Application of Control Volume Energy Analysis
Most thermodynamic devices consist of a series of
components operating in a cycle, e.g., steam power plant
Main components of the above cycle are:
1) Boiler (steam generator) – heat exchanger
2) Turbine – generates work
3) Condenser – heat exchanger
4) Pump
Others components include:
nozzles, diffusers, throttling devices
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Nozzles and Diffusers
Devices that increase or decrease the flow velocity by
passing the flow through a variable area duct, A1  A2
1
2
Applying conservation of mass assuming steady flow:
dM CV
 1 A1V1   2 A2V2
dt
1 A1V1   2 A2V
For low subsonic flow (1 = 2)
V2 A1

V1 A2
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Subsonic Nozzle: A2 < A1  V2 > V1
Subsonic Diffuser: A2 > A1  V2 < V1
Aircraft gas turbine
diffuser
nozzle
Applying the energy equation (assuming steady, no heat
loss, PE=0):
dE
 q  w s  (hi  Vi 2 / 2  gZ i )  (he  Ve2 / 2  gZ e )  0
dt
h1  V12 / 2  h2  V22 / 2
V22  V12  2(h1  h2 )
For a rocket nozzle V2 >> V1
V22  2(h1  h2 )
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Turbine
A device in which shaft work is generated as a result of
gas passing through a set of blades attached to a freely
rotating shaft
The rotating blades redirect the flow off axis, so you need
a set of fixed blades that straighten out the flow before the
next set of rotating blades
Rotating
blades
Fixed
blades
Rotating
blades
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In this course we are not interested in the details of the
flow through each blade, or row of blades. We are
interested in the overall energy balance
W
1
Flow
T
2
Applying First Law (steady-state, neglect heat transfer)
0  q  w  (h1  V12 2)  (h2  V22 2)
w  (h1  h2 )  (V12 2  V22 2)
work per unit mass
Often the change in KE is small compared to change in h
i.e., h1  h2  V12 2  V22 2
w  h1  h2
Note:
work output ( w  0)  h1  h2
Power is work output per unit time
  m w  m (h1  h2 )
W
87
EXAMPLE
Steam enters a turbine operating at steady-state with a mass flow rate of
4600 kg/h. The turbine develops a power output of 1000 kW. At the
inlet the pressure is 60 bars, the temperature is 400C and the velocity is
10 m/s. At the exit the pressure is 0.1 bar, the quality is 0.9 and the
velocity is 50 m/s. Calculate the rate of heat transfer between the turbine
and the surroundings, in kW.
steam
m = 4600 kg/hr
P1= 60 bar
T1= 400C
V1= 10 m/s
1
W
T
liquid/vapor
x2= 0.9
P2= 0.1 bar
V2= 50 m/s
2
T
400C
276C
60 bar
1
0.1 bar
2
v
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Assume steady-state and PE is negligible
0  q  w  (h1  V12 2)  (h2  V22 2)
q  w  (h2  h1 )  (V22 2  V12 2)
Need enthalpy at states 1 and 2
State 1:
From saturated water Table A-3 Tsat (60 bar)= 275.6C
since T1>Tsat at same pressure have superheated vapor
From superheated water vapor Table A-4
h(60bar, 400C)= 3177.2 kJ/kg
State 2:
From saturated water Table A-3
hf(0.1 bar)= 191.8 kJ/kg hg(0.1 bar)= 2584.7 kJ/kg
h2= hf + x2(hg-hf) = 191.8 + 0.9(2392.8)= 2345.4 kJ/kg
so h2- h1= 2345.4 - 3177.2 = -833.8 kJ/kg
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0.5V22 V12  = 0.5(502 - 102 ) = 1200 m 2 /s 2
m 2  1 N  1 J 
J
 1200 2 

1200



kg
s  kg  m/s 2  N  m 
Collecting terms:
q  w  (h2  h1 )  (V22 2  V12 2)
Q  W  m (h2  h1 )  m (V22 2  V12 2)
 1000 kW  4600
kg  1 hr  
kJ
kJ 

  833.8  1.2 
hr  3600 s  
kg
kg 
Q  61.3 kW
Negative sign implies heat loss from turbine
Note:
1) Difference in magnitude between h and ke
2) Magnitude of heat loss Q (61 kW) compared to
magnitude of power output W (1000 kW)
90
Compressor/pump
A device in which shaft work input is used to raise the
pressure of a fluid (liquid or vapor)
91
Again we are not interested in the details of the flow
through each blade or row of blades. We are interested in
the overall energy balance
Flow
1
W
C
2
Applying First Law (steady-state, neglect heat transfer
and PE)
0  q  w  (h1  V12 2)  (h2  V22 2)
w  (h1  h2 )  (V12 2  V22 2)
work per unit mass
Often the change in KE is small compared to change in h
i.e., h1  h2  V12 2  V22 2
w  h1  h2
Note:
work input ( w  0)  h2  h1
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Throttling Device
A device that generates a significant pressure drop via a
flow restriction, e.g., partially closed valve.
1
2
Applying First Law (steady-state, neglect heat transfer
and PE)
0  q  w  (h1  V12 2)  (h2  V22 2)
h1  V12 / 2  h2  V22 2
If the state 2 is taken far downstream from the blockage
the change in velocity is negligible, i.e., V1  V2
h1  h2
Throttling process is characterized by constant enthalpy
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Heat Exchangers
These are devices that transfer energy between fluid
streams at different temperatures cool or heat one of the
fluids. The following is a tube-in-tube heat exchanger
4
2
1
3
Can have cross-flow or parallel-low type
Applying First law to above cross flow heat exchanger
assuming steady flow, no heat loss to the environment and
KE and PE is negligible
0  Q  W  m 1 (h1  V12 2  gZ1 )  m 3 (h3  V32 2  gZ 3 )
 m 2 (h2  V22 2  gZ 2 )  m 4 (h4  V42 2  gZ 4 )
Steady flow so m 1  m
 2 and m
3  m
4
0m
 1 (h1  h2 )  m
 3 (h3  h4 )
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Solving we get
m 1 h4  h3

m 3 h1  h2
To get the rate of heat transfer from one stream to the
other perform CV analysis on only the inner-tube (assume
inner-stream is hotter than outer-stream)
Qi
2
1
Again Applying First Law with same assumptions
0  (Q i )  m 1 (h1  h2 )
Q i
 (h1  h2 )
m 1
Since Q i  0  h1  h2 , so T1 > T2 (fluid cools down)
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A CV analysis of the outer-stream would give
4
3
Qo
0  (Q o )  m 3 (h3  h4 )
Q o
 (h4  h3 )
m 3
Since Q o  0  h4  h3 , so T4 > T3 (fluid heats up)
Note, the magnitude of the energy transfer rate from the
inner stream Q i equals the magnitude of the energy
transfer rate into the outer-stream Q o
Q i  Q o
m 1 h1  h2   m 3 h4  h3 
m 1 h4  h3

m 3 h1  h2
Note we recover the same relationship obtained using the
global energy balance
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Another common type of heat exchanger is a direct
contact heat exchanger, e.g., open feed water heater.
Hot stream, m
1
Warm stream, m
3
Cold stream, m 2
This type of heat exchanger consists of a vessel where a
hot stream and cold stream of the same fluid are mixed
and exit at an intermediate temperature through a single
outlet.
Apply conservation of mass and First Law to the CV and
assuming steady state, negligible KE and PE change to
get:
dM CV
 m 1  m 2  m 3
dt
0m
1  m
2 m
3
dECV
 Q  W  m 1 (h1 )  m 2 (h2 )  m 3 (h3 )
dt
0m
 1h1  m
 2 h2  m
 3 h3
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Transient Control Volume Analysis
Applies when the CV has only one inlet or one exit
Reservoir:
constant T
constant P
valve
CV
Considering the filling of a rigid tank of volume Vcv with
a gas supplied at a constant pressure and temperature
Applying conservation of mass
dM CV
 m i  m e
dt
Applying First Law, neglecting heat transfer to
environment, KE and PE
dECV dU CV

 Q  W  m i (hi  Vi 2 / 2  gZ i )  m e (.....)
dt
dt
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Note: constant enthalpy across the valve (throttling
device)  gas specific enthalpy, hi , into the CV equals
the gas specific enthalpy in the reservoir, hR
dU CV
 m i hi  m i hR
dt
Substituting
dU CV
dM CV
 hR
dt
dt
Integrating from the initial state " i" to the final state " f"
Uf
Mf
Ui
Mi
 dU hR  dM
U f  U i  ( M f  M i ) hR
M f u f  M i u i  ( M f  M i ) hR
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If the tank is initially empty (vacuum) mi = 0
M f u f  M f hR
cV T f  c PTR
Tf 
cP
TR  kTR
cV
Work done getting gas into the CV results in a final tank
gas temperature higher than the reservoir temperature
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