1. Thomas J. Crawford, PhD, PE November 08, 2013 A. Economics: Pay Back Period or Pay Back Time. ( MAY HAVE ERRORS, TYPOS !!) If we make this Investment to improve Efficiency or Decrease Costs, how much Time is required to recover the Investment Costs and Expenditures? Example: Coal–Fired Fossil Power Plant. Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082/kwhr, (Neglect Fuel and O&M costs) Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is 3 months. (boiler tubes replaces, condenser tubes cleaned, turbine upgrade, precipitators, pulverizers, etc. ) Solution: Analogous to the Tortoise and the Hare running a race. Considering only the costs of power, neglecting replacement power, Fuel Savings and Capital Investment Costs, etc. Amount = Rate * time We set Amount (1) = Amount (2)  R1 * ( t + Δ t ) = R2 * t; Δ t = outage time t = R1 * Δ t we need R1 and R2, the rates that money is generated. R2 – R1 R1 = (0.38 * 2000) * 1000 * 0.082 = $62, 320 / hr. or 62, 320 $ / hr. R2 = (0.40 * 2000) * 1000 * 0.082 = $65, 600 / hr. 65, 600 $ / hr. t = 62,320 * ( 3 * 30 * 24 )  t = 41,040 hrs = 4.68 years ( 65,600 – 62,320 ) Check: A1 = R1 * ( t + Δ t ) = 62,320 * ( 41,040 + 3*30*24) = $ 2.692E09 = $ 2.692 billion A2 = R2 * ( t ) = 65.600 * ( 41,040 ) = $ 2.692E09 = $ 2.692 billion Lecture Data Analysis 04: Page 1 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

2. Thomas J. Crawford, PhD, PE November 02, 2013 A. Economics:, continued Pay Back Period or Pay Back Time, continued. Solution: Analogous to the Tortoise and the Hare running a race. Both are trying to cover some distance in the same time; the tortoise runs steadily, while the Hare or Rabbit runs rapidly, but is also shutdown for periods. Note: the cost per day of Outage is below. Lecture Data Analysis 04: Page 2 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt R1 ( Tortoise ) t + Δ t R2 (Hare) Outage t ΔtΔt Find: Extra Cost for One Additional Day of Outage. Solution: t = R1 * ( Δ t + 1*24 ) = 62,320 * ( 2160 + 24 ) = 41,496 hrs R2 – R1 ( 65,600 – 62,320 ) Extra time to catch up = 41,496 – 41,040 hrs = 456 hrs Extra cost to catch up = 456 * ( 65,600 – 62,320 ) = $1.496E06 = $ 1.5 million This is 456 hrs (19 days) at the old rate R1 instead of the New Rate R2.

3. Thomas J. Crawford, PhD, PE November 02, 2013 A. Economics:, continued Pay Back Period or Pay Back Time, continued. Example: Coal–Fired Fossil Power Plant. continued Notice: Total time = t + t = 41,040 + 3*30*24 = 41,040 + 2160 = 43200 hrs and ( 41,040 / 43,200 ) = ( 0.38 / 0.40 ) or R1 / R2 = eff1 / eff2 because R1 / eff1 = R2 / eff2 Δ t = outage time = 3 months = 3*30*24 = 2160 hrs Lecture Data Analysis 04: Page 3 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

4. Thomas J. Crawford, PhD, PE November 11, 2013 A. Economics:, continued Pay Back Period or Pay Back Time. If we make this Investment to improve Efficiency or Decrease Costs, how much Time is required to recover the Investment Costs and Expenditures? Example: Coal–Fired Fossil Power Plant. Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082 /kwhr, Cost of Replacement Power = $0.07 /kwhr (Neglect Fuel, O&M, and Capitol costs) Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is again 3 months.. Solution: Similar to previous Example. Considering only the costs of power and replacement power; neglecting Capital Investment Costs, etc. Amount = Rate * time We set Amount (1) = Amount (2)  R1 * (t + Δ t ) = R2 * t + Rp * Δ t t = R1 * Δ t – Rp * Δ t we have R1 and R2, need Rp; Rates of Money. R2 – R1 R1 = $ 62, 320 $ / hr. R2 = $65, 600 / hr. Rp = 0.38 * 2000 * 1000 * ( 0.082 – 0.070 )  Rp = $ 9120 / hr. t = ( 62,320 – 9120 ) * ( 3 * 30 * 24 )  t = 35,034 hrs = 4.00 years ( 65,600 – 62,320 ) Check: ( 3*30*24) = 2160 hrs A1 = R1 * ( t + Δ t ) = 62,320 * ( 35,034 + 2,160 ) = $ 3.0665 E09 = $ 2.318 billion A2 = R2 * t + Rp * Δ t = 65,600 * 35,034 + 9120 * 2160 = $ 2.318 E09 OK Lecture Data Analysis 04: Page 4 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

5. Thomas J. Crawford, PhD, PE November 11, 2013 A. Economics:, continued Pay Back Period or Pay Back Time. Example: Coal–Fired Fossil Power Plant. Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082/kwhr, Cost of Replacement Power = $0.07 / kwhr Capital Costs = $18.4 E06 (Neglect Fuel and O&M costs) Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is again 3 months.. Solution: Similar to previous Examples. Considering only the costs of power, replacement power, and Capital Investment Costs. Amount = Rate * time We set Amount (1) = Amount (2)  R1 * (t + Δ t ) = R2 * t + Rp * Δ t – Cp t = R1 * Δ t – Rp * Δ t + Cp we have R1, R2, and Rp; Rates of Money. R2 – R1 R1 = $ 62, 320 $ / hr. R2 = $65, 600 / hr. Rp = $ 9120 / hr. Cp = $ 18.4E06 t = ( 62,320 – 9120 ) * ( 3 * 30 * 24 ) + 18.4E06  t = 40,644 hrs = 4.64 years ( 65,600 – 62,320 ) Check: ( 3*30*24) = 2160 hrs A1 = R1 * ( t + Δ t ) = 62,320 * ( 40,644 + 2,160 ) = $ 2.668 E09 A2 = R2 * t + Rp * Δ t – Cp = 65,600 * 40,644 + 9120 * 2160 – 18.4E06 = $ 2.668 E09 OK Lecture Data Analysis 04: Page 5 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

6. Thomas J. Crawford, PhD, PE November 11, 2013 A. Economics:, continued Pay Back Period or Pay Back Time. Example: Coal–Fired Fossil Power Plant. ( MAY HAVE ERRORS, TYPOS ) Given: Power = 2000 MWth in the Boiler, current efficiency = eff = 38.0 % Cost of Power Output = $0.082/kwhr, Cost of Replacement Power = $0.07 / kwhr Capital Costs = $18.4 E06, Fuel Costs = $1.92 / million Btu (Neglect O&M costs) Find: Pay Back Time if upgrades are made to increase eff to 40.0%, and outage time is again 3 months.. Solution: Similar to previous Examples. Considering only the costs of power, replacement power, Capital Investment Costs, and Fuel Costs (Savings). Amount = Rate * time We set Amount (1) = Amount (2)  R1 * (t + Δ t ) = R2*t + Rp * Δ t +Rf * Δ t – Cp t = R1 * Δ t + Rp * Δ t + Cp – Rf * Δ t we have R1, R2, Rp; Rates of Money. R2 – R1 R1 = $ 62, 320 $ / hr. R2 = $65, 600 / hr. Rp = $ 9120 / hr. Rf = $13,104 / hr ( explained page 7) Cp = $ 18.4E06 t = ( 62,320 – 9120 – 13,104 ) * ( 3 * 30 * 24 ) + 18.4E06  t = 32,014 hrs = ( 65,600 – 62,320 ) t = 3.65 years Check: ( 3*30*24) = 2160 hrs A1 = R1 * ( t + Δ t ) = 62,320 * ( 32,014 + 2,160 ) = $ 2.130 E09 A2 = R2 * t + Rp * Δ t + Rf * Δ t – Cp = = 65,600 *32,014 + ( 9120 + 13,104 ) * 2160 – 18.4E06 = $ 2.130 E09 OK Lecture Data Analysis 04: Page 6 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

7. Thomas J. Crawford, PhD, PE November 11, 2013 A. Economics:, continued Pay Back Period or Pay Back Time. Example: Coal–Fired Fossil Power Plant. Given: Same Example Find: Fuel Rate for equation on page 6 if coal costs $1.92 / million Btu. Solution: 38% efficiency  the Heat Rate. 9700 Btu / kwhr is a reasonable rate. The conversion factor is 3412 Btu / kwhr and the definition of efficiency. 38% eff.  38 kwe / 100 kwth 3412 Btu * 100 kwth = 8980 Btu / kwhr (electric) (very low) 1.0 kwhr 38 kwe Rf = 2000 * 0.38 * 1000 * 8980 * $1.92 / 1.00E06 = $ 13,104 / hr units: ---- kwe -------- ( Btu / kwhr ) ( $ / Btu ) = $ / hr Lecture Data Analysis 04: Page 7 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

8. Thomas J. Crawford, PhD, PE November 02, 2013 A. Economics: continued Comparing Fuel Costs, the Costs of different types of Fuels Example: Electric Power. Given: Coal Cost = $1.94 / million Btu, Natural gas = $4 / million Btu Find: ??? Costs per tone; still working on this problem. Solution: Example: Electric Light Bulbs. Given: Cost of Electricity = 0.12 / kwhr Incandescent Bulb = $1.02, life = 800 hrs, Florescent = $12.50, life = 8000 hrs Find: Payback period for changing over the light bulbs STILL WORKING ON THIS PROBLEM Solution: WORKING HERE NOV 05, 2013 Lecture Data Analysis 04: Page 6 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

9. Thomas J. Crawford, PhD, PE November 09, 2013 A. Economics: continued Costs of Stand-by Equipment: Compare the Inventory cost of various parts or pieces of equipment vs. the cost of replacement power and lost revenue due to equipment failure. Example: Electric Power. WORKING TO DEVELOP SOMETHING HERE Given: Find: ??? Costs of Each Solution: Example: House and Furnishings or replacement Parts. Given: Replacement Lights, Garbage Disposal, various sizes of Batteries, Candles, Bottled Water, canned food needed in Emergency situations. Find: Cost of Inventory; compare to consequences of not having Replacements and backups. STILL WORKING ON THIS PROBLEM Solution: Example: Automobile and Replacement Parts. Given: Replacement Lights, Battery, Air Filter, Extra Gasoline, Oil, Water, Tires, Fuel Pump, Water Pump, Alternator, Spark Plugs, Seats, Windows, etc. Find: Cost of Inventory; compare to consequences of not having Replacements and backups. STILL WORKING ON THIS PROBLEM Solution: Lecture Data Analysis 04: Page 6 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt

10. Thomas J. Crawford, PhD, PE November 02, 2013 B. Thermodynamics:, continued Thermodynamic Efficiencies, Moran 4 th ed., Chapter 5, pp 215 - 218. 1. Heat Engine eff = η = desired output = W = Q H – Q C required input = Q H Q H Q H = W + Q C  W = Q H – Q C Carnot efficiency for a Reversible Cycle eff = η = T H – T C η = 1.0 – T H / T C T H 2. Refrigerator. C.O.P = β = output = Q C = Q C input W Q H – Q C Q H = W + Q C  W = Q H – Q C For a Reversible Cycle C.O.P = β = __ T C _ T H – T C 3. Heat Pump. C.O.P = γ = output = Q H = Q H input W Q H – Q C Q H = W + Q C  W = Q H – Q C For a Reversible Cycle C.O.P = γ = __ T H _ T H – T C Lecture Data Analysis 04: Page 7 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt THTH QHQH W TCTC QCQC THTH QHQH W TCTC QCQC THTH QHQH W TCTC QCQC Eng.

11. Thomas J. Crawford, PhD, PE November 02, 2013 B. Thermodynamics:, continued WORKING HERE NOV. 02, 2013 Example: Coal–Fire Fossil Power Plant. continued Notice: Lecture Data Analysis 04: Page 8 Data Analysis: Economics, Thermodynamics ENGR1181_Lect_Data_Anal04.ppt