Exam 2

General Comments:  Score range: 32 - 77; Class Average, 63 .  Approximate letter grades: A,  69 up; B, 55-66.

Some have asked about the possibility of additional work to improve the grade.  It's very late in the term for this, but if you want to write a term paper [worth 40 points] see me ASAP about a topic.  No papers accepted without previous approval of the topic and no two persons may write on the same topic.
 
 

Problem Solutions:

P1.  3.016066 u + 2.014102 u = 5.030158 u.  Subtracting (4.002603 + 1.08665) leaves 0.018900 u, the net mass difference.
        0.0189 u x 931 MeV/u  = 17.6 MeV
 

P2.  The BE (binding energy) of U-235 is about the same as that of U-238.  From the graph on the 'NuclearEnergy' hand-out, the BE/nucleon for U is about 7.4 MeV.
235 x 7.4 = 1739 MeV, total BE for U-235
235 - 3 = 232 bound nucleons remain after fission.   232 x 8.3  MeV/nucleon [given in the question] = 1926 MeV
1926 - 1739 = 187 MeV energy yield
 

P3.  632 Btu/ft²  x 700 ft²  = 442,400 Btu
    x 50% = 221,200 Btu heating from the collector .
Daily requirement is 40,000 Btu/hr x 24 hr/d = 960,000 Btu
Fraction supplied is thus 221200/960000 = 0.23 or 23%

P4.  Average monthly energy costs are 2000 kWh x $0.08 = $160
    The PV panels supply:  1 kW x0.3 x 24 hr/d x 30 d/mo =  216 kWh per month
    Monthly saviing:  216 kWh x $0.08  = $17.28

Payback time is thus $8000/ $17.28  = 463 months, or about 38 years!
[About half of this system's cost is for controls and storage, so 'scaling up' to 2 or 3 kW with more panels would shorten the payback time]