Unit 12 – Refrigeration and air standard cycles
November 30, 2010
Outline
Unit Twelve – Refrigeration
and Air Standard Cycles
• Two somewhat related topics this week
• Use assumptions for cycle analysis
developed in the unit on Rankine cycles
• Refrigeration cycle fluids that exist in
both a liquid and a gas during the cycle
• Air-standard cycles model combustion
engines gases as air (an ideal gas) as
the working fluid
Mechanical Engineering 370
Thermodynamics
Larry Caretto
November 30, 2010
– Internal vs. external combustion
2
Review Cycle Analysis Basics
• Basic assumptions
Unit Twelve Goals
• As a result of studying this unit you
should be able to
– No line losses (output state of one
device is input to the next device)
– Work devices are isentropic (w = h)
– Heat transfer has no work and P = 0
– Exit from two-phase device is saturated
– calculate heat, work and coefficient of
performance for refrigeration cycles
– understand the assumptions for airstandard cycles
– compute heat, work, and efficiency for
otto, brayton, diesel and similar cycles
• Use actual data if available
• Account for different mass flow rates
3
Refrigeration Cycle
Refrigeration Calculations
• Only required input is pressure (or temperature) of evaporator and condenser
Evaporator outlet is
saturated vapor
Isentropic compressor
No work and P = 0 in
condenser and
evaporator
h = 0 for throttling
valve
3
• Pevaporator = P1 = P4 = Psat(T4 = T1)
• Pcondenser = P2 = P3 = Psat(T3 < T2)
2
Condenser
Throttling
Valve
Compressor
Evaporator
4
1
5
ME 370 – Thermodynamics
4
State 1 is saturated
vapor at Pevaporator
State 2, gas, s2 = s1
and P2 = Pcondenser
State 3 is saturated
liquid at Pcondenser
State 4, mixed, h4 = h3
and P4 = Pevaporator
3
2
Condenser
Throttling
Valve
Compressor
Evaporator
4
1
6
1
Unit 12 – Refrigeration and air standard cycles
November 30, 2010
R-134a Refrigeration Cycle
Refrigeration Cycle for R-134a
310
Tcondenser
10,000
Saturation
290
Pcondenser
Saturation
Compressor
Condenser
Throttle
Evaporaor
100
Temperature (K
Pressure (kPa)
300
1,000
T2 >
Tcondenser
280
Throttling
Valve
Compressor
270
Condenser
260
Evaporator
Pevaporator
250
Tevaporator
10
0
50
100
150
200
250
300
350
400
450
500
240
Enthalpy (kJ/kg)
1
1.1
1.2
7
1.4
1.5
1.6
1.7
1.8
Entropy (kJ/kg-K)
Refrigeration Calculations
•
•
•
•
•
•
•
1.3
Refrigeration Calculations II
Given Pevaporator and Pcondenser
h1= hg(Pevaporator) and s1= sg(Pevaporator)
h2 = h(Pcondenser, s2 = s1)
h4 = h3 = hf(Pcondenser)
|qL| = qevaporator = h1 – h4
|w| = |wcompressor| = h2 – h1
COP = |qL| / |w|
•
•
•
•
•
•
•
•
Given Tevaporator and Tcondenser
h1 = hg(Tevaporator)
s1= sg(Tevaporator)
P2 = P3 = Pcondenser = Psat(Tcondenser)
h2 = h(Pcondenser, s2 = s1)
Do not use
Tcondneser to
h4 = h3 = hf(Tcondenser)
find h2. T2 >
|qL| = qevaporator = h1 – h4
Tcondenser
|w| = |wcompressor| = h2 – h1
COP = |qL| / |w|
9
R-134a Example Calculation
R-134a Example Answers
• Pevaporator = 100 kPa; Pcondenser =1 MPa
• Find: Coefficient of Performance, COP
h1 = hg(100 kPa) = 234.44 kJ/kg
s1 = sg(100 kPa) = .95183 kJ/kgK
h2 = h(P2 = Pcond = 1 MPa, s2 = s13)
Condenser
= 282.53 kJ/kg
h3 = hf(1 MPa) = 107.32 kJ/kg
Throttling
h4 = h3= 107.32 kJ/kg
Valve
• |w| = |wcomp| = |h1 – h2| = |231.35 kJ/kg – 279.14
kJ/kg = 48.09 kJ/kg
• |qL| = |qevap| = |h1 – h4| = |231.35 kJ/kg – 105.29
kJ/kg = 129.15 kJ/kg
2
Compressor
Evaporator
4
COP 
COP 
qevap
wcomp
3

129.15 kJ
48.09 kJ
h1 - h4
h2 - h1
kg
kg
2
Condenser
Throttling
Valve
 2.69
Compressor
Evaporator
4
1
1
11
ME 370 – Thermodynamics
10
12
2
Unit 12 – Refrigeration and air standard cycles
November 30, 2010
Pevap= 140 kPa; Pcond = 0.9 MPa
Air-Standard Cycles
• Otto cycle is model for spark-ignition
(gasoline powered) engine
• Diesel (compression-ignition) cycle
• Brayton cycle for gas turbines
• Stirling cycle efficient but problems in
implementation
• h1 = hg(Pevaporator)
s1= sg(Pevaporator)
• If Tevaporator is given, Pevaporator = Psat at T
= Tevaporator so that h1 = hg(Tevaporator) and
s1= sg(Tevaporator)
• h2 = h(Pcondenser, s2 = s1)
• Note if Tcondenser is
239given,
.16 kJ you
101must
.61 kJ find

Pcondenserh1=Phsat
(T
).
kg
3 kg
4
qevap h3.-35

h
= h3=hhf(P
) kJCOP239
• h4COP
 .16 kJ  1 4
280.21
h condenser
2
1
kg
COP = ?

– Suggested use with solar collectors
• Others: Ericksson, Atkinson, Dual
• All have similar analysis
h2 - h1
w
kgcomp
13
14
Air-Standard Cycle Analysis
Otto Cycle Definition
• Use air properties as ideal gas with
variable or constant heat capacity
• Model chemical energy release as heat
addition (~1,200 Btu/lbm or 2,800 kJ/kg
for Otto cycle engine)
• Heat addition at constant pressure,
volume or temperature
• Isentropic work
• Closed system except Brayton Cycle
• Start at initial point with piston at bottom
of cylinder filled with air at v = v1
• Isentropic compression to v2
– Compression ratio, CR = v1/v2
• Constant volume heat addition at v2 = v3
• Isentropic expansion to v4 = v1
• Model exhaust as constant volume heat
rejection
15
16
Otto Cycle
14
12
10
Compression
Ratio = 10:1
8
6
Efficiency =
60.2%
4
2
0
0
0.2
0.4
0.6
0.8
Specific volume (m3/kg)
Compression
Expansion
ME 370 – Thermodynamics
Heat addition17
Heat rejection
1
•
•
•
•
•
•
•
•
•
T3 = T2 + qH/cv
v3 = v2
P3 = RT3/v3
v4 = v1 = (CR)v3
P4 = P3/(CR)k
T4 = P4 v4/R
|qL| = cv|T1 – T4|
 = 1 - |qL| / |qH|
 = 1 – (CR)1-k
16
14
12
Presure (MPa)
16
Presure (MPa)
• Given: CR, P1,
T1, qH
• v1 = RT1/P1
• v2 = v1/CR
• Isentropic compression to P2
• P2 = P1(CR)k
• T2 = P2v2/R
Otto Cycle Continued
10
Compression
Ratio = 10:1
8
6
Efficiency =
60.2%
4
2
0
0
0.2
0.4
0.6
0.8
1
Specific volume (m3/kg)
Compression
Expansion
Heat addition18
Heat rejection
3
Unit 12 – Refrigeration and air standard cycles
Otto Cycle Calculations
November 30, 2010
Otto Cycle Calculations II
• Given: P1 = 100 kPa; T1 = 300 K;
CR = 10; qH = 2,800 kJ/kg
• For air, R = 0.287 kPam3/kgK
• v1 = RT1/P1 = (0.287 kPam3/kgK) (300
K)/(100 kPa) = 0.861 m3/kg
• v2 = v1 / CR = 0.0861 m3/kg
• Continue with parallel computations for
constant and variable heat capacity
• Constant cv
P2 = P1(CR)k = (100)
(10)1.4 = 2,510 kPa
T2 = P2v2/R = (2510)
(0.0861) / (0.287) =
753.1 K
• Air tables
vr(300 K) = 621.2
vr(T2) = 621.2/10
Find: T2 = 730.0 K
u(T2) = 536.10 kJ/kg
P2 = RT2/v2 =
(0.287) (730.0) /
(0.0861 ) = 2,433
kPa
19
20
Otto Cycle Calculations III
Otto Cycle Calculations IV
• Air tables
• Constant cv
u(T3) = u(T2) + qH=
T3 = T2 + qH/cv =
753.1 K + (2800
536.10 kJ/kg +
kJ/kg) / (0.718
2800 kJ/kg =
kJ/kgK) = 4653 K
3336.10 kJ/kg
v3 = v2 (both
Find: T3 = 3512 K
calculations)
P3 = RT3/v3 =
P3 = RT3/v3 = (0.287) (0.287) (3512) /
(4653) / (0.0861 ) =
(0.0861 ) = 11 707
15 510 kPa
kPa
• Constant cv
• Air tables
P4 = P3/(CR)k = (100) vr(3512 K) = .09918
/(10)1.4 = 617.4 kPa v (T ) = .09918*10
r 4
v4 = v1 = 0.861 m3/kg
Find: T4 = 1854 K
(both calculations)
T4 = P4v4/R = (617.4) u(T4) = 1537.42
kJ/kg
(0.861)/(0.287) =
1852 K
u(T1) = u(300 K) =
214.36 kJ/kg
qL = cv(T4 – T1)
21
22
Otto Cycle Calculations V
Quiz Solution
• Constant cv
• Air tables
qL = cv(T4 – T1) =
qL = u(T4) – u(T1) =
(0.718 kJ/kgK)
1537.42 kJ/kg –
(1852 K – 300 K) =
214.36 kJ/kg =
1114.3 kJ/kg
1323.06 kJ/kg =
 = 1 - |qL|/|qH| = 1 –
 = 1 - |qL|/|qH| = 1 –
(1114.3 kJ/kg) /
(1323.06 kJ/kg) /
(2800 kJ/kg)
(2800 kJ/kg)
 = 60.2%
= 52.7%
– Given: Rankine cycle with
P3 = 1000 psia
data shown
3
Turbine
T3 = 800oF
– Find: Efficiency for (a)
ideal cycle, (b) cycle with
4
s,turbine = 85% and s,pump
Steam
Generator
= 80%
Condenser
– Point 1is saturated liquid;
P = 2 psia
h1 = 94.12 Btu/lbm, v1 =
2
0.016224 ft3/lbm
1
Pump
– h2 = h1 + |wP| = 94.12
0.016224 ft 3
1000
w
v
P
P
psia






Btu/lbm + 3.00 Btu/lbm = P 1 2 1
lbm
97.12 Btu/lbm
1 Btu
3.00 Btu
23
ME 370 – Thermodynamics
 2 psia 
5.40395 psia  ft 3

lbm 24
4
Unit 12 – Refrigeration and air standard cycles
November 30, 2010
Quiz Solution II
Quiz Solution III
h(800oF,
• h3 = h(T3,P3) =
1000 psia) = 1388.3
Btu/lbm and s3 = s(T3,P3) = 1.5662 Btu/lbm·R
• h4 = h(Pcond,s4 = s3) is in mixed region
h4  h f ( Pcond )  x 4 h fg ( Pcond )  h f ( Pcond ) 
•
•
•
•
s 3  s f ( Pcond )
h fg ( Pcond ) 
s fg ( Pcond )
1.5662 Btu 0.1751 Btu

lbm  R
lbm  R 1021.8 Btu 908.92 Btu
94.12 Btu


1
.
7445
Btu
lbm
lbm
lbm
lbm  R

With s  100% h1 and h3 do not change
Turbine work is 85% of work found above
Pump work = 3.00 Btu/lbm/80%
h2 = h1 + 3.00 Btu/lbm/80%
wP , s
 s ,T h3  h4  
wturb  wP

qSG
1382.4 Btu 908.92 Btu 3.00 Btu


w  wP h3  h4  wP
lbm
lbm
lbm
  turb

 36.6%

1382.4 Btu 97.12 Btu
qSG
h3  h2

lbm
lbm
 s,P

wP , s
h3   h1 

s,P






 1382.4 Btu 908.92 Btu 
1 3.00 Btu
 

0.85
lbm
lbm

 0.80 lbm
1382.4 Btu  94.12 Btu
1 3.00 Btu 

 

lbm
0.80 lbm 
 lbm
 = 31.0%
25
26
Quiz Eleven Solution
– Given: Rankine cycle
with data shown
– Find: (a) Efficiency for
ideal cycle
– Point 1 is saturated
liquid; h1 = 191.78
kJ/kg; v1 = 0. 001010
m3/kg
– h2 = h1 + |wP| = 191.78
kJ/kg + 8.07 kJ/kg +
199.85 kJ/kg
Quiz Eleven Solution II
45 MW Net Output
P3 = 8 MPa
3
Turbine
T3 = 500oC
4
Steam
Generator
• h3 = h(T3,P3) = h(500oC, 8 MPa) = 3397.8
kJ/kg; s3 = s(8 MPa, 500oC) = 6.7231 kJ/kg·K
• h4 = h(Pcond,s4 = s3) is in mixed region
x4 
Condenser
P = 10 kPa
2
1
Pump
0.001010 m3
8000 kPa
kg
1 kJ
8.070 kJ
 10 kPa 

kPa  m3
kg 27
wP1  v1 P2  P1  
s4  s f (10 kPa )
s fg (10 kPa )

 0.6487 kJ
kg  K
kg  K
 0.8099
7.5006 kJ
kg  K
6.7231 kJ
h4 = hf + x4hfg = 191.78 kJ/kg + (0.8099)(2392.4 kJ/kg) = 2129.3 kJ/kg
qH = h3 – h2 = 3397.8 kJ/kg – 199.85 kJ/kg = 3198.0 kJ/kg
|qL| = |h1 – h4| = |191.81 kJ/kg – 2129.3 kJ/kg| = 1937.5 kJ/kg
Net work, w = qH - |qL| = 3212.5 kJ/kg - 1961.8 kJ/kg = 1260.5
kJ/kg
The efficiency = w / qH = (1260.5 kJ/kg) / (3198.0 kJ/kg)
 = 39.4%
28
Quiz Eleven Solution III
• Find: (b) Mass flow rate of steam in cycle
m 
W
45 MW

w 1260.5 kJ
kg
1000 kJ 35.70 kg

MW  s
s
• Find: (c) Cooling water temperature rise if its
mass flow rate is 2000 kg/s
Q cw  Q L  m cw hcw ,out  hcw ,in   m cw c p ,cw Tcw ,out  Tcw ,in   m cw c p ,cw Tcw
Tcw 
Q L
m cwc p ,cw
m
q
 steam L 
m cw c p ,cw
35.702 kg 1937.5 kJ
s
kg
 8.27 o C
2000 kg 4.18 kJ
s
kg  K
29
ME 370 – Thermodynamics
5