topic1-2 refrigeration cycle

Topic1-2 Refrigeration Cycle1. Carnot Refrigeration cycle2. Coefficient of performance (COP)3. Conditions for highest COP4. Temperature limitations5. Carnot heat pump6. Using vapor as a refrigerant7. Wet versus dry compression

8. Expansion process 9. Properties of refrigerants 10. Standard vapor-compression cycle

11. Performance of the SVC cycle12. System Efficiency Index (SEI)

13. Condenser and evaporator temperature differences14. Liquid Subcool Heat exchanger 15. Subcooling and superheating

1

16. Analysis of Carnot refrigeration cycle17. Comparison between Carnot & standard VCRS 18. Actual vapor-compression cycle19. Multi-stage cycles20. Non vapor-compression cycles20.1 Trans-critical carbon dioxide cycle

20.2 Expendable refrigerant cooling system20.3 Absorption cycle20.4 Air cycle

20.5 Stirling cycle20.6 Thermoelectric cooling20.7 Magnetic refrigeration20.8 Steam jet refrigeration20.9 Vortex tube21. Natural Refrigeration

1. Carnot Refrigeration cycle

2

Carnot cycle: thermodynamically reversible processes

Net work

2

1-2 Adiabatic compression

Heat from low-Temp. source

Heat to high-Temp. sink

Tem

per

atu

re, K

Entropy, kJ/kgK

1

23

4

1

23

4WorkCompressor

2-3 Isothermal rejection of heat

3-4 Adiabatic expansion4-1 Isothermal addition of heat

Work

Turbine

1-2 Adiabatic reversible compression = isentropic compression

3-4 Adiabatic reversible expansion = isentropic expansion

-Standard of comparison- Guide of T to max eff.

Refrigeration

2. Coefficient of performance (COP)

3

Efficiency = Output / Input → Heat Engine = Wout/Qin

Net work

3

Tem

per

atu

re, K

Entropy, kJ/kgK

1

23

4

reversible process: qrev = ∫Tds :

Refrigeration cycle: input = net workOutput = q2-3 – wasteDesired output = q4-1 – useful refrigeration

HE = (TH - TL)/ TH

TH

TL

Coefficient of performanceCOP = desired output/input

COPRefrigeration Carnot

= T1(s1 – s4)/[(T2 – T1)(s1 – s4)]= T1 / (T2 – T1)= TL / (TH - TL)

s1 = s2s3 = s4

3. Conditions for highest COP

low T2 → high COP

4

COPR Carnot = T1 / (T2 – T1)

Refrigeration

4

Net work

4

Tem

pe

ratu

re, K

Entropy, kJ/kgK

1

23

4

T2

T1

s1 = s2s3 = s4high T1 → high COP

T1 → more effect upon COP than T2

4. Temperature limitations

Ex: a cold room at -20C & reject heat to the atm. at 30C

5

Tem

per

atu

re, K

Entropy, kJ/kgK

1

23

4

T2

T1

253 K cold room

303 K atmosphere

2-3 heat rejection processT2 > 303 K

4-1 refrigeration processT1 < 253 K

high COP → low T2 but T2 > 303 K

high COP → high T1 but T1 < 253 K

→ T2 – 303 = T1

→ 253 – T1 = T2

T1

T2

What can we do on keeping t as small as possible?

Heat exchanger: Condenser → qc = (UA)c T1

Heat exchanger: Evaporator → qe = (UA)e T2 = (UA)e Tlm

= (UA)c Tlm

5. Carnot heat pump

Heat pump: same equipment as a refrigeration system- delivering heat at high Temp.

6

Coefficient of performanceCOP = desired output/input

Net workTe

mp

era

ture

, K

Entropy, kJ/kgK

1

23

4Heat rejected

COPHeat Pump Carnot = T2 / (T2 – T1) = COPR + 1Performance factor = COPR + 1

Performance factor- vary from 1 to

6. Using vapor as a refrigerant

Refrigerant: gas, i.e. air

7

Tem

per

atu

re, K

Entropy, kJ/kgK

1

2

3

4

x

Cold room

Atmosphere

y

2-3 constant-pressure heating4-1 constant-pressure cooling

Different from Carnot cycle by area x and y

area x and y → decrease COP

6. Using vapor as a refrigerant

8

To keep rectangle shape as Carnot cycle:

Refrigerant condenses during heat-rejection at const. T, Pand boil during refrigeration at const. T,P

Isothermal heat-rejection and refrigeration

→ Condenser

→ EvaporatorRefrigerant operates between liquid and vapor states.

Tem

per

atu

re, K

Entropy, kJ/kgK

Cold room

Atmosphere

1

23

4

Saturated liquid

Saturated vapor

7. Wet compression versus dry compression

Wet compressionLiquid refrigerant may damage valves or cylinder head.Droplets of liquid may wash the lubricating oil from cylinder wall

9

Tem

per

atu

re, K

Entropy, kJ/kgK

Superheat horn

1

2

3

4

Dry compressionSuperheat horn – more work required for dry compression

8. Expansion process Constant-enthalpy throttling -- Isenthalpic process

10

Tem

per

atu

re, K

Entropy, kJ/kgK

1

2

3

4

Expansion engine: 1. Small work compared to compression work2. Difficulties of lubrication intrude when a fluid of two-phase drives the engine3. Not economic for work done compared to cost of expansion engine

Throttling device: reducing pressure by valve or other restriction

9. Properties of refrigerants : low boiling point

11

Pre

ssu

re,

kPa

Enthalpy, kJ/kg

Critical point

t = constant

Saturated liquid

Saturated vapor

Mollier (Pressure-enthalpy) diagram of a Refrigerant

12

1

23

4

Isobaric

Isobaric

Net work

Refrigeration effectTe

mp

erat

ure

, K

Entropy, kJ/kgK

1

23

4

T2

T1

Carnot refrigeration cycle

high T1, low T2 → high COP


T1 → more effect on COP than T2

12

10. Standard vapor-compression cycle

13

Tem

per

atu

re, K

Entropy, kJ/kgK

1

2

3

4

Process 1-2: Isentropic compression of saturated vapour in compressor Process 2-3: Isobaric heat rejection in condenser Process 3-4: Isenthalpic expansion of saturated liquid in expansion device Process 4-1: Isobaric heat extraction in the evaporator

Pre

ssu

re,

kPa

Enthalpy, kJ/kg

1

23

4

Condensation

Exp

ansi

on

Evaporation

Isen

thal

pic

Isobaric

Isobaric


The standard vapor-compression cycle

14

Mollier (Pressure-enthalpy) diagram

Pre

ssu

re,

kPa

Enthalpy, kJ/kg

1

23

4

Condensation

Exp

ansi

on

Evaporation

1

2

3

4

PowerQc

Qe

Power= ሶ𝑚(h2-h1)Qe = ሶ𝑚(h1-h4)

Qc = ሶ𝑚(h2-h3)

11. Performance of the standard vapor-compression cycle

15

Pre

ssu

re,

kPa

1

23

4

Condensation

Exp

ansi

on

Evaporation

Enthalpy, kJ/kg

Work of compression = wc = h2 – h1

Pc = ሶ𝑚(h2-h1)

Heat rejection rate = qc = h2 – h3

Qc = ሶ𝑚(h2-h3)

Refrigeration effect = qe = h1 – h4

Qe = ሶ𝑚(h1-h4)

COP = qe/wc

Volume flow rateper kW of refrigeration = ሶ𝑚vsuc /Qe

Power per kW of refrigeration = P/Qe = 1/COP

Example 1 A standard vapor-compression cycle developing 50 kW of refrigeration using refrigerant 22 operates with a condensing temperature of 35C and an evaporating temperature of -10C. Calculate (a) the refrigerating effect in kilojoules per kilogram, (b) the circulation rate of refrigerant in kilograms per second, (c) the power required by the compressor in kilowatts, (d) the coefficient of performance, (e) the volume flow rate measured at the compressor suction, (f) the power per kilowatts of refrigerant, and (g) the compressor discharge temperature.

Data: VCC, Qe, R22, tcond, tevap

Assume: standard VCC; isentropic compression, isobaric, isenthalpic processes

Method: evaporator, energy balance, (a) qe = h1-h4

evaporator, mass balance, (b) ሶ𝑚 = Qe/qe

compressor, energy balance, (c) Pc = ሶ𝑚(h1-h2)(d) COP = Qe/Pc

(e) Vsuc = ሶ𝑚vg@pe, pe= psat@Te

(f) P/Qe = 1/COP(g) Tdischarge = T2 = T@(pc, s2=s1), pc= psat@Tc

Pre

ssu

re,

kPa

1

23

4

Enthalpy, kJ/kg

35C= Tc

-10C= Te

(a)qe (b) ሶ𝑚 (c) Pcomp (d) COP (e)vsuc (f)P/Qe (g)Tdischarge

16

17

-10 C = tevap

35 C= tcond

1

2

h1=402 h2=435h4=243

T2=57C

3

4

17

Calculation: (a) qe = h1-h4 = 59 kJ/kg(b) ሶ𝑚 = Qe/qe = 0.315 kg/s(c) Pc = ሶ𝑚(h1-h2) = 10.3 kW(d) COP = Qe/Pc = 4.86(e) Vsuc = ሶ𝑚vg@pe = 20.4 L/s(f) P/Qe = 1/COP = 0.207(g) Tdischarge = T2 = 57C

h@0 C =200 kJ/kg

Example 1 VCC R22 at 35C and -10C

pe=355 kPa

pc=1355 kPa

Example 2 A standard vapor-compression refrigeration cycle using R22 operates with a condensing temperature of 35C and an evaporating temperature of -10C. If efficiency of compression is 70%, calculate (h) the actual COP, (i) actual discharge temperature T2’ and (j) System Efficiency Index, SEI = COPact/COPCarnot.

SEI is the ratio between the actual COP and the Carnot COP with reference to the cooling load and outside air temperatures, i.e. when the heat exchanger temperature differences, ΔT = 0, Tload = Te , To = Tc .

Assume: actual compression, isobaric, isenthalpic processes

Method: COPact = (h1-h4)/Pc,act = (h1-h4)/(h1-h2’)compression efficiency, c = (h2-h1)/(h2’-h1) = Pc/Pc,a

(h2’-h1) = (h2-h1)/c = (435-403)/0.7 = 47 kJ/kg

(h) COPact = (402-243)/47 = 3.40

T2’ = T@(pc, h2’), h2’ = 402 + 47 = 449 kJ/kg

T2’ = 74C

SEI = COPact/COPCarnot

COPCarnot = Te/(Tc-Te) = (-10+273)/(35-(-10)) = 5.84

(j) SEI = 3.40/5.84 = 0.582

Pre

ssu

re,

kPa

1

23

4

Enthalpy, kJ/kg

35C= Tc

-10C= Te

18

2’

12. System Efficiency Index (SEI) = COPactual/COPCarnot@ΔT=0

19

13. Condenser and evaporator temperature differencesCOP values for R22 cooling at -10C = Tload

with an outside air temperature of 35C = To

Varying Te and Tc forT1 = Tload - Te = T2 = To - Tc = T

Pre

ssu

re,

kPa

1

23

4

Enthalpy, kJ/kg

45C= Tc

-20C= Te

1

23

4

Enthalpy, kJ/kg

40C= Tc

-15C= Te

Tload

T = 5 C

To

Tload

T = 10 C

To

2’

2’

Pre

ssu

re,

kPaT at evaporator and condensor, °C 0

Evaporating temperature, °C -10

Condensing temperature, °C 35

Temperature lift = T c - Te, °C 45

Evaporating pressure, pe, kpa 355

Condensing pressure, Pc, kPa 1355

Pressure ratio = pc/pe 3.82

Carnot COP (refrigeration cycle) 5.84

h1 = hg,@Te, kJ/kg 402

s1 = sg,@Te, kJ/kgK 1.767

h2, kJ/kg 435

Discharge temperature, T2, °C 57

h3 = hf,@Tc kJ/kg 243

Refrigerating effect, kJ/kg 159

work for compressor, kJ/kg 33

COP, standard vapor compression cycle 4.86

compression efficiency, % 70

actual work for compressor, kJ/kg 47

COP, with 70% compression efficiency 3.40

System efficiency index, SEI 0.582

actual h2', kJ/kg 449

Actual discharge temperature, T2', °C 74

10

-20

45

65

244

1729

7.08

3.89

398

1.784

447

77

257

142

48

2.92

70

69

2.04

0.350

467

95

5

-15

40

55

296

1534

5.19

4.69

400

1.775

441

67

250

150

41

3.70

70

58

2.59

0.443

458

82

20

80 kPa = pe

1000 kPa = pc

1

3

h1=1413 h3=1783h@0 C =200 kJ/kg

h6=317

Example 3 Calculate the power needed to compress 1.2 kg/s of ammonia to saturated vapor at 80 kPa to 1000 kPa

Assume: standard VCCCalculation: Pc = ሶ𝑚(h1-h2) = -444 kWQe = ሶ𝑚(h1-h4) = 1315 kWCOP = Qe/Pc = 2.96COPCarnot = Te/(Tc-Te) = 6.18SEI = COP/COPcarnot = 0.77Tdischarge = T2 = 143C

14. Liquid Subcool Heat exchanger (LSHX)

21

Subcool liquid from condenser with suction gas from evaporator

Heat balance: h3-h4 = h1-h6

Refrigerating effect: h6-h5 = h1-h3

qe = h6-h5

21

22

15. Subcooling and superheating

1

23

4 1’

2’3’

4’

Superheating by heat -transfer from cold room →qe

-exchange with pipeline → loss-transfer from subcool liquid → 0→wc in some refrigerant, i.e.→Tdischarge affects lubrication→ Dry vapor to prevents entry of liquid droplets into compressor

cold room Subcooling by heat transfer to

surrounding or to suction vapor→qe by reduce throttling loss→ Reduce flash gas → lower

pressure drop in evaporator- ensures that only liquid enters into throttling device leading to its efficient operation.

atmosphere

Example 4 ระบบท ำควำมเยน็แบบอดัไอใช้ R22 มอุีณหภมูทิี่ Evaporator Te = 0C และที่Condenser Tc = 40C ใหห้ำ a) sketch เสน้ refrigeration cycle บน p-h diagram, b) COP ของ Carnot cycle, c) Refrigeration effect, 2d) Work of compression, e) อุณหภมูขิอง discharge, f) COP ของ standard vapor-compression refrigeration cycle, g) ถำ้น ำ Liquid-to-suction heat exchanger มำใชเ้พือ่ superheating เพิม่ 10C แลว้ subcool คำ่อุณหภมูทิีข่ำออก Condenserเป็นเทำ่ใด, h) Refrigeration effect เมือ่ใช ้LSHX และ i) Work of compression เมือ่ใช ้LSHX

Data: VCC, R22, Te, Tc ,

Assume: standard VCC; isentropic compression, isobaric, isenthalpic processes

Method: (b) COPcarnot = T1 / (T2 – T1) = 6.83 (T, K)

Evaporator (c) qe = h1-h4 = 406-250 = 156 kJ/kg

compressor (d) wc = h1-h2 = 406-433 = -27 kJ/kg

(e) Td = 58C

(f) COP = qe/wc = 5.82

LSHX, h3-h3’ = h1’-h1 = h = 8 kJ/kg

(g)Tsp = T1’-T1 = 10C, Tsc 6C

T3’ = 40-6 = 34C

(h) qe,LSHX = qe +h = 156+8 = 164 kJ/kg

(i) Te = 0+10C, wc = h2’-h1’ = 443-414 = 29 kJ/kg

Pre

ssu

re,

kPa

1

23

4

Enthalpy, kJ/kg

3’

1’

(a)p-h (b) COPcarnot (c) qe (d) wc (e)Td (f)COP (g) t3’ (h) ሶ𝑚

40C= Tc

0C= Te

23

3

3’ 1

1’

2’

24

0 C = Te

40 C= Tc

1

2

h1=406 h2=433h4=250

T2=58C

3

4

s1=s2=1.752

1’

3’2’

h2’=443

h3’=242

LSHX, h3-h3’ = h1’-h1 = h = 8 kJ/kg

qe,LSHX = qe +h = 156+8 = 164 kJ/kg

h1’=414

wc = h2’-h1’ = 443-414 = 29 kJ/kg

high T1, low T2 → high COP

25


Refrigeration

25

Net work

25

Tem

pe

ratu

re, K

Entropy, kJ/kgK

1

23

4

T2

T1

s1 = s2s3 = s4

T1 → more effect on COP than T2

16. Analysis of Carnot refrigeration cycle

T1

T2

2626

17. Comparison between Carnot & standard VCRS

COPStandard VCRS =COP = qe / wcomp = (h1 – h4)/(h2 – h1)

R = SEI = COPStandard VCRS / COPR Carnot

COP = (hg@Te – hf@Tc)/(hs1=s2 – hg@Te)

COP asTevap or Tsuction

COP as Tcond

SEI =System Efficiency Index

27

1

23

4

Isen

thal

pic

Isobaric

Isobaric

Standard VCRS cycle

As Tevap

1. RE(qe)2. wcomp then COP 3. Tdischarge

4. qcond

5. %flash gas bcoz %x

6. ሶ𝑚 bcoz v (L/kg)

1’

2’

4’

27

Refrigeration effect

Net work

Refrigeration effect

28

Tem

per

atu

re, K

Entropy, kJ/kgK

1

23

4


Pre

ssu

re,

kPa

Enthalpy, kJ/kg

Standard VCRS cycle

1

2

3

4

T2

T1

Irreversibility due to 1) non-isothermal heat rejection (process 2-3) 2) isenthalpic throttling (process 3-4)

Throttling loss

Superheat horn

Superheat horn – more work required for dry compression, high Td

Throttling loss - flash gas, reduction in refrigeration effect and increase more work because irreversibility

qewc

17. Comparison between Carnot & standard VCRS

Good for heat pumps

wc

18. Actual vapor-compression cycle

29

Discharge line: gas, p is a penalty on compressor power

Liquid line: liquid, smaller diameter, more p → liquid flashes in to vapor →expansion device will not work properly

Suction line: gas, p is a penalty on efficiency, reduce ps to comp.


30

Irreversibility due to 1. Pressure drops in evaporator, condenser and LSHX 2. Pressure drop across suction and discharge valves of the compressor 3. Pressure drop and heat transfer in connecting pipe lines 4. Heat transfer in compressor

4-1d → pressure drop in evaporator1d-1c → Superheat of vapor in evaporator 1c-1b → Useless superheat in suction line

1b-1a → Suction line pressure drop 1a-1 → Pressure drop across suction valve

1-2 → Non-isentropic compression 2-2a → Pressure drop across discharge valve

2a-2b → Pressure drop in the delivery line

2b-2c → Desuperheating of vapor in delivery pipe

2b-3 → Pressure drop in the condenser

3-3a → Subcooling of liquid refrigerant

3a-3b → Heat gain in liquid line

Actual systems differ from standard cycles -- foreign matter such as lubricating oil, water, air, particulate matter→ pressure drop, heat transfer coefficient in evaporator

31

1

2

3

4

4-1d → pressure drop in evaporator

1d-1c → Superheat of vapor in evaporator

1c-1b → Useless superheat in suction line

1b-1a → Suction line pressure drop

1a-1 → Pressure drop across suction valve

1-2 → Non-isentropic compression

2-2a → Pressure drop across discharge valve






32

Discharge line: vapor line

Liquid line: liquid

Suction line: vapor line

Standard cycle


2-2a → Pressure drop across discharge valve


→wc (compressor work)→ Tdischarge – compressor life

1a-1 → Pressure drop across suction valve

1-2 → Non-isentropic compression

suction pressure (ps) →specific volume → compressor volumetric efficiency (v)→wc

→wc

→ liquid flashes into vapor qewc

→ expansion device will not work properly

4-1d → pressure drop in evaporator

1d-1c → Superheat of vapor in evaporator

1c-1b → Useless superheat in suction line 1b-1a → Suction line pressure drop



→ efficient operation→ qe



→wc but smaller than that in vapor line

need cooling

Pipe design for small pdrop

Reduce pdrop by velocity flow or pipe size→ heat transfer coefficient in evaporator*minimum velocity is required to carry lubricating oil back to compressor

19. Multi-stage cycles

33

1. High pr; ratio of suction to discharge pressure is high enough to cause a serious

2. Low v or high Td; drop in volumetric efficiency or an unacceptably high discharge temp.

Two-stage cycle

a) with intercooling

b) with flash-gas removal and intercooling

c) with flash-gas removal

a) with intercooling and liquid-subcooler

20. Non vapor-compression cycles

34

20.1 Trans-critical carbon dioxide cycleTcr = 31C; heat rejection near Tcr – low refrigerating effectGas cooler as condenser without condensation

Mollier diagram for R744 showing transcritical cycle with evaporation at –10°C, compression to 100 bar and gas cooling to 40°C

120C40C

-10C

Liquid formation only takes place duringexpansion to the lower pressure level.

Actual compression process with compression efficiency 70-80%Isentropic processAt lower Td = 100°C

35

Total loss refrigerant: liquid nitrogen, carbon dioxide

20.2 Expendable refrigerant cooling system

Transportation refrigerationof perishable goods. Base system uses liquid, non-toxic, low temperature refrigerant. Low-cost liquid can be used as a refrigerant is then dumped into the atmosphere. This is called chemical or open cycle cooling.

Carbon dioxide is below triple point: snow or gasLiquid nitrogen released at -199Ccarbon dioxide released at -78C

36

20.3 Absorption cycle

removing evaporated refrigerant vapor without compressor

Lithium bromide - water

1810 J. Leslie - H2SO4 -absorbent, water-refrigerant

1860 F. Carre - aqua-ammonia absorption system. Water is a strong absorbent of NH3

Solar energy based absorption refrigeration systems in 1950s

GeneratorAbsorberPump

37

20.4 Air cycle

Air cycle refrigeration works on the reverse Brayton or Joule cycle. Air is compressed and then heat removed; this air is then expanded to a lower temperature.Expansion turbine can be useful energy to run compressors.Main application; air conditioning and pressurization of aircraft.Low COP

38

20.5 Stirling cycle

The Stirling cycle is an ingenious gas cycle which uses heat transferred from the gas falling in temperature to provide that for the gas rising in temperature.A detailed explanation of the cycle is referred to Gosney (1982) and Hands (1993). The Stirling cycle has been successfully applied in specialist applications requiring low temperatures at very low duties.

Portable refrigeration; The Free Piston Stirling Cooler (FPSC

invented in 1964) is a completely sealed heat transfer system that has only two moving parts (a piston and a displacer), and which can use helium as the working fluid.

39

20.6 Thermoelectric cooling

1821-- T.J. Seebeck - an electro motive force (emf)1834 -- J. Peltier - the reverse effect observedThe passage of an electric current through junctions of dissimilar metals causes a fall in temperature at one junction and a rise at the other.

1838-H.F.E. Lenz froze a drop of water using antimony and bismuth 1857– W. Thomson -- Thomson effect1949-A. F. Ioffe : thermoelectric refrigeration systems using semiconductors

1960s- mainly used for storing medicines, vaccines, in electronic cooling Some large refrigeration capacity systems such as a 3000 kcal/h air conditioner and a 6 ton capacity cold storage were also developed.

40

20.7 Magnetic refrigeration

Magnetic refrigeration depends on the magnetocaloric effect, which is the temperature change observed when certain magnetic mater-ials are exposed to a change in magnetic field. Magnetic refrigeration is a research topic, and historically has been used at ultra-low temperatures. Only recently has it been seen as a possible means of cooling at near room temperatures. An overview of magnetic refrigeration is given by Wilson et al. (2007)Ref. Wilson N , Ozcan S , and Sandeman K , Overview of magnetic refrigeration, IOR , 2006–7.

41

20.8 Steam jet refrigeration

water is sprayed into a low pressure chamber - evaporate - required good vacuum- low- grade energy process steam in chemical plants or a boiler 1838 F. Pelletan – patent of steam compression by a jet of motive steam. 1907 – ejectors - high velocity steam jet (≈ 1200 m/s)

42

20.9 Vortex tube

1931 G. Ranque – “vortex effect” tangential injection of air into a cylindrical tube Temperatures as low as −46°C and as high as 127°C are possible.

21. Natural Refrigeration

43

1.1 Use of Ice :Natural convection of air

1.3 Evaporative cooling :evaporation of water

1.4 Salt solution:Water + saltNaCl → -20C CaCl2 → -50 C

1.2 Dry ice : CO2 -78C

topic1-2 refrigeration cycle

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