(10!4!3) NPTEL - Gas Liquefaction and Refrigeration Systems

8/22/2019 (10!4!3) NPTEL - Gas Liquefaction and Refrigeration Systems

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1

1010


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Earlier LectureEarlier Lecture

For a real gas, J T coefficient depends on

TINV.

The isentropic expansion of a gas always results in

h

T

p

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

.

J T expansion is normally used where phasechanges are required, while an isentropic

expansion is used for single phase fluids.

The isentropic expansion coefficient iss

T

p


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Earlier LectureEarlier Lecture The gases like Air, N2, show J T cooling when

expanded at room temperature while He, H2, Neon

are required to be precooled to result in J T cooling.

In a thermodynamic ideal system, all the gas that is


.

Using the ideal thermodynamic cycle, one cancalculate the ideal work requirement for liquefaction

of unit mass of a given gas.

This Ideal Work requirement depends on the initial

condition of the gas (p and T).


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Outline of the LectureOutline of the LectureTopic : Gas Liquefaction and Refrigeration

Systems (contd)

Parameters of Gas Liquefaction systems


Linde Hampson system Liquid yield

Work requirement

Optimization of liquid yield


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IntroductionIntroductionm

1 2 As seen earlier, the schematic of

an Ideal system and its T s

diagram are as shown.

The processes of compression


fm

2frespectively.

The initial condition 1 of the gas

determines the position of point f.The point 2 determines the finalstate of the gas after thecompression process.


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IntroductionIntroductionm

1 2 Lets us take an example of N2 and

the initial condition at point 1 be

ambient (1 bar (0.9869 atm), 300K).


fm

to follow an Ideal cycle is morethan 70000 bar (690846.3 atm).

Such high pressures areimpractical and hence there is aneed to modify the system tolower the maximum pressures.


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IntroductionIntroduction As stated earlier, devices like

heat exchangers, J T valve,

turbo expanders can be used tomodify the systems.


conserve cold and J T devicesare used to achieve lowertemperatures.

The following slides explainvarious cycles that are used forgas liquefaction.

eW

HighPr.

Low

Pr.


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In the refrigeration systems, the Carnot COP isoften used as a benchmark to compare the

performances.

On the similar lines, there is a need to compare

Gas Liquefaction ParametersGas Liquefaction Parameters


.

In liquefaction systems, an ideal cycle is used as abenchmark to compare the performances.

Different ratios and functions are defined to give aqualitative and quantitative information of different

liquefaction systems.


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Performance Parameters

Work/unit mass of gas compressed

Work/unit mass of gas liquefied

1

W

m

f

W

m

fm

m m

mRQ

cW1

2

Gas Liquefaction ParametersGas Liquefaction Parameters


Compressor isothermal efficiencyCompressor mechanical efficiency

,c iso

,c mech

g

fm

Figure of Merit (FOM)

Fraction of total gas liquefied

iW

W

1/fy m m=


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FundamentalsFundamentalsSign Convention

The work done by the system is taken as positive.

The heat transferred to the system is taken aspositive.


Pressure Measurement

Bar or Pascal is the S.I. unit. The conversion tableis as follows.

Pressure1 Pa = 1 N/m2

1 bar = 105 Pa

1 atm = 1.01325 bar


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LindeLinde Hampson SystemHampson System The salient features of this system

are as follows.

Linde Hampson cycle consists ofcompressor, heat exchanger and a

m

mRQ

1 2

cW

Makeup gas


.

Only a part of the gas that iscompressed, gets liquefied.

Being an open cycle, the massdeficit occurring is replenished bya Makeup Gas connection.

3

4g

fm m

fm


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LindeLinde Hampson SystemHampson System All the processes are assumed to

be ideal in nature and there are

no irreversible pressure drops inthe system.

m

mRQ

1 2

cW

Makeup gas


isothermal while the J Texpansion is isenthalpic.

The system incorporates a Two-Fluid heat exchanger which isassumed to be 100% effective.

3

4g

fm m

fm


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LindeLinde Hampson SystemHampson System The heat exchange process is an

isobaric process and it is used to

conserve cold in the system.

That is, the stream of gas (23)

m

mRQ

1 2

cW

Makeup gas


(g1).

The J T expansion device is

used for phase change of gasstream to liquid stream bylowering the temperature.

3

4g

fm m

fm


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LindeLinde Hampson SystemHampson System

T=const

12

m

mRQ

1 2

cW

Makeup gas

14Prof. M D Atrey, Department of Mechanical Engineering, IIT Bombays

h=const3

4gf

3

4g

fm m

fm


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m

mRQ

1 2

cW

Makeup gas Consider a control volume for this

system as shown in the figure.

It encloses the heat exchanger, J T device and the liquid


3

4g

fm m

fm

.

The 1st Law of Thermodynamics isapplied to analyse the system.

The changes in the velocities anddatum levels are assumed to be

negligible.


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m

mRQ

1 2

cW

Makeup gas

IN OUTm @ 2 (m mf) @ 1

The quantities entering andleaving this control volume are as

given below.


3

4g

fm m

fm

Using 1st Law , we get

( )2 1f f fmh m m h m h= +

in out E=


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m

mRQ

1 2

cW

Makeup gas Rearranging the terms, we have

The fraction of gas liquefied or

1 2

1

f

f

m h hm h h

=


3

4g

fm m

fm

liquid yield is defined as

y depends on the initialconditions and the compressionpressure.

1 2

1

f

f

m h hy

m h h

= =


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The values ofh1 and hfare governed by theinitial conditions, which are often ambient.

1 2

1 f

h hy

h h

=


In order to maximize y, the value ofh2 shouldbe as small as possible.

To have a minimum h2, the change in enthalpyfor a given change in pressure should be zero attemperature T1.


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Mathematically,

Using calculus, for variables enthalpy (h),pressure (p) and temperature (T), we have seen

1 2

0

T T

h

p=

=


Substituting the J T coefficient and the Cp for

the second and third terms respectively, we have

1h pT

h p T

p T h =

0JT pC =


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Cp is a positive quantity and hence cannot bezero.

0JT pC =


,

It implies that, in order to maximize y, the state2 should lie on the inversion curve for a

particular gas at the temperature of compressionprocess.

JT


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LindeLinde Hampson SystemHampson System Consider three constant

enthalpy lines on the T P

chart as shown in the figure.

Here, h1> hi > h2.T1


From the figure, it is clearthat (h1 - h2) is maximum,when the point 2 lies on the

inversion curve, so that they is maximum.

Pressure

1 2


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m

mRQ

1 2

cW

Makeup gas The work requirement for a Linde

Hampson system can be

derived by considering a controlvolume enclosing the compressor.


3

4g

fm m

fm

IN OUT

m @ 1 m @ 2-Wc -QR

leaving this control volume are asgiven below.


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m

mRQ

1 2

cW

Makeup gas

IN OUTm @ 1 m @ 2

-Wc -QR

Using 1st Law for the followingtable, we get


3

4g

fm m

fm

Rearranging the terms, we have

in out E=

1 2c Rmh W mh Q =

( )2 1R cQ W m h h =


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m

mRQ

1 2

cW

Makeup gas

The expression for QR can be

obtained by using 2nd Law for anisothermal compression. It isgiven by,

( )2 1R cQ W m h h =


3

4g

fm m

fm

Combining the above equations,the work required for a unit mass

of gas compressed is

( ) ( )1 1 2 1 2cW T s s h h

m =

( )1 2 1RQ mT s s=


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m

mRQ

1 2

cW

Makeup gas

The liquid yield y is given by

1 2fm h h

y

= =

( ) ( )1 1 2 1 2cW T s s h h

m =


3

4g

fm m

fm

Combining the above equationsthe work required for a unit massof gas liquefied is

c

c c

f

WW Wm

ym ym

= =

1 f


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TutorialTutorial 11 Determine the liquid yield, the work per unit

mass compressed and work for unit mass

liquefied for a Linde Hampson cycle with air asworking fluid. The system is operated between1.013 bar (1 atm) and 202.6 bar (200 atm) at300 K. 200 1


Step 1

The T s diagram for

a Linde HampsonCycle is as shown.

T=const

h=const

12

3

4

s

300


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Step 2

The state properties atdifferent points are asgiven below.

TutorialTutorial 11

T=const12

200 1

300


The properties are takenfrom NIST.

1 2 f

p (bar) 1.013 202.6 1.013T (K) 300 300 78.8

h (J/g) 28.47 -8.37 -406

s (J/gK) 0.10 -1.5 -3.9

=cons

4g

s


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Liquid yield

TutorialTutorial 11

1 2

1 f

h hyh h

=

T=const12

200 1

300


p (bar) 1.013 202.6 1.013T (K) 300 300 78.8

h (J/g) 28.47 -8.37 -406s (J/gK) 0.10 -1.5 -3.9

1 2

1 f

h hy

h h

=

28.47 8.37

28.47 406

+ =

+

36.84

434.47

=

0.085=

=cons

4g

s


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Work/unit mass of gascompressed

TutorialTutorial 11

( ) ( )1 1 2 1 2cW T s s h h

m =

T=const12

200 1

300


p (bar) 1.013 202.6 1.013T (K) 300 300 78.8

h (J/g) 28.47 -8.37 -406s (J/gK) 0.10 -1.5 -3.9

( ) ( )300 0.1 1.5 28.47 8.37cW

m = + +

443.16 / g=

=cons

4g

s


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Work/unit mass of gasliquefied

TutorialTutorial 11

443.16cW

m =

0.085y =

T=const12

200 1

300


c c

f

W W

m ym =

443.16

0.085= 5213.64 /

g=

=cons

4g

s


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TutorialTutorial 22 Determine the liquid yield for a Linde Hampson

cycle with Nitrogen as working fluid for thefollowing operating conditions. Comment on theresults.

Point 1 Point 2

T=const12


,

1 bar

,

50 barII 200 K,

1 bar200 K,50 bar

III 300 K,

1 bar

300 K,

100 barIV 200 K,

1 bar200 K,100 bar

h=const3

4g

s


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TutorialTutorial 22Step : 1

Assuming the heatexchange process to100% effective, the T

T=const12


Hampson Cycle is asshown.

In this tutorial , weassume that 1 atm = 1bar.

h=const3

4g

s


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1 2 fp (bar) 1 50 1

TutorialTutorial 22T=const

12

50 1

300

Point 1 Point 2I 300 K,

1 bar

300 K,

50 bar


.

h (J/g) 464 454 35s (J/gK) 4.35 3.25 0.42

1 2

1

I

f

h hyh h

=

464 454

464 35 =

10

429 =

0.023=

=cons

4g

s


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TutorialTutorial 22

1 2 fp (bar) 1 50 1

T=const12

50 1

200

Point 1 Point 2II 200 K,

1 bar

200 K,

50 bar


1 2

1

II

f

h hyh h

=

357 332

357 35 =

25

325 =

0.076=

.

h (J/g) 357 332 35s (J/gK) 4.0 2.8 0.42

=cons

4g

s


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TutorialTutorial 22

1 2 fp (bar) 1 100 1

T=const12

100 1

300

Point 1 Point 2III 300 K,

1 bar

300 K,

100 bar


1 2

1

III

f

h hyh h

=

464 445464 35

=

19429

=

0.044=

.

h (J/g) 464 445 35s (J/gK) 4.35 3.1 0.42

=cons

4g

s


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TutorialTutorial 22

1 2 fp (bar) 1 100 1

T=const12

100 1

200

Point 1 Point 2IV 200 K,

1 bar

200 K,

100 bar


1 2

1

IV

f

h hyh h

=

357 312357 35

=

45322

=

0.14=

.

h (J/g) 357 312 35s (J/gK) 4.0 2.5 0.42

=cons

4g

s


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TutorialTutorial 22Point 1 Point 2 y

I 300 K,

1 bar

300 K,

50 bar

0.023

II 200 K,1 bar

200 K,50 bar

0.076

III 300 K 300 K 0.044


As the compression pressure increases, the liquidyield y increases at a given compressiontemperature.

1 bar 100 barIV 200 K,

1 bar200 K,100 bar

0.14


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TutorialTutorial 22Point 1 Point 2 y

I 300 K,

1 bar

300 K,

50 bar

0.023

II 200 K,1 bar

200 K,50 bar

0.076

III 300 K 300 K 0.044


As the compression temperature decreases, theliquid yield y increases at a given compressionpressure.

1 bar 100 barIV 200 K,

1 bar200 K,100 bar

0.14


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Liquid Yield : NitrogenLiquid Yield : Nitrogen

0.15

0.3

0.2

0.25

y

Nitrogen

Point 2 y

Plotting the values forthe followingconditions, we have


0.05

T2 K10 0 150 200 250 300 350

0

0.1

, .

II 200 K, 50 bar 0.076

III 300 K, 100 bar 0.044

IV 200 K, 100 bar 0.14


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Liquid Yield : NitrogenLiquid Yield : Nitrogen

0.15

0.3

0.2

0.25

y

Nitrogen Summarizing, we have

As the compressionpressure increases, theliquid yield y increases


0.05

T2 K10 0 150 200 250 300 350

0

0.1

temperature.

As the compression

temperature decreases,the liquid yield yincreases at a givencompression pressure.


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AssignmentAssignment1. Consider a Linde Hampson cycle with Nitrogen

as working fluid. The system is operated between1.013 bar (1 atm) and 101.3 bar (100 atm) at300 K. Determine

1. Li uid ield

2. Work per unit mass compressed3. Work for unit mass liquefied.

2. Calculate the above parameters if the same cycle

is operated with air as working fluid.



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3. Determine the liquid yield for a Linde Hampson cycle with Nitrogen as working fluid forthe following operating conditions. Comment onthe results.

Point 1 Point 2I 250 K,

250 K,

AssignmentAssignment


Verify your answers from the yield versustemperature chart for Nitrogen.

1 bar 50 barII 300 K,

1 bar300 K,200 bar


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SummarySummary The Ideal cycle demands very high pressure

which is impractical and hence modified cyclesare proposed to lower the maximum pressure.

An ideal cycle is used as a benchmark to compare


systems.

In a Linde Hampson system, only a part of the

gas that is compressed, gets liquefied.


44/46

SummarySummary A heat exchanger is used in a Linde Hampson

system to conserve cold in the system. Thisprocess is an isobaric process and is assumed to

be 100% effective.


system, the state 2 should lie on the inversioncurve at the temperature of compression process.

The work required for a unit mass of gascompressed for a Linde Hampson system is

( ) ( )1 1 2 1 2cW T s s h h

m =


45/46

SummarySummary For a Linde Hampson system following hold

true.

As the compression pressure increases, theliquid yield y increases at a given compression


.

As the compression temperature decreases,the liquid yield y increases at a given

compression pressure.


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(10!4!3) NPTEL - Gas Liquefaction and Refrigeration Systems