topic1-2 the vapor-compression cycle

Topic1-2 The Vapor-Compression Cycle

The vapor-compression cycle: the most widely used refrigeration in practice

Carnot cycle: the highest efficiency cycle

Net work

1-2 Adiabatic compression

Heat to low-Temp. sink

Heat from high-Temp. source

Tem

per

atu

re, K

1

2 3

4

1

2 3

4

Work

Compressor

2-3 Isothermal addition of heat

3-4 Adiabatic expansion4-1 Isothermal rejection of heat

Work

Turbine? ?

Entropy, kJ/kgK

Carnot cycle: an ideal-Standard of comparison- Guide of T to max eff.

1

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

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

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 (cont.)

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= w(h2-h1)Qe = w(h1-h4)

Qc = w(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

Heat rejection rate = qc = h2 – h3

Refrigeration effect = qe = h1 – h4

COP = qe/wc

Volume flow rateper kW of refrigeration = wvsuc /Qe

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

Example 10-1 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) w = Qe/qe

compressor, energy balance, (c) Pcomp = w(h1-h2)(d) COP = Qe/Pcomp

(e) Vsuc = wvsat vapor@pevap, pevap= psat@tevap

(f) P/Qe = 1/COP(g) tdischarge = t@(pcond, s2=s1), pcond= psat@tcond

Pre

ssu

re,

kPa

1

23

4

Enthalpy, kJ/kg

35C= tcond

-10C= tevap

(a)qe (b)w (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

(b) w = Qe/qe

(c) Pcomp = w(h1-h2)(d) COP = Qe/Pcomp

(e) Vsuc = wvg@pevap

(f) P/Qe = 1/COP(g) tdischarge = t@(pcond, s2=s1)

h@0 C =200 kJ/kg

Example 10-1 VCC R22 at 35C and -10C

p-h diagram R717

18

80 kPa = Psuction

1000 kPa = Pdischarge

1

3

h1=1410 h3=1800h@0 C =200 kJ/kg

h6=316

Example VCC ammonia at 80 kPa to 1000 kPa

12. Liquid Subcool Heat exchanger (LSHX)

19

Subcool liquid from condenser with suction gas from evaporator

Heat balance: h3-h4 = h1-h6

Refrigerating effect: h6-h5 = h1-h3

qe = h6-h5

19

13. Actual vapor-compression cycle

20

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 pi to comp.

Exam10-2 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 และ subcool คำ่อุณหภมูทิีข่ำออก Condenser เป็นเทำ่ใด, h) ถำ้เพิม่ อุณหภมูิ Evaporator Te เป็น 10C โดยทีป่รมิำตร suction เทำ่เดมิ จะไดอ้ตัรำกำรไหลเชงิมวล เพิม่ขึน้หรอืลดลง เพรำะData: VCC, R22, te, tc ,

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

Method: (b) COPcarnot = T1 / (T2 – T1)=6.825CV evaporator, energy balance, (c) qe = h1-h4

CV compressor, energy balance, (d) w = (h1-h2)(e) tdischarge =(f) COP = qe/wc=5.85

CV liq-to-suc HE, energy balance, h3-h3’ = h1’-h1

(g)tsp = 10C, t3’= 40-5.5C(h) te w

Pre

ssu

re,

kPa

1

23

4

Enthalpy, kJ/kg

3’

1’

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

40C= tcond

0C= tevap

21

22

0 C = tevap

40 C= tcond

1

2

h1=405 h2=432h4=250

T2=60C

3

4

s1=s2=1.752

1’

3’

high T1, low T2 → high COP

23


Refrigeration

23

Net work

23

Tem

pe

ratu

re, K

Entropy, kJ/kgK

1

23

4

T2

T1

s1 = s2s3 = s4

T1 → more effect on COP than T2

14.1 Analysis of Carnot refrigeration cycle

T1

T2

2424

14.2 Comparison between Carnot & standard VCRS

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

R = COPStandard VCRS / COPR Carnot

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

COP asTevap

or Tsuction

COP as Tcond

25

1

23

4

Isen

thal

pic

Isobaric

Isobaric

Standard VCRS cycle

asTevap → RE(qe) but wcomp then COP

→ qcond and %flash gas

1’

2’

4’

25

Refrigeration effect

Net work

Refrigeration effect

26

Tem

per

atu

re, K

Entropy, kJ/kgK

1

23

4


Tem

per

atu

re, K

Entropy, kJ/kgK

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

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

qewc

14.2 Comparison between Carnot & standard VCRS

Good for heat pumps

wc

27

14.3 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 → -→wc in R717 but not in R22→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

14.4 Actual vapor-compression refrigeration cycle

28

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

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

29

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






30

Discharge line: vapor line

Liquid line: liquid

Suction line: vapor line

Standard cycle

14.4 Actual vapor-compression refrigeration 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

topic1-2 the vapor-compression cycle

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