Transcript Lecture 18: Control Volumes: System Analysis

EGR 334 Thermodynamics
Chapter 4: Section 10-12
Lecture 18:
Integrated Systems and
System Analysis
Quiz Today?
Today’s main concepts:
• Be able to explain what an integrated system is
• Be able to describe the components of some common integrated
systems
• Apply mass balance, energy balance, and continuity to streams of
flow through integrated systems.
Reading Assignment:
• No New Reading Assignment for Wed.
• Read Chap 5 for Friday.
Homework Assignment:
Problems from Chap 4: 95, 98, 102
3
Terms for Chap 5 Class Discussion :
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Spontaneous Heat Transfer
Clausius Statement
Kelvin-Planck Statement
Entropy Statement
Irreversible vs. Reversible
Internally Reversible Process
Carnot Corollaries
Carnot Efficiency
Max. heat pump efficiency
Max. refrigeration cycle COP
Carnot Cycle
Clausius Inequality
You may want to create a
summary sheet to help you
discuss each of the
concepts.
System Integration
►Engineers creatively combine components to
achieve some overall objective, subject to
constraints such as minimum total cost. This
engineering activity is called system integration.
►The simple vapor
power plant of Fig 4.16
provides an
illustration.
Sec 4.11: System Integration
5
Integrated Thermodynamics Systems:
System Integration : Combine components to make a useful cycle
Some common systems:
- Power Plant
- Refrigerator
Components:
- Heat Pump
Pipes
Nozzles/Diffusers
Turbines
Compressors/Pumps
Heat Exchanger
Throttling
Sec 4.11: System Integration
6
Power Plant Cycle
1. Boil Water
Burn something
Nuclear Reaction
Geothermal
2. Use the steam as it
rises to turn a turbine.
Sec 4.11: System Integration
7
Power Plant Cycle
3.Condense Steam
(to recycle water)
using a heat exchanger
4. Pump water back to boiler.
Sec 4.11: System Integration
8
Power Plant Cycle
2. Turbine
3. Condenser
4. Pump
1. Boiler
Sec 4.11: System Integration
9
Power Plant Cycle
QCV
1. Boiler
2. Turbine
.
m
tixe
WCV
.
m
ni
.
WCV
4. Pump
QCV
3. Condenser
Sec 4.11: System Integration
10
Refrigeration cycle
1.Condenser
We know that condensing
something will remove heat
from the fluid.
3. Evaporator
To have something to
condense, we must have
evaporated something.
(Both are heat exchangers)
Sec 4.11: System Integration
11
Refrigeration cycle
1. Condenser
4. Compressor
3. Evaporator
2. Throttling valve/
Expander
Sec 4.11: System Integration
12
Refrigeration Cycle
1. Condenser
QCV
WCV
4. Compressor
2. Throttling valve/
Expander
3. Evaporator
QCV
Sec 4.11: System Integration
13
Example: (4.103) A simple gas turbine power cycle operating at steady
state with air as the working substance is shown in the figure. The cycle
components include an air compressor mounted on the same shaft as the
turbine. The air is heated in the high-pressure heat exchanger before
entering the turbine. The air exiting the turbine is cooled in the lowpressure heat exchanger before returning to the compressor. KE and PE
effects are negligible. The compressor and turbine are adiabatic. Using
the ideal gas model for air, determine the
(a) Power required for the
compressor, in hp,
(b) Power output of the
turbine, in hp,
(c) Thermal efficiency of
the cycle
Sec 4.11: System Integration
Example: (4.103)
state
state
14
(a) Power required for the compressor, in hp,
(b) Power output of the turbine, in hp,
(c) Thermal efficiency of the cycle
11
22
33
44
11
>p11
pp22>p
=p22
pp33=p
11
p4=p
520
520
650
650
2000
2000
980
980
AV (ft
(ft33/min)
/min) 30,000
30,000
AV
(atm)
pp (atm)
(°R)
TT (°R)
(BTU/lb)
hh (BTU/lb)
124.27
155.51 504.71
236.02
From Table A-22E : Ideal Gas Properties of Air
state
Q
W
Compressor
0
?
Heat Ex 1
Turbine
Heat Ex 2
0
0
?
0
Assumptions
•Steady State
•KE=  PE = 0
•Turbine and compressor are
Adiabatic (QCV= 0)
•No work in Heat Ex. (WCV = 0)
•Air is modeled as an ideal gas
Sec 4.11: System Integration
15
Example: (4.103) (a) Power required for the compressor, in hp,
(b) Power output of the turbine, in hp,
(c) Thermal efficiency of the cycle
state
1
AV (ft3/min)
T (°R)
p2>p1
p3=p2
1
520
650
2000
980
124.27
155.51 504.71
state
Q
W
Compressor
0
?
Heat Ex 1
0
Turbine
236.02
0
?
Heat Ex 2
0
Ideal Gas Eq.
 A V 1
and
v1
3
m 
4
1
Using Continuity Eq.
m 
3
30,000
p (atm)
h (BTU/lb)
2
v
R T1
P1
so
m 
 A V 1
p1 M
R T1
2
(3 0, 0 0 0 ft / m in )(1 4 .7 lb f / in )( 2 8 .9 7 lb m / lb m o l ) 1 4 4 in 2 6 0 m in
(1 5 4 5 ft  lb f / lb m o l  R )(5 2 0 R )
1
ft
2
1
 137, 394
hr
lb
hr
Sec 4.11: System Integration
16
Example: (4.103) (a) Power required for the compressor, in hp,
(b) Power output of the turbine, in hp,
(c) Thermal efficiency of the cycle
state
1
AV (ft3/min)
p (atm)
T (°R)
h (BTU/lb)
2
3
4
30,000
1
p2>p1
p3=p2
1
520
650
2000
980
124.27
155.51 504.71
236.02
state
Q
W
Compressor
0
-1687
?
Heat Ex 1
Turbine
0
0
Heat Ex 2
?
0
Energy Balance (Compressor):
dE CV
dt
 QCV  WCV 

W C V   137 394 
2


V
mh
 gz 
2


lb m
hr
124.27  155.51 

BTU
lb m
W C V  m  h1  h 2 
1 hp
2545 B tu / hr
  1687 hp
Sec 4.11: System Integration
17
Example: (4.103) (a) Power required for the compressor, in hp,
(b) Power output of the turbine, in hp,
(c) Thermal efficiency of the cycle
state
AV (ft3/min)
p (atm)
T (°R)
h (BTU/lb)
1
2
3
4
30,000
1
p2>p1
p3=p2
1
520
650
2000
980
124.27
155.51 504.71
W C V  (137 394
lb m
hr
Q
W
Compressor
0
-1687
Heat Ex 1
Turbine
236.02
Energy Balance (Same as Compressor):
state
0
0
Heat Ex 2
14,505
0
W CV  m  h 3  h 4 
)  504.71  236.02 
BTU
lb m
1 hp
2545 B tu / hr
 14, 505 hp
Sec 4.11: System Integration
18
Example: (4.103) (a) Power required for the compressor, in hp,
(b) Power output of the turbine, in hp,
(c) Thermal efficiency of the cycle
state
1
AV (ft3/min)
h (BTU/lb)
3
4
30,000
p (atm)
T (°R)
2
p2>p1
p3=p2
1
520
650
2000
980
155.51 504.71
Q
W
Compressor
0
-1687
18,851
0
0
14,505
Heat Ex 1
1
124.27
state
Turbine
Heat Ex 2
236.02
0
Find Q23 Energy Balance around heat exchanger:
dE CV
dt
 Q CV  W CV 

Q C V  137 394
 

lb m
hr
2


v
m  h 
 gz 


2


  504.71  155.51 
W Turbne  W Compressor
Q HeatEx

Q C V  m  h3  h 2 

BTU
lb m
hp
2545 B tu / hr
14 ,505  1687
18 ,851
 18, 851 hp
 0 . 68
Sec 4.9: Heat Exchangers (Revisited)
19
Example: (4.78) As sketched in the figure, a condenser using river water to
condense steam with a mass flow rate of 2x105 kg/h from saturated vapor
to saturated liquid at a pressure of 0.1 bar is proposed for an industrial
plant. Measurements indicate that several hundred meters upstream of the
plant, the river has a volumetric flow rate of 2x105 m3/h and a
temperature of 15°C. For operation at steady state and ignoring changes in
KE and PE, determine the river-water temperature rise, in °C, downstream
of the plant traceable to use of such a condenser, and comment.
Sec 4.9: Heat Exchangers (Revisited)
20
Example: (4.78)
state
state
state
Pl,i
Pl,i
Pl,i
Pl,e
Pl,e
Pl,e
AV
Av(m
(m333/h)
/h)
Av
(m
/h)
m
m(kg/h)
(kg/h)
m
(kg/h)
R,i
R,i
R,i
R,e
R,e
R,e
2x10
2x10555
2x10
2x10
2x10555
2x10
000
000
???
2x10
2x10555
2x10
2x10
2x10555
2x10
111
000
pPP(bar)
(bar)
(bar)
0.1
0.1
0.1
0.1
0.1
0.1
TTT(°C)
(°C)
(°C)
45.81
45.81
45.81
45.81
15
15
15
hhh(kJ/kg)
(kJ/kg)
(kJ/kg)
2584.7
191.83
62.99
xxx
Look up intensive properties
for water from Tables based
on known property values
From Table A-2 : Properties of Saturated Water
mR 
 AV R
vR

2  10
5 m3
1 .0 0 0 9  1 0
 2  1 0 kg / h r
8
h
3 m 3
kg
Sec 4.9: Heat Exchangers (Revisited)
21
Example: (4.78)
state
Pl,i
Pl,e
Av (m3/h)
m (kg/h)
x
P (bar)
T (°C)
Using the energy balance
simplified for a heat exchanger h (kJ/kg)
R,i
R,e
2x105
2x105
2x105
2x105
2x108
2x108
1
0
0
0
0.1
0.1
45.81
45.81
15
?
2584.7
191.83
62.99
From Table A-2 : Properties of Saturated Water (p817 & 819)
dE CV
dt
 Q CV  W CV 

2


v

m h 
 gz 


2


0  m P l  h P l ,i  h P l , e   m R  h R ,i  h R , e 
h R , e  h R ,i 
m Pl
mR
h
P l ,i
 hP l ,e 
Sec 4.9: Heat Exchangers (Revisited)
22
Example: (4.78)
state
Pl,i
Pl,e
Av (m3/h)
m (kg/h)
h R ,e  h R ,i 
h R , e  6 2 .9 9
m Pl
m R
kJ
kg
h
Pl , i
5
8
2x105
2x105
0
0
?
2x105
1
0
P (bar)
0.1
0.1
T (°C)
45.81
45.81
15
h (kJ/kg)
2584.7
191.83
62.99
 h Pl , e 
 2  10

 2  10
R,e
2x105
x
Plugging known values:
R,i
From Table A-2 : Properties of Saturated Water (p817 & 819)
kg / h r 
kg / h r 
 2 5 8 4 .7  1 9 1 .8 3 
kJ
 6 5 .3 8
kg
Using this value, the exit temperature may be found from Table A-2.
TR,e = 15.6°C 
giving a temperature rise of 0.6 °C
kJ
kg
23
end of Lecture 18 Slides