Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Chapter 4
The First Law of Thermodynamics: Control Volumes
The conservation of mass and the conservation of energy principles for open
systems or control volumes apply to systems having mass crossing the system
boundary or control surface. In addition to the heat transfer and work crossing the
system boundaries, mass carries energy with it as it crosses the system
boundaries. Thus, the mass and energy content of the open system may change
when mass enters or leaves the control volume.
Wnet

VCM

m i , Vi
Zi
Q net
Control Surface
ZCM
Ze

m e , Ve
Reference Plane
Typical control volume or open system
Thermodynamic processes involving control volumes can be considered in two
groups: steady-flow processes and unsteady-flow processes. During a steadyflow process, the fluid flows through the control volume steadily, experiencing no
change with time at a fixed position.
Mass Flow Rate
Mass flow through a cross section area per unit time is called the mass flow rate,
m . It is expressed as
z

m  Vn dA
A

where Vn is the velocity normal to the cross sectional flow area.
Chapter 4 -1
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
If the fluid density and velocity are constant over the flow cross sectional area, the
mass flow rate is


V A
  Vav A  av
m
v

where  is the density, kg/m3 ( = 1/v), A is the cross-sectional area, m2, and Vav is
the average fluid velocity normal to the area, m/s.
Example 4-1
Refrigerant-134a at 200 kPa, 40% quality flows through a 1.1 cm inside diameter,
d, tube with a velocity of 50 m/s. Find the mass flow rate of the refrigerant-12.
At P = 200 kPa, x = 0.4
v  v f  xv fg
 0.0007532  0.4(0.0993  0.0007532)
m3
 0.0402
kg


Vav A Vav  d 2
m 

v
v 4
50m / s
 (0.011m) 2

0.0402m3 / kg
4
kg
 0118
.
s
Chapter 4 -2
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
The fluid volume flowing through a cross section per unit time is called the
volume flow rate V . The volume flow rate is given by integrating the product of
the velocity normal to the flow area and the differential flow area over the flow
area. If the velocity over the flow area is a constant, the volume flow rate is given
by (note we are dropping the “av” subscript on the velocity)


V  VA
(m3 / s)
The mass and volume flow rate are related by
V

  V 
m
v
( kg / s)
Example 4-2
Air at 100 kPa, 50oC flows through a pipe with a volume flow rate of 40 m3/min.
Find the mass flow rate through the pipe.
Assume air to be an ideal gas, so
RT
kJ (50  273) K m3 kPa
v
 0.287
P
kg  K 100kPa
kJ
m3
 0.9270
kg
V
40m3 / min 1 min
m  
v 0.9270m3 / kg 60s
kg
 0.719
s
Conservation of Mass for General Control Volume:
The conservation of mass principle for the open system or control volume is
expressed as
Chapter 4 -3
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Sumof rate
Sumof rate
Time rate changeO
L
O
L
O
L
M
P
of mass flowing P
M
of mass flowing
M
of massinside P
M
P
M
P
M
P
M
P
M
P
M
P
into
control
volume
fromcontrol
volume
control
volume
N
QN
QN
Q
or
 m   m
in
out
 system
 m
( kg / s)
Conservation of Energy for General Control Volume:
The conservation of energy principle for the control volume or open system has
the same word definition as the first law for the closed system. Expressing the
energy transfers on a rate basis, the control volume first law is
Sumof rate
Sumof rate
Time rate changeO
L
O
L
O
L
M
P
M
P
M
of energy flowing P
M
of energy flowing P
M
of energy inside P
M
P
M
into control volumeP
fromcontrol volumeP
control volume P
N
QM
N
QM
N
Q
or
E in  E out

Rate of net energy transfer
by heat, work, and mass
E system

 


( kW )
Rate change in internal, kinetic,
potential, etc., energies
Considering that energy flows into and from the control volume with the mass,
energy enters because net heat is transferred to the control volume, and energy
leaves because the control volume does net work on its surroundings, the open
system, or control volume, first law becomes
dE
Q   m i i  W   m e e  CV
dt
for each inlet
for each exit
4
:
( kW )
Where  is the energy per unit mass flowing into or from the control volume. The
energy per unit mass, , flowing across the control surface that defines the control
volume is composed of four terms, the internal energy, kinetic energy, potential
energy, and the flow work.
Chapter 4 -4
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
As the fluid upstream pushes mass across the control surface, work done on that
unit of mass is
A
 FdL  FdL  PdV  Pvm
A
Wflow

 Pv
m
Wflow
w flow
The term Pv is called the flow work done on the unit of mass as it crosses the
control surface. The total energy carried by a unit of mass as it crosses the control
surface is
2
V
  u  Pv 
 gz
2
2
V
 h
 gz
2
Here we have used the definition of enthalpy, h = u+ Pv.


F
I
F
I
V
V
Q   m G
h
 gz J W   m G
h 
 gz J E
H2K 
H2K

2
2
e
i
i
i
i
e
for each inlet
e
e
CV
( kW )
for each exit
Where the time rate change of the energy of the control volume has been written
as E CV .
Steady-state, Steady-flow Processes (SSSF)
Chapter 4 -5
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Most energy conversion devices operate steadily over long periods of time. The
rates of heat transfer and work crossing the control surface are constant with time.
The states of the mass streams crossing the control surface or boundary are
constant with time. Under these conditions the mass and energy content of the
control volume are constant with time.
dmCV
 m CV  0
dt
dECV
 E CV  0
dt
Steady-state, Steady-flow Conservation of Mass:
Since the mass of the control volume is constant with time during the steadystate, steady-flow process, the conservation of mass principle becomes
Sum of rate
Sum of rate
L
O
L
O
M
P
M
P
of mass flowing P
M
of mass flowing
M
P
M
into control volumeP
fromcontrol volumeP
N
QM
N
Q
or
 m   m
in
out
( kg / s)
Steady-state, Steady-flow Conservation of Energy:
Since the energy of the control volume is constant with time during the steadystate, steady-flow process, the conservation of energy principle becomes
Sum of rate
Sum of rate
L
O
L
O
M
P
M
P
of
energy
flowing

of
energy
flowing
M
P
M
P
M
P
M
P
into
control
volume
fromcontrol
volume
N
QN
Q
Chapter 4 -6
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
or
0
E in  E out

E system

 


Rate of net energy transfer
by heat, work, and mass
( kW )
Rate change in internal, kinetic,
potential, etc., energies
or
E
1
:E

in
Rate of net energy transfer
by heat, work, and mass
into the system
out
Rate of energy transfer
by heat, work, and mass
from the system
Considering that energy flows into and from the control volume with the mass,
energy enters because heat is transferred to the control volume, and energy leaves
because the control volume does work on its surroundings, the steady-state,
steady-flow first law becomes
Qin

F
I
V
 W   m G
h 
 gz J Q
H2K

2
e
in
e
e
e
out
 Wout

F
I
V
  m G
h
 gz J
H2K

2
i
i
for each exit
i
for each inlet
Most often we write this result as
Q net  Wnet


F
I
F
I
V
V
  m G
h 
 gz J  m G
h
 gz J
2
H2K H
K




2
e
e
2
i
e
e
for each exit
i
i
for each inlet
Where
Q net   Q in   Q out
Wnet   Wout   Win
Chapter 4 -7
i
i
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Steady-state, Steady-flow for One Entrance and One Exit:
A number of thermodynamic devices such as pumps, fans, compressors,
turbines, nozzles, diffusers, and heaters operate with one entrance, one exit. The
steady-state, steady-flow conservation of mass and first law of thermodynamics
for these systems reduce to
m 1  m 2
( kg / s)
1 
1 
V1 A1  V2 A2
v1
v2
2 2
V
 V1
Q  W  m h2  h1  2
 g ( z2  z1 )
2
L
M
N
O
P
Q
( kW )
where the entrance is state 1 and the exit is state 2 and m is the mass flow rate
through the device.
When can we neglect the kinetic and potential energy terms in the first law?
Consider the kinetic and potential energies per unit mass.
2
V
ke 
2

m
(45m / s) 2 1kJ / kg
kJ
For V = 45
ke 

1
s
2
1000m2 / s2
kg

m
(140m / s) 2 1kJ / kg
kJ
V = 140
ke 

10
s
2
1000m2 / s2
kg
pe  gz
For z  100m
z  1000m
m
1kJ / kg
kJ
100
m

0
.
98
s2
1000m2 / s2
kg
m
1kJ / kg
kJ
pe  9.8 2 1000m

9
.
8
s
1000m2 / s2
kg
pe  9.8
Chapter 4 -8
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
When compared to the enthalpy of steam (h  2000 to 3000 kJ/kg) and the
enthalpy of air (h  200 to 6000 kJ/kg), the kinetic and potential energies are often
neglected.
When the kinetic and potential energies can be neglected, the conservation of
energy equation becomes
 (h2  h1 )
Q  W  m
( kW )
We often write this last result per unit mass flow as
q  w  (h2  h1 )
( kJ / kg )
Q
W
Where
q
and w 


m
m
Some Steady-Flow Engineering Devices
Below are some engineering devices that operate essentially as steady-state,
steady-flow control volumes.
Chapter 4 -9
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Nozzles and Diffusers
For flow through nozzles, the heat transfer, work, and potential energy are
normally neglected and have one entrance and one exit. The conservation of
energy becomes
m in  m out
m 1  m 2  m
E in  E out
2
2
V
V
m h1  1  m h2  2
2
2
F
I
F
I
G
H J
K G
H J
K

solving for V2

2
V2  2(h1  h2 )  V1
Example 4-3
Steam at 0.4 MPa, 300oC enters an adiabatic nozzle with a low velocity and
leaves at 10 kPa with a quality of 90%. Find the exit velocity, in m/s.
Control volume: The nozzle
Property relation: Steam tables
Chapter 4 -10
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Process: Assume steady-state, steady-flow
Conservation Principles:
Conservation of mass: for one entrance, one exit, the conservation of mass
becomes
 m
in
 out
 m
1  m
2  m

m
Conservation of energy:
According to the sketched control volume, mass crosses the control surface, but
no work or heat transfer crosses the control surface. Neglecting the potential
energies, we have
E in  E out
2
2
V
V
m h1  1  m h2  2
2
2
F
I
F
I
G
H J
K G
H J
K
Neglecting the inlet kinetic energy, the exit velocity is

V2  2(h1  h2 )
Now, we need to find the enthalpies from the steam tables.
U
V
W
T1  300o C
kJ
h1  3066.8
kg
P1  0.4 MPa
P2  0.2 MPa
x2  0.90
U
h
V
W
2
At 0.2 MPa hf = 504.7 kJ/kg and hfg = 2201.9 kJ/kg
h2 = h f + x2 h fg
= 504.7 + (0.90)(22019
. ) = 2486.4
Chapter 4 -11
kJ
kg
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles

kJ 1000 m2 / s 2
V2  2(3066.8  2486.4)
kg kJ / kg
1077.4
m
s
Control
Surface
Turbines
Wout
m in
1
m out
2
Turbine control volume.
If we neglect the changes in kinetic and potential energies as fluid flows through
an adiabatic turbine having one entrance and one exit, the conservation of mass
and the SSSF (steady state, steady flow) first law becomes
m in  m out
m 1  m 2  m
E in  E out
m 1h1  m 2 h2  Wout
Wout  m (h1  h2 )
Example 4-4
High pressure air at 1300 K flows into an aircraft gas turbine and undergoes
Chapter 4 -12
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
a steady-state, steady-flow, adiabatic process to the turbine exit at 660 K.
Calculate the work done per unit mass of air flowing through the turbine when
(a) Temperature dependent data is used.
(b) Cp;av at the average temperature is used.
(c) Cp at 300 K is used.
Control volume: The turbine
Property relation: Assume air is an ideal gas and use ideal gas relations.
Process: Steady-state, steady-flow, adiabatic process
Conservation Principles:
Conservation of mass:
 m
in
 out
 m
1  m
2  m

m
Conservation of energy:
Qin

F
I
V
  m G
h 
 gz J W
H2K

2
e
e
e
e
out

F
I
V
  m G
h
 gz J
H2K

2
i
i
for each exit
i
i
for each inlet
According to the sketched control volume, mass and work cross the control
surface. Neglecting kinetic and potential energies and noting the process is
adiabatic, we have
0  m 1h1  Wout  m 2 h2
 (h  h )
W  m
out
1
2
The work done by the air per unit mass flow is
wout 
Wout
 h1  h2

m
Notice that the work done by a fluid flowing through a turbine is equal to the
enthalpy decrease of the fluid.
Chapter 4 -13
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
(a) Using the air tables, A.17
at T1 = 1300 K,
h1 = 1395.97 kJ/kg,
at T2 = 660 K,
h2 = 670.47 kJ/kg
wout  h1  h2
 (1395.97  670.47)
 7255
.
kJ
kg
kJ
kg
(b) Using Table A.2 at Tav = 980 K, Cp = 1.138 kJ/(kg K)
wout  h1  h2  C p , av (T1  T2 )
kJ
(1300  660) K
kg  K
kJ
 728.3
kg
 1138
.
c. Using Table A.2 at T = 300 K, Cp = 1.005 kJ/(kg K)
wout  h1  h2  C p , av (T1  T2 )
kJ
(1300  660) K
kg  K
kJ
 643.2
kg
 1005
.
Compressors and Fans
Win
m in
1
m out
2
Steady-Flow Compressor
Chapter 4 -14
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Compressors and fan are essentially the same devices. However, compressors
operate over larger pressure ratios than fans. If we neglect the changes in kinetic
and potential energies as fluid flows through an adiabatic compressor having one
entrance and one exit, the SSSF (steady state, steady flow) first law or the
conservation of energy equation becomes.
Wnet  m (h2  h1 )
 ( W )  m (h  h )
in
2
1
Win  m (h2  h1 )
Example 4-5
Nitrogen gas is compressed in a steady-state, steady-flow, adiabatic process from
0.1 MPa, 25oC. During the compression process the temperature becomes 125oC.
If the mass flow rate is 0.2 kg/s, determine the work done on the nitrogen, in kW.
Control volume: The compressor (see the compressor sketched above)
Property relation: Assume nitrogen is an ideal gas and use ideal gas relations.
Process: Steady-state, steady-flow
Conservation Principles:
Conservation of mass:
 m
in
 out
 m
1  m
2  m

m
Conservation of energy:
Q net

F
I
V
  m G
h 
 gz J W
H2K

2
e
e
e
e
net
for each exit

F
I
V
  m G
h
 gz J
H2K

2
i
i
i
for each inlet
Chapter 4 -15
i
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
According to the sketched control volume, mass and work cross the control
surface. Neglecting kinetic and potential energies and noting the process is
adiabatic, we have for one entrance and one exit
 1 (h1  0  0)  ( Win )  m 2 (h2  0  0)
0m
Win  m (h2  h1 )
The work done on the nitrogen is related to the enthalpy rise of the nitrogen as it
flows through the compressor. The work done on the nitrogen per unit mass flow
is
Win
win 
 h2  h1

m
Assuming constant specific heats at 300 K, we write the work as
win  C p (T2  T1 )
kJ
(125  25) K
kg  K
kJ
 103.9
kg
 1039
.
Chapter 4 -16
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Throttling Devices
Consider fluid flowing through a one-entrance, one-exit porous plug. The fluid
experiences a pressure drop as it flows through the plug. No net work is done by
the fluid. Assume the process is adiabatic and that the kinetic and potential
energies are neglected, then the conservation of mass and energy equations
become
m i  m e
m i hi  m e he
hi  he
This process is called a throttling process. What happens when an ideal gas is
throttled?
hi  he
he  hi  0
z
e
i
C p (T )dT  0
or
Te  Ti
Chapter 4 -17
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
When throttling an ideal gas, the temperature does not change. We will see later
in Chapter 10, that the throttling process is an important process in the
refrigeration cycle.
Chapter 4 -18