Lecture 38: Vapor-compression refrigeration systemscc.sjtu.edu.cn/Upload/20160505155144117.pdf ·...

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1.1 ME 200 –Thermodynamics I Lecture 38: Vapor-compression refrigeration systems Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 200240, P. R. China Email : [email protected] Phone: 86-21-34206056; Fax: 86-21-34206056

Transcript of Lecture 38: Vapor-compression refrigeration systemscc.sjtu.edu.cn/Upload/20160505155144117.pdf ·...

1.1

ME 200 –Thermodynamics I

Lecture 38: Vapor-compression refrigeration systems

Yong Li

Shanghai Jiao Tong University

Institute of Refrigeration and Cryogenics

800 Dong Chuan Road Shanghai, 200240, P. R. China

Email : [email protected]

Phone: 86-21-34206056; Fax: 86-21-34206056

1.2

Carnot Refrigeration Cycle

1.3

Coefficient of Performance β

Coefficient Of Performance ::: Ratio of the refrigeration

effect to the net work input required to achieve that effect.

(COP)

Represents the maximum theoretical coefficient of performance of any refrigeration cycle operating between regions at TC and TH.

1.4

Departures from the Carnot Cycle

Compression two-phase liquid–vapor mixture. Wet compression.

Temperature difference T

.

In actual systems, compressor handles vapor only. Dry compression.

The work output of the turbine is normally

sacrificed by using a simple throttling

valve for the expansion turbine, saving in

initial and maintenance costs.

1.5

Departures from the Carnot Cycle

T

Dry compression

Throttling valve

1.6

Cycle Analysis

P

h

2 3

4 1

pcond

pevap

T

s

2

1

3

4

Tcond

Tevap

TH

TL

subcooling

Tsc

superheat: Tsh

Compressor Throttling

Valve

Condenser

1

2 3

4

Evaporator

W

Qcond

Qevap

T-s diagram P-h diagram

1.7

Ideal Vapor-Compression Refrigeration Cycle

Process 1–2s: Isentropic compression of the

refrigerant from state 1 to the condenser

pressure at state 2s.

Process 2s–3: Heat transfer from the

refrigerant as it flows at constant pressure

through the condenser. The refrigerant exits as

a liquid at state 3.

Process 3–4: Throttling process from state 3 to

a two-phase liquid–vapor mixture at 4.

Process 4–1: Heat transfer to the refrigerant as

it flows at constant pressure through the

evaporator to complete the cycle.

All of the processes in the

above cycle are internally

reversible except for the

throttling process.

1.8

Analyzing Rankine Cycle---I

Assumptions: 1) Steady state. 2)

ΔKE=ΔPE=0. 3)stray Q=0

Turbine

Condenser

Pump

e

ee

ee

i

ii

iicvcvcv gz

Vhmgz

VhmWQ

dt

dE

22

22

Boiler

1.9

Vapor-compression refrigeration cycle

Evaporator:

The heat transfer rate is referred to

as the refrigeration capacity. ( kW).

» Another unit for the refrigeration

capacity is the ton of refrigeration, =

211 kJ/min.

Compressor

Condenser

Throttling process

Turbine

Condenser

Pump

Boiler

e

ee

ee

i

ii

iicvcvcv gz

Vhmgz

VhmWQ

dt

dE

22

22

1.10

Example 1: Ideal Vapor-Compression Refrigeration Cycle

Known: R-134a as the working fluid and operates on an

ideal vapor-compression refrigeration cycle between 0.14

and 0.8 MPa. mass flow rate of the refrigerant is 0.05 kg/s,

Find: Qin, Win, mass flow rate, Qout

Assumptions: 1) Steady state. 2) ΔKE=ΔPE=0.

Analysis:

1.11

Actual cycle

The heat transfers between the

refrigerant and the warm and

cold regions are not

accomplished reversibly:

Tevaporator < TC, Tcondenser > TH.

» The COP as Tevaporator and COP

as TH.

Adiabatic irreversible

compression

1.12

Cycle Analysis

Second Law Efficiency:

actual

Carnot

COP

COP

Irreversibilities

T

s

2

Tcond

Tevap

TH

TL

3

4 1

1.13

Compressor Analysis

Overall isentropic Efficiency:::Ratio of isentropic

compressor power input to actual compressor power input:

r 2s 1

o,is

comp

m h h

W

1.14

Household Refrigerator

1.15

Vapor Compression Cycles

Best possible performance for a heat pump cycle:

» Carnot Cycle!

What is the difference between

a.) Refrigeration/Air Conditioning, and

b.) Heat Pumping?

1.16

Refrigerator & Heat pump

The objective of a

refrigerator is to remove heat

(QL) from the cold medium;

The objective of a heat pump

is to supply heat (QH) to a

warm medium

1.17

A heat pump can be used to heat a house in winter and to cool it in summer

1.18

Example 2:Vapor Compression Cycles

Consider a 300 kJ/min refrigeration system that operates on an vapor-compression

cycle with R134a as the working fluid. The refrigerant enters the compressor as

superheated vapor at 140 kPa and is compressed to 800 kPa. If the isentropic

efficiency of the compressor is 0.85, and the superheat and subcooling are 5 oC,

determine: a) the quality of the refrigerant at the end of the throttling process, b) the

coefficient of performance, c) the power input to the compressor (kW).

Given: p1 = p4 = 140 kPa, Tsuperheat = 5 oC, p2 = p3 = 800 kPa, Tsupcooling = 5 oC,

QL = 300 kJ/min Find: x4 , COPR , Win

Assumptions:

» Steady state, steady flow, Neglect KE and PE

» Adiabatic compression

» Constant pressure heat transfer processes

» No work or heat interactions in the expansion process

1.19

Example 2

Solution:

» Quality at the end of the throttling process:

» From R134a Tables: h3 hf (T3) = 87.85 kJ/kg

hf(p4) = 27.08 kJ/kg

hfg(p4) = 212.08 kJ/kg

3 4

4 4

4

4

3 3 3

3

3

,

800

26.31

f

fg

sat subcooling

o

h h

h h px

h p

h h p T

T T kPa T

T C

4 0.287x

p1 = p4 = 140 kPa, Tsuperheat = 5 oC, p2 = p3 =

800 kPa, Tsupcooling = 5 oC, QL = 300 kJ/min

1.20

Continue Solution:

» Coefficient of Performance cooling:

– Compressor analysis:

1 4

2 1

1 1 1

1 1 superheat

1

1

,

18.77 5 13.77

242.12

R

sat

o o o

kJkg

h hCOP

h h

h h p T

T T p T

T C C C

h

2 2 2

2 1

2

,

0.9604

280.41

s s

kJkg Ks

kJkg Ks

h h p s

s s

h

Example 2 p1 = p4 = 140 kPa, Tsuperheat = 5 oC, p2 = p3 =

800 kPa, Tsupcooling = 5 oC, QL = 300 kJ/min

1.21

Continue Solution:

» Coefficient of Performance cooling:

» Power input:

2 1

2 1

1 4

2 1

3.425

sC

C

R

s

R

h h

h h

h hCOP

h h

COP

12 hhmWin

41 hhmQin

Example 2 p1 = p4 = 140 kPa, Tsuperheat = 5 oC, p2 = p3 =

800 kPa, Tsupcooling = 5 oC, QL = 300 kJ/min

1.22

Continue Solution:

» Power input:

1 4

2 1

1 4

min300 1min

3.425 60

1.46

in

inin

kJin

in

R

in

Qm

h h

QW h h

h h

QW

COP s

W kW

Example 2 p1 = p4 = 140 kPa, Tsuperheat = 5 oC, p2 = p3 =

800 kPa, Tsupcooling = 5 oC, QL = 300 kJ/min

1.23

Cycle Analysis

Compressor Inlet State

p1 = pevap h1 = h (Tevap+Tsh , pevap)

Compressor Outlet State

p2 = pcond

h2s = h (p2 , s2 = s1)

isentropic specific work

2s 12 1

s

isentropic efficiency

(h h )h h

T

s

2

Tcond

Tevap

p2

1

2s

1.24

Reading material

1.25

Cycle Analysis

Condenser Outlet State

p3 = pcond

h3 = h (Tcond+Tsc , pcond)

Evaporator Inlet State

p4 = pevap

h4 = h3

P

h

3

4

Pcond

Pevap

1.26

Cycle Analysis

Coefficient of Performance

Volumetric Capacity

P

h

2 3

4 1

pcond

pevap

q = v

refrigerating effect

compressor inletspecific volume

h h

v1 4

1

COPh h

h h =

refrigerating effect

specific work

1 4

2 1

indicator of the compressor

displacement requirement

1.27

Cycle Analysis

T

s

2

1

Tcond

Tevap

TH

TL

Sources for Irreversibilities

3

4

Throttling Losses

T

s

Tcond

Tevap

TH

TL

3

4

Compressor Losses

1

2 3s

2s

1.28

Cycle Analysis

Sources for Irreversibilities

» also includes losses due to pressure drops in heat exchangers and

distribution piping

T

s

2s

1

Tcond

Tevap

TH

TL

3

4

Desuperheating Losses

T

s

2s

1

Tcond

Tevap

TH

TL

3

4

Heat Exchange Losses

3s