj

tP

ffjfj VGqVtGE

dtVdGq

=)(

)( (9)

where Gis energy per fission and )(tPo the total reactor power. The balance equation for decay heat groupjcan

then be written

jjojj

tPEdt

d

= )( (10)

and the total decay heat source is

= jjj ttP )()( (11)

In principle, Equation 10 can be solved for any operating history. Consider the special case of an infinite operating

time, such that the system has reached equilibrium and

0=dt

d j

Then from Equation 10,

j

ojj

PE

=)( (12)

If the system is then shut down, then for any shutdown time st , j is the solution of

jjj

dt

d

= (13)

subject to the initial condition


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j

ojjsj

PEt

=== )()0( (14)

Solution of Equation 13 gives

)exp()( sjj

ojsj t

PEt

= (15)

and the decay heat source is

=j

sjjsdo

s tEtP

tP)exp()(

)(

Typical values for the yields and decay constants are given in Table 3 below.

Group jE j (sec-1)

1 0.00299 1.772

2 0.00825 0.57743 0.01550 6.743 x 10-2

4 0.01935 6.214 x 10-3

5 0.01165 4.739 x 10-4

6 0.00645 4.810 x 10-5

7 0.00231 5.344 x 10-6

8 0.00164 5.726 x 10-7

9 0.00085 1.036 x 10-7

10 0.00043 2.959 x 10-8

11 0.00057 7.585 x 10-10

Table 3 Decay Heat Group Constants (From RETRAN code manual)


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HEAT GENERATION IN REACTOR STRUCTURE

Gamma and neutron radiation emanating from the reactor core interacts with and is absorbed by structural

materials such as core barrels, pressure vessels, etc. The absorbed radiation is converted into heat which must be

removed. As core barrels and reactor pressure vessels are relatively thin compared to their diameter, they can be

accurately approximated as slabs. Consider a monodirectional, monoenergetic gamma flux incident upon a slab

wall as illustrated below.

x=0 x=L

Figure 1: Gamma Radiation Incident on a Slab Wall

The interaction rate within the slab is given by ( )x where is the total attenuation coefficient. Photon

interactions inevitably lead to the production of short range electrons through Compton scattering, pair production,

and the photoelectric effect. We can again assume the deposition of the electron energy occurs locally such that the

heat generated by photons is proportional to the photon interaction rate. We therefore write the heat generation rate

as

=q x E xa( ) ( ) (1)

where a is the energy absorption coefficient for the slab material at photon energyE.Values of attenuation

coefficient and energy absorption coefficient are given in Table 1. The energy absorption coefficient accounts for

the fraction of the incident photon energy carried by the electrons after an interaction. For the simple example of amonodirectional beam, the uncollided gamma flux within the slab is

o

x x( ) exp( )= 0 (2)

To account for the contribution of scattered photons we introduce the concept of a Buildup Factor B E x( , ) where

the Buildup Factor is defined as

B E x x

x( , )

Total Energy Absorbed at from Scattered and Unscattered Photons of Incident Energy E

Energy Absorbed at from Unscattered Photons of Incident Energy E (3)

The volumetric heat generation rate for this example would then be

= q x E x B E xa( ) exp( ) ( , ) 0 . (4)

The Buildup Factor is an empirical fit to data obtained from detailed radiation transport calculations and is available

in most standard shielding texts.


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62

Attenuation Coefficients (cm-1)

Photon Energy

(Mev)

Water Iron Lead Concrete

0.5 = 0.0966a= 0.0330

= 0.651a= 0.231

= 1.640a= 0.924

= 0.2040a= 0.0700

1.0 = 0.0706

a= 0.0311 = 0.468

a= 0.205 = 0.776

a= 0.375 = 0.1490

a= 0.06501.5 = 0.0574

a= 0.0285 = 0.381a= 0.190

= 0.581a= 0.285

= 0.1210a= 0.0600

2.0 = 0.0493a= 0.0264

= 0.333a= 0.182

= 0.518a= 0.273

= 0.1050a= 0.0560

3.0 = 0.0396a= 0.0233

= 0.284a= 0.176

= 0.477a= 0.284

= 0.0853a= 0.0508

5.0 = 0.0301a= 0.0198

= 0.246a= 0.178

= 0.483a= 0.328

= 0.0674a= 0.0456

10.0 = 0.0219a= 0.0165

= 0.231a= 0.197

= 0.554a= 0.419

= 0.0538a= 0.0416

Table 1: (Photon Attenuation and Energy Absorption Coefficients,from Todreas and Kazimi)

For energy distributions other than monoenergetic, the volumetric heat generation rate is somewhat more

complicated. We consider two cases: one where the incident photon flux consists of a finite number of discrete

photon energies, and the second where the incident photon flux is a continuous spectrum of energies.

Multiple Discrete Photon Energies

For the case of multiple discrete photon energies, the volumetric heat generation rate is the sum of the heat produced

by each incident photon, i.e.

= q x E x B E xi a i i ii i i( ) exp( ) ( , ) 0 (5)

where the subscript idenotes the individual energies.

Continuous Spectrum of Photon Energies

If the incident photon flux is a continuous spectrum of energies, the volumetric heat generation rate is obtained by

integrating over all incident energies, i.e.

=

q x E E x B E x dE a( ) ( )exp( ) ( , ) 00

(6)

where 0( )E contains the incident photon spectrum.


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63

HEAT GENERATION FROM RADIOISOTOPE SOURCES

Radioisotope power sources produce heat as a result of exothermic decay reactions. This heat is then converted

to electricity through thermoelectric devices or other similar means. Charged particles emitted during the decay

process have kinetic energies corresponding to the mass defect of the decay reaction. We again assume that the

kinetic energy of these particles is deposited locally at the point of decay, such that the heat generation rate is

proportional to the spatial distribution of the radioisotope within the power source. This is generally uniform. IfEis the energy associated with the mass defect of the decay reaction, then the volumetric heat generation rate dueto radioactive decay is

=q r E N r ( ) ( )v v

(1)

where: = Decay constant of the radioisotope

N r( )v

= Radioisotope number density as a function of position

Example: (Adapted from Example 4-6,El-Wakil)

A radioisotope power source is fueled with 475 gm of Pu238

C, 100 % enriched in Pu238

. If the density of PuC is

12.5 gm/cm3, calculate the volumetric heat generation rate and total thermal output of the radioisotope source.

SOLUTION

Pu238decays to U234via alpha emission with an 86 year half life, i.e.

94238

92234

24

Pu U He + .

U234has a half life of 2.47 x 105years and relative to Pu238can be considered stable. The mass defect is given by

m =

=

238 0495 234 0409 4 0026

0 006

. . .

. amu

or in terms of energy

E m=

=

=

931

931 0 006

5 586

Mev

amu

Mev/reaction

( )( . )

.

The decay constant is related to the half life through

= = =

= 0 693 0 693

86

0 693

2 7156 102 5519 10

12

9

10 1. . .

. sec. sec

t yr.

The Pu238

number density is given by

N= =

+ =

PuC

PuC

3Av

Mnuclei/cm

( . )( . )

. ..

12 5 6 023 10

238 041 12 013 01 10

2322

such that the volumetric heat generation rate is


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= =

=

q E N ( . )( . )( . )

. sec.

5 586 2 5519 10 3 01 10

4 291 10

10 22

13 Mev/cm3

The total thermal output of the device is

Q q V q m

= =

=

=

=

( . )( . )

.

4 291 10 475 12 5

1 631 10

261

13

15Mev sec

W

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