Lecture 38: Vapor-compression refrigeration systemscc.sjtu.edu.cn/Upload/20160505155144117.pdf ·...
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
1.3
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.
1.6
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
1.15
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
1.18
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
1.25
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