report on r22

Thermodynamic properties of R22

REPORT

THERMODYNAMIC PROPERTY TABLES OF REFRIGERANT R22

Submitted to

DR. T.P. ASHOK BABUProfessor,

Department of Mechanical EngineeringNational Institute of Technology, Karnataka

Surathkal.

Submitted by,

Nagendra 10TH09F

1st Sem, M.Tech

Thermal Engineering

Department of Mechanical Engineering ( M.Tech)


ABSTRACT:

The report consists of the development of thermodynamic properties of refrigerant 22.

The equations for the development of the refrigerant properties are taken from the technical

paper and ASHRAE data hand book. Critical temperature and critical pressure are taken from

the ASHRAE data hand book. Manual calculations are done and the program is developed

using Microsoft Excel. The enthalpy of saturated liquid at 00c is taken from the ASHRAE

data hand book and is taken as the reference. The graphs of (Psat v/s Vg), (Psat v/s hg),

(Psat v/s Tsat), (Psat v/s Sg) are plotted using Microsoft excel. The graphs for the calculated

values is then compared with the graphs taken from the ASHRAE data hand book.



CONTENTS:

Abstract………………………………………………………………………. 2

1. Introduction…………………………………………………………............... 3

2. General properties of Refrigerant R22...……………………………….......... 4

3. Notations……….............................................................................................. 5

4. Experimental data /correlation required to establish thermodynamic

properties of Refrigerant ……………………………………...…..................... 6

5. Procedure

5.1 To calculate saturation pressure.................................................................... 7

5.2 To caluculate dpsatdt

...................................................................................... 7

5.3 To calculate specific volume of saturated vapour...................................... 8

5.4 Enthalpy of Vapourisation............................................................................ 9

5.5 To calculate enthalpy departure term............................................................ 9

5.6 To calculate h4.............................................................................................10

5.7 To calculate enthalpy departure temperature...............................................11

5.8 To calculate h6.............................................................................................12

5.9 To calculate entropy....................................................................................12

6. Calculations…………………………………………………………...................13-18

7. Thermodynamic table for R22 …………………………………............................ 19

8. Graphs………………………………………………………………….................20-23

9. References………………………………………………………...…......................24



1. INTRODUCTION

A refrigerant is a substance used in a heat cycle usually including, for enhanced

efficiency, a reversible phase change from a gas to a liquid. Traditionally, fluorocarbons,

especially chlorofluorocarbons were used as refrigerants, but they are being phased out

because of their ozone depletion effects. Other common refrigerants used in various

applications are ammonia sulphur dioxide, and non-halogenated hydrocarbons such as

methane.

There are different types of refrigerants namely

In-Organic Refrigerants Designation

a)Water (H20) R 718

b) Ammonia (NH 3) R717

c) Carbon Dioxide ( CO₂) R744

d) Nitrous Oxide (N2O) R764

.Organic Refrigerants

a) Monofluoro-Tetrachloro Methane(CFCl3¿ R11

b) Difluoro-Dichloro Methane (CF₂Cl₂) R12

c) Difluoro-Monochloro Methane (CHF₂Cl) R22

Refrigerant22(CHClF₂) is commonly used in reciprocating compressors in window

type air conditioners and large units such as package units and central air conditioning

plants.It is also used for low temperature refrigeration applications, cold storages,food

freezing and storage etc, with reciprocating and often with centrifugal compressors in very

large capacity air conditioning plant



2. GENERAL PROPERTIES OF REFRIGERANT 22

From ASHRAE table the following data was obtained

TABLE 1: General properties of Refrigerant 22.

Designation R22

Chemical name Difluro-Monochloro Methane

Chemcial formula CHCLF2

Molecular Weight 86.48 Kg/mole

Normal Boiling point -40.8 0C

Critical Temparature 369.14 K

Critical Pressure 4990 Kpa



3. NOTATIONS

Psat Saturation pressure

Tsat Saturation temperature

ρ density

vf specific volume of liquid

vg specific volume of vapour

Cf specific heat of liquid

Cpo constant pressure specific heat at zero pressure

Cvo constant volume specific heat at zero pressure

R gas constant

Ru Universal gas constant (8314.3 J/kmol K)

Pc Critical pressure

Tc Critical Temperature

H enthalpy

U internal energy

S entropy

ω accentric factor

T r reduced temperature

M Molecular weight



4. EXPERIMENTAL DATA / CORRELATIONS REQUIRED TO ESTABLISH

THERMODYNAMIC PROPERTIES OF REFRIGERANTS

The following minimum experimental data / correlations are required to establish the Thermodynamic properties of Refrigerants.

Psat versus Tsat or Tsat versus Psat. P-v-T data or equation of state for gaseous phase. Liquid density ρ or specific volume vf. Liquid specific heat Cf. Zero-pressure constant pressure specific heat Cpo or constant volume specific

heat Cvo (=Cpo – R) of the gaseous phase.

P-H diagram



5. PROCEDURE

5.1 To find Psat

The correlation between Psat and Tsat is given by

ln (PsatPc )=( 11−x ) [Ax+B x1.5+C x3+D x6 ] → (1)

Where x=1− TTc

A = -6.99913B = 1.23014C = -2.49377D = -2.21052

5.2 dPsatdT

is obtained by differentiating equation (1)

dPsatdT

= ddT [ ln( PsatPc )]= d

dT [( 11−x )[ Ax+B x1.5+C x3+Dx6 ] ]

dPsatdT

=−Psat [ ( 11−x ) ( A+1.5 Bx0.5+3C x2+6 D x5 )× 1

Tc

+¿+( Ax+B x1.5+C x3+Dx6 )× 1

(1−x )2×

1Tc ] → (2)

5.3 To find Vg – specific volume of saturated vapour

The Soave modified Redlich – Kwong equation of state is given by

P= RTV g−b

− aαV g (V g+b ) → (3)



Where

a=0.42748R2T C

2

PC

→ 3(a)

b=0.08664 RTC

PC

→ 3(b)

α=[1+ (0.48508+1.55171ω−0.15613ω2 ) (1−√T r ) ]

2 → 3(c)

Where

ω=¿ accentric factor (which is taken as zero)

T r=TT C

Ru= 8314.3 J/kmol K, R=8314/86.48=96.14 J/KgK

Rearranging the terms

(Vg−b ) (Vg+b )Vg= RTP

Vg (Vg+b )−aαP

(Vg−b )

Vg3−b2Vg=RTP

Vg (Vg+b )−aαP

(Vg−b )

Vg3+(−RTP )Vg2+( aαP − RTb

P−b2)Vg+(−aαb

P )=0 → 3(d)

Solving by taking roots, V g is obtained

Since Vf is very less it can be neglected. Therefore Vfg = Vg

5.4 To find h fg – Enthalpy of vapourisation

hfg is obtained by using Clasius – Clapeyron equation which is written as

dPsatdT

=h fg

T ×V fg

→ (5)



5.5 To find the Enthalpy Departure Term

The Enthalpy departure term is h3-h2

h3−h2=(U 3−U 2 )+(P3V 3−P2V 2 ) → (6)

Where,

U 3−U 2=∫2

3

[T ( ∂P∂T )V

−P ]dV → (7)

Using equation (3)

P= RTV g−b

− aαV g (V g+b )

Where , ∝=(1+m−m√T r )2

m=0.48508

( ∂ P∂T )V

=R

V−b+am (1+m−m√T r )V (V +b )T C √T r

T ( ∂P∂T )V

−P=RT

V−b+amT (1+m−m√T r )V (V +b )TC √Tr

−RTV−b

+aα

V (V +b )

Therefore , using equation (6)

U 3−U 2=∫2

3 [ a (m√T rα+α )V (V +b ) ]dv

U 3−U 2=∫2

3 a (m√T rα+α )b [ 1

V−

1(V +b ) ]dv

U 3−U 2=a (m√T rα+α )

bln ( V 3

V 3+b×V 2+bV 2

) → (8)

Substituting equation (8) in equation (6) i.e.,

h3−h2=(U 3−U 2 )+(P3V 3−P2V 2 )



h3−h2=a (m√Trα+α )

bln( V 3

V 3+b×V 2+bV 2

)+[ RT3

V 3−b−

aαV 3 (V 3+b ) ]V 3−[ RT 2

V 2−b−

aαV 2 (V 2+b ) ]V 2

→(9)

Substituting V 3=∞at P3=0, we obtain


bln(V 2+b

V 2)+RT 3−

RT 2V 2

V 2−b+

aαV 2

V 2 (V 2+b )

→ (10)

Equation (10) is the expression for enthalpy departure h3−h2 . From this equation the value of h3 is found.

5.6 To find h4 :

h4−h3=∫3

4

CP0dT → (11)

Where CP0=A+BT +CT 2+DT 3

a=8.675 E−1

b=23.96 E−1

c=−1.457 E−4

d=3.394 E−8

Therefore,

h4−h3=∫3

4

( A+BT+CT 2+DT3 )dT

h4−h3=A T 4+BT 4

2

2+CT 4

3

3+DT 4

4

4−[AT 3+

BT 32

2+CT3

3

3+DT 3

4

4 ] → (12)

From equation (12) the value of h4 is found.

5.7 To find the Enthalpy Departure Term

The Enthalpy departure term h5-h4 is found using equation (9)



h5−h4=a (m√T rα+α )

bln( V 5

V 5+b×V 4+bV 4

)+[ RT 5

V 5−b−

aαV 5 (V 5+b ) ]V 5−[ RT4

V 4−b−

aαV 4 (V 4+b ) ]V 4

→ (13)

Substituting V 4=∞at P4=0, we obtain


bln( V 5

V 5+b )+RT 5V 5

V 5−b−

aαV 5

V 5 (V 5+b )−RT 4 → (14)

From equation (14) the value of h5 is found.

5.8 To find h6 :

h5−h6=hfg → (15)

Where h fg is found by using Clasius-Clayperon equation at a given temperature. Therefore

h6=h5−hfg → (16)

5.9 To find entropy :

Entropy is found by using the following equations :

S fg=hfg

T sat

Sg=hg

T sat

6. SAMPLE CALCULATION :

The sample calculation is done for -40 0C.

To start with -40 0C =233.15 K

The value of Psat is obtained by using equation (1)

ln (PsatPc )=( 11−x ) [Ax+B x1.5+C x3+D x6 ]

Where x=1− TTc

A = -6.99913B = 1.23014C = -2.49377



D = -2.21052

SubstitutingT=233.15 KTc=96.145+273=369.14 K

x=1− TTc

=x=1− 233369.14

=0.34022

Pc=4990×103N /m2

ln ( Psat

4990×103 )=( 11−0.34022 )[ (−6.99913 ) ( 0.34022 )+ (1.23014 ) 0.34022

1.5+(−2.49377 )0.340223+(−2.21052 ) 0.34022

6 ]

Psat =105861.5 pa

To find dPsatdT

: Using equation (2)

dPsatdT

=Psat [( 11−x ) ( A+1.5 B x0.5+3C x2+6 D x5 )× 1

Tc

+¿ (Ax+Bx1.5+C x3+D x6 )× 1

(1−x )2×

1Tc ]

Substituting the values,

dPsatdT

=2.753 [ ( 11−0.2932 ) ((−6.99913 )+1.5 (1.23014 )0.340220.5+¿3 (−2.49377 ) (0.34022 )2+6 (−2.21052 ) 0.340225 )

×1

369.14

+¿ ((−6.99913 ) 0.34022+(1.23014 )0.340221.5+¿ (−2.49377 ) 0.340223+ (−2.21052 ) 0.340226 )× 1

(1−0.34022 )2

×1

369.14

]dPsatdT

=4920 .783 pa/K

To find Vg – specific volume of saturated vapour using equation (3)

P= RTV g−b

− aαV g (V g+b )



The above equation is rearranged as

Vg3+(−RTP )Vg2+( aαP − RTb

P−b2)Vg+(−aαb

P )=0

Where

a=0.42748R2T C

2

PC

b=0.08664 RTC

PC

α=[1+ (0.48508+1.55171ω−0.15613ω2 ) (1−√T r ) ]

2

Where ω=¿ accentric factor (which is taken as zero)

T r=TT C

R = 8314.3/ 86.48 = 96.14Kj/kg K

a=0.42748R2T C

2

PC

=0.42748 ( 96.14 )2 (369.14 )2

4990×103

a=107.89

b=0.08664 RTC

PC

=0.08664 (96.14 ) (369.14 )❑

4990×103

b=6.16×10−4

T r=TT C

=233.15368.14

=0.63160

α=¿

α=1.209054

Substituting the values

Vg3+(−96.13×233105125.5 )Vg2( 107.8913×1.21

105125.5−125.8788×233×6.168×10−4

105125.5−(6.16×10−4 )2)Vg+(−107.8913×1.21×6.16×10−4

105125.5 )=0

Solving the above equation,



V g=0.2076 m3

kg

V fg=V g−V f=0.2076−0.00687=0.20076m3

kg

To find h fg- enthalpy of vapourisation

h fg is found by using Clasius – Clapeyron equation

¿dPsatdT

=h fg

T ×V fg

Substituting,

4920.78=h fg

233.15×0.20076

h fg=238.1754 KJ /¿kg¿

Assuming h f=200 KJ /¿kg¿ at 0 0C , hfg=214.69 at 0 0C

hg=h f+h fg=200+214.694=412.69KJ /Kg

Where hg=h2 at 0 0 C

To find h3−h2 , enthalpy departure term

Using equation (10)


bln(V 2+b

V 2)+RT 3−

RT 2V 2

V 2−b+

aαV 2

V 2 (V 2+b )

Where V 2=V g=0.2076

Substituting,

h3−h2=107.89 (0.48508√0.6311×1.21+1.21 )

6.16×10−4

ln ¿

h3−h2=5.63 KJ /¿kg¿

h3=5.63+h2=5.63+414.69



h3=420.32 KJ /¿kg¿

To find h4,

h4−h3=∫3

4

CP0dT

Where CP0=a+bT +c T2+d T3

a=1.73 E+1

b=23.96 E−1

c=−1.457 E−4

d=3.394 E−8

Substituting and Integrating

h4−h3=aT 4+bT 4

2

2+c T4

3

3+dT 4

4

4−[aT 3+

bT 32

2+c T3

3

3+dT 3

4

4 ] T 4=−40+273.15=233.15 K

T 3=0+273.15=273.15 K

Substituting the values,

h4−h3=(8130.289−10235.29)/1000KJ /¿kg¿

h4=¿ 417.3189KJ /¿kg¿

To find the enthalpy departure term h5−h4

By using equation (14)


bln( V 5

V 5+b )+RT 5V 5

V 5−b−

aαV 5

V 5 (V 5+b )−RT 4

At -400C,

T r=TT C

= 233369.14

=0.63



α=1.21

V 5=V gat point5

T 5=T 4=233 K

To find V 5 find Psat at T 5=T 4=233 K

To find h5-h4


bln( V 5

V 5+b )+RT 5V 5

V 5−b−

aαV 5

V 5 (V 5+b )−RT 4

h5-h4=-5.2677KJ/kg h5=412.0512Kj/kg

h5=hg at -40 0C

h6=h5-hfg=412.051234-238.1754=173.87kj/kg

h6=hf at -40 0C

To calculate entropy Sf and Sfg

Sf=hf/Tc

=173.87369.14

=0.7457 KJ/Kg K

Sfg=hfg/Tc

238.175369.14

=¿

Sfg1 .02155Kj/Kg K

Sg=hg/Tc

412.05369.14

=

Sg= 1.7673 Kj/Kg K



7. Table of Thermodynamic properties of R22

T in ⁰C

Psat in Pa

Vg in m3/Kg

hfg in KJ/Kg

hg in KJ/Kg

hf in KJ/Kg

sf in KJ/KgK

sfg in KJ/Kg

Ksg in

KJ/KgK

-4010586

2 0.2076 238.175 412.0512 173.8758 0.745768 1.022 1.767322

-3811606

9 0.1907 237.23 412.1271 174.8969 0.743767 1.009 1.752614

-3612703

5 0.1738 234.045 412.1986 178.1538 0.751228 0.987 1.738134

-3413879

6 0.1569 228.352 412.2639 183.9124 0.769025 0.955 1.723872

-3215139

1 0.1489 233.847 412.3452 178.4984 0.740197 0.97 1.709912

-3016485

9 0.1375 232.669 412.4144 179.7451 0.739235 0.957 1.696132

-2817924

1 0.1272 231.573 412.482 180.909 0.737952 0.945 1.68257

-2619457

8 0.1178 230.406 412.5473 182.1413 0.736967 0.932 1.669218

-2421091

1 0.1092 229.151 412.6104 183.4597 0.736343 0.92 1.656072

-2222828

2 0.1014 227.985 412.6714 184.6862 0.735362 0.908 1.643127

-2024673

4 0.094 226.154 412.7282 186.5742 0.73701 0.893 1.63037

-1826631

2 0.0878 225.753 412.7864 187.0336 0.733034 0.885 1.617819

-1628705

8 0.0818 224.504 412.8396 188.3352 0.732394 0.873 1.605443

-1430901

9 0.076 222.385 412.8869 190.5023 0.735105 0.858 1.593235

-1233223

8 0.0713 222.178 412.938 190.7597 0.73046 0.851 1.581229

-1035676

3 0.0666 220.762 412.9816 192.2197 0.730457 0.839 1.569377

-838263

9 0.0624 219.788 413.0238 193.2362 0.728781 0.829 1.557698

-640991

5 0.0584 218.346 413.0604 194.7146 0.728859 0.817 1.546174

-443863

6 0.0548 217.262 413.0951 195.8329 0.727598 0.807 1.534814



-246885

2 0.0514 215.879 413.1244 197.2458 0.727442 0.796 1.5236010 50061

0 0.0483 214.694 414.0539 199.3596 0.729854 0.786 1.515848

253396

1 0.0454 213.379 415.913 202.5336 0.736084 0.776 1.511587

456895

3 0.0402 199.597 416.2105 216.6133 0.781574 0.72 1.501752

660563

7 0.0395 207.003 416.3568 209.3537 0.749969 0.742 1.491516

864406

3 0.0379 209.46 416.5441 207.0838 0.73656 0.745 1.481572

1068428

3 0.0355 206.736 416.7785 210.0425 0.741807 0.73 1.471935

1272634

9 0.0337 206.631 416.9964 210.3653 0.737736 0.725 1.462375

1477031

20.0327

6 211.325 417.1731 205.848 0.716866 0.736 1.452805

1681622

5 0.03 203.443 417.479 214.0356 0.740223 0.704 1.443815

1886414

2 0.0294 209.444 417.647 208.2027 0.715105 0.719 1.434474

2091411

6 0.0268 200.423 417.9895 217.5662 0.742167 0.684 1.425855

2296620

30.0254

6 199.741 418.2455 218.5048 0.740318 0.677 1.41706124 1E+06 0.024 197.391 418.5319 221.1414 0.744208 0.664 1.40848726 1E+06 0.0225 193.877 418.8467 224.9699 0.75203 0.648 1.40012328 1E+06 0.0216 194.875 419.1007 224.2261 0.744566 0.647 1.39166830 1E+06 0.0205 193.532 419.3997 225.8679 0.74507 0.638 1.38347332 1E+06 0.0194 191.533 419.7202 228.1869 0.747786 0.628 1.37545534 1E+06 0.0184 189.872 420.0455 230.1737 0.749385 0.618 1.36755836 1E+06 0.0175 188.645 420.3727 231.7281 0.749565 0.61 1.35976938 1E+06 0.0166 186.832 420.7235 233.8919 0.751701 0.6 1.35215740 2E+06 0.0158 185.573 421.0736 235.5003 0.752037 0.593 1.3446398. GRAPHS



8.1 Psat v/s Vg

0 0.0500000000000001 0.1 0.15 0.2 0.2500000000000010

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

ASHRAER22 values obtained

Vg in m3/kg

Psat

in M

pa

Figure 1: Graph of Psat v/s vg

8.2 P sat V/S hg



385 390 395 400 405 410 415 420 4250

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

R22 values obtainedASHRAE VALUES

hg in KJ/Kg

Psat

in M

Pa

Figure 2: Graph of Psat v/s hg

8.3 Psat V/S sg



1.3 1.4 1.5 1.6 1.7 1.8 1.90

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

R22 values obtainedASHRAE values

Sg in KJ/Kg K

Psat

in M

pa

Figure 3: Graph of Psat v/s Sg

8.4 Psat v/s T



220 230 240 250 260 270 280 290 300 310 3200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

R22 values obtainedASHRAE VALUES

T sat in K

Psat

in M

pa

Figure 4: Graph of Psat v/s Tsat



9 References

1. C.P Arora Refrigeration and airconditoning.

2. ASHRAE data hand book 2005


report on r22

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