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Engines

Engines. Physics 202 Professor Lee Carkner Lecture 16. PAL #15 Entropy. 1 kg block of ice (at 0 C) melts in a 20 C room D S = S ice + S room S ice = Q/T = mL/T = [(1)(330000)]/(273) S ice = S room = Q/T = -330000/293 S room = D S = +1219.8 – 1136.5 =

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Engines

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  1. Engines Physics 202 Professor Lee Carkner Lecture 16

  2. PAL #15 Entropy • 1 kg block of ice (at 0 C) melts in a 20 C room • DS = Sice + Sroom • Sice = Q/T = mL/T = [(1)(330000)]/(273) • Sice = • Sroom = Q/T = -330000/293 • Sroom = • DS = +1219.8 – 1136.5 = • Entropy increased, second law holds

  3. PAL #15 Entropy • Entropy of isobaric process, 60 moles compressed from 2.6 to 1.7 ms at 100 kPa • DS = nRln(Vf/Vi) + nCVln(Tf/Ti) • Ti = (100000)(2.6)/(60)(8.31) =521 K • Tf = (100000)(1.7)/(60)(8.31) = 341 K • DS = -530 J/K • Something else must increase in entropy by more than 530 J/K

  4. Engines • General engine properties: • A working substance (usually a gas) • A net output of work, W • Note that we write Q and W as absolute values

  5. The Stirling Engine • The Stirling engine is useful for illustrating the engine properties: • The input of heat is from the flame • The output of heat makes the cooling fins hot

  6. Heat and Work • How does the work compare to the heat? • Since the net heat is QH-QL, from the first law of thermodynamics: DEint=(QH-QL)-W =0 W = QH - QL

  7. Efficiency • In order for the engine to work we need a source of heat for QH e = W/QH • An efficient engine converts as much of the input heat as possible into work • The rest is output as QL

  8. Efficiency and Heat e = 1 - (QL /QH) QH = W + QL • Reducing the output heat means improving the efficiency

  9. The Second Law of Thermodynamics (Engines) • This is one way of stating the second law: It is impossible to build an engine that converts heat completely into work • Engines get hot, they produce waste heat (QL) • You cannot completely eliminate friction, turbulence etc.

  10. Carnot Efficiency eC = 1 - (TL / TH) • This is the Carnot efficiency • Any engine operating between two temperatures is less efficient than the Carnot efficiency e < eC There is a limit as to how efficient you can make your engine

  11. The First and Second Laws • You cannot get out more than you put in • You cannot break even • The two laws imply: • W < QH • W  QH

  12. Dealing With Engines W = QH - QL e = W/QH = (QH - QL)/QH = 1 - (QL/QH) eC = 1 - (TL/TH) • If you know TL and TH you can find an upper limit for e (=W/QH) • For individual parts of the cycle you can often use the ideal gas law: PV = nRT

  13. Engine Processes • We can find the heat and work for each process • Net input Q is QH • Net output Q is QL • Find p, V and T at these points to find W and Q

  14. Carnot Engine • A Carnot engine has two isothermal processes and two adiabatic processes • Heat is only transferred at the highest and lowest temperatures

  15. Next Time • Read: 20.8

  16. When water condenses out of the air onto a cold surface the entropy of the water, • Increases, since entropy always decreases • Decreases, but that is OK since the 2nd law does not apply to phase changes • Decreases, but that is OK since the entropy of the air increases • Increases, since phases changes always increase entropy • Remains the same

  17. Water is heated on a stove. Which of the following temperature changes involve the greatest entropy change of the water? • T increased from 20 to 25 C • T increased from 40 to 45 C • T increased from 80 to 85 C • T increased from 90 to 95 C • All are equal

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