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Economics of power production

Economics of power production. What are the cost factors in Power Production? Fuel Costs Capital Costs Operation and Maintenance Waste Related Costs Decommissioning. Fuel Costs: Comparing Coal and Nuclear.

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Economics of power production

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  1. Economics of power production • What are the cost factors in Power Production? • Fuel Costs • Capital Costs • Operation and Maintenance • Waste Related Costs • Decommissioning

  2. Fuel Costs: Comparing Coal and Nuclear • For a nuclear plant, these tend to be lower even though the following steps occur in the production of the fuel assemblies used in the reactor: • mining of the uranium ore, • conversion to U3O8 (uranium oxide - yellowcake form), • conversion to uranium hexafluoride, • enrichment from 0.7% U235 to 2-5% U235, • conversion to uranium dioxide (UO2) pellets, • loading of the pellets into rods, then into fuel assemblies • Transportation costs are high for coal because of the amount of material needed to generate the same energy as the nuclear fuel. • Mining of coal • Transportation

  3. Capital Costs: Coal and Nuclear • Coal plants are required to include scrubbers to remove airborne pollutants as sulfur dioxide, nitrous oxides, and particulates • Nuclear plant have buildings used for containment and safety-related equipment must meet higher standards than the traditional structures. Also, safety-related systems are redundant. • Capital costs are usually amortized over a period of time as allowed by IRS regulations

  4. Operation and Maintenance Costs • labor and overheads (e.g. medical and pension benefits), • expendable materials, • NRC (e.g. license changes, on-site and regional inspectors, and headquarters staff) and state (e.g. health department, emergency planning) fees, • local property taxes (varies from state to state).

  5. Waste Related Costs • The costs associated with the byproduct waste. For a coal plant this is ash. For a nuclear plant, these costs include the surcharge levied by the Department of Energy for ultimate storage of the high level waste. The DOE charge is a flat fee based on energy use.

  6. Decommissioning • The costs associated with restoration of the plant site back to "greenfield" status. Usually restoration would occur over a long period of time, e.g. 20 years. Parts of the plant could be used for energy generation by other sources.

  7. Nuclear versus Coal for similar age and size plants for $ per Megawatt-hour (10 $/Mw-hr = 1 cent/kw-hr):

  8. Cash Flow Diagrams Nuclear Power Plant Cost: 1 Billion Dollars, Rate income: 0.70 billion/year, Interest Payments/year = 0.3 billion, Labor Costs 0.200 billion/yr, Refueling costs 0.1 billion/yr, Plant Lifetime 30 years. Decommissioning after 30 years 1 Billion.

  9. Types of Cash Flow • single payment uniform series gradient series • exponential gradient • Amount of time = t = Et = Eo (1+g)t • t =1,2,3,….n • g = exponential growth rate

  10. Assumptions • Year-end convention is applicable • There is no inflation, nor will there be any during the lifetime • A before-tax analysis is needed • The effective interest rate is constant • Non-quantifiable factors can be disregarded • Funds invested in a project are available • Excess funds earn interest

  11. Equivalence Suppose the bank has an interest rate of 5%. You can put $100 in the bank at t = 0. After 1 year the money increases to $105. After 2 years the money increases to $110.

  12. Single Payment Equivalence The equivalence of any present amount P at t = 0 to any future amount F at t = n is called future worth. • (1+i)-n = single payment present worth factor • i is the effective rate over the period.

  13. Example If you have a savings account with an effective annual rate of 10%, how much would you put in to have $10,000 after 5 years? P = $6,209

  14. Table 1. Conversion

  15. Example • What factor will convert a gradient cash flow ending at t=8? (What is (F/G, i%, n). The effective annual interest rate is 10%.

  16. Uniform Series A cash flow that repeats each year for n years without change in the amount is called an annual amount (A). Example, Your car’s warranty has run out. In your experience, your maintenance costs have averaged $100/month. After five years what is the future value of you car maintenance costs?

  17. Sinking fund • a fund or account into which annual deposits A are made in order to accumulate F at t=n in the future. • Example • I want to buy a car for $25,000 in five years. If I can get a CD rate of 6%, what should my monthly deposit be over 5 years to get a future value of $25,000?

  18. Annuity • – a series of equal payments (A) made over a • period of time. In the simplest case of an • annuity that starts at the end of the first year • and continues for n years, the purchase price • P is • P = A (P/A, i%, n) • A $10,000,000 lottery prize is typically a • thirty-year annuity. Suppose for argument sake that • you use an annual interest rate of 10%. What would • the yearly payment, A, be to the prizewinner.

  19. Equivalent uniform annual cost or EUAC. A machine has a maintenance cost of $250/yr. What is the present worth of these maintenance costs over a 12 year period if i = 8%?

  20. Past Value I purchased a new car in 1984. My monthly payment is $179 for 60 months at an interest rate of 8%. What is the current value of the purchase price?

  21. Doubling Time Tripling Time

  22. Gradient cash flow • This factor finds the present worth of a uniform • gradient cash flow that starts in year 2 not in year 1. • 1) this assumes that the increasing cost begin • at the end of year one because of the year end • convention. • Then after year one, costs begin to increase.(P/G) find • the relative worth of onlyt he increasing part of the • annual expenses. • The present worth of the base expense incurred • at t = 1 and must be found separately with (P/A). • The sign convention used with a gradient cash • flow is as follows: if an expense increases each • year, the gradient is negative. If the revenue • increases each year, the gradient is positive.

  23. Example: Maintenance on an old machine is $100 this year but is expected to increase $25 per year thereafter. What is the present worth of 5 year of maintenance? (i=10%)

  24. Stepped cash flow Example: An investment costing $1,000 returns $100 for the first five years, and returns $200 for the following five years. Find present worth. P = -1000 + 200(P/A, i%, 10) – 100(P/A, i%, 5)

  25. Non Standard Model You may run into problems where the cash flow does not match the standard models. An example, how do you handle a missing or extra parts of cash flow as shown in the figure below? how do you handle a missing or extra parts of cash flow as shown in the figure below? P = A(P/A,i%,10) – A(P/F,i%,9)

  26. The same series can be represented by taking finding the present worth with an annual series over 8 years and then adding a finding the present worth of a single payment in the 10th year. P = A(P/A,i%,8) + A(P/F,i%,10)

  27. Delayed and premature cash flow In the delayed and premature cash flow problem, the present worth is first referenced to year 2 P’ = -75 (P/A, i%, 7)

  28. Find present worth of future value P’ referenced to t = 0P = P’ (P/F, i%, 2) = -75 (P’/A, i%, 7) (P/F, i%,2)

  29. Superposition Example: In 10 years you deposit $300 at the beginning of each year for a total of 10 payments. 1st payment occurs at t = o, there are no payments at t = 10. F = 300 (F/P, i%,10) + 300 (F/A, i%, 10) – 300

  30. Another Example • Suppose that someone asks you to loan them • 100$ and that at the end of one year they will • pay you $5 in interest. Is this a good deal? • P = -100 + 105 (P/F, 5%, 1) • = -100 + 105 (0.9524) = 0 In this example, the present worth stays the same. You loan $100 and at the end of 1 year your investment plus interest is returned. You would do just as well if you put the money in a 5% bank account and would assume no risk. This is not a good deal.

  31. Simple & Compound Interest Compound—if you put $100 in a bank account with 5% interest, Year 0 $100 Year 1 ($100)(1.05) = 105$ Year 2 ($105) (1.05) = $110.25 Year 3 ($110.25) (1.05) = $115.76 You are earning interest on principal + on interest from previous years. Simple – if you earn interest only on the principal,

  32. Rate of Return (ROR): Effective annual interest rate at which an investment accrues income. A working definition of ROR would be the effective annual interest rate that makes the present worth of the investment zero. Suppose you have a $100 investment that pays back at $65 at the end of each of the first two years. • This is a quadratic equation. • If n were large then you would need to find the • root of an nth order polynomial. • The solution to this problem is, • (i = 31.87%)

  33. Rate of ReturnUsually the ROR is found by trial and error. Example: What is the ROR on a capitol investment of $1,000 at t = 0 when $500 is returned in year 4, and $1,000 returned in year 8? Set up as a present worth calculation, iteration 1, use i = 5% P = -1000 + 500 (P/F 5%,4) + 1000 (P/F, 5%, 8) = -1000 + (500)(0.8227) + (1,000) (0.6768) = $88 5% is too low now try 10% P = -1000 + (500 (P/F 10%, 4) + 1000 (P/F, 10%, 8) = -1000 + (500)(0.6830) + (1,000) (0.4665) = -$162 10% is too high, now try 6% You will find that 6% is too low but close. So the next step is to try 7%. You then try 6.1%. Continuing this iteration procedure, you find that the solution approaches 6.37%.

  34. Rate on Investment (ROI) ROR is the effective annual interest rate %/year. ROI is the dollar amount • Minimum Attractive ROR (MARR): • If a company establishes a minimum ROR on al of • its investment, this is known as the Minimum • Attractive Rate of Return. • If a bank ROR is 5% in a simple account, the • company may decide that 5% is the MARR. • Characteristics of an Engineering Economy Problem • An interest rate will be given • Two or more alternatives will compete for funding • Each alternative will have its own cash flow • It is the best alternative that must be found

  35. Example: Investment A cost $10,000 today and pays back $11,500 in two years. Investment B costs $8,000 today and pays back $4,500 each year. Which is best? • Present worth method • P(A) = -10,000 + 11,500 (P/F, 5%, 2) • P(A) = -10,000 + 11,500 (0.9070) • P(A) = $431 • P(B) = -8,000 + 4,500 (P/F, 5%, 2) • P(B) = -8,000 + 4,500 (1.8594) • P(B) = $367 • A is better than B

  36. Capitalization Cost Method Capitalized cost is the amount of money at t = 0 needed to perpetually support the project or the earned interest only. One assumes infinite life in making comparisons. Capitol Costs = initial Cost + (Annual Costs/i) This assumes a constant annual cost. If the costs vary from year to year, you must determine a cash flow of equal annual amounts (EAA) that is equivalent. EAA --- Capital Costs = initial cost + (EAA/i) = initial cost + present worth of all expenses

  37. Example What is the capitalized cost of a public works project that will cost $25,000,000 now and will require $2,000,000 in maintenance annually? • Assume i = 12% • Use the units of Millions of dollars • Capital Costs = 25M + 2M (P/A, 12%, ¥) • = 25M +(2M/0.12) = $41.67M

  38. Annual Cost Method Suppose that two projects have unequal lives. You can use the annual cost method to examine the alternatives. (Also called the annual return method or capital recovery method) If one alternative has a life of 10 years and another 5 years

  39. Equivalent Uniform Annual Cost (EUAC) Example : over 30 years and i = 7% which alternative is superior? A B Type brick wood Life 30 years 10 years Initial cost $1800 $450 Maintenance $5/year $20/yr EUAC (A) = 1800 (A/P, 7%, 30) + 5 = 1800 (0.0806) + 5 = $150 EUAC (B) = 450 (A/P, 7%, 10) + 20 = 450 (0.1424) + 20 = $84 B is cheaper than A. You assume over 30 years you replace the wood at 10 years and 20 years.

  40. Benefit - Cost Ratio Method This method is typically used in municipal projects where benefits and costs accrue for different segments of the community. The present worth of all benefits is divided by the present worth of all costs B/C ³1.0 Disbursements by initiators – costs Disbursements by users – disbenefits B/C =

  41. Example: you want to build a bridge over a ravine in order to take time off of a route that goes through the mountains. Benefits (decreased travel time, fewer accidents, reduced gas usage, etc.). Should the bridge be built? Since B/C is less than one, the benefit to cost ratio of the project does not justify the costs of the project.

  42. Rate of Return Method You can use the minimum attractive ROR (MARR) for comparison purposes. • Ranking Mutually Exclusive Multiple Projects If you have two projects and need to rank their relative economic benefits, you can rank the projects by the following method, In this case project 2 has a higher benefit to cost ratio than project 1.

  43. Alternatives with different lifetimes Suppose that you have a choice between two cars. Car A has a 3-year projected lifetime Car B has a 5-year projected lifetime The first question you ask, is how long do you need a car? The length of need is the parameter that will allow you to evaluate the alternatives properly.

  44. Opportunity Costs An opportunity cost would be something like a trade-in allowance. In the case of a car, you can negotiate a trade-in value for your used car when you buy a new car. • Replacement Studies • If you decide to replace some equipment, you • need to investigate the costs of retiring the • equipment. In these types of studies, the existing • equipment is called the “defender” and the new • piece of equipment is called the “challenger.” • Salvage Value • Since most defenders have a market value when • you decide to retire them, they will have a salvage • value that needs to be considered. • By convention, the salvage value is subtracted from • the defender’s present cost.

  45. The Equivalent Uniform Annual Cost of the defender is next year’s maintenance cost plus interest x (current salvage value) plus the current salvage value-next year’s salvage value. EUAC = Next Year’s Maintenance Cost + (1+ i) x Current Salvage Value – Next Year’s Salvage Value

  46. Economic Life: As an asset grows older, the maintenance value increases. Additionally, the amortized cost of the purchase price decreases. An example would be your car. When does it make sense to replace your car? The maintenance cost and the salvage value should be evaluated and the point where this sum is minimum should be the economic life of the equipment.

  47. Example As an example, lets assume that the city of Columbia has purchased a bus for $120,000. The projected maintenance cost and salvage value of the bus is known from prior experience with this model of bus. Find the present worth and amortize the maintenance costs. Assume an interest rate of 8%.

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