ch9-

Chapter 9

Simple Vapour Compression  Refrigeration Systems 9.1 Introduction A vapour compression refrigeration system system is an improved type of air refrigeration system system in in which a suitable working substance, termed as refrigerant, is used. It condenses and evaporates at temperatures a nd pressures close to the atmospheric conditions. conditions. The refriger refrigerants, ants, usuall usually y, used for t his p urpose are ammoni a (N H!, carbon dio"ide (#$%! and sulphur dio"ide (&$2!. The refrigerant used, does. not leave the system, but is circulated throughout the system alternately alterna tely condensing and evaporatin evaporating. g. In evaporating, evaporating, the refrigerant refrigerant absorbs its its latent heat from from the brine (salt water! wh ich is used u sed for circulating it around the cold chamber. 'hile condensing, it gives out its latent heat to the circulating water of the cooler. cooler. The vapour compression refrigeration system is, therefore a latent heat pump, as it pumps its latent heat from the brine and delivers it to the coole r. The vapour compression refrigeration refrigeration system is nowadays used for all  purpose  purp ose refrigera refri geration tion.. It I t is general gen erally ly used for all industria indu striall purpo p urposes ses from a small s mall domestic refrigerator refrigerator to a big air conditioning plant.

9.2 Advantages and Disadvantages of Vapour Vapour Compression and Air Refrigeration Systems )ollowing are the advantages and disadvantages of the vapour compression refrigeration refrigeration system over air air refrigeration refrigeration system* Advantages 1. It has smaller si+e for the given capacity of refrigeration. 2. It has less running cost. . It can be employed over a large range of temperatures. coefficient of performance performance is uite high. !. The coefficient

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Disadvantages 1. The initial cost is high. 2. The prevention of leakage of the refrigerant is the ma-or problem in vapour compression system.

9. $echanism of a Simple Vapour Vapour Compression Refrigeration System

%ig. 9.1. Simple vapour compression refrigeration system .

)ig. ./ shows the schematic diagram of a simple vapour compression refrigeration system. It consists of the following five essential parts* 1. Compressor . The low pressure and temperature vapour refrigerant from evaporator is drawn into the compressor through the inlet or suction valve A, where it is compressed to a high pressure and temperature. This high pre high pressu ssure re and temperatu temperature re vapour vapour refrigeran refrigerantt is dischar discharged ged into the condense cond enserr throug t hrough h the delivery valve B. valve B.

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2. Condenser . The condenser or cooler consists of coils of pipe in which the high pressure and temperature vapour refrigerant is cooled and condensed. the refrigerant, refrigerant, while, while, passing through the th e condenser, gives up its latent heat to the surrounding condensing condensing medium which is normally air or water. . Receiver . Thee condensed liuid refrigerant from the condenser is stored in a vessel known as receiver from where it is supplied to the evaporator through the e"pansion, valve or refrigerant refrigerant control valve. !. &'pansion valve . It is also called throttle valve or refrigerant control valve. The function of the e"pansion valve is to allow the liuid refrigerant under high pressure and temperature to pass at a controlled rate after reducing its pressure and temperature. &ome of the liuid refrigerant evaporates as it passes through the e"pansion valve, but the greater portion is vaporised in the evaporator at the low pressure and temperature. (. &vaporator . An evaporator consists of coils of pipe in which the liuid vapour. refrigerant at low pressure and temperature is evaporated and changed into vapour vapou r refrigerant refrig erant at low pressure pressure and temperature. temperature. In evaporating, evaporating, the liuid vapour vapour refrigerant absorbs0 its latent heat of vaporisation from the medium (air, water or brine! which is to be cooled.

)ote* In any compression, refrigeration system, there are two different pressure conditions. $ne is called the high pressure side and other is known as low pressure side. side. The high pressure side includes the discharge line ( i.e. piping i.e. piping from from delivery delivery valve valve 1 to the condenser!, condenser, receiver and e"pansion valve. The low pressure side includes the evaporator, piping from the e"pansion valve to the evaporator and the suction line (i.e. piping from the evaporator to the suction valve A). valve A).

9.! +ressure,&nthalpy -p,h Chart The most convenient convenient chart for studying study ing the behavior behav ior of a refrigerant refrige rant is the pthe p-h h chart, in which the vertical ordinates represent pressure and hori+ontal ordinates represent enthalpy (i.e. total heat!. A typical chart is shown in )ig. .%. in which a few important importan t lines of the complete chart are drawn. drawn. The The saturated saturated liuid liuid line and the saturated vapour line merge into one another at the critical point. A saturated liuid is one which has a temperature eual to the saturation temperature corresponding to its  pressure. The space to the left of the saturated liuid line will, therefore, therefore, be sub sub cooled liuid region. The space between the liuid and the vapour lines is called wet vapour region and to the right of the saturated vapour line is a superheated vapour region.

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%ig. 9.2. +ressure,enthalpy (p-h) chart.

9.( 0ypes of Vapour Compression Cycles 'e have already discussed that vapour compression cycle essentially consists of compression, condensation, throttling and evaporation. 2any scientists have focussed their attention to increase the coefficient of performance of the cycle. Though there are many cycles, yet the following are important from the sub-ect point of view 1. #ycle with dry saturated vapour after compression, 2. #ycle with wet vapour after compression, . #ycle with superheated vapour after compression, !. #ycle with superheated vapour before compression, and (. #ycle with undercooling or subcooling of refrigerant.

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9.# 0heoretical Vapour Compression Cycle ith Dry Saturated Vapour after Compression A vapour compression cycle with dry saturated vapour after compression is shown on T-s and p-h diagrams in )ig. 9.3 (a) and (b) respectively at point /, let T  ,l  p l  and  s1, be the temperature, pressure and entropy of the vapour refrigerant respectively. The four processes of the cycle are as follows %ig. 9.. 0heoretical vapour compression cycle ith dry saturated vapour after

compression.

1. Compression process . The vapour refrigerant at low pressure p1 and temperature T l   is compressed isentropically to dry saturated vapour as shown by the vertical line /% on T-s diagram and by the curve /% on p-h diagram. The pressure and temperature rises from pl  to p2 and T 1 to T 2 respectively.

The 'ork done during isentropic compression per kg of refrigerant is given by w

h2

h1

(./!

'here h1 3 4nthalpy of vapour refrigerant at temperature T  I  , i.e. at suction of the compressor, and h2 = 4nthalpy of the vapour refrigerant at temperature T 2, is i.e. at discharge of the compressor. 2 . Condensing process . The high pressure and temperature vapour refrigerant from the compressor is passed through the condenser where it is completely condensed at

1/

constant pressure p2 and temperature T 2, as shown by the hori+ontal line % on T-s and  p-h diagrams. The vapour refrigerant is changed into liuid refrigerant. The refrigerant, while passing through the condenser, gives its latent heat to the surrounding condensing medium. . &'pansion process . The liuid refrigerant at pressure p3 = p2 and temperature

T 3 = T 2 is e"panded by throttling process through the e"pansion valve to a low  pressure p4 = pl  and temperature T 4 = T 1 , as shown by the curve 3-4 on T-s diagram and by the vertical line 3-4 on p-h diagram. 'e have already discussed that some of the liuid refrigerant evaporates as it passes through the e"pansion valve5 but the greater portion is vaporised in the evaporator. 'e know that during the throttling  process, no heat is absorbed or re-ected by the liuid refrigerant.

)otes* -a In case an e"pansion cylinder is used in place of throttle or e"pansion valve to e"pand the liuid refrigerant, then the refrigerant will e"pand isentropically as shown  by dotted vertical line on T-s diagram in )ig. . (a!. The isentropic e"pansion reduces the e"ternal work being e"panded in running the compressor and increases the refrigerating effect. Thus, the net result of using the e"pansion cylinder is to increase the coefficient of performance. &ince the e"pansion cylinder system of e"panding the liuid refrigerant is uite complicated and involves greater initial cost, therefore its use is not -ustified for small gain in cooling capacity. 2oreover, the flow rate of the refrigerant can be controlled with throttle valve which is not possible in case of e"pansion cylinder which has a fi"ed cylinder volume. -3 In modern domestic refrigerators, a capillary (small bore tube! is used in place of an e"pansion valve. !. Vaporising process . The liuidvapour mi"ture of the refrigerant at pressure p4 = pl  and temperature T 4 =T l  is evaporated and changed into vapour refrigerant at constant  pressure and temperature, as shown by the hori+ontal line 6/ on T-s and p-h diagrams.

7uring evaporation, the liuidvapour refrigerant absorbs its latent heat of vaporisation from the medium (air, water or brine! which is to be cooled. This heat which is absorbed by the refrigerant is called refrigerating effect and it is briefly

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written as 8 4. The process of vaporisation continues up to point I which is the starting  point and thus the cycle is completed. 'e know that the refrigerating effect or the heat absorbed or e"tracted by the liuid vapour refrigerant during evaporation per kg of refrigerant is given by  R E 

h1

h4

h1

h f 3

(.%! ...( h f3

h4  )

h f  3

'here

 &ensible heat at temperature T 3 , i.e. enthalpy of 9iuid refrigerant leaving the condenser.

It may be noticed from the cycle that the liuidvapour refrigerant has e"tracted heat during evaporation and the work will be done by the compressor for isentropic compression of the high pressure and temperature vapour refrigerant. #oefficient of performance, C  .O . P  .

 Re  frigeratin g effect 

h1

h4

Work  done

h2

h1

h1

h f 3

h2

h1

(.!

&'ample 9.1 In an ammonia vapour compression system, the pressure in the evaporator is % bar. Ammonia at e"it is :.;< dry and at entry its dryness fraction is :./. 7uring compression, the work done per kg of ammonia is /<: k=. #alculate the #.$.>. and the volume of vapour entering the compressor per minute, if the rate of ammonia circulation is 6.< kg?min. The latent heat and specific volume at % bar are /%< k=?kg and :.<; m?kg respectively.

Solution  @iven*  p1 h fg 

 p 4

% bar 5  x1

:.;< 5  x 4

:./ 5 w

/<: k=?kg 5 ma

6.< kg?min 5

/%< k=?hg 5 v g   :.<; m?kg

C..!. The T-s and p-h diagrams are shown in )ig. . (a) and (b) respectively. &ince the ammonia vapour at entry to the evaporator (i.e. at point 6! has dryness fraction (x4 ) eual to :./, therefore enthalpy at point 6, h4   x4  h fg   '.19  132&  2&1.%& "#$"g 

1/2

&imilarly, enthalpy of ammonia vapour at e"it i.e. at point /, h1   x1  h fg   '.&  132&  112 .2& "#$"g 

 Heat e"tracted from the evaporator or refrigerating effect,  * +   h1  h4  112 .2&  2&1 .%&  %4 .& "#  $  "g 

'e know that work done during compression, w   1&' "#$"g 

C .  !  . .   * +  $  w  %4 .& $ 1&'  &.3  A,s.



olume of vapour entering the compressor per minute 'e know that volume of vapour entering the compressor per minute 3 2ass of refrigerant? min   &pecific volume  m a  v g   4.&  '.&  2.1 m 3  $  mi  As.

&'ample 9.2 The temperature limits of an ammonia refrigerating system are %
5i6uid heat -787g

5atent heat -787g

5i6uid entropy -787g :

2&

29.9

11.94

1.1242

-1'

13&.3%

129%.

'.&443

0emperature -4C

Solution @iven* T 2  T 3  %
The T-s and p-h diagrams are shown in )ig. .6  (a) and (b) respectively.  x1   7ryness fraction at point /. 9et 'e know that entropy at point /,  x1 h fg 1  x  129% .  s1  s f 1   '.&443  1 T 1

23

1/

 '.&443  4.934 x1 ...( i )

&imilarly, entropy at point %,  s 2  s f  2 

 x1 h fg 2 T 2

 '.&443 

11 .94 29

 &.'4

...( ii )

&ince the entropy at point I is eual to entropy at point %, therefore euating euations (i) and (ii),

%ig. 9.!

'e know that enthalpy at point /, h1  h f 1  x1 h fg1  13&.3%  '.91  129% .  131 .2 "#$"g 

and enthalpy at point %, h2  h f 2  h fg 2  29.9  11 .94  14&.4 "#$"g 

  #oefficient of performance of the cycle h1  h f  3 131 .2   29 .9     .  A,s. h2  h1 14&.4  131 .2 

&'ample 9. A vapour compression refrigerator works between the pressure limits of F: bar and %< bar. The working fluid is -ust dry at the end of compression and there is no undercooling of the liuid before the e"pansion valve. 7etermine * 1. #.$.>. of the cycle 5 and 2. #apacity of the refrigerator if the fluid flow is at the rate of < kg?min

1/!

 Entha&p' (kkg)

 Pressre (!ar)

 "atration #e$peratre (%)

'

29&

1&1.9

2&

21

&.32

 *i+id

Entrop' (kkg %)

,apor

 

*i+id

,apor 

293.29

'.&&4

1.'332

322.&

'.22

1.244

Solution @iven *  p 2   p 3  F: bar    p1   p4  %< bar   T 2  T 3  %< G 5 T 1  T 4  %F/ G 5 h f  3  h4  /
%ig. 9.(

 

9et

 x1   7ryness fraction of the vapour refrigerant enterin the

compressor at point /. 'e know that entropy at point /,  s1  s f 1  x1  s fg1  s f 1  x1 (  s g 1  s f 1 )

...(  s g 1   s f 1  s g 1 )

1/(

 '.22   x1 ( 1.244  '.22  )  '.22   1.'2'4 x1



(i)

and entropy at point %,  s 2  1.'332 "#  $ "g  ...( ive ) ...(ii)

&ince the entropy at point / is eual to entropy at point %, therefore euating euations (i) and (ii), '.22   1.'2'4  x1  1.'332    x1  '.%91 'e know that enthalpy at point /, h1  h f 1  x1 h fg1  h f 1  x1 ( h g 1  h f 1 )

...(  h g 1  h f 1  h fg 1  )

 & .32  '.%91( 322.&  & .32 )  2 .93 "#  $ "g 

 #.$.> of the cycle 

h1  h f 3 h2  h1



2 .93  1&1.9  293.29  2 .93

 4.3   As.

%. #apacity of the refrigerator  'e know that the heat e"tracted or refrigerating effect produced per kg of refrigerant h1  h f 3  2 .93  1&1.93  114 .9% "#$"g 

&ince the fluid flow is at the rate of < kg?min, therefore total heat e"tracted  &  114 .9%   &%4.& "#$mi  #apacity of the refrigerator  

&%4.& 21'

 2.%4 T*  As.

9.". 0heoretical Vapour Compression Cycle ith ;et Vapour after Compression A vapour compression cycle with wet vapour after compression is shown on T-s and p- h diagrams in )ig. .F (a) and (b) respectively. In this cycle,

1/#

the enthalpy at point % is found out with the help of dryness fraction at this point. The dryness fraction at points / and % may be obtained by euating entropies at points 1 and 2.

%ig.9.# 0heoretical vapour compression cycle ith et vapour after compression.

Now the coefficient of performance may be found out as usual from the relation, C  .O . P  .

 Re  feigeratin g effect  Work  done

h1 h2

h f  3

(.6!

h1

)ote* The remaining cycle is the same as discussed in the last article. &'ample 9.!

)ind the theoretical #.$.>. for a #$% machine working between the temperature range of %
 *i+id

,apor 

#e$peratre

 *atent heat  kkg 

C 

 Entha&p' kkg 

 Entrop' kkg.% 

 Entha&p' kkg 

 Entrop' kkg.% 

2&

14.%%

'.&9%

22.23

'.991

11%.4  

-&

%2.&%

'.22

321.33

1.214

24.%  

 

1/"

Solution @iven* T 2  T 3  2&5 C = 2&62%3 = 29   T 1  T 4   &5 C = -&62%3 = 2   x1  '.  h f 3  h f 2  14.%% "#$"g  h f 1  h f 4  %2.&% "#$"g   s f 2  '.&9% "#$"g     s f 1  '.22 "#$"g   h2 7  22.23 "#$"g  h17  321.33 "#$"g   s 2 7  '.991 "#$"g    s17  1.214 "#$"g   h fg 2  11%.4 "#$"g  h fg 1  24.% "#$"g   

The T-s and p-h diagrams are shown in )ig. .Ea and .Eb respectively. )irst of all, let us find the dryness fraction at point %, i.e.  x 2 . 8e know that the entropy at point /,  x1 h fg1 '.  24.%  ...( i )  s1  s f 1   '.22   '.431 T 1

2

&imilarly, entropy at point %,  s 2  s f 2 

 x 2 h fg2 T 2

 '.&9% 

 '.&9%  '.3941 x 2 ...( ii )

%ig.9."

&'ample 9.(

1//

 x 2  11% .4 29

An ammonia refrigerating machine fitted with an e"pansion valve works between the temperature limits of/:B# and :B#. The vapour is < dry at the end of isentropic compression and the fluid leaving the condenser is at :B#. Assuming actual #.$.>. as F: of the theoretical, calculate the kilograms of ice produced per k' hour at :B# from water at /:B#. 9atent heat of ice is < k=?kg. Ammonia has the following properties #e$peratre  *i+id heat (hf) C  kkg 

 *atent heat ( h fg  )

 *i+id entrop'( s f  )

#ota& entrop' of dr' satrated /apor 

kkg 

3'

323.'

114&.'

1.2'3%

4.942

-1'

13&.3% 

129%.

'.&443

&.4%%'

Solution @iven * T 1  T 4   /:B # 3  1' D%E 3 %F G 5 T 2  T 3  :B # 3 :D%E 3 : G5  x 2  :.< 5 h f 3  h f  2  %.:; k=?kg 5 h f 1  h f 4  /<.E k=?kg 5 h fg 2  //6<.; k=?kg 5 h fg 1  /%E.F; k=?kg 5  s f 2  /.%:E   s f 1  :.<66   s 2 7  6.;6% 5  s17  3<.6EE: The Ts and ph diagrams are shown in )ig. .; (a! and (b! respectively.  x1   7ryness fraction at point /. 9et

 

'e know that entropy at point /,  x1 h fg1  x  129% .  s1  s f 1   '.&443  1 T 1

23

 '.&443  4.934 x1 ...( i )

&imilarly the entropy at point %  s 2  s f 2 

 x 2 h fg2 T 2

 1.2'3% 

'.9&  114& . 3'3

 4.%9 

...( ii )

&ince the entropy at point / (  s1 )  is eual to entropy at point % (s2!, therefore euating euations (i) and (ii), '.&443  4.934  x1  4.%9     x1  '. 

1/9

%ig. 9./  4nthalpy at point /,

h1  h f 1  x1 h fg1  13&.3%  '.  129% .  12&1.4 "#$"g 

and enthalpy at point %, h2  h f 2  x 2 h fg2  323.'  '.9&  114& .  1411 .  "#$"g 

'e know that theoretical #.$.>. h1  h f 3 12&1.4  323 .'    & . h2  h1 1411 .   12&1.4 

Actual #.$.>.3 '.  &.  3.4

'ork to be spent corresponding to / k' hour, 8   3'' "#   Actual heat e"tracted or refrigeration effectproduced per k' hour   8   A0/ al C..!.  3''  3.4  12&2 "# 

'e know that heat e"tracted from / kg of water at /:B# for the formation of / kg of ice at :B #  1  4.1%  1'  33&  3% .% "# 

19

 Amount of ice produced 

12&2 3% .% 

 33.2 "g$"8  h  As

9./ 0heoretical Vapour Compression Cycle ith Superheated Vapour after Compression

%ig. 9.9. 0heoretical vapour compression cycle ith superheated vapour after

compression

A vapour compression cycle with superheated vapour after compression is shown on Ts and p-h diagrams in )ig. . (a) and (b) respectively. In this cycle, the enthalpy at point 2 is found out with the help of degree of superheat. The degree of superheat may be found out by euating the entropies at points / and 2. Now the coefficient of performance may be found out as usual from the relation C  .O . P  .

 Re  feigeratin g effe Work  done

h1 h2

h f  3 h1

(.
A little consideration will show that the superheating increases the refrigerating effect and the amount of work done in the compressor. &ince the increase in refrigerating effect is less as compared to the increase in work done, therefore, the ne t effect of supe rheating is to have low coefficient of performance.

)ote * In this cycle, the cooling of superheated vapour will take place in two stages. )irstly, it will be condensed to dry saturated stage at constant pressure (shown by graph %%0! and secondly, it will b e conden sed at constant temp erature (show n by

191

graph %0!. The remaining cycle is same as discussed in the last article.

&'ample 9.# A vapour compression refrigerator uses methyl chloride (86:! and operates  between temperature limits of /:B# and 6. of the refrigerator. The relevant properties of methyl chloride are as follows  "atration te$peratre in  C 

 Entha&p' in kkg  *i+id

,apor

Entrop' in kkg %  *i+id

,apor 

-1'

4&.4

4'.%

'.13

1.3%  

4&

133.'

43.

'.4&

1.&%  

Solution @iven * T 1  T 4   /:B # 3  1' D%E 3 %F G 5 T 2 7  T 3  6
 

The T-s and p-h diagrams are shown in )ig. 9.1' (a) and (b) respectively.

192



%ig. 9.1

 

9et 0 p = &pecific heat at constant pressure for superheated vapour. 'e know that entropy at point 2,   T     s 2  s 2 7 2.3 0 p lg  2   T 2 7  

 333     31  

1.3%   1.&%   2.3 0 p lg 

 1.&%   2.3 0 p  '.'2  1.&%   '.'4 0 p



0 p  1.'9

and enthalpy at point 2, h2  h27 0 p  :egee f  spehea/  h27 0 p ( T 2T 27  )

 43.  1.'9( 333  31 )  &'' "#$"g 



C..!. f efigea/ h1  h f  3 4' .%  133    3.%%  A,s. h2  h1 &''  4' .% 

&'ample 9." A refrigeration machine using 8/% as refrigerant operates between the pressures %.<  bar and  bar. The compression is isentropic and there is no undercooling in the condenser. The vapour is in dry saturated condition at the beginning of the compression. 4stimate the theoretical coefficient of performance. If the actual coefficient of performance is :.F< of theoretical value, calculate the net cooling  produced per hour. The refrigerant flow is < kg per minute. >roperties of refrigerant are  Pressre0 !ar 

 "atration

 En th a& p'0 k k g 

te$peratre0 C 

 *i+id

,apor 

9.'

3

%'.&&

2'1.

2.&

-%

29.2

14.&

Take c p for superheated vapour at  bar as :.F6 k=?kg G.

Solution

19

 Entrop' of satrated  /apor0 kkg. % 

'.3 '.%''1

 

 @iven * T 2 7  T 3  35 C = 362%3 = 3'9   T 1  T 4   %5 C =-%62%3 = 2 .  )a0/al   '.&( C . !  . .)1h  m & "g $mi  h f 3  h4  %'.&& "#$"g    ( C . !  h f 1  h f 4  29.2 "#$"g  h2 7  2'1. "#$"g  h1  14.& "#$"g   s 2 7  '.3 "#$"g     s1  s 2  '.%''1 "#$"g   0 p   '.4 "#$"g  

The Ts and p-h diagrams are shown in )ig. .//a and .//b respectively. Thee/i0al 0effi0ie/ f pefma0e )irst of all, let us find the temperature at point % ( T 2 ! 'e know that entropy at point %,   T     s 2  s 2 7 2.3 0 p lg  2   T 2 7  

  T 2     3'9  

'.%''1  '.3  2.3  '.4 lg 

  T 2   '.%''1  '.3   '.'112  2.3  '.4  3'9   T 2  1.'2  3'9

lg 

...( Ta"ig  a/i lg f  '.'112)



T 2  1.'2  3'9  31%   

%ig. 9.11

19!

'e know that enthalpy of superheated vapour at point %, h2  h2 7 0 p ( T 2  T 2 7  )

 2'1.  '.4( 31%  3'9 )  2' .92 "#$"g 

  Theoretical coefficient of performance, h1  h f  3 14 .&  %' .&&   &.1  A,s. ( C . !  . .)/h  h2  h1 2' .92  14 .&

 Net cooling produced per hour  'e also know that actual #.$.>. of the machine, ( C . !  . .)a0/al   '.&  ( C . !  . .)/h  '.&  &.1  3.31& and actual work done, w a0/al   h2  h1  2' .92  14 .&  22 .42 "#$"g  'e know that net cooling (or refrigerating effect! produced per kg of refrigerant  wa0/al   ( C . !  . .)a0/al   22.42  3.31&  %4.3 "#$"g    Net cooling produced per hour   m  %4.3  &  %4.3  3%1.& "#$mi 

3%1.& 21'

 1.%% T*  As

...( 1T*  21' "#$mi)

&'ample 9./ A water cooler using 8/% works on the condensing and evaporating temperatures of %FB# and %B# respectively. The vapour leaves the evaporator saturated and dry. The average output of cold water is /:: kg ? h cooled from %FB# to FB#. Allowing %: of useful heat into water cooler and the volumetric efficiency of the compressor as ;: and mechanical efficiency of the compressor and the electric motor as ;< an d < respectively, find*. volumetric displacement of the compressor 5 and %. power of the motor. 7ata for 8/% is given below

#e$peratre 4C 

 Pressre !ar 

 Entha&p'0 kkg  *i+id

,apor

Entrop'0 kkg %  *i+id

,apor

 "pecific heat  kkg %  *i+id

,apor 

 "pecific /o&$e of  /apor  m kg 

19(

2

.9

'.4

.19.1'

'.22%'

'.&

'.99

'.%4

'.'2  

2

3.29%

3%.92

1.39

'.14%

'.9&

1.'%

'.2'

'.'&2

Solution @iven ; T 2 7  T 3  25 C = 262%3 = 299   T 1  T 4 = 25 C = 262%3 = 2%&   m w  1'' "g$h  T w1  25 C = 262%3 = 299   Tw 2  5 C = 62%3 = 2%9     v  '< = '.'   m1  &< = '.&   m 2  9&< = '.9&  h f 3  '.4 "#$"g  h f 1  3%.92 "#$"g  h2 7  19.1' "#$"g  h 1  1.39 "#$"g    s f 3  '.22%' "#$"g   s f 1  '.14% "#$"g    s 2 7  '.& "#$"g    s1 s 2 '.9& "#$"g   



0 p 3  '.99 "#$"g   0 p 4  1.'% "#$"g   0 p 2 7  '.%4 "#$"g   0 p 1  '.2'

"#$"g   v 2 7  '.'2 m3 $"g  v1  '.'&2 m3 $"g  1. lme/i0 >ispla0eme/ f /he 0mpess    The T-s and p-h diagrams are shown in )ig. ./%a and ./ %b respectively. &ince %: of the useful heat is lost into water cooler, therefore actual heat e"tracted from the water cooler, h +   1 .2 m w  0 w ( T w1  T w 2  )

 1.2  1''  4.1% ( 299  2%9 )  1''&' "#$h  1%.& "#$mi

...( ?p. hea/ f  wa/e, 0 w  4.1% "#$"g  )

'e know that heat e"tracted or the net refrigerating effect per kg of the refrigerant  h1  h f 3  1.39  '.4  12% .%& "#$"g 

  2ass flow of the refrigerant, m * 

1% .& 12% .%&

 1.3 "g$mi

19#

and volumetric displacement of the compressor 



m *  v1



1.3  '.'&2 '.'

v

 '.'& m 3  $  mi,  A,s.

2. pwe f /he m/   )irst of all, let us find the temperature at point % (T 2 ). 'e know that entropy at point %

  T 2    T  7    2  

 s 2  s 2 7 2.3 0 p2 7 lg 

%ig. 9.12

  T 2     299  

'.9&   '.&  2.3  '.%4 lg 

  T 2     299 

'.''91  1.&& lg 

T 2   T 2    1.'134   '.''&%    299  299  

lg 

...( Ta"ig  a/i lg f  '.''&%)

19"

 T 2  299  1.'134  3'3   

'e know that enthalpy at point %, h2  h2 7 0 p 2 7 ( T 2  T 2 7  )

 19.1'  '.%4( 3'3  299 )  2''. "#$"g 

'e also know that work done by the compressor per kg of the refrigerant h2  h1  2'' .  1 .39  12 .41 "#$"g  and work done per minute  m *  12.41  1.3  12.41  1 .133 "#$mi

 '.2% "#$s  '.2% "8    >ower reuired for the compressor  

'.2% 



 m1

and power of the motor



'.31%  m 2

'.2%  '.&



 '.31% "8 

'.31%  '.9&

 '.334 "8  As

9.9 0heoretical Vapour Compression Cycle ith Superheated Vapour 3efore Compression

19/

-a 0,s diagram

-3 +,h diagram

%ig. 9.1. 0heortical vapour compression cycle ith superheated vapour 3efore compression.

A vapour compression cycle with superheated vapour before compression is shown on T-s and p-h diagrams in )ig../a and ./b respectively. In this cycle, the evaporation starts at point .6 and continues upto point /0, when it is dry saturated. The vapour is now superheated before entering the compressor upto the point /. The coefficient of performance may be found out as usual from the relation, C  .O . P  .

 Re  frigeratin g effect  Work  done

h1 h2

h f  3

(.F!

h1

)ote* In this cycle, the heat is absorbed (or e"tracted! in two stages. )irstly from point 6 to point /0 and secondly from point /0 to point /. The remaining cycle is same as discussed in the previous article.

&'ample 9.9 A vapour compression refrigeration plant works between pressure limits of <. bar and %./ bar. The vapour is superheated at the end of compression, its temperature being EB # The vapour is superheated by
 "atration te$peratre0C 

 *i+id heat0 kkg

199

*atent heat0 kkg 

 

&.3

1&.&

&.1&

144.9

 

2.1

-14.'

2&.12

1&.%  

Solution @iven *  p 2  &.3 ba   p1  2.1 ba  T 2  3%5C = 3%62%3 = 31'   T 1  T 17  = &5C  0 p  '.3 "#$"g   T 2 7  1&.&5C = 1&.&62%3 = 2.&   T 17   145C = -1462%3 = 2&9   h f 3  h f  2 7  &.1& "#$"g  h f 17  2&.12 "#$"g  h fg 2 7  144.9 "#$"g  h fg 17  1&.% "#$"g  The Ts and ph diagram are shown in )ig. ./6 (a! and (b! respectively. 'e know that enthalpy of vapour at point /, h1  h17 0 p ( T 1  T 17  )  ( h f 17 h fg 17  )  0 p ( T 1  T 17  )

 ( 2&.12  1&.%  )  '.3  &  1 .9% "#$"g 

&imilarly, enthalpy of vapour at point %, h2  h2 7 0 p ( T 2  T 2 7  )  ( h f  2 7  h fg 2 7  )  0 p ( T 2  T 2 7  )  ( & .1&  144.9 )  '.3( 31'  2.& )  214.  "#$"g 

 #oefficient of performance of the plant, h1  h f 3 1 .9%   & .1& 13'.2 C . !  . .    4.%3&  A,s. h2  h1 214 .   1 .9%  2% .3

%ig. 9.1!

2

9.1 0heoretical Vapour Compression Cycle ith
-a 0,s diagram

-3 +,h diagram

%ig. 9.1(. 0heoretical vapour compression cycle ith undercooling or su3cooling of the refrigerant.

&ometimes, the refrigerant, after condensation process %00 , is cooled below the saturation temperature (T 37) before e"pansion by throttling. &uch a process is called undercooling or subcooling of the refrigerant and is generally done along the liuid line as shown in )ig. ./
and work done , 

h1

w C  .O . P  .

h4

h2

h1

...( h4  h f 3  )

h f  3

h1

 Re  frigeratin g effect  Work  done

21

h1 h2

h f  3 h1

(.E!

)ote * The value of hf may be found out from the relation, h f  3

h f 3   c p  egree of  ndercoo&i ng 

&'ample 9.1 A vapour compression refrigerator uses 8/% as refrigerant and the liuid evaporates in the evaporator at  /
 *i+id

,apor

"pecific entrop' in k  kg %  *i+id

,apor 

-1&

22.3

1'.

'.'9'4

'.%'&1

61'

4&.4

191.%

'.1%&'

'.921

Solution  @iven * T 1  T 4   1&5C = -1&62%3 = 2&   T 2  1&5C = 1&62%3 = 2   T 2 7  1'5C = 1'62%3 = 23   0 pv  '.4 "#$"g   0 p1  '.94 "#$"g   h f 1 

22.3 "#$"g  h f 3 7  4&.4 "#$"g  h17  1'. "#$"g  h2 7  19.% "#$"g   s f 1  '.'9'4 "#$"g   s f 3  '.1%&' "#$"g    s g 1  '.%'&1 "#$"g    s 2 7  '.921 "#$"g   (i) Ceffi0ie/ f pefma0e if /hee is  >e0lig    The T-s and p-h diagrams, when there is no undercooling, are shown in )ig../Fa and ./Fb respectively. 9et

 x1  7ryness fraction of the refrigerant at point /.

'e know that entropy at point /,  s1  s f 1  x1  s fg1  s f 1  x1 (  s g 1  s f 1 ) ...(  s g 1  s f 1  s g 1 )

22

 '.'9'4  x1 ( '.%'&1  '.'9'4 )  '.'9'4  '.14%  x1 ...( i )

and entropy at point %,

  T 2    T  7    2  

 s 2  s2 7 2.3 0 pv lg 

 2   '.921  2.3  '.4 lg     23   '.921  2.3  '.4  '.''%%   '.%'34 ...( ii )

&ince the entropy at point / is eual to entropy at point %, therefore euating euations (i! and (ii!,

'.'9'4  '.14%   x1  '.%'34

 

 x1  '.99% 

%ig.9.1#

'e know that the enthalpy at point /, h1  h f 1  x1 h fg1  h f 1  x1 ( h g 1  h f 1 )

 22.3  '.99% ( 1'  22.3 )  1'.4 "#$"g  ...( h g 1  h17  )

2

and enthalpy at point %, h2  h2 7 0 pv ( T 2  T 2 7  )  191.%  '.4( 2  22.3 )  194.9 "#$"g 



C..!. 

h1  h f 37  h2  h1



1'.4  4&.4 194.9  1'.4

 9.2%  As.

(ii) Ceffi0ie/ f pefma0e whe /hee is a >e0lig f & 5C    The T-s a> p-h diagrams, when there is an undercooling of < o#, are shown in )ig. ./E (a! and (b! respectively. 'e know that enthalpy of liuid refrigerant at point  h f 3  h f 3 7 0 p1  :egee f  >e0i g  



4&.4

C..!. 

 '.94  & 

h1  h f 37  h2  h1



4'.% "#$"g 

1'.4  4'.%  194.9  1'.4

 9.&9  As

%ig. 9.1"

&'ample 9. 11 &imple NH vapour compression. system has compressor with piston displacement of % 2?2in, a condenser pressure of /% bar and evaporator pressure of %.< bar. The liuid is subcooled to %:B# by soldering the liuid line to suction line. The temperature of vapour leaving. the compressor is /::B#, heat re-ected to compressor

2!

cooling water is <::: k=?hour, and volumetric efficiency of compressor is :.;. #ompute * #apacity 5 Indicated power 5 and #.$.>. of the system.

Solution  @iven  v p  2 m 3  $ mi -  p 2   p 2 7   p3 7   p3  12 ba   p1   p4  %.< bar 5 T 3 # 3 %:D%E 3 % G 5 T 2  /:: # 3 /::D%E 3 E G 5  v  :.;

 %:

Capa0i/ f /he ss/em   The T-s and p-h diagrams are shown in )ig. ./;a and ./;b respectively.   )rom p-h diagram, we find that the evaporating temperature co rresponding to 2.&  bar is T 1  T 4  14 C   -14  2%3  2&9   

#ondensing temperature corresponding to 12 bar is T 2 7  T 2  3' C  3'  2%3  3'3   &pecific volume of dry saturated vapour at 2.& bar ( i.e. at point / !, v1  '.49 m 3  $  "g  4nthalpy of dry saturated vapour at point /, h1 = 142 "#$"g  4nthalpy of superheated vapour at point 2, h2 =13' k=?kg  and enthalpy of subcooled liuid at %:: # at point at  h f 3  h4  2%' "#$"g 

2(

%ig.9.1/ m *   2ass flow of the refrigerant in kg?min. 9et 'e know that piston displacement, v p 

m *  v1 v

 

m * 

v p   v v1



2  '.   3.2& "g$mi, '.49

'e know that refrigerating effect per kg of refrigerant h1  h f 3  142  2%'  11& "#$"g 

and total refrigerating effect  m * ( h1  h f 3  )  3.2&( 142  2%' )  3%1 "#$mi



#apacity of the system  3%1 $  21'  1 T*  A,s.

 I>i0a/e> pwe  , f /he ss/em 'e know that work done during compression of the refrigerant  m * ( h2  h1 )  3.2&( 13'  142 )  &9.&3 "#$mi, Heat re-ected to compressor cooling water   &''' "#$#   &'''$'  3.33 "#$mi, ...( ive )

 Total work done by the system  &9 .&3  3 .33  %42 .  "#$mi,

2#

and indicated power of the system  %42 .  $ '  12 .3 "8  A,s. C..!. f /he ss/em 'e know that #.$.>. of the system 

T/al  efigea/ ig effe0/  T/al  w"  >e



3%1 %42. 



&.1 As.

&'ample 9.12 A vapour compression refrigerator uses methyl chloride (86:! and operates  between pressure limits of /EE.6 k>a and FE.< k>a. At entry to the compressor, the methyl chloride is dry saturated and after compression has a temperature of /:%B#. The compressor has a bore and stroke of E< mm and runs at 6;: r.p.m with a volumetric efficiency of ;:. The temperature of the liuid refrigerant as it leaves the condenser is .5 %. mass flow rate of refrigerant5 and . cooling water reuired by the condenser if its temperature rise is limited to /%B#. &pecific heat capacity of water3 6./;E k=?kg G. The relevant properties of methyl chloride are as follows*

 "at. te$p.  C 

 Pressre kPa

-1' 4&

 "pecific entha&p' kkg 

 "pecific /o&$e

 "pecific entrop' kkg % 

 *i+id

,apor

*i+id

,apor

*i+id 

,apor 

1%%.4

'..''1'2

'.233

4&.3

4'.%

'.13

1.%2

9%.&

'. .''11&

'.'4  

132.9

6;.F

:.6;<

1.&% 

Solution

2"

@iven *  p1   p4  /EE.6 k>a 5  p2   p3  FE.< k>a 5 T 2  /:% # 3 /:%D%E 3 E< G 5  :  @  E< mm 3 :.:E< m 5     6;: r.p.m 5  v  ;: 3 :.; 5 T 3  <# 3 
 



The T-s and p-h diagrams are shown in )ig. ./ (a! and (b) respectively.

%ig. 9.19

1. *efigea/ C..!. 'e know that enthalpy at point %, h2  h2 7 0 pv ( T 2  T 2 7  )

 43.  1.24( 3%&  31 )  &% .2 "#$"g 

and enthalpy of liuid refrigerant at point , h f 3  h f 3 7 0 pl ( T 3 7 T 3  )

 132.9  1.24( 31  3' )  11 .%4 "#$"g 

'e know that refrigerator #.$.>. h1  h f  3 4' .%   11 .%4    2.9  A,s. h2  h1 &% .2  4' .% 

2/

2. ass flw a/e f efigea/  m * 3 2ass flow rate of refrigerant in kg?min.

9et

'e know that suction volume or piston displacement per minute,   pis/



 

4

aea  ?/"e  *.!.B.

( '.'%& ) 2 '.'%&  4'  '.1  m 3  $  mi,

...( i )

'e also know that piston displacement per minute

 m *  v1 

1 v

 m *  '.233 

1  '.29 m * '.

...( ii )

4uating euations (i) and (ii), m *  '.1  $ '.29  '.&& "g$mi,

3. Clig wa/e eie> b /he 0>ese   

9et

m w   #ooling water reuired by the condenser in kg?min.

'e know that heat given out by the refrigerant in the condenser   m * ( h2  h f 3  )  '.&&( &% .2  11 .%4 )  2&2.% "#$mi ...( iii )

and heat taken by water in the condenser   m w  0 pw  *ise i /empea/ e

 m w  4.1%  12  &'.244 m w ...( iv )

4uating euations (iii) and (iv), mw  2&2.%  $ &'.244  &.'3 "g$mi As.

&'ample 9.1

29

A commercial refrigerator operates with 8/% between /.%F; bar and /.FE% bar. The vapour is dry and saturated at the compressor inlet. Assuming isentropic compression, determine the theoretical #.$.>. of the plant. The isentropic discharge temperature is F6.;FB#. If the actual #.$.>. of the plant is ;: of the theoretical, calculate the power reuired to run the compressor to obtain a refrigerating capacity of / T8. If the liuid is subcooled through /:B# after condensation, calculate the power reuired The properties of 8/% are given below*

 "atration te$p. (C)

 "atration pressre (!ar)

 Entha&p' (kkg)

Entrop' (kkg %)

 *i+id

,apor 

 *i+id

,apor 

-2 &

1.23

13.33

1%.4

'.'&&2

'.%12  

&&

13.%2

9'.2

2'%.9&

'.319%

'.%%4

>roperties of superheated 8/% #e$peratre (  C) Pressre (!ar)  Entha&p' (kkg) Entrop' (kkg %)

4.

13.%2

22'.  

'.%12 

Assme spe0ifi0 hea/ f lii> / be 1.'&& "#$"g .

Solution @iven * p1   p 4  /.%F; bar 5  p 2   p3   /.FE% bar 5 T 2  F6.;F # 3 F6.;FD%E 3 . .)a0/al   ;: ( C . !  . .)/h  5 D  / T8 5 T 1   %< # 3 E.;F G 5 ( C . !  %


%:E.< k=?kg 5  s f 3  :./E k=?kg 5  s 2 7   :.FEE6 k=?kg G 5 h2  %%:.F k=?kg 5 0 p 1  /.:<< k=?kg G.

21

%ig. 9.2

 

The T-s and  p-h diagrams are shown in )ig* .%:a and .%:b respectively.

Thee/i0al C..!. f /he pla/ 'e know that theoretical #.$.>. of the plant 

h1  h f 37  h2  h1



1% .4  9'.2 22'.  1% .4

 1.9& As.

 !we eie> /  /he 0mpess  &ince the actual #.$.>., of the plant is ;: of the theoretical, therefore ( C . !  . .)a0/al   '.  1.9&  1.& and actual work done by the compressor, wa0/al   h2  h1  22' .   1% .4  44 .12 "#$"g   Net refrigerating effect produced per kg of refrigerant . .)a0/al   wa0/al   ( C . !  

44.12  1.&   .3 "#$"g 

'e know that refrigerating capacity 1 T*  21' "#$mi,   2ass flow of refrigerant,

211

m *  21' $ .3  3.'& "g$mi,

and work done during compression of the refrigerant  m * ( h2  h1 )  3.'&( 22'.  1% .4 )  134.&% "#$mi,  >ower reuired to run the compressor  134 .&%  $ '  2.24 "8  A,s.

 !we eie> if /he lii> is sb0le> /hgh 1' EC  The Ts and ph diagrams with subcooling of liuid are shown in )ig. .%/a and .%/b respectively. 'e know that enthalpy of liuid refrigerant at point , h f 3  h f 3 7 0 p1  :egee f  sb0li g  

 Theoretical C . . p . 

And

9'.2  1.'&&  1'

h1  h f  3 h2  h1





%9.%3 "#$"g 

1% .4  %9 .%3  2 .2 22' .   1% .4

( C . !  . .)a0/al   '.  2.2  1.% 

 A0/al w" >e b /he 0mpess , wa0/al   h2  h1  22' .   1% .4  44 .12 "#$"g 

 Net refrigeration effect produced per kg of refrigerant . .)a0/al   wa0/al   ( C . !  

44.12  1.%   %% .& "#$"g 

212

%ig. 9.21

'e know that refrigerating capacity  1 T*  21' "#$mi  2ass flow of the refrigerant, m *  21' $ %% .&  2.% "g$mi,

and work done during compression of the refrigerant  m * ( h2  h1 )  2.% ( 22'.  1% .4 )  119 .12 #$mi,  >ower reuired  119 .12 $ '  1.9& "8   A,s.

&'ample 9.1! A vapour compression refrigerator works between the pressures 6. bar and /.;F  bar. The vapour is superheated at the end of compression, its temperature being %
"atration te$p.0 C

#ota& heat (&i+id)0 kkg

*atent heat0 kkg 

1.

-1&

21.%

11.41

4.93

14.4&

49.'%

14%.'

The specific heat at constant pressure for he superheated vapour is :.F6< k=?kg G and for the liuid is :.F k=?kg G.

21

Solution  @iven *  p 2   p 3  4.93 ba   p1   p4  1. ba  T 2  2&E C = 2&62%3 = 29    T 3   9E C = 962%3 = 22    x1  9&< = '.9&  T 3 7  T 2 7  14.4&E C = 14.4&62%3 = 2%.4&   T 1  T 4   1&E C = -1&62%3 = 2&   h f 1  21.% "#$"g  h f 3 7  h f 2 7  49.'% "#$"g  h fg 1  11.41 "#$"g  h fg 2 7  14%. "#$"g  0 pv  '.4& "#$"g   0 pl   '.93 "#$"g   The Ts and ph diagrams are shown in )ig. .%%a and .%%b respectively. Ceffi0ie/ f pefma0e 'e know that enthalpy at point /, h1  h f 1  x1 h fg1  21.%  '.9&  11.41  1%& "#$"g 

&imilarly, enthalpy at point %, h2  h2 7 0 pv  :egee f  spehea/ 

 ( h f 2 7  h fg 2 7  )  0 pv ( T 2  T 2 7  )

 ( 49.'%  14% . )  '.4&( 29  2% .4& )  2'3.% "#$"g 

and enthalpy of liuid refrigerant at point , h f 3  h f 3 7 0 p 1  :gee f  >e0lig 

 h f  3 7 0 p1 ( T 3 7 T 3  )

 49.'%  '.93( 2% .4&  22 )  43.2 "#$"g  

#oefficient of performance, h1  h f 3 1%&  43.2 C . !  . .   4.&%  As. h2  h1 2'3.%  1%&

8efrigerating effect per kg of the working substance 'e know that the refrigerating effect  h1  h f 3  1%&  43.2  131.1 "#$"g   As.

21!

%ig.9.22

&'ample 9.1( In a /< T8 ammonia refrigeration plant, the condensing temperature is %
 Entha&p'0 kkg

Entrop' 0 kkg %

"pecific heat0 kkg % 

te$peratre  *i+id

,apor

*i+id

,apor

*i+id

,apor 

2&

29.9

14&.4

1 .124 2

&.'391

4.

2.

-1'

13&.3%

1433.'&

'.&443

&.4%%'

-

-

 C 

Solution @iven * D  /< T8 5 T 2 7  T 3 7   %< # 3 %
<.:/ k=?kg G 5 0 p 1  6.F k=?kg G 5 0 pv  %.; k=?kg G 5 h f 1  /<.E k=?kg 5 h17  /6.:< k=?kg 5  s f 1  :.<66 k=?kg 5  s g 1  <.6EE: k=?kg G 

21(

The Ts and ph diagrams are shown in )ig..%a and .%b respectively. )irst of all, let us find the temperature of refrigerant at point % (T%!. 'e know that entropy at point l,  s1  s f 1  x1  s fg1  s f 1  x(  s g 1  s f 1 )

...(  s g 1   s f 1  s fg 1 )

 '.&443  '.9% ( &.4%%'  '.&443 )  &.329 "#$"g    ...( i )

and entropy at point %,

  T 2     T      &.'391  2.3  2. lg  2   29   T 2 7  

 s 2  s 2 7 2.3 0 pv lg 

  T     &'.'391   .44 lg  2   29   ...( ii )

&ince entropy at point / is eual to entropy at point %, therefore euating euations (i) and (ii!,   T    &.329  &.'391   .44 lg   2   29  

  T 2   &.329  &.'391  '.'4&   .44  29  

lg 

T 2

or

29

 1.1'9

...( Ta"ig  a/ilg f  '.'4&)

T 2  29  1.'9  33'   



1. Ceffi0ie/ f pefma0e 'e know that enthalpy at point / h1  h f 1  x1 h fg1  h f 1  x1 ( h g 1  h f 1 ) ...( h g 1  h f 1  h fg 1  )

 13&.3%  '.9% ( 1433.'&  13&.3%  )  1394.12 "#$"g  ...( h g 1  h17  )

4nthalpy at point %, h2  h2 7 0 pv  :egee f  spehea/   h2 7 0 pv ( T 2  T 2 7  )

21#

 14&.4  2.( 33'  29 )  1&&&.44 "#$"g 

%ig.9.2

and enthalpy of liuid refrigerant at point , h f 3  h f 3 7 0 pl   :egee f  >e0lig 

 29 .9  4.  &  2%& .9 "#$"g 

 #oefficient of performance, h1  h f 3 1394.12  2%& .9    .93  A,s. C . !  . . h2  h1 1&&&.44  1394.12

2. !we eie>  'e know that the heat e"tracted or refrigerating effect produced per kg of refrigerant  * +   h1  h f 3  1394.12  2%&.9  111 .22 "#$"g 

and refrigerating capacity of the system, D  1& T*  1&  21'  31&' "#$mi

...( ive )

  2ass flow of the refrigerant, D 31&' m *    2.1 "g$mi,  * +  111 .22

'ork done during compression of the refrigerant  m * ( h2  h1 )  2.1( 1&&&.44  1394.12 )  4&3 .3 "#$mi,   >ower reuired  4&3 .3 $ '  % .&& "8  A,s.

21"

&'ample 9.1# The following data refers to a %: T8 ice plant using ammonia as refrigerant * The temperature of water entering and leaving the condenser are %:B# and %EB# and temperature of brine in the evaporator is /
*i+id

,apor

-1&

112.34

142.&4

2&

29.9'

14&.4

Entha&p'0 kkg

Entrop'0 kkg %

"pecific heat0 kkg %  

,apor

*i+id

,apor  

'.4&%2

&.&49'

4,39

2.3'3

1.1242

&.'391

4.'

2.'&

*i+id  

Solution @iven ; D  2' T*  T w 2  2'E C = 2'62%3 = 293  T w1 2%E C = 2%62%3 = 3''   T 2 7  T 3 7  2& EC = 2&62%3 = 29   T 1  T 4   1& EC = -1&62%3 = 2&    T 3  2'EC = 293   h f 1  112.34 "#$"g  h f 3 7   29.9' "#$"g  h1  142.&4 "#$"g  h2 7  14&.4 "#$"g  s f 1  '.4&%2 "#$"g    s f 3  1.1242 "#$"g    s1 s 2 &.&49' "#$"g    s 2 7  &.'391 "#$"g   0 pl 4  4.39 "#$"g   0 pl 3  6.:F k=?kg G 5 0 pv 1  %.: k=?kg G 5 0 pv 2 7   %.;:< k=?kg G 



The Ts and ph diagrams are shown in )ig. .%6a and .%6b respectively. )irst of all, let us find the temperature of refrigerant at point % (T%!. 'e know that entropy at  point %,   T     s 2  s 2 7 2.3 0 pv 7 lg  2   T 2 7  

  T 2     29  

&.&49'  &.'391  2.3  2.'& lg 

  T 2   &.&49'  &.'391  '.'%9  2.3  2.'&  29  

lg 

21/

T 2 29

 1 .2

...( Ta"ig  a/ilg f  '.'%9)

 

T 2  1.2  29  3&% .   

 

4.  C 

+/halp a/ pi/ 2, h2



h2 7  0 pv 2 7 ( T 2

 T  2

7  )

 14&.4  2.'&( 3&% .  29 )  133.'2 "#$"g 

and enthalpy of liuid refrigerant at point , h f 3  h f 3 7 0 pl 3  :egee f  >e0lig 

 h f 3 7 0 pl 3 ( T 3 7 T 3  )  29.9  4.' ( 29  293 )  2%&.% "#$"g 

%ig. 9.2!

'e know that heat e"tracted or refrigerating effect produced per kg of the refrigerant,  * +   h1  h f 3  142 .&4  2%&.%   11&' .% "#$"g 

and capacity of the ice plant, D  2' T*  2'  21'  42'' "#$mi

 2ass flow of the refrigerant,

219

m * 

D 42''   3.& "g$mi,  * +  11&' .% 

'ork done by the compressor per minute

 m * ( h2  h1 )  3.&( 133.'2  142 .&4 )  %&3 .& "#$mi,  >ower e"pended per T8  

%&3.& '  2'

 '.2 "8$T*  As.

Amount of cooling water in the condenser  9et m w  = Amount of cooling water in the condenser. 'e know that heat given out by the refrigerant in the condenser   m * ( h2  h f 3  )

 3.&( 133.'2  2%&.%  )  49&3. "#$mi ...( i )

&ince the specific heat of water, 0 w  4.1% k=?kg G, therefore heat taken by water in the condenser   m w  0 w ( T w1  T w 2  )  mw  4.1% ( 3''  293 )  29.3 mw "#$mi ...( ii )

&ince the heat given by the refrigerant in the condenser is eual to the heat taken by water in the condenser, therefore euating euations (i) and (ii!, 29.3 mw  49&3. 

 

mw  19 "g$mi As.

&'ample 9.1" A vapour compression refrigeration machine, with )reon/% as refrigerant, has a capacity of %: tonne of refrigeration operating between  %;B# and %FB#. The refrigerant is subcooled by 6B# before entering the e"pansion valve and

22

the vapour is superheated by ., . olumetric efficiency 5 and !. 1ore and stroke of cylinder. The speed of co mpressor is /::: r.p.m. The following properties of )reon/% may be used *  "at. te$p.0  C   Pressre0 !ar 

 " p. /o&$e of  /apor0 $3 kg   

-2

1.'93

'.14%&

2

.9%

'.'22

 Entha&p'0 kkg

Entrop'0 kkg % 

*i+id

,apor

*i+id

,apor  

1'.4

1%&.11

'.'444

'.%1&3

19.11

'.22%1

'.&

'.%  

&pecific heat of liuid refrigerant 3 :.F k=?kg G and specific heat of superheated vapour 3 :.F/< k= ?kg G.

Solution @iven * D  12 T*  T 17   2E C = -262%3 = 24&   T 2 7  T 3 7  2E C = 262%3 = 299   T 3 7 T 3  4E C  T 3  22E C = 2262%3 = 29&   T 1  T 17  &E C  T 1   23E C = -2362%3= 2&'    @  1.2& :  Cleaa0e vlme = 3< ?/"e vlme      1''' .p.m.  v17  '.14%& m3 $"g  v2 7  '.'22 m3 $"g  h f 1  1'.4 "#$"g  h f 3 7  '.% "#$"g  h17  1%&.11 "#$"g  h2 7  19.1 "#$"g   s f 1  '.'444 "#$"g    s f 3  '.2%1 "#$"g    s17  '.%1&3 "#$"g    s 2 7  '.& "#$"g   0 pl   '.93 "#$"g   0 pv  '.1& "#$"g    

The T-s and p-h diagrams are shown in )ig. 9.2& (a) and (b) respectively.

1. Thee/i0al pwe eie>  )irst of all, let us find the temperature of superheated vapour at point 2 (T 2 ). 'e know that entropy at point /,

221

  T 1    T  7    1  

 s1  s17 2.3 0 pv lg 

 2&'   '.%1&3  2.3  '.1& lg    '.%2%%    24&  ...( i )

and entropy at point %,

  T 2     T      '.&  2.3  '.1& lg  2   299   T 2 7     T     '.&  1.414& lg  2   299  

 s 2   s 2 7 2.3 0 pv lg 

...( ii )

%ig. 9.2(

&ince the entropy at point / is eual to entropy at point %, therefore euating euations (i) and (ii ),

  T 2     299  

'.%2%%   '.&  1.414& lg 

  T 2   '.%2%%  '.&  '.'291  1.414&  299  

lg 

222

T 2 299

 1.'93

...( Ta"ig  a/ilg f  '.'291)

T 2  299  1.'93  319 .%   



'e know that enthalpy at point /,

h1  h17 0 pv ( T 1  T 17  )  1%&.1  '.1&( 2&'  24& )  1%.1 "#$"g 

4nthalpy at point %, h2  h2 7 0 pv ( T 2  T 2 7  )  19.11  '.1&( 319.%  299 )  21'.4 "#$"g 

and enthalpy of liuid refrigerant at point , h f 3





h f 3 7  0 pl ( T  3 7  T  3  ) 





'.% 



'.93( 299



29& )

4.&2 "#$"g 

'e know that heat e"tracted or refrigerating effect per kg of the refrigerant,  * +   h1  h f 3  1%.1  4.&2  113 .  "#$"g 

and refrigerating capacity of the system,

D  12 T*  12  21'  2&2' "#$"g 

...( ive )

 2ass flow of the refrigerant, D 2&2' m *    22 .1%  "g$mi,  * +  113 . 

'ork done during compression of the refrigerant  m * ( h2  h1 )  22.1% ( 21'.4  1%.1 )  %24 "#$mi,  Theoretical power reuired  %24  $ '  12 .'%  "8  A,s.

2. C..!.

22

'e know that #.$.>.



h1  h f  3 h2  h1



1%.1  4 .&2  3.4  A,s. 21' .4  1%.1

3. lme/i0 effi0ie0  

v 2 3 &pecific volume at point %, and

9et

C  3 #learance 3 3< = :.:.

F

(ive) )irst of all, us find the specific volume at suction to the compressor, i.e. at point /. Applying #harles0 law to point / and /0,

v1 T 1



v17  T 17 

 

v1  v17 

T 1 T 17 

 '.14%& 

2&'  '.1&'& m 3  $ "g  24&

Again applying #harles0 law to points % and %0, v2 T 2



v 2 7  T 2 7 

 

v 2  v 2 7 

T 2 T 2 7 

 '.'22 

319 .%   '.'2 m 3  $ "g  299

'e know that volumetric efficiency, v

  v    '.1&'&   1  C   C  1   1  '.'3  '.'3    '.'2     v 2    '.%    %<  As.

4. Be a> s/"e f 0li>e   @ = 9ength of cylinder 3 /.%< >, and   3 &peed of compressor 3 /::: r.p.m.

J (@iven! J (@iven!

'e know that theoretical suction volume or piston displacement per minute 1 1  m *  v1   22 .1%  '.1&'&   3.4 m 3  $  mi, v '.%  &ince the machine has si" cylinder, single acting compressor, therefore, theoretical suction volume or piston displacement per cylinder per minute

22!



3.4  '.4 m 3  $ mi,  

...( iii )

'e also know that suction volume or piston displacement per minute = !is/ aea   ?/"e   *.!.. 

 

4

 : 2  @    

 

4

 : 2  1.2& :  1'''  92 : 3 m 3  $ mi,

...( iv )

4uating euations (iii) a> (iv), :3 = '.4$92 = '.''' &2 : = '.'% m = .% mm Ans.  @ = 1.2&

  .% = 1'.4 mm Ans.

&'ample 9.1/ A food storage locker reuires a refrigeration capacity of /% T8 and works between the evaporating temperature of  ;B# and condensing temperature of :B#. The refrigerant 8/% is subcooled by
 Entha&p'0 kkg

Entrop'0 kkg % 

!ar   *i+id

,apor

*i+id

, apor 

 "pecific /o&$e of  /apor0 $3 kg 

-

2.3&4

2.%2

14.'%

'. 1149

'.%''%

'.'%9'

3'

%.4&1

4.&9

199.2

'. 24''

'..&3

'.'23&

The specific heat of liuid 8/% is /.%< k=?kg G, and of vapour 8/% is :.E k=?kg G.

22(

Solution @iven D = 1 2 T*  T 17 = -5C = -  6 2%3 = 2&   T 27 = 3'5 C = 3' 6 2%3 =3'3   T37  -T 3=&5 C T 1 = -25 C = -262%3 = 2%1   h f1=2.%2"#$"g h f37 =4.&9 "#$"g  hl 7 = 14.'% "#$"g  h27 = 199.2 "#$"g  s fl  = '.1149 "#$"g   s f3 = '.24'' "#$"g    sl 7 = '.%''% "#$"g  s27='.&3 "#$"g   vl 7 = '.'%9 m3 $"g v27 = '.'23& m3 $"g  0 p/ = 1.23& "#$"g   0 pv = '.%33 "#$"g   The Ts and ph diagrams are shown in )ig. .%F (a! and (b! respectively, 1. Ceffi0ie/ f pefma0e )irst of all, let us find the temperature of superheated vapour at point % (T 2 ). 'e know that entropy at point /,   T     s1  s17 2.3 0 pv lg  1   T 17     2%1   '.%''%  2.3  '.%33 lg    '.%1%1  2&  ...( i )

And entropy at point %,

  T 2     T      '.&3  2.3  '.%33 lg  2   3'3   T 2 7  

 s 2  s 2 7 2.3 0 pv lg 

  T     '.&3  1.  lg  2   3'3  ...( ii )

&ince the entropy at point / is eual to entropy at point %, therefore euating euations (i) and (ii),   T    '.%1%1  '.&3  1.  lg  2   3'3 

  T 2   '.%1%1  '.&3  '.'1  1.   3'3  T 2  1.'444 3'3

lg 

or

...( Ta"ig  a/ilg f  '.'1)



T 2  1.'444  3'3  31 .4   

22#

%ig. 9.2#

'e know that enthalpy at point /, h1  h17 0 pv ( T 1  T 17  )

 14.'%  '.%33( 2%1  2& )  1.4% "#$"g 

4nthalpy at point %, h2  h2 7 0 pv ( T 2  T 2 7  )

 199.2  '.%33( 31 .4  3'3 )  2'9.44 "#$"g 

and enthalpy of liuid refrigerant at point , h f 3  h f 3 7 0 pl ( T 3 7 T 3  ) 



C . !  . .

4.&9  1.23&  &

h1  h f 3 h2  h1





&.42 "#$"g 

1 .4%   & .42 2'9 .44  1 .4% 



13' .'& 2' .9% 

  .2  A,s.

2. Thee/i0al pwe pe /e f efigea/i 'e know that the heat e"tracted or refrigerating effect per kg of the refrigerant,  * +   h1  h f 3  1.4%  &.42  13'.'& "#$"g 

and the refrigerating capacity of the system

22"

D  12 T*  12  21'  2&2' "#$mi

...( ive )

 2ass flow of the refrigerant, D 2&2'   19 .4 "g$mi, m *   * +  13' .'&

'ork done during compression of the refrigerant  m * ( h2  h1 )  19.4( 2'9 .44  1.4%  )  4' .2 "#$mi,  Theoretical power per tonne of refrigeration



4' .2 '  12

 '.&& "8$T*  As.

3. Be a> s/"e f 0mpess   

9et

 : 3 1ore of compressor   @ 3 &troke of compressor 3 /.< :, and

...( ive )

   3 &peed of compressor 3 /::: r.p.m. ...( ive )

)irst of all, let us find the specific volume at suction to the compressor, i.e. at  point /. Applying #harles0 law , v1 T 1



v17  T 17 

 

v1  v17 

T 1 T 17 

 '.'%9' 

2%1 2&

 '.'1 m 3  $ "g 

(a) 8he /hee is  0leaa0e 'e know that theoretical suction volume or piston displacement per minute  m *  v1  19.4  '.'1  1.&% m 3  $ mi

and theoretical suction volume or piston displacement per cylinder per minute  1.&%  $ 2  '.%& m 3  $ mi

...( iii )

22/

Also theoretical suction volume or piston displacement per minute = !is/ aea   ?/"e   *.!..



 

4

 

 : 2  @     3

4

 : 2  1.& :  1'''

3

 11% .2& : m  $  mi ...( iv )

4uating euations (iii) and (iv!, 

and

:3 = '.%&$11%.2& = '.'''    : 4 '.'% m = % mm As.  @ 4 /.< : = /.<   ;E 3 /:.< mm Ans.

(b) 8he /hee is a 0leaa0e f 2< 9et 2 3 &pecific volume at point %, and C =Cleaa0e = 2< = '.'2 ...( ive )

Applying charles law to point % and %0 v2 T 2



v 2 7  T 2 7 

v 2  v 2 7 

 

 '.'23& 

31 .41 3'3

T 2 T 2 7 

 '.'24& m 3  $ "g 

'e know that volumetric efficiency of the compressor, v

 v     '.'1    1  C  C  1   1  '.'2  '.'2  v ' . '24&       2    1.'2  '.'  '.9&4

 >iston displacement per cylinder per min m *  v1 1 19.4  '.'1



2





v

2



...( v )

4uating euations (iv) and (v), 11% .2& : 3  '.23 

or

 : 3  '.23  $ 11% .2&  '.'''%'

229

1 '.9&4

 '.23 m 3  $ mi,



  and

 :  '.'% m  .% mm  As.

 @  1.& :  1.&  .%   133 mm  A,s.

9.11 Actual Vapour Compression Cycle .

%ig. 9.2" Actual vapour compression cycle.

The actual vapour compression cycle differs from the theoretical vapour compression cycle in many ways, some of which are unavoidable and cause losses. The main deviations between the theoretical cycle and actual cycle are as follows* 1. The vapour refrigerant leaving the evaporator is in superheated state. 2. The compression of refrigeration is neither isentropic nor polytropic. . The liuid refrigerant before entering the e"pansion valve is subcooled in the condenser. !. The pressure drops in the evaporator and condenser.

The actual vapour compression cycle on T-s diagram is shown in )ig. .%E. The various processes are discussed below

2

-a !0ess 1-2-3. This process shws the flow of refrigerant in the evaporator. The point / represents the entry of refrigerant into the evaporator and the point  represents the e"it of refrigerant from evaporator in a superheated state. The point  also represents the entry of refrigerant into the compressor in a superheated condition. The superheating of vapour refrigerant from point % to point  may be due to -i automatic control of e"pansion valve so that the refrigerant leaves the evaporator  as the superheated vapour. -ii picking up of larger amount of heat from the evaporator through pipes located within the cooled space. -iii picking up of heat from the suction pipe, i.e., the pipe connecting the evaporator delivery and the compressor suction valve.

In the first and second case of superheating the vapour refrigerant, the refrigerating effect as well as the compressor work is increased. The coefficient of  performance, as compared to saturation cycle at the same suction pressure may be greater, less or unchanged. The superheating also causes increase in the reuired displacement of compressor and load on the compressor and condenser. This is indicated by % on T-s diagram as shown in )ig. .%E. -3 !0ess 3-4-&--%-. This process represents the flow of refrigerant through the compressor. 'hen the refrigerant enters the compressor through the suction valve at  point , the pressure falls to point 6 due to frictional resistance to flow. Thus the actual suction pressure (p s ) is lower than the evaporator pressure ( !  +  ). 7uring suction and  prior to compression, the temperature of the cold refrigerant vapour rises to point < when it comes in contact with the compressor cylinder walls. The actual compression of the refrigerant is shown by <F in )ig. .%E, which is neither isentropic nor  polytropic. This is due to the heat transfer between the cylinder walls and the vapour refrigerant..

The temperature of the cylinder walls is somewhat in between the temperatures of cold suction vapour refrigerant and hot discharge vapour refrigerant. It may be assumed that the heat absorbed by the vapour refrigerant from the cylinder walls during the first  part of the compression stroke is eual to heat re-ected by the vapour refrigerant to the cylinder walls. 9ike the heating effect at suction given by 6< in )ig. .%E, there is a cooling effect at discharge as given by FE. These heating and cooling effects take place at constant pressure. 7ue to the frictional resistance of flow, there is a pressure drop i.e., the actual discharge pressure.( p7! is more than the condenser pressure ( pC !. -c !0ess -9-1'-11. This process represents the flow of refrigerant though the condenser. The process ; represents the cooling of superheated vapour refrigerant to the dry saturated state. The process /: shows the removal of latent heat which

21

changes the dry saturated refrigerant into liuid refrigerant. The process /:// represents the subcooling of liuid refrigerant in the condenser before passing through the e"pansion valve. This is desirable as it increases the refrigerating effect per kg of the refrigerant flow. It also reduces the volume of the refrigerant partially evaporated from the liuid refrigerant while passing through the e"pansion valve. The increase in refrigerating effect can be obtained by large uantities of circulating cooling water which should be at a temperature much lower than the condensing temperatures. -d >rocess ///. This process represents the e"pansion of subcooled liuid refrigerant by throttling from the condenser pressure to the evaporator pressure.

9.12 &ffect of Suction +ressure 'e have discussed in, the previous article that in actual practice, the suction  pressure (or evaporator pressure! decreases due to the frictional resistance of flow of the refrigerant. 9et us consider a theoretical vapour compression cycle /0 %060 when the suction pressure decreases from p& to ps0 as shown on p-h diagram in )ig. .%;. It may be noted that the decrease in suction pressure 1. decreases the refrigerating effect from ( h1   h4  ) to ( h17  h4 7  ) and

 

2. increases the work reuired for compression from (h2 G h 1 ) to (h 27 G h17).

%ig. 9.2/. &ffect of suction pressure

22

&ince the #.$.>. of the system is the ratio of refrigerating effect to the work done, therefore with the decrease in suction pressure, the net effect is to decrease #.$.>. of the refrigerating system for the same amount of refrigerant flow. Hence with the decrease in suction pressure, the refrigerating capacity of the system decreases and the refrigeration cost increases.

9.1 &ffect of Discharge +ressure 'e have already discussed that in actual practice, the discharge pressure (or condenser pressure! increases due to frictional resistance of flow of the refrigerant. 9et us consider a theoretical vapour compression cycle /%0060 when the discharge  pressure increases from  !   :  to  !   : 7  as shown on p-h diagram in )ig. .%. It may be noted that the increase in discharge pressure 1. decreases the refrigerating effect  from ( h1   h4  ) to ( h1  h4 7  ) . 2. increases the work reuired for compression from ( h2  h1 ) to ( h2 7 h1 ) .

)rom above, we see that the effect of  increase in discharge pressure is similar to the effect of decrease in suction pressure. 1ut the effect of increase in discharge pressure is not as severe on the refrigerating capacity of the system as that of decrease in suction pressure.

%ig. 9.29. effect of discharge pressure

&'ample 9.19 &imple ammoniacompression system operates with a capacity of /<: tonnes. The condensation temperature in the condenser is 
2

1. >ower 5 2. Heat transferred to cylinder  . >iston displacement 5 !. Heat transfer in condenser5 and (. #oefficient of performance.

Solution @iven * D   /<: T8 5 T 2  T 2 7 7  T 3 7  < # 3 
The. T-s and p-h diagrams are shown in )ig. .: (a! and (b) respectively.   )rom p-h diagram, we find that the pressure corresponding to evaporation temperature of  %




 .

a 

&ince there is a suction pressure drop of :.//; bar due to wire drawing, therefore  pressure point /0,  p17  1.&1  '.11  1.4 ba  1.4  1' &  $m 2 >ressure corresponding to condensation temperature of 
%ig. 9.

2!

&ince there is a discharge pressure drop of :.% bar due to wire drawing, therefore  pressure at point %0,  p 2 7  13.&  '.23  13.%3 ba  13.%3  1' &  $m 2

)ote* In )ig. .: point / represents the inlet of the suction valve and point /0 is the outlet of the suction valve. The point %0 represents the inlet of discharge valve and point % is the outlet of the discharge valve.   )rom p-h diagram, we also find that enthalpy of superheated ammonia vapours at  point I or /0, h1  h17  144' "#$"g  &pecific volume at point /0, v17  '. m 3  $ "g 

Temperature at point /0, T 17  9 C   -9  2%3  24     

9et

v 2 7  &pecific volume at point %0.

&ince the compression is according to  pv 1.22



C   ,  therefore

 p17 ( v17  ) ,   p 2 7 ( v 2 7  ) , 1   p17   

or

v 2 7  v17  



   p 2 7  

1

  1.4   1.22  ' .   '.123 m 3  $  "g    13.%3 

Now plot a point %0 on the p-h diagram corresponding to  p2 7  13.%3  bar and v 2 7  '.123 m 3  $ "g . )rom the p-h diagram, we find that 4nthalpy of superheated ammonia vapours at point % or %0, h2  h2 7  12' "#$"g  Temperature at point %0, T 2 7  9' C  and enthalpy of liuid ammonia at point , h f 3  h4  32' "#$"g 

1. !we 

2(

'e know that refrigerating effect per kg,  * +   h1  h f 3  144'  32'  112' "#$"g 

and refrigerating capacity  1&' T*  1&'  21'  31&'' "#$mi, ...( ive )

  2ass flow of the refrigerant, 31&''  2 .12 "g$mi m *  112'

'e know that work done by the compressor per minute  m * 

  1

 2.12 

(  p 2 7 v2 7  p17 v17  )

1.22 ( 13 .%3  1' &  '.123 1.22  1

 1.4  1' &  '. ) #  $ mi  9  1' &  #$mi  9'' "#$mi

 

!we= 9'' $ '  14 .3 "8  A,s.

2. ea/ /asfee> / 0li>e wa/e Ja0"e/  'e know that actual work done by the compressor 

 m * ( h2 7  h17  )  2.12( 12'  144' )  &'2 "#$mi,  Heat transferred to cylinder water -acket

=9''-&'2=33 "#$mi As. 3. !is/ >ispla0eme/  'e know that piston displace ment

2#



m *  v17 



2 .12  '.

v

'.%&

 3' m 3  $  mi,  A,s.

4. ea/ /asfe i 0>ese  'e know that heat transfer in condenser   m * ( h2  h f 3  )  2.12( 12'  32' )  3&&  "#$mi  As.

&. Ceffi0ie/ f pefma0e 'e know that coefficient of performance, C . !  . .

 *efigea/ ig 0apa0i/ 8"  >e



31&'' 9''

 3.&4  As.

9.1! Improvements in Simple Saturation Cycle The simple saturation cycle may be improved by the following methods 1. 1y introducing the flash chamber between the e"pansion valve and the evaporator. 2. 1y using the accumulator or precooler. . 1y subcooling the liuid refrigerant by the vapour refrigerant. !.1y subcooling the liuid refrigerant leaving the condenser by liuid refrigerant from the e"pansion valve. The effect of the above mentioned methods on the simple saturation cycle are discussed, in detail, in the following pages

9.1( Simple Saturation Cycle ith %lash Cham3er 'e have already discussed that when the high pressure liuid refrigerant from the condenser passes through the e"pansion valve, some of it evaporates. This  partial evaporation of the liuid refrigerant is known as flash. It may be noted that the vapour formed during e"pansion is of no use to the evaporator in producing refrigerating effect as compared to the liuid refrigerant which carries the heat in the form of latent heat. This formed vapour, which is incapable of producing any refrigeration effect, can be by passed around the evaporator and supplied directly

2"

to the suction of the compressor. This is done by introducing a flash chamber  between the e"pansion valve and the evaporator as shown in )ig. ./ The flash chamber is an insulated container and it separates the liuid and vapour due to centrifugal effect. Thus the mass of the refrigerant passing through the evaporator reduces.

%ig. 9.1. Simple saturation cycle ith flash cham3er.

9et us consider that a certain amount of refrigerant is circulating through the condenser. This refrigerant after passing through the e"pansion valve, is supplied to the flash chamber which separates the liuid and vapour refrigerant. The liuid refrigerant from the flash chamber is supplied to the evaporator and the vapour refrigerant flows directly from the flash chamber to the suction of the compressor. The p-h diagram of the cycle is shown in )ig. .% 9et

m1   2ass of liuid refrigerant supplied to the evaporator,

and m 2  2ass of refrigerant (liuid and vapour! circulating through the condenser, or leaving the e"pansion valve.

 2ass of vapour refrigerant flowing directly from the flash chamber to the suction of the compressor  m 2  m1

Now considering the thermal euilibrium of the flash chamber. &ince the flash chamber is an insulated vessel, therefore, there is no heat e"change between

2/

the flash chamber and the atmosphere. In other words, the heat taken and given out  by the flash chamber are same. Heat taken by the flash chamber = ea/ give / b /he flash 0hambe  $ 2 h4

or $ 2 ( h1



$1

$2

h4  )

h1 h1

$ 1 h f  4   ( $ 2 $ 1 ( h1

h4 h f 4  

$2

$ 1  )h1

h f 4   )

h1

h f  3

h1

h f  4 

 ...( h4

h f  3  )

(.;!

'e know that the heat e"tracted or refrigerating effect,  R E 

$ 1 ( h1

h f  4   )

h1

h f  3

h1

h f  4  

$ 2 ( h1

h f  3  )

$2

(h1

h f 4   )

 ...9 8ro$ e+ation (5.6)7 

%ig. 9.2. p,h diagram of simple saturation cycle ith flash cham3er.

And work done in compressor, W 

$ 2 ( h2

h1  )

( . !

29



C  .O . P  .

$ 2 ( h1

 R E  W 

h f 3  )

$ 2 ( h2

h1  )

h1 h2

h f 3 h1

(./:!

and power reuired to drive the compressor,  P 

$ 2 ( h2 :;

h1  )

kW 

(.//!

)rom above, we see that the refrigerating effect, coefficient of performance and the  power reuired are same as that of a simple saturation cycle when wh en the flash chamber is not used. Thus the use of flash chamber has no effect on the thermodynamic cycle. The only effect resulting from the use of flash chamber is the reduction in the mass of refrigerant flowing through the evaporator and hence the reduction in si+e of evaporator.

9.1# Simple Saturation Cycle ith Accumulator or +re,cooler &ometimes, the liuid refrigerant passing through the evaporator is not completely evaporated. If the compressor is supplied with liuid along with vapour refrigerant, then the compressor has to do an additional work of evaporating and raising the temperature of liuid refrigerant. It will also upset the normal working of the compressor which is meant only for compressing the pure vapour refrigerant.

%ig.9.. Simple saturation cycle ith accumulaor or pre,cooler. pre,cooler.

In order to to avoid this this difficulty, difficulty, an insulated vessel, known as accumulator or  precooler,  preco oler, is used use d in the system, as shown in )ig. ) ig. .. . . The accumulator accu mulator receives rec eives the discharge (a mi"ture of liuid and vapour refrigerant refrigerant ! from the e"pansion valve and supplies the liuid refrigerant only to the evaporator, evaporator, as in the case of flash

2!

chamber. The discharge discharge from the evaporator is sent again again to the accumulator which helps to keep off the liuid from entering the compressor. Thus the accumulator supplies dry and saturated vapour to the compressor. compressor. A liuid pump is provided in the system in order to maintain circulation of the refrigerant in the evaporator.

%ig. 9.!. p-h 9.!. p-h diagram  diagram of simple saturation cycle ith accumulator.

 

9et

m1 3 2ass of liuid refrigerant circulating through the evaporator, m2 3 2ass of refrigerant flowing in the condenser.

'hen all the liuid refrigerant does not evaporate in the evaporator, it is represented by b y point /0 on p on p-h -h diagram as shown in )ig. .6. 9et the mass of refrigerant that leaves the evaporator at point /0 is same i.e., ml . #onsider the thermal euilibrium of the accumulator. &ince the accumulator is an insulated vessel, therefore there is no heat e"change between the accumulator and the atmospher atmosphere. e. In other other words, the heat taken in and given out by the accumulator is eual. 2athematically, 2athematically, Heat taken in by the accumulator = ea/ give / b /he a00mla/ $ 2 h4

$ 1 h1  $ 2 h1

$ 1 ( h1  h f  4   )

$ 2 ( h1

$ 1 h f  4  

h4  )

2!1

$1

h1

$2

h4

$2

h1  h f  4  

h1

h f  3

h1  h f  4  

 

(./%!

'e know that the heat e"tracted or refrigerating effect,  R E 

$ 1 ( h1  h f  4   )

$2

h1

h f  3

h1  h f  4 

 ...9 8ro$ e+ation (5.12)7  $ 2 ( h1

(./! (h1  h f  4   )

h f  3  )

and work done in compressor, W 



C  .O . P  .

$ 2 ( h2

h1  )

(./6!

$ 2 ( h1

 R E  W 

h f  3  )

$ 2 ( h2

h1  )

h1 h2

h f  3 h1

(./
and power reuired to drive the compressor,  P 

$ 2 ( h2 :;

h1  )

kW 

(./F!

)rom above, we see that when the accumulator is used in the system, system, the the refrigerating refrigerating effe effect ct,, coefficient of performance, and power po wer reuired is same as the simpl simplee saturation cycle. The accumulator is used only to protect the liuid refrigerant to flow into the compressor and thus dry compression is always ensured.

9.1" 9.1" Simple Saturation Cycle ith Su3,cooling of 5i6uid Refrigerant 3y Vapour Refrigerant 'e know that the liuid refrigerant leaving the the condenser isat a higher temperature than the vapour refrigerant leaving the evaporator. evaporator. The liuid refrigerant leaving the condenser can be subcooled by passing it through a heat e"changer which is supplied with saturated vapour from the evaporator as shown in )ig. .<. In the heat e"changer, the liuid refrigerant gives heat to the vapour refrigerant. The pThe p-h h diagram of the cycle is shown in )ig. .F.

2!2

%ig. 9.(. Simple saturation cycle ith su3,cooling of li6uid refrigerant 3y vapour refrigerant.

9et m1 3 2ass of the vapour refrigerant,

m 23 2ass of the liuid refrigerant, T 33 Temperature of liuid refrigerant entering the heat e"changer, T 37 3 Temperature of liuid refrigerant leaving the heat e"changer, T 13 Temperature of vapour refrigerant entering the heat e"changer, T 103 Temperature of vapour refrigerant leaving the heat e"changer, 0 pv3 &pecific heat of vapour refrigerant, and 0 pl 3 &pecific heat of liuid refrigerant. #onsidering the thermal euilibrium of heat e"changer, Heat lost by liuid refrigerant= ea/ gaie> b vap efigea/  i.e.

$2

c p&  ( # 3

# 3   )

$1

c p/ ( # 1   # 1  )

2!

%ig. 9.#. p-h diagram of simple saturation cycle ith su3,cooling of li6uid refrigerant 3y vapour refrigerant.

0   &ince m 2  m1 and  pl  is more than 0 pv , ,therefore (T 3 - T 37) will be less than ( T 17 T 1  ). . )or energy balance, h f  3

h f  3   h1  h1

(./E!

In the ideal case, the liuid and vapour refrigerant leave the heat e"changer at the same temperature, say (T m ). 

# 1  # 3   # $

(./;! Gnowing the condition of points / and , the condition of points /0 and 0 can be obtained by trial and error method on p-h chart. A little consideration will show that subcooling increases the refrigerating effect from ( h1  h f 3  ) to ( h1   h f 3 7  )  per kg of refrigerant as compared with the simple saturation cycle. If D tonnes of refrigeration is the load on the evaporator, then the mass of refrigerant ( m * ) reuired to be circulated through the evaporator for sub coole d cycle is given by

2!!

$ R

21; < h1

h f  3  

kg$in

(./!

 >ower reuired to drive the compressor,  P 1

$ R ( h2   h1  ) :;

kW 

h 2   h1 

21; < :;

h1

h  f  3  

kW 

(.%:!

and power reuired to drive the compressor without heat e"changer ( i.e. for simple saturation cyc le !,  P 2

$ R ( h2

h1 )

:; h2

21; < :;

h1

kW  h1 h f  3  

kW 

(.%/!

  4"cess power reuired to drive the compressor as compared to simple saturation cycle,  P e=cess

 P 1

P 2

21; < :;

h2   h1  h1

h f  3 

h2 h1

h1 h f  3

kW 

(.%%!

)rom above, we see that subcooling the liuid refrigerant by vapour  refrigerant, the coefficient of performance of the cycle is reduced. 4ven with theoretical loss resulting from the above type of subcooling, there are many actual installations which adopt this process.

9.1/ Simple Saturation Cycle ith Su3,cooling of 5i6uid Refrigerant 3y 5i6uid Refrigerant 'e know that the liuid refrigerant leaving the condenser is at a higher temperature than the liuid refrigerant leaving the e"pansion valv e. The liuid refrigerant leaving the condenser can be subcooled by passing it through a heat e"changer which is supplied with liuid refrigerant from the e"pansion valve, as shown in )ig. .E. In the heat e"changer, the liuid refrigerant from the condenser gives heat to the liuid refrigerant from the e"p ansion valve. The  p-h diagram of the cycle is shown in )ig..;.

2!(

%ig. 9." Simple saturation cycle ith su3,cooling of li6uid refrigerant 3y li6uid refrigerant.

9et m1   2ass of refrigerant leaving the evaporator, m 2   2ass of liuid refrigerant passing through the condenser, m3   2ass of liuid refrigerant supplied to the heat e"chang er, from the e"pansion valve. #onsidering the thermal euilibrium of the heat e"changer, Heat lost by liuid refrigerant from condenser3Heat gained by liuid refrigerant from e"pansion valve $ 2 ( h f  3 h f  3   ) $ 3 ( h1 h f 4   ) i.e., 

$3

$2

h f  3 h1

h f  3 

$2

h f 4  

h f  3 h1

h f  3   h f 3  

 

(.%!

'e know that refrigerating effect,  R E 

$ 1 ( h1

$2

h f  4   )

$2

( $ 2

h f 3 h1

h f 3  h f 4  

2!#

$ 3  )( h1

(h1

h f 4   )

h f 4   )

(.%6! (.%
$ 2 ( h1

h f 4   )

$ 2 ( h f  3

$ 2 h1

$ 2 h f  4   $ 2 h f  3

$ 2 h1

$ 2 h f  3

$ 2 ( h1

h f  3   )

$ 2 h f  3   h f  3  )

...( m 2 h f 4 7   m 2 h f 3 7  )

and work done in compressor, W 



C  .O . P  .

$ 2 ( h2

h1  )

 R E 

$ 2 ( h1

W 

(.%F!

$ 2 ( h2

h f  3  )

h1

h f  3

h1  )

h2

h1

(.%E!

If D tonnes of refrigeration is the load on the evaporator, then the mass of refrigerant (m ) reuired to be circulated through the evaporator is given by l  $1

$2

or

$3

21; < h1

(.%;!

h f  4  

21; < h1

(.%!

h f 4  

&ubstituting the value f m3 from euation (i), $2

$2

$2 1

$2

h1

h f  3

h f  3 

h1

h f 4  

h f  3

h f  3  

h1

h f 4  

h f  4   h f  3

h f  3  

h1

h f 4  



21; < h1

h f 4  

21; < h1

h f  4  

21; < h1

$2

h f  4   21; <

h1

h f  3

...(  h f 4 7  h f 3 7  )  

(.:!

'e know that power reuired to drive the compressor,  P 

$ 2 ( h2 :;

h1  )

kW 

&ubstituting the value f m 2 the above euation,

2!"

(./!

 P 

21; < :;

h2 h1

h1 h f  3

kW 

(.%!

%ig. 9./. p-h diagram of simple saturation cycle ith su3,cooling of li6uid refrigerant 3y li6uid refrigerant.

9.19 +ro3lems 9.19.1. An ammonia refrigerator works between  .%5 # and 2.%5 #, the vapour being dry at the end of isentropic compression. There is no undercooling of liuid ammonia and the liuid is e"panded through a throttle valve after leaving the condenser . &ketch the cycle on the T-s and p-h diagram and calculate the refrigeration effect per kg of

2!/

ammonia and the theoretical coefficient of performance of the unit with the help of the  properties given below*

 Entha&p'0 kkg 

#e$peratre0 C 

 *i+i d

,apo r

*i+i d

,apo r 

-.%

1&2.1

143%.'3

'.'1

&.43'

2.% 

3'%.1

14%.'3

1.1&1&

&.'2'3

 Entrop'0 kkg % 

KAns. 1'2.3 k=?kg 5 %.2 L 9.19.2. An ammonia refrigerator produces 3' tonnes of ice from and at :B # in 24 hours. The temperature range of the compressor is from 2&5 # to  1&5 #. The vapour is dry saturated at the end of compression and an e"pansion valve is used. Assume a coeficient of performance to be '< of the theoretical value. #alculate the power reuired to drive the compressor. 9atent heat of ice5 3 33& k=? kg. >roperties of ammonia are*

 Entha&p' kkg 

#e$peratre0 C 

 Entrop'0 kkg % 

 *i +i d

,apor

*i+i d

,apor 

2&

29.9

14&.4

1.1242

&.'391

-1&

112.34

142.&4

'.4&%2

&.&49'

KAs. 33.24"wL 9.19.. An ammonia refrigerating machine fitted with an e"pansion valve works  between the temperature limits of  /:B # and :B #. The vapour is < dry at the end of isentropic compression and the fluid leaving the condenser is at  :B #. If the actual coefficient of performance is F: of the theoretical, find the ice  produced per k' hour at :B # from water at /:B #. The latent heat of ice is < k=?kg. The ammonia has the following properties *i +id heat0 kkg 

*a tent heat0 kkg 

 *i+id

,apor

*i+id

,apor 

:

%.:;

//6<.E

/.%:E

6.;6%

/:

/<.E

/%E.F;

:.<66

<.6EE:

#e$peratre0 C 

2!9

 Entrop'0 kkg % 

KAs. 33.24 "g$"8hL 9.19.!. A 8/% refrigerating machine works on vapourcompression cycle. The temperature of refrigerant in the evaporator is  %:B #. The vapour is dry saturated when it enters the compressor and leaves it in a superheated condition. The condenser temperature is : B #. Assuming specific heat at constant pressure for 8 /% in the superheated condition as /.;;6 k=?kg G, determine 1. condition of vapour at the entrance to the condenser 5 2. condition of vapour at the entrance to the evaporator 5 and . theoretical #.$.>. of the machine. The properties of 8 /% are  Entha&p'0 kkg

Entrop'0 kkg % 

#e$peratre0 C 

 *i+id

,apo r

*i+id

,apo r 

%:

/E.;%

/E;.E

:.:E/

:.E:;E

:

F6.<

/.F%

:.%6::

:.F;6

KAs. 33.5 C  29< >  4.'% L 9.19.(.  A #$% refrigerating plant fitted with an e"pansion valve, works between the pressure limits of <6.;/ bar and %:. bar. The vapour is compressed isentropically and leaves the compressor cylinder at %B #. The condensation takes place at /;B # in the condenser and there is no undercooling of the liuid. 7etermine the theoretical coefficient of performance of the plant. The properties of #:% are *   Pressre0  !ar 

 "atration te$peratre0 C 

<6.;/ %:.

 Entha&p'0 kkg

Entrop'0 kkg % 

 *i +id 

,apor

*i+id

,apor  

D/;

/E.6;

:%.<<

:.<:F<

/.:E;

/;

6.%E

%.:F

:./E

/.%F%

KAs. 4.92 L 9.19.#. A single stage NH refrigeration system has cooling capacity of <:: k'. The evaporator and condenser temperatures are  /:B # and :B # respectively. Assuming saturation cycle, determine * /. mass flow rate of refrigerant 5 %. adiabatic discharge temperature 5 . compressor work in k' 5 6. condeser heat re-ection 5 <. #.$.>. 5 and F. compressor swept volume in m?min, if volumetric efficiency is E:. The following values may be taken*

2(

h g ( 1'  C)  1431.  "#$"g h f  ( 3' C)  322. "#$"g v g ( 1'  C)  '.41& m 3  $  "g  -

 s g ( 1'  C)  &.4%1% "#$"g  .

The properties of superheated NH at condenser pressure of //.FF bar (:B #! are as follows* At & C,h  121. "#$"g - s  &.&44 "#$"g   .    At 9'  C, h  134.& "#$"g - s  &.43 "#$"g  KAs. '.4& "g$s  .35 C  9.& "8  &9' "8  &.&&  1.2 m3 $miL 9.19.". The evaporator and condenser temperatures in an NH refrigeration system are  /:B # and 6:4 # respectively. 7etermine per T8 basis * /, mass flow rate 5 %. compressor work 5 . condenser heat re-ection5 6. #.$.>. 5 and <. refrigerating efficiency. Cse only the properties of NH given below*  Pressre (p) !ar 

 "pecific /o&$e of  /apor (/ g  ) $  k g 

 *i + id

,apor

*i + id

,apor 

-1 '

%.:;

:.6/;

/6.<

/6/.6/

:.<6<

<.6E/%

6:

/<.<<

:.:;

E/.6E

/6E%.:%

/.
6.;E%;

 "atra tion te$peratre (t) C 

 Entha&p'0 kkg



Entrop'0 kkg % 

)or superheated  3 at /<.<< bar, the following values may be taken  "perheat   % 

 "pecific /o& $e(/) $  k g 

 Entha& p' (h) k&kg 

 Entrop' (s) k&kg % 

F:

:./:;

/F6E.

<.;;

;:

:.//F

/E::.

<.<%<

3

KAs. '.19 "g$mi  '.2 "8 4.32 "8  4.2% 1.297' L 9.19./. In a vapour compression refrigeration system using 8/%, the evaporator  pressure is /.6 bar and the condenser pressure is ; bar. The refrigeranat leaves the condenser subcooled to :B#. The vapour leaving the evaporator is dry and saturated. The compression process is isentropic. The amount of heat re-ected in the condenser is /.6% 2=?min. 7etermine * /. refrigerating effect in k=?kg 5 %. refrigerating load in T85 . compressor input in k' 5 and 6. #.$.>. &how the cycle on a p-h diagram. KAs. 114 "#$"g  49 T*  &1.4 "8  3.3&.L

2(1

9.19.9. A vapour compression refrigerator works between the temperature limits of  %:B# and %. Assume specific heat of the refrigerant as 6.;. The properties of the refrigerant are

#e$peratre0 C 

 Entha&p'0 kkg

Entrop'0 kkg % 

 *i+id

,apor

*i+id

,apor 

%:

;.E;

/6%:.:%

:.F;6

<.F%66

%<

%;.:

/6F<.;6

/./%6%

<.:/

KAs. 19.% "#$"g  99'.2 "#$"g  &.'1 L 9.19.1. A food storage chamber reuires a refrigeration system of /% T8 capacity with an evaporator temperature of  ;B# and condenser temperature of :B# . The refrigerant 8/% is subcooled by
 Entha&p'0 kkg

Entrop'0 k kg % 

 *i + id

,apor

*i + id

,apor 

;

%;.E:

/;6.:F

:.//6;

:.E::E

:

F6.<

/.F%

:.%6::

:.F;<

 KAns. /:.:
3/:: mm

&troke

3/<: mm

&peed

3%:: r. p. m.

Indicated mean effective pressure

3.% bar  

2(2

#ondenser pressure

3/: bar  

4vaporator pressure

3 bar  

Temperature of water at entry to condenser

3<
Temperature of water at e"it from condenser

3%:B#

8ate of cooling water flowing in the condenser

3/%.< kg?min

Inlet water temperature

3/%.
$utlet water temperature

3 %:.
If the mass of ice produced per hour from water at /<# is <: kg and the latent heat of  ice is < k=?kg, find 5 (a) coefficient of performance 5 (b) mass flow of ammonia per minute 5 and (c! condition of ammonia entering the compressor. The relevant properties of ammonia are given below*  Pressre0 !ar 

 "atration te$peratre0  C 

 Entha&p'0 kkg

"pecific heat0 kkg % 

 *i+id 

,apor

*i+id

,apor  

/%

/

%E.

/6F.<

6.F

%.;

%.

/ :

/<.6

/6





KAs. 4.4  '.32 "g  '. L

9.19.12. A free+er of %: T8 capacity has evaporator and condenser temperatures of  :B# and %
:

 Pressre !ar 

/.::66

 Entrop'0 kkg % 

 Entha&p'0 kkg 

 "pecific /o&$e0 $3 kg 

 *i+id

,apor

*i+id

/apor 

 *i+id

,apor 

;.;F

/E6.%:

:.:E/

:.E/E/

:.::FE

:./<F

2(