Phasor Analysis of a Pulse Tube Cryocooler: Mass Flow Rate and Pressure Vector Phase Angle, Lecture notes of Mechanical Engineering

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1

Earlier Lecture

• In the earlier lecture, we have seen a Pulse Tube (PT) cryocooler in which the mechanical displacer is removed and an oscillating gas flow in the thin walled tube produces cooling.
• This gas tube is called as Pulse Tube and this phenomenon is called as Pulse Tube action.
• PT systems can be classified based on the
  • Stirling type or GM type
  • Geometry and Operating frequency
  • Phase shift mechanism

Topic : Cryocoolers

• Phasor Analysis (contd)
• Phasor Diagrams
• PT Classification based on Phase Shift Mechanism

Outline of the Lecture

Introduction

Regenerator Pulse Tube

AC CHX HHX

Q^ AC ,TAC Qc ,Tc Qh ,Th

m^ c m^ pt m^ h mo

• In the earlier lecture, Phasor analysis of an Orifice Pulse Tube Cryocooler (OPTC) working on a monatomic gas was explained.
• The pressure ( p ) and the temperature ( T ) variations in the PT are assumed to be sinusoidal.

p = p 0 + p 1 cos( ω t) T = T 0 +T 1 cos( ω t)

Introduction

(^1) cos 1 1 cos ( ) 2

pt (^) h c c c

p V (^) T m t C p t RT T

= ^ + +

Regenerator Pulse Tube

AC CHX HHX

Q^ AC ,TAC Qc ,Tc Qh ,Th

m^ c m^ pt m^ h mo

• The mass flow rate at the cold end ( m (^) c ) is as given below.
• It is clear that vector m (^) c is a sum of two vectors which are at 90 o^ to each other.

Phasor Analysis

θ

m^ h^ Pressure h h c

T (^) m T

     



1 pt c

p V RT

ω γ

m c

(^1) cos 2

pt (^) h c h c c

p V (^) T m t m RT T

= ^ + +

Regenerator Pulse Tube

AC CHX HHX

Q^ AC ,TAC Qc ,Tc Qh ,Th

m^ c m^ pt m^ h mo

• Plotting these two vectors, we have the figure as shown.
• This diagram is called as the Phasor diagram of Pulse Tube.

Phasor Analysis

• Going ahead with the analysis, consider a control volume enclosing the Cold End Heat Exchanger (CHX) as shown in the figure.
• Let and denote the time averaged enthalpy flow and heat flow respectively, across the control volume.

Regenerator Pulse Tube

AC CHX HHX

Q^ AC ,TAC Qc ,Tc Qh ,Th

Phasor Analysis

• Let us assume the following.
  • <H (^) r> be the average enthalpy flow in the Regenerator.
  • <H (^) pt > be the average enthalpy flow in the Pulse Tube.
  • be the average heat flow or the heat lifted at the Cold End (refrigeration effect).

Regenerator Pulse Tube

AC CHX HHX

Q^ AC ,TAC Qc ,Tc Qh ,Th

H^ r Hpt

Phasor Analysis

• Applying 1 st^ law of thermodynamics to this control volume, we have
• Assuming a perfect regeneration in the regenerator, we have
• Therefore, the heat lifted ( Qc ) is the enthalpy flow into the Pulse Tube.

Regenerator Pulse Tube

AC CHX HHX

Q^ AC ,TAC Qc ,Tc Qh ,Th

H^ r Hpt

H (^) r = 0

Q^ c = Hpt

Q^ c = H^ pt − Hr

Phasor Analysis

• At any cross section of the OPTC, the definition of enthalpy is as given below.
• Let τ be the total time period of one cycle. Let Cp be a constant and the time average enthalpy is
• Substituting, temperature T into the above equation, we get

H^ = mC T p

0

H C^ p mTdt

τ

τ

 = ∫  T = T 0 +T 1 cos( ω t)

( 0 1 (^ ))

0

p cos

C

H m T T t dt

τ

Phasor Analysis

• From the 1 st^ law, the heat lifted ( Q (^) c ) at the cold end of the PT is the enthalpy flow of the PT.
• This is obtained by substituting m (^) c and Tc in the following equation.

Regenerator Pulse Tube

AC CHX HHX

Q^ AC ,TAC Qc ,Tc Qh ,Th

H^ r Hpt

Q^ c = Hpt

0 0

cos 5

H C^ p m p T t dt p

τ

Phasor Analysis

c^1 pt^ sin^ (^ )^ h 1 1 cos(^ )

c c

p V (^) T m t C p t RT T

ω ω ω γ

(^0 )

cos 5

c p c

T C p H m t dt p

τ

(^0 0 )

sin 2 cos 5 2

c p pt (^) h c c

T C p p V (^) T H t dt C p t dt p RT T

• In the above equation, the first term is zero. This is because, the sine function when integrated over one full cycle vanishes.

Phasor Analysis

• Therefore, the time averaged enthalpy is given as
• We know that the mass flow rate at the hot end ( m (^) h ) is a vector and it is as given below.
• The magnitude of this vector is given by

2 1 1 (^50)

H C C T p^ p^ h p

mh =C p 1 1 cos ( ωt)

mh =C p 1 1

Phasor Analysis

• Using and |m (^) h | equations, we have.

1 1 1 (^50)

H C p C T p^ p^ h p

m^ h =C p 1 1  =^1 (^50)

H m^ h^ C T pp^ h p

θ m^ h^ Pressure h h c

T (^) m T

     



1 pt c

p V RT

ω γ

m c

cos h^ h c c

adj T^ m hyp T m

c c^ cos h h

T m m T

1 0

cos 5

c c^ p^ h h

T m C T p H T p

^ 

1 0

cos 5

p c c

C (^) p H T m p

θ