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Gross Energy Available through Light Water Reactors
by John W. Gofman, Ph.D.
C.N.R. Report 1977-2, May 1977
This CNR Report addresses the question of the maximum possible contribution
(gross) to the U.S. energy supply via light water nuclear reactors. The
nuclear promotional industry, inside and outside of the Carter Administration,
has made it appear that light water reactors are sorely required, if we are
to meet our energy "demand". The trivial energy contribution possible via
light water technology is evident in the tabulations of this report.
Considered here are:
Low, medium and high estimates of workable-grade
uranium available in the U.S.
The electrical yield (kwhrs/e) gross per short ton U3O8
mined, with all supporting assumptions and calculations.
Lifetime U3O8 requirement per 1000-megawatt light water
reactor.
Quads of energy (thermal and electrical) available
through LWR technology; also in barrels of oil-equivalent.
The number of light water reactors fuelable within the
estimated U3O8 supply.
The energy which would be lost to the economy if no
further nuclear plants were initiated.
Except for explicit use of various estimates of the fuel supply, the
calculations throughout this report are based upon the optimistic assumptions
of the nuclear industry, in order to present the most favorable case for
nuclear power. Even with this approach, nuclear power via light water
technology is a trivial source of energy.
Following the tables is Appendix #1 which shows the calculations and
assumptions from which the tables are derived.
Appendix #2 discusses some fundamental terms in the fuel issue, like ppm,
MTU, yellowcake, etc.
Committee for Nuclear Responsibility, Inc.
P.O.B. 11207, San Francisco, California 94101
-2
References:
1.
Starr, Chauncey, "On the Distribution of Fissions in Light
Water Reactors", Letter, dated May 12, 1976, to Conunittee for
Nuclear Responsibility. (Dr. Starr is President of the Electric
Power Research Institute, POB 10412, Palo Alto, CA 94303).
2.
Perry, A.M., Letter, dated February 2, 1976 to Conunittee for
Nuclear Responsibility. For extensive details, see the
comprehensive report, "Net Energy from Nuclear Power",
Rotty, Ralph M; A.M. Perry, D.B. Reister, Institute for Energy
Analysis, IEA-75-3, November, 1975, Oak Ridge Associated
Universities, Oak Ridge, TN.
3.
Wilde, R.M., Nuclear Regulatory Conunission Staff Testimony
(quoted by P.M. Wood) in Hearings before the Atomic Safety and
Licensing Board in the Matter of Gulf States Utilities Company
(River Bend Station, Units 1 and 2), Docket Nos. 50-458,
50-459, second half of 1976.
4.
Nuclear Regulatory Conunissian Staff Tes.l:irnony, ibid. (Presented
l:5y P.M. Wood).
5.
Lieberman, M.A., "United States Uranium Resources ---An Analysis
of Historical Data",
Number 4238, 431-436,
Science 192,
April 30, 1976.
6.
"The National Energy Plan", April 29, 1977, Executive Office of
the President, Energy Policy and Planning, Washington, D.C. 20500.
(See also "Uranium Reserves, Resources, and Production", Energy
Resources Council, June 15, 1976).
No permission is requi.red to reprint this report in whole or in part.
-3
Table No. 1
Number of 1000-Mega..att/e Plants Fuelable
INCLUDING those already built
Because the actual size of nuclear plants varies so much, for clarity one should discuss
a standard size like 1000-megawatts (electrical).
Based on Assured Fuel Supply of 640,000 tons U 3 Ow
Plants at 70% Plants at 55%
Capacity Factor Capacity Factor*
Fuel performance at 100% of
its "theoretical" yield**
105.5 plants 134.3 plants
Fuel performance at 75% of
its "theoretical" yield
79.1 plants 10-0. 8 plants
Fuel performance at 50% of
its "theoretical" yield I 52.8 plants I 67.2 plants
Probable Fuel Supply of 1,130,000 tons
(Ref. 5).
U 3 01
Fuel performance at 100% of 186.3 plants 237.1 plantsits "theoretical" yield
Fuel performance at 75% of 139.7 plants I 178.0 plantsits "theoretical" yieldá I
Fuel performance at 50% of
93.2 plants 118.5 plants
its "theoretical" yield
Exaggerated Fuel Supply of 1,840,000 tons UJo(Ref. 6).
i
Fuel performance at 100% of I 303.3 plants 386.1 plantsits "theoretical" yield
Fuel performance at 75% of 227.4 plants 289.8 plantsits "theoretical" yield
Fuel performance at 50% of
151.7 plants 193.0 plants
its "theoretical" yield
*
At 55% capacity factor, more plants (highly inefficient) can be built than at 70%
capacity factor, provided that refueling schedules are altered to prevent premature
unloading of "unburnedá" fuel.
** "Theoretical" yield = 3.033 x 107 kwh(e) per short ton U30g mined.
NUMBERS IN BOXES:
See reverse side of this sheet.
-4-
Table No. 1, continued
NUMBERS IN BOXES
52.8:
Since approximately 50,000 megawatts are already built, if fuel performance remains at 50% of theoretical yield, then
no more nuclear plants can be fueled with the known 640,000-ton
reserve.
U308
139.7=
1
This number turns out to correspond closely with the number of nuclear megawatts which the National Energy Plan (Ref. 6) proposes to have in operation by 1985. See Table 3. The Pian is obviously risky with respect to fuel for 140,000 Mw. If fuel performance and capacity factors remain as they are now (50% and 55% respectively), then only 118,700 megawatts will be fuelable, even if 1,130,000 tons of fuel are found.
303.3
1
:I This number corresponds closely to recent statements by ,James Schlesinger that he is thinking of 300 presum..bly fuelableLWR's on line by the end of this century. This table shows that such talk is based on wildly optimistic assumptions about the fuel supply and fuel performance.
NOTE:
Not all of the uranium discovered in this country, and to be discovered in this country, belongs to the United States!
According to the General Accounting Office (GAO) report to Congress of July 7, 1976, entitled "Certain Actions that Can Be Taken to Help Improve This Nation's Uranium Picture", no one knows what fraction of U.S. uranium is already owned and will be owned by foreign investors. In1974, at least 10% of all uranium exploration in this country was done
by companies which were wholly owned by foreign companies or countries, with the right to export the fuel. {p.18).
Additionally, some domestic companies are conducting joint ventures with foreign companies for uranium exploration. Under one such agreement, half the uranium discovered is to be controlled by the foreign investors
(p.19).
"According to an ERDA official, foreign countries are exploring for uranium in this country because they believe there is more opportunity to recover and export uranium from the U.S. than from other countries which have rigid export requirements" (p.19).
Table No. 2 -5-
Energy Contributions from Nuclear Power
Based on Assured Fuel Supply of 640,000 tons U 3 O 8
Total Quads nuclear Average Quads nuclear
over 30 years per year
Fuel performance at 100% of 199.4 66.3 6.6 2.2
its "theoretical" yield thermal elec. * thermal elec.
Fuel performance at 75% of 149.4 49.7 5.0 1. 66
its "theoretical" yield thermal elec. thermal elec.
Fuel performance at 50% of 99.8 33.2 3.3 1.11
its "theoretical" yield thermal elec. theármal elec.
Share Possible from Nuclear, if U.S. Energy Consumption :_ 91.65 Quads per Year
In 1976. !0181 U.S. primary energy demand: 75 Ouads. Carter's prediction 10< 1985: 91.65 Quads.
Thermal Quad basis Elec. Quad basis
Fuel performance at 100% 7.2% 2.4%
Fuel performance at 75% 5.5% 1. 8%
Fuel performance at 50% 3.6% 1.2%
Probable Fuel Supply of 1, 1 30,000 tons U 3 0 8
Fuel performance at 100% of
its "theoretical" yield
Fuel performance at 75% of
its "theoretical" yield
Fuel performance at 50% of
its "theoretical" yield
Total Quads nuclear
over 30 years
352.l 117.0
thermal elec.
264.0 87.7
thermal elec.
176.1 58.5
thermal elec.
Average Quads nuclear
per year
11. 7 3.9
thermal elec.
8.8 2.9
thermal elec.
5.9 1. 95
thermalá elec.
Share Possible from Nuclear, if U.S._...!'.ergy ..onsumptlon: 91.65 Quads per Year
In 1976: !0181 U.S. primary energy demand .a:......- Carte<'s pre<11ct1on 10< 1985 :=, 91.65 Quads.
Fuel performance at 100%
Fuel performance at 75%
Fuel performance at 50%
Thermal Quad basis
12.8%
9.6%
6.4%
Elec. Quad basis
4.3%
3.2%
2.1%
áProducing etectncity t0< purposes whef9 electricity 1s reQU1red ;...ritl cnlditing as "the!Tnaf'.' Q.iada. since olhet'_ fl!els would olhefwise be YSed at their them11t valYe.
Producing etec1ricity fa l)Ufl)(9el wh- ellciJicity is NOT reQUired. but is mef9fy IJS8d fa il9 hetlt '1111"9 to replace Olhef fuels, deser,es crediting ONLY as "electrical" Quads
*
-6- Table No. 2, continued
Exaggerated Fuel Supply of 1,840,000 tons U 3
07
Total Quads nuclear Average Quads nuclear
over 30 years per year
Fuel performance at 100% of
its "theoretical" yield
573.0
thermal
190.4
elec.
19.l
thermal
6.3
elec.
Fuel performance at 75% of
its "theoretical" yield
429.8
thermal
142.8
elec.
14.3
thermal
4.8
elec.
Fuel performance at 50% of
286.7
9.6
3.2
its "theoretical" yield thermal elec. thermal elec.
Share Possible from Nuclear, If U.S. Energy Consumption: 91.65 Quads per Year
In 1976, total U.S. primary energy demond :. 75 Quads. Carte(s predicllon lat 1955: 91.65 Quads.
Thermal Quad
Fuel performance at 100% 20.8%
Fuel performance at 75% 15.6%
Fuel performance at 50% 10.5%
basis Elec. Quad
6.9%
5.2%
3.5%
basis
Table No. 3
Jimmy Carter's Energy.Plan proposes (p.71) to have a total of 138 nuclear
plants (of various sizes) in operatiort as of 1985 without specifying the
combined generating capacity in megawatts. However, the Plan does reveal
the expected thermal energy (gross) to be yielded in 1985 by these plants:
7.71 Quads. From that figure, one can derive the total number of nuclear
megawatts (or number of 1000-megawatt plants) which are proposed (below).
Relationship of Thermal Quads
to Megawatt Design-Capacity of Nuclear Plants
Capacity Factor Megawatt Annual Quads Number 1000-meg (e) Plants
Rating thermal from a Required to Produce
sin9:le plant 7.71 Quads (t) per Year
70% 1,000 Mw 0.063 Quads 122.4 Plants
60% 1,000 Mw 0.054 Quads 142.8 Plants
55% 1,000 Mw 0.0495 Quads 155.8 Plants
At recent Nuclear Regulatory Commission hearings, expected capacity
factor has been stated by the NRC as 60%,
.-
Table No. 4 -7
Energy to Be Lost to the Economy
if no ADDITIONAL nuclear plants are initiated
Assumptions:
1.) 1,130,000 tons U309 are the ultimate U.S. reserve from suitable ore.
2.) Fuel performs at 75% of its theoretical yield in LWR's.
3.) Plants operate at 70% of their rated capacity.
4.) Enrichment "tails" are 0.2% instead of 0.3%.
Total number fuelable 1000-Mw plants = 139.7, or 140,000 Mw (Table 1).
BUT many of them have already been built. We now have 63 plants
"operable" with a combined capacity of about 50,000 Mw, plus about
30,000 additional Mw underway = 80,000 Mw.
140,000 Mw minus 80,000 Mw operable and underway = 60,000 Mw.
Thus only 60 ADDITIONAL plants @1000 Mw can be considered at all. And
even these are risky with respect to fuel supply.
Since 140 plants @1000 Mw would consume the entire 1,130,000 tons of
fuel, .. decision NOT to build 60 of them (a "ban" on new nuclear
¥
construction) would mean a maximum loss of 43% (60.,-140) of the
á energy which 140 plants could deliver. 140 plants could deliver
8.8 thermal Quads per year, because that is the yield from 1,130,000
Thus Quads per year lost by a "ban". = 43% of 8.8 Quads = 3.8 Quads.
Annual U.S. energy consumption in 1985 = 91.65 Quads (Carter Plan).
3.8 Quads/yr lost by a nuclear "ban" = 4.1% of total (3.8+.91.65).
Conclusion:
Loss of 3.8 Quads/yr from initiating NO additional nuclear plants is
a negligible 4.1% loss.
Increment of 3.8 Quads/yr from new nuclear construction would solve
only 4.1% of the country's supply problem.
This utterly trivial amount of energy can not possibly justify the
common scaremongering about the effects of a ban on new nuclear
constructi0n! In fact, even this trivial increment of energy will be
non-existent if fuel-performance remains poor, or if less than
1,130,000 tons of fuel are found.
Appendix 1 -8
The Electrical Yield in Kwhrs (e) per Short Ton U1 Mined:á
0 0
There are two methods for estimating the "theoretical" yield
of electrical power (via light water reactors) per ton of U30a mined.
is based upon assuming that some fraction of
U-235 in the reactor is utilized in one pass through the reactor
(..reprocessing). That fraction "burned" is generally quoted as
between 70% and 73.3%. We shall utilize the mid-range value of 71.7%.
Then it is necessary to know how many fissions ("bonus fissions")
are contributed by nuclides other than U-235. The other potential
contributors are U-238, Pu-239, Pu-241. Starr (Ref. 1) has estimated
that under the best conditions of fuel burn-up, the distribution of
fissions is:
52 from U-235
40 from Pu-239 and Pu-241 combined
8 from U-238.
This means that the energy per U-235 fission must be supplemented
by 48 -+ 52 , or O. 923 for energy from the combined fissions of
,U-238
plus plutonium nuclides.
The second method is more empirical and is based upon the best
yield of energy observed for light water reactors
their usual
operating yield). Perry (Ref. 2) has suggested 30,000 megawatt-days
thermal per metric ton of 3.0% enriched fuel, barring premature
fuel-discharge due to fuel failures. This value can be translated
directly to the "theoretical" or "design" expectation of energy,yield.
We shall develop both methods here.
Method 1.
Step 1 : Losses in milling
Ore is mined containing one short ton u309. Wilde (Ref. 3) has
The first method
(not
-9
suggested that "mill losses may conservatively be taken
at 10%". The lost uranium-235 ends up in the mill "tailings".
Therefore, per short ton mined, we end up with 0.9
u3o8
short tons U309 at the end of the milling step.
Step i: Conversion to UF and back to UO 2.
_
It is estimated that the combined losses in these steps is
1.5% of the fuel.. Therefore, we end up with 98.5% of initial
uranium.
:. {0.985) x {0.9) = 0.8865 short tons equivalent.
u3o
8
Step 3: Losses to "tails" during enrichment
In preparing 3.0% U-235, some U-235 is left in the "tails" of
the enrichment plant. Currently as much as 0.3% U-235 is the
concentration in the "tails". Since this wastes uranium, we shall
assume {to give nuclear power every advantage) that additional
enrichment facilities will be built if the U.S. goes on with
nuclear expansion, that we will then leave the "tails" at 0.2%
rather than 0.3% {which produces a 19.8% saving of fuel), and on
thatá'.basis, we calculate the loss of U-235 in enrichment tails:
By the law of conservation of mass, the U-235 comes out of
the enrichment plant either in the enriched product or in the tails.
Let us assume 100 tons of uranium go into the enrichment step.
Let x = tons of enriched uranium produced {3% U-235)
100 -x = tons of depleted uranium in tails (0.2% U-235)
0.0071 = fraction of feed uranium that is U-235 .
.. á¥ 100 (O. 0071) = 0. 03x + (0. 002) (100 -x). Conservation of U-235.
0..71 = 0.03x + 0.2 -0.002x
0.028x = 0.51
100 -x = 81.8 tons.
We started with 100 x 0.0071, or 0.71 tons U-235.
In the enriched fraction, we now have:
(18.2) (0.03) = 0.546 tons U-235
So, 0.546 = 0.769, or 76.9% of the U-235 survives the enrichment
0.71 step.
x = 18.2 tons;
-10
From step 2, we had left 0.8865 short tons of with its
U308
U-235 content. Since we say that 76.9% of the U-235 survives the
enrichment step, it is the same as saying:
(0.769) x (0.8865), or 0.6817 short tcms u309 equivalent
get into the reactor cycle.
Step 4: Energy per short ton U ::.Oin the reactor step
r
l short ton U309 .. 2000 pounds U309
1 pound U309 ---:,, 454 grams U309
¥
..
..
1 short ton U308 .. 908,000 grams U309
.. .
.
..
The uranium fraction of U309 is 0.848
1 short ton U309 .. (908,000) (0.848), or
769,984 grams uranium.
U-235 is 0.0071 of natural uranium.
.
. . l short ton u309 .. (7.69,984) (0.0071), or
5466.9 grams U-235
from step 3, our original short ton of 0309 is down to
0.6817 short tons after enrichment and reconversion to uo2 fuel
(3.0% U-235). Therefore:
U-235 entering reactor cycle = (0.6817) (5466.9), or
3726.8 grams U-235
Step 5: u 23s Utilization In the reactor
The estimate is (in the introduction of this Appendix) that
71.7% of the U-235 gets utilized in the reactor, if there is no
premature unloading of the fuel. Therefore:
(3726.8) (0.717)
=
2672.1 grams U-235 utilized in the reactor
per original short ton of u309 mined.
Step 6: Fraction of U235 undergoing fission
Some U-235 is utilized in a non-productive manner, by
capturing neutrons to produce non-fissionable U-236. The remainder
But,
-11
is fissioned. Starr (Ref. 1) estimates for the U-235
useaáup , 81% of the U-235 fissions, while 19% does £2!,
because it goes to U-236. Therefore:
(0.81) (2672.1), or 2164.4 grams U-235 undergo fission per
short ton of U309 mined.
Step 7: Energy from fission
The fission of one gram U-235 yields 0.92 megawatt-days of
thermal energy (from first principles of physics).
1 megawatt-day----.:,. 24,000 kilowatt-hours (thermal)
¥¥ Fission of 1 gram U-235 .. (0.92) (24,000), or 22,080 kwhrs (t)
Therefore:
Fission of 2164.4 grams U-235 --.:..) (2164.4) (22,080), or
4.779 x 107 kwhrs, thermal.
But, for
every U-235 fission, we have 0.923 "bonus fissions".
:. (0.923) (4.779 x 107) = 4.411 x 107 kwhrs thermal from
bonus fissions.
Final total thermal yield
= 4.779 x 107 + 4.411 x 107
= 9.19 x 10
7
kwhrs
thermal
from our original short ton
of mined.
U308
Step_ 8: Thermal to electrical conversion
Generally a value of 0.33 is taken as the electrical energy
obtained per thermal energy-unit produced in light water reactors.
:. (9.19 x 107) (0.33) = 3.033 x 107 kwhrs
from our original short ton
of u3oa mined.
This value, 3.033 x 107 , or 30. 33 million kilowatt hours
electrical per short ton u3oa mined is what we shall call the
"theoretical" yield in the light water reactor.
electrical
-12
Theoretical Energy Yield per Ton vs. Actual Yield:
The figure 30.33 million kwhrs (e) is "theoretical" in the sense
that this is the energy yield to be obtained if all design criteria
are fulfilled.
Thus if 7 1.7% of the U-235 is not utilized or "burned", the
yield will fall. If the "tails" in enrichment are 0.3% instead of
0.2%, the yield will fall. If the "bonus fissions" have been
overestimated (and they may be), the yield will fall.
It appears that the actual fuel performance on the average
to date has been approximately 50% of its theoretical yield,
according to testimony and tables presented by the Nuclear
Regulatory Staff in the second half of 1976 (Ref. 4).
Method 2.
Direct Estimate from Claimed Empirical Results_
Perry has estimated 30,000 megawatt-days, thermal, per metric
ton 3% enriched uranium, barring premature discharge of fuel.
1 metric ton.. 1,000 Kg.
At 3% U-23 5, (1,000) (0.03)= 30 kilograms U-235, or
30,000 grams u-235
In step 4 of Method 1 (above), we estimated that 3762.8 grams
U-235 at 3% enrichment enter the reactor cycle per short ton
U308
mined.
Therefore, if 30,000 grams U-235 --..') 30,000 megawatt-days
thermal, it follows that
3762.8 grams U-235 -..:> 3762. 8 x 30,000 = 3762á.8 megawatt-days
30,000
thermal,
per original short ton mined.
u3o 8
1 megawatt-day .. 24,000 kwhrs
.'. (3762.8) (24,000) = 9.03 x 107
kwhrs thermal
per original short ton mined
U308
Utilizing the factor of 0.33 for conversion of thermal to
electrical, we have
(0.33) (9.03 x 107) = 2.98 x 107
kwhrs electrical per
original short ton uo
mined ..¥
38
in good agreement with Method 1.
-13
Confusion over the term "megawatt-days of burn-up"
Some observers carelessly use megawatt-days of "burn-up" to
describe fuel performance in reactors. It is extremely important
to point out that 30,000 megawatt-days thermal is a design value
which applies if there is no premature discharge of fuel and if we
start with 1 metric ton of 3.0% enriched uranium. It should be
self-evident that if lower or higher enrichment is utilized (and
both occur frequently in commercial light water reactors), the
design expectation is different from 30,000 megawatt-days thermal
per metric ton áof fuel metal.
It is therefore far preferable to speak of "fuel duty" or
performance in terms of kwhrs per original short ton of mined,
u3o8
than to speak of megawatt-days per metricá ton of enriched fuel, the
latter being so dependent upon extent of enrichment ¥
. Reconciliation. with Our erevious Estimates:
In 1976, C.N.R. presented some estimates of the gross electrical
yield, kilowatt-hours (e), per short ton u3og. Four parameters have
been revised in this report.
(1) ¥ The fraction of U-235 going to fission (vs. "duds"ágoing to
U-236) is reduced from 85% to 81%, áwhich confo:t'!llls with the "dud"fraction
(19%) used by the Electric Power Research Institute (Ref. 1).
(2). The bonus fissions from U-238 plus plutonium nuclides are
here taken as 92.3% of U-235 fissions instead of 43% of U-235 fissions,
because that appears to conform with the values observed for burn-up
at design levels (Ref. 1).
(3). The loss of U-235 in enrichment tails is here taken as 0.2%,
whereas earlier we used 0.3% tails. Actually, 0.3% corresponds better
with current practice, but. to present nuclear power prospects most
favorably again, we are assuming sufficient additional enrichment
capacity will be'built to permit going to 0.2% tails---a development
which should increase energy-yield per short ton
u
3
o
8 mined by a
fuLl 19 .,8%.
(4). An estimated loss of 10% of the u3o
in mining-plus-milling
8
is used in this report, whereas this loss was not included earlier.
-14
Lifetime U 0Requirement per 1 COO-megawatt Reactor:
31
If we are to calculate the number of plants fuelable by any
estimated supply, it is necessary to know the
u
requirement
u
3
o
8 3
o
8
for the lifetime of one plant.Three factors are essential to specify:
(1). The expected average capacity factor. If a plant could
operate at full power every hour of every day, it would be operating
at 100% "capacity factor". Expected capacity factors are discussed
as percents, which makes it possible to confuse these values with
fuel performance at some percent of its theoretical energy yield.
These are not the same.
(2). The lifetime of the plant. In 1976, we assumed a 40-year
lifespan for nuclear plants; the nuclear industry no longer expects
them to last 40 years. So this report uses 30 years. Therefore each
plant will use only 7 5% as much fuel as calculated earlier. (However,
the effective capital cost of nuclear power increases by 33% with this
adjustment) .
(3). Fuel performance relative to its theoretical yield.
a.) 30-year lifetime; 70% capacity factor ; fuel performance 100% of theoretical
(0.7) (1000 Mw) (1000 kw) (24 hrs) (36 5' days) (30 years)
.
Mw day year
(0. 7) (106 ) (8.760 X 103) (30) )'
1.84 x 1011 kilowatt-hours output in 30 years.
Theoretical yield per short ton mined = 3.033 x 107 kwhrs (e)
u3o
8
Therefore under these conditions, each plant requires
1.84 X 1011
or 6,06 7 short tons
u3o
8
b.) 30-year lifetime; 55% capacity factor ; fuel performance 10 0% of theoretical
Obviously if the nuclear generating plants operate much more
poorly than designed, e.g., at 55% capacity factor instead of 70%, the
U309 requirement per plant for 30 years will be less, provided the
refueling schedule is revised to take the much poorer performance
into account.
3.033 x 107
-15
At 55% capacity factor, electrical output in 30 years is
(0.55)
(1000 Mw) (1000 kw) (24 hours) (36 5 days) (30 years) >
Mw day year
(0.55 X 106 ) (8.76 X 103) (30) .)
1011
1.445 x kilowatt-hours output in 30 years.
Therefore under these conditions, each plant requires
1 . 445 X 1011 or 4,76 4 short tons uo8 mined.
3
3.033 X 10' Number
of Plants
Fuelable:
Illustrative Calculations
a.) If nuclear plants operate with fuel performance at 100% of
theoretical yield, how many plants (operating at 70% capacáity factor)
can be fueled with the assured -!U. s. reserve of 6 40 ,.000 tons of fuel?
How many,if plants operate at 55% capacity factor?
At 70% capacity factor,. it takes 6 ,067 tons u3o.gáminQd to
supply one plant for a 30-year lifetime. Therefore, for 6 40,000
tons reserve of u3o8:
Number of plants fuelable = 640,000 , or 105.5 plants.
6 ,067
At 55% capacity factor, it t..es 4,76 4 short tons u3_oa mined
to supply one plant for a 30-year lifetime.á Therefore, for a
6 40,000-ton reserve of u3o8:
Number of plants fuelable = 6 40,000 , or 134.3 plants.
4,76 4
b.) Nuclear plants have thus far 0perated with fuel performance at
about 50% of "theoretical" energy yield. With that performance, how
many plants would be fuelable from a reserve of 640,000 tons u3Qg?
For operation with fuel performance at 50% of its theoretical
energy yield, a plant at 70% capacity factor requires much more U30g.
Indeed, one plant requires 6 ,06 7 x 2, or 12,134 tons u3
oa for a
30-year lifetime. Therefore, from 6 40,000 tons u.08,
3
Number of plants fuelable = 6 40,000 , or 52.7 plants.
12,134
-16
c.) If there are fewer fuel failures in the future, fuel performance
in nuclear plants will rise. If we assume that fuel performance on
the average reaches 75% of its theoretical energy yield, how many I
plants would be fuelable with the assured supply of 640,000 tons U3Qg?
For fuel performance at 75% of its theoretical energy yield
instead of 100%, a plant at 70% capacity factor requires 100 ...;..75
as much fuel, or 1. 3 3 x 6,067, or 8,069 tons uo
for a 30-yr. lifetime.
38
Therefore, from 640,000 tons of u
3 o 8
Number of plants fuelable =
Quads (and oil-equivalent) from Nuclear Power
In describing an energy economy, one finds two units in common
use. The first is the "Quad", which is the abbreviation for
1015 British Thermal Units, or 1015 BTU's. The second unit is the
barrel of 9il equivalent.
In order to express the energy contribution of nuclear electric
power plants in a set of units comparable to those for other energy
sources, it is necessary to convert kilowatt-hours to Quads or to
barrels of oil equivalent.
one Quad represents 180,000,000 barrels of oil. (That is
the conversion from a General Elec..ric handbook; a Federal Energy
Administration handbook uses 172,000,000 barrels per Quad. The
difference is not great considering variations in the energy-content
of crude oil).
Thus, when the National Energy Plan (April 1977) reports the
1976 U.S. primary energy demand as 3 7 million barrels of oil equivalent
per day, we convert to Quads as follows:
= 0.2056 Quads per day
180,000,000
Annual energy use = (0.2056 Quads) (3 65 days)
day year
= 75.0 Quads per year
Next, a 1000-rnegawatt plant at 70% capacity factor produces
(0.7) (1000 Mw) (1000 kw) (24 hrs) (3 65 days) ,or (0. 7 x 106 kw) (8760 hours),
Mw day year
or 6.13 2 x 109 kilowatt-hours electrical per year.
640,000 , or 79. 3 plants.
8,069
First,
37,000,000
C:
á -17
1 Kilowatt-hour is equivalent to 34 15 BTU.
\ Therefore, a 1000-megawatt plant at 70% capacity factor has an
annual energy output, electrical,of (6 .132 x 109 kwhrs) (3,4 15 BTU)
Kwhr
or, 2.094 x 1013 BTU electrical per year.
But 1 Quad= 1 x 1015 BTU
:. Annual Output, electrical= 2.094 x 1013 = 0.02094 Quads.
1 X 1015
Annual output, thermal= 1 x Annual output, electrical
0.33
(2/3 of annual output is waste heat).
¥
¥ ¥ Annual output, thermal= 1 x 0.02094= 0.063 Quads thermal .
0.33
Lifetime outputs per plant (30 years):
Quads, electrical, (30) (0.02094 ) = 0.628 Quads
Quads, thermal, (30) (0.063) = 1.89 Quads
In barrels of oil equivalent
Annua,l output,. electrical = (O. 02094 Quads) (180,000,000 barrels)
Quad
=
3.77 x 106 barrels of oil
Daily output, electrical= 3.77 x 10.. barrels
3 .. 65 JC 10 days
= 1.033 x 104 barrels or oil per day,
.or about 10,000 barrels per day.
Annual output, thermal= (0..06 3 Quads) (180,000,000 barrels)
Quad
106
=
11.34 x barrels of oil
6
Daily output, thermal= 11.34 x 10
barrels
3. 65 x 10.. days
=
3.11 x 104 barrels of oil per day,
or about 30,000 barrels per day.
-18
Appendix 2
Some realities about uranium ore:
There is a very great deal of uranium on earth, including
uranium in sea water. This fact is often used to obscure the
fundamental truth that there is an extremely small amount of uranium
in discovered deposits which are worth working for nuclear plants of
the types currently at hand ..
Obviously, in general the richer an ore is in its uranium
content, the more likely it is to be worth working. On the other
side, for ores poor in uranium, there are limits to what can be used
for nuclear power: (a) energy limits, and (b) absurdity limits.
(a) Energy Limits: If an ore is of too low a grade in uranium
content, the energy required to mine, mill, convert, and enrich the
uranium can exceed the energy obtainable from such uranium in light
water reactors. Probably most of the uranium in the world falls into
this class. Use of such uranium sources would represent a drain on
the nation's energy supply, not an increment. Nevertheless, unless
there is careful monitoring, such a practice could occur!
(b) Absurdity limits: In some instances, it can be shown that
a low-grade uranium ore still could.yield more energy out in LWR's
than the combined energy inputs required for mining, milling,
enrichment, etc.
In principle, such an ore could be considered as a potential
source of fuel for light water reactors. In practice, however, it
could pe an absurdity to consider such a source.
One of the ostensible advantages of nuclear power over coal is
the elimination of the requirement for mining and transporting
"billions of tons" of coal. Rarely mentioned by nuclear salesmen is
the fact that this advantage is true only for the high-grade
uranium ores, which are rare indeed. The poor but abundant ores (like
Tennessee Shale, at 60 to 70 ppm) could require about TWICE as much
mining as would coal to get the same amount of net energy via LWR's.
No rational society would tolerate a mining industry both huge and
toxic by comparison with coal mining for such a poor energy yield
compared with coal.
-19
,,
Ore.. , ((
ft
""
The most meaningful estimate of grade is the term "parts per
million" by weight, or p.p.m. With rare exceptions, any other grading
system can be reduced to the p.p.m. system, and of course the p.p.m.
system is directly translatable to the percentage system. Thus, we
can tabulate some uranium contents of ores in both systems:
Ore grade in ppm Uranium Ore grade in % Uranium
á10 I 000 ppm 1%
1,000 ppm 0.1%
100 ppm 0.01%
10 ppm 0.001%
An ore containing 1,000 ppm uranium means exactly that: out of
every 1,000,000 units of ore mixture by weight, 1,000 units are
uranium.
High grade uranium ore deposits are those above 1,000 ppm of
uranium. Ore deposits over 5,000 ppm are virtually unknown. Ores
between 100 and 500 ppm have not been found in this country. Below
100 ppm, we are approaching the level where one mines as much uranium
ore as coal to get the same amount of net :energy.
It is not commonly understood that a very large amount of toxic
waste is created in the process of converting uranium ore to those
neat little "fuel pellets" of uo2 featured so often in nuclear
commercials, side by side with a carload of coal.
For ores of 1,000 ppm, from ground to reactor, a concentrationfactor
of about 5,000 is required for uranium, as shown below:
1 ton uo2 pellets loaded into reactor.. 0.88 tons of uranium1 the
rest is oxygen.
The enrichment step (to 3.0%, with 0.2% tails) has concentrated the
uranium 5.49-fold (100-;.-18.2). So before enrichment, we had
(5.49) (0.88) = 4.83 tons of uranium.
At 1,000 ppm, the ore mined.is 1,000-fold the final uranium content.
(1,000) (4.83) = 4,830 tons of ore mined.
-20
With 10% loss in mining and milling, and 1.5% loss in conversion steps:
(4, 830) (10) (100) = (4,830) (1.11) (1.015)
9 98.5
=
5,442 tons of ore mined per ton
of uopellets fed into a reactor.
2
When the nuclear industry is using ores
ppm, there will
of 500
be nearly 11,000 tons of radioactive waste (mostly in the mill
tailings) for every single ton of uo2 loaded into a reactor. Though
they don't get moved, one could "picture" them in carloads too.
In the mining and milling of uranium ores as generally practiced
in the U.S., the final product shipped from the mill is a
uranium oxide (one of several possible oxides) which is bright yellow,
which accounts for the trade name "yellowcake".
Yellowcake is essentially pure u3
o8. Uranium reserves are often
referred to in short tons of uo8, whether or not the uranium is
3
actually present in that oxide form.
Two other oxides of uranium are encountered in discussions:
uo2 and uo3. uo2 is the actual final form of uranium used in the
manufacture of nuclear fuel pellets. uo3 is not commonly encountered.
For most discussions, the key point is the ability to convert
quantities between uranium metal and its two common oxides, namely
uoand uo2.
38
Uranium constitutes 84.8% of uby weight; oxygen is 15.2%
.
3
o
8
Therefore, to convert utons to uranium, multiply by 0.848 .
3
o8
To convert uranium to uo8, multiply by 1.18.
3
Uranium constitutes 88.1% of uo2. Therefore, to convert U02 tons
to uranium, multiply by 0.881. To convert uranium to uo2, multiply
by 1.14.
Uranium Hexafluoride (UF6) is only of importance in that it is
the gaseous compound actually used in the current form of diffusion
enrichment plants. Following enrichment, the hexafluoride is
converted back to uoin general.
2
-21
It is important to be sure in a discussion to specify whether
one is referring to ore, or to a purified uranium compound derivable
from such an ore.
Thus a "find" of an ore body ofáan estimated 10 million tons
sounds like the solution to the uranium shortage, but it can be simply
trivial. Suppose the ore discovered is 200 ppm uranium! Then
10,000,000 tons of ore means only 2,000 tons of uranium, a miniscule
amount.
Sometimes a discovery is even reported in pounds! But what seems
huge in pounds is quickly reduced by a factor of 2,000 when converted
to short tons.
While on the subject, we must give attention to the variety of
"tons" encountered in the nuclear fuel literature.
1 short ton = 2,000 pounds
1 long ton = 2,240 pounds
1 metric ton = 1,000 kilograms
Since 1 kilogram = 2.2 pounds
1 metric ton = 2,200 pounds, close to the long ton.
One metric ton is labeled as l tonne.
'< M. T. U. >>
The abbreviation will commonly be encountered, 1 M.T.U., meaning
1 metric ton of uranium. It is important to ascertain whether a
specific reference is to 1 MTU of natural uranium (meaning the
uranium is 99.3% U-238, and 0.7% of the fissionable U-235), or to
enriched uranium (meaning uranium enriched to 2%, 3%, or more in U-235).
The literature is exceedingly sloppy in specifying enriched uranium
when this is meant.
But if a meaningful discussion of nuclear fuel is t.. take place,
we must have specified the degree of enrichment of the uranium in
U-235 content. Trying to figure out the energy which can be yielded
by 11
1 M.T.U." is simply hopeless unless it is clearly specified
that the uranium is not enriched at all (contains the 0.71% U-235
found in nature), or is enriched to 2%, 3%, 3.7%, or whatever.
There is no single "standard" degree of enrichment. However,
one commonly encounters 3.0% or 3.2% enrichment in discussions of
fuels for Pressurized Water Reactors.