10. Heat devices: heat engines and refrigerators (Hiroshi Matsuoka)

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1 10 Heat devices: heat engines and refrigerators (Hiroshi Matsuoka) 1 In this chapter we will discuss how heat devices work Heat devices convert heat into work or work into heat and include heat engines such as internal combustion engines in automobiles and steam engines in power plants as well as refrigerators and air conditioners We will idealize processes in these devices to be quasi-static so that we can calculate the quasi-static heat flowing into a system during these processes in terms of changes in state variables such as the internal energy enthalpy and entropy of the system Heat devices based on cyclic processes Each heat device is based on a cyclic process of a substance such as a gas mixture of air and vaporized gasoline in an internal combustion engine so that the device including the substance inside returns to the same macroscopic state after each cycle and by repeating the same cyclic process it continuously converts heat into work or work into heat A special feature of the cyclic process is that the device returns to the same macroscopic state after each cycle so that the value of the internal energy of the device at the end of the cycle is the same as that at the beginning of the cycle: U f = U i In other words there is no net change in the internal energy after each cycle:!u = U f "U i = 0 The first law applied to a cyclic process find Applying the first law of thermodynamics to the cyclic process in the heat device we then!u = Q + W = 0 where Q is the total amount of heat that flows into the device in each cycle while W is the total amount of work that is done on the device in each cycle We have therefore obtained a very useful relation between Q and W:!W = Q 10 1 Heat engines A heat engine is a heat device that converts heat into work Clearly it would be ideal if we could construct a heat engine that could absorb some heat from a heat source and convert it completely into a work output to the outside It turns out that we cannot construct such a heat

2 2 engine As we will discuss later this statement of the impossibility of a perfect heat engine is nothing but the second law of thermodynamics Efficiency of a heat engine Any real heat engine must therefore discard some energy as heat to the outside so that a part of the input heat must be wasted as the exhaust heat discarded to the outside Note that the input heat Q in is positive or Q in > 0 as it flows into the engine whereas the exhaust heat Q ex is negative or Q ex < 0 as it flows out of the engine The net total amount of heat Q for each cycle of the engine is then the sum of Q in and Q ex : Q = Q in + Q ex Note that as Q ex < 0 we find Q < Q in As the total amount W of work done on the engine is related to Q by!w = Q the work done by the engine!w is related to Q in by!w = Q = Q in + Q ex < Q in To improve engine we must then aim at increasing the efficiency! defined by! " #W Q in which measures how much of the input heat is converted to the work output In other words! measures the efficiency of the conversion of the input heat into the work output As!W < Q in the efficiency of the engine must be always less that 1! <1 and the closer to 1! gets the more efficient the engine gets To improve a heat engine we must therefore understand what determines the efficiency of the heat engine so that we can find a way to increase its efficiency toward 1 The efficiency! can be also expressed in terms of Q in and Q ex as

3 ! = "W Q in = Q in + Q ex Q in ( = 1" "Q ex ) Q in 3 so that we can calculate the efficiency directly from Q in and Q ex This equation indicates that to increase the efficiency we must make the discarded heat!q ex smaller compared to the input heat Q in Q in!w! Q ex Efficiency is not everything Clearly the efficiency of a heat engine is not the only factor that dictates the design of the particular engine for a particular purpose which requires for example a particular size for work output Once we have incorporated all the specifications required for the particular engine we then try to maximize its efficiency Ideal heat engines based on quasi-static cyclic processes Applying the second law of thermodynamics we will show later that the efficiency of a heat engine reaches a maximum value when the cyclic process for the heat engine becomes completely quasi-static so that the quasi-static cyclic process provides an upper limit for the efficiency of the hat engine We will therefore examine quasi-static cyclic processes as idealized models for heat engines 1011 The Carnot cycle: an idealized model for a heat engine As an idealized model for a heat engine a Carnot cycle is a quasi-static cyclic process in which a substance inside the engine exchanges heat with two heat reservoirs at two different temperatures T c where T c < T h These heat reservoirs are much larger than the engine so that their heat capacities are much larger than that of the substance inside the engine (recall that a heat capacity of a system is extensive so that it is proportional to the mole number of the system) and any heat exchange between the substance and the reservoirs hardly changes the temperatures of the reservoirs

4 In the Carnot cycle the substance absorbs heat from the hotter reservoir through a quasistatic isothermal process at temperature T h and discards heat to the colder reservoir through another quasi-static isothermal process at temperature T c Following the isothermal process at temperature T h the substance decreases its temperature to T c through a quasi-static adiabatic process before it goes through the quasi-static isothermal process at temperature T c followed by another quasi-static adiabatic process that increases the temperature back to T h 4 Temperature changes as volume changes in a quasi-static adiabatic process To construct a Carnot cycle for a given substance we must be able to change the temperature of the substance by changing its volume during a quasi-static adiabatic process To see that we can actually do this we can use the first ds equation Since there is no heat flowing into the substance during a quasi-static adiabatic process we find no change in entropy of the substance: ds =!Qqs T = 0 Using the first ds equation we then find so that ds = C V T dt +! " T dv = 0 1 T dt =! " C V # T dv

5 which indicates that as long as! " 0 the temperature of the substance changes as its volume is changed so that by either expanding or compressing the volume we can change the temperature from a given value to any other value we like For example if! > 0 then! ( C V " T ) > 0 so that when the volume expands or dv > 0 dt < 0 or the temperature decreases By solving this differential equation we can also find the relation between T and V along the quasi-static adiabatic process For example as we have found in Sec81 along a quasi-static adiabatic process of a lowdensity gas with C V = const T and V satisfy TV = const so that we can sketch a Carnot cycle with a low-density gas on the TV diagram as shown below Both the quasi-static process from state 1 to state 2 and the quasi-static process from state 3 to state 4 are adiabatic 5 The universal efficiency for the Carnot cycles between reservoirs at T c The Carnot cycle occupies a special place in the study of heat engines because the value of its efficiency! depends only on T c so that all the Carnot cycles that operate with two heat reservoirs at T c share the same universal efficiency:! Carnot = 1" T c T h To derive this universal efficiency we need to express the quasi-static heat flowing into a system in an isothermal process in terms of an entropy change in the system Quasi-static heat and entropy change along a quasi-static isothermal process As we have found in Sec82 the quasi-static heat Q T qs flowing into a system in an isothermal process at temperature T is simply given by Q T qs = T!S

6 6 where!s is a change in the entropy of the system after the isothermal process:!s = S( TV f n) " S( T V i n) = S( TP f n) " S( T P i n) so that if we can find the initial and final values of the entropy S through an analytical expression for S or in a numerical table for S in some reference handbook we can find the quasi-static heat Q T qs Deriving the universal efficiency for the Carnot cycles between reservoirs at T c In a Carnot cycle the quasi-static heat Q qs in hotter reservoir at T h is then that the substance in an engine absorbs from a Q in qs = T h!s h where!s h is a change in the entropy of the substance in this isothermal process while the quasistatic heat!q qs ex that the substance in the engine discards to a colder reservoir at T c is!q qs ex =!T c "S c where!s c is a change in the entropy of the substance in this isothermal process In contrast along the quasi-static adiabatic processes there is no heat flowing into the substance in the engine so that the entropy of the substance remains constant and there is no change in the entropy of the substance in these processes As the Carnot cycle is a cyclic process the entropy of the substance must return to its initial value after the cycle so that the net change in the entropy after the cycle is zero: 0 =!S =!S h +!S c from which we obtain! "S c = "S h We then find the efficiency of the Carnot cycle to be! Carnot = 1" "Q qs ( ex ) qs Q in ( =1 " "T #S c c) = 1" T #S c h = 1" T c T h #S h T h #S h T h This expression for the efficiency is universal for all the Carnot cycles that operate between two heat reservoirs at temperatures T c

7 The output work of a Carnot cycle As the efficiency of a cyclic process is defined by! = "W qs Q in qs we can calculate the output work of a Carnot cycle by 7 #!W qs = " Carnot Q qs in = 1! T & c % ( $ ' T )S = T h h (! T h c ))S h T h For example for a low-density gas with C V = const we find #!W qs = ( T h! T c )"S h = nr( T h! T c )ln V 3 % $ & ( ' so that to increase the output work!w qs we must increase the temperature difference T h! T c and/or the volume ratio V 3 The Carnot cycle in the TS diagram and its universal efficiency We can understand the above derivation of the efficiency of the Carnot cycles between two heat reservoirs at T c by sketching a Carnot cycle in the TS diagram shown below In this diagram the quasi-static input heat Q in qs = T h!s h corresponds to the sum of the two shaded areas below the line from state 2 to state 3 while the quasi-static discarded heat qs =!T c "S c corresponds to the area of the shaded area below the line from state 1 to state 4!Q ex The efficiency of the cycle depends on the ratio between Q qs ex and Q qs in which is equal to the ratio between the two rectangular areas whose heights along the T-direction are T c respectively A crucial point here is that the entropy changes along the two isothermal processes are equal in magnitude:!s h =!S c This point is also the reason why the efficiency of the

8 Carnot cycles between two heat reservoirs at T c is universal despite that the values of entropy at states 1 through 4 may vary from one Carnot cycle to another The shaded area enclosed by the Carnot cycle in this TS diagram also corresponds to the work output because 8!W qs = Q qs in + Q qs ex = ( T h! T c )"S h Input heat exhaust heat and work output in the TS diagram In general for a quasi-static cyclic on the TS diagram shown below the area underneath the upper portion of the curve for the cyclic process represents the input heat Q in qs of the process while the area underneath the lower portion of the curve for the cyclic process represents the exhaust heat!q qs ex of the process As! W qs = Q qs in + Q qs ex the area enclosed by the curve for the cyclic process represents the work output! W qs Therefore for a given upper portion of the curve for the cyclic process the smaller the area underneath the lower portion of the curve for the cyclic process becomes the higher the efficiency gets T! W qs qs!q ex S 1012 The Otto cycle: an idealized model for an internal combustion engine As an idealized model for a cyclic process in an internal combustion engine we will examine an Otto cycle of a low-density gas or a mixture of air and vaporized gasoline for which we assume: (i) the ideal gas law is a good approximation for the equation of state of the gas (ii) the heat capacity at constant volume of the gas is constant: C V = const

9 The Otto cycle consists of two quasi-static adiabatic processes and two quasi-static isochoric or isovolumic processes Before we describe the Otto cycle in more detail we need to remind us how we can find the quasi-static heat flowing into a system during a quasi-static isochoric process 9 Quasi-static heat in an isochoric or isovolumic process: V = const As we have found in Sec77 the quasi-static heat Q V qs flowing into a system in an isochoric process at volume V is simply given by Q V qs =!U where!u is a change in the internal energy of the system after the isochoric process:!u = U( T f Vn) "U( T i Vn) = U( T f P f n) " U( T i P i n) so that if we can find the initial and final values of the internal energy U through an analytical expression for U or in a numerical table for U in some reference handbook we can find the quasi-static heat Q V qs The Otto cycle of the low-density gas The Otto cycle of the low-density gas consists of the following four quasi-static processes: (i) A quasi-static adiabatic compression of the gas proceeds from state T 1 n ( T 2 n) For this process we find T 1 = T 2 ( ) to state (ii) A quasi-static isochoric ignition of the gas proceeds from state ( T 2 n) to state ( T 3 n) In this process an input heat Q in qs flows into the gas and is given by Q qs in =!U = C V ( T 3 " T 2 ) (iii) A quasi-static adiabatic expansion of the gas proceeds from state T 3 n ( T 4 n) For this process we find T 4 = T 3 ( ) to state ( ) back to the initial (iv) A quasi-static isochoric exhaust of the gas proceeds from state T 4 n state ( T 1 n) In this process an exhaust heat!q qs ex flows out of the gas and is given by

10 !Q qs ex =!"U =!C V ( T 1! T 4 ) = C V ( T 4! T 1 ) 10 We can now sketch this Otto cycle on the TV diagram as shown below The efficiency of the Otto cycle The efficiency of the Otto cycle is then given by! Otto = 1" "Q qs ( ex) qs =1 " C T " T V 4 1 Q in C V T 3 " T 2 ( ) ( ) = 1" T " T 4 1 T 3 " T 2 Subtracting T 1 = T 2 from T 4 = T 3 we obtain so that T 4! T 1 = V R cv " % 2 $ ' T 3! T 2 # &! Otto = 1" T 4 " T # 1 = 1" V & 2 T 3 " T % ( 2 $ ' ( ) = 1" T 1 T 2 = 1" T 4 T 3 The efficiency of the Otto cycle is determined solely by the compression ratio so that to improve its efficiency we must decrease the compression ratio The compression ratio for an automobile engine is typically 1 8 and for air c v! ( 5 2)R so that! Otto = 1" V R c # & v # 2 % ( =1 " % 1& $ ' $ 8' 2 5 ( = 056

11 The output work of the Otto cycle As the efficiency of a cyclic process is defined by! = "W qs Q in qs we can calculate the output work of the Otto cycle by ) + #!W qs = " Otto Q qs in = C V 1! V & 2 * % $ V ( + 1 ' - + / + T! T 3 2 ( ) 11 so that to increase the output work!w qs we must increase the temperature difference T 3! T 2 and/or decrease the compression ratio The Otto cycle in the TS diagram and its efficiency compared with that of the Carnot cycle The efficiency of the Otto cycle is also less than that of the Carnot cycle between reservoirs at T 1 and T 3 as! Otto = 1" T 1 T 2 = 1" T 4 T 3 <1 " T 1 T 3 =! Carnot which can be seen in the sketch of an Otto cycle in the TS diagram shown below 1013 The Rankine cycle: an idealized model for a steam engine in a power plant As an idealized model for a cyclic process in a steam engine in a power plant we will examine a Rankine cycle of water The Rankine cycle consists of two quasi-static adiabatic processes and two quasi-static isobaric processes at constant pressure Before we describe the Rankine cycle in more detail we need to remind us how we can find the quasi-static heat flowing into a system during a quasi-static isobaric process Quasi-static heat in an isobaric process: P = const As we have found in Sec77 the quasi-static heat Q P qs flowing into a system in an isobaric process at pressure P is simply given by

12 Q P qs =!H 12 where!h is a change in the enthalpy of the system after the isobaric process:!h = H( T f V f n) " H( T i V i n) = H( T f P n) " H( T i P n) so that if we can find the initial and final values of the enthalpy H through an analytical expression for H or in a numerical table for H in some reference handbook we can find the quasi-static heat Q P qs The Rankine cycle of water The Rankine cycle of water consists of the following four quasi-static processes: (i) An quasi-static adiabatic pumping of liquid water to increase its pressure proceeds from state ( n) to state ( P 2 n) (ii) A quasi-static isobaric evaporation of liquid water into steam inside a boiler proceeds from state P 2 n ( ) In this process an input heat Q qs in flows into the gas and is given by ( ) to state V 3 P 2 n Q qs in =!H = H( V 3 P 2 n) " H( P 2 n) # H 3 " H 2 (iii) A quasi-static adiabatic expansion of the steam into a mixture of steam and liquid water inside a turbine proceeds from state ( V 3 P 2 n) to state ( V 4 n) (iv) A quasi-static isobaric condensation of the mixture of steam and liquid water into liquid water proceeds from state V 4 n ( ) back to the initial state ( n) In this process the temperature of the mixture of steam and liquid water also remains constant at T 1 In this qs process an exhaust heat!q ex flows out of the mixture of steam and liquid water and is given by!q ex qs =!"H =! H n { ( )! H( V 4 n)} #!( H 1! H 4 ) = H 4! H 1 We can now sketch this Rankine cycle on the PV diagram as shown below

13 13 The efficiency of the Rankine cycle The efficiency of the Rankine cycle is then given by! Rankine = 1" "Q qs ( ex) qs =1 " H " H 4 1 Q in H 3 " H 2 Using a table for saturated liquid water and steam we can look up the value of H 1 while using a table for superheated steam we can also look up the value of H 3 We can also use the table for saturated liquid water and steam to calculate the value of H 4 Although we can estimate the value of H 2 it turns out that in many applications we can approximate H 2 by H 1 without losing too much accuracy for the efficiency Example: As an example consider a typical fossil-fuel power plant for which P 1 = 0023 atm and P 2 = 300 atm (1 atm! 10 "10 5 Pa ) The temperature for the mixture of liquid water and steam is set at T 1 = 20 C = 293 K while the superheated steam temperature is set at T 3 = 600 C = 873 K Using a table for saturated liquid water and steam we find the value of H 1 for 1 kg of liquid water to be H 1 = 84 kj whereas using a table for superheated steam we find the value of H 3 for 1 kg of steam to be H 3 = 3444 kj We can also use the table for saturated liquid water and steam to calculate the value of H 4 for 1 kg of mixture of liquid water and steam as detailed below and find it to be

14 H 4 = 1824 kj 14 We can estimate the value of H 2 for 1 kg of liquid water as described below and find it to be roughly H 2 = 114 kj Note that H 1 and H 2 are much smaller than H 3 and H 4 H 1 H 2 << H 3 H 4 and H 1 and H 2 are practically on the same order of magnitude: O( H 1 ) ~ O( H 2 ) ~ 100 kj so that we can approximate H 2 by H 1 without losing too much accuracy for the efficiency Using these values of the enthalpy we find the efficiency of this steam engine to be! Rankine = 1" H 4 " H 1 H 3 " H 2 ( 1800 kj) " 100 kj =1 " ( ) ( 3400 kj) "( 100 kj) = 1 " = 048 Finding H 4 To calculate H 4 for 1 kg of mixture of liquid water and steam we need to know the mass fraction x of steam in the mixture at state ( V 4 n) As the adiabatic expansion takes the system from state V 3 P 2 n ( ) to state ( V 4 n) and the entropy of the system must remain constant along a quasi-static adiabatic process the entropy of the system at these states must be equal: S( V 4 n) = S( V 3 P 2 n) We can look up the value for S( V 3 P 2 n) in the table for superheated steam and find it to be S( V 3 P 2 n) = 6233 kj K whereas S( V 4 n) is expressed as so that S( V 4 n) = ( 1! x)s lquid water ( T 1 n) + xs steam ( T 1 n)

15 ( )! S lquid water ( T 1 P 1 n) ( )! S lquid water ( T 1 n) x = S V 4 P 1 n S steam T 1 n ( )! ( 0297 kj K ) ( )! ( 0297 kj K) = kj K = 8667 kj K 15 and H( V 4 n) = ( 1! x)h lquid water ( T 1 n) + xh steam ( T 1 P 1 n) = ( 1! 0709) ( 84 kj) + ( 0709) ( 2538 kj) = 1824 kj Finding H 2 To calculate H 2 for 1 kg of liquid water we first need to find for 1 kg of liquid water: # = +!V = + "V & % ( $ "P ' = 10 )3 m 3 Sn!P = ) * S!P = ( 1 ) * S!P) { } ( ) 1) ( )10 Pa )1 )( Pa) ( )( 1 ) )2 ) = 10 )3 m 3 10 )3 m 3 We can calculate H 2 by H 2 = H 1 + ( H 2! H 1 ) = H 1 + "H = H 1 + "( U + PV) = H 1 + "U + "( PV) where and!u = Q qs + W qs = W qs = " # PdV $ " 1 ( 2 P + P 2 1)!V $ 1 ( %10 5 Pa) ( 12 %10 "5 m 3 ) $180 J!( PV ) " P 2 " ( 300 #10 5 Pa) ( 10 $3 m 3 ) " 30 kj We therefore find H 2! ( 84 kj) + ( 180 J) + ( 30 kj)!114 kj

16 The output work of the Rankine cycle We can calculate the output work of the Rankine cycle by 16!W qs = Q qs in + Q qs ex = ( H 3! H 2 ) + ( H 1! H 4 ) " H 3! H 4 where we have used H 2! H 1 so that to increase the output work!w qs we must increase the enthalpy difference H 3! H 4 The Rankine cycle in the TS diagram and its efficiency compared with that of the Carnot cycle The efficiency of the Rankine cycle is less than that of the Carnot cycle between reservoirs at T 1 and T 3 as can be seen in the sketch of the Rankine cycle in the TS diagram shown below In the previous example for the Rankine cycle in a power plant T 1 = 20 C and T 3 = 600 C so that the efficiency of a Carnot cycle between reservoirs at T 1 and T 3 is indeed higher than the efficiency for the Rankine cycle:! Carnot = 1" 293 K 873 K = 066 > 048 =! Rankine