Lecture note refrigeration cycle

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Refrigeration Cycle

12/11/2013Refrigeration Cycle 1

Chapter Objective

To carry out first law analysis on a refrigeration

cycle in which the working fluid undergoes

changes of phase as it completes the cycle.


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What is Refrigerator?

� A reversed heat engine.

� Absorb heat QL from a low-temperature medium

and reject the heat QH to a high-temperature

medium.

� A working fluid called refrigerant flows through

components of the refrigerator, forming a

thermodynamic cycle called refrigeration cycle.

� To perform the heat absorption and rejection

processes, the refrigerator requires a work input,

W net, in.


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Environment

Reverse

Heat

Engine

COLD

refrigerated

space

W net, in = required input

QL = desired output

QH

WQQWQQ LHHL =−→−=−

LH

LLref

QQ

Q

W

QCOP

−==


Total heat transfer to a cycle = Total work done by the cycle

∑ ∑= WQ

Coefficient of performance of a refrigerator,

(1)

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The most efficient and

ideal refrigerator is the

one in which the

refrigerant undergoes a

reversed Carnot cycle

while working between

2 specified

temperatures.

Q23

condenser

engine compressor

evaporator

1

3 2

4

W12W34

Q41

Components of a refrigerator working on

reversed Carnot cycle


Q41

Q23

W

s

T

T2 = T3

T4 = T1

s4 =s3 s1 =s2

14

3 2

T-s diagram for a refrigerator

working on reversed Carnot cycle

Q23

condenser

engine compressor

evaporator

1

3 2

4

W12W34

Q41

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The processes shown on the T-s

diagram are as follows:

1-2 Wet vapor at state 1 enters

the compressor and is

compressed isentropically to

state 2. The required input is

denoted W12.

2-3 The vapor at state 2 and is

condensed at constant

pressure and temperature to

state 3, when it is completely

liquid. The heat rejected from

the refrigerant is denoted by

Q23.

Q41

Q23

W

s

T

T2 = T3

T4 = T1

s4 =s3 s1 =s2

14

3 2


3-4 The liquid refrigerant at state 3

expands isentropically behind

the piston of an engine, doing

work of amount denoted by

W34.

4-1 The refrigerant at state 4 enters

the evaporator where it absorb

heat denoted by Q41 from the

cold space, until it reaches state

1.

Q41

Q23

W

s

T

T2 = T3

T4 = T1

s4 =s3 s1 =s2

14

3 2

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4123

41

QQ

Q

W

QCOP L

ref−

==∑

dsTdQT

dQds =→=


Cycle’s Performance

The performance of the cycle is

measured by the Coefficient of

Performance (COP), defined as,

Q41

Q23

W

s

T

T2 = T3

T4 = T1

s4 =s3 s1 =s2

14

3 2

The definition of entropy change is,

( )41141 ssTQ −=∴

( )32223 ssTQ −=

( ) ( )3241 ssss −=−


Q41

Q23

W

s

T

T2 = T3

T4 = T1

s4 =s3 s1 =s2

14

3 2

and

and

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( )( )( )4112

411

ssTT

ssTCOPref

−−

−=∴

( )12

1

TT

TCOPref

−=


Q41

Q23

W

s

T

T2 = T3

T4 = T1

s4 =s3 s1 =s2

14

3 2

i.e.

(2)

Note: - where T1 = evaporator temperature

and T2 = condenser temperature

- only valid for reversed Carnot

cycle (for refrigerator)

- COP value is the maximum value

for a cycle

A vapor-compression refrigeration system operates on a

Carnot cycle, uses R-12 as the working fluid. The refrigerant

is dry saturated at entry to the condenser at 28°C, and

saturated liquid leaving the condenser. The evaporator

temperature is -10°C. Determine,

(a) The compressor input work (kJ/kg);

(b) The heat transfer in the evaporator (kJ/kg);

(c) The COP

Show the cycle on a T-s diagram.


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Q41

Q23

W

s

T

T2 = T3 = 28oC

T4 = T1 = -10oC

s4 =s3 s1 =s2

14

3 2


Main components of

a household refrigerator

Evaporator coils

Freezer

compartment

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Q23

condenser

engine compressor

evaporator

1

3 2

4

W12W34

Q41

T2

T1

T2’

T1’T2’ = T atmosphere

T1’ = T cold space

T2’ < T 2

T1’ > T 1

Therefore, Q23 can be

transferred from the

refrigerant to the atmosphere

and Q41 can be transferred

from the refrigerated space to

the refrigerant.


T2’T2’ < T 2

Q41

Q23

s

T

T2

T11

4

3 2

T1’T1’ > T 1

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Reversed Carnot cycle is not

practical since,

• All the reversible processes

cannot be achieved.

• Process 1-2 is bad for

compressor – low efficiency and

two phase compression is

impractical.

• It is difficult to stop evaporation

process at point 1 and

compressed it just to state 2.

To get practical cycle, modification

to the reversed Carnot cycle is

made.

Q41

Q23

W

s

T

T2 = T3

T4 = T1

s4 =s3 s1 =s2

14

3 2


Carnot Cycle/

Ideal

Replacing the turbine

with a throttling

Valve

(Ideal/actual)

Throttling valve

+

Dry saturated or

superheated vapor at

entry to the

Compressor

(Ideal/actual)

Throttling valve

+

Dry saturated or

superheated vapor

at

entry to the

compressor

+

Under-cooling at

the condenser exit

(Ideal/actual)

Two or more

throttling valves

+

Flash chamber

+

2-stage

compression

(Ideal/actual)

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Process 3-4 is replaced with a throttle valve.

s

Replacement of expansion cylinder with a throttle valve

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2

Q in

Wc

Expansion

valve

Evaporator

Condenser

Compressor

Q out

Saturated or

superheated vapor

1

23

4 The expansion of the liquid refrigerant

through the throttle valve is represented

by a dashed line 3-4 since the process is

not reversible.


s

4-1 Constant Pressure

Evaporation

Heat from a cold space is

absorbed by the refrigerant.

As a result, the refrigerant

evaporates at a constant

evaporator pressure, from

state 4 to become a drier

liquid-vapor mixture at state 1.

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2

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s

1-2 Isentropic compression

The liquid-vapor mixture is

compressed from the

evaporator pressure to the

condenser pressure, in a

reversible adiabatic manner.

The refrigerant exits the

compressor as a saturated

vapor at state 2.

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2


s

2-3 Constant Pressure

Condensation

Heat is rejected from the

refrigerant to a warm space. As a

result, the refrigerant condenses

at a constant condenser

pressure until it becomes a

saturated liquid at state 3.

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2

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s

3-4 Constant Enthalpy

Expansion

The refrigerant expands

through the throttle valve

adiabatically. As a result, it’s

pressure drops from the

condenser to the evaporator

pressure. The enthalpy is

constant during the process,

i.e. h3 = h4.

Note: The expansion process is highly irreversible, thus making

the vapor-compression cycle an irreversible cycle.

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2


s

Effects of using an expansion

valve

i) A throttling process is an

expansion process with

constant enthalpy, (h3 = h4).

Therefore, no work done by

the cycle, i.e. w34 = 0.

ii) The refrigeration effect, q41

decreases.

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2

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s

Analysis of the Cycle

The cycle operates as steady flow.

So, each component of the vapor-

compression refrigeration cycle

applies the steady flow energy

equation (SFEE) to analyze the

energy interaction. The changes in

the kinetic and potential energy are

ignored.

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2

1212 hhw −=

4141 hhq −=


s

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2


Evaporation process (4-1):

(3)

(Refrigerating effect /

heat absorbed by the refrigerant)

Compression process (1-2):

(4)

(input work to the compressor)

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43 hh =

3223 hhq −=


s


Condensation process (2-3):

(5)

(heat rejected from the refrigerant)

Expansion process (3-4):

(6)

(constant enthalpy process)

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2

12

41

12

41

hh

hh

w

qCOPref

−

−==


s

Performance of the Cycle

The performance of the vapor-


is measured by the coefficient of

performance of refrigerator,

COPref, defined as,

(7)

q41

q23

T

T2

T1

s4 s1 =s2

14

3 2

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An ideal vapor-compression refrigeration cycle uses R-134a as the

working fluid. The evaporator and condenser operate at –20oC and

1000 kPa respectively. The refrigerant is dry saturated at entry to the

compressor and saturated liquid at the condenser outlet. The mass

flow of the refrigerant is 3 kg/min. Determine,

(a) The COPref;

(b) The refrigerating effect (kW);

(c) The COPhp;

(d) The heat transfer to the cooling water in the condenser (kW).

Show the cycle on a T-s diagram.



Q41

Q23

T

T2

T1 = -20oC

s4 s1 =s2

14

3

2

P2 = P3 = 1000 kPa

s

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To make full use of the specific enthalpy of evaporation of the

refrigerant, the evaporation process is continued until the refrigerant

becomes a saturated vapor at 1.

q41

q23

T

T2

T11

4

3

2

sThe refrigerant exits the evaporator

as a saturated vapor


In actual condition, the

refrigerant evaporates until it

becomes superheated vapor at

state 1. This is to prevent the

carry-over of liquid refrigerant

into the compressor. However,

the amount of superheat is

usually kept to minimum.

q41

q23

T

T2

T11

4

3

2

s

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q41

q23

T

T2

T1 14

3

2

s

Effects of This Process

1. The refrigerant becomes

superheated vapor as it exits

the compressor.

2. More work needs to be

supplied to the compressor to

compress the refrigerant.

3. The condensation process still

occurs at a constant condenser

pressure, but not at a constant

temperature.

4. More heat needs to be rejected

from the refrigerant to the

warm region.


The refrigerant is condensed until it’s temperature is lower than

the saturation temperature at the condenser pressure, when it

exits the condenser.

Under-cooling of the refrigerant in the

condenser

q41

q23

T

Tsat 2

1

4

3

2

s

Saturation temperature

at condenser pressure

P evaporator

P condenser

Degree

of sub-cooling

Degree

of superheat

Tsat 1

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Effects of This Process

• The line 3-4 representing the expansion process is shifted to the left,

thus line 4-1 representing the evaporation process increases in length.

• Therefore, under-cooling of the refrigerant increases the refrigerating

effect of the cycle, which is a desirable effect.

q41

q23

T

Tsat 2

1

4

3

2

s



P evaporator

P condenser

Degree

of sub-cooling

Degree

of superheat

Tsat 1

( )41 1 4q h h= −

( )12 2 1w h h= −



Evaporation process (4-1):

(8)

Compression process (1-2):

(9)

q41

q23

T

Tsat 2

1

4

3

2

s



P evaporator

P condenser

Degree

of sub-cooling

Degree

of superheat

Tsat 1

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( )23 2 3q h h= −

43 hh =



Condensation process (2-3):

(10)

Expansion process (3-4):

(11)

q41

q23

T

Tsat 2

1

4

3

2

s



P evaporator

P condenser

Degree

of sub-cooling

Degree

of superheat

Tsat 1

( )( )1 441

12 2 1

ref

h hqCOP

w h h

−= =

−


Performance of the Cycle

Performance of the vapor-compression refrigeration

cycle is measured by the Coefficient of Performance of

refrigerator, COPref, defined as,

(12)

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The p-h diagram

Another diagram frequently used

in the analysis of vapor-

compression refrigeration cycles is

the P-h diagram.

Three of the four processes appear

as straight lines.

The heat transfer in the condenser

and the evaporator is proportional

to the lengths of the corresponding

process curves.

P, (bar)

h

1

2 2’3

4

T sat condenser

T sat evaporator

The air temperature inside a cold room is controlled at a constant

value using a vapor compression refrigeration system with R-134a as

the refrigerant. The refrigerant leaves the evaporator dry saturated

and is compressed to 1 MPa. The refrigerant leaves the condenser at

35oC. The refrigerant is then throttled to the evaporator pressure of

240 kPa. The isentropic efficiency of the compressor is 85%. The

refrigeration load is 100 kW. Determine,

(a) The temperature of the refrigerant which leaves the compressor

(oC);

(b) The mass flow rate of the refrigerant (kg/s);

(c) The coefficient of performance of the refrigeration system.

Show the cycle on a p-h diagram.


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P, (bar)

h

10.84

2.610

Ts = 45oC

1

2 2’3

4

It is proposed to use a heat pump working on the ideal vapor-compression cycle for the purpose of heating the air supply to anincubation room from 25oC to 37oC at a rate of 1.06 kg/s. The supplyof heat is taken from a refrigerated room at 7oC. For the air, take cp

=1.005 kJ/kgK.

The refrigerant is R12 which is dry saturated leaving the evaporator.A temperature difference of 17 K is necessary for the transfer of heatfrom the refrigerated room to the refrigerant in the evaporator. Thedelivery pressure of the compressor is 10.84 bar and there is 5 K ofunder-cooling of refrigerant in the condenser.


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Show the schematic diagram of the components and the cycle on a

T-s diagram and p-h diagram. Calculate,

(a) The heat load (kW);

(b) The mass flow of the refrigerant (kg/s);

(c) The refrigerating effect (kW);

(d)The motor power required to drive the compressor if the

mechanical efficiency is 87%;

(e) The COPhp;

(f) The COPref.



Q41

Q23

T

Tsat

14

3

2

s

5K

17K

-10oC

10.84 bar

P

h

4 1

23

2’

2’

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• The most important quantity in refrigeration or

freezing application is the total of heat required to be

moved from a cold space.

• This heat is called refrigerating load.

• Measured in `ton’ or kilowatt or Btu/min.

• Unit conversion: 1 ton = 200 Btu/min = 3.516 kW.


( )414141hhmQxmQ refref −==

•••

∴

( )( )skg

hh

Qmref /

41

41

−=

••


(13)

Refrigerating load is,

The mass flow rate of the refrigerant is,

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Features of the diagram

It is more convenient to

represent the vapor-

compression refrigeration

cycle on a pressure-enthalpy

(p-h) diagram, because the

enthalpies required for the

calculation can be read off

directly from the diagram.

Enthalpy (kJ/kg)

Pressure (bar)


Figure shows the vapor-


shown on a p-h diagram. Note that

the cycle shown is with under-

cooling of the refrigerant at the

exit of the condenser.

Q1 = (h1 – h4)

W =

(h2 – h1)

Vapor-compression cycle on a p-h diagram

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What is Flash Chamber?

• Flash chamber is a device used

to separate vapor refrigerant

from the liquid refrigerant.

• When a flash chamber is used,

the compression process needs

to be carried out in two stages.

Schematic diagram of a refrigeration

plant with two-stage compression

and a flash chamber.

Wc

Expansion

valve 1

Condenser

Compressor

Q out

3

5

6

WcExpansion

valve 2

Evaporator

Flash

chamber

Compressor

1

2

(1)

8

Direct contact

heat exchanger

9

(1)

(1)

(x)

(1-x)27(1-x)

4

(1-x)

Qin


The refrigerant leaving the

expansion valve 1 as a mixture of

vapor and liquid (wet vapor). The

vapor refrigerant has no more

capability to absorb heat in the

evaporator. Thus, it is no use to pass

the vapor through the evaporator.

It is more practical to separate the

vapor from the mixture at some

intermediate pressure, pi and allow

only the liquid refrigerant to flow

through the evaporator. The liquid

refrigerant has full capability to

absorb the heat in the evaporator.

Wc

Expansion

valve 1

Condenser

Compressor

Q out

3

5

6

WcExpansion

valve 2

Evaporator

Flash

chamber

Compressor

1

2

(1)

8

Direct contact

heat exchanger

9

(1)

(1)

(x)

(1-x)27(1-x)

4

(1-x)

Qin

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Process

Suppose 1 kg of refrigerant flowing

through the condenser. At the flash

chamber, x kg of dry saturated vapor at

pressure pi and enthalpy hgi is bled off

to the inter-sage of the compressor.

The remaining mass of (1-x) kg of liquid

with enthalpy hfi passes through the

throttle valve 2 and then to the

evaporator.

At the intermediate pressure pi , (1-x)

kg of vapor at state 2 is mixed with x kg

of flash vapor of enthalpy hgi . The

resultant mixture at state 3 is

compressed in the second stage

compressor, to state 4.

Wc

Expansion

valve 1

Condenser

Compressor

Q out

3

5

6

WcExpansion

valve 2

Evaporator

Flash

chamber

Compressor

1

2

(1)

8

Direct contact

heat exchanger

9

(1)

(1)

(x)

(1-x)27(1-x)

4

(1-x)

Qin


• 2 compressors, 1 evaporator, 1

condenser, and 1 flash chamber.

• Point 2, 3, 6, 7 and 9 are located on

the same intermediate pressure pi . pi

is chosen so that the compressor

work is minimum.

pi = (p1 . p4)1/2

• The flash chamber pressure = The

intermediate pressure, pi .

Vapor-compression refrigeration cycle

with two-stage compression and a flash

chamber on a p-h diagram

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

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• Saturated liquid enter the 2nd

expansion valve at point 7.

• The flash chamber contains

mixture of saturated water and

vapor.

• The dry saturated vapor from

the flash chamber at point 9 mix

with the superheated vapor

from the 1st stage compressor at

point 2 to yield superheated

vapor at point 3.

• The superheated vapor at point 3

is compressed to point 4

through the 2nd stage

compressor.

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

fgifi hxhhh )(56 +==


The amount of dry saturated

vapor bled off is given by the

dryness fraction x at state 6, at

the intermediate pressure pi .

Since the enthalpy of the

refrigerant at state 6 is equal to

the enthalpy at state 5, we

have,

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

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29

fgi

fi

h

hhx

−=

5


Thus,

(14)

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

( )[ ]293 .1..10 hxhxh

HHWQ inletexit

−+−=

−=/−/

( )9223 . hhxhh −−=


Thus,

(15)

9

(x) kg

2

3

(1-x) kg

(1) kg

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

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30

∑ += 3412 WWWin

( )( ) ( )34121 hhhhxWin −+−−=∑


a) Total work input to the

compressors

Since the compression process is

carried out in two stages, the total

work input to the cycle is,

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

i.e.

(16)

( )( )8181 1 hhxQ −−=


b) Refrigerating effects

The refrigerating effects of the

cycle is

(17)

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

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( )5445 hhQ −=


c) Heat rejected in the condenser

The heat rejected from the

refrigerant is

(18)

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

( )( )( )( ) ( )3412

81

1

1

hhhhx

hhxCOPref

−+−−

−−=

∑=

in

refW

QCOP 81


d) Coefficient of performance

The coefficient of performance of the

cycle is

(19)

P

h

1

4

3

9

267

5

8

(1-x) kg

(x) kg

(1) kg

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• Increase in refrigerating effects of the cycle.

• Lower amount of compression work.

• Increase in the coefficient of performance.


• The refrigeration system becomes more complex

since there are more equipments and piping

system required.

• Higher capital costs for setting up the plant

• Higher maintenance costs.


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A two-stage vapor-compression refrigeration system uses Refrigerant-

134a as the working fluid. Saturated vapor enters the first compressor

stage at -30oC and saturated liquid leaves the condenser at 11.595 bar.

The system is equipped with a flash chamber, which operates at a

pressure of 4.1459 bar. If each compressor stage operates

isentropically and the refrigerating capacity of the system is 10 tons,

determine,

(a) the mass flow rate of the refrigerant through the evaporator

kg/s;

(b) the power input to each compressor (kW);

(c) The coefficient of performance.

Show the cycle on a T-s diagram. (1 ton refrigeration = 3.516 kW)


4

T oC

s

1

5

3

2

6 97

8

(1)

(x)

(1-x)

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A vapor-compression refrigeration plant uses refrigerant R134a and

has a suction saturation temperature of -5oC and a condenser

saturation temperature of 45oC. The vapor is dry saturated on entering

the compressor and there is no under-cooling of the condensate. The

compression is carried out isentropically in two stages and a flash

chamber is employed at an intersage saturation temperature of 15oC.

Determine:

a) the amount of vapor bled off at the flash chamber;

b) the state of vapor at inlet to the 2nd compressor stage;

c) the refrigerating effect;

d) the work input;

e) the COPref.

A two-stage compression refrigeration system is to operate

between the pressure limits of 1 MPa and 0.1 MPa. The working

fluid is Refrigerant-134a. The refrigerant leaves the condenser as

a saturated liquid and is throttled to a flash chamber operating at

0.32 MPa. Part of the refrigerant evaporates leaving the low

flashing process, and this vapour mixed with the refrigerant

leaving the low-pressure compressor. The mixture is then

compressed to the condenser pressure by the high-pressure

compressor. The liquid in the flash chamber is throttled to the

evaporator pressure and cools the refrigerated space as it

vaporizes in the evaporator. Assuming the refrigerant leaves the

evaporator as a saturated vapour and both compressors are

isentropic, determine:


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(a) The fraction of the refrigerant which evaporatesas it throttled to the flash chamber;

(b) The amount of heat removed from therefrigerated space (kJ/kg);

(c) The compressor work (kJ/kg);

(d) The COPref.

Illustrated the schematic layout of the refrigerationsystem showing clearly the respective componentsand also its p-h diagram.



4

T oC

s

1

5

3

2

6 97

8

(1)

(x)

(1-x)

Wc

Expansion

valve 1

Condenser

Compressor

Q out

3

5

6

WcExpansion

valve 2

Evaporator

Flash

chamber

Compressor

1

2

(1)

8

Direct contact

heat exchanger

9

(1)

(1)

(x)

(1-x)27(1-x)

4

(1-x)

Qin

Lecture note refrigeration cycle

Education

Transcript of Lecture note refrigeration cycle