Earlier Lecture. This gas tube is called as Pulse Tube and this phenomenon is called as Pulse Tube action.

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2 Earlier Leture In te earlier leture, we ave seen a Pulse Tube (PT) ryoooler in wi te meanial displaer is removed and an osillating gas flow in te tin walled tube produes ooling. Tis gas tube is alled as Pulse Tube and tis penomenon is alled as Pulse Tube ation. PT systems an be lassified based on te Stirling type or GM type Geometry and Operating frequeny Pase sift meanism Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 2

3 Earlier Leture In te year 1990, Ray Radebaug of NIST, proposed Pasor analysis of a Pulse Tube Cryoooler. Tere exists a pase angle between te mass flow rate at te old end and te pressure vetor. Tis pase angle depends on te dimensions, frequeny, p 1 and te oter operating parameters. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 3

4 Outline of te Leture Topi : Cryooolers Pasor Analysis (ontd) Pasor Diagrams PT Classifiation based on Pase Sift Meanism Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 4

5 Introdution m m pt m m o Regenerator Pulse Tube Q, T Q, T Q, T In te earlier leture, Pasor analysis of an Orifie Pulse Tube Cryoooler (OPTC) working on a monatomi gas was explained. Te pressure (p) and te temperature (T) variations in te PT are assumed to be sinusoidal. p = p + p os( ωt) T = T + T os( ωt) Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 5

6 Introdution m m pt m m o Regenerator Pulse Tube Q, T Q, T Q, T At any ross setion of te Pulse Tube, using te adiabati law, we ave T1 2 p 1 = T0 5 p0 Mass flow rates at te Hot end and te Orifie are equal. m = mo m = Cp ωt Mass flow rate troug te orifie is 6 Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 1 1 os ( )

7 Introdution m m pt m m o Regenerator Pulse Tube Q, T Q, T Q, T Te mass flow rate at te old end (m ) is as given below. ω pv os π os ( ω ) 1 pt m = ωt+ + Cp 1 1 t γ RT 2 T It is lear tat vetor m is a sum of two vetors wi are at 90 o to ea oter. T Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 7

8 Pasor Analysis m m pt m m o m Regenerator Pulse Tube Q, T Q, T Q, T ω pv 1 pt RT γ ω pv π 1 pt m = os ωt+ + m γ RT 2 T Plotting tese two vetors, we ave te figure as sown. T θ m Pressure T T m Tis diagram is alled as te Pasor diagram of Pulse Tube. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 8

9 Pasor Analysis m m pt m m o θ m m Pressure T T Regenerator Pulse Tube Q, T Q, T Q, T ω pv m 1 pt RT γ ω pv π 1 pt m = os ωt+ + m γ RT 2 T Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay It is lear tat tere exists a pase angle between m and pressure vetor. Te importane of pase angle is explained in tis leture. T 9

10 Pasor Analysis Regenerator Pulse Tube Q, T Q, T Q, T Going aead wit te analysis, onsider a ontrol volume enlosing te Cold End Heat Exanger () as sown in te figure. Let <H> and <Q> denote te time averaged entalpy flow and eat flow respetively, aross te ontrol volume. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 10

11 Pasor Analysis Regenerator Pulse Tube Q, T Q, T Q, T Let us assume te following. H r <H r > be te average entalpy flow in te Regenerator. H pt <H pt > be te average entalpy flow in te Pulse Tube. <Q > be te average eat flow or te eat lifted at te Cold End (refrigeration effet). Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 11

12 Pasor Analysis Regenerator Pulse Tube Q, T Q, T Q, T H r Applying 1 st law of termodynamis to tis ontrol volume, we ave Q = H H If te net eat energy stored by te regenerator in a yle is zero, it is alled as perfet regeneration. H pt pt r Tat is, te eat energy lost by te gas during pressurization is same as te eat energy gained by te gas during depressurization. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 12

13 Pasor Analysis Regenerator Pulse Tube Q, T Q, T Q, T H r Applying 1 st law of termodynamis to tis ontrol volume, we ave Q = H H Assuming a perfet regeneration in te regenerator, we ave H r = 0 Terefore, te eat lifted (Q ) is te entalpy flow into te Pulse Tube. Q = H H pt pt pt r Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 13

14 Pasor Analysis At any ross setion of te OPTC, te definition of entalpy is as given below. Let τ be te total time period of one yle. Let C p be a onstant and te time average entalpy is C p τ 0 H H = mtdt τ T = T + T os( ωt) 0 1 = mc T p Substituting, temperature T into te above equation, we get τ C p H = m ( T ( )) 0 + T1os ωt dt τ 0 Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 14

15 Pasor Analysis C p τ 0 ( ω ) H = m ( T ) 0 + T1os t dt τ From te adiabati law, we ave T T 2 p = 5 p Eliminating T 1 from bot te above equations, we get τ C p 2 p 1 H = m T0 + T0os ( ωt) dt τ 5 p 0 0 τ τ Cp Cp 2m p1 H = mt 0dt + T0os ( ωt) dt τ τ 5 p Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 15

16 Pasor Analysis Regenerator Pulse Tube Q, T Q, T Q, T H r From te 1 st law, te eat lifted (Q ) at te old end of te PT is te entalpy flow of te PT. Q Tis is obtained by substituting m and T in te following equation. C = H τ p 2m p1 H = T0 os( ωt) dt τ 5 p 0 0 H pt pt Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 16

17 2TC p ω pv T = + Pasor Analysis τ p 1 H = m os( ωt) dt 5τ p0 0 ( ω ) os( ω ) 1 pt m sin t Cp 1 1 t γ RT T τ τ 2TC p p pv 1 1 pt T H ω = sin ( 2ωt ) dt + C 2 1p 1 os ( ωt) dt 5τ p0 2γRT T 0 0 In te above equation, te first term is zero. Tis is beause, te sine funtion wen integrated over one full yle vanises. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 17

18 2TC p C pt 5τ p 2T Pasor Analysis τ p H = + t dt 0 0 Rearranging, we ave ( ω ) ( 1 os 2 ) 2 τ τ CC 1 ptp1 H = dt + os( 2ωt ) dt 5τ p0 0 0 Here again, te seond term is zero. Tis is beause, te osine funtion wen integrated over one full yle vanises. 2 2 CC 1 ptp1 H CC 1 ptp1 = [ τ ] = 5τ p 5p 0 0 Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 18

19 Pasor Analysis Terefore, te time averaged entalpy is given as H = CC T p 2 1 p 1 5p 0 We know tat te mass flow rate at te ot end (m ) is a vetor and it is as given below. m = Cp t os 1 1 ( ω ) Te magnitude of tis vetor is given by m = Cp 1 1 Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 19

20 Pasor Analysis Using <H> and m equations, we ave. H C pc T p 1 1 p 1 p 1 = Cp 1 1 H m = 5 p 0 m CTp = 5p 0 adj osθ = = yp T m T m m = T m T osθ m ω pv 1 pt RT γ H T m osθ CTp p 1 = T 5 p0 θ m Pressure T T m H C p T m p 1 = 5 p0 osθ Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 20

21 Pasor Analysis For a monatomi gas, γ=5/3. Terefore, te speifi eat at onstant pressure (C p ) is given by te following equation. C p Rγ 5R = = γ 1 2 θ m m Pressure T T ω pv m 1 pt RT γ H C p T m p 1 = 5 p0 1 p Q = H = RT m 1 2 p0 osθ Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay osθ 21

22 Q RTp m 1 = 2 p0 Pasor Analysis From te adjaent equation, it is osθ lear tat te eat lifted at te old end (Q ) is dependent on m θ m m Pressure T T ω pv m 1 pt RT γ p 1 /p 0 T pase angle. For a given design and operating parameters, te Q is maximum wen pase angle is minimum. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 22

23 Q RTp m 1 = 2 p0 Pasor Analysis Various pase sifting osθ meanisms ave been developed in order to minimize te pase angle. m ω pv 1 pt RT γ It is important to note tat te pase angle an be minimized, and in ertain ases it an be made zero. θ m Pressure T T m Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 23

24 PT Classifiation Pulse Tube Cryoooler Frequeny Stirling Gifford MMaon Geometry Pase Sift Inline U Type Coaxial Annular Basi Orifie Inertane Tube Double Inlet Valve Low Frequeny Hig Frequeny Very Hig Frequeny Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 24

25 PT Classifiation Basi Regenerator Pulse Tube Orifie Regenerator Pulse Tube Double Inlet Valve Regenerator Pulse Tube Inertane Tube Regenerator Pulse Tube Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 25

26 Pasor Diagram BPTC Regenerator Pulse Tube Q, T Q, T Q, T Te semati of a Basi Pulse Tube Cryoooler (BPTC) is as sown above. 1 pt m ωt m γ RT 2 T Te mass flow rate at te Hot end eat exanger is zero. ω pv π T = os + + Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 26

27 Pasor Diagram BPTC Regenerator Pulse Tube Q, T Q, T Q, T Te Pasor diagram for a BPTC is as sown in te figure. m ω pv RT 90 1 pt γ Pressure 1 p Q = RT m 1 2 p0 osθ It is lear tat te pase angle is 90 o, rendering te net eat lifted at te old end as zero. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 27

28 Pasor Diagram OPTC Regenerator Pulse Tube Q, T Q, T Q, T As seen earlier, in an Orifie Pulse Tube Cryoooler (OPTC), an orifie and a ompliane volume is used as pase sift meanism. Tere exists a finite mass flow rate at te Hot end. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 28

29 Pasor Diagram OPTC Regenerator Pulse Tube Q, T Q, T Q, T Te Pasor diagram for an OPTC is as sown in te figure. m ω pv Te relative opening or losing of te orifie alters m osθ. 1 pt RT γ θ Pressure T m T For an optimum orifie opening, te produt m osθ reaes a maximum to maximize Q. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 29

30 Pasor Diagram DIPTC Regenerator Pulse Tube Q, T Q, T Q, T Te semati of a Double Inlet Pulse Tube Cryoooler (DIPTC) is as sown in te figure. Some portion of gas is bypassed from te After Cooler and is fed at te Hot end Heat Exanger. Tis is alled as Double Inlet Line. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 30

31 Pasor Diagram DIPTC m Q Regenerator, T Q, T Pulse Tube Q, T On te double inlet line, an orifie is used to ontrol/regulate te flow of working fluid. m DI m o Tis double inlet orifie togeter wit an Hot end orifie alter te pase angle. Mass flow rates are as sown. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 31

32 Pasor Diagram DIPTC m Q Regenerator, T Q, T Pulse Tube Q, T For te mass flow rates, we ave m = m + m o DI m DI m o T m T Pressure o T T T m = m + m T T T o DI It implies tat m o is te vetor sum of m and m DI. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 32

33 Pasor Diagram DIPTC m DI m Q Regenerator, T Q, T Pulse Tube Q, T m o T T T m = m + m T T T o DI T m T Pressure o We know tat m o is always in pase wit te pressure vetor. Te pasor diagram for tis vetor is as sown in te figure. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 33

34 Pasor Diagram DIPTC m Q Regenerator, T Q, T Pulse Tube Q, T m and m DI are vetorially added to yield m o as sown in te figure. m DI m o T T m ω pv 1 pt RT γ θ T m o T Pressure T m T m DI From te equation of m, we ave te oter sides of te triangle, as sown in te figure. It is lear tat te pase angle is redued due to te modifiation. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 34

35 Pasor Analysis Regenerator Pulse Tube Q, T Q, T Q, T In te pasor analysis, we ave assumed an adiabati proess in te Pulse Tube. In te oter elements, for example in onneting tubes,,,, an isotermal proess is assumed. For te sake of understanding, let us analyze te Cold End Heat Exanger in an OPTC. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 35

36 Pasor Analysis m r m Regenerator Pulse Tube Q, T Q, T Q, T As done before, p and T are assumed as follows. p = p + p os( ωt) 0 1 T = T0 Let m r and m be te mass flow rates at te inlet and outlet as sown. We ave m = m m Following te earlier derivation, we ave x r ω pv 1 x mr = sin ( ωt) + m RT Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 36

37 ω pv 1 x π mr = os ωt+ + m RT 2 Pasor Analysis Combining te above equations, we ave ω pv ω pv ω pv 1 x 1 pt π T mr = + os ωt+ + m RT γ RT 2 T π 1 pt m = os ωt+ + m γ RT 2 T Terefore, te mass flow rate (m r ) is te sum of two vetors wi are at 90 o to ea oter. T Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 37

38 Pasor Analysis ω pv ω pv π 1 x 1 pt mr = + os ωt+ + m RT γ RT 2 T T From te above equation, it is lear tat te first term is at 90 o to te seond term. θ m m r m Pressure T T ω pv RT m 1 x ω pv 1 pt RT γ Plotting tese two vetors, we ave te figure as sown. m r lies above te m vetor. Tis is due to te vetor addition as sown in te figure. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 38

39 Pasor Analysis Regenerator Pulse Tube ω pv 1 p RT0 ω pv 1 a m m a RT p 0 ω pv 1 r RTm m ra ω pv 1 x m r RT ω pv 1 pt m RTγ m x ω pv 1 x θ RT Pressure T m T m Following a similar proedure for te rest of te elements, we ave a pasor diagram as sown. From te figure, it is lear tat m p vetor is almost double te lengt of m vetor. Terefore, for a given m, we need a very large m p. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 39

40 Summary Tere exists a pase angle between mass flow rate at te old end and pressure vetor. Q RTp m 1 = 2 p0 osθ Heat lifted at te old end (Q ) is dependent on m, p 1 /p 0, T, pase angle. In te pasor analysis, adiabati proess is assumed in PT and isotermal proess is assumed in all oter parts. Te relative lengt of te vetors indiate te mass flow rate in tose parts. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 40

41 A self assessment exerise is given after tis slide. Kindly asses yourself for tis leture. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 41

42 Self Assessment 1. In a perfet regeneration, te net eat energy stored by regenerator is. 2. For a given design and operating parameters, te Q is maximum wen pase angle is. 3. Te eat exange during pressurization and depressurization is. 4. Pase sifting meanisms are used to te pase angle. 5. In a Basi PT Cryoooler, te pase angle is. 6. In Pasor Analysis, PT undergoes proess. 7. Apart from PT, all oter parts undergo proess. 8. In, double inlet orifie and ot end orifie alter te pase angle. Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 42

43 1. Zero 2. Minimum 3. Same 4. Minimize Answers o 6. Adiabati 7. Isotermal 8. DIPTC Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 43

44 Tank You! Prof. M D Atrey, Department of Meanial Engineering, IIT Bombay 44