22.812 NUCLEAR ENERGY ECONOMICS POLICY ANALYSIS S’04

Classnote

Th e Economic s o f th e Nuclea r Fue l Cycle : (1 ) Once-Throug h Fue l Cycle

1 . ​Introduction

 A complex cycle of industrial operations is required to prepare and manufacture fresh fuel for nuclear power reactors and to manage ‘spent’ (irradiated) fuel after it is discharged. The particular characteristics of the nuclear fuel cycle depend on the type of reactor that is being supported. Here we will concentrate mainly on the fuel cycle for light water reactors (LWRs)

 One of the objectives of this module is to develop a simple model for estimating the contribution of the nuclear fuel cycle to the overall cost of nuclear energy. We will not discuss each of the stages of the cycle in great detail, but in each case we will provide some background on current costs and likely trends.

 The basic flowsheet for the LWR fuel cycle is shown below:

Deutsch, John, Ernest Moniz et al. "The Future of Nuclear Power: An Interdisciplinary MIT Study." Massachusetts Institute of Technology, 2003 (ISB N 0 -615-12420-8). Available at http://web.mit.edu/nuclearpower/. p. 101.

 A key distinction is between ‘open’ and ‘closed’ fuel cycles. In the open or once-through fuel cycle, the spent fuel discharged from the reactor is treated as waste. In the closed fuel cycle, the spent fuel is reprocessed, and the products are partitioned into uranium, plutonium, and the residual material, mostly fission products, which is treated as high level waste.

 We begin by considering the once-through fuel cycle.

2. Stage s o f th e nuclea r fue l cycle

 The nuclear fuel cycle can be divided into three stages: the front-end, which extends from the mining of uranium ore to the delivery of fabricated fuel assemblies to the reactor; at-reactor; and the back-end, which starts wiith the shipping of spent fuel offsite and ends with the disposal of high level waste.

Mining:

 Uranium mining is the first stage of the fuel cycle. Uranium ore deposits are found in many parts of the world. The main producing nations today are the United States, Australia, South Africa, Canada, Russia, and other nations of the former Soviet Union.

 Large deposits of uranium ore typically contain only a few tenths of a percent of uranium, although a few very rich deposits in Canada and Australia contain 10-20% uranium. The ore is processed in a uranium mill to produce ‘yellowcake’, a concentrate containing 85-90% by weight of uranium oxide (U 3 O 8 ) . The mill is typically located close to the mine site in order to minimize the cost of transporting the ore. The non-uraniferous material which constitutes the vast bulk of the ore is rejected at the mill. This material, known as the mill tailings , contains most of the radioactive daughter products of uranium that were present in the ore, and must be stabilized to prevent the release of these radioisotopes (including radon gas) into the environment.

​

Conversio n an d Enrichment

 In the next stage of the cycle, the yellowcake is purified and converted to uranium hexafluoride (UF 6 ), the only stable compound of uranium that is volatile at temperatures close to ambient. UF 6 is the feed material for the uranium enrichment stage, in which the weight fraction of the fissile isotope 235 U is increased from 0.711% up to about several percent -- the fissile concentration needed for LWR fuel. Different isotopes of the same element exhibit identical chemical behavior, so considerable ingenuity is needed to devise physical separation means. The isotopic enrichment of uranium is one of the most technically challenging stages of the fuel cycle.

 The two main enrichment technologies in commercial use today are gaseous diffusion and the gas centrifuge process. For several decades gaseous diffusion plants produced almost all of the enriched uranium used in nuclear power reactors, and still today account for most of the world’s enrichment capacity. The process relies on the slight (less than 1%) mass difference between molecules of 235 UF 6 and 238 UF 6 . Gaseous UF 6 is pumped under pressure across a semi-porous diffusion barrier. The lighter 235 UF 6 molecules have a slightly higher probability of diffusing through the barrier, and the gas

on the downstream side is thus slightly enriched in the fissile isotope, while the undiffused gas is slightly depleted (see Figure 1).

x W

Figur e 1 : Schemati c o f a gaseou s diffusio n stage

 The ratio of U-235 to U-238 in the downstream gas rises only by a very small amount, and more than 1000 stages are needed to achieve a U-235 enrichment of 3%.

 The performance of each enrichment stage is described by the separation factor,  , given by the expression:

x P

  1  x P

x W

1  x W

where x P and x W are the weight fractions of U-235 in the enriched and depleted product streams respectively. The stage separation factor for a gaseous diffusion stage is 1.00429. (An analogous separation factor is used to characterize other isotope separation processes too.)

 The stages are arranged in a ‘cascade’, in which the enriched product from one stage becomes the feed to the next highest stage, while the depleted product becomes the feed to the next lowest one. The feed stream is

introduced into a central stage of the cascade, while the enriched product and depleted ‘tails’ streams are withdrawn from each end (see Figure 2).

Enriched product, P kg x p

Tails, W kg x x

Figure 2: An enrichment cascade An overall material balance on the cascade yields:

F = P + W eq. (1)

and a material balance on the U-235 isotope leads to

Fx F = Px P + Wx w eq. (2)

where F, P, and W are the masses of uranium in the feed, product, and tails streams respectively, and x F , x P , and x w are the weight fractions of U-235 in the three streams.

In these equations, P and x P are determined by the in-core fuel management scheme; x F is given by the U-235 content of natural uranium (0.711%); and x W is set to optimize enrichment plant operations. Thus we have two equations in two unknowns (F and W). Solving for F, we get that:

F  P  x p  x W ˘

  x F  x W  

Example: For a cascade enriching natural uranium to 3% in U-235 at a tails assay, x w , of 0.2%, solving equations (1) and (2) gives:

 x P  x W 

F  P    5.48P

 x F  x W 

i.e. , to produce one kilogram of 3% enriched uranium product requires about 5.5 kilograms of natural uranium feed.

 Gaseous diffusion plants are extremely large and very capital intensive, and use large amounts of energy. A full-scale gaseous diffusion plant can consumes 2000-3000 megawatts of electric power, enough to meet the needs of a city of half a million or more people. Commercial-scale plants are today operating in the United States, Russia, and France.

 The gas centrifuge process, the other leading enrichment technology, also relies on the small mass difference between molecules of 235 UF 6 and 238 UF 6 . In this case the separation is achieved in ultra-high-speed centrifuges. UF 6 gas introduced into the centrifuges is subject to centrifugal acceleration thousands of times greater than gravity. The heavier 238 UF 6 molecules tend to congregate at the centrifuge wall, while the gas at the axis is enriched in 235 UF 6 . The overall separation factor in an optimally designed centrifuge is roughly 1.4 – much higher than in a gaseous diffusion stage. However, the throughput of each machine is small, because of materials and mechanical constraints that limit the size of the centrifuge and its rotation speed. To produce commercial-scale quantities of enriched uranium tens or hundreds of thousands of centrifuges must therefore be piped together in a cascade. Gas centrifuge cascades are even more capital- intensive than gaseous diffusion plants, but only consume about 5% of the energy. An Anglo-Dutch-German consortium operates the only full-scale gas centrifuge enrichment plants in service today. USEC, the American enrichment corporation, has announced its intention to build a new gas centrifuge plant to replace its aging gaseous diffusion plant at Paducah, Kentucky.

Fue l Fabrication

 In the fabrication stage, the UF 6 is first converted to UO 2 and the UO 2 is then formed into pellets, the pellets are sintered, and then stacked into zircaloy tubes. Sufficient space is left in the tube for the fission product gases to accumulate without overpressurizing the tube, an end-cap is added, and the tube is sealed. The tubes (or rods) are then fastened together to make

assemblies. In a typical PWR, the rods are assembled into a 17x17 square array.

[Note: A useful technical presentation on the stages of the front-end of the cycle prepared by Argonne National Laboratory can be found at: http://web.ead.anl.gov/uranium/guide/prodhand/sld001.cfm ]

Reacto r Operatio n - - Irradiation : Batches , Cycle s an d Energ y Generation

 A typical PWR core has

 ~ 200 assemblies (~ 12 feet long)

 ~ ~300 rods/assembly (~ 0.5” diameter)

 ~ 200 fuel pellets/rod  ~ 8,000,000 pellets

 Nuclear fuel is typically loaded in staggered ‘batches’ consisting of 1/n th of the total number of in-core assemblies (the ‘batch fraction’). Typical values of n are 3 or 4.

 The period between refueling outages is called the ‘refueling cycle length’, T c .

 The energy extracted from a given batch of fuel is expressed in terms of the fue l burnup , B, reported in MWD (thermal) per Metric Ton of Initial Heavy Metal (or MWD(th)/MTIHM)

 Example: n = 3

​

Cycl e No. Batc h #

I II III IV

Startup core

2	3	4 (new)

3	4	5 (new)

4	5	6 (new)

V

Discharged after 1 cycle B d ~B c *

Discharged after 2 cycles B d ~2 B c *

Discharged after 3 cycles B d ~3 B c *

* Useful approximation. In practice, need to track B d with computer codes

Utilities prefer to have cycles that are multiples of 1/2 year in length, so as to be able to match their refueling outages with periods of low power demand (spring and fall in most parts of the country.) For many plants, T c has been 1 year in the past, but utilities are rapidly switching to longer cycles (18 months, or two years.)

 If the batch fraction is 1/n, under steady state conditions each batch remains in the core for n cycles.

 Similarly, at steady state the energy produced by all n batches in the core during one cycle is equal to the energy produced by one batch during its total residence time in the core (i.e., n cycles.) .

 Thus the total electrical energy produced by a given batch during its in-core lifetime at steady state is:

E b (kwhr (e)/batch) = 8766 (hrs/yr) x CF x K (kwe) x T c (yr)

where:

CF = cycle average capacity factor (including refueling downtime) K = plant rating (kwe)

T c = cycle length (yrs) including downtime

 We can also write that the energy produced per batch is:

E b (kwhr (e)/batch) = B d (MWD(th)/MT) x 24 (hr/day) x 1000 (kw/MW) x  x P (MT) where:

B d = discharge burnup of the fuel (MWD(th)/MT of heavy metal)

 = thermodynamic efficiency (Mwe/MW(th)) P = batch fuel inventory (MT of heavy metal)

Also,

T b = n T c B d = n B c

And P, the fuel inventory per batch = Total core inventory/n

 Note, for a batch fraction of 1/n, steady state is achieved after n cycles, to a first approximation.

3. Materia l Balanc e o n th e Front-En d o f th e Nuclea r Fue l Cycle

 Three key functions:

 Electric power system manager

 In-core fuel manager

 Out-of-core fuel manager

 The system manager specifies the required output from the power plant (capacity factor, refueling interval)

 The in-core fuel manager provides the specifications for each fuel batch, using in- core physics and fuel management codes to meet target energy production.

 The out-of-core fuel manager is responsible for timely delivery of each fuel batch, which involves purchasing fuel and fuel cycle services.

 Example : Assume the following steady-state specifications.

K = 1000 MWe

T c = 1.5 years CF = 90%

 = 0.33

n = 3

With this information, we can calculate the size of a steady-state fuel batch, since as noted previously the energy output per batch over its in-core lifetime = energy output of entire core over one cycle.

Energy output per batch = 1000 (MWe) x 365 x 1.5 x 0.9 (days/cycle) x 1/0.33 (MWD( th)/MWD(e))

= 1.49 x 10 6 MWD(th) per batch

And if B d = 50,000 MWD( th)/MTHM, we have that the mass of heavy metal in each batch, P, is

P = 1.49 x 10 6 (MWD( th))/50,000 (MWD (th)/MTHM)= 29. 8 MTHM

The fuel manager’s physics codes will also calculate the initial enrichment of

uranium-235, x p , that is required to attain the desired discharge burnup, B d . This can be usefully approximated by the following correlation (presented in Zhiwen Xu’s doctoral thesis (2003)), which is valid for enrichments up to 20%:

 n  1   n  1 

x p  0.41201  0.11508   2 n  B d   0.00023937   2 n  B d 

   

And for B d = 50,000 MWD( th) /MT and n=3, x p =4.51%.

With this information, we can work back through each of the stages in the fuel cycle to obtain the amount of uranium ore that must be mined for each batch (see Figure 3).

29.8 MTHM of L EU

4.51% U-235

1000 MWe PWR

Cap. fact.= 90%

Spent fuel

1% losses

29.8/0.99 = 30.1MTHM of U

Therm. eff.=33% Discharge burnup =

50,000 MWD/MTHM

30.1 x (270/238)

= 34.15 MT UO 2

34.15 352

.995  270  44.74 MT UF 6

UF 6

Product, P X P =4.51%

0.5% losses

Feed, F X F =0.711%

Nat.UF 6

F  4.5 1  0.3  44.74

.711  0.3

 458.3 MT nat. UF 6

Yellowcake

458.3



842

0.5% losses

3  367.3 M T U O  367.3  238

 311.4 M T U

Tails, W X W = 0.3%

(U 3 0 8 )

.995

352

842

3

458.3 - 44 .74 =413.6 MT UF 6

Mill tailings

Uranium ore

Figure 3: Materia l balanc e o n th e fron t en d o f th e PW R fue l cycle (Basis: 1 steady state batch; 1000 MWe PWR; thermal efficiency = 33%; 90% capacity factor; 18- month refueling cycle; batch fraction = 1/3; discharge burnup = 50,000 MWD(th)/MT)

4 . ​Simpl e Cos t Mode l fo r A Singl e Fue l Cycl e Batch

 The operations associated with each batch of fuel typically extend over many years, from mining the uranium ore to finally disposing of the high level waste. Payments for these various operations are made at widely differing times. Thus it is important to take into account the time value of money in calculating the overall fuel cycle cost.

 Recall also that nuclear fuel is not permitted to be expensed for tax purposes, but must be capitalized and depreciated (like buildings or machinery)

 The task is to calculate the revenue stream that is equivalent in a present worth sense to the series of payments on the fuel batch.

 To make this easier, we will assume that the revenues are received as a single cash flow, R b , occurring at the midpoint of the in-core fuel irradiation lifetime. We will also assume that the taxes, T, are paid at the midpoint of fuel irradiation.

R b

	0	T b

I 1 T

I 3

I 2

 We can transform this into the equivalent ‘tax-implicit’ problem:

R b (1-  )

 D

	0	T b

I 1

I 3 Discount

rate , x

I 2

 Next, calculate the revenue requirement, R i , for each I i .

R i (1-  )

0  I i  (1   )R i e  x  T i   De  x  T i

where  T i is the time from cash outlay to irradiation midpoint

Note also that D  I i

 R i

 I ˘

 1    e

x  T i     ˘

1  

   

 I i  1  x  T    

1   i

 I  x I i  T

i 1   i

 I i  I i    T i

==========

i.e. , the revenue requirement = direct cost + carrying charges for  T I years.

Next, for each fuel cycle transaction, I i , we can write:

I i = M i x C i

where

M i = mass processed at stage i C i = unit cost of transaction i

we can write that the total cost

 M i C i   M i C i     T i

and the total batch cost

  M i C i    M i C i     T i

i i

===================

Question : Ho w accurat e i s thi s approximation?

We can compare it with the more exact expression for the levelized annual revenue requirement for capital charges we derived previously.

Recall:

And the levelized annual revenue requirement, R L is given by

R L   I o

where, for straight line depreciation

    1    ( A / P , x %, N ) 

  ( 1 

  I N  A / F , x %, N  ˘

SLD

 1     



 I o  

Compare this with the approximate expression derived above:

We can write that the revenue, R, required to balance this investment if received in a lump sum at N/2 is given by:

R = R o + R N

Where, from the above approximation, we can write that

R o ≈ I o + I 0   (N/2)

R n ≈ -I N + (-I N )   (-N/2)

Therefore,

R  I

   1 

  

( N / 2 )   1   ˘

o   

  

  

The annual revenue requirement can therefore be approximated by R/N,

R I o



   1 

  

( N / 2)   1   ˘

N N   

  

  

For x=0.1; I o /I N = 0.1; and  = 0.4, we have

	Exact expression,  SLD I 0	Approximate expression, R/N	Error
N=20	0.1629 I o	0.1367 I o	~ 19%
N=5	0.293 I o	0.2717 I o	~ 8%

Fue l Cycl e Cos t fo r Once-Throug h Cycle

We can use the approximate cost model for a single fuel cycle batch to estimate the fuel cycle cost for a PWR operating on the once-through fuel cycle according to the material balance shown in Figure 3.

NOTE: Some minor transactions such as chemical conversion of UF 6 to UO 2 and transportation have been included in the price assigned to a contiguous major transaction.

Calculatio n o f Once-Throug h Fue l Cycl e Cost :

Basis: 1 kg of 4.51% enriched uranium (see fuel cycle material balance)

Transaction	Unit Cost, C i ($/kg)	Mass Flow, M I (kg)	 T I (years)	Direct Cost, M I C I ($)	Carrying Charge, M I C I    T I
Ore purchase	30	10.45	4.25	313.5	133
Yellowcake	8	10.45	4.25	83.6	35.5
conversion
Enrichment	100 ($/kg SW)	6.23 kg SW	3.25	623	202.5
Fabrication	275	1.01	2.75	277.8	76.4
Interim SF	100	1.0	-2.25	100	-22.5
storage
Final disposal	400	1.0	-2.25	400	-85
TOTAL				1797.9	339.9
GRAND $2137.8/kg 4.51% U TOTAL

(Note: We have assumed   = 0.1/yr)

We can obtain the fuel cycle cost in c/kwh (e) as follows

Fuel cycle cost (cents/kwh(e) = 2137.8 ($/kg U) x 1000 (kg/MT) x 1/50,000 (MTHM/MWD) x

1/24 (days/hr) x 1/ 1000 (MW/kw) x 1/0.33 (kwh(th)/kwh(e))

= 0.5 4 cents/kwh(e)

( Note: This is not a levelized cost over the reactor lifetime; it is the fuel cycle cost for the specified batch.)