Lecture 17: Control Volumes: Applications 2

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EGR 334 Thermodynamics
Chapter 4: Section 9-10
Lecture 17:
Control Volume Applications:
Day 2
Quiz Today?
Today’s main concepts:
• Be able to set up mass and energy balance models for
Turbines
Pumps
Compressors
Boilers
Heat Exchangers
Nozzles
Diffusers
Throttle
Reading Assignment:
• Read Chapter 4, Sections 11-12
Homework Assignment:
Problems from Chap 4: 75, 83, 87, 90
3
Modeling applications with Control Volumes:
Many important applications involve one inlet, one exit control volumes at
steady state. Other systems include feature additional inlet and outlet
streams. One very common device, the heat exchanger, requires multiple
streams of mass flow to be modeled.
Mass Rate Balance: multi-path, steady state
dm C V
0  m i m e
mi
dt
me
Energy Rate Balance: 1 path, steady state
1


2
2
0  Q cv  W cv   m  ( hi  h e )  (V i  V e )  g ( z i  z e ) 
2


W
Q
dECV
 m i ei
dt
 m e ee
4
Control Volume Applications:
Nozzles
Compressor
Diffuser
Boiler
Turbine
Pump
Throttling Valve
Heat Exchanger
Heat Exchangers
►Direct contact: A mixing chamber in which hot and
cold streams are mixed directly.
►Tube-within-a-tube counterflow: A gas or liquid
stream is separated from another gas or liquid by a
wall through which energy is conducted. Heat
transfer occurs from the hot stream to the cold
stream as the streams flow in opposite directions.
Sec 4.9: Heat Exchangers
6
Direct Contact Heat Exchangers:
In direct contact heaters, mass streams combine
and exchange heat by mass transport as well as by
convection and conduction.
m
Examples:
H ot _ in
m
C old _ in
 m exit
Swamp Coolers
Hot-Cold water
faucet
Cooling Towers
Sec 4.9: Heat Exchangers
7
Separate Flow (Tube) Heat Exchangers:
Heat transfer occurs by conduction
through a material that separates
individual flow streams
Parallel Flow: Streams enter and
leave along same directions
Cross Flow: Stream pass along
direction perpendicular to each
other
Counter Flow: Streams enter and leave
opposite each other.
8
Tube Heat Exchanger are all around you.
Still
Radiator
Home air conditioning
Industrial
Heating Boiler
Geothermal
Heating
Refrigerator
condenser
Industrial
Systems
Power Plant
Sec 4.9: Heat Exchangers
A heat exchanger attempts to maximize the heat transfer between
separate fluid streams.
Designed to maximize “contact” between the two fluids
Maximize time of contact
and
Maximize area of contact
9
Sec 4.9: Heat Exchangers
10
For Separate Stream Heat Exchangers:
Mass Balance Model:
m
F luid 1 _ in
m
F luid 1 _ exit
and
m
F luid 2 _ in
m
F luid 2 _ exit
Energy Balance Model:
dE C V
dt
 QCV  WCV
2
2




Vi
Ve
  m i  hi 
 gz i    m e  he 
 gz e 
2
2




Steady Insulated
No work
State
from
surroundings
V0
Horizontal Section
(or very short vertical)
0  m f 1_ i  h f 1_ i  h f 1_ e   m f 2 _ i  h f 2 _ i  h f 2 _ e 
First Fluid
Second Fluid
Sec 4.9: Heat Exchangers
11
Example: (4.81) A feedwater heater operates at steady state with liquid water
entering at inlet #1 at 7 bar, 42 °C and a mass flow rate of 70 kg/s. A separate
stream of water enters at inlet 2 as a two phase liquid-vapor mixture at 7 bar with a
quality of 98%. Saturated liquid at 7 bar exits the feedwater heater at #3. Ignoring
heat transfer with surroundings and neglecting KE and PE effects, determine the
mass flow rate in kg/s at inlet #2.
state
1
m (kg/s)
70
x
2
3
0.98
0
p (bar)
7
7
7
T (°C)
42
165
165
175.9
2722.17
697.22
h (kJ/kg)
Look up h values on
Table A-2 and A-3
#2
7 bar
98%
#1
7 bar
42 °C
70 kg/s
#3
7 bar
Sat. Liq.
Assumptions
•Steady State
•KE=  PE = 0
•QCV= 0
Sec 4.9: Heat Exchangers
12
Example: (4.81) Determine the mass flow rate in kg/s at inlet #2.
state
1
m (kg/s)
70
2
x
3
0.98
0
p (bar)
7
7
7
T (°C)
42
165
165
175.9
2722.17
697.22
h (kJ/kg)
Mass Balance:
h3  h f ( a t 7 b a r )  6 9 7 .2 2
kJ
kg
h2  h f  x ( h g  h f )
 697.22
kJ
kg
 2722.17
h1  h
f
 0.98[2763.5  697.22] kJ
kg
kJ
kg
 a t T1   1 7 5 .9 kJkg
Energy Balance:
m 1  m 2  m 3
Combining
From Table A-3
In
Out
0  m 1 h 1  m 2 h 2  m 3 h 3
0  m 1 h 1  m 2 h 2   m 1  m 2 h 3
 h 1h 3 
 175.9  697.22 
m 2  m1 
 18.0 kg / s
   70 kg / s  

 697.22  2722.2 
 h 3h 2 
Sec 4.9: Heat Exchangers
13
Example: (4.85) A parallel flow heat exchanger has separate streams of air and
water. Streams undergo no significant change of pressure. Ignore heat loss to the
surroundings and changes in KE and PE. Apply ideal gas to the air stream. If both
streams leave at the same temperature, determine the exit temperature.
state
state
11
22
33
H
H22O
O
44
Air
Air
m
m (kg/s)
(kg/s)
10
10
10
10
55
55
pp (bar)
(bar)
11
11
11
11
TT
99.63 oC
hh (kJ/kg)
(kJ/kg)
1200
1200 KK
2675.5
from Table A-3 at psat = 1 bar, T1= Tsat = 99.63 C and h1=hg=2675.5 kJ/kg
Using energy balance: for the heat exchanger
dE C V
dt
 QCV  WCV
2
2




Vi
Ve
  m i  hi 
 gz i    m e  he 
 gz e 
2
2




Reducing this with assumptions
0  m H 2 O ( h1  h 2 )  m a ir ( h3  h 4 )
Sec 4.9: Heat Exchangers
14
Example: (4.85) A parallel flow heat exchanger has separate streams of air and
water. Streams have no significant change of pressure. Ignore heat loss to the
surroundings and changes in KE and PE. Apply ideal gas to the air stream. If both
stream leave at the same temperature, determine the exit temperature.
state
1
2
3
H 2O
4
Air
m (kg/s)
10
10
5
5
p (bar)
1
1
1
1
T
99.63oC
h (kJ/kg)
2675.5
1200 K
Using ideal gas model
0  m H 2 O ( h1  h 2 ( Texit , 1 b a r ))  m a ir c p (T3  T exit )
Note: since h2 depends upon T2=T4 and p2 = 1 bar, it could be found from Table
A-4 if the exit temperature were known. This problem can be solved by
picking a value of T2 and then looking up h2 and checking to see if it satisfies
the equation. It can also be easily solved using IT’s iterative solver.
Sec 4.9: Heat Exchangers
15
Solution using IT
Exit Temperature
T2  T4  5 6 1 K  2 8 8 C
o
Then the heat transfer
between the two fluids is
Q a ir  H 2 O  m H 2 O ( h 2  h1 )
 (1 0 kg / s )(3 0 4 9  2 6 7 5) kJ / kg
 3 7 4 0 kJ / s
state
1
2
3
H 2O
4
Air
m (kg/s)
10
10
5
5
p (bar)
1
1
1
1
T
99.63oC
288
1200 K
288
h (kJ/kg)
2675.5
3049
Throttling Devices
►Throttling Device: a device that achieves
a significant reduction in pressure by
introducing a restriction into a line
through which a gas or liquid flows. Means
to introduce the restriction include a
partially opened valve or a porous plug.
Sec 4.10: Throttling Devices
17
Throttling Device: Reduces Pressure
Mass Balance:
 m
in

 m
i
exit
 m in  m exit  m
i
Typical Energy Balance simplifications,
dE CV
dt
Steady
State
 Q CV  W CV
2
2




vi
ve
 m i  hi 
 gz i   m e  h e 
 gz e 
2
2




No heat No work
transfer
v0
hi  h e
Horizontal Section
(or very short vertical)
Sec 4.10: Throttling Devices
18
Example: (4.92) Refrigerant 134a enters the expansion valve of an air
conditioning unit at 140 psi, 80 °F and exits at 50 psi. If the refrigerant
undergoes a throttling process, what are the temperature, in °F, and the
quality at the exit valve?
state
inlet
x
Assumptions
• Steady State
• KE= PE = 0
• QCV= WCV = 0
exit
?
P (psi)
140
50
T (°F)
80
?
h (BTU/lb)
37.27
Enthalpy of saturated liquid at 80°F
140 psi
80 °F
50 psi
Sec 4.10: Throttling Devices
19
Example: (4.92) Refrigerant 134a
state
inlet
x
exit
?
P (psi)
T (°F)
h (BTU/lb)
140
80
50
40.27
At 140 psi, the saturation T is 100.56 °F, so this
is a super-cooled liquid. There are no tables for
super-cooled liquid, so neglect pressure effects.
h1  h f ( T at 80 F )  37.27
o
37.27
h 2  h1 = 3 7 .2 7
From Table A-10E
h2  h f  x ( h g  h f )  x 
he  h f
hg  h f
x 
BTU
lb
BTU
lb
37.27  24 . 14
83 . 29
 0 . 158
20
end of Lecture 17 Slides
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