College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 496ALT
Alternative Energy
Spring 2009 Number: 18650 Instructor: Larry Caretto
March 5 Homework Solutions
1
Use the atomic mass data shown in the table below in solving the various parts of this
problem.
Nucleus
235
U
n
139
Xe
95
Sr
Mass (amu)
235.04394
1.008665
138.9187869
94.9193582
Reference
Fay and Golomb, Energy and the Environment, p 121.
Class notes (taken from NIST web site)
www2.bnl.gov/CoN/nuc/X/Xe139.shtml
www2.bnl.gov/ton/cgi-bin/nuclide?nuc=Sr95
(a) Calculate the mass deficit (m) in atomic mass units (amu) of the following fission
reaction. (Use literature values for the exact masses of the isotopes and neutrons.)
U + n → 139Xe + 95Sr + 2n
235
The mass deficit for the reaction given is 138.9187869 + 94.9193582 + 1.008665 - 235.04394.
m = –0.19713 amu
(b) Calculate the energy (MeV) released per one fission.
The annihilation of 1 amu of mass produces 931.495 MeV. Thus, the mass deficit computed here
would produce 183.6 MeV .
(c) Calculate the energy released per kilogram of
released in the combustion of 1 kg of carbon.
235
U and compare it to the energy
(One MeV = 1.602176462x10-19 MJ and one amu = 1.66053873x10-27 kg. So the energy release
per unit mass of the 235U atom for the fission reaction shown above is found as follows.
183.6 MeV 1.6022 x10 19 MJ
amu
MJ
 7.54 x107
235.0439 amu
MeV
1.6605 kg
kg
The energy released in the combustion of 1 kg of carbon is given in various references. I used
the text Energy and the Environment by Fay and Golomb, which gives a value of 33 MJ/kg on
page 121. The fission energy release is over two million times this amount.
(d) If the uranium feed to a reactor has a U235 concentration of 3% and the spend fuel has a
U235 concentration of 0.8%, what weight of UO2 fuel is required for a 1000 MWe plant
that runs for one year at a 95% capacity factor and an efficiency of 33%?
For each kg of uranium loaded (which is a mixture of U235 and U238), the U235 concentration will
decrease from 3% to 0.8%. This means that 0.03 – 0.008 = 0.022 kg of U235 per kg of loaded
uranium will undergo fission, producing something like the 7.54x107 MJ/kg computed above.
Thus, on the basis of uranium loaded, the fission produces (0.022)( 7.54x107) = 1.66x106 MJ/kg.
A 1000 MWe power plant, operating at an efficiency of 33% requires a heat input of
(1000 MW)/33% = 3000 MW = 3000 MJ/s. If the plant operates at a 95% capacity factor, the total
Jacaranda (Engineering) Room 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
March 5 homework solutions
ME496ALT, L. S. Caretto, Spring 2009
Page 2
heat energy input during the year will be (95%)(3000 MJ/s)(3600 s/h)(24 h/day)(365 d/yr) =
9.08x1010 MJ. The mass of uranium required then is (9.08x1010 MJ/yr) / (1.66x106 MJ/kg) =
5.47x104 kg/yr of uranium. The amount of UO2 is found by multiplying by the molar mass ratio of
UO2 to U which is (238 + 32)/238 = 1.143. Thus the required amount of UO 2 is (1.143)(5.47x104
kg/yr) = 6.21x104 kg/yr of UO2 .
2
Use the following data in answering the questions here.
Nucleus
2
D
n
3
T
4
He
Mass (amu)
2.014102
1.008665
3.0160493
4.002603
Reference
http://www.sisweb.com/referenc/source/exactmaa.htm
Class notes (taken from NIST web site)
http://www2.bnl.gov/ton/cgi-bin/nuclide?nuc=H3
http://www.sisweb.com/referenc/source/exactmaa.htm
(a) Calculate the mass deficit (m) in atomic mass units (amu) of the following fusion
reaction.
D + 3T → 4He + n
2
The mass deficit for the reaction given is 4.002603 + 1.008665 – 2.014102 – 3.0160493.
m = –0.01888 amu
(b) Calculate the energy (MeV) released per one fusion.
The annihilation of 1 amu of mass produces 931.495 MeV. Thus, the mass deficit computed here
would produce 17.59 MeV .
(c) Calculate the energy released per kilogram of deuterium.
One MeV = 1.602176462x10-19 MJ and one amu = 1.66053873x10-27 kg. So the energy release
per unit mass of deuterium for the fusion reaction shown above is found as follows.
3. The isotope 129I has a half-life of 15.7 years. In a nuclear power plant accident, 1 kg of the
isotope is dispersed into the surroundings of the plant. How much of the iodine isotope
will remain in the surroundings after 1, 10, and 100 years.
The lecture notes for nuclear energy have the following equation that relates the current
concentration, N, the initial concentration, N0, the time, t, and the half-life, t1/2.
N  N0e
 ln(2 ) t / t1
2
We can solve this equation for the ratio, N/N 0, which is independent of the measure of the isotope
present. Thus, we can also apply this equation to determine the mass of 129I that will be present
for the various times shown. If we replace N and N0 by m and m0 = 1 kg, we obtain the following
results for the mass remaining after 1, 10, and 100 years.
m1 year  (1 kg)e
l yr

ln( 2)
15.7 yr
m10 years  (1 kg)e
l 0 yr

ln( 2)
15.7 yr
 0.957 kg
 0.643 kg
March 5 homework solutions
ME496ALT, L. S. Caretto, Spring 2009
m100 years  (1 kg)e
l 00 yr

ln( 2)
15.7 yr
Page 3
 0.0121 kg
Of course, all these calculations assume that there is no movement of the 129I, which would dilute
its local concentration, during the time periods used here.
4. The lecture presentation on nuclear energy discussed separative work units, SWU, defined
The lecture presentation on nuclear energy discussed separative work units, SWU, defined
by the following equation: SWU = PV(N P) + TV(NV) – FV(NF), where V(N) is the value
function defined in the notes. The meanings of there terms and the following equations
were defined in the notes: F = P + T; FNF = PNP + TNV. Use the three equations in this
paragraph to show that the product-to-feed ratio, the SWU-to-feed ratio, and the SWU-toproduct ratio are given by the following equations.
P
T N  NT
 1  F
F
F N P  NT
F
T N  NT
 1  P
P
P N F  NT
SWU P
T
 V N P   V N T   V N P 
F
F
F
SWU
T
F
 V N P   V N T   V N P 
P
P
P
Use these equations to verify the SWU calculations on slide 36 of the presentation that for
NP = 3%, 3.8 SWU required if NT = 0.25%, or 5.0 SWU if NT = 0.15%. Recall that the value of
NF for natural uranium is 0.71%. Note that the results on that slide are for the SWU-toproduct ratio, SWU/P.
We can rearrange the given equations by dividing by the feed rate, F.
F  P T
 1
P T

F F
FN F  PN P  TN T
 NF 
P
T
N P  NT
F
F
Combining the second and fourth equation above gives the desired result for P/F.
NF 
P
 P
N P  1   NT
F
 F

P N F  NT

F N P  NT
This is the first desired result. Taking the reciprocal of the right-hand side of the equation for P/F
gives the result for F/P.
F
T N  NT
 1  P
P
P N F  NT
To get the result for SWU/F we start by dividing the equation for SWU by F to and use the P/F
equation just derived to get the N terms in the equation for SWU/F.
 N  NT
N  NT
SWU P
T
 V N P   V N T   V N P   F
V N P   1  F
F
F
F
N P  NT
 N P  NT

V NT   V N F 

The entire right hand side can be placed over a common denominator of NP – NF as follows.
 N  NT  N F  NT
SWU N F  N T

V N P    P
F
N P  NT
N P  NT


N  NT
V N T   P
V N F 
N P  NT

March 5 homework solutions
ME496ALT, L. S. Caretto, Spring 2009
Page 4
SWU N F  N T V N P   N P  N F V N T   N P  N T V N F 

F
N P  NT
To get the equation for SWU/P we divide the SWU definition by P and use the equation for P/F to
introduce the fraction terms.

N  NT
SWU
T
F
 V N P   V N T   V N P   V N P   1  P
P
P
P
 N F  NT

N  NT
V N T   P
V N F  T
N F  NT

he formula for the value function, shown below, is taken from slide 35 of the nuclear energy
presentation. We can use this formula to compute the value of V(N) for the various
concentrations used in this problem. Note that the concentrations must be expressed as a
fraction, not as a percentage, in this formula.
 N 
V ( N )  (2 N  1) ln 

1 N 
 0.03 
V (0.03)  2(0.03)  1ln 
  3.268
 1  0.03 
 0.0071 
V (0.0071)  2(0.0071)  1ln 
  4.780
 1  0.0071 
 0.0025 
V (0.0025)  2(0.0025)  1ln 
  5.959
 1  0.0025 
 0.0015 
V (0.0015)  2(0.0015)  1ln 
  6.481
 1  0.0015 
Using the appropriate fractions and results of the value function for NF = 0.0071, NP = 0.03, and
NT = 0.0025 in the equation for SWU/F gives.
SWU N F  N T V N P   N P  N F V N T   N P  N T V N F 


F
N P  NT
0.0071  0.00253.268  0.03  0.00715.989  0.03  0.00254.780  0.638
0.03  0.0025
This does not match the figure of 3.8 SWU given in the notes. Perhaps the ratio not expressed in
the notes is really SWU/P. We can get this ratio by dividing SWU/F by P/F, which we can
compute from the equation derived above.
P N F  NT 0.0071  0.0025


 0.1673
F N P  NT
0.03  0.0025
F
1

 5.978
P 0.1673
This matches the figure of 6.0 kg of feed for 1 kg of enriched uranium given in the lecture notes.
We can now compute SWU/P.
SWU SWU F

 0.6385.978  3.82
P
F P
This matches the value of 3.8 given in the notes. Similar agreement with the results in the slide is
found for the tails fraction, NT = 0.15% by repeating the computations above with this new value
of NT, and its corresponding value function result.
March 5 homework solutions
ME496ALT, L. S. Caretto, Spring 2009
Page 5
SWU N F  N T V N P   N P  N F V N T   N P  N T V N F 


F
N P  NT
0.0071  0.00153.268  0.03  0.00716.481  0.03  0.00154.780  0.979
0.03  0.0015
P N F  NT 0.0071  0.0015


 0.1965
F N P  NT
0.03  0.0015
F
1

 5.089
P 0.1965
SWU SWU F

 0.9795.089  4.98
P
F P
5. For many years now there has been only little development of new nuclear power plants
worldwide and none in the US. Recently the issue of global warming has prompted a
reconsideration of nuclear power as a source of power that is free of CO 2 emissions. Find
a website of either (a) an environmental organization or (b) a company or industrial
organization that favors nuclear power and summarize the arguments you find there. Give
a brief critique of the arguments you find. Do this in one to two pages.
As usual there is no correct answer to questions like these. The grade on your answers will be
based on how well you provide information to present your point of view.