Chapter 4
Control Volume Analysis
Using Energy
(continued)
Learning Outcomes
►Distinguish between steady-state and
transient analysis,
►Distinguishing between mass flow rate
and volumetric flow rate.
►Apply mass and energy balances to
control volumes.
►Develop appropriate engineering models
to analyze nozzles, turbines, compressors,
heat exchangers, throttling devices.
►Use property data in control volume
analysis appropriately.
Mass Rate Balance
time rate of change of
mass contained within the
control volume at time t
time rate of flow of
mass in across
inlet i at time t
dmcv
= å mi - å me
dt
i
e
time rate of flow
of mass out across
exit e at time t
(Eq. 4.2)
Determine the amount of water
In tank after 1 hour
dmcv
= å mi - å me
dt
i
e
Energy Rate Balance
time rate of change
of the energy
contained within
the control volume
at time t
net rate at which
energy is being
transferred in
by heat transfer
at time t
net rate at which
energy is being
transferred out
by work at
time t
net rate of energy
transfer into the
control volume
accompanying
mass flow
dEcv
Vi2
Ve2
= Qcv -W + mi (ui +
+ gzi ) - me(ue +
+ gze ) (Eq. 4.9)
dt
2
2
Evaluating Work for a Control Volume
The expression for work is
 e ( peve )  m
 i ( pi vi )
W  Wcv  m
(Eq. 4.12)
where
cv accounts for work associated with rotating
► W
shafts, displacement of the boundary, and electrical
effects.
► m
 e ( pe ve ) is the flow work at exit e.
►
 i ( pi vi ) is the flow work at inlet i.
m
Control Volume Energy Rate Balance
(One-Dimensional Flow Form)
Using Eq. 4.12 in Eq. 4.9
2
2
dEcv 
V
V
 Qcv  Wcv  m i (ui  pi vi  i  gzi )  m e (ue  peve  e  gze )
dt
2
2
(Eq. 4.13)
For convenience substitute enthalpy, h = u + pv
2
2
dEcv 
V
V
 Qcv  Wcv  m i (hi  i  gzi )  m e (he  e  gze )
dt
2
2
(Eq. 4.14)
Control Volume Energy Rate Balance
(One-Dimensional Flow Form)
In practice there may be several locations
on the boundary through which mass enters
or exits. Multiple inlets and exits are
accounted for by introducing summations:
2
2
dEcv 
V
V
 Qcv  Wcv   m i (hi  i  gzi )   m e (he  e  gze )
dt
2
2
i
e
(Eq. 4.15)
Eq. 4.15 is the accounting balance for the
energy of the control volume.
Turbines
►Turbine: a device in which power is
developed as a result of a gas or liquid
passing through a set of blades attached to
a shaft free to rotate.
Determine the Velocity
At each exit duct
AV
m 
v
where
V is velocity
v is specific
volume
2
2


(V

V
)
2  g(z  z )
0  Q cv  Wcv  m (h1  h2 )  1
1
2 
2


Turbine Modeling
2
2

 Eq.
(V

V
)
2  g(z  z )
0  Q cv  Wcv  m (h1  h2 )  1
1
2  4.20a
2


►If the change in kinetic energy of flowing matter is
negligible, ½(V12 – V22) drops out.
►If the change in potential energy of flowing matter is
negligible, g(z1 – z2) drops out.
►If the heat transfer with surroundings is negligible,
Q cv drops out.
 (h1  h2 )
Wcv  m
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
(Eq. 4.18)
► Wcv  0.
►If the kinetic energies of the flowing streams are
negligible, m i(Vi2/2) and m e (Ve2/2) drop out.
►If the potential energies of the flowing streams are
negligible, m igzi and m e gze drop out.
►If the heat transfer with surroundings is negligible,
Q cv drops out.
0   m i hi   m ehe
i
e
4
3
Vi2
Ve2
0 = Qcv -Wcv + å mi (hi +
+ gzi ) - å me(he +
+ gze )
2
2
i
e
m3 (h4 - h3 ) = m1 (h1 - h2 )
m1
h4 = h3 + (h1 - h2 )
m