Control Volume Analysis
Used for flow through systems
HEAT
CONTROL
VOLUME
WORK
Use closed system analysis (fixed system mass M) to
derive expressions for conservation of mass and energy
Time t
MS(t) = MCV(t) + mi
Time t+t
MS(t+t) = MCV(t+t) + me
Note: mi doesn’t have to be equal to me since MCV(t)
doesn’t have to be equal to MCV(t+t)
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For a closed system MS(t) = MS(t+t), so
MCV(t) + mi = MCV(t+t) + me
MCV(t+t) - MCV(t) = mi - me
Divide through by t to get time rate quantities
M CV (t  t   M CV (t) mi me


t
t t
Taking limit as t0, closed system and control volume
boundaries coincide
 M CV (t  t   M CV (t)
lim
 m i  m e

Δt 0 

t


Note: m is called the mass flow rate with units kg/s,
subscript i is for inlet and e is for exits
dM CV
 m i  m e
dt
Time rate of change of mass contained within the control
volume equals the net mass flow rate m into the control
volume
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For multiple inlets and outlets
dM CV
  m i   m e
i
e
dt
For steady-state (dMCV/dt=0)
 i   m e
m
i
e
Mass flow rate in terms of local properties
Consider
a small quantity of matter flowing with velocity

V across an incremental area dA, in a time interval t

V

Vt
A
Vnt
dA
Vn is the normal component of the velocity vector
The volume of matter in the cylinder is VntdA
Mass of matter crossing dA in t is m=VntdA
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Divide through by t and take the limit as t0 defines
the mass flow rate through the area dA,
 m 
m dA  lim 
  Vn dA
t  0 t 
Integrating over the entire area A yields the total mass
flow rate
m   Vn dA
A
If the flow is one-dimensional
across area A, i.e.,:

(i)
Flow velocity V is normal to area A
(ii) Intensive properties are uniform across area A
VA
v
Substituting into mass conservation
m  VA 
dM CV
  AV   AV
i
e
dt
Mass flux defined as mass flow rate per unit area
m
V
units kg/m 2 s 
e.g.,  V 
A
v
Volumetric flow rate defined as
m

 VA
units (m 3 /s)
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Conservation of Energy for a Control Volume
Perform closed system analysis where the system at time t
includes the fluid in the CV and inlet region i
Time t
Time t+t
At time t, the energy of the system is:
ES (t)  ECV (t)  mi(ui  Vi 2 / 2  gZi )
At time t+t, the energy of the system is:
ES (t  t)  ECV (t  t)  me(ue  Ve2 / 2  gZe )
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First law applied to the closed system gives:
ES (t  t)  ES (t)  QS  WS
ECV (t  t)  me(ue  Ve2 / 2  gZ e )
 [ ECV (t)  mi(ui  Vi 2 / 2  gZ i )]  QS  WS
ECV (t  t)  ECV (t)  QS  WS  mi(ui  Vi 2 / 2  gZ i )
 me(ue  Ve2 / 2  gZ e )
The change in energy in the CV is equal to the energy
transferred to the system by heat and work plus the energy
entering into the CV by mi less the energy leaving the CV
by me
Divide through by t, take limit as t0, closed system
and control volume boundaries coincide, note
Q s  Q CV , W s  W CV
dECV (t)
 Q  W  m i(ui  Vi 2 / 2  gZ i )
dt
 m e(ue  Ve2 / 2  gZ e )
where ui, Vi, Zi are evaluated at the inlet
ue, Ve, Ze are evaluated at the exit
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Consider the work term
Work consists of shaft type work and flow work
Wshaft
Work to push
mi into CV
Work to push
me out of CV
Consider the inlet of cross-sectional area A
Fluid mi enters CV in time t
Pi
x
Flow work is the energy used to get fluid into CV in time
t
W  Fx  ( Pi Ai )x
Dividing through by t and taking limit as t0
W
x
lim
 ( Pi Ai ) lim
t  0 t
t  0 t
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This yield the rate of work done, note m
 i   i AiVi
Wi  ( Pi Ai )Vi  Pi ( AiVi )  Pi (m i /  i )
Flow work at the inlet is Wi  m i ( Pi vi )
Similarly, flow work at the exit W e  m e ( Pe ve )
Substituting into the energy conservation equation
dECV (t)
 Q  W shaft  m i Pi vi  m e Pe ve
dt
 m i(ui  Vi 2 / 2  gZ i )  m e(ue  Ve2 / 2  gZ e )
Generalizing for multiple inlets and exits and substituting
for h = u + pv
dECV (t)
 Q  W shaft   m i(hi  Vi 2 / 2  gZ i )
i
dt
  m e(he  Ve2 / 2  gZ e )
e
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Common engineering applications involve one inlet and
one exit and the flow is steady, dE dt  0, m
i  m
e  m

dE dt
Q W shaft


 (hi  Vi 2 / 2  gZi )  (he  Ve2 / 2  gZ e )
m
m
m
0  q  w s  (hi  Vi 2 / 2  gZi )  (he  Ve2 / 2  gZe )
W shaft
Q
where q  , w 
are heat added and shaft work
m
m
done per unit mass, units are J/kg
Note q , w are not rates, i.e., J/s
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