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1.To go to all coal, using Table 1.1 for 2003, when we used 98.3 QBtu/year (Table 1.1)

Monday’s CAPA #2. 1.To go to all coal, using Table 1.1 for 2003, when we used 98.3 QBtu/year (Table 1.1) Now (+10%) we use 1.10 * 98.3 Quads = 108.13 *10 15 Btu/year. Table 1.1, upper right corner, gives 1QBtu=47.8*10 6 tons of coal

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1.To go to all coal, using Table 1.1 for 2003, when we used 98.3 QBtu/year (Table 1.1)

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  1. Monday’s CAPA #2 1.To go to all coal, using Table 1.1 for 2003, when we used 98.3 QBtu/year (Table 1.1) Now (+10%) we use 1.10 * 98.3 Quads = 108.13 *1015 Btu/year. Table 1.1, upper right corner, gives 1QBtu=47.8*106 tons of coal Tons = 108.13 QBtu *47.8 * 106 tons/QBtu =5168.6*106 tons=5.168*109 tons =5.168E9 for CAPA OR—if you used from the front cover: Front cover, 1 ton (2000 pounds) of coal holds 2.668*107 Btu/ton Tons = Btu / (Btu/ton)  108.13 *1015 Btu / (2.66*107Btu/ton)=40.65 * 10 15-7 =40.65*108 tons = 4.065*109 tons 4.065E9 for CAPA Which is correctly done, but CAPA will mark it wrong. Coal is too variable! (compare to Fig. 2.7, with about 1050*106 tons/year; 1.050*109/4.065*109 = 26%, much as this problem) 2. Your share=( 5.168*109 tons *2000 pounds/ton) / (320*106 people) = 32.3*109-6 =32,300 pounds/person each year = 32,300/365 =88.4 pounds / day= 88.4 for CAPA

  2. TodayHeat engines Use heat energy to do work. Chapter 3 All power point images are only for the exclusive use of Phys3070/Envs3070 Spring term 2014

  3. The Industrial Revolution Convert millions of years of collected solar energy stored in fossil fuels for heat, into mechanical work= force times distance. By ‘heat engines’—many kinds Diesel, gas turbine, steam pistons, ……. To lift, push, spin (electricity)….

  4. Think about a water mill. Water at height Hhigh * work Falls to Hlow CHANGE in potential energy = m g (Hhigh-Hlow) With g =9.8 m/sec2

  5. Reminder! Be orthodox in your units. If m is in kg, heights in meters, and g=9.8 m/sec2, then the energy will be in Joules.

  6. Because water falls down spontaneously. And we can use that fall to do useful work.

  7. Ditto for heat • And heat (a form of energy) also ‘falls’ spontaneously from high temperature to low. Either as heat alone conducting through material, or by moving the hot stuff. (old idea—heat as an invisible fluid, ‘caloric’) • As water falling, we may extract useful energy from this process. • Called a ‘heat engine’. Forget the details.

  8. A Law and a Definition • Energy is conserved Heat in = heat out + work done Qin = Qout + W Or--- QHot = QCold + W • Efficiency = e = Work done/Heat energy in = W/Qin ( a definition) (careful! Heat in Calories or Btu, work in Joules or kW-hr)

  9. Example I burn one ton of coal to generate 2.66 x 107 Btu of heat energy (text), and use this energy to do 10 10 Joules of useful work. A) how much heat was exhausted? Qout = Qin – W = 2.66 x 107 Btu – 1010 J =2.66*107Btu * 1055 J/Btu – 1010 J = 2.81 x 1010 J – 1 x 1010 J = 1.81 x 1010 J OR Qout=2.66*107 Btu-1010J / (1055 J/Btu) =2.66*107 Btu - 0.948*107 Btu = 1.71*107 Btu B) What is the efficiency of my engine? e=W/Qin = 1010 J / 2.66 x 107 Btu =1010J / 2.81 x 1010J = 0.3559 = 35.59%.

  10. Ditto for the RATE of doing work Heat Power in (tons of coal per day) = Heat power out (Btu/second)+(work) Power out (kWatt). Efficiency =work power out/Rate of heat energy in, in the same units!

  11. I burn one bbl of oil per hour at 25% efficiency. How many kW of electricity can I sell? I bbl/hr=6.12x109 J/3600 sec =1.7x106 J/sec =1.7 x 106 watts = 1700 kW= heat power in Useful power out = 0.25 x 1700 kW = 425 kW

  12. Or-- One bbl/hr =1700kWh/hour=1700 kW At 25%425 kW

  13. How many Btu must I dump in that hour? • The other 75%, or 0.75 x 1700 kWh=1275 kWh • 1 kwh=3413 Btu • So I must dump 1275 kWh x 3413 Btu/kwh =4,351,575=4.35 million Btu each hour into a cooling stream or the air.

  14. Two ways to specify power • Power plants may be rated by their thermal power GWt • Or their electrical power output GWe with GWe = efficiency * GWt • A typical thermal power plant has e = 30-35%. • The rest of the thermal energy is dumped somewhere. • Why so poor?

  15. There is a limit to the efficiency of any heat engine. • Since we cannot extract all the heat energy, since to do so would have to exhaust heat into some place at absolute zero, equal to -273 deg C.

  16. Change in potential energy = m g H = work energy available ALL of the gravitational potential energy? ALL the way down.

  17. Qin Work=Qin-Qout Qout ALL of the heat energy? ALL the way down, to absolute zero = -273 deg C

  18. Best possible efficiency? e = W / Qin And best (Carnot) e= 1-Tcold/Thot With T in degKelvin (K) = deg C + 273 0 deg K = -273 deg C Room temperature = 20 deg C = 293 deg K

  19. Example of the best possible efficiency Make a fire at T hot = 500 deg C Exhaust waste heat at Tcold = 10 deg C. ecarnot= 1 – Tc/Th Thot =500+273 = 773 deg K Tc =10+273 =283 deg K eCarnot = 1 – 283/773 = 1 – 0.366 =0.634 =63.4%

  20. e=100% ? ecarnot = 1 – Tc/Th = 1.00000 • Tc = zero deg K (absolute zero) In practice, about 60 % of ecarnot

  21. Text pp138-145

  22. Generating electricity Spin a coil of wire in a magnetic field. The work done to rotate the coil appears as electrical energy in the wires of the coil. Can be 40% or so efficient, turning heat energy into mechanical rotational energy into electrical energy. More on Feb. 19

  23. Exam 1 • In class, Monday Feb. 17, 0900-0950. • Odd seats, please, or sit in the front three rows. • Text and calculator OK • Covers Ch 1-3, and the extra as in the posted page. • Some short multiple choice numerical problems. • Some longer problems, with parts. Some may ask for analysis, commentary, or a conclusion from the numbers. (HW #3 is also due) More on Friday

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