The Ideal thermodynamic cycle for gas liquefaction is impractical and hence modified cycles are proposed.

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2 Earlier Lecture The Ideal thermodynamic cycle or as liqueaction is impractical and hence modiied cycles are proposed. An Ideal cycle is used as a benchmark and in eect, dierent ratios and unctions are deined to compare various liqueaction systems. A Linde Hampson cycle consists o a compressor, heat exchaner and a J T expansion device. In this system, only a part o the as that is compressed, ets liqueied. 2

3 Earlier Lecture The isobaric heat exchane process occurrin in the heat exchaner is used to conserve cold and J T expansion device is used or producin lower temperatures. The work required or a unit mass o as compressed or a Linde Hampson system is Wc T s s 2 h h 2 = ( ) ( ) The yield y is maximum when the state 2 (ater compression) lies on the inversion curve at the temperature o the compression process. 3

4 Earlier Lecture For a Linde Hampson system ollowin hold true. As the compression pressure increases, the liquid yield y increases or a iven compression temperature. As the compression temperature decreases, the liquid yield y increases or a iven compression pressure. 4

5 Outline o the Lecture Topic : Gas Liqueaction and Rerieration Systems (contd) Basics o Heat Exchaners Eect o heat exchaner eectiveness on Linde Hampson system Fiure o Merit (FOM) 5

6 Heat Exchaner A heat exchaner is a device in which the coolin eect rom cold luid is transerred to precool the hot luid. It can either be a two luid type or a three luid type dependin upon the number o inlets and outlets attached to the heat exchaner. The process o heat exchane occurs at a constant pressure and hence, it is an isobaric process. 6

7 Heat Exchaner Fluid A, in, hot Fluid A, out, cold Fluid B, out, hot Fluid B, in, cold A, in, hot B, out, hot A, out, cold B, in, cold Lenth, x 7

8 Heat Exchaner In order to speciy the perormance o the heat exchaner in terms o actual heat exchane occurrin, a term " ε" is deined which is called as Heat Exchaner Eectiveness. It is deined as a ratio o actual heat transer that is occurrin to the maximum possible heat transer that can occur theoretically. Mathematically, ε = Q act Q max ε is a dimensionless number between 0 and. 8

9 Heat Exchaner Eectiveness Fluid A, in, hot Fluid B, out, hot A, in, hot B, out, hot Fluid A, out, cold Fluid B, in, cold A, out, cold The actual heat exchane is The maximum possible heat exchane is (,, ) (, Aout, ) Q = C T T = C T T act B PB Bin Bout A PA Ain ( ɺ ) (,, ) Q = mc T T max P min Bin Ain Lenth, x Q Eectiveness is B, in, cold Q ε = act Q max 9

10 Linde Hampson System 2 The isothermal compression process or a Linde Hampson system on T s diaram is rom 2. The Isenthalpic J T expansion is rom h=const The as and liquid states are iven by and. s 0

11 Linde Hampson System 2 ' I the ε is assumed to be 00%, the isobaric heat exchane would be 2 3 3' 3 4 h=const s In actual, the heat exchane is not perect and hence, these processes are rom 2 3

12 Linde Hampson System 3' h=const 4' ' In such a case, the as is always compressed rom the state to state 2. The J T expansion occurs rom 3 4 as shown in the iure. s 2

13 Linde Hampson System 2 ' It is clear that the actual heat exchane is rom, where as the maximum possible heat exchane is rom. 3' 3 4 h=const 4' On the other pressure line, the actual heat exchane is rom 2 3, where as the maximum possible heat exchane is rom 2 3. s 3

14 Linde Hampson System 2 In an ideal system, chane in enthalpy on these two isobaric lines is equal. 3' 3 4 h=const 4' ' Thereore, the eectiveness is iven as h' h ε = h h OR s ε = h h 3' 2 h h 3 2 4

15 Linde Hampson System 2 Makeup as Q R 2 ' W c ( m ɺ m ɺ ) ' 3' 3 4 h=const 4' s 3' 4' 5

16 ( m ɺ m ɺ ) Linde Hampson System Makeup as Q R 2 W c 4' Consider a control volume or this system as shown in the iure. The quantities enterin and leavin this control volume are as iven below. ' IN OUT 2 (m m 3' Usin the st Law, we have ( ) mh ɺ = h + h 2 ' 6

17 ( m ɺ m ɺ ) Linde Hampson System Makeup as Q R 2 W c ' 3' Rearranin the terms, we have liquid yield y as y h h h' h ' 2 = = As seen earlier, the eectiveness is h' h ε = h h Rearranin the terms, we have 4' ( ) h = ε h h + h ' 7

18 Linde Hampson System y = h h h h ' 2 ' ( ) h = ε h h + h ' From the above two equations, substitutin one into another, we have y as y = ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h 8

19 Linde Hampson System y = ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h The second term bein neative, it should be minimum to maximize the yield y. All other parameters bein constant or a iven cycle, the eectiveness ε should be very close to. The next tutorial depicts the eect o the heat exchaner eectiveness on the liquid yield. 9

20 Tutorial Determine the liquid yield or a Linde Hampson cycle with air as workin luid when the system is operated between.03 bar ( atm) and bar (200 atm) at 300 K. The eectiveness o HX is 00%, 95%, 90% and 85%. Comment on the results. 200 The T s diaram o Linde Hampson system i assumed the heat exchaner to be 00% eective is as shown. 300 h=const T=const s 20

21 y = Tutorial ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h 2 p (bar) T (K) h (J/) s (J/K) ε = y = ( ) ( )( ) ( ) ( )( ) =0.085 y =

22 y = Tutorial ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h 2 p (bar) T (K) h (J/) s (J/K) ε 2 = 0.95 y 2 = ( ) ( 0.95)( ) ( ) ( 0.95)( ) = =0.060 y 2 =

23 y = Tutorial ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h 2 p (bar) T (K) h (J/) s (J/K) ε 3 = 0.90 y 3 = ( ) ( 0.90)( ) ( ) ( 0.90)( ) = =0.034 y 3 =

24 y = Tutorial ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h 2 p (bar) T (K) h (J/) s (J/K) ε 4 = 0.85 y 4 = ( ) ( 0.85)( ) ( ) ( 0.85)( ) = =0.006 y 4 =

25 y Workin Fluid : Air Pressure : bar 200 bar Temperature : 300 K Linde Hampson System Tutorial ε y ε Plottin the values or the above conditions, we have the trend as shown. 25

26 y Workin Fluid : Air Pressure : bar 200 bar Temperature : 300 K Linde Hampson System Tutorial Joinin the points, we have the trend as shown in the iure. It is clear that as the eectiveness decreases, the yield y decreases drastically ε Furthermore, the eectiveness should be more than 85% in order to have a liquid yield. 26

27 Fiure o Merit (FOM) In the earlier lecture, we have seen that the ideal cycle is used as a benchmark to compare various liqueaction systems. In eect, a parameter called Fiure o Merit (FOM) is deined. It is the ratio o the ideal work input to the actual work input o the system per unit mass o as liqueied. Mathematically, W i W 27

28 Tutorial 2 Determine the ollowin or a Linde Hampson system with Nitroen as workin luid when the system is operated between.03 bar ( atm) and bar (200 atm) at 300 K. The eectiveness o HX is 00%. Ideal Work requirement Liquid yield Work/unit mass compressed Work/unit mass liqueied FOM 28

29 Tutorial 2 Ideal Work Requirement ɺ = Wi T s s h h ( ) ( ) T 2 p (bar) T (K) h (J/) s (J/K) s W c ( ) ( ) = = 76 J/ 29

30 Liquid yield y h h h h 2 = Tutorial 2 2 p (bar) T (K) h (J/) s (J/K) T=const h=const 4 s y h h h h 2 = = = 433 =

31 Tutorial 2 Work/unit mass o as compressed Wc T s s 2 h h 2 = ( ) ( ) 2 p (bar) T (K) h (J/) s (J/K) T=const h=const 4 s W c ( ) ( ) = = 463 J/ 3

32 Tutorial 2 Work/unit mass o as liqueied T=const W c = 463 y= h=const W Wc y c = 463 = = J/ s 32

33 Tutorial 2 Fiure o Merit (FOM) T=const W c = W i = h=const FOM W m = ɺ Wc i 767 = = s 33

34 Tutorial 3 Determine the ollowin or a Linde Hampson system with Aron as workin luid when the system is operated between.03 bar ( atm) and bar (200 atm) at 300 K. The eectiveness o HX is 00%. Ideal Work requirement Liquid yield Work/unit mass compressed Work/unit mass liqueied FOM 34

35 Tutorial 3 Ideal Work Requirement ɺ = Wi T s s h h ( ) ( ) T 2 p (bar) T (K) h (J/) s (J/K) s W c ( ) ( ) = = 476 J/ 35

36 Liquid yield y h h h h 2 = Tutorial p (bar) T (K) h (J/) s (J/K) T=const h=const 4 s y h h h h 2 = = = =

37 Tutorial 3 Work/unit mass o as compressed Wc T s s 2 h h 2 = ( ) ( ) p (bar) T (K) h (J/) s (J/K) T=const h=const 4 s W c ( ) ( ) = = 326 J/ 37

38 Tutorial 3 Work/unit mass o as liqueied T=const W c = 326 y= h=const W Wc y c = 326 = = J/ s 38

39 Tutorial 3 Fiure o Merit (FOM) T=const W c = W i = h=const FOM W m = ɺ Wc i = = s 39

40 Perormance o L H System Fluid Boil. Pt y c c ɺ N Ar Air O W m W m ɺ FOM The above table is or a Linde Hampson system when the pressures are rom bar to 200 bar at 300K. The heat exchaner eectiveness is 00%. 40

41 Tutorial 4 For the conditions speciied in Tutorial 2 or the Linde Hampson system with Nitroen as workin luid, calculate the ollowin when the eectiveness o HX is 90%. Liquid yield Work/unit mass compressed Work/unit mass liqueied FOM Comment on the results 4

42 Liquid yield y = Tutorial 4 ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h 2 p (bar) T (K) h (J/) s (J/K) ε = 0.90 y = ( ) ( 0.90)( ) ( ) ( 0.90)( ) =

43 Tutorial 4 Location o h and Additional Work ε = h h ' h h h ( ) = ε h h + h ' 2 p (bar) T (K) h (J/) s (J/K) ' 0.9( ) 230 h = + = J/ Wc add = h h ' = = 23.2 J/ 43

44 Tutorial 4 Work/unit mass o as compressed Wc T s s 2 h h 2 ( ) ( ) 2 p (bar) T (K) h (J/) s (J/K) W c ( ) ( ) = Wc Wc Wc m = + m m ɺ ɺ ɺ total h h add = 463 J/ = = J/ 44

45 Tutorial 4 Work/unit mass o as liqueied W c = c = y=0.025 W Wc = = J/ y Fiure o Merit (FOM) Wc Wi = = 76 Wi Wc 76 FOM = / = =

46 y W c ɺ Wc m ɺ FOM Tutorial 4 ε = ε = 0.9 % chane The above table hihlihts the siniicance o heat exchaner eectiveness or a Linde Hampson system 46

47 Assinment. Determine the ollowin or a Linde Hampson system with Air as workin luid when the system is operated between.03 bar ( atm) and bar (200 atm) at 300 K. The eectiveness o HX is 00%. Ideal Work requirement Liquid yield Work/unit mass compressed Work/unit mass liqueied FOM 47

48 Assinment 2. Determine the ollowin or a Linde Hampson system with Oxyen as workin luid when the system is operated between.03 bar ( atm) and bar (200 atm) at 300 K. The eectiveness o HX is 00%. Ideal Work requirement Liquid yield Work/unit mass compressed Work/unit mass liqueied FOM 48

49 Assinment 3. Repeat the Problem and Problem 2 the Linde Hampson system when the heat exchaner eectiveness is 90%. Compare and comment on the results. 49

50 Summary A heat exchaner is a device in which the cold is transerred rom cold luid to hot luid. Eectiveness ε is deined as a ratio o actual heat transer that is occurrin to the maximum possible heat transer that can occur theoretically. It is a dimensionless number between 0 and. In a Linde Hampson cycle, the heat exchaner eectiveness ε is h or ' h h3' h2 ε = ε = h h h h

51 Summary The liquid yield y or a Linde Hampson system is iven by y = ( h ) ( )( ) h2 ε h h ( h ) ( )( ) h ε h h As the eectiveness decreases, the yield y decreases drastically. Furthermore, the eectiveness should be more than 85% in order to have a liquid yield in Linde Hampson cycle. 5

52 Thank You! 52