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```									                        Learning Objectives
1. Develop the conservation of mass principle.
2. Apply the conservation of mass principle to various systems
3. Apply the first law of thermodynamics as the statement of the
conservation of energy principle to control volumes.
4. Identify the energy carried by a fluid stream crossing a control
surface as the sum of internal energy, flow work, kinetic energy,
and potential energy of the fluid and to relate the combination of
the internal energy and the flow work to the property enthalpy.
5. Solve energy balance problems for common steady-flow devices
such as nozzles, compressors, turbines, throttling valves, mixers,
heaters, and heat exchangers.
5-1 Conservation of Mass
Mass Flow Rate
• The amount of mass flowing through a
cross section per unit time
• Denoted, m 
General Analysis
• The mass flow rate through a
differential area dAc is
 m  rVn dAc
• r → density of fluid
Vn → velocity of fluid normal to the
cross-sectional area
• Mass flow rate through entire cross-
sectional area
m   rVn dAc
Ac
5-1 Conservation of Mass
Uniform Flow
• In many practical applications, we can
make the following approximations:
1. One-dimensional flow
2. All properties are uniform at any
cross-section normal to the flow
direction
3. Properties have the bulk average
values over the cross-section

m  rVavg Ac

• r → density of fluid
Vavg → mean velocity of fluid normal to the cross-sectional area
Ac → cross-sectional area normal to flow direction
5-1 Conservation of Mass
Volumetric Flow Rate
• The volume of a fluid flowing through a cross-sectional area per
unit time is

V   Vn dAc
Ac

V  Vavg Ac

• The mass and volume flow rates are related by

m  rV
V
m
v
5-1 Conservation of Mass
Conservation of Mass Principle
• One of the most fundamental principles in nature
• The net mass transfer to or from a control volume (CV) during a
process is equal to the net change (increase or decrease) in the
total mass of the control volume during that process

• Or                      min  mout  mCV

d
• In rate form           min  mout    mCV
dt
• These expressions are referred to as mass balances
• The expressions are valid for any system undergoing any process
5-1 Conservation of Mass
Mass Balance Expressions for Multiple Inlets and Outlets

 m   m  m
in       out
CV

d
 m   m  dt mCV
in    out

• When the properties at the inlets and exits as well as within the
control volume are not uniform, the mass flow rate can be
expressed in the differential form
 m  rVn dAc
5-1 Conservation of Mass
Mass Balance for Steady Flow Processes
• For a steady flow process, mCV = constant
• The conservation of mass principle simplifies to

m  m
in     out

Incompressible Flow (r = constant)
• For an incompressible fluid, the conservation of mass principle
simplifies to

V  V
in     out
5-2 Flow Work and the Energy of a Flowing Fluid
Flow Work
• The work required to push a mass into or out of a control volume
• Necessary for maintaining a continuous flow through a control
volume

• Force applied on the fluid element by the imaginary piston

F = PA
5-2 Flow Work and the Energy of a Flowing Fluid

• To push the entire fluid element into the control volume, this
force must act through a distance L
• The work done in pushing the fluid element across the boundary
(flow work) is then
Wflow = FL = PAL = PV
• The flow work per unit mass is then
wflow = Pv
• The flow work is the same whether the fluid is pushed into or out
of the control volume
5-2 Flow Work and the Energy of a Flowing Fluid
Flow Work (cont.)
• Unlike other work quantities, flow work is expressed in terms of
properties
• For this reason, flow work can be viewed as a combination
property, convected energy, or transport energy
• As a result, the flow energy can be considered part of the energy
of a flowing fluid
• This greatly simplifies energy analysis of control volumes
5-2 Flow Work and the Energy of a Flowing Fluid
Total Energy of a Flowing Fluid
• The total energy of a stationary simple compressible system
consists of three parts: internal, kinetic, and potential energies
1 2
e  u  ke  pe  u  V  gz           non flowing fluid
2
• A fluid entering or leaving a control volume possesses an
additional form of energy, the flow energy, Pv
• The total energy of a flowing fluid is then,

  Pv  e  Pv  u  ke  pe

• This expression can be simplified by recalling that the enthalpy h
is given by h = u + Pv
1
  h  ke  pe  h  V 2  gz          flowing fluid
2
5-2 Flow Work and the Energy of a Flowing Fluid
Total Energy of a Flowing Fluid (cont.)
• By using the enthalpy instead of the internal energy to represent
the energy of a flowing fluid, one does not need to be concerned
• The energy associated with pushing the fluid into or out of a
control volume is automatically taken care of by the enthalpy
5-3 The First Law of Thermodynamics
Energy Balance

Ein  Eout  Esystem

• This relation is referred to as
the energy balance
• The relationship is valid for
any system undergoing any
process
5-3 Energy Balance for Steady-Flow Systems
• A process during which a fluid flows through a control volume
• No intensive or extensive properties within the control volume
change with time
• The total volume V, the mass m, and the total energy content E of
the control volume remain constant

Wb = 0


Q  constant

W  constant

5-3 Energy Balance for Steady-Flow Systems
• The boundary work is zero for steady flow systems (VCV = const.)
• The total mass or energy entering the control volume must be
equal to the total mass or energy leaving it (since, mCV = const.
and ECV = const.)
• The heat and work interactions between a steady-flow system and
its surroundings do not change with time

Wb = 0


Q  constant

W  constant

5-3 Energy Balance for Steady-Flow Systems

m  m
in       out

 rVA   rVA
in            out

For a single stream (one inlet and one outlet),

m1  m2
r1V1 A1  r2V2 A2
5-3 Energy Balance for Steady-Flow Systems
• During a steady-flow process, the total energy content of a
control volume remains constant, and thus ECV = 0
• Therefore, the total amount of energy entering a control volume
must be equal to the amount of energy leaving it

Ein  Eout  ECV  0

Ein  Eout
     
5-3 Energy Balance for Steady-Flow Systems
Energy Balance for Steady-Flow Systems (cont.)

Qin  Win   m  Qout  Wout   m
in                      out

  h  ke  pe
Qin  Win   m  h  1 V 2  gz   Qout  Wout   m  h  1 V 2  gz 
2                                      2
in                                             out

Using the formal sign convention,

Q  W   m  h  1 V 2  gz    m  h  1 V 2  gz 
2                        2
out                      in

For a single stream (one inlet (1) and one outlet (2)),
           V22  V12                 
Q  W  m  h2  h1             g  z2  z1  
               2                     
5-3 Energy Balance for Steady-Flow Systems
Energy Balance for Steady-Flow Systems (cont.)

Q  W   m  h  1 V 2  gz    m  h  1 V 2  gz 
2                        2
out                     in


•Q      → Rate of heat transfer between the control volume and its
surroundings
negative → CV is losing heat


• W → Power
positive → CV is doing work on surroundings
• h → Enthalpy of flowing fluid
2
• 1 V → Kinetic energy of flowing fluid
2
• gz → Potential energy of flowing fluid
Nozzles
• Devices that increase the
velocity of a fluid at the
expense of pressure

Diffusers
• Devices that increase the
pressure of a fluid by
decreasing its velocity

General Characteristics
• Typically, W  0

• Usually, Q  0 , pe  0


• Large changes in velocity
ke  0
Turbine
• In gas, steam, or hydroelectric power plants, the
device that drives the electric generator
• A turbine produces work, W  0


Compressors, Pumps, and Fans
• Devices used to increase the pressure of a fluid
• Work is supplied to these devices, W  0

Fan
•Increases the pressure of a gas slightly
•Mainly used to mobilize a gas
Compressor
•Capable of compressing a gas to very high
pressures
Pumps
•Like compressors except that they handle
General Characteristics of Turbines and Compressors
• Usually, Q  0 , pe  0


• Often, ke ≈ 0, except turbines and fans, however often
|h| >> |ke| in turbines and fans making ke negligible
Throttling Valves
• Any kind of flow-restricting device that causes a significant
pressure drop in the fluid
• The pressure drop is often accompanied by a large drop in
temperature
• The following assumptions can often be made,
q ≈ 0,       w ≈ 0,        pe ≈ 0,        ke ≈ 0
• Based on these assumptions, the conservation of energy equation
for a single-stream steady-flow device reduces to
h2 ≈ h1
• For this reason, a throttling valve is sometimes called an
isenthalpic device
Mixing Chamber
• The section where a mixing process takes place
• The following assumptions can often be made,
q ≈ 0,      w ≈ 0,         pe ≈ 0,       ke ≈ 0
Heat Exchangers
• Devices where two moving fluid streams exchange heat without
mixing
• Typical characteristics of heat exchangers,
w ≈ 0,       pe ≈ 0,         ke ≈ 0

Double Shell (tube and shell)
•Composed of two concentric pipes of different diameters
Problem 5-38
500 oC with a mass flow rate of 6000 kg/h and leaves at 100 kPa
and 450 m/s. The inlet area of the nozzle is 40 cm2. Determine (a)
the inlet velocity and (b) the exit temperature.
Problem 5-54
Argon gas enters an adiabatic turbine steadily at 900 kPa and 450 oC
with a velocity of 80 m/s and leaves at 150 kPa and 150 m/s. The
inlet area of the turbine is 60 cm2. If the power output of the turbine
is 250 kW, determine the exit temperature of the argon.
Problem 5-66
Refrigerant-134a is throttled from the saturated liquid state at 700 kPa
to a pressure of 160 kPa. Determine the temperature drop during this
process and the final specific volume of the refrigerant.
Problem 5-81
Refrigerant-134a at 1 MPa and 90 oC is to be cooled to 1 MPa and
30 oC in a condenser by air. The air enters at 100 kPa and 27 oC
with a volume flow rate of 600 m3/min and leaves at 95 kPa and
60 oC. Determine the mass flow rate of the refrigerant.
Problem 5-91
A thin-walled double-pipe counter-flow heat exchanger is used to cool
oil (cp = 2.20 kJ/kg · °C) from 150 to 40 °C at a rate of 2 kg/s by water
(cp = 4.18 kJ/kg · °C) that enters at 22 °C at a rate of 1.5 kg/s.
Determine the rate of heat transfer in the heat exchanger and the exit
temperature of water.

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