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Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Some Steady-Flow Engineering Devices
Below are some engineering devices that operate essentially as steady-state,
steady-flow control volumes.
Chapter 4 -1
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 -2
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   m
in
out
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
kJ
h  3066.8
V
kg
P  0.4 MPa W
T1  300o C
1
1
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 -3
kJ
kg
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles

kJ 1000 m2 / s2
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  E
in
out
m 1h1  m 2 h2  Wout
W  m (h  h )
out
1
2
Example 4-4
High pressure air at 1300 K flows into an aircraft gas turbine and undergoes
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.
Chapter 4 -4
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
(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
 1h1  Wout  m
 2 h2
0m
 (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.
(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
Chapter 4 -5
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
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
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
Chapter 4 -6
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
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

F
I
V
  m G
h
 gz J
H2K

for each exit
2
i
i
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 for one entrance and one exit
Chapter 4 -7
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
 1 (h1  0  0)  ( Win )  m
 2 (h2  0  0)
0m
 (h  h )
W  m
in
2
1
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 -8
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
 i  m e
m
 i hi  m e he
m
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 -9
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.
Example 4-6
One way to determine the quality of saturated steam is to throttle the steam to a
low enough pressure that it exists as a superheated vapor. Saturated steam at
0.4 MPa is throttled to 0.1 MPa, 100oC. Determine the quality of the steam at
0.4 MPa.
Throttling orifice
1
2
Control
Surface
Control volume: The throttle
Property relation: The steam tables
Process: Steady-state, steady-flow, no work, no heat transfer, neglect kinetic and
potential energies, one entrance, one exit
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

F
I
V
  m G
h
 gz J
H2K

for each exit
2
i
i
i
i
for each inlet
According to the sketched control volume, mass crosses the control surface.
Neglecting kinetic and potential energies and noting the process is adiabatic with
no work, we have for one entrance and one exit
Chapter 4 -10
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
0  m 1 (h1  0  0)  0  m 2 (h2  0  0)
m 1h1  m 2 h2
h1  h2
U
kJ
h  2676.2
V
kg
P  01
. MPa W
T2  100o C
2
2
therefore
h1  h2  2676.2
d
 h f  x1h fg
x1 
kJ
kg
i
@ P1  0.4 MPa
h1  h f
h fg
2676.2  604.74
21338
.
 0.971

Mixing chambers
The mixing of two fluids occurs frequently in engineering applications. The
section where the mixing process takes place is called a mixing chamber. The
ordinary shower is an example of a mixing chamber.
Chapter 4 -11
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Example 4-7
Steam at 0.2 MPa, 300oC enters a mixing chamber and is mixed with cold water
at 20oC, 0.2 MPa to produce 20 kg/s of saturated liquid water at 0.2 MPa. What
are the required steam and cold water flow rates?
Steam
1
Mixing
Chamber
Cold Water
2
Saturated Water
3
Control
Surface
Control volume: The mixing chamber
Property relation: Steam tables
Process: Assume steady-state, steady-flow, adiabatic mixing, with no work
Conservation Principles:
Conservation of mass:
Chapter 4 -12
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
 m
in
  m out
m 1  m 2  m 3
m 2  m 3  m 1
Conservation of energy:
Q net

F
I
V
  m G
h 
 gz J W
H2K

2
e
e
e
e

F
I
V
  m G
h
 gz J
H2K

2
i
net
for each exit
i
i
i
for each inlet
According to the sketched control volume, mass crosses the control surface.
Neglecting kinetic and potential energies and noting the process is adiabatic with
no work, we have for two entrances and one exit
m 1h1  m 2 h2  m 3h3
m 1h1  (m 3  m 1 )h2  m 3h3
m 1 (h1  h2 )  m 3 (h3  h2 )
(h  h )
m 1  m 3 3 2
(h1  h2 )
Now, we use the steam tables to find the enthalpies:
U
kJ
h  30718
.
V
kg
P  0.2 MPa W
T  20 C U
kJ
h

h

83
.
69
V
kg
P  0.2 MPa W
P  0.2 MPa U
kJ
h h
 504.7
V
Sat . Liquid W
kg
T1  300o C
1
1
o
2
2
f @ 20o C
3
f @ 0.2 MPa
2
3
Chapter 4 -13
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
m 1  m 3
(h3  h2 )
(h1  h2 )
kg (504.7  83.69) kJ / kg
s (30718
.  83.96) kJ / kg
kg
 2.82
s
 20
m 2  m 3  m 1
 (20  2.82)
 17.18
kg
s
kg
s
Heat Exchangers
Heat exchangers are normally well-insulated devices that allow energy exchange
between hot and cold fluids without mixing the fluids. The pumps, fans, and
blowers causing the fluids to flow across the control surface are normally located
outside the control surface.
Chapter 4 -14
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Example 4-8
Air is heated in a heat exchanger by hot water. The water enters the heat
exchanger at 45oC and experiences a 20oC drop in temperature. As the air passes
through the heat exchanger its temperature is increased by 20oC. Determine the
ratio of mass flow rate of the air to mass flow rate of the water.
1
Air Inlet
Control
Surface
1
Water Inlet
2
Water Exit
2
Air Exit
Control volume: The heat exchanger
Property relation: air: ideal gas relations
water: steam tables or incompressible liquid results
Process: Assume steady-state, steady-flow
Conservation Principles:
Chapter 4 -15
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Conservation of mass:
0(steady)
 in  m
 out  m
 system
m
( kg / s)
for two entrances, two exits, the conservation of mass becomes
m in  m out
m air ,1  m w,1  m air , 2  m w, 2
For two fluid streams that exchange energy but do not mix it is better to conserve
the mass for the fluid streams separately.
 air ,1  m
 air , 2  m
 air
m
 w ,1  m
 w,2  m
w
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 kinetic and
potential energies, we have for steady-state, steady-flow
0(steady)
E in  E out

Rate of net energy transfer
by heat, work, and mass

E system

 

Rate change in internal, kinetic,
potential, etc., energies
Chapter 4 -16
( kW )
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
m air ,1hair ,1  m w ,1hw ,1  m air , 2 hair , 2  m w , 2 hw , 2
m air (hair ,1  hair , 2 )  m w (hw , 2  hw ,1 )
(hw , 2  hw ,1 )
m air

m w (hair ,1  hair , 2 )
We assume that the air has constant specific heats at 300 K ( we don't know the
actual temperatures, just the temperature difference). Because we know the initial
and final temperatures for the water, we can use either the incompressible fluid
result or the steam tables for its properties.
Using the incompressible fluid approach for the water, Cp, w = 4.184 kJ/(kg K) .
C p , w (Tw , 2  Tw ,1 )
m air

m w C p , air (Tair ,1  Tair , 2 )
b g
b g
kJ
20 K
kg w  K

kJ
1005
.
25 K
kgair  K
4.184
kgair / s
 3.33
kgw / s
A second solution to this problem is obtained by determining the heat transfer rate
from the hot water and noting that this is the heat transfer rate to the air.
Considering each fluid separately for steady-state, steady-flow, one entrance, one
exit, and neglecting the kinetic and potential energies, the first law, or
conservation of energy equations become
Chapter 4 -17
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
E in  E out
m air ,1hair ,1  Q in , air  m air , 2 hair , 2
m w ,1hw ,1  Q out , w  m w , 2 hw , 2
Q in , air  Q out , w
Pipe and Duct Flow
The flow of fluids through pipes and duct is often steady-state, steady-flow. We
normally neglect the kinetic and potential energies; however, depending on the
flow situation the work and heat transfer may or may not be zero.
Example 4-9
In a simple steam power plant, steam leaves a boiler at 3 MPa, 600oC and enters a
turbine at 2 MPa, 500oC. Determine the in-line heat transfer from the steam per
kilogram mass flowing in the pipe between the boiler and the turbine.
Q out
1
Steam from
Boiler
Steam to
Turbine
2
Control
Surface
Control volume: Pipe section in which the heat loss occurs.
Property relation: Steam tables
Process: Steady-state, Steady-flow
Chapter 4 -18
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Conservation Principles:
Conservation of mass:
0(steady)
 in  m
 out  m
 system
m
( kg / s)
for one entrance, one exit, the conservation of mass becomes
m in  m out
 1  m 2  m

m
Conservation of energy:
According to the sketched control volume, heat transfer and mass cross the
control surface, but no work crosses the control surface. Neglecting the kinetic
and potential energies, we have for steady-state, steady-flow
0(steady)
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
We determine the heat transfer rate per unit mass of flowing steam as
m 1h1  m 2 h 2  Q out
Q out  m (h1  h2 )
qout
Q out

 h1  h2
m
We use the steam tables to determine the enthalpies at the two states as
Chapter 4 -19
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
U
kJ
h  3682.3
V
kg
P  3 MPa W
T  500 C U
kJ
h  3467.6
V
kg
P  2 MPa W
T1  600o C
1
1
o
2
2
2
qout  h1  h2
kJ
 ( 3682.3  3467.6)
kg
kJ
 214.7
kg
Example 4-10
Air at 100oC, 0.15 MPa, 40 m/s flows through a converging duct with a mass
flow rate of 0.2 kg/s. The air leaves the duct with at 0.1 MPa, 113.6 m/s. The
exit-to-inlet duct area ratio is 0.5. Find the required rate of heat transfer to the air
when no work is done by the air.
Q in
Air Exit
2
1
Air inlet
Control
Surface
Control volume: The converging duct
Property relation: Assume air is an ideal gas and use ideal gas relations.
Process: Steady-state, steady-flow
Chapter 4 -20
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Conservation Principles:
Conservation of mass:
0(steady)
 in  m
 out  m
 system
m
( kg / s)
for one entrance, one exit, the conservation of mass becomes
m in  m out
 1  m 2  m

m
Conservation of energy:
According to the sketched control volume, heat transfer and mass cross the
control surface, but no work crosses the control surface. Here keep the kinetic
energy and still neglect the potential energies, we have for steady-state, steadyflow
0(steady)
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
2
2
V
V
m 1 h1  1  Q in  m 2 h 2  2
2
2
2 2
V  V2
Q in  m (h2  h1 )  1
2
F
I
G
H J
K
F
G
H
F
G
H
In the first law equation, the following are known, P1, T1 (and h1),
Chapter 4 -21
IJ
K
IJ
K
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
 
 , and h2 (or T2). We use the first
V1 , V2 , m , and A2/A1. The unknowns are Q
in
law and the conservation of mass equation to solve for the two unknowns.
m 1  m 2
( kg / s)
1 
1 
V1 A1  V2 A2
v1
v2


P1
P
V1 A1
 V2 A2 2
RT1
RT2
Solving for T2

P A V
T2  T1 2 2 2
P1 A1 V1
 (100  273) K
F
F
01
. MPa I
113.6 m / s I
0.5g
b
G
J
G
J
. MPa K H40 m / s K
H015
 3531
. K
Assuming Cp = constant, h2 - h1 = Cp(T2 - T1)
2 2
V1  V2

Qin  m C p (T2  T1 ) 
2
F
G
H
IJ
K
kg
kJ
 0.2 (1005
.
(3531
.  373) K
s
kg  K
(113.62  402 )m2 / s 2 kJ / kg

)
2
2
2
1000m / s
kJ
 2.87
 2.87 kW
s
Chapter 4 -22
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
Looks like we made the wrong assumption for the direction of the heat transfer.
The heat is really leaving the flow duct. (What type of device is this anyway?)
Q out  Qin  2.87 kW
Liquid Pumps
Fluid exit
2
Pump
W
h
in
Fluid inlet
1
Liquid flow through a pump.
The work required when pumping an incompressible liquid in an adiabatic steadystate, steady-flow process is given by
2 2
V  V1
 h2  h1  2
Q  W  m
 g ( z2  z1 )
2
L
M
N
O
P
Q
( kW )
The enthalpy difference can be written as
h2  h1  (u2  u1 )  ( Pv ) 2  ( Pv )1
For incompressible liquids we assume that the density and specific volume are
constant. The pumping process for an incompressible liquid is essentially
isothermal, and the internal energy change is approximately zero (we will see this
more clearly after introducing the second law). Thus, the enthalpy difference
reduces to the difference in the pressure-specific volume products. Since v2 = v1
= v the work input to the pump becomes
Chapter 4 -23
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles

L
V
M
W  m
v( P  P ) 
N
2
2
2
1
2
 V1
 g( z2  z1 )
2
O
P
Q
( kW )
W is the net work done by the control volume, and it is noted that work is input to
  W
the pump; so, W
in , pump
If we neglect the changes in kinetic and potential energies, the pump work
becomes
 v ( P2  P1 )
( Win , pump )  m
( kW )
 v ( P2  P1 )
Win , pump  m
We use this result to calculate the work supplied to boiler feedwater pumps in
steam power plants.
Uniform-state, Uniform-flow Problems
During unsteady energy transfer to or from open systems or control volumes, the
system may have a change in the stored energy and mass. Several unsteady
thermodynamic problems may be treated as uniform-state, uniform-flow
problems. The assumptions for uniform-state, uniform-flow are



The process takes place over a specified time period.
The state of the mass within the control volume is uniform at any instant of
time but may vary with time.
The state of mass crossing the control surface is uniform and steady. The
mass flow may be different at different control surface locations.
To find the amount of mass crossing the control surface at a given location, we
integrate the mass flow rate over the time period.
z
z
t
t
 i dt Exits: me  m
 edt
Inlets: mi  m
0
0
The change in mass of the control volume in the time period is
Chapter 4 -24
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
(m2  m1 ) CV 
z
dm
dt
0 dt
CV
t
The uniform-state, uniform-flow conservation of mass becomes
m  m
i
e
(m2  m1 ) CV
The change in internal energy for the control volume during the time period is
(m2 u2  m1u1 ) CV 
z
dU
0 dt
t
dt
CV
The heat transferred and work done in the time period are
z
t
Q
0
z
t

Qdt
and

W  Wdt
0
The energy crossing the control surface with the mass in the time period is
m 
j
2
Vj
F
I
z
m G
h 
 gz J
dt
H 2 K
t
j
0
j
j
j
where
j=
i, for inlets
e, for exits
The first law for uniform-state, uniform-flow becomes


F
I
F
I m e me g
V
V
Q W  m G
h 
 gz J  m G
h
 gz J b
H 2 K H 2 K
2
e
e
e
2
i
e
i
i
i
2 2
1 1 CV
When the kinetic and potential energy changes associated with the control volume
and the fluid streams are negligible, it simplifies to
Chapter 4 -25
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
b
Q  W   mehe   mi hi  m2u2  m1u1
g
CV
( kJ )
Example 4-11
Consider an evacuated, insulated, rigid tank connected through a closed valve to a
high-pressure line. The valve is opened and the tank is filled with the fluid in the
line. If the fluid is an ideal gas, determine the final temperature in the tank when
the tank pressure equals that of the line.
Control volume: The tank
Property relation: Ideal gas relations
Process: Assume uniform-state, uniform-flow
Conservation Principles:
Conservation of mass:
m  m
i
e
(m2  m1 ) CV
Or, for one entrance, no exit, and the initial mass is zero, this becomes
mi  (m2 ) CV
Conservation of Energy:
Chapter 4 -26
Chapter 4 Lecture notes for Thermodynamics: An Engineering Approach, 3rd Ed by Cengel and Boles
For an insulated tank Q is zero and for a rigid tank with no shaft work W is zero.
For one inlet mass stream and no exit mass stream and neglecting changes in
kinetic and potential energies, the uniform-state, uniform-flow conservation of
energy reduces to
b
Q  W   me he   mi hi  m2 u2  m1u1
g
CV
( kJ )
0  mi hi  (m2 u2 ) CV
Or,
mi hi  (m2 u2 ) CV
hi  u2
ui  Pv
i i  u2
u2  ui  Pv
i i
Cv (T2  Ti )  Pv
i i
Cv (T2  Ti )  RTi
T2 
Cp
Cv  R
Ti 
Ti
Cv
Cv
 kTi
If the fluid is air, k = 1.4 and the absolute temperature in the tank at the final state
is 40% higher than the fluid absolute temperature in the supply line. The internal
energy in the full tank differs from the internal energy of the supply line by the
amount of flow work done to push the fluid from the line into the tank.
Rework the above problem for a 10 m3 tank initially open to the atmosphere at
25oC and being filled from an air supply line at 90 psig, 25oC until the pressure
inside the tank is 70 psig.
Chapter 4 -27