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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 . 

.. ᥠ100 (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.