Chapter 1Refrigeration

Air conditioniong (A/C) – control of temperature, humidity and air circulation (+ purity)

Refrigeration For A/C, temperature requirement 5 oC or higher Food processing/storage requires – 40 oC Industrial refrigeration requires – 100 oC Cryogenics; below – 150 oC

1.0 Revision

1.1 Definitions Work done by a system is positiveHeat transferred to a system is positive

1.2 First Law

1.2.1 Closed system

where U = total internal energy = mu

and for a cycle;

1.2.2 Open systems

(a) Conservation of mass;

(b) Conservation of energy

1.2.3 Applications(a) Compressors, fans, pumps : power requirement is

(b) Condensors, evaporators, heat exchangers. Rate of heat absorption is

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(c) Throttling – a restriction which causes a temperature drop is assumed as isenthalpic

(d) Turbines – the power output is:

1.3 The second law

(a) cycle – a thermodynamic cycle is said to occur whenever a system undergoes a number of prosesses and finally returns to its intial state.

(b) heat engine – produces work as a result of heat transfer from a high temperature body to a low temperature body.

(c) thermal efficiency,

=

where subscripts C and H refer to heat transfer to a cold reservoir and heat transfer from a hot reservoir, respectively.

1.3.1 Statements of the second law

Clausius – “It is impossible to construct a system which will operate in a cylce and transfer heat from a cooler to a hotter body without work being done on the system by the surrounding”

Kelvin-Plank statement : < 1

1.3.2 Corollaries to the second law

1st corollary – “It is impossible to construct an engine which operates between two given reservoirs and is more efficient than a reversible engine” ie ………………..

2nd corollary : = f(TH, TC) and we can show that resulting in the efficiency of a

reversible heat engine =

2

1.3.3 Entropy

For a reversible process, and can also be shown thatTds = du + p.dvTds = dh – v.dp

1.4 Perfect gas

Equation of state : PV = mRT

1.5 Specific heat

at constant volume ; cv and at constant pressure ; cp

and ratio of specific heats,

For perfect gases, du = cv dT, dh = cp dT and cp – cv = R which are applicable for any process.

1.6 Reversible adiabatic (isentropic) process for perfect gas obeys

constant

1.7 Reversed Carnot cycle

Transfers heat from the cold reservoir to hot reservoir as a result of work transfer to the system.

1.7.1 Refrigerator – Coefficient of Performance (COP) is

COP =

Maximum theoretical COP is , COPmax =

1.7.2 Heat Pump – Performance Factor (PF) or COPHP is

PF = COPHP =

Energy Efficiency Ratio (EER) is equivalent to COP

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Example

An ideal refrigerator provides cooling at (i) 0 C (ii) – 20 C (iii) – 40 C and rejects heat at 40 C. Determine the COP and work input to provide 100 kW of cooling in each case.

Example

Heat pump rejects heat at 40 C and takes in heat at (i) 0 C (ii) – 20 C (iii) – 40 C. Determine PF and work input for 100 kW of heating.

(Ans. Ex 1.1 (i) 6.83, 14.6 kW (ii) 4.22, 23.76 kW (iii) 2.91, 34.3 kW Ex 1.2 (i) 7.83, 12.8 kW (ii) 5.22, 19.2 kW (iii) 3.91, 25.6 kW)

1.8 Production of low temperatures

1.8.1 Expansion of a liquid with flashing

Flashing occurs when part of a liquid is vaporized in an expansion process such as throttling. The vaporized liquid cools the remaining liquid, thus lowering its temperature. During the expansion process, the pressure drops, volume increases and entropy increases. Consider a saturated liquid undergoing an irreversible expansion process,

T p1

1 T1

T2 p2

2 s1.8.2 Reversible adiabatic expansion of a gas

Permanent gas such as air initially at low temperature and high pressure is expanded adiabatically and reversibly in an expander, doing work on the surroundings. For an ideal gas, for a pressure drop from p1 to p2, the expansion is isentropic, it obeys

C and = C

Thus, the final temperature is,

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1.8.3 Irreversible adiabatic expansion (throttling ) of a real gas

A real gas is initially compressed and cooled. It is then expanded irreversibly through a throttle device (e.g. porous plug). For an adiabatic process, h2 = h1. For a perfect gas T2 = T1, but for a real gas substantial temperature decrease is produced when its pressure is dropped via throttling. The Joule-Thomson coefficient is defined as,

=

< 0 Heating ; T2 > T1

= 0 T2 = T1

> 0 Cooling ; T2 < T1

P2, T2 P1, T1 real gas

T maximum inversion temperature (eg N2= 620K, He= 40K, air = 600K). T2 2 pressure drop through plug 1 T1 h = constant >0 < 0

p2 p1 p1.8.4 Thermoelectric cooling

Peltier in 1834 discovered that cooling is produced at one junction of two dissimilar metals, if current is passed through them. At the same time heat is produced at the other junction. QC

TC Cold end

. material A material B QH QH

Hot end TH

current

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Chapter 2Vapour Compression Refrigeration

2.1 The ideal cycle

T

3 2 1 – 2 isentropic compression 2 – 3 isothermal condensation 3 – 4 isentropic expansion 4 – 1 isothermal evaporation

4 1

s

2.1.1 Selection of operating temperatures

COP = =

(a) Selection of TC (operating or evaporating temperature)

Application TC (oC)Air conditioningCold storageDomestic refrigerationFrozen foodFreeze dry

0 to 10– 10– 15– 35

– 35 to – 45

Typical A/C system T coil

Room 25 C Q 25 C

. air

. 15 C coil fan supply air refrigerant 15 C (typical). TC 10 C

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(b) Selection of TH (condensing temperature)

Dependent on environment in which heat is rejected from the condenser, usually temperature difference between TH and the environment is about 10 C.

(i) Air-cooled condensers can be found in refrigerators, water coolers, window-type A/C and most package type A/C.

(ii) Water-cooled condensers are prevalent in large systems where large quantities of heat are being rejected. The benefits of water include

Available at much lower temperature than air Cp of water is about 4 times that of air; for the same condenser heat rejection rate

and the same require ¼ mass flow rate. Convection heat transfer coefficient, h is greater for water than air. For same ,

need smaller heat exchanger (A = heat rejection rate/h* ).

T

. TH 10 C

. condenser T higher than ambient T of cooling medium

.evaporator T lower than cold T of cold room .room temperature TC

. 5 C

s2.2 Limitations of ideal cycle2.2.1 Throttling vs isentropic expansion

Isentropic expansion in turbine is impractical. Turbine is replaced by throttling device, resulting in loss of refrigerating effect (h1 – h4) < (h1 – h4s) and loss of turbine work.

T. 3

. 4s 4 s

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2.2.2 Dry vs wet compression.

Wet compression (1’ – 2’) is avoided because Liquid refrigerants trapped in the head of the cylinder Liquid refrigerant droplets may wash away lubricating oil from walls of compressor,

increasing wear.With dry compression (1 – 2) additional compressor work is required.

T

. 2’ 2

. 1’ 1 s

2.3 Theoretical vapour compression cycle (1, 2, 3, 4)

T. qH 2 w = qH - qC

. 3

qC

.

. 4s 4 1 s

2.4 The pressure-enthalpy diagram

p T1 T2 s1 s2

v1

v2

y2 > y1

y=T,s,v h

2.5 Analysis of the theoretical cycle

p . 3 2

8

Pcond

Pevap . 4 1 compressor work

. refrigerating effect h

. heat rejected

Refrigerating effect, qC = h1 – h4

Heat rejected, qH = h2 – h3 per unit of mass, e.g. kJ/kgCompressor work, w = h2 – h1

COP = refrigerator PF = heat pump

Refrigerant mass flow rate, =

Refrigerant volumetric flow rate, = (also known as theoretical piston

displacement rate).Example 1

A standard vapour compression cycle develops 50 kW of refrigeration, using R134a and operating with a condensing temperature of 35.51 oC and an evaporating temperature of –10 oC. Using a p-h diagram, for R134a, determine:

(a) refrigerating effect(b) refrigerant mass flow rate(c) specific work and power required by compressor(d) COP(e) volumetric refrigerant flow rate(f) compressor discharge temperature(g) the heat rejected in the condenser

p . 3 35.51 C 2 Pcond

Pevap

9

. 4 – 10 C 1

. h h1 = hg,-10C = 244.51 kJ/kg, s1 = 0.93766 kJ/kgK

h2 = 274.17 + (284.77 – 274.17) = 275.75 kJ/kg

h3 = h4 = hf 35.51C = 101.61 kJ/kg ; v1 = vg,-10C = 0.099516 m3/kg

(a) qref = h1 – h4 = 244.51 – 101.61 = 142.90 kJ/kg

(b) = = 0.350 kg/s

(c) w = h2 – h1 = 275.75 – 244.51 = 31.24 kJ/kg ; = = 0.350(31.24) = 10.9 kW

(d) COP = = = 4.59

(e) = = 0.350(0.099516) = 0.0348 m3/s = 34.8 L/s

(f) t2 = 40 + (0.93766 – 0.9327)(50 – 40)/(0.9660 – 0.9327) = 41.5 C(g) = (h2 – h3) = 0.350(275.75 – 101.61) = 60.9 kWRepeat example above with these data: The refrigerant leaves the condenser with 2 K subcooling and exits the evaporator with 3K superheat. The compressor has an isentropic efficiency of 90%.2.5.1 Effect of evaporator pressure

p . 3, 3” 2 2” Pcond

. 4 1 pe pe” . 4” 1”

. hA decrease in evaporator pressure (pe to pe”) results in:

(i) decrease in refrigerating effect; (h1” – h4”) < (h1 – h4)(ii) increase in specific volume of suction vapour; v1” > v1

(iii) decrease in volumetric efficiency*, due to increase in pressure ratio(iv) increase in compressor work, due to increase in pressure ratio, as well as change from

steeper isentropic 1 – 2 to flatter isentropic 1” – 2”(v) decrease in COP (refrigerating effect decreases, compressor work increases).

* volumetric efficiency;

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=

valid for positive displacement compressors only.

2.5.2 Effect of condenser pressure

. pc” 3” 2” pc

. 3 2

. pe . 4 4” 1.

. hAn increase in condenser pressure from pc to pc” results in:

(i) decrease in refrigerating effect; (h1 – h4” ) < (h1 – h4)(ii) increase in compressor work(iii) decrease in volumetric efficiency(iv) decrease in COP

2.5.3 Effect of suction vapour superheat

Superheating of suction vapour is desirable in practice because:(i) it ensures complete vaporization of liquid in evaporator, before it enters the

compressor(ii) the degree of superheat is often used as a means of actuating and modulating the

expansion valve’s capacity

p . 3 2 2” Pcond

. pevap . 4 1 1”

. hThe superheating a vapour from t1 to t1” results in:

(i) increase in specific volume of suction vapour; v1” > v1

(ii) increase in refrigerating effect; (h1” – h4) > (h1 – h4)(iii) inrease in specific compressor work; (h2” – h1”) > (h2 – h1)(iv) COP may increase, decrease or remains unchanged

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In R12 systems, suction temperature in practice is between 15 to 25 C.

2.5.4 Effect of liquid subcooling

p . 3” 3 2 Pcond

. pevap . 4” 4 1

. hSubcooling results in

(i) reduced flashing of liquid during throttling process; x4” < x4

(ii) increase in refrigerating effect; (h1 – h4”) > (h1 – h4)(iii) increase in COP(iv) reduced piston displacement and horsepower per ton for all refrigerants

In some systems, a subcooler is placed between the condenser and expansion valve. Cooling water at temperature lower than condensing temperature is passed through the subcooler to achieve subcooling of the refrigerant.

2.5.5 Using liquid-vapour regenerative heat exchanger

Combination of vapour superheating and liquid subcooling results in a liquid-vapour regenerative heat exchanger.

. 2”

. condenser

. 3 1”

Liquid vapour regenerative heat exchanger

. 3”

.

. 4” 1

. evaporator

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p . 3” 3 2 2” pcond

. pevap . 4” 4 1 1”

. hRefrigerant vapour from evaporator is superheated in the heat exchanger with consequent subcooling of the liquid from the condenser. Energy balance of the heat exchanger gives,

h1” – h1 = h3 – h3”

i.e. heat gained by refrigerant vapour equals heat loss from liquid refrigerant in the heat exchanger. The refrigerating effect is,

qref = h1 – h4” or qref = h1 – h3”

and

COP =

Compared to the standard cycle, refrigerating effect increases but specific work also increases because compression is carried out further into the superheat region. Consequently, COP improvement if any is marginal. The heat exchanger probably offers no thermodynamic benefits, but it is justified because:

(i) superheated vapour ensures that no liquid enters the compressor(ii) subcooled liquid prevents bubbles of vapour from impeding flow of refrigerant

through expansion valve.

2.6 Actual Vapor-Compression Refrigeration Cycle (2nd meeting)

In real vapor-compression refrigeration cycle, irreversibilities are present due to fluid friction (causes pressure drops) and heat transfer to or from the surroundings that occur across a finite temperature difference.

In real cycles: The refrigerant leaves the evaporator (and enters the compressor) as superheated vapor.

This superheating increases power input to compressor since steady-flow work is proportional to the specific volume.

The compression process is not isentropic. The entropy may increase or decrease depending on the direction of heat transfer during the compression process. It is desirable to cool the refrigerant during the compression process since it lowers win

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The refrigerant leaves the condenser as compressed liquid (subcooled). This is desirable since the refrigerant can absorb more heat from the refrigerated space

Pressure drops in the condenser and the evaporator

Example 2Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and –10 oC at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50 oC. The refrigerant is cooled in the condenser to 26 oC and is throttled to the evaporator pressure. Neglecting pressure drops in the evaporator and condenser, determine (a) the rate of heat removal from the refrigerated space, (b) the power input to the compressor, (c) the isentropic efficiency of the compressor, and (d) the COP of the refrigerator.

State 1 (0.14 MPa, – 10 C)h1 = 246.36, s1 = 0.9724

State 2 (0.8 MPa, 50 C)h2 = 286.69, s2 = 0.9802

s1 = 0.9724

= 284.21

State 3 (0.8 MPa, 26 C) = 87.83

State 4 = 87.83

(a) The refrigeration load, = 7.93 kW

(b) = 2.02 kW

(c) = 0.939 or 93.9 %

(d)

2.7 Multistage Compression Refrigeration System

Consider the following two-stage compression refrigeration system with a flash chamber

Warm environment

QH T 4 Condenser 5 4

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Expansion high-pressure 5 1valve compressor 6 9 2 Flash . ... x 9Chamber 3 Pi

7 2 7 6 3Expansion low-pressureValve compressor (1-x) 8 1 evaporator 8 1

QL

Cold environment s

The intermediate pressure Pi is chosen so that the total compressor work required is minimum, Assuming an adiabatic flash chamber, we can write the following energy balance eqn.

0 – 0 =

where hf and hg are evaluated at Pi

In other words, mass fraction of saturated vapor extracted from the flash chamber, x is equal to the quality of mixture at point 6.

The specific enthalpy at point 9 is obtained from the adiabatic mixing eaquation,

The rerfrigerating effect, Total compressor work (required),

Example 3A two-stage compression refrigeration system operates between the pressure limits of 0.8 and 0.14 MPa. Refrigerant-134a leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.32 MPa. Part of the refrigerant evaporates during this flashing process, and this vapor is 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 vapor and both compressors are isentropic, determine (a) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) the amount of heat removed from the refrigerated space

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and the compressor work per unit mass of refrigerant flowing through the condenser, and (c) the coefficient of performance.

h1 = hg,0.14MPa = 239.16s1 = 0.94456

P2 = 0.32 MPa, s2=s1 = 0.94456

= 255.93

h5 = hf,0.8MPa = 95.47, h6 = h5 = 95.47h7 = hf,0.32MPa = 55.16, h8 = h7 = 55.16

h3 = hg,0.32MPa = 251.88

(a) = = 0.2049

= 0.2049(251.88) + (1-0.2049)(255.93) = 255.10 kJ/kg

(b) = 146.3 kJ/kg

At state 9 , P9 = 0.32 MPa, h9 = 255.10 . Therefore

= 0.9416

At state 4, P4 = 0.8 MPa, s4 = s9 = 0.9416, Thus

= 274.48

= (1 – 0.2049)(255.93 – 239.16) + (274.48 – 255.10) = 32.71 kJ/kg

(c) = 4.47

2.8 Absorption Refrigeration

Low-pressure vapour can be transformed into high-pressure vapour in refrigeration system by the conventional vapour compression method or the absorption method.

Vapour compression High-pressure vapour 1. Compressor condenser

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Absorption1. Absorb vapour in liquid while removing expansion valve

heat 2. Elevate pressure of liquid with pump3. Release vapour by applying heat evaporator

. Low-pressure vapour

In an absorption system, the low pressure vapour is first absorbed in an appropriate absorbing liquid. During the absorption process, conversion of vapour into liquid (akin to condensation) results in heat being rejected. The pressure of the liquid is then elevated with a pump. Finally, heat addition causes the vapour to be released from the absorbing liquid.

A vapour compression is known as a work-operated cycle because pressure elevation of refrigerant is accomplished by a compressor which requires work.

The absorption cycle is referred to as a heat-operated cycle, because most of the operating cost is associated with the heat required to drive off vapour from the high-pressure liquid. The pump work is negligibly small compared to that needed by a compressor in a vapour-compression system.

2.8.1 The absorption cycle . High-pressue vapour. lean solution Generator Condenser heat heat

. rich

. solution Throttling valve Expansion valve

. Low-pressure vapour

Absorber Evaporator heat heat pumpHigh-temperature heat enters the generator while low-temperature heat from the substance being refrigerated enters the evaporator. Heat rejection from the cycle occurs at the absorber and condenser at temperatures that allow heat rejection to atmosphere.

2.8.2 Ideal absorption cycle

The absorption cycle can be thought of as a combination of a power cycle and refrigeration cycle.

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qg Ts Ta qc

W qa Ta Tr qe

Ts source temperature, Ta ambient temperature, Tr refrigerating temperature. For an ideal power cycle,

……………………..……. (4.1)

and for an ideal refrigerator,

………………………….….(4.2)

Using the definition of COPabs for an absorption cycle,

COPabs = ……… (4.3)

Thus, for an ideal absorption cycle,

COP = ……...……(4.4)

ExampleWhat is the COP of an ideal heat-operated refrigeration system that has a source temperature of heat of 100 oC, a refrigerating temperature of 5 oC, and an ambient temperature of 30 oC ?

COP = = 2.09

These trends are detectable from eq. 4.4:1. As Ts increases, the COP increases2. As Tr increases, the COP increases3. As Ta increases, the COP decreases

Typical absorbent-refrigerant pair includes LiBr-water which is most popular, while aqua-ammonia pair (water as absorbent, ammonia as refrigerant) was used before LiBr-water was discovered.

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2.9 Gas cycle refrigeration

The gas cycle is used exclusively in air-conditioning of aircrafts; military and commercial. Often referred to as air cycle refrigeration because air is the working fluid in the cycle. Its COP is lower than that of conventional vapour compression cycle. It can work as an open or closed cycle. The benefits of an air cycle includes:

(i) Small amount of leakages are tolerable since air is the refrigerant.(ii) The simplest open system requires only one heat exchanger(iii) Readily available refrigerant(iv) Cabin pressurization and air-conditioning can be combined into one operation(v) Initial compression of the air is obtained by the ram effect (conversion of the high

kinetic energy of ambient air relative to aircraft into enthalpy and ultimately pressure rise). Refrigerated air capacity ranges from 3 to 70 kg a minute.

2.9.1 Simple air-cycle sytem for aircraft refrigeration

to combustion chamber cooling air exit

compressed air cooling air fan

jet compressor air cooler

ram air cooling turbine

cooling air to cabin

Compressed air after cooling in air-cooler is passed through a cooling turbine Turbine work is used to drive a fan which draws cooling air through the heat exchanger; ie

to overcome friction in air-cooler Air is dischaged from the turbine at a pressure slightly above cabin pressure Fan is located downstream, avoiding unnecessary temperature rise of cooling air Turbine work is not available for the compressor (compressor is driven by power turbine

downstream of compressor after the combustion chamber)

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