44 ■ A Textbook of Refrigeration and Air Conditioning
. , th n 8 Carnot refrigerator used for air-conditiont<t for making ice at 0°C (273 K) will huve Jess C.O.P. 8
• 4ooc In other words, we can say th th )spheri c temperature ts · lit plant in summer at 20°C when _e ut,m_ .
1 than the Carnot C.0 .P. of a domestic air,
the Carnot C.O.P. of a domesuc refrigerator is ess
conditioner.
C t Cycle operates between 305 K and 260 ~ 2 A I · working on a arno · E,arnple 2. . mac ,me .~ . ting machine,· 2. a heal pump; and 3
Dett!rmine tire C.O.P when ;, is operated as: ]. a reJrzgera .
a heat engine.
SoJution. Given : T2 = 305 K; Ti = 260 K
I. C 0. p of O refrigerating machi11e We know that C.O.P. of a refrigerating machine,
Ti - 260 = 5.78 Ans. (C.O.P.)R = r
2 _ J; - 305 - 260
2. C. O.P. of a heat pump
*We know that C.O.P. of a heat pump,
T2 (C.O .P.)p =
Tz - Ti 305
= 305 -260 = 6.78 Am,.
3. C. O.P. of a heat engine
**We know that C.O.P. of a heat engine,
T,. - T. 305 - 260 (c o p ) = 2 I = ---- = 0. )47 \11",.
. . . E T,. 305 2
ExampJe 2.3. A Carnot refrigeration cycle absorbs heat at 270 Kand rejects it at 300 K. I . Calculate the coefficient of performance of this refrigeration cycle.
2. If the cycle is absorbing 1130 kJ/min at 270 K, how many kl of work is required per second?
3. If the Carnot heat pump operates between the same temperatures as the above refrigeration cycle, what is the coefficient of performance ?
4. How many kJlmin will the heat pump deliver at 300 Kif it absorbs 1 J 30 kl/min at 2 70 K. Solution. Given : T, = 270 K ; T
2 = 300 K
1. Coefficient of performance of Carnot refrigeration cycle
We know that coefficient of performance of Carnot refrigeration cycle,
7j 270 (C.O.P.)R = -- = --- = 9 \n..,
T2 - Tj 300 - 270 .
We know that C.O.P. of a heat pump (C O p) - {C O p) , . . . P - . • • R + I ::: 5.78 + I = 6.78 A ns.
We know th t C O p a • .. of a heat engine, (C.O. P. )E == ____ I
(C.O.P.)p =~ = 0.147 Ans.
z. 1
3.
Cll'-'lir It llrltll&HIIII t,1111 ■ • 2.-''qaM,_,.,_
-'· c~"' of ~rjon,,anc, of Carnot INtll ,,.., We know that coefficient of perfonnancc of a Carnot heat pump,
7i 300 (C.O.P.)p = r. = --- = 10 Ans. 2-1i 300-270
4. ff,ol ,uUt11rtd by heat pump at 300 K Let Q2 = Heat delivered by heat pump at 300 K.
Heat absorbed at 270 K (i.e. T1),
Q, = 1130 kJ/min ... (Given)
We know that
Ql 0"' 10 -- Q2 (C.O.P.)p = 'I
Q2 - QI Q2 - 1130
10Q2
- 11 300 = Q2 or Q2 = 1256 kJ/min An~. IJMplt 2.4. A cold storage is to be maintaintd at -5°C while w s11rro,wli,ag1 a,r at
iJS°C. 77te 1teat leakagt from the surroundings into the cold storage is estimaud to be 29 UV. ~ «-1 C.0.P. of tht refrigeration plant is one-third of an itual plant worling bttwtn tM SIJIM
ltlllf#rtllll~S. Find the power required to drfre the plant.
Solution. Given : rl = -5°C = -5 + 273 = 268 K;
T2
:::: 35°C = 35 + 273 = 308 K ; Q1 = 29 kW ;
l ~ (C.O.P.)..,_ = 3(C.O.P.)itkoi
The refrigerating plant operating between the temperatures T1 WR -~ R Refrigerating
and T2
is shown in Fig. 2.3. plant Let WR = Work or power requir~d to drive the plant We know that the coefficient of performance of an ideal
4o • A Textbook of Refrigeration and Air Conditioning
2.4 Difference Between a Heat Engine, Refrigerator anct
Heat Pump ln h . 'ed h ~ engine is converted into
a eat engme, as shown in Fig. 2. 1 (a). the heal suppli to l c . f h •
useful w k If . . h heat reJected rom t e eng1nc
or · Q2 ts the heat supplied to the engine and Q1 1s t e '
then the net work done by the engine is given by
Hot body ....._...,.__, (High temp.)
Heat engine
a,
WE = Q2 - QI
Refrigerator
r, Cold body '---~ (Low temp.) ~ T ~< ~ ,, > a
Hot bOdy Hot body
Heat pump
Cold body Cold body
(c)
(a) (b)
Fig. 2.1. Difference between a heat engine, refrigerator and heat pump. .
The performance of a heat engine is expressed by its efficiency. We know that the efficiency
or coefficient of performance of an engine,
Work done _ WE = "E or (C.O.P.)E = 1. d - Q •1
• Heat supp 1e 2
A refrigerator as shown in Fig. 2. 1 (b), is a reversed he_a t engi ne which eithe~ ~ool or
maintain the temperature of a body (T1
) lower than the atmospheric temperature (T0 ) . :n,1s ts done
by extracting the heat (Q ) from a cold body and del ivering it to a hot body (Q2). In domg so, work I
•
WR is required to be done on the system. Accordi ng to First Law of Thermodynam1cs,
WR = Q2 - Q,
The performance of a refrigerator is expressed by the ratio of amount of heat taken from the
cold body (Q1) to the amount of work required to be done on the system ( WR). This ratio is called
coefficient of performance. Mathematically, coefficient of performance of a refrigerator,
Q, Q, (C.O.P. )R = - = --
WR Q2 - Q,
Any refrigerating system is a heat pump as shown in Fig. 2. 1 (c). which extracts heat (Q1)
from a cold body and delivers it to a hot body. Thus there is no difference between the cycle of
operations of a heat pump and a refrigerator. The main difference between the two is in their
operating temperatures. A refrigerator works between the cold body temperature (T,) and the
atmospheric temperature (T) whereas the heal pump operates between the hot body temperature
C'i ) and the atmospheric temperature (Ta>, A refrigerator used for cooling in summer can be used
as a heat pump for heating in winter.
In the simi lar way, as discussed for refrigerator. we have
wr = Q2 - Q,
Ctmptor l ; Air Refrlg•ratlon Cyclt • • 41
,i1,,nnt11h.'l' nl It hi.' 111 pur11p ,~ l1xp,e,!l{'d 1 1 lhl' ll{f (n) 10 the omount ol wo,~ H'llll l. • l >) ,I It· 1111111 ol lht· 11 1111111111 t, f lwa, ,1,•ll vc·u.,t
I p(ll' v, •u In w douc < r ti ,, iftl' hl' 1.1 ii.:nt ,)f pc1'fonnnt1~t' or <'llN11y ll . j '
b ve we; sec that the C.O.P. may he lcs , II proni a O . , . , , , , • s 1an one or greater thun one depending on
f -efrigerat1on system ust:d. But the C.O.P. of a hear pump i 1 t'ine o I s a wuys greater than one.
then·-open Air Refrigeration Cycle
2,d · f · · tion · cl th · · d' In an open air re n_gera cy e, e uir ,s rrec tly led to lhe space to be cooled (i.e. a
) al lowed to circulate through the cooler and then returned to the comp . i:..icrerator • · . . , . , . ressor to start
reu~ cle Since the air ts supplted to the re1 ngerator at atmospheric pressure th f (her cy ' , , a I ere ore,
aOO of air handled by the compr~ssor and expa nder is large. Thus the size of compressor and
volutne hould be large. Another disadvantage of the open cycle system is that the moisture is
eipandt :arried away by the ai r ci rculared through the cooled space. This leads to the formation
regular y t the end of expansion process and clog the line. Thus in an open cycle system a drier of (rost a
,
should be used.
2•6 Closed or Dense Air Refrigeration Cycle
In a closed or dense air refrigerati on cycle, the air is passed through the pipes and
nent parts of the system at all times. The air, in this system, is used for absorbing heat from
compo d h. I d b . . . l d . the otller fluid (say brine) an t 1s c?o e nne _,s c1rcu ~te mto the space to be cooled. The air
in the closed system does not come 111 contac t directly with the space to be cooled.
The closed air refrigeration cyc le has the fo llowing thermodynamic advantages :
1, Since it can work at a suction pressure higher than that of atmospheric pressure.
therefore the volume of ai r handled by the compressor and expander are smaller as
compared to an open air refrigerati on cycle system.
2. The operating pressure ratio can be reduced, which results in higher coefficient of
performance.
2.7 Air Refrigerator Working on Reversed Carnot Cycle
In refrigerating systems, the Carnot cycle considered is the reversed Carnot cycle. We know
that a heat engine worki ng on Carnot cyc le has the highest possible efficiency. Similarly, ti
refrigerating system working on the reversed Carnot cyc le. wi II have the maximum possible
coefficient of performance. We also know that it is not possible to make an engi ne working on the
Carnot cycle. Similarly, it is also not possi hlc to make a refrigerating machine working on th?
reverse{! Carnot cycle. However, it is used as the ultimate standard of comparison.
r. . A reversed Carnot cycle, usi ng air as working medium (or refrigerant) is shown on p-v and
·s diagram . F' I
an<1 te s m ,g. 2.2(a) and (b) respccrively. At point I, let p 1• ,· 1
• T1 be the pressure.vo ume
mperature of air respectively.
42 ll A Textbook of Refrigeration and Air Conditioning
i ~ lsen. exp.
l T,• T,
2
Wso 3
l I ' • "o ~
! P-i I ~/J. 2 ~ i i
, , ....
I ! ! r4 lsen. f: T1 := T4 4 0.. p• Cl)
I comp. t-ls0 ,
I 2' I ' &,t l : 3' P1 --,-1 lJ,1 , 1
51 = 5-1 I I 54 = 53
V3 V4 V2 v, Entropy Volume
(b) T-s diagram. (a) p-v diagram.
F. 2 2 Reversed Ca rnot cycle. 1g . ..
f h le are as follows : b h The four processes o t e eye . . d . ntropically as shown y t e . . . , The alf is compresse ise
t. /se11tropic compre.n 1011 p,m~s\. . the ressure of air increases from Pi to P2, curve 1-2 onp-v and T-s diagrams. Dunng this procests, e i:creases from T1 to Tz· We know that
d from v to v and tempera ur specific volum~ ecreases . J ~ absorbed or rejected by the air. during isentrop,c compress10n, no heat ,s . . w com ressed isothermally (i.e. at
2. l wthermal compre\·sio11 proce~·s. The air is2 n3o :nd T-s diagrams. During this t T - T ) as shown by the curve - on p-v
constant tempera ure, f 2 .- . 3 from p top and specific volume decreases from Vz to V3· We process, the pressure o atr increases . 2 • J . f . know that the heat rejected by the air dunng isothermal compression per kg o air,
q -q = Area 2-3-3'-2' R - 2 - J
= T3 (s2 - s) = T2 (s2 - s3)
J . fsentropic expansion proceH. The air is now expanded isentropically as shown by the curve 3-4 on p-v and T-s diagrams. The pressure of air decreases from p3 to p4, specific volume increases from v3 to v4 and the temperature decreases from T3 to T4• We know that during isentropic expansion, no heat is absorbed or rejected by the air.
4. l wtltermal expau!J io11 proceu. The air is now expanded isothermally (i. e. at constant temperature, T4 = Ti) as shown by the curve 4-1 on p-v and T-s diagrams. The pressure of air decreases from p4 to pl' and specific volume increases from v
4 to v
1. We know that the heat
absorbed by the air (or heat extracted from the cold body) during isothermal expansion per kg of air,
qA = q4 _ 1 = Area 4- 1- 2'- 3'
= T4 (s 1 - s4) = T4 (s2 - sJ) = T1 (s
2 - s
1)
We know that work done during the cycle per kg of ai r
w H = Heat rejected - Heal absorhed = q _ q = 0 _ q • R A r: ~ .J - 1 lli
In a refrigerating machine, heal rejected is more than heat absorbed.
Chapter 2 ; Air Refrigeration Cycle, 43
:: --' I ( .\ 1 - \ 1 ) ::: '/ L_
(7 , - r )( , > 1· 1 .. I ., 2 . \ l .' - I
1-i-ough 1he reversed Carnot cycle is the 111 > 1 ,·,·· 11 • t !'I c I1: Ien1 betwe th fi d 1· ·
frigera lor has been made us.ing th • , ·I . en e 1xe temperature 1m1ts •
. t 110 re . is 1:yc e. This I\ due to the ~, ' th t th · ·
)e f the cycle require high speed h .1 h . reason a e 1sentrop1c
ocesses o . w , et e isothermal processes req . t I l
pr ed. This variation in speed of air is not practicable. . u1re an ex reme y ow
spt . We have already discussed that C.O.P. of a heat p
~ 01e . ump,
(C.O.P.)p = (C.O.P.)R + I = __Ii_+ l = __!]__ ~-Ti Ti - Ti
C o.P. or efficiency of a heat engi ne, and ·
(C.O.P. )E = wR = (T2 - Ti )(s2 - s3) = T2 - Ti - 1
qR T2(s2 -s3) T2
- (C.O.P.)p
2•8 Temperature Limitations for Reversed Carnot Cycle
We have seen in the previous article that the C.O.P. of the refrigeration system working on
..cPd Carnot cycle is given by revev--
(C.O.P.)R = 7;
where r, = Lower temperature, and
T2 = Higher temperature.
The C.O.P. of the reversed Carnot cycle may be improved by
1. decreasing the higher temperature (i.e. temperature of hot body, T2), or
2. increasing the lower temperature (i. e. temperature of cold body, Ti).
This applies to all refrigerating machines, both theoretical and practical. It may be noted that
temperatures T1 and T2 cannot be varied at will , due to
certain functional limi tations. It should be kept in mind
that the higher temperature (T2) is the temperatu re of
cooling water or air available for rejection of heat and the
lower temperature (T1
) is the temperature to be
maintained in the refrigerator. The heat transfer will take
place in the right direction only when the higher
temperature is more than the temperature of cooling water
or air to which heat is to be rejected, while the lower
temperature must be less than the tem perature of
~ub~tance to be cooled.
Thus, if the temperature of cooling water or air (i.e.
T2) available for heat rejection is low. the C.O. P. of the
Carnot refrigerator will be high. Since T, in winter is les~
than r · -h' 2 in summer. therefore, C.O. P. in winter wi ll be
igf~er than C.O.P. in summer. In other words, the Carnot
re ngerators k . . . . . . t . wor more etf1c1en1l y 111 winter I u 111 111
surnmer s· . . .. ref . · . 1m1larly, 1f the lower temperature fix ed by th~
ngerat1on 1 · . refn app 1cat1on is high, the C.O.P. of the Carnot
gerator will be high. Thus a Carnot refrigerator used
,.
•
Dorncst1c A,r Cond1t10m•r
~
I
I I ~
cHAprER
Simple Vanour Compression Refrigeration svstems
4~ troduction A vapour compression refrigeration
S}Stem is an improved type of air refrigeration
sy~tem in which a suitable working substance.
termed as refrigerant, is used. It condenses
and evaporates at temperatures and pressures
close to the atmospheric conditions. The
refrigerants, usually, used for this purpose are
ammonia (NH3), carbon dioxide (CO~) and
sulphur dioxide (S02
). The refrigerant used,
does not leave the system. but is circulated
Since low pressure vapour refrigerant from
the evaporator is changed into high pressure
vapour refrigerant in the compressor,
therefore it is named ru, vapour compression Ii ~ . e ngeration system.
1
2
3
4
Introduction.
Advantages and Disadvantages of Vapour
Compression Refrigeration System over Air
Refrigeration System.
Mechanism of a Simple Vapour
Compression Refrigeration System.
Pressure-Enthalpy ( p-h) Chart.
S. Types of Vapour Compression Cycles.
6 Theoretical Vapour Compression Cycle with
Dry Saturated Vapour after Compression.
7 Theoretical Vapour Compression Cycle with
Wet Vapour after Compression.
8. Theoretical Vapour Compression Cycle with
Superheated Vapour after Compression.
! . Theoretical Vapour Compression Cycle with
Superheated Vapour before Compression.
10 Theoretical Vapour Compression Cycle with
Undercooling or Subcooling of Refrigerant.
1 Actual Vapour Compression Cycle.
12 Effect of Suction Pressure.
13 Effect of Discharge Pressure.
1-1 Improvements in Simple Saturation Cycle.
15 Simple Saturation Cycle with Flash
Chamber.
l ~ Simple Saturation Cycle with Accumulator
or Precooler.
17 Simple Saturation Cycle with Subcooling of
Liquid Refrigerant by Vapour Refrigerant.
lF Simple Saturation Cycle with Subcooling of
Liquid Refrigerant by Liquid Refrigerant.
\'
121 G A Textbook of Refrigeration and Air conditioning
throughout the system alternately condensing and evaporating. In evaporating, the refrigerant absorbs its latent heat from the Heal in brine (salt water) which is used for circulating it around the cold chamber. While condensing, it gives out its latent beat lo the circulating water of the coo!er. The vapour compressron refrigeration system is, theref~re a latent heat pump, as it p~mps its latent heat from the bnne and delivers it to the cooler.
The vapour compression refrigeration system is now-adays used for all purpose refrigeration. It is generally used for all industrial purposes from a small domestic refrigerator to a
big air conditioning plant. d . 1834 by Jacob Perkins using hand ope . · t was develope in ra1ton
1\ ote: The first vapour cornpress10n sys em · .
d Disadvantages of Vapour Compression 4 .2 RAd;'~ntagt~snaSnystem over Air Refrigeration System
e, r1gera 10 . . . d disadvantages of the vapour compression refngeration
Following are the advantages an system over air refrigeration system :
td, llllfll),;l' . .
I. It has smaller size for the given capacity of refngeratt0n.
2. It has less running cost. 3. It can be employed over a large range of temperatures. 4. The coefficient of performance is quite high.
J)iwd, anlagn
I . The initial cost is high. 2. The prevention of leakage of the refrigerant is the major problem in vapour compression
system.
4.3 Mechanism of a Simple Vapour Compression Refrigeration System
Fig. 4.1 shows the schematic diagram of a si mple vapour compression refrigeration system.
Ct consis~f the fo llowing five essential parts :
vf.' Compre\·sor. The low pressure and temperature vapour refrigerant from evaporator is drawn into the compressor through the in let or suction valve A, where it is compressed to a high pressure and temperature. This high pressure and temperature vapour refrigerant is discharged into the condenser through the delivery or discharge valve B.
v1. CmulnHtr. The condenser or cooler consists of coil s of pi pe in whil: h the high pres!-~re and temperature vapour refrigerant is cooled and condensed . The refrigerant, while passin_g through the condenser, gives up its latent heat to the surrounding condensi ng medium which 15
normally air or water.
>i< Brine is used as it has a very low freezi ng temperature.
J. Receiver. The co~densed_ liquid refrigerant from the condenser is s~red in a vessel ~own . r from where it 1s supplied to the evaporator through the expansion valve or refngerant
as receive control valve.
µ xpansion valve. It is also call:d ~ottle_ valve or refrig~rant control valve. The function
f th expansion valve is to allow the hqmd refngerant under high pressure and temperature to 0
:ta controlled rate after reducing its pressure and temperature. Some of the liquid refrigerant pass h . l b h . . "sed . th evaporates as it passes through t e expansion va ve, ut t e greater portion is vapon m e
evaporator at the low pressure and temperature.
0 Evaporator. An evaporator consists of coils of pipe in which the liquid-vapour refrigerant at low pressure and temperature is evaporated and changed into vapour refrigerant at low pressure and temperature. In evaporating, the liquid vapour refrigerant absorbs its latent heat of vaporisation
from the medium (air, water or brine) which is to be cooled.
Note: In any compression refrigeration system, there are two different pressure conditions. One is called the high presmre .1ide and other is known as !011 pressure .1ide. The high pressure side includes the discharge line (i.e. piping from delivery valve B to the condenser), condenser, receiver and expansion valve. The low pressure side includes the evaporator, piping from the expansion valve to the evaporator and the suction line (i.e. piping from the evaporator to the suction valve A).
4.4 Pressure-Enthalpy (p-h) Chart
. The most convenient chart for studying the behaviour of a refrigerant is the p-h chart, in
wtuch the vertical ordinates represent pressure and horizontal ordinates represent enthalpy (i.t.
: heat). A typical chart is shown in Fig. 4.2, in which a few important lines of the complete
the ~drawn.The saturated liquid line and the saturated vapour line merge into one anodler at -::•I point. A saturated liquid is one which has a temperature equal to the saturation lherefore : corresponding to its pressure.The space to the left of the saturated liquid line will. let v~ su_b-cooled liquid region. The space between the liquid and the vapour lines is called
region and to the right of the saturated vapour line is a superheated vapour repon.
11%1•
d Air conditioning . b k of Refrigeration an
128 ■ A Text 00 h hart along with the T-s diagram . e shall draw the p- c . Of the
f Vapour Compression Cycles . 4 .5 Types O h ur compression cycle essentially consists of
We have already discussed1. t at dvap:aporation Many scient ists have focussed their
· densation thrott mg an e · compress10n, con ' . f rformance of the cycle. Though there are many cycles, yet attention to increase the coefficient O pe . .
b' t of view · the following are important from the su ~ect porn .. I. Cycle with dry saturated vapour after compression,
2_ Cycle with wet vapour after compression, 3. Cycle with superheated vapour after compression, 4. Cycle with superheated vapour before compression, and
5. Cycle wi th undercooling or subcooling of refrigerant. Now we shall discuss all the above mentioned cycles, one by one, in the following pages.
4.6 Theoretical Vapour Compression Cycle with Dry Saturated Vapour after Compression
A vapour compression cycle with dry saturated vapour after compression is shown on T-5
and p-h diagrams in Fig. 4.3 (a) and (b) respectively. At point I , let T1, p
1 and sl' be tbe
temperature, pressure and entropy of the vapour refrigerant respecti vely. The four processes of the cycle are as follows :
, l fanpressi~n process . The vapour refrigerant at low pressure p1
and temperatu~ 71 ~ compr(ssed isentropically to dry saturated vapour as shown by the vertical line 1-2 on T-s diagram and by th~ curve l-2 on p-h diagram. The pressure and temperature ri ses from p to P2 and T110 r2 respectively. I
The work done during isentropic compression per k t· f · · · by go re ngerant 1s given w = h, - h
/.ensing process. The high pressure and temperature vapour --1. • fr: ),/40~11~ . . nangerant om the ---c.W- is passed through the condenser where 1t 1s completely condensed at
CIJll'f" "'-- . h b h h . . COllltant preaure and temperature T2,_ as s. o~n Y ~ e onzontal h~e 2-3 on T-s and p-h diagrams. The vapour
~gerant is changed into hqu1d re~ngerant. Th~ refngerant, while passing through the condenser, pves its latent heat to the surrounding condensing medium .
• 4 nsion process. The liquid refrigerant at pressure p = p and •"""-h•- T T • ~ ~I"- 3 2 --....--.... 3:: 2 11
expanded by • throttling process through the expansion valve to a low pressure p4
= Pi and remperature r4 = T" as shown _by the curve 3-4 on T-s diagram and by the vertical line 3-4 00 ,-A diagram. We have already discussed that some of the liquid refrigerant evaporates as it passes dll'ough the expansion valve, but the greater portion is vaporised in the evaporator. We know that !hiring the throttling process, no heat is absorbed or rejected by the liquid refrigerant.
Neta:(,) Jo case an expansion cylinder is used in place of throttle or expansion vaJve to expand the liquid lffligennt, then the refrigerant will expand isentropically as shown by dotted vertjcal line on T-s diagnm i1 Fig. 4.3 (a). The isentropic expan\ion reduces the external work being expanded in running the ~ and increases the refrigerating effect. Thus, the net result of using the expansion cylinder is to
icraae the coefficient of performance.
Since the expansion cylinder system of expanding the liquid refrigerant is quite complicated and hives greater initial cost, therefore its use is not justified for small gain in cooling capacity. Moreover, die flow rate of the refrigerant can be controlled with throttle valve which is not possible in case of Gpllaioii cylinder which has a fixed cylinder volume.
f ) In modern domestic refrigerators, a capillary (small bore tube) is used in place Man ex..-,n Vlhe.
~ rising process. The liquid-vapour mixture of the refrigerant at pressure P, = Pa and -..=me T4 = r, is evaporated and changed into vapour refrigerant at constant preuun, • :. •e. as sh?wn by the horizontal line 4-1 on T-s an~ p-~ diagrams. Durin~ ev~ die liliia} ~ refngerant absorbs its latent heat of vaponsat1on fro!° the ~um (m, .., • .:-.18 t~ be cooled. This heat which is absorbed by~ refrig~t 15 ~ tr/rllt~ ...... • 11.briefly written as RE. The process of vaporisatJon conbnues UplO poa l wllic:II •
PGint and thus the cycle is completed. .. .. ii t
Ii PfOCess is an irreversible process.
' .. , ••.. 4' .... , ........ ,u ..... dll adll effect or the hell .-.bed or extraetcd by the • .,.. ..... ~ ~-"~tit is given by NIia ., .... mpuadoll per.._ nanr-~ = ,., - "· = ,., - "'3 ... (-.• .,, •
1a13
= Sensible heat at temperature T3, i.e. enthalpy of refrigerant leaving the condenser.
k be noticed from the cycle that the liquid-vapour refrigerant has extracted heat~ ....: ..i mo work will be done by the compressor for isentroPic compression of die liiiii preame and temperature vapour refrigerant.
:. Coefficient of perfonnance, Refrigerating effect h1 - h4 _ hi - h13
C.O.P. = Work done = lti - h1 - hi - hi Nott: The ratio of C.O.P. of vapour compression cycle to the C.O.P. of Carnot cycle is known 11
re.frigaatio11 efficiency (1lR) or performance index (P.l.). Eumple 4.1. In an ammonia vapour compression system, the pressure in the evapo,_,
ii 2 bar. Ammonia al ail is 0.85 dry and at entry its dryness fraction is 0. 19. During comprr--. ,,w· worl doM ptr kg of ammonia is 150 Id. Calculate the C. O.P. and the volume of V"POI, •ring the comprrssor per minute, if the rate of ammonia circulation is 4.5 kg/min. The latew ..,_ sptcific volume at 2 bar are 1325 Id/kg and 0.58 m3/kg respectively.
c.o.r. The T-s and p-h diagrams are shown in Fig. 4.3 (a) and (b) respectively. Since the ammonia vapour at entry to the evaporator (i.e. at point 4) has dryness fraction
(x4) equal to 0.19, therefore enthalpy at point 4,
h4 = X4 X ",g = 0.19 X 1325 = 251.75 kJ/kg
Similarly, enthalpy of ammonia vapour at exit i.e. at point 1,
hi = x, x ",g = 0.85 x 1325 = 1126.25 kJ/kg
:. Heat extracted from the evaporator or refrigerating effect,
RE = h1 - h4 = 1126.25 -151.75 = 874.5 kJ/kg
We know that work done during compression, w = 150 kJ/kg
:. C.O.P. = REI w = 874.5 / 150 = 5.83 Ans. Volumt of vapour mitring lht compressor per mi,iult
We know that volume of vapour entering the compressor per minute = Mass of refrigerant / min x Specific volume = max vg = 4.5 x 0.58 = 2.61 ml/min Ans.
-JO"C. Example 4~2. The temperature limits of an ammonia refrigerating system art i,•c qe# .:;::asn:,s dry at the_ end of co~ pr~ssion, calculate the coefficient of perf onnallU a •• : g undercoolmg of the liquid ammonia. Use the following table for p
~IIIIIW Liquid heat Latent heat
re, (kl/kg) (kl/kg)
25 298.9 1166.94
-ltl 135.37 1297.68
J
f
as
Or
If! ,
11 ,
flt
Chapter 4 : Simple Vapo.,, Comp . ress,on R f .
. . e nge,ation S
.~11luri1HI, Given . Tl= T3 = 2soc:::: 25 + 273 - Y•tems SJ 131
_ I ::: J,4 = 298.9 kJ/kg ; h == 1166 9 - 298 K . T _ _
1 ,.. : l;i 68 kJ/k . fg'l • 4 kJ!k . , I - T4::: - Joo C
:ti I == 1297. g , s11 == 0.5443 kJn. g , s/2 .::: l 1242 ::: - IO+ 273 -
~i! . 'r~ , . , "'g K ~ · kJtk K . -
~]1 • The T-s and p-h diagrams are shown in p· g ' ht, ::: 135.37
1g. 4.4 (a)
I.,et xi == Dryness frac . and (b) respectiv I
We kJlOW that entropy at point J, tion at point 1. e y.
- s + x, h/gl
s, - f l T==0.5443+~
= 0.5443 + 4.934 X 263
Similarly, entropy at point 2, ,
hfg ' S2 = s/2 +T==l. I242+~ 66.9_i _
2 298 -5.04 ..
Since the entropy at point 1 is equal to entro . · · · (u)
. PY at point 2, therefore .
and (11 ),
equating equations (i)
0.5443 + 4.934 x, = 5.04 or x, ::: 0.91
.. . (i)
1; t 298 g
3-+-------1 2 i /
3 /
lO
I, ~ I
:, I
~ I
8263 ;.- - +
~ f._ 1 4 1
I s,, : s, = ., -+I \
~ - Entropy --..
(a) T-s diagram.
We know that enthalpy at point 1. f ig. 4.4
I I
I I
I
I I
I I
h, --., I
~ --i':....._---L,I h I
I I " 2 ~
hf3 = h4
- Enthalpy--►
( b) p-h diagram.
and hi = h,, + x, hfrtl = 135.37 + 0.91 x 1297.68 = 131 6.26 kl/kg
enthalpy at point 2, h2 = hi 2 +hf,(!.·= 298.9 + 11 66.94:: 1465.84 kl/kg
:. Coefficient of performance of the cycl~
h1 -hn 1316.26-298.9
= h, -h1 1465.84-1316.26 = 6·8 Anli.
bar ~ ~ample -l.J. A vapour compressio~ ref rigerator works between the pressure limits of 60
co0/i 5
bar. nze working flu id is just dry at the end of compression and there is no u11der-
1tg of the /' ·d b 2. Ca,,ac . iqui efo re the expansion valve. Determine : I. C. O.P. of the cycle : and
,.. uy of the refrigerator if the fluid flow is at the rate of 5 kg/min.
