Today: Estimating our Energy Predicament

1. Today:Estimating our Energy Predicament Prof. Tom Murphy U C San Diego All power point images are only for the exclusive use of Phys3070/Envs3070 Spring term 2014

2. Read for Friday • R and K -10.2 –ozone --- pp 343-344 • www.kyotoprotocol.com, and click on a few interesting links • www.epa.gov/ozone/intpol • Read about the Montreal Protocol (NOT the Convention)

3. Friday FCQ International agreements (the Montreal Protocol) (the Kyoto Protocol) Wednesday May 7 7:30-10 PM

4. 3070 end game What is to be done? Please create drafts of a single specific step to improve our energy/environment future, large or small, that you would take as world dictator. Send to me, and volunteer to make a five minute presentation on our last day. (double extra credit if selected!)

5. (15) If I invest $5billion to build a nuclear power plant that will operate for and be paid off in 40 years, how much per year will my ‘mortgage’ payments be to pay the initial capital cost at an annualized interest at a rate of 5%? If I produce 1010 kW-hr of electrical power each year, how much must I charge my customers per kW-hr to service my debt? If I just let the debt grow, without payments, the annualized total cost of the investment is $5 billion *(1.05)40 =$5 billion * 7.04 =$35.3 billion = $35.2 *109, which must be paid to the investors at the end. In that 40 years I produce 40*1010 kW-hr. I must charge my customers at the rate of $35.2*109 / 40 * 1010 kW-hr = $0.088/kW-hr. But I do not do it that way. I will make monthly payments, using the site www.mortgagecalculator.com. I entered a house value and loan amount of $500,000, at 5% for 40 years, with no taxes or the like. The output from that site was a payment of $2410.98 per month, and a total cost (principal + interest) of $1,157,271.84. My $5B plant costs more by a factor of 104 than the $500,000 input. My monthly payment must be $2411*104=$24.11 million, and for the year, $289 million. So the total loan amount for my plant is $1.157*106 * 104 (scale factor) = $1.157*1010 =$11.57 billion. The rate I must charge is $1.157*1010 / 40*1010 kW-hr (over the 40 years) = $0.0289 / kW-hr, or 2.89cents/kW-hr to pay for my financing. The same rate comes out for one year at a time.

6. (10) A nuclear power plant operates at 80% of the best thermal efficiency (Carnot) of 31%. • Transmission losses are 10% before the power gets to the customers. How much fission heat power must the plant generate to give 60,000 customers each 5 kW? • How much heat power must be extracted from the plant by its cooling facilities? • The power that gets to the customer is 0.80 * 0.31 * (1-0.10)=0.2232 of the thermal heat. • The power to the customers is 60,000 * 5000 watts = 3*108 W =0.30 GWe. • This requires 0.30 GW/0.2232 = 1.34 GWt of heat power, Joules/sec. • The plant (not the transmission system) has an efficiency • of 0.80*0.31 = 0.248, so 1-0.248 = 0.752* 1.34 GWt =1.008 GW of heat power • must be handled by the cooling system.

7. A typical fission reaction for uranium is • 1on1 +23592U 14313153I78 +AZ XX N +3 10n1, where I is the element iodine. • (5) What are the atomic weight A, the atomic number Z and the • neutron number N for the element XX in the specific fission reaction above, one of the many possible? • Protons: 0 + 92 = 53 + Z +3*0, so Z = 39 (the element we call yttrium (Y) • Neutrons: 1 + 143 = 78 + N +3 , so N = 63. • Total nucleons: 1 + 235 = 131 + A +3, so A = 102. • Check: A = N + Z, or 102 = 63 + 39, yes indeed

8. (10) If we fission I kg of 235U completely in one year, at a plant efficiency of 30%, how many electrical GWe can we sell? • Each 235U fission releases 1.8*108 eV, as in HW #8. One gram of 235U holds 6.02*1023/235 nuclei, using the ‘mole’, as in ‘not in the text’ notes. I kg=1000 grams that fissions holds 6.02*1023 *1000/235= 25.6*1023235U nuclei, each releasing the 1.8*108 eV. • The fission heat power is then 25.6*1023(fissions)*1.8*108 (eV/fission), but we prefer to use our Joule standard, so we also multiply by • *1.60*10-19 (Joule/eV) to get heat energy of 25.6*1.8*1.6 * 10(23+8-19) =73.728*1012 Joules of fission hear energy in one year. • But the question is about power=energy/time, so divide by seconds in a year, so divide by this time • /(365 days * 24 hours/day*3600 sec/hour) • = (73.728*1012 Joule/year)/(3.16*107 sec/year) = (73.7*1012 Joules)/(3.16*107 sec) = 23.4*105 Joule/sec = 2.34 MWt. • Then we can sell only 30% of this heat power as electricity, 0.30*2.34 = 0.702 MWe. • (the graders will be generous for partial credit on these steps)

9. (10) What is the greatest single reason to enhance US nuclear electricity, and why? • Examples could be: electricity without CO2, declining coal and gas, other nasty features of coal, a known and familiar nuclear industry, having lots of uranium, being able to breed new plutonium fuel for a very long time. Each would need a few well-written words. • (10) What is the greatest single reason NOT to continue to rely on nuclear power for the US, and why? • Examples could include: the need to dispose of radioactive spent fuel, proliferation of nuclear weapons, the danger of radioactive releases from accidents, cheap coal and gas, public fears, excessive regulations. Each topic would again need a few well-written words to explain.

10. (10) Why is the Arctic such an important source of feedback to accelerate global climate change? • Ice and snow have a very high albedo, reflecting sunlight. If these melt, the average albedo of the earth would be lower, capturing more solar energy, leading to more melting, a classic positive feedback. There are also concerns that thawed permafrost would release methane, as another positive feedback.