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1 Earlier Lecture • A Cryocooler is a mechanical device operating in a closed cycle, which generates low temperature.
• It eliminates cryogen requirement, offers reliable operation and is also cost effective.
• Heat exchangers can either be regenerative or recuperative type depending upon heat exchange.
• Recuperative Type: J – T, Brayton, Claude.
• Regenerative Type: Stirling, GM, Pulse Tube.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 2 Outline of the Lecture Topic : Cryocoolers
• Ideal Stirling cycle
• Working of Stirling Cryocooler
• Schmidt's Analysis
• Conclusions
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 3 History • A well developed and a most commonly used Cryocooler is the Stirling Cycle Cryocooler.
• This cycle was first conceived by Robert Stirling in the year 1815. It was an engine cycle and was aimed to produce work (engine).
• The important events that occurred in the history of cryocoolers are as given in the next slide.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 4 The Chronology Year Event 1815 Robert Stirling – Stirling Engine 1834 John Herschel – concept of using as a cooler 1861 Alexander Kirk – The concept into practice 1873 Davy Postle – Free Piston system 1956 Jan Koehler – First commercial machine for air liquefaction 1965 Jan Koehler – Nitrogen Liquefaction
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 5 An Ideal Stirling Cycle • Consider a p – V chart as shown in the figure.
2 QC • 1 2: Isothermal p compression at TC. – 1 pV11= pV 2 2
Pressure TTT12= = C
V2 dQ= dW = −ℜ TC ln V Volume – V 1
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 6 An Ideal Stirling Cycle • 23: Constant volume heat rejection.
2 VV23=
QC p dQ=+− CVE( T T C) QRT – 1 • 34: Isothermal expansion. 3 pV33= pV 44 Pressure
Q E 4 TTT34= = E
Volume – V V4 dQ= dW = −ℜ TC ln V3
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 7 An Ideal Stirling Cycle • 41: Constant volume heat absorption. =−− 2 VV41= dQ CVC( T T E)
QC p QRT
– Q 1 COP = E QQ− 3 CE
QRT Pressure V +ℜ 4 Q TE ln E 4 V = 3 VV −ℜ24 − ℜ Volume – V TTCEln ln VV13
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 8 An Ideal Stirling Cycle V V 2 = 3 VV14 2 V4 QC +ℜTE ln p QRT V3 – 1 = VV24 3 −ℜTTCEln − ℜ ln VV13 QRT Pressure
Q TE E 4 COP = TTCE−
Volume – V COP(Stirling) = COP(Carnot)
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 9 Stirling & Carnot Cycles Stirling Cycle
2 T TC 2 1 – p
– 1 3 Pressure Temperature TE 3 4 4
Volume – V Entropy – s
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 10 Stirling & Carnot Cycles Carnot Cycle Stirling Cycle
2 T TC 2 5 1 – p 5 – 1
3 . Pressure Temperature 6 TE 3 6 4 4
Volume – V Entropy – s
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 11 Ideal Stirling Cycle
Expander Regenerator Piston
1 Tc 2 Compression QC QC p QRT 2 – 1 Regenerative Cooling 3 3 QRT Pressure Q Expansion Q E E 4 4 TE
Volume – V Regenerative Heating
1 Tc
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 12 Ideal Stirling Cycle V Expander Regenerator E,max VVEC= = 0 VC,max Piston 1 1 1 Tc
2 2 Compression 2 3 3 Regenerative Cooling Time 4 4 3
Expansion 4 TE 1 1 Displacement Regenerative Heating
1 Tc
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 13 Ideal Stirling Cycle • As mentioned in the earlier VE,max VV= = 0 V EC C,max lecture, the characteristics 1 1 of a Stirling cycle are 2 2 • High frequency.
3 3 • Regenerative heat Time 4 4 exchanger.
• Phase difference between the piston and the 1 1 Displacement displacer motions.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 14 Actual Stirling Cycle • In actual Stirling cycle the VE,max VV= = 0 V EC C,max discontinuous motion can 1 1 not be achieved. In view of 2 2 this sinusoidal motion may be implemented. 3 3
Time • This motion is realistic and 4 4 can be achieved using a Crank or gas spring mechanism. 1 1 Displacement
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 15 Actual Stirling Cycle • In reality, the actual working cycle will be different from Ideal Stirling Cycle in following ways.
• Discontinuous motion, difficult to realize in practice.
• Presence of void volume or dead space (not swept by piston or displacer), pressure drop.
• Ineffectiveness in heat transfer or regeneration.
• Non isothermal compression and expansion.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 16 Stirling Cryocooler – Types Regenerator • Depending upon the relative arrangements of piston and displacer/piston, various types of Stirling Cryocoolers are
Compressor Expander possible, namely α Type • α type Stirling Cryocooler.
• β type Stirling Cryocooler.
• γ type Stirling Cryocooler.
β Type γ Type Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 17 Stirling Cryocooler – Types Regenerator • Two Piston arrangement (α type) • whose drive mechanisms may be mounted on same crank Compressor Expander shaft. α Type • Integral Piston & Displacer arrangement (β type) • The piston and displacer are housed inside same cylinder.
β Type γ Type Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 18 Stirling Cryocooler – Types Regenerator • Split Piston & Displacer arrangement (γ type) • The compression space is divided. Compressor Expander α Type • These systems have variable dead volume in compression space due to the movement of displacer.
β Type γ Type Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 19 Design Parameters • The various design parameters of a Stirling Cryocooler are as
Regenerator follows.
VR TC VC TE • Evaporator temperature (TE) VE • Condenser temperature (TC)
Compressor Expander • Compression Volume (VC) α Type • Expansion Volume (VE)
• Regenerator Volume (VR)
• Pmax, Pmin, Pavg. • Phase angle (α) • Crank angle (ø)
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 20 Schmidt’s Analysis • In the year 1861, Gustav Schmidt, a German scientist, presented a Stirling Cryocooler analysis.
• This analysis is based on a realistic cycle and is assumed to provide a first guess of dimensions. The following are the assumptions.
• Perfect isothermal compression, expansion.
• Harmonic motion of piston and displacer.
• Perfect regeneration.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 21 Schmidt’s Analysis • The non – dimensional parameters in the Schmidt’s analysis are
V • Swept volume ratio : k = C VE
T • Temperature ratio : τ = C TE
V • Dead volume ratio : X = D VE
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 22 Schmidt’s Analysis • Expansion volume variation : 1 α VVeE=(1 + cosφ ) 2 VE • Compression volume variation : VC V 1 VC = – VVcC=(1 +− cos(φα )) k 2 VE
1 Volume V= kV (1 +− cos(φα )) cE2
Angle – ø
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 23 Schmidt’s Analysis
pVee pV cc pV dd KVE MT =++ = RTecd RT RT 2RTC
• Let the instantaneous pressure in the system be same throughout the system, p.
• Also, Te and Tc are assumed to be constants as TE and TC respectively.
KVE • Let MT be given as shown. MT = 2RTC
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 24 Schmidt’s Analysis
pV (1+ cosφ)TkC ( 1 +− cos(φα )) VTKV E+ +=DE E RTC22 TE VED T 2 RT C
T V + τ τ = C = D TTEC 2X X Td = S = TE VE 2 τ +1 K =τ(1cos + φ) ++kS( 1cos( φα −+ )) 2 p A Ak=++(τα cos )22 ( k sin α ) B=++τ kS2 δ = B k sinα tanθ = τα+ k cos
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 25 Schmidt’s Analysis • Substituting, A, B, θ and δ in the mass equation and rearranging, we get K p = B[δcos( θφ−+ ) 1] K p = @ φθ= min B[1+δ ] K p = @ φθπ= − max B[1−δ ] [1+δ ] p = ratio [1−δ ]
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 26 Schmidt’s Analysis • Mean pressure 1 2π =θφ − pm ∫ pd() 2π 0
1−δ pp= m max 1+δ
πδpVsin θ πpVmE δsin( θα− ) k mEQ= pdV = QEe= pdV = 0.5 Cc∫ 0.5 ∫ 2 +−δ 2 11+−δ 11
Q Q T COP = E COP = E = E WT QQCE− TTCE−
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 27 Losses • In the earlier slide, we saw the cooling effect based on Schmidt's analysis.
• But, in an actual system, there are many losses. Few of them are as listed below.
• Ineffectiveness of regenerator. • Pressure drop in system. • Solid conduction losses. • Shuttle conduction losses. • Losses in power input.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 28 Losses • Considering the above mentioned losses, the net cooling effect and gross power required is given by the following correlations.
• Qnet = QE – Σ(losses).
• Wtotal = WT + Σ(losses).
• In general, QE calculated from Schmidt's analysis, in which 60 – 70% are considered as losses, while losses in power input is due to mechanical efficiency.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 29 Summary • A Stirling Cycle was first conceived by Robert Stirling in the year 1815.
• COP(Stirling) = COP(Carnot).
• In reality, the actual working cycle has discontinuous motion, pressure drop, ineffectiveness and non isothermal processes.
• Depending upon the relative arrangements of piston and displacer/piston, α, β, γ are the different types of Stirling cryocooler.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 30 Summary • Gustav Schmidt presented a Stirling Cryocooler analysis in the year 1861, it is assumed to provide a first guess of dimensions.
• The net cooling effect and gross power required is given by the following correlations.
• Qnet = QE – Σ(losses).
• Wtotal = WT + Σ(losses).
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 31 • A self assessment exercise is given after this slide.
• Kindly asses yourself for this lecture.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 32 Self Assessment 1. A Stirling cycle consist of two ______processes. 2. In an isothermal process, dQ is given by ______. 3. In a constant volume process, dU is given by ____.
4. COPCarnot and COPStirling are _____. 5. COP of Stirling cycle is ______. 6. In an actual Stirling cycle, the discontinuous motion is approximated to ______motion. 7. The volume not swept by piston/displacer is ____. 8. In a _____ type unit, the piston and displacer are housed inside same cylinder. 9. In Schmidt's analysis, instantaneous pressure is assumed to be _____.
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 33 Answers 1. Isothermal and Constant volume
2. mdQ= dW = −ℜ TC ln[ V21 / V ] 3. R dU=+− CVE( T T C) 4. Equal.
5. MTE/ ( TT CE− ) 6. Sinusoidal 7. Void volume 8. Beta 9. Constant
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 34 Thank You!
Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay 35