27 - NPTEL

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27

2Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay

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.

Topic : Cryocoolers

• Ideal Stirling cycle

• Working of Stirling Cryocooler

• Schmidt's Analysis

• Conclusions


Outline of the Lecture


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.

The Chronology


Year Event1815 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


An Ideal Stirling CyclePr

essu

re –

p

Volume – V

• Consider a p – V chart as shown in the figure.

• 12: Isothermal compression at TC.

CQ2

11 1 2 2p V p V=

1 2 CT T T= =

2

1

lnCVdQ dW TV

= = −ℜ



essu

re –

p

Volume – V

• 23: Constant volume heat rejection.

• 34: Isothermal expansion.

CQ

EQ

RTQ

2

3

4

1

2 3V V=

( )V E CdQ C T T= + −

3 3 4 4p V p V=

3 4 ET T T= =

4

3

lnCVdQ dW TV

= = −ℜ



essu

re –

p

Volume – V

• 41: Constant volume heat absorption.

CQ

EQ

RTQ

2

3

4

1RTQ

4 1V V= ( )V C EdQ C T T= − −

E

C E

QCOPQ Q

=−

4

3

2 4

1 3

ln

ln ln

E

C E

VTV

V VT TV V

+ℜ

=

−ℜ −ℜ


An Ideal Stirling Cycle

4

3

2 4

1 3

ln

ln ln

E

C E

VTV

V VT TV V

+ℜ

=

−ℜ −ℜ

32

1 4

VVV V

=

Pres

sure

–p

Volume – V

CQ

EQ

RTQ

2

3

4

1RTQ

E

C E

TCOPT T

=−

COP(Stirling) = COP(Carnot)


Stirling & Carnot CyclesPr

essu

re –

p

Volume – V

2

3

4

1

Tem

pera

ture

–T

Entropy – s

2

3 4

1TC

TE

Stirling Cycle


Stirling & Carnot CyclesPr

essu

re –

p

Volume – V

2

3

4

1

Tem

pera

ture

–T

Entropy – s

2

3 4

1TC

TE

5

6

5

6

Carnot Cycle Stirling Cycle

.


Ideal Stirling Cycle

Tc

TE

Tc

Compression

Pres

sure

–p

Volume – V

Regenerative Cooling

Expansion

Regenerative Heating

CQ2

1RTQ

3

EQ 4

RTQ

PistonRegeneratorExpander

1

2

3

4

1

CQ

EQ


Ideal Stirling CycleTi

me

Displacement

Tc

TE

Tc

Compression

Regenerative Cooling

Expansion

Regenerative Heating

PistonRegeneratorExpander

1

2

3

4

1

1

2

3

4

1

1

2

3

4

1

0E CV V= =,maxEV ,maxCV


Ideal Stirling Cycle• As mentioned in the earlier

lecture, the characteristics of a Stirling cycle are• High frequency.

• Regenerative heat exchanger.

• Phase difference between the piston and the displacer motions.

Tim

e

Displacement

1

2

3

4

1

1

2

3

4

1



Actual Stirling Cycle• In actual Stirling cycle the

discontinuous motion can not be achieved. In view of this sinusoidal motion may be implemented.

• This motion is realistic and can be achieved using a Crank or gas spring mechanism.

Tim

e

Displacement

1

2

3

4

1

1

2

3

4

1



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.


Stirling Cryocooler – Types• Depending upon the relative

arrangements of piston and displacer/piston, various types of Stirling Cryocoolers are possible, namely• α type Stirling Cryocooler.

• β type Stirling Cryocooler.

• γ type Stirling Cryocooler.

ExpanderTypeα

Compressor

Regenerator

Typeβ Typeγ


Stirling Cryocooler – Types• Two Piston arrangement

(α type)• whose drive mechanisms may

be mounted on same crank shaft.

• Integral Piston & Displacer arrangement (β type)

• The piston and displacer are housed inside same cylinder.

ExpanderTypeα

Compressor

Regenerator

Typeβ Typeγ


Stirling Cryocooler – Types• Split Piston & Displacer

arrangement (γ type)• The compression space is

divided.

• These systems have variable dead volume in compression space due to the movement of displacer.

ExpanderTypeα

Compressor

Regenerator

Typeβ Typeγ


Design Parameters• The various design parameters of

a Stirling Cryocooler are as follows.

• Evaporator temperature (TE)• Condenser temperature (TC)• Compression Volume (VC)• Expansion Volume (VE)• Regenerator Volume (VR)• Pmax, Pmin, Pavg.• Phase angle (α)• Crank angle (ø)

ExpanderTypeα

Compressor

Regenerator

TC TEVE

VC

VR


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.


• The non – dimensional parameters in the Schmidt’s analysis are

• Swept volume ratio :

• Temperature ratio :

• Dead volume ratio :

Schmidt’s Analysis

C

E

VkV

=

C

E

TT

τ =

D

E

VXV

=


• Expansion volume variation :

• Compression volume variation :


( )1 1 cos2e EV V φ= +

( )1 1 cos( )2c CV V φ α= + − C

E

VkV

=

( )1 1 cos( )2c EV kV φ α= + −

Volu

me

–V

Angle – ø

VC

VE

α


• Let the instantaneous pressure in the system be same throughout the system, p.

• Also, Te and Tc are assumed to be constants as TEand TC respectively.

• Let MT be given as shown.

Schmidt’s Analysise e c c d d

Te c d

p V p V p VMRT RT RT

= + +

2E

TC

KVMRT

=

2E

C

KVRT

=


Schmidt’s Analysis( ) ( )1 cos 1 cos( )

2 2 2CE D E E

C E E D C

T kpV V T KVRT T V T RT

φ φ α+ + − + + =

C

E

TT

τ = D

E

VXV

=2

E Cd

T TT +=

21

XS ττ

=+

( ) ( )1 cos 1 cos( ) 2K k Sp

τ φ φ α= + + + − +

2 2( cos ) ( sin )A k kτ α α= + + 2B k Sτ= + +AB

δ =

sintancos

kk

αθτ α

=+



[ ]cos( ) 1Kp

B δ θ φ=

− +

[ ]min 1Kp

B δ=

+

[ ][ ]11ratiop

δδ

+=

−

@ φ θ=

[ ]max 1Kp

B δ=

−@ φ θ π= −

• Substituting, A, B, θ and δ in the mass equation and rearranging, we get


Schmidt’s Analysis• Mean pressure

2

0

1 ( )2mp pd

π

θ φπ

= −∫

max11mp p δ

δ−

=+

0.52

sin

1 1m E

E ep VQ pdV π δ θ

δ= =

+ − ∫ 0.52

sin( )

1 1m E

C cp V kQ pdV π δ θ α

δ

−= =

+ − ∫

E

T

QCOPW

= E

C E

TT T

=−

E

C E

QCOPQ Q

=−


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.


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.


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.


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).


• A self assessment exercise is given after this slide.

• Kindly asses yourself for this lecture.

Self Assessment1. 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 _____.33Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombay

Answers


1. Isothermal and Constant volume

2. m

3. R

4. Equal.

5. M

6. Sinusoidal

7. Void volume

8. Beta

9. Constant

[ ]2 1ln /CdQ dW T V V= = −ℜ

( )V E CdU C T T= + −

( )/E C ET T T−


Thank You!

27 - NPTEL

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