Chapter 04 the Second Law of Thermodynamics (Pp 82-99)

CH4 SECOND LAW OF THERMODYNAMICS

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CHAPTER 04 THE SECOND LAW OF THERMODYNAMICS 4-1 EFFICIENCY OF A HEAT ENGINE Example 4-1 A diesel engine performs 2200 J of mechanical work and discards 4300 J of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine? Solution (a) QIN = QOUT + W = 4300 + 2200 = 6500 J W 2200 (b) η = = = 0.34 or 34 % QIN 6500 Example 4-2 An engine with 20 % efficiency does 100 J of work in each cycle. (a) How much heat is absorbed in each cycle? (b) How much heat is rejected in each cycle? Solution (a) The efficiency of a heat engine is given by W η= QIN Q 100 QIN = IN = = 500 J η 0.20 (b) QOUT = QIN − W = 500 − 100 = 400 J

CH4 SECOND LAW OF THERMODYNAMICS 83 Example 4-3 An engine absorbs 100 J and rejects 60 J in each cycle. (a) What is its efficiency? (b) If each cycle takes 0.5 s , find the power output of this engine in watts. Solution (a) The efficiency of the heat engine is defined as Q − QOUT 100 − 60 W η= = IN = = 0.40 or 40 % QIN QIN 100 W QIN − QOUT 100 − 60 (b) P = = = = 80 W t t 0 .5 Example 4-4 Find the efficiency of a Carnot engine working between the temperatures 225 0 C and 25 0 C . B.U. B.Sc. (Hons.) 1983A Solution The efficiency of a Carnot engine is given by  T  η = 1 − L  × 100 %  TH   

η = 1 −

25 + 273   × 100 % = 40 % 225 + 273 

Example 4-5 What is the maximum efficiency of a steam engine that utilizes steam from a boiler at 480 K and exhausts at 373 K . K.U. B.Sc. 2003 Solution The efficiency of a Carnot engine is given by  T  η = 1 − L  × 100 %  TH   

η = 1 −

373   × 100 % = 22.3 % 480 

CH4 SECOND LAW OF THERMODYNAMICS 84 Example 4-6 A heat engine performs 200 joules of work and at the same time rejects 3000 joules of heat to the cold body. Calculate the efficiency. B.U. B.Sc. (Hons.) 1989A Solution The efficiency of a Carnot engine is given by  Q  η = 1 − OUT  × 100 % QIN    

η = 1 −

3000   × 100 % = 6.25 % 3000 + 200 

Example 4-7 A Carnot engine is operated between two heat reservoirs at temperatures of 400 K and 300 K . If the engine receives 1200 joules of energy from the reservoir at 400 K in each cycle, how many joules of energy does it reject to the reservoir at 300 K ? B.U. B.Sc. (Hons.) 1986A Solution The efficiency of a Carnot engine is given by   T   Q η = 1 − L  = 1 − OUT  QIN   TH  

TL QOUT = TH QIN T   300  QOUT =  L  QOUT =   (1200) = 900 J  400   TH  Example 4-8 A steam engine, operating between 115 0 C and 60 0 C , develops 2500 hp . How much heat must be supplied if its overall efficiency is 10 percent? What is the maximum efficiency that an ideal heat engine can have when it operates between the stated temperatures? (1 hp = 746 W). B.U. B.Sc. 1999A

CH4 SECOND LAW OF THERMODYNAMICS Solution Let QIN be the supplied heat, then

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W = 10 % of QIN = 0.1 QIN 2500 × 746 = 0.1 QIN 2500 × 746 = 1.865 × 10 7 J QIN = 0 .1 The efficiency of ideal heat engine is given by  T  60 + 273 η = 1 − L  = 1 − = 0.142 or 14.2 % 115 + 273  TH  Example 4-9 A heat engine absorbs 52.4 kJ of heat and exhausts 36.2 kJ of heat per cycle. Calculate the efficiency and the work done by the engine per cycle. P.U. B.Sc. 2003, 2009 Solution The efficiency of a Carnot engine is given by  Q  η = 1 − OUT  × 100 % QIN    36.2   × 100 % = 30.9 %  52.4  The work done by the engine per cycle is W = QIN − QOUT = 52.4 − 36.2 = 16.2 kJ Example 4-10 A Carnot’s engine takes 3.36 × 10 6 joules of heat from a reservoir at 527 0 C and gives some heat to sink at 27 0 C . What is the efficiency of engine? How much work does it perform in joules? K.U. B.Sc.2006 Solution The efficiency of heat engine is given by  T  η = 1 − L   TH 

η = 1 −

CH4 SECOND LAW OF THERMODYNAMICS 86 27 + 273 η = 1− = 0.625 or 62.5 % 527 + 273 The work done by the engine is W = η QIN = (0.625)(3.36 × 10 6 ) = 2.1 × 10 6 J Example 4-11 A Carnot engine extracts 240 J of heat from a high temperature reservoir during each cycle. It rejects 100 J of heat at reservoir at 15 0 C . (i) How much work does the engine do in one cycle? (ii) What is its efficiency? (iii) What is the temperature of the hot reservoir? K.U. B.Sc. 2001 Solution (i) The work done by the engine is given by W = QIN − QOUT = 240 − 100 = 140 J The efficiency of the given Carnot engine is (ii)  Q   100  η = 1 − OUT  × 100 % = 1 −  × 100 % = 58.3 % QIN   240   (iii) The efficiency of Carnot engine in terms of temperatures of reservoirs is given by  T  η = 1 − L  × 100 %  TH 

 15 + 273   × 100 58.3 = 1 − T H   58.3 288 = 0.583 = 1 − 100 TH 288 = 1 − 0.583 = 0.418 TH 288 TH = = 691 K or 4180 C 0.418

CH4 SECOND LAW OF THERMODYNAMICS 87 Example 4-12 A heat engine operates between two reservoirs at 300 K and 500 K . During each cycle it absorbs 200 cal of heat from the hot reservoir. (a) What is maximum efficiency of the engine? (b) Determine the maximum work the engine performs during each cycle. B.U. B.Sc. 2008A Solution (a) The efficiency of heat engine is given by  T  η = 1 − L   TH  300 = 0.40 or 40 % η = 1− 500 (b) The heat absorbed by the engine from hot reservoir in each cycle is QIN = 200 cal = (200)(4.186) J = 837.2 J The work done by the engine in each cycle is W = η QIN = (0.40)(837.2) = 334.88 J Example 4-13 A Carnot engine absorbs heat from a reservoir at a temperature of 1000 0 C and rejects heat to a cold reservoir at a temperature of 0 0 C . If the engine absorbs 1000 joules from the high temperature reservoir, find heat rejected and efficiency of heat engine. B.U. B.Sc.(Hons.) 1988A, 1989A Solution Now QOUT TL = QIN TH T  (0 + 273)(1000) QOUT =  L  QIN = = 214 J (1000 + 273)  TH  The efficiency of heat engine is given by

CH4 SECOND LAW OF THERMODYNAMICS  T  η = 1 − L  × 100 %  TH   

η = 1 −

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0 + 273   × 100 % = 78.6 % 1000 + 273 

Example 4-14 In a Carnot cycle, the isothermal expansion of an ideal gas takes at 140 0 C and isothermal compression at 25 0 C . During the compression, 2090 J of heat energy is transferred to the gas. Calculate (i) The work performed by the gas during isothermal expansion. (ii) Heat rejected from the gas during isothermal compression. (iii) Work done on the gas during isothermal compression. K.U. B.Sc. 2002, 2005 Solution (a) W = −QIN = −2090 J The negative sign indicates that the work is done by the gas. (b) The efficiency of Carnot engine is given by  T  η = 1 − L  × 100 %  TH  25 + 373    × 100 % = 27.8 %  140 + 273   Q  η = 1 − OUT  × 100 % QIN  

η = 1 −

Now

 Q  27.8 = 1 − OUT  (100)  2090  27.8  Q  = 0.278 = 1 − OUT  100  2090  QOUT = 1 − 0.278 = 0.722 2090

CH4 SECOND LAW OF THERMODYNAMICS 89 QOUT = (0.722)(2090) = 1509 J (c) Work done on the gas during isothermal compression W = QOUT = 1509 J Example 4-15 A Carnot cycle has efficiency 75 % working between highest and lowest temperatures. What is the lowest temperature of its high temperature is 927 0 C ? B.U. B.Sc. 2007S Solution The efficiency of a Carnot engine is given by  T  η = 1 − L   TH  TL 75 = 1− 100 927 + 273 T 0.75 = 1 − L 1200 TL = 1 − 0.75 = 0.25 1200 TL = (0.25)(1200) = 300 K = 27 0 C Example 4-16 A Carnot engine has an efficiency of 22 % . It operates between heat reservoirs differing in temperature by 75 0 C . Find the temperature of the reservoirs. K.U. B.Sc. 2004, 2008 Solution Let TL and TH be the temperatures of cold and hot reservoirs (i.e. sink and source) respectively, then the efficiency of Carnot engine is given by T η = 1− L TH

Now

η = 22 % = 0.22

CH4 SECOND LAW OF THERMODYNAMICS TH = (TL + 75) K Hence TL T + 75 − TL 0.22 = 1 − = L TL + 75 TL + 75 75 0.22 = TL + 75

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0.22 TL + 16.5 = 75 0.22 TL = 75 − 16.5 = 58.5 58.5 TL = = 266 K or − 7 0 C 0.22 TH = TL + 75 = 266 + 75 = 341 K or 680 C Example 4-17 A Carnot heat engine with an efficiency of 0.20 operates between two temperatures that differ from each other by 100 K . What are the temperatures between which the cycle operates? Solution The efficiency of a Carnot engine is given by T η = 1− L TH Now η = 0.20 TH = (TL + 100) K Hence TL T + 100 − TL 0.20 = 1 − = L TL + 100 TL + 100 100 0.20 = TL + 100 0.20 TL + 20 = 100 0.20 TL = 100 − 20 = 80 80 TL = = 400 K 0.20

CH4 SECOND LAW OF THERMODYNAMICS 91 TH = TL + 100 = 400 + 100 = 500 K The given engine operates between 400 K and 500 K. Example 4-18 The exhaust temperature of a heat engine is 220 0 C . What must be the high temperature if the Carnot efficiency is to be 36 % ? Solution T Now η = 1− L TH 36 220 + 273 = 1− 100 TH 493 0.36 = 1 − TH 493 = 1 − 0.36 = 0.64 TH 493 TH = = 770 K or 497 0C 0.64 Example 4-19 In one cycle of operation a heat engine takes in 2200 J of heat and performs 620 J of work. (a) What is the efficiency of this engine? (b) How much heat is exhausted in each cycle? (c) If the engine completes a cycle each 0.033 s , then at what rate is heat added, heat exhausted, work done? Solution (a) The efficiency of heat engine is defined as W 620 η= = = 0.28 or 28 % QIN 2200 (b) The heat exhausted in each cycle is QOUT = QIN − W = 2200 − 620 = 1580 J Q 2200 (c) PIN = IN = = 6.67 × 10 4 W = 66.7 kW t 0.033

CH4 SECOND LAW OF THERMODYNAMICS 92 Q 1580 POUT = OUT = = 4.79 × 10 4 W = 47.9 kW t 0.033 W 620 P= = = 1.88 × 10 4W = 18.8 kW t 0.033 Example 4-20 The low temperature reservoir of a Carnot engine is at 10 0 C and has an efficiency of 50 % . How much the temperature of the high temperature reservoir is increased to measure the efficiency to 50 % ? K.U. B.Sc. 2007 Solution The efficiency of a Carnot engine is given by  T  η = 1 − L   TH  0.50 = 1 −

10 + 273 TH

283 = 1 − 0.50 = 0.50 TH 283 TH = = 566 K 0.50 Let η1 be the new efficiency of heat engine with new source T temperature as TH* , then η1 = 1 − L* TH 283 0.60 = 1 − * TH 283 = 1 − 0.60 = 0.40 TH* 283 TH* = = 707.5 K 0.40 The desired increase in temperature of hot reservoir is ∆T = TH* − TH = 707.5 − 566 = 141.5 K

CH4 SECOND LAW OF THERMODYNAMICS 93 Example 4-21 A heat engine utilizes a heat source at 580 0 C and has a Carnot efficiency of 29 percent. To increase the efficiency to 35 percent, what must be the temperature of the heat source? Solution The temperature of the cold reservoir i.e. sink is calculated as under T η = 1− L TH TL 0.29 = 1 − 580 + 273 TL = 1 − 0.29 = 0.71 853 TL = (0.71)(853) = 606 K Let η1 be the new efficiency of heat engine with new source temperature as TH* , then T η1 = 1 − L* TH 606 0.35 = 1 − * TH 606 = 1 − 0.35 = 0.65 TH* 606 TH* = = 932 K or 6590 C 0.65

CH4 SECOND LAW OF THERMODYNAMICS 94 Example 4-22 A Carnot heat engine operates between a reservoir at 950 K and has heat transfer of 400 J exhausted to another reservoir at 300 K during each cycle. What is the amount of work done by the heat engine during each cycle? Solution TL QOUT Now = TH QIN T   950  QIN =  H  QOUT =   (400) = 1267 J T 300    L W = QIN − QOUT = 1267 − 400 = 867 J The given heat engine performs 867 J of work in each cycle.

CH4 SECOND LAW OF THERMODYNAMICS

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4-2 COEFFICIENT OF PERFORMANCE OF A REFRIGERATOR ENGINE Example 4-23 The low temperature of a freezer cooling coil is − 15 0 C and the discharge temperature is 30 0 C . What is the maximum theoretical coefficient of performance? Solution The coefficient of performance is defined as TL K= TH − T L (−15 + 273) K= = 5.733 (30 + 273) − (−15 + 273) Example 4-24 A heat pump acting as a refrigerator is used to heat a house. The temperature outside the house is − 10 0 C and the interior to be kept at 22 0 C . Find the maximum coefficient of performance of heat pump. P.U. B.Sc. 2005, 2006 Solution The coefficient of performance of heat pump acting as a refrigerator is given by TL K= TH − T L

TL = −10 0 C = (−10 + 273) K = 263 K TH = 22 0 C = (22 + 273) K = 295 K TH − TL = 295 − 263 = 32 K 263 Hence K= = 8.218 32 Example 4-25 What is the coefficient of performance of a refrigerator that operates with Carnot efficiency between temperatures − 3 0 C and 27 0 C ? Now

CH4 SECOND LAW OF THERMODYNAMICS 96 The coefficient of performance of heat pump acting as a refrigerator is given by TL K= TH − T L

TL = −30 C = (−3 + 273) K = 270 K TH = 27 0 C = (27 + 273) K = 300 K TH − TL = 300 − 270 = 30 K 270 Hence K= =9 30 Example 4-26 A household refrigerator, whose coefficient of performance is 4.7, extracts heat from the cooling chamber at rate of 250 J per second. How much work per cycle is required to operate the refrigerator? P.U. B.Sc. 2002 Solution The coefficient of performance K is defined as QL QL K= = QH − QL W Now

W =

QL 250 = = 53.2 J / cycle K 4.7

Example 4-27 A refrigerator does 153 J of work to transfer 568 J of heat from its cold compartment. (a) Calculate the refrigerator’s coefficient of performance. (b) How much heat is exhausted to the kitchen? Solution (a) The coefficient of performance K is defined as QL 568 K= = = 3.71 W 153 (b) Heat exhausted to kitchen will be QL + W = 568 + 153 = 721 J

CH4 SECOND LAW OF THERMODYNAMICS 97 Example 4-28 Apparatus that liquefies helium is in a laboratory at 296 K . The helium in the apparatus is at 4 K . If 150 mJ of heat is transferred from helium, find the minimum amount of heat delivered to the laboratory. Solution The coefficient of performance is defined as TL K= TH − T L 4 K= = 1.37 × 10 − 2 296 + 4 Now QL QH = QL + W = QL + K 1   QL = (150 × 10 −3 )1 + = 11.1 J −2   1.37 × 10 

CH4 SECOND LAW OF THERMODYNAMICS 98 ADDITIONAL PROBLEMS (1) A Carnot engine is operated between two reservoirs at temperatures 227 0 C and 127 0 C . Evaluate the efficiency of the engine. B.U. B.Sc.(Hons.) 1988S (2) A heat engine performs 1000 joules of work and at the same time rejects 4000 joules of heat energy to the cold reservoir. What is the efficiency of the engine? B.U. B.Sc.(Hons.) 1991A (3) A heat engine receives heat 120 joules and the work done by it is 90 joules. Calculate the efficiency in percentage. B.U. B.Sc. 1997A (4) A heat engine performs 2000 joules of work and at the same time rejects 6000 joules of heat energy to the cold reservoir. What is the efficiency of the engine? B.U. B.Sc. 1998A (5) A Carnot engine whose heat source is at 127 0 C takes 100 calories of heat at this temperature in each cycle and gives up 80 calories to the heat sink. Find the temperature of heat sink. B.U. B.Sc. 1988S (6) A Carnot engine absorbs heat from a reservoir at a temperature of 100 0 C and rejects heat to a cold reservoir at a temperature of 0 0 C . If the engine absorbs 1000 joules from the high temperature reservoir, find heat rejected and efficiency of heat engine. B.U. B.Sc. 1986S (7) What is the maximum efficiency of a steam engine that utilizes steam from a boiler at 480 K and exhausts at 373 K ? K.U. B.Sc. 2003 (8) A heat engine performs 200 joules of work and at the same time rejects 300 joules of heat to the cold body. Calculate its efficiency. B.U. B.Sc.(Hons.) 1989A (9) Calculate the efficiency of a Carnot engine working between the temperatures 927 0 C and 27 0 C . P.U. B.Sc. 1989

CH4 SECOND LAW OF THERMODYNAMICS 99 (10) A Carnot engine’s working substance is water. It uses steam at 300 0 C and condenses to water at 40 0 C . What is the maximum theoretical efficiency of this engine? B.P.S.C. 1995 (11) A Carnot engine operates between the temperatures 850 K and 300 K . The engine performs 1200 J of work each cycle, which takes 0.25 s . Calculate its efficiency and its average power. What are the rates of heat input and heat exhaust per cycle? F.P.S.C. 2009

Answers (1) 20 % (2) 20 % (3) 75 % (4) 25 % (5) 320 K (6) QOUT = 732 J , η = 26.8 % (7) 22.3 % (8) 6.25 % (9) 75 % (10) 37 % (11) 65 %, 4800 W, 13600 W and 8800 W