ME 200 L18: Conservation Laws: Heat Exchangers
HW 7 Posted Due in One Week:
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Common Steady-flow Energy Devices
Nozzles
Diffusers
Heat Exchangers and Mixers
Water, Steam, Gas Turbines
Compressors
Pump
Throttles
2
Rotating Machinery
A turbine is a steady-flow
device used to produce
mechanical work (W) by
reducing the internal &/or
kinetic &/or potential energy
of the working fluid.
• For gas turbines, the fluid
drives rotating blades
while the υ increases from
inlet to exit as the working
fluid expands (or the p
drops).
3
Steam Turbine Example
6 Kg/s of steam at an inlet velocity of 75 m/s enter a turbine stage at 3 MPa and
400 oC and exit at a velocity of 125 m/s at a pressure of 2 Mpa at 360 oC. Find: (a)
the power developed by this steam turbine stage, (b) the percentage change in
the steam density across the turbine and (c) the percentage change in the flow
area. The heat loss through the casing is 33 kW.
Q cv  33 kW
m 1  6 kg / s
Wcv
m 1  6 kg / s

V12
V22 
Q Turbine  WTurbine  m  h1 
 h2 

2
2 

V
AV
AV
me  m 2  m  2  2 2  1 1
v2
v2
v1
2
2

75 
125  


33  WTurbine  6  3230.9 
 3159.3 



2000
2000


(a ) WTurbine  33  6 (71.6  5)  366.6 kW
(b) (2  1 ) / 1  (v1  v2 ) / v2  (0.094  0.141) / 0.141  33%
(c ) A2V2 / v2  AV
1 1 / v1  125 A2 / 0.141  75 A1 / 0.094
A2 / A1  (75 /125) *(0.141/ 0.094)  0.9
4
Air Compressors
A pump is a steady-flow
device that consumes
shaft work from rotating
blades that compress the
fluid.
• Compressors are used for
gas systems, pumps for
liquids so operating
assumptions are similar.
September 17th, 2010
ME 200
5
Example
At steady state, a well-insulated compressor takes in
air at 60 ºF, 14.2 psi, with a volumetric flow rate of
1200 ft3/min, and compresses it to 500 ºF, 120 psi.
Kinetic and potential energy changes from inlet to
exit can be neglected. Determine the compressor
power, in hp, and the volumetric flow rate at the
exit, in ft3/min.
6
Example
Find
Assumptions
– Wcv = ? in hp
– A2V2 = ? in ft3/min
System (air flowing through
compressor)
• The control volume is at steady
state; the flow is steady
• Q, Δke, and Δpe are negligible.
• The air is an ideal gas.
Basic Equations
1
air
P1 = 14.2 psi
T1 = 60 ºF
A1V1 = 1200 ft3/min
compressor
2
P2 = 120 psi
T2 = 500 ºF
dmcv
  m i   m e
dt
i
e
P  RT
2
2




dEcv 
V
V
i
e




 e he 
 Qcv  Wcv   m i hi 
 gzi   m
 gze 

 e 

dt
2
2
i




7
Example
Solution
dmcv
  m i   m e
dt
i
e
m 1  m 2  m




dEcv 
Vi2
Ve2




 e he 
 Qcv  Wcv   m i hi 
 gzi   m
 gze 

 e 

dt
2
2
i




 h1  h2 
Wcv  m

m
AV


RT
P
AV P
m  1 1 1
RT1
8
Example
14.2 psi 1200 ft 3 min  60 min 144in 2
m
2
1540 ft  lbf
1
h
1
ft
 520R 
28.9 lbm  R
  5310 lbm h
m
 h1  h2 
Wcv  m
From Table A-22E
h1  124 Btu lbm
Wcv  5310
h2  231 Btu lbm
lbm
Btu
1hp
124

231


h
lbm 2540 Btu h
Wcv  223hp
9
Example
A2V2  m 2
m RT2
A2V2 
P2
5310
A2V2 
lbm  1540 lbf  ft 

 960 R 1 ft 2
1h
h  28.9 lbm  R 
120 psi
144in 2 60 min
A2V2  262 ft 3 min
10
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.
Heat Exchanger Modeling
2
2
V
V
0  Qcv  Wcv   m i (hi  i  gzi )   m e (he  e  gze )
2
2
i
e
► Wcv  0 if there is no stirring shaft or moving boundary.
► ΔKE = (Vi2/2-Ve2/2) negligible unless specified.
 e gze  m i gzi ) negligible unless specified.
► ΔPE = (m
► Qcv  0 If Heat transfer with surroundings is negligible.
► Control Volume includes both hot and cold flows. The
“heat exchange,” between them is internal!
0   m i hi   m ehe
i
e
Example Problem: Heat Exchanger
Given: Air and Refrigerant R-22 pass through separate streams through
an insulated heat exchanger. Inlet and exit states of each are defined.
Find:
(a) Mass flow rates, (b) Energy transfer from air to the refrigerant.
3
R-22
Air
2
1
Assumptions: Flow work only, insulated casing,
steady state, steady flow, no leaks.
Data: Av1= 40m3/min,1=27 C=300K, P1= 1.1 bars
T2=15 C= 288 K, P2= 1bar
P4= 7 bars, T4=15 C, P3=7 bars, x3=0.16
4
0  m R 22 h3  m Air h1  m R 22 h4  m Air h2
 h1  h2 

m R 22  m Air 
 h4  h3 
AV1 P1 AV1
(1.1bars)( 40m3 / min)


v1
RT1
(0.287 kJ / kg  K )(300 K )
 51.11kg / min
m Air 
R22 properties: Table A-9, A-8. P=7 bars, Tsat=10.91 C. Therefore, 4 is
superheated and h4=256.86 kJ/kg. Table A-8 h3=hf3+x3hfg3= 58.04+(.16)(195.6)
=89.34 kJ/kg.
Example Problem: Heat Exchanger
Given: Air and Refrigerant R-22 pass through separate streams through
an insulated heat exchanger. Inlet and exit states of each are defined.
Find:
(a) Mass flow rates, (b) Energy transfer from air to the refrigerant.
3
R-22
Air
2
1
4
Assumptions: Flow work only, insulated casing,
steady state, steady flow, no leaks.
Data: Av1= 40m3/min,1=27 C=300K, P1= 1.1 bars
T2=15 C= 288 K, P2= 1bar
P4= 7 bars, T4=15 C, P3=7 bars, x3=0.16
h h 
 300 .19  288 .15 
m R 22  m Air  1 2   51.11

h

h
256
.
86

89
.
34


 4 3
 3.673 kg / min

 R 22 ( h4  h3 )  3.673(256 .86  89.34)  615 .3 kJ / min
Q
R 22  m

 Air (h2  h1 )  51.11288 .15  300 .19   615 .3 kJ / min
Q
Air  m
Summary
• Control volume energy and mass conservation
equations that we learned are applicable to many
practical energy devices and equipment.
• Important learning comes from application of
appropriate assumptions, considering the
appropriate working substances and their
properties in the proper range of operation to
estimate different energy quantities.
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