119

CHAPTERS

EXERGY ANALYSIS OF AN ORIFICE PULSE TUBE

REFRIGERATOR

5.1 Introduction

Exergy or the available work energy function is a useful concept to analyse

thennodynamic cycles. By definition, for a heat engine, exergy of a state 1 is the

maximum work that would be obtained in bringing the system to the dead state (normally

ambient). For a refrigerator, exergy is the minimum work required to take state 1 to 'dead

state'.

Energy (first law) analysis is still the most commonly used method in the analysis

of thermal systems. The first law is concerned only with the conservation of energy, and

it gives no information on how, where, and how much the system performance is

degraded. Exergy analysis is a powerful tool in the design, optimization and performance

evaluation of energy systems[1,2,31. The principles and methodologies of exergy analysis

are well established. Exergy analysis is usually aimed to determine the maximum

performance of the system and identify the sites of exergy destruction. Identifying the

main sites of exergy destruction shows the direction for potential improvements. The

requirement is to obtain quantitative information that will lead to a better understanding

of the process irreversibility and their distribution among the components of a thermal

system. ~y determining the sources and magnitudes of the exergy losses, and minimizing

them, an optimal refrigeration cycle emerges.

For systems above ambient temperature, higher the temperature at which quantity

of energy is released, more valuable is that quantity of energy. Conversely, for cryogenic

systems, lower the temperature at which a quantity of energy is absorbed, greater the total

degradation incurred and thus the greater the financial cost of that energy. Thus, it has

become desirable to characterize energy in terms of quality, and so the concept of exergy

120

has been introduced. The exergy of a system is defined as maximum available power

from that system. In general, eXergy, remains invariant during reversible process, where

as it is reduced as a result of irreversible phenomenon.

5.2 Exergy transfer by heat, work, and mass

Exergy, like energy, can be transferred to or from a system in three forms: heat,

work and mass flow. Exergy transfer is recognized at the system boundary as energy

crosses it, and it represents the exergy gained or lost by a system during a process. Heat

transfer and work are the only two forms of exergy interactions associated with a fixed

mass or closed system.

5.2.1 Exergy of heat

It is known that, for an engine when the heat supplied is Q, the availability or

exergy of this amount of heat varies as the temperature at which the heat supplied varies.

Thus,

(5.1)

In the case of a refrigerator, the minimum work required or exergy to remove the

given amount of heat increases as the temperature decreases. In this case the exergy given

is negative.

5.2.2 Exergy flow of mass flux

This is the useful work due to the energy content of the mass flux.

It can be shown that, for a flow process between the state point I and ambient 0 the useful

work is given by,

(5.2)

121

5.2.3 Exergy of work

Exergy of work is equal to that work itself, since there is no thennodynamic restriction in

its availability.

5.2.4 Exergy balance equation

For any system working between state points 1& 2 or for any specific component,

the exergy balance equation can be written as,

(5.3)

where, ef is the exergy of flow ofmass, eq exergy ofheat and !1e exergy loss[4] .

5.3. Model development

Enthalpy, entropy and exergy are good tools for understanding the

thennodynamics of a system. Here, thennodynamic analysis ofpulse tube refrigerator has

been perfonned, by analyzing these quantities. Through the exergy analysis, exergy losses

of each component of the orifice pulse tube refrigerator can be more effectively

understood. The exergy analysis of the orifice pulse tube refrigerator provides the

infonnation about how the input exergy provided by the power input to the compressor is

utilized, as the working fluid goes through each component of the pulse tube refrigerator.

For calculating the exergy losses in each component of the pulse tube refrigerator,

exergy balance equations are developed for all components.

122

5.3.1 Compressor

Figure 5.1 Control volume for compressor

For low-pressure ratios and good cooling arrangement, isothermal compression

can be assumed [4]. Helium gas is compressed isothermally at ambient temperature. So,

there is no temperature oscillation and hence there is no enthalpy flow through the

compressor. It is not reasonable to use an adiabatic transfer line with the isothermal

compressor, as this defeats any advantage of the isothermal compression by having the

losses associated with the isothermal to adiabatic transitions. An isothermal transition is

difficult to implement without the significant viscous losses and is not considered here.

Thus an integral compressor, which incorporates an after cooler, is considered. Hence, for

the analysis purpose, here the compressor and after cooler are considered as one unit [5].

By exergy balance equation (5.3), for any component of a system

For isothermal compression at ambient temperature To, w= ef2 -efl

Since, an amount of heat q IS evolved during compression, its exergy eq = 0, the

compression being at To.

Since, the process is assumed to be reversible, there is no loss, i.e. l:1e =0

Hence, W =e2 - el (5.4)

123

The exergy supplied is equal to the compressor work. If the compressor is 100%

effective there will be no losses inside the compressor. In a real compressor, there will be

some losses in the compressor due to its inefficiency. In practical cases, we can assume

that the isothermal efficiency of the compressor to be 0.6.

So, Actual work required = exergy supplied / l1iso

Loss of exergy in the compressor, ~ecp = actual work supplied - ideal work supplied

5.3.2 Valve model

Figure 5.2 shows the valve model, which can be used for rotary valve and orifice.

For throttling process, due to high speed and short duration, the heat exchanged with the

environment is very small. Hence the process can be treated as an adiabatic process [6].

The exergy loss of control volume can be derived as follows:

1

r---------..,I II I

III IL J

2

Figure 5.2 Control volumes for rotary and orifice valve

Since, there is no heat transfer, eql = eq2 = 0;

The exergy change is due to mass flux only.

Since, no work interaction, w = 0,

The exergy loss, ~e = e2 - e j(5.5)

124

5.3.3 Regenerator and pulse tube

Figure 5.3 shows the model of a regenerator and pulse tube. For the process 1-2, the

exergy loss of the control volume can be derived as follows:

Exergy loss = (exergy lost by low pressure stream - exergy gained by

high pressure stream)

4

1

3

2

Figure 5.3 Control volume for regenerator and pulse tube

Since, no heat transfer and work done, eq =0 and w =O.

By exergy balance equation, _ + + Ae\ + e3 - e2 e4 ue

5.3.4 Cold and hot end heat exchangers

(5.6)

(5.7)

Let 8q be the amount of heat transfer with the heat exchanger. Exergy transfer

due to the heat transfer is given by, eq~oqx(T:' ) (5.8)

125

oq

4..

~. 3· :+--

····1 · 2--.;

Figure 5.4 Control volumes for the cold and hot end heat exchangers

Since, no work transfer, WI = W2 = 0

The equation reduces to, + + _ + + Ae, e3 eq - e2 e4 tie

5.4 Exergy calculations

(5.9)

(5.10)

Working of the OPTR is already explained in chapter 2. The diagram showing the

temperature history of a single stage orifice pulse tube refrigerator is shown in figure 5.5.

During the compression part of the cycle, high-pressure gas at a pressure PH and ambient

temperature To is cooled to the cold end temperature of the regenerator. The cold gas is

admitted to the pulse tube, which already contains the gas at low pressure PL due to the

previous cycle. Gas in the pulse tube gets adiabatically compressed and its temperature

rises. The hot gas is cooled to the ambient temperature in the HHX and expands to the

average pressure in the reservoir as it flows through the orifice. During the return flow the

gas expands adiabatically and gets cooled to a temperature lower than the CHX

temperature and provides the refrigeration load and while going out through the

regenerator gets warmed back. If the regenerator is 100% efficient, the gas will get

warmed back to the temperature at which it entered the regenerator.

126

- Forward flow- - - Return flow

L- 7,9 8Q)0>m c.r:t> ax +-'Q) cu

L-+-' Q)m 0>Q)

L-

~.r:0"'0 C L-

C Q) +-'Q) t> Q) Q)

+-' l.i= rJ) Z0 ·C Q)

I 0 c::

t

610

2

L- Q)L-

a a> ......en coen co L-Q) > Q)L-

~C0- Q)

E co 0>a ...... Q)a() c:: c::

60 K 4_________ .1

54 K ----------

308 K

300 K 1

Figure 5.5 Temperature history of a single stage orifice pulse tube refrigerator

In an OPTR, input power to the compressor provides the source of exergy and the

input exergy's magnitude diminishes as the working fluid passes through each component

of the pulse tube. The exergy destruction is mainly due to the irreversibility of each

component in the pulse tube refrigerator. In an OPTR, the only useful product is exergy

associated with the refrigerating power.

Exergy values of state points with reference to figure 5.5 are taken from the

Enthalpy -Exergy chart [4] shown in figure 5.6 and are tabulated in Table 5.1.

127

Figure 5.6 Exergy - enthalpy chart for helium

Table 5.1 Exergy values at various state points (To = 300 K)

State Exergy (callg) Enthalpy (cal/g) Entropy (calg-) K-))

1 320 375 6.5

2 425 377 6.2

3 422 376 6.3

4 620 105 5.1

5 620 105 5.1

6 421 395 6.4

7 422 377 6.0

8 385 378 6.2

9 360 375 6.5

10 350 375 6.5

11 540 99 5.3

12 520 103 5.4

13 310 375 6.5

5.4.1 Exergy supplied

Exergy supplied is equal to the compressor work.

Isothennal compressor work = e2 - e)= 425 - 320 =105 callg = 439.53 J/g.

For an isothennal efficiency of 0.6,

. d 439.53 7325 J/Actual compressor work reqUIre = =. g.0.6

128

129

5.4.2 Exergy of refrigeration produced

Refrigeration produced qc = (h t2 - hIt) = (103 - 99) callg == 4 callg

f fri o (T-T ) (60-300)Exergy 0 re geratlOn = eq = qc x ~ = 4x 60 = - 16 cal/g = - 66.97 Jig

This is the exergy utilized. The negative sign indicates that, work is to be supplied to

produce refrigeration.

5.4.3 Exergy losses

a) Loss in the compressor

~ecp = actual work - ideal work = (732.5 - 439.53) = 292.97 Jig

b) Loss in rotary valve

~eRV = e2 - e3 = (425 - 422) callg =3 callg = 12.56 Jig

c) Loss in regenerator

~erg = (e3

+ e12

) - (e4+ en) = [(422+520) - (620+310)] callg = 17 callg = 71.16 Jig

d) Loss in CHX

~eCHX = [(e4+ell) - (es+en) + eq ] = [(620+540) - (620+520) -16] = 4callg = 16.744J/g

e) Loss in pulse tube

~e == (e + e ) - (e + e ) = [(620+350) - (421 +540)] callg = 37.67 JigpI s 10 6 II

130

f) Loss in HHX

Heat removed in the HHX = h6-h7 = 395 - 377 = 18 callg = 75.35 Jig

(308-300)eq = 18 x = 0.48cal/g = 2 Jig

308

l1eHHX = (e6 +e9 )-(e7+elO) + eqh = ( 421+360) - ( 422 +350) + 0.48) cal/g = 9.48 callg

= 39.69 Jig

g) Loss in orifice valve

l1eOR =e7 -eg = (422 -385) = 37callg = 154.88 Jig

Exergy balance for a single stage orifice pulse tube refrigerator is

shown in Table 5.2 and the corresponding exergy flow diagram in figure 5.7. In an orifice

pulse tube refrigerator, exergy is provided by input power to the compressor and leaves

the CHX as the product of the system transferred to the cold reservoir. In a real OPTR,

exergy loss takes place in all the components of the system. The most significant are

exergy destruction in regenerator, compressor, and orific [8]. Exergy always depends upon

the pressure and temperature gradient. The process becomes irreversible if it occurs

through a finite temperature and pressure difference or if there is a dissipative effect like

friction, viscosity, fluid mixing etc.

The exergy loss in the compressor is about 40%. In the compressor, helium gas is

compressed isothermally. The irreversibility of the operation of the compressor results

from the internal irreversibility of compression, mechanical losses within the compressor

and electromechanical losses in the driving electric motor [9]. Shaft power wastage in

compressor can be attributed to fluid throttling, irreversible fluid mixing, friction and

irreversible heat transfer. All such defects can be described as exergy destruction rates.

131

The location and mechanism of the defects can be identified. For instance, the

exergy destruction at the suction valve can be compared with the exergy destruction

at the discharge valve, or with that due to heat transfer between the compressor and the

surroundings. Even leakage can be characterized in terms of exergy destruction, as it

involves throttling and irreversible fluid mixing [9].

Table 5. 2 Exergy Balance for Single Stage Orifice Pulse Tube Refrigerator

JigPercentage

Parameterofexergy supplied

1. Exergy supplied 732.5

2. Exergy utilized 66.97 9.14

3. Exergy losses

a) Compressor 292.97 40

b) Rotary valve 12.56 1.71

c) Regenerator 71.16 9.71

d) Cold end heat exchanger 16.74 2.2

e) Pulse tube 37.67 5.1

'f) Hot end heat exchanger 39.69 5.4

g) Orifice 154.88 21.14

h) Unaccounted 39.86 5.4

The total instantaneous exergy destruction rate in a compressor can be written in

the form [9],

(5.11)

132

Exergysupplied

100Compressor

Pulse tuhe 40

5.1

HHX Rotary valve

5.4 1.71

Regenerator

Orifice valve 9.71

21.14CHX

2.2

Unaccounted

5.4Exergy utilized

9.14

Figure 5.7 Exergy flow diagram ofa single stage OPTR

The most significant part of the exergy destruction is due to the throttling in

suction and discharge valves and the next significant part is due to the internal

convection. The performance of the compressor could have been improved significantly

133

by increasing the areas of the suction and discharge valve ports. This could have been

done as these ports occupied only a small part of the area of the cylinder head. The steps

necessary to optimize the exergy destruction due to internal convection are less obvious

and would require detailed consideration [9]. The exergy destruction by this mechanism

can be reduced by providing an insulated cylinder liner with a very low heat capacity and

to make modifications to reduce the internal convective heat transfer coefficient, by

reducing the swirl and turbulence with in the cylinder. However these modifications can

also increase the work of compression.

Exergy loss in a regenerator is mainly due to pressure drop in the regenerator,

thermal loss due to ineffectiveness of the regenerator, conduction heat transfer through the

finite temperature difference and fluid friction. The physical exergy at the regenerator exit

is reduced not only by the temperature difference at the exit related to its effectiveness,

but also due to conduction heat transfer in the regenerator and its wall. This factor

together with the pressure drop due to fluid flow in the regenerator is responsible for the

high level of exergy destruction. Therefore optimization of the regenerator is an

essential aspect of the design of PTRs [10].

The overall goal of an OPTR is to deliver the thermal exergy at the CHX by

cooling the load. Exergy transfer due to the heat transfer to the CHX is given by equation

(5.8). The above equation represents the heat transfer to CHX from the load. The exergy

utilized from the CHX is 9.14% of exergy supplied. The exergy transfer to the cold end

heat exchanger is the only product of the OPTR.

The main reasons for exergy destruction in the pulse tube are not very well

understood [8]. The losses in the pulse tube are mainly due to mixing of hot and cold fluid,

temperature distribution at both ends of the pulse tube, fluid friction and conduction heat

transfer.

Since, the hot end heat exchanger has low temperature gradient with no pressure

drop; the exergy loss in the HHX is small. In most applications, the irreversibility at the

cold and hot heat exchangers is small [10].

134

The orifice is used as a flow resistance, and flow through a resistance is an

irreversible process. In the orifice, since the gas expands at high speed and short time, the

heat exchanged with the environment is very small. Hence the process can be treated as

adiabatic process. Major portion of the exergy loss in the orifice is due to the throttling

effect and fluid friction [11]. Orifice has higher exergy loss similar to the compressor. It is

about 22%.

5.5 Exergy, entropy and enthalpy flows in OPTR

Exergy, entropy, and enthalpy flows are studied from the exergy-enthalpy chart

and are shown in figures 5.8 - 5.1 O.

5.5.1 Exergy flow in various components of OPTR

Flow of exergy in different components of a pulse tube refrigerator is shown in the

figure 5.8. Since the compressor in an OPTR is isothermal, the heat of compression is

assumed to be transferred to the environment temperature. For the combined control

volume of the compressor and aftercooler with no exergy destruction; the input work to

the compressor is equal to the input exergy to the regenerator. As the temperature of gas

in compressor increases at the compressor outlet, resulting in thermal exergy input to

aftercooler. This exergy is lost in the aftercooler. Exergy loss in the aftercooler is due to

the fluid friction and heat transfer through finite temperature difference [8]. Due to the

flow resistance of the rotary valve, there is a small loss in the exergy. In the regenerator it

is interesting to note that the exergy of the working fluid increases as it moves forward in

the regenerator. This is due to the heat transfer out of the working fluid in the forward

direction, resulting in input exergy to the fluid. The source of the exergy is the

regenerator matrix. The slope of the line in a regenerator is lower mainly due to the fluid

friction and heat transfer irreversibility [8].

The exergy of the working fluid decreases substantially in the backward direction in

the regenerator. Not only the working fluid transfer exergy back to the matrix, but also

there is irreversibility due to fluid friction and heat transfer. This results in a greater deal

of exergy destruction in the regenerator of an OPTR in a cycle [8].

135

2600 --ForwardBackward

2400

-- 2200C>- A \-,..- 2000 I \

>- I \

e> I \Q) 1800 I \><

UJ I \

I \

1600 I \

I \

I \

1400 I \

"\

- - - 1- - - -I

1200

COMP RV REG CHX PT HHX ORV

Figure 5.8 Exergy flow for forward and backward flow

5.5.2 Enthalpy flow in OPTR

Above 30 K, helium gas in the pulse tube refrigerators, can be considered as an

ideal gas[l21. Then enthalpy of an ideal gas at a given state depends upon the temperature

at that particular state as given by the following equation,

(5.14)

The nature of variation of enthalpy of an orifice pulse tube refrigerator for forward

and backward flow has been shown in figure 5.9. In an isothermal compressor, there are

no temperature oscillations and therefore no enthalpy flow. An isothermal transfer line is

difficult to implement without significant viscous losses and is not considered here.

Hence the compressor and aftercooler are considered as a single unit and there is no

enthalpy flow across the compressor.

136

The enthalpy flow in the regenerator originates from heat conduction via matrix

and gas, heat transfer between matrix and gas and from gas flow [12]. In a regenerator the

gas gets cooled from the Th to Tc as it flows from the hot end to the cold end, hence the

enthalpy decreases significantly. In the subsystem containing pulse tube and its two heat

exchangers, the temperature of the gas in the tube is high, when it moves to the right and

low, when it moves to the left. The enthalpy of the working fluid inside the pulse tube

increases due to the compression of the gas.

1800--Forward

1600 - - - Backward

1400..-..C)--, 1200'-'"

>-a.ro 1000..c:+Jc:W

800

600

400

COMP RV REG CHX PT HHX ORV

Figure 5.9 Enthalpy flow for forward and backward flow

Thus enthalpy flows from the cold end heat exchanger to the hot end heat

exchanger. Enthalpy decreases in the HHX due to cooling of the gas to the ambient

temperature. Since, the orifice is at room temperature, helium flowing through the orifice

can be considered as an ideal gas. Hence there is no heat exchange between the gas and

the environment so, there is no enthalpy difference across the orifice valve.

137

5.5.3 Entropy flow in OPTR.

The variation of the entropy of an orifice pulse tube refrigerator has been obtained

and is shown in figure 5.10. The flow of gas over a pressure difference in the rotary valve

creates entropy. During first half of the cycle a flow of gas enters PTR from the

compressor at high pressure and during the other half of the cycle gas flows at a low

pressure to the compressor.

The entropy flow in the regenerator is directed from the cold to the warm end of

the regenerator, since, the gas moves to the right, when the pressure is high (low entropy)

and to the left, when the pressure is low (entropy high). The total entropy production rate

in the pulse tube consists of; entropy production due to the heat exchange of the gas with

the tube walls, entropy production due to the heat conductivity in the gas with the tube

walls and entropy production due to turbulence of gas [12]. The entropy production due to

turbulence occurs due to cross sectional area changes (e.g. from the connecting tube into

the pulse tube) or large Reynolds numbers of the gas flow. As a result, the cold end and

hot end gases get mixed and entropy is produced.

The entropy flow through the right border of the control volume of the orifice

towards the reservoir is zero. The reason is that both pressure and temperature are

constant in the reservoir. But the entropy flow through the left border is not zero. This is

because; the gas is flowing to the right with high pressure (low entropy) and to the left

with low pressure (high entropy). So there is a net entropy flow from the orifice to HHX

[12]. In the HHX the entropy decreases due to cooling of the gas to the temperature of the

environment.

138

18 ...L--_--'--_--J-__----'-__'--__'-------''------'

COMP RV REG CHX PT HHX ORV

Figure 5.10 Entropy variations for forward and backward flow

5.6 Exergetic efficiency of OPTR

The exergetic efficiency of an OPTR is defined based upon the input exergy and

exergy of the product of the system.

The exergetic efficiency of OPTR,

Exergy of cold produced

17e = Exergy supplied(5.15)

The exergetic efficiency of OPTR has been determined for different cold end

temperatures of the pulse tube and is shown in figure 5.11. From the figure it is clear that,

as the cold end temperature increases the exergetic efficiency of the orifice pulse tube

refrigerator increases. This is due to the fact that, the exergy utilized increases as the cold

end temperature increases and hence the exergetic efficiency.

139

12070 80 90 100 110

Cold end temperature (K)

o.0 +----r-~-.____r_-_r____r-_r_____r--r--___r--r--__f

60

0.7

0.8

>- 0.6(,)c.~ 0.5(,)

~ 0.4(,)

:,j:j

& 0.3L..

~w 0.2

0.1

Figure 5.11 Variation of exergetic efficiency of an OPTR

The exergy analysis of a single stage OPTR has been perfonned and the exergetic

efficiency at different cold end temperatures is calculated. The exergy analysis is

conducted by using the exergy- enthalpy (e-h) chart. Exergy losses can be directly read

off as ordinates in the e-h chart, whereas in other commonly used charts like P-V or T-S

charts, work and heat have to be read as areas. The exergy analysis provides quantitative

measures of deviations from ideality. Moreover, the exergy losses allow the location, type

and cause of a loss to be clearly identified. This infonnation is critical for efforts to

increase exergetic efficiency.

140

5.7 References

1. Gaggioli R A, Available energy and exergy. International J. Applied

Thennodynamics, 1 - 4, 1998.

2. Bejan A, Advanced Engineering Thennodynamics. Wiley, New York, 1988

3. Bejan A, Entropy generation through heat and fluid flow. Wiley, New York, 1982

4. Thirumaleshwar M, Exergy method of analyses and its application to a Helium

cryorefrigerator, Cryogenics 19, 1979.

5. Kittel P, Enthalpy, entropy, and exergy flows in ideal pulse tube cryocoolers.

Cryocoo1ers 13, 333-341, 2004.

6. Ya Ling He, Jiang Huang, Chung Feng Zhao, and Ying Wen Liu, First and second

law analysis of pulse tube refrigerator. Applied Thermal Engineering, 26, 2301­

2307,2006.

7. Thirumaleshwar M, Design, fabrication and performance of Gifford-McMahon

Cryorefrigerator operating at 20 K. Ph D Thesis, Indian Institute of Science,

Bangalore, India, 1985.

8. Razani A, Flake Band Yarbrough S, Exergy flow in pulse tube refrigerators and

their perfonnance evaluation based on exergy analysis. Adv. in Cryog. Engg., 49

1508-1518,2004

9. J A Mcgovern and S Harte, An exergy method for compressor perfonnance

analysis. IntI. J. Refrig, 18,421-433,2002.

10. Razani A, T. Roberts, B Flake, A thennodynamic model based on exergy flow for

analysis and optimization of pulse tube refrigerators. Cryogenics 47, 166-173,

2007.

11. Ya-Ling He, Jing Huang, Chung'-Feng Zhao, Ying-Wen Liu, First and second law

analysis of pulse tube refrigerator. Applied Thennal Engg. 26, 2301-2307, 2006.

12. Peter Paul Steijaert, Thenno dynamical aspects of pulse tube refrigerators. Ph D

Thesis, Eindhoven University ofTechnology, Netherlands.

13. Razani A, Flake B and Yarbrough S, Exergy flow in pulse tube refrigerators and

their perfonnance evaluation based on exergy analysis. Adv. in Cryog. Engg. 49

1508-1518, 2004.