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               Chapter 3                     Hence (3.9) becomes
               The Second Law of Thermodynamics
                                                                   dq    0   Carnot cycle, perf. gas        (3.12)
                                                                T

                                         We have
                                                               dq      2    dq      3    dq      4    dq      1    dq  (3.13)
                                                             T       T        T       T        T
                                                                   1       2        3       4
                                         Since processes 2 → 3 and 4 → 1 are adiabatic with dq   0, the second and fourth inte-
                                         grals on the right side of (3.13) are zero. For the isothermal process 1 → 2, we have T
                                                                                                           2
                                                                                            2
                                         T . Since T is constant, it can be taken outside the integral:   T   1  dq   T H  1    dq
                                          H
                                                                                            1
                                                                                                           1
                                                         4
                                         q /T . Similarly,   T  1  dq   q /T . Equation (3.12) becomes
                                          H
                                             H
                                                                    C
                                                         3
                                                                       C
                                                            dq     q H     q C    0   Carnot cycle, perf. gas  (3.14)
                                                           T    T H  T C
                                             We now find e , the maximum possible efficiency for the conversion of heat to
                                                         rev
                                         work. Equations (3.3) and (3.14) give e   1   q /q and q /q   T /T . Hence
                                                                                  C  H     C  H      C  H
                                                                    T C   T   T C
                                                                           H
                                                           e rev    1               Carnot cycle            (3.15)
                                                                    T H     T H
                                         We derived (3.15) using a perfect gas as the working substance, but since we earlier
                                         proved that e rev  is independent of the working substance, Eq. (3.15) must hold for any
                                         working substance undergoing a Carnot cycle. Moreover, since the equations e rev    1
                                         q /q and e rev    1   T /T hold for any working substance, we must have q /q
                                                                H
                                                             C
                                                                                                              H
                                                                                                           C
                                          C
                                             H
                                          T /T or q /T   q /T   0 for any working substance. Therefore
                                            C
                                                               H
                                                            H
                                                       C
                                                    C
                                               H
                                                                 dq     q C     q H    0   Carnot cycle     (3.16)
                                                               T    T C  T H
                                         Equation (3.16) holds for any closed system undergoing a Carnot cycle. We shall use
                                         (3.16) to derive the state function entropy in Sec. 3.3.
                                             Note from (3.15) that the smaller T is and the larger T is, the closer e  ap-
                                                                            C                H              rev
                                         proaches 1, which represents complete conversion of the heat input into work output.
                                         Of course, a reversible heat engine is an idealization of real heat engines, which in-
                                         volve some irreversibility in their operation. The efficiency (3.15) is an upper limit to
                                         the efficiency of real heat engines [Eq. (3.7)].
                                             Most of our electric power is produced by steam engines (more accurately, steam
                                         turbines) that drive conducting wires through magnetic fields, thereby generating elec-
                                         tric currents. A modern steam power plant might have the boiler at 550°C (with the
                                         pressure correspondingly high) and the condenser at 40°C. If it operates on a Carnot
                                         cycle, then e    1   (313 K)/(823 K)   62%. The actual cycle of a steam engine is
                                                    rev
                                         not a Carnot cycle because of irreversibility and because heat is transferred at temper-
                                         atures between T and T , as well as at T and T . These factors make the actual effi-
                                                       H      C             H     C
                                         ciency less than 62%. The efficiency of a modern steam power plant is typically about
                                         40%. (For comparison, James Watt’s steam engines of the late 1700s had an efficiency
                                         of roughly 15%.) River water is commonly used as the cold reservoir for power plants.
                                         A 1000-MW power plant uses roughly 2 million L of cooling water per minute
                                         (Prob. 3.29). About 10% of the river flow in the United States is used by power plants
                                         for cooling. A cogeneration plant uses some of the waste heat of electricity generation
                                         for purposes such as space heating, thereby increasing the overall efficiency.
                                                                 14
                                             The annual intake of 10 gallons of cooling water from rivers, lakes, and coastal
                                         waters by United States electric-power plants and industrial plants kills billions of fish.
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