HE SECOND LAW OF HERMODYNAMICS 9 EXERCISES Setions 9. and 9.3 e Seond Law of ermodynamis and Its Appliations 3. INERPRE is problem requires us to alulate te effiieny of reversible eat engines tat operate between te given temperatures. DEELOP e maximum effiieny of a reversible engine, operating between two absolute temperatures, >, is given by Equation 9.3, e Carnot /. Apply tis to ea part of tis problem to find te orresponding effiienies. EALUAE (a) e ( 73 K) ( 373 K) 6.8%. (b) e Δ ( K) ( 98 K) 7.05%. () Wit room temperature at e ( ) ( ) ASSESS 93 K 73 K 77.0%. e engine wit te largest differene in reservoir temperature as te largest effiieny. 4. INERPRE is problem is an exerise in alulating te termal effiieny of a eat engine, given te temperature of te ot and old reservoirs. DEELOP Assuming tis osmi eat engine were a reversible engine, its effiieny would be tat of a Carnot engine, given by Equation 9.3: ecarnot EALUAE Substituting te values given in te problem, we obtain e Carnot.7 K 4 4.66 0 99.95% 5800 K ASSESS e engine effiieny is almost 00%. is is too good to be true. 5. INERPRE is problem involves a reversible Carnot engine tat runs between te boiling and melting point of He. We are given te engine s effiieny and are asked to find te temperature of its old reservoir (i.e., te melting point of He). DEELOP Apply Equation 9.3, wi gives te effiieny of a Carnot engine. We are given e Carnot 0.777 and 4.5 K, so we an solve for, wi will be te melting point of He. EALUAE e melting point of He is ecarnot e (.000 0.777)(4.5 K) 0.948 K ASSESS ( ) Carnot is agrees wit te melting point of He found in te literature, so it seems to be a reasonable result. 6. INERPRE is problem is about a Carnot engine tat operates via te Carnot yle. We are given te termal energy absorbed per yle and te work done per yle by te engine, and we are asked to find its effiieny. 9- Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9- Capter 9 DEELOP By definition, te effiieny of an engine is W e Q were W is te work done by te engine and Q is eat absorbed from te ot reservoir per yle. For part (b), te work W is defined as te work done by te gas, so W 350 J in tis ase. e eat transferred is Q Q Q. Beause te tere is no net ange in te internal energy of te engine over a yle, te first law of termodynamis gives (note tat W in Capter 9 is defined ontrary to W in Capter 8!) W Q Q For part (), Equation 9. gives Q Q so we an solve for given tat 0 C 83 K. EALUAE (a) From te equation above, te effiieny of te engine is (b) e eat rejeted ea yle is W 350 J e 38.9% Q 900 J Q Q W 900 J 350 J 550 J () e temperature of te ot reservoir is Q 900 J ( 83 K ) 463 K 90 C Q 550 J ASSESS e maximum temperature is greater tan, as our alulation onfirms. Note tat Carnot s teorem applies to te ratio of absolute temperatures. 7. INERPRE We are to find te oeffiient of performane of a reversible refrigerator tat operates between 0 C and 30 C. DEELOP For a refrigerator, te oeffiient of performane is given by Equation 9.4, Q COPrefrigerator Q Q Use Equation 9., Q /Q / to onvert tis to an expression involving temperature: Q Q COPrefrigerator Q Q Q Q Q wi we an solve given tat 0 C 73 K and 30 C 303 K. EALUAE Inserting te given quantities into te expression for te COP gives 73 K COP refrigerator 9.0 303 K 73 K ASSESS Notie tat te temperatures are absolute temperatures (i.e., Kelvin). 8. INERPRE is problem requires us to find te work done by a refrigerator to freeze te given quantity of water. e eat of transformation (Capter 7) is involved in te liquid-to-solid pase ange. DEELOP e amount of eat tat must be extrated in order to freeze te water is (Equation 7.5) ( )( ) Q mlf 0.67 kg 334 kj/kg 4 kj e work onsumed by te refrigerator wile extrating tis eat is given by Equation 9.4, W Q /COP. EALUAE Inserting te given values, we obtain Q 4 kj W 53 kj COP 4. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9-3 ASSESS A COP of 4. means tat ea unit of work an transfer 4. units of eat from inside te refrigerator. A smaller COP would mean tat more work is required to freeze te water. 9. INERPRE We want to know if te uman body an be onsidered as a eat engine, wi as stringent limits on its effiieny. DEELOP If te uman body were a eat engine, te maximum effiieny it ould attain would be given by te effiieny for a Carnot engine (Equation 9.3): e /. EALUAE Body temperature is 37 C 30 K. So if we assume te ambient temperature is 0 C 93 K, te maximum effiieny would be 93 K e 5% 30 K So under normal irumstanes, te uman body is far too effiient at 5% to be a eat engine. ASSESS We often say te body burns alories, wi sounds like it s just releasing random eat from te food we eat. But atually te proess is more speifi. Energy-storing moleules interat wit oter moleules to ause preise emial reations tat result in, for example, a musle ontrating or a neuron produing a urrent. Not all of te stored energy is onverted to useful work, owever. Some of it ends up as eat. Setion 9.4 Entropy and Energy Quality 0. INERPRE is problem requires us to alulate te entropy ange involved wit melting te given quantity of ie at 0 C. DEELOP Beause te ie is at 0 C, no temperature ange is involved in melting it. erefore, Equation 9.6 takes te form ΔQ Δ S e ange in eat is just te latent eat of water, wi is given by Equation 7.5 (ΔQ ml f ) and able 7. (L f 334 kj/kg). e temperature is 0 C 73 K. EALUAE Inserting te given quantities into te expression for entropy ange gives ΔQ mlf (.0 kg)( 334 kj/kg) Δ S. kj/k 73 K ASSESS us te water as more entropy tan te ie, so it as a greater apaity to do work.. INERPRE e problem onerns te entropy inrease assoiated wit metabolizing a amburger. DEELOP We ll assume te energy in te burger, Q, flows into te body as eat. erefore, te entropy ange from state (burger ingested) and state (burger metabolized) is given by Equation 9.6: Δ S dq/. EALUAE e body temperature, 37 C 30 K, remains onstant trougout, so Q 650 kal 4.84 kj Δ S dq 8.8 kj/k 30 K kal ASSESS Altoug we assumed te burger s energy went into eat, te answer would be te same if te body used some of te energy to do work. In eiter ase, te burger s energy is no longer available to do work one it as been metabolized.. INERPRE is problem asks us to alulate te entropy inrease upon eating up a given mass of water from 0 C to 95 C. DEELOP e ange in eat ontent of te water is given by Equation 6.3, Q mδ, were 4.84 kj/(kg K) for water (see able 6.). For a substane like water wit onstant speifi eat (in tis ase at onstant pressure), we an differentiate tis expression to find dq md, so (from Equation 9.6) te ange in entropy is Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-4 Capter 9 dq d Δ S m mln e initial and final temperatures are 0 C 83 K and 95 C 368 K, respetively. EALUAE Inserting te given values, we ave 368 K Δ S mln ( 0.5 kg) 4.84 kj/ ( kg K) ln 80 J/K 83 K to two signifiant figures. ASSESS e final entropy of te system as inreased. 3. INERPRE is problem requires us to find te mass of a blok of lead given its entropy inrease assoiated wit its solid-to-liquid pase transition (i.e., melting it). DEELOP e lead starts at its melting-point temperature, so tere is no ange in temperature assoiated wit te solid-to-liquid pase ange. erefore, Equation 9.6 for entropy ange takes te form ΔQ Δ S e eat ange is given by Equation 7.5, ΔQ ml f, were L f 4.7 kj/kg (see able 7.). Insert tis into te expression for entropy ange and solve for te mass m. EALUAE From te above equation, we find te mass of lead to be ASSESS mlf Δ S f ( 600 K)( 900 J/K) ΔS m.9 kg L 4.7 kj/kg As expeted, te mass of te blok is proportional to te ange in entropy. 4. INERPRE We are to find te energy tat beomes unavailable during an irreversible isotermal proess given te entropy inrease. DEELOP e energy tat beomes unavailable is given by E minδ S (see setion Entropy and te Availability of Work), were min is te lowest temperature available to te system. e ange in entropy ΔS is Δ S 5 J/K, and te temperature is min 440 K. EALUAE Inserting te given quantities in te expression for unavailable work gives E minδ S ( 440 K)( 5 J/K) kj ASSESS Sine tis is an isotermal proess, te minimum temperature is te maximum temperature is te temperature. 5. INERPRE We re asked to find te probability tat 6 moleules are distributed in different ways inside a box. is as relevane to te statistial interpretation of entropy. DEELOP Considering a single moleule, te probability tat it is loated on te left-side or te rigt side of te box is ½. Considering moleules, te probability for one partiular left-rigt arrangement (mirostate) is ¼. Anoter way to say tis is tat tere are 4 different ways to sort te moleules between te two sides. For 6 moleules, tere are 6 64 ways to sort, so te probability for one partiular arrangement (mirostate) is /64. We now ave to ount ow many of tese mirostates mat te following marostates (see Figures 9.8 and 9.9). EALUAE (a) ere s one mirostate in wi all of te moleules are found on te left side of te box, and anoter one were all are found on te rigt side of te box. So te probability of tis marostate is /64 + /64 /3. (b) It s a bit arder to find te number of mirostates wit alf te moleules on one side, alf on te oter. So let s label te moleules A, B, C, D, E, and F, and let s identify a mirostate by te 3 moleules on te left-and side. So for example, (ABC) is te mirostate wit A, B, C on te left-and side, and te oters on te rigt. We an Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
PROBLEMS e Seond Law of ermodynamis 9-5 swit out C in tree different ways: (ABD), (ABE) and (ABF). Similarly we an swit out B in tree different ways and swit out A in tree different ways. at gives us 0 mirostates. We ount anoter 0 mirostates if we start wit (DEF), and swit out D, ten E, ten F. e total number is 0 mirostates. So te probability of te marostate wit te moleules split evenly between te sides is 0/64. ASSESS It is 0 times more likely tat te 6 moleules will be spread out evenly between te two sides of te box vs. all on one side. In general, if tere are n moleules, te probability tat k of tem will be on one side and (n k) on te oter side is given by te oeffiients from te binomial teorem: n n! P n n k k! ( n k)! In te ase above, n 6 and k 3, so te probability is P 6!/ ( 3! 3! ) 6 0 / 64, as we found. 6. INERPRE is problem is about a Carnot engine, its work, effiieny and power output. DEELOP For a yli operation, te ange in internal energy is zero, Δ U 0. From te first law of termodynamis, we ave. e W/ Q. For a Carnot engine, Q/ Q /. EALUAE (a) e work done by te engine during ea yle is W Q Q 745 J 458 J 87 J (b) e effiieny of te engine is () For a Carnot engine, W 87 J e 38.5% Q 745 J Q 458 J (59 K) 364 K Q 745 J (d) e work from (a) is per yle, so te meanial power output of te engine is tis divided by te yle period, wi is equivalent to multiplying by te frequeny ( f / ) : W Pout Wf ( 87 J/yle)( 8.6 yles/s) 5.34 kw ASSESS e ool reservoir temperature is ooler tan te ot reservoir temperature ( ) te effiieny an be verified by using Equation 9.3: e / 38.5%. < as expeted. And, 7. INERPRE is problem requires us to find te termodynami effiieny of a nulear power plant in winter and in summer, wen te temperature of its old reservoir is 0 C and 5 C, respetively. DEELOP From Equation 9.3, te termodynami effiieny of a Carnot engine is ecarnot were te temperatures are in Kelvin. EALUAE Inserting te given temperatures for summer and winter gives 98 K e summer 47.7% 570 K 73 K e winter 5.% 570 K ASSESS e plant is more effiient in winter tan in summer beause tere is a greater eat differene. However, as explained in Setion 9.3, irreversible proesses, transmission losses, and so on make atual effiienies less tan te teoretial maxima. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-6 Capter 9 8. INERPRE You want to alulate te minimum oeffiient of performane for a eat pump tat s to be installed in a new ouse. DEELOP For a eat pump, wat you want is eat, and wat you put in is work in te form of eletriity, so COP Q / W. In tis ase, you are dealing wit rates, tat is, te rate tat te ouse needs to be eated, H, and te eletri power, P, tat te solar voltai system supplies to te eat pump. e oeffiient of performane an terefore be written as COP H / P. EALUAE e minimum COP needed to keep te ouse warm on a old day is H 6.85 kw COP.95 P.3 kw ASSESS is is entirely reasonable for a eat pump. It migt seem impossible to obtain more eat energy tan you put in, but te eat pump is not generating eat but only moving it from outside to inside. (see Figure 9.). 9. INERPRE is problem involves a nulear power plant and asks us to alulate te rate of energy extration, te effiieny, and te igest temperature te plant attains. DEELOP From Equation 6.3, Q mδ, we see tat to raise te temperature of te ooling water by 8.5 K, eat must be exausted to it at a rate of dq dm 4 4.84 kj/ ( kg K) (.8 0 kg/s)( 8.5 K) 996 MW dt Δ dt We take tis to be te rate of all te eat rejeted by te power plant. Sine te rate of work output dw/dt is also given, te eat input to te plant (extrated from its fuel) is dq dq dw + dt dt dt were we ave used te first law of termodynamis (see Problem 9.6). In terms of te rates, te effiieny of te plant is dw dt e dq dt If we onsider te plant to operate like a Carnot engine, ten its igest temperature an be alulated using Q Q (from Equation 9.). EALUAE (a) Substituting te values given, we obtain dq dq + dw 996 MW + 750 MW.75 GW dt dt dt were te negative sign orresponds to te energy being extrated from te fuel. (b) e plant s effiieny (from te definition of effiieny in terms of rates) is dw dt 750 MW e 43.0% dq dt.75 GW () Wit te assumption tat te plant operates like an ideal Carnot engine, ten Q dq dt.75 GW.75 Q dq dt 996 MW (Note tat te energy rate per yle and te energy rate per seond are proportional.) If 5 C 88 K, ten ( ).75.75 88 K 505 K 3 C ASSESS e atual igest temperature would be somewat greater tan tis, beause te atual effiieny is always less tan te Carnot effiieny. 30. INERPRE is problem requires us to alulate te effiieny of an eletrial power plant, given te temperature of its ot and old reservoirs. From tis, we are to find te power disarged as waste eat and te number of ouses we ould eat wit tis waste eat. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9-7 DEELOP e maximum effiieny of te plant as a funtion of te temperature of its ot and old reservoirs is given by Equation 9.3 ecarnot wi we are to ompare wit te atual effiieny wi is W e 0.9 Q e atual effiieny may also be expressed as e Q /Q (Equation 9.). In tis apter, W is defined as te work done by te system (as ompared to W in te first law of termodynamis, Equation 7., wi is te work done on te system). us, W is equal to te net eat, W Q Q Q Q + W were Q is te eat rejeted by te system to te old reservoir and Q is te eat absorbed by te system from te ot reservoir. Using tis result to eliminate Q in te expression for te plant s atual effiieny gives ( e) Q W e Differentiating tis wit respet to time gives te waste power dq /dt in terms of te power output dw/dt 800 MW and te atual effiieny of te plant e: dq dw e dt dt e Finally, te number of ouses tat an be warmed wit tis waste power is simply te waste power divided by te power requirement of a single ouse, or dq dt Nouses Pouse were P ouse 3 kw. EALUAE (a) e maximum possible effiieny for te power plant is 73 K + 30 C 303 K ecarnot 45.% > e 73 K + 80 C 553 K (b) e waste power is dq dw e 0.9 ( 800 MW). GW dt dt e 0.9 () e number ouses tat ould be eated by te waste power is 9 dq dt. 0 W 4 Nouses 3 9.4 0 P 3 0 W ASSESS As expeted, te maximum effiieny e Carnot is greater tan te atual effiieny e. ouse 3. INERPRE is problem asks us to find te rate (i.e., kg/s) at wi all te power plants in te USA use ooling water. We are given te atual effiieny of te power plants and te temperature rise in te ooling water. DEELOP For a yli operation, te ange in internal energy is zero, Δ U 0. From te first law of termodynamis, we ave W Q Q, were W is te work done by te system (ontrary to te definition of W in Capter 8), Q is te eat absorbed by te system, and Q is te eat rejeted by te system. erefore, te total rate at wi eat is exausted by all te power plants is dq d dw ( Q W) ( ) dt dt dt e 33% 0 W 4.06 0 W Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-8 Capter 9 e mass rate of flow at wi water ould absorb tis amount of energy, wit only a 5 C temperature rise, is dq dm water Δ dt were water 484 J/(kg K) (see able 6.). is equation an be solved to give te mass rate of ooling water used. EALUAE Solving te equation above for dm/dt, we obtain dm dq water dt 4.06 0 W 7 Δ 0 kg/s dt dt 484 J/ ( kg K) ( 5 K) or about Mississippi (a self-explanatory unit of river flow). ASSESS o absorb te power output of 0 W wit only an inrease of temperature of 5 C, we expet te mass flow rate to be large. 3. INERPRE For tis problem, we are to sow tat te overall effiieny of a two-stage eat engine is te same as a single-stage engine operating between and of te two-stage engine. DEELOP Let te eat exausted Q i by te first engine equal te eat input to te seond. en, Q W+ Qi and Qi W + Q, were Q is te eat exausted to te environment. For a single-stage engine operating between and, te atual effiieny is te total work W + W divided by Q, (Equation 9.), wi we an evaluate to ompare wit te Carnot effiieny of te two-stage engine, wi is e / (Equation 9.3). EALUAE e atual effiieny of te single-stage engine is W+ W Q Qi + Qi Q Q e Q Q Q wi is te Carnot effiieny of te two-stage engine. ASSESS In te last equality, we used Equation 9., / Q /Q. 33. INERPRE is problem involves finding te COP of a freezer for wi te igest and te lowest temperatures are 3 C 305 K and 0C 73K. In addition, we are to find ow mu water at 0 C te freezer an freeze in one our. DEELOP e oeffiient of performane (COP) of a reversible freezer is given by Equation 9.4: COP Q Q W Q Q were we ave used Equation 9., / Q /Q for te last equality. One te COP is known, we an solve for Q and te amount of water te freezer an freeze in one our, wi is m Q Lf wit Lf 334 kj/kg (see Equation 7.5 and able 7.). EALUAE (a) e COP of te freezer is 73 K COP 8.53 305 K 73 K (b) e eat rejeted in one our is Q W ( ) COP 8.53 kw 369 MJ, so te water we an freeze is Q 369 MJ 3 m.0 0 kg. L f 334 kj/kg ASSESS ypial freezers ave a COP lower tan 8.53. us, more eletrial energy is needed to freeze te same amount of water. 34. INERPRE We are to use an energy-flow diagram to analyze weter using a refrigerator to ool te lowtemperature reservoir of a Carnot an inrease its overall effiieny. DEELOP In order to lower te temperature of te old reservoir, te ombination must remove more eat from te old reservoir tan it puts into te reservoir. Sine bot te engine and te refrigerator are at te Carnot effiieny, tey are reversible. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9-9 EALUAE As sown in te figure below, te best tat tis an do is zero work output wile te temperatures stay te same. If te temperature of te old reservoir is to beome lower, ten te eat extrated from te old reservoir must be larger, wi will require more work tan te engine produes. ASSESS Migt as well do noting at all! 35. INERPRE is problem requires us to find te montly ost of using all te inoming eletrial energy to power a eat pump wit COP 3.4 to eat a ouse. We are given tat te eletrial energy osts $30 per mont in te winter. DEELOP e same eletrial energy W used for diret onversion in eletri eating would produe eat Q W + Q. Using Equation 9.4 allows us to express tis as ( COP) Q W + Q W + Q Q W( COP) us, te eat pump an produe a fator COP more eat tan if te eletrial energy is onverted diretly to eat. EALUAE Beause te eat pump is a fator COP more effiient, te ost will be redued by tis same fator, so te montly eating bill would be $30 $30 $68 COP 3.4 ASSESS e savings are signifiant, wi is wy eletrial eating is not reommended. 36. INERPRE is problem requires us to find te power needed to run a refrigerator tat leaks eat at te given rate to its old reservoir (i.e., te environment). e refrigerator operates as a reversible eat engine, so we know its COP. DEELOP e rate at wi eat leaks from te refrigerator is dq dt 340 W. For a reversible refrigerator, te COP is given by Equation 9.4: Q 77 K COP refrigerator 0.65 W 6 K Q W 0.65 Differentiating tis expression wit respet to time gives te power needed to run te refrigerator. EALUAE e power needed to run te refrigerator is dw dq 340W 3 W dt dt 0.65 0.65 ASSESS e ooling apaity of te refrigerator is not given, so 340 W is te minimum eat leakage, wi means tat tis power is te minimum power required to run te refrigerator. 37. INERPRE We are to find te minimum COP required to save money if we swit from an oil furnae to an eletrially powered eat pump, onsidering te ost of oil and of eletriity. We will do tis by alulating te ost of te eat delivered by bot te oil-burning eater and te eletri eat pump. DEELOP e oeffiient of performane (COP) is te relationsip between te eat sent to te old reservoir and te work done. Set te eat Q to be te same for bot eating meanisms, and solve for COP. e ost of oil $.75 is $ oil $0.0583 kw, and te ost of eletriity is $ 30 kw eletri $0.65 kw. e eat delivered is Q W COP Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-0 Capter 9 (see derivation of Equation 9.4), and we are paying for te work done, so te COP must exeed te ratio of te osts (COP > $ eletri /$ oil ). EALUAE e COP must satisfy COP > $ eletri /$ oil.83 So in order to be ost-effetive, te eat pump must ave a COP of greater tan.83. ASSESS Most eat pumps ave a COP mu iger tan tis value, so it s probably a good idea to swit. 38. INERPRE For tis problem, we are to sow tat te Clausius statement (i.e., te seond law of termodynamis) would be violated by te existene of a perfet eat engine, wi would allow te onstrution of a perfet refrigerator. DEELOP If it were possible to onstrut a perfet eat engine (one wi would extrat eat and perform an equivalent amount of work), ten it ould be oupled to a real refrigerator in su a way tat te work output of te engine equals te work input to te refrigerator, as sown in te figure below. EALUAE e net effet of tis arrangement is to produe a perfet refrigerator (a yli devie wose sole effet is te transfer of eat, Q + Q, Q Q from a old reservoir to a ot one), in violation of te Clausius statement of te seond law. ASSESS is ompletes te proof of te equivalene of te Kelvin-Plank and Clausius statements in Setion 9.. 39. INERPRE We are asked to find te COP, power usage, and operating ost ompared to tat of an oil-burning eater of a eat pump. We will assume tat te eat pump is a Carnot eat pump. DEELOP e maximum COP of a eat pump (wen its eating, not ooling) is given in Equation 9.4b: COP / ( ). In tis ase, 0 C 83 K and 70 C 343 K. In general, te COP for a eat pump is te eat supplied divided by te work input. In terms of rates, tat an be written as COP H / P, were H is te supplied eat rate and P is te eletri power onsumption. EALUAE (a) Assuming te eat pump is maximally effiient, 343 K COP 5.7 5.7 343 K 83 K (b) e power onsumption needed to supply eat at 0 kw is H 0 kw P 3.5 kw COP 5.7 () Given te utility rate for eletri power, te eat pump s ourly operating ost is ( Pt) R ( )( )( ) Cpump utility 3.5 kw 5.5 / kw 54. In omparison, an oil furnae, supplying te same eat, would ave an ourly operating ost of ( ) ( ) 0 kw C oil $3.60 / gal $.40 30 kw/gal ASSESS e ost per kw of oil is atually less tan tat of eletriity: R oil 8.7 / kw. But te eat pump as su a ig COP tat it ost tree times less to eat te ouse. 40. INERPRE is problem involves a reversible engine tat ontains a given volume of monatomi gas. e system goes troug te four termodynamis proesses indiated in te problem statement, and we are to find net Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9- work done by te system and te net eat added to te system over a omplete yle, and te engine s effiieny as defined by te ratio of te work done by te engine to te eat absorbed over a omplete yle. DEELOP e p diagram for te yle is as sown below. Let us alulate te work and eat absorbed for ea stage of te yle. For te isotermal expansion AB te ange in internal energy of te system is zero (ΔU 0), so te first law of termodynamis gives 4.84 L QAB WAB nraln ( B A ) ( 0.35 mol) 8.34 J/ ( mol K) ( 586 K) ln 8 J.4 L were Q AB is te eat absorbed and W AB is te work done by te system (note tat tis definition of W is ontrary to te definition used in Capter 8). e seond equality above is from Equation 8.4. For te isovolumi ooling BC, te volume does not ange so no work is done. e eat absorbed is given by Equations 8.5 and 8.3, wi give WBC 0and 3 3 QBC nc Δ BC nr( C B ) ( 0.35 mol ) 8.34 J/ ( mol K ) ( 9 K 586 K ) 83.3 J For te isotermal ompression CD,.4 L QCD WCD nrc ln ( D C ) ( 0.35 mol) 8.34 J/ ( mol K) ( 9 K) ln 589 J 4.84 L and for te isovolumi eating DA, WDA 0and 3 QDA nr( A D) QBC For tese proesses, we are given tat B A C D 4.84 L, A B 586 K, and C D 9 K, so we an sum up te ontributions to work and eat to find te total for ea for a omplete yle. EALUAE (a) e net eat added to te system is e net work done by te system is 0 ( ) Q Q + Q + Q + Q 8 J+ 589 J 593 J ABCDA AB BC DA CD 0 ABCDA AB BC DA CD ( ) W W + W + W + W 8 J+ 589 J 593 J (b) e ratio of te work done to te eat absorbed is WABCDA 593 J 4.0% Q + Q 8 J+ 83 J AB DA A Carnot engine operating between 586 K and 9 K as effiieny 9/586 50.%. is is not ASSESS a ontradition of Carnot s teorem, beause te engine in tis problem does not absorb and exaust eat at onstant temperatures. 4. INERPRE Our engine yle onsists of four pats, two of wi are isoori and two of wi are isobari. We are to determine te effiieny, defined as te work done per unit eat absorbed, and ompare te result wit te effiieny of a Carnot engine operating between te same temperatures. Finally, we need to explain any differene between te two effiienies. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9- Capter 9 DEELOP Label te states in Fig. 9. A, B, C, and D going lokwise from te upper left orner. e work done and te eat absorbed during te isobari segments AB and CD are and ( ) ( )( ) ( ) ( )( ) WAB pa B A 6 atm 6 L L 4 L atm W p 3 atm L 6 L L atm CD C D C 5 p B B p A A 5 QAB ncp ( B A ) n R ( 36 ) L atm 60 L atm nr nr 5 5 QCD ncp ( D C ) ( pdd pcc ) ( 6 8 ) L atm 30 L atm were we ave assumed an ideal monatomi gas (see Equation 8.3). For te isoori segments, we ave 3 QBC nc ( C B ) ( 8 36 ) L atm 7 L atm 3 QDA nc ( A D ) ( 6 ) L atm 9 L atm and W W 0. e net eat added for one yle is terefore BC DA ( 60 7 30 9 ) L atm ( L atm)( 0.3 J/L atm) QABCDA QAB + QBC + QCD + QDA +. kj and te net work done is W ( 4 + 0 + 0 ) L atm L atm. kj. Note tat te first law of termodynamis, applied to a yli proess, requires tat W Q wen using te definition tat W is te work done by te system (wi is opposite to te definition used in Capter 8). EALUAE (a) Sine te eat absorbed is Q + ( 60 + 9 ) L atm 69 L atm, te effiieny is W L atm e 7.4% Q + 69 L atm (b) e maximum and minimum temperatures are B pbb nr and D pdd nr so te effiieny of a Carnot engine operating between tese temperatures is D pdd 6 L atm ecarnot 83.3% B pbb 36 L atm is is not a ontradition of Carnot s teorem, beause te given engine does not operate between two eat reservoirs at fixed temperatures. ASSESS e effiieny of a real engine is always less or equal to tat of a Carnot engine. 4. INERPRE For te given Carnot yle, we are to find te eat absorbed, te eat rejeted, and te work done per yle. We are ten to find te effiieny of te engine and te maximum and minimum temperatures, and sow tat te effiieny as given in Equation 9. equates to te Carnot effiieny of Equation 9.3. DEELOP See Figure 9.5 and te aompanying disussion of te Carnot engine. e eat absorbed is in te isotermal expansion is B B Q nraln P A Aln A A were we ave used te ideal-gas law (Equation 7.) p nr for te last equality, state A (8.000 atm,.000 L), and state B (4.000 atm,.000 L). e eat rejeted during isotermal ompression is C Q P C Cln D Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9-3 were state C (.050 atm, 3.4 L) and state D (4.00 atm,.6 L). Beause te internal energy of te engine does not ange, ΔU 0, so te first law of termodynamis states tat W Q Q were W is te work done by te system. From tese results, we an alulate te effiieny using Equation 9., e W/Q. e maximum and minimum temperatures may be found from te ideal-gas law (Equation 7.), EALUAE (a) e eat absorbed is P / nrand P / nr A A A C C C B 0.3 J Q paaln ( 8.000 atm)(.000 L) 8.34 J/ ( mol K) ln( ) 56.7 J A atm L (b) e eat rejeted is C 3.4 L 0.3 J Q pccln (.050 atm)( 3.4 L) 8.34 J/ ( mol K) ln 464.J D.6 L atm L () e work done by te engine is (d) e effiieny is W Q Q 56.7 J 464.J 97.66 J W 97.66 J e 7.39% Q 56.7 J (e) e maximum and minimum temperatures are ( 8.000 atm)(.000 L) ( 0.0 mol) 8.34 J/ ( mol K) (.050 atm)( 3.4 L) ( 0.0 mol) 8.34 J/ ( mol K) p A A A 487.4 K 490 K nr p C C C 40.6 K 400 K nr to two signifiant figures. e Carnot effiieny is tus 40.6 K e Carnot 7.39% 487.4 K wi is te same result as for part (d). ASSESS For a Carnot engine, te atual effiieny is te Carnot effiieny. ese imply a Carnot effiieny of e C/ A 0.739, exatly as before. Equations 9. and 9.3 are idential beause Q / Q / 0.86, expliitly. C A 43. INERPRE is problem is about te inrease in entropy as te ie is melted and eated up. DEELOP e entropy inrease is given by Equation 9.6: Δ S dq/. We onsider te entropy inrease in two steps. First, te eat needed to melt te lake ie is, f L f 334 kj/kg from able 7.. e temperature is onstant during te melting, 0 73 K. In te seond step, te eat input raises te water temperature aording to dq md, were 4.84 kj/kg K from able 6.. Here, te temperature is not onstant, so we will ave to integrate. EALUAE e entropy inrease during melting is dq ml ( 94 Mg)( 334 kj/kg) ( 73 K) Δ f S melt 0 5 MJ/K e entropy inrease during warming is md 88 K Δ Swarm mln ( 94 Mg)( 4.84 kj/kg K) ln.0 MJ/K 0 0 73 K Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-4 Capter 9 So te total entropy inrease is ASSESS Δ S Δ S +Δ S + tot melt warm 5 MJ/K.0 MJ/K 36 MJ/K.4 0 MJ/K As expeted, te entropy ange is positive in bot melting and warming proesses. 44. INERPRE You want to know te rate at wi your body s entropy inreases during normal metabolism. DEELOP e normal alorie intake is about 000 kal/day for women and about 500 kal/day for men. We ll split te differene and assume a metaboli rate of 50 kal/day. emperature of metabolism rate (or basal metaboli rate) for males is about 300 kal/day and for females is about 00 kal/day. Q flows into te body as eat. erefore, te entropy ange from state (burger ingested) and state (burger metabolized) is given by Equation 9.6: Δ S dq/. EALUAE e body onsumes food energy and onverts it to work and eat at a onstant body temperature of 37 C 30 K, so te rate of entropy inrease is ΔS dq ΔQ/ Δt 50 kal/day 4.84 kj Δt Δt 30 K kal ASSESS One ould say tis is te rate at wi we reate disorder. 0.4 mw/k 45. INERPRE We are to derive te formula given in te problem statement tat desribes te entropy ange for n moles of an ideal gas tat undergoes an isovolumi temperature ange from to. DEELOP From te first law of termodynamis ( dq du + dw) and te properties of an ideal gas ( du nc d and p nr), an infinitesimal entropy ange is dq d p d d ds nc + d nc + nr Integrate from state (, ) to state (, ), and apply te isovolumi onstraint to obtain te given formula. EALUAE Integrating te expression above gives Δ S nc ln + nrln For an isovolumi proess so Δ S nc ln ( ). ASSESS Of ourse, we ould ave started wit dq nc d at onstant volume, but we wanted to display ΔS for a general ideal-gas proess, for use in oter problems. 46. INERPRE We are to derive te formula given in te problem statement tat desribes te entropy ange for n moles of an ideal gas tat undergoes an isobari temperature ange from to. DEELOP From te first law of termodynamis ( dq du + dw) and te properties of an ideal gas ( du nc d and P nr), an infinitesimal entropy ange is dq d p d d ds nc + d nc + nr Integrate from state (, ) to state (, ), and use Equation 8.9, C P C + R, to obtain te given formula. Note tat wen te pressure is onstant, te ideal-gas law gives. EALUAE Substituting te seond equation into te first one yields ln ln ( ) ln S nc nr n C R ncpln Δ + + ASSESS e same expression an also be obtained by using dq ncpd at onstant pressure. Note tat Δ S > 0 if, > as expeted. 47. INERPRE is problem involves te entropy ange in an ideal diatomi gas eated under tree different onditions: onstant volume, onstant pressure, and adiabatially. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9-5 DEELOP From Problem 9.45, te entropy ange at onstant volume is S nc ( ) Δ ln, were C 5R/ for a diatomi gas (see disussion after Equation 8.3). From Problem 9.46, te entropy ange at onstant pressure is Δ S nc ln n( C + R)ln ( ) ( ) P P For an adiabati proess, onsider te disussion aompanying Figure 9.6. e entropy ange is γ nr Δ S nrln nrln ln γ γ γ were we ave used te relation for an adiabati proess (Equation 8.b). For a diatomi gas, CP C + R 5R + R 7 γ C C 5R 5 EALUAE (a) At onstant volume, te entropy ange is 5 55K Δ S ncln ( ) ( 6.36 mol) 8.34 J/ ( mol K) ln 86.0 J/K 88K (b) At onstant pressure, te entropy ange is 7 55K Δ SP ncpln ( ) n( C + R) ln ( ) ( 6.36 mol) 8.34 J/ ( mol K) ln 0.4 J/K 88K () For an adiabati proess, Q 0 (no eat flow between system and environment) and te entropy ange is Δ S 0. ASSESS e entropy ange is greater in isobari (onstant pressure) proess tan isoori (onstant volume) proess sine CP > C. 48. INERPRE is problem involves te entropy ange tat results from mixing te given amount of ot and old water. We are to find te entropy ange for te ot water, te old water, and te entire system. DEELOP Wen mixing two liquids tat are initially at different temperatures, te termal energy ange in bot liquids must be te same, assuming all te termal energy lost by te ot liquid is gained by te old liquid. us, Δ Qot +Δ Qold 0 mototδ ot + moldoldδ old 0 For tis problem, m ot m old and ot old, beause te ot and old liquids are bot water. us, Δ ot Δ old Beause te equilibrium temperature of te mixed liquid is te same for te (previously) ot and old portions, we also ave ot +Δ ot old +Δold ot old Δold Δot Δold Δ ot ot old 70 K Combined wit te previous expression gives Δ old Δ ot 35 K and te equilibrium temperature is old + Δ old (0 C + 73 K) + 35 K 38 K. For tis problem, te entropy ange given by Equation 9.6 takes te form Δ S dq From te argument above, we ave dq md, so Equation 9.6 gives d Δ S m mln were m 0.50 kg and 484 J/kg, so we an solve for te entropy ange for ea ase. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-6 Capter 9 EALUAE (a) For te ot water, 353 K and 38 K, so te entropy ange is 38 K Δ Sot ( 0.50 kg) 484 J/ ( kg K) ln 09 J/K 353 K (b) For te old water, 83 K and 38 K, so te entropy ange is 38 K Δ Sold ( 0.50 kg) 4.84 kj/ ( kg K) ln J/K 83 K () e total entropy ange is te sum of tese two, so Δ S tot.97 J/K 09. J/K.7 J/K ASSESS Note tat te results are given to tree signifiant figures beause we assumed te data were valid to tree signifiant figures. e total entropy of te system as inreased, wi onforms wit te seond law of termodynamis. 49. INERPRE You want to find te entropy ange in going from one state to anoter state tat lie on te same adiabat. DEELOP We re told tat te system goes from state p, to state p,, were te two states are related by te adiabati equation: γ p γ. Our first inlination would be tat te entropy ange would be zero, sine tere is no eat exange in te adiabati proess tat onnets tese two states. However, we re told tat te system goes between te two states in two segments: one a onstant pressure proess (going from p, to p, ) and te oter a onstant volume proess (going from p, to p, ). See te figure below. is is an ideal gas, so te temperatures of te tree endpoints are: /, p nr /, p nr and p /. nr From able 8., te differential eat flows for tese two proesses are: dq ncpd for onstant pressure dq nc d for onstant volume Reall tat γ Cp / C. EALUAE For te onstant pressure proess, te entropy ange is ncpd Δ Sp ncpln ncpln nγ C ln For te onstant volume proess, te entropy ange is e total entropy is te sum: γ nc d p Δ S nc ln nc ln nc ln γ p a were we ave used te matematial identity: ( ) Δ S Δ Sp +Δ S nc γ ln γ ln 0 ln x aln x. e total entropy ange is zero as we expeted, sine te system ould ave gone from state to state by an adiabati proess for wi Q 0. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9-7 ASSESS As explained in te text, te entropy is a state variable, wi doesn t depend on ow a system arrived at a partiular state. Note tat tis separates entropy from te eat, Q, absorbed or expelled by a system. You an t say tat a system ontains a partiular amount of eat, but you an say tat it ontains a partiular amount of entropy. 50. INERPRE is problem asks about te energy quality resulting from a termodynami proess (an adiabati free expansion in tis ase) during wi entropy as inreased. e system onsists of a given quantity of ideal gas, and we are given te initial and final volume oupied by te gas. DEELOP From te disussion aompanying Figure 9.6, we know tat te ange in entropy during te adiabati free expansion is S nrln Δ were 5. e energy made unavailable is E Δ S. EALUAE Substituting te values given in te problem statement, te energy tat beomes unavailable to do work in te free expansion of an ideal gas ( remains onstant), is E Δ S nrln ( 6.36 mol ) 8.34 J/ ( mol K ) ( 305 K ) ln ( 5) 43.7 kj ASSESS is is te work tat ould ave been reovered from a reversible isotermal expansion. However, due to te irreversible nature of te proess, we give up te possibility of extrating tis work. 5. INERPRE is problem asks for te entropy ange of te pan-water system, wen termal equilibrium as been reaed after a ot pan as been plunged into te given amount of old water. DEELOP Assume all te eat lost by te pan is gained by te water. e equilibrium temperature is given by Equation 6.4, or (.4 kg) 900 J/ ( kg K) ( 48 K) + ( 3.5 kg) 484 J/ ( kg K) ( 88 K) (.4 kg) 900 J/ ( kg K) + ( 3.5 kg) 484 J/ ( kg K) eq 306 K Using te result of Exerise 48, te ange in entropy for te pan is Similarly, te ange in entropy for te water is eq Δ Span mpanpan ln pan eq Δ Swater mwaterwater ln water e sum of tese two terms is te ange of entropy of te pan-water system. EALUAE e entropy ange of te pan and water togeter is eq eq Δ S Δ Span +Δ Swater mpanpan ln + mwaterwater ln pan water 306 K 306 K (.4 kg) 900 J/ ( kg K) ln + ( 3.5 kg) 484 J/ ( kg K) ln 60 J/K 48 K 88 K to two signifiant figures. ASSESS e entropy ange for te pan is negative, wile tat of te water is positive. e total entropy ange is positive, in aordane wit te seond law of termodynamis. 5. INERPRE is problem is about te effiieny of an engine operating between two temperatures. DEELOP We assume te engine is reversible and operates between te two given temperatures ( 40 K and 73 K). e effiieny an ten be omputed using Equation 9.3, ecarnot. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-8 Capter 9 EALUAE (a) Substituting te values given in te problem statement, we find te effiieny to be e Carnot 73 K 35.0% 40 K (b) e total eat te blok of ie an absorb as it melts at 73 K is 3 ( )( ) Q ml f.00 0 kg 334 kj/kg 334 MJ en te melt-water temperature will rise and te engine s effiieny will drop. Wile running at te original effiieny, te engine exausts eat at te rate dq dq dw dw ( 8.5 kw) 5.8 kw dt dt dt ecarnot dt 0.35 (ombine te first law wit te definition of effiieny). us, it an operate between te original temperatures for a time Q 334 MJ 4 t. 0 s 5.88 dq dt 5.8 kw ASSESS For real engines in wi e < e, Carnot eat is exausted at a greater rate. is sortens te duration for wi te engine an maintain its effiieny. 53. INERPRE is problem is about te inrease in entropy as water is vaporized at 00 C. DEELOP e entropy inrease is given by Equation 9.6: Δ S dq/. e eat needed to vaporize te water is, v L v 57 kj/kg from able 7.. e temperature is onstant during te vaporization proess, 373 K. EALUAE e entropy inrease during te vaporization proess is ΔQ mlv (.0 kg )( 57 kj/kg ) Δ S. kj /K 373 K ASSESS As expeted, te entropy ange is positive in bot melting and warming proesses. 54. INERPRE We will alulate te effiieny of te Otto yle, on wi gasoline engines are modeled. DEELOP e engine absorbs eat ( Q ) during ombustion, and expels eat to te environment ( Q ) during te exaust segment. Bot tese proesses are at onstant volume, so Q nc Δ, and te effiieny is: e Otto W Q Δ Q Q Δ We an find te respetive temperature anges assuming te gas mixture in te engine is ideal: p / nr. EALUAE (a) e ot temperature ange is between point and point 3 in Figure 9.4: ( 5 ) ( 5 ) p Δ 3 3p p nr 5nR were we use te values for te pressure and volume given in te figure. e old temperature ange is between point and point 4 in te figure, but te pressures aren t given in tis ase. However, point and point are on te p γ γ, so: p p( 5 ), and similarly for point 4 and point 3: p4 p( 5 ) same adiabat ( onstant) te old temperature ange an be written: p Δ 4 p 4 p nr 5 nr From tese temperature anges, te effiieny of te Otto yle is: e Otto Δ γ 5 Δ γ γ 3. erefore, Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
min γ Carnot 3 max e Seond Law of ermodynamis 9-9 (b) e minimum temperature ours at point at te end of te exaust segment: p min γ 5 nr e maximum temperature ours at point 3 at te end of ombustion: 3p γ max 3 35 min 5nR () A Carnot yle working between minimum and maximum temperatures would ave an effiieny of (Equation 9.3): e 5 So, te Carnot yle effiieny is greater tan tat of te Otto yle: e, Carnot > e Otto as we d expet sine te Carnot yle as te maximum effiieny for an engine. ASSESS If we assume γ.4, just for argument sake, ten e Otto 47%, wile e Carnot 8%. In tis ligt, gasoline engines are woefully ineffiient. Mu of te ombustion energy is lost as exaust eat. 55. INERPRE is problem is about te effiieny of te Otto yle as a funtion of te ompression ratio. DEELOP As argued in te previous problem, te effiieny of te Otto yle is: eotto Δ / Δ. We will write tis in terms of te ompression ratio, r /, were and are te volumes before and after te adiabati ompression in Figure 9.4. In tis ase 5, but we ll derive te expression for te general ase first. EALUAE e ot temperature ange is between point and point 3 in te figure: Δ 3 ( p 3 3 p ) ( p3 p) nr nr e old temperature ange is between point and point 4 in te figure: Δ 4 ( p 4 4 p ) ( p4 p) nr nr Sine p γ onstant for points and and for points 3 and 4, we ave p p r γ, and p4 p3 r γ. erefore, te effiieny redues to γ γ ( ) Δ p r p r 3 γ eotto r Δ ( p3 p) ASSESS Sine te effiieny inreases for larger r, one migt assume engineers would try to maximize te ompression ratio. In pratie, owever, te ompression ratio annot be too large, oterwise te fuel pre-ignites, wi results in knoking tat redues engine performane. 56. INERPRE is problem is about te work done by te Diesel engine as a funtion of te ompression ratio. DEELOP As in te previous problem, te yle onsists of 4 steps. We alulate te work done in ea step in terms of te ompression ratios, r /, α 3 / r( 3 / ). e work done by gas and eat absorbed during te four proesses omprising te diesel yle are: (adiabati) P P W, Q 0, γ 3 (isobari) W3 P ( 3 ), Q3 ncp( 3 ) Qin, 3 4 (adiabati) ( P 3 3 P 4 4) W34, Q34 0, γ 4 (isovolumi) W 0, Q nc ( ) Q. 4 4 4 Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9-0 Capter 9 For te wole yle, te work done is p p + p p γ ( p p ) p + p W + p p ( γ ) ( γ ) γ EALUAE From p p 3 3 4 4 3 3 4 4 3 3 γ and p γ p γ, 3 3 4 4 p p pr γ 3 in terms of p, te ompression and utoff ratios, we ave γ γ γ γ 4 3 3 4 3 3 3 3 4, p p ( / ) p ( / ) p ( α / r) pα r αr wi, upon substituting into te expression for W, gives p W ( p 3 3 p ) p 4 4 p r γ γ + γ ( α ) α γ + ( γ ) γ ASSESS In a diesel engine, te isobari proess (from to 3) is were ombustion takes plae. 57. INERPRE is problem is about te effiieny of te Diesel yle as a funtion of te ompression ratio. DEELOP As in te previous problem, te yle onsists of 4 steps. We alulate te work done in ea step in terms of te ompression ratios, r /, α 3 / r( 3 / ). e eat added is Qin nc ( ) P 3 We use te ideal-gas law p nr, and te relations C / R C /( C C ) γ /( γ ) to alulate te effiieny P P P in terms of r and α. EALUAE (a) Combining te expressions above, we find e eat flowing into te engine to be nγr p p γ p p γ pr γ pr α Qin ncp( 3 ) γ nr nr γ γ γ (b) Using te result obtained in Problem 9.56 for W, we find te effiieny to be γ γ 3 3 ( 3 3 ) ( 3 ) ( ) W ( 4 4 ) ( ) ( ) diesel p p r γ γ r γ γ + γ Qin γ( p 3 3 p ) γr ( α ) γ( α ) e γ α α α ASSESS Due to te ig ompression ratio, diesel engines are more effiient tan gasoline engines (Otto yles). 58. INERPRE In tis problem we ompare te effiieny of a Diesel engine to a gasoline engine (Otto yle) of a ar. DEELOP From te previous two problems, te effiienies of te two engines an be written in terms of te ompression and utoff ratios as γ eotto r γ γ r ( α ) ediesel γα ( ) EALUAE Combining te expressions above, wit γ.4 for diatomi gas, r 8.3 for Otto yle, and r 9 and α.4 for diesel yle, we obtain γ.4 eotto r (8.3) 0.57 57.% γ γ.4.4 r ( α ) (9) (.4 ) ediesel 0.378 0.6 6.% γα ( ).4(.4 ) e diesel engine is more effiient. ASSESS Even toug diesel engine is more effiient, diesel fuel is more expensive tan regular gasoline. So you need to fator te fuel prie into onsideration wen you make your purase. 59. INERPRE Find te maximum effiieny of a power plant, given te temperature range of its yle. We will alulate te Carnot effiieny, and ompare tis wit te atual effiieny. DEELOP e ig temperature 950 F 783 K. e low temperature is 90 F 305 K. e Carnot effiieny is given by e. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
e Seond Law of ermodynamis 9- EALUAE e maximum effiieny is e 6%. ASSESS e atual effiieny of tis plant is given as 5%, wi is onsiderably lower due (at least in part) to aving to evaporate moisture out of te wood-ip fuel. 60. INERPRE We are to find te final temperature and entropy ange for a system in wi two objets at different temperatures are brougt into termal ontat and allowed to ome to termal equilibrium. DEELOP o find te final temperature, use Equation 6.4, wi may be expressed as mww0,w + mcucu0,cu f mww + mcucu e entropy ange is ds dq (Equation 9.6), so we an integrate to find te ange in entropy for te water and for te opper. e initial temperature and te mass of te opper are 0,Cu 80 C 353 K and m Cu 0.5 kg. e initial temperature and te mass of te water are 0,W 0 C 83 K and m W.0 kg. e speifi eats W and Cu an be found in able 6.. EALUAE (a) Inserting te given quantities gives m f m + m W W 0,W Cu Cu 0,Cu + m W W Cu Cu (.0 kg) 484 J/ ( kg K) ( 83 K) + ( 0.50 kg) 386 J/ ( kg K) ( 353 K) (.0 kg) 484 J/ ( kg K) + ( 0.50 kg) 386 J/ ( kg K) 86 K (b) e ange in entropy for te water and opper are, respetively: ds W dq m d W W W W W W Δ S m m ds Cu f d f 86 K W W W W Wln (.0kg) 484J/ ( kg K) ln 44. J/K 0,W 0,W 83 K dq m d Cu Cu Cu Cu W Cu Δ S m m f d f 86 K Cu Cu Cu Cu Cu ln ( 0.50 kg) 386 J/ ( kg K) ln 40.6 J/K 0,Cu 0,Cu 353 K e total ange in entropy is Δ S Δ SW +Δ SCu 3. J/K. ASSESS e entropy of te opper atually dereases, but tis derease is more tan offset by te inrease in entropy of te water. is is an irreversible proess, and entropy always inreases in irreversible proesses. 6. INERPRE We are asked to alulate te entropy ange in an objet wose eat apaity is inversely proportional to its temperature. DEELOP By definition, te eat apaity relates te eat flowing into an objet to te ange in its temperature: dq Cd. In tis ase, C ( ) Δ S C00 d /. /. 0 0 We an plug tis into Equation 9.6 for te entropy ange: EALUAE Performing te integration from 0 to : Cd 0 0 0 0 0 0 0 0 0 Δ S C C ASSESS For >, te entropy ange is positive. For, te entropy ange beomes onstant: ΔS C. 0 0 0 0 6. INERPRE is problem deals wit a Carnot engine for wi te temperature of te eat reservoir varies wit time. We are to express instantaneous temperature of te ot reservoir as a funtion of time and find te time it takes for te engine s power to rea zero. Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
9- Capter 9 DEELOP In time dt, te engine extrats eat dq md from te blok, and does work dw Pdt. Equation 9. gives te definition of te atual effiieny e: dw edq ( md) Pdt were for te seond equality we ave used e e Carnot /. e power is also assumed to be proportional to, so tis equation beomes d Pdt m 0 0 Integrating tis expression yields as a funtion of time t. For (b), note tat te power output beomes zero wen. EALUAE (a) Integrating from t 0 and 0 to t and gives or 0 t d Pdt 0 Pt 0 ln 0 0 m( 0 ) m( 0 ) () t (b) e power output is zero for. is ours at time Pt 0 0exp m( 0 ) ( ) m 0 0 t0 ln P0 ASSESS We find te instantaneous temperature of te ot blok to derease exponentially wit time. At t 0, ( 0 ) 0. However, for Pt 0 >> m( 0 ) beomes very small. Note tat te expression for P was originally assumed to be valid for, or for times t t. If we allow 0 <, or t > t0, ten dw Pdt < 0 beomes work input to an engine wi ats like a refrigerator tat ools te blok. 63. INERPRE You ave an infinite eat reservoir, but a finite ool reservoir. e question is ow mu work an you obtain wit an engine plaed between te reservoirs before te ool reservoir is exausted. DEELOP e infinite eat reservoir will maintain its temperature,, no matter ow mu eat, Q, you extrat from it. e ool reservoir, on te oter and, will not maintain its initial temperature,, 0 as eat from te engine is expelled into it. e temperature will rise in te ool reservoir aording to dq. Cd But one te ool reservoir temperature is equal to, no more work an be extrated. EALUAE You an assume tat te engine yles fast enoug tat during a single yle te ool reservoir temperature is approximately onstant. o maximize te amount of work tat you extrat, plae a Carnot engine between te reservoirs so tat te work extrated during one yle is: dw d ( Q Q ) dq Cd were Q / Q / for a Carnot engine. o find te total work, integrate tis expression from 0 to, W Cd Cln( ) C Cln C( 0) 0 0 0 If we let x / 0, ten W C ( ln x + / x). ASSESS e work is proportional to te eat apaity, as you migt expet. e eat apaity is a measure of ow mu eat te ool reservoir an aept, so te larger te eat apaity, te more work tat an be extrated. You migt worry tat te work ould be negative for some value of x / 0. For x, te work is approximately W C ln x, wi is positive. For x, ln x x, and te work is approximately Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.
( ) e Seond Law of ermodynamis 9-3 W C x + / x, wi is positive as well. erefore, te work extrated is positive for all possible temperature differenes. 64. INERPRE You want to know ow mu te temperature of a river will inrease wen eat exaust from a power plant is absorbed by te water. DEELOP Sine te plant as an effiieny of e W / Q 35%, te rate at wi it is generating waste eat must be: Q Q W W 750 MW 390 MW e 0.35 were te dots ere signify tat tese are rates, for example, Q /. dq dt e rate at wi water is flowing past 3 te plant is given by te volume rate: 0 m /s. o better understand ow te eat flows into te flowing water, it migt elp to freeze bot flows and just imagine wat is appening over a sort period of time, Δt. In tis ase, te plant expels a finite amount of eat, Q Q Δt, into a volume of water Δt. See te figure below. is volume of water will rise in temperature aording to: Q mδ ρδ, were ρ is te density of water and is te speifi eat. You an use te equations formulated ere to ek tat te temperature rise is below te environmental regulation. EALUAE Solving for te temperature ange of te water: ( ) Δ Q Q 390 MW o 3.0 C 3 3 ρ ρ 000 kg/m 0 m /s 484 J/kg K ( )( )( ) e temperature rise is below te regulated limit. ASSESS e river makes for a good reservoir, sine te flow will onstantly bring ool water tat as yet to be eated by te plant s exaust. 65. INERPRE We re asked to find te entropy ange wen ot and old water are mixed togeter. Sine te ot and old water an t be unmixed, Equation 9.6 doesn t apply diretly, but you an find a reversible proess tat mimis tis irreversible mixing. DEELOP Before mixing, imagine ooling te ot water and warming te old water until tey are bot at, wi f is te final temperature wen te water volumes are irreversible mixed. After te temperatures are equilibrated, te two water samples an simply be added togeter. is is a reversible proess, sine we ould easily divide te mixed water in alf and re-warm one sample and re-ool te oter to te original temperatures. But as te final state is te same as in te irreversible mixing ase, we laim tat te entropy ange applies to bot system pats. EALUAE e differential eat flow in bot te warming and ooling proesses is dq md. e entropy ange, terefore, in ooling te ot water from to f is f dq f Δ Sf mln Copyrigt 06 Pearson Eduation, In. All rigts reserved. is material is proteted under all opyrigt laws as tey urrently exist. No portion of tis material may be reprodued, in any form or by any means, witout permission in writing from te publiser.