SSL8 Mass & Energy Analysis of Control Systems

MASS & ENERGY
ANALYSIS OF CONTROL
SYSTEMS
Lecture 8

Keith Vaugh BEng (AERO) MEng

Reference text: Chapter 6 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition
Yunus A. Cengel, Robert H. Turner, John M. Cimbala
McGraw-Hill, 2008
KEITH VAUGH

OBJECTIVES
}

KEITH VAUGH

OBJECTIVES
Develop the conservation of mass principle.
}

KEITH VAUGH

OBJECTIVES

Apply the conservation of mass principle to various
systems including steady- and unsteady-ﬂow control
}
volumes.

KEITH VAUGH

OBJECTIVES

}
volumes.

Apply the ﬁrst law of thermodynamics as the statement of
the conservation of energy principle to control volumes.

KEITH VAUGH

OBJECTIVES

}
volumes.

Apply the first law of thermodynamics as the statement of
the conservation of energy principle to control volumes.

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.

KEITH VAUGH

Solve energy balance problems for common steady-ﬂow
devices such as nozzles, compressors, turbines, throttling
valves, mixers, heaters, and heat exchangers.
}

KEITH VAUGH

Solve energy balance problems for common steady-flow
devices such as nozzles, compressors, turbines, throttling
valves, mixers, heaters, and heat exchangers.
}
Apply the energy balance to general unsteady-flow
processes with particular emphasis on the uniform-flow
process as the model for commonly encountered charging
and discharging processes.

KEITH VAUGH

CONSERVATION OF MASS
Conservation of mass
}
Mass, like energy, is a conserved property, and it cannot
be created or destroyed during a process.

Closed systems
The mass of the system remain constant during a
process.

Control volumes
Mass can cross the boundaries, and so we must keep
track of the amount of mass entering and leaving the
control volume.

KEITH VAUGH

Mass is conserved
even during chemical
reactions

KEITH VAUGH

Mass is conserved
even during chemical
reactions

Mass m and energy E can be converted to each other according to

E = mc 2

where c is the speed of light in a vacuum, which is c = 2.9979 × 108 m/s.
The mass change due to energy change is absolutely negligible.

KEITH VAUGH

MASS & VOLUME FLOW RATES
&
δ m = ρVn dAc
& &
m = ∫ δ m = ∫ ρVn dAc
Ac Ac
}
1 Deﬁnition of
Vavg = ∫ Vn dAc
Ac Ac average velocity
The average velocity Vavg is deﬁned as the
average speed through a cross section.

KEITH VAUGH

&
δ m = ρVn dAc
& &
Ac Ac
}
1 Deﬁnition of
Vavg = ∫ Vn dAc

&
m = ρVavg Ac ( s)
kg

KEITH VAUGH

&
δ m = ρVn dAc
& &
Ac Ac
}
1 Deﬁnition of
Vavg = ∫ Vn dAc

&
m = ρVavg Ac ( s)
kg

&
V
& &
m = ρV = Mass ﬂow rate
v

KEITH VAUGH

&
δ m = ρVn dAc
& &
Ac Ac
}
1 Deﬁnition of
Vavg = ∫ Vn dAc

&
m = ρVavg Ac ( s)
kg

&
V
& &
v
Volume ﬂow rate

&
V = ∫ Vn dAc = Vavg Ac = VAc
Ac ( s)
m3

KEITH VAUGH

&
δ m = ρVn dAc
& &
Ac Ac
}
1 Definition of
Vavg = ∫ Vn dAc

&
m = ρVavg Ac ( s)
kg

&
V
& &
v
Volume flow rate

&
V = ∫ Vn dAc = Vavg Ac = VAc
Ac ( s)
m3 The volume flow rate is the
volume of fluid flowing through
a cross section per unit time.
KEITH VAUGH

CONSERVATION OF MASS
PRINCIPLE

The conservation of mass principle for a control volume
The net mass transfer to or from a control volume during a time
interval Δt is equal to the net change (increase or decrease) in
}
the total mass within the control volume during Δt.

⎛ Total mass entering ⎞ ⎛ Total mass leaving ⎞ ⎛ Net change in mass ⎞
⎜ the CV during Δt ⎟ − ⎜ the CV during Δt ⎟ = ⎜ within the CV during Δt ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠

KEITH VAUGH

min − mout = Δmcv ( kg )
dmcv
& &
min − mout =
dt ( s)
kg

Conservation of mass principle for an ordinary
bathtub

KEITH VAUGH

dmcv
& &
min − mout =
dt ( s)
kg

General conservation of mass

d r r
dt ∫
cv
ρ dV + ∫
cs
( )
ρ V ⋅ n dA = 0

bathtub

KEITH VAUGH

dmcv
& &
min − mout =
dt ( s)
kg


d r r
dt ∫
cv
ρ dV + ∫
cs
( )
ρ V ⋅ n dA = 0

in rate form

d
∫ & &
ρ dV = ∑ m − ∑ m
dt cv
in out
bathtub dmcv
or &
= ∑m − ∑m &
dt in out

KEITH VAUGH

MASS BALANCE FOR STEADY
FLOW PROCESSES
During a steady-ﬂow process, the total amount of mass
contained within a control volume does not change with time
(mCV = constant).
}
Then the conservation of mass principle requires that the total
amount of mass entering a control volume equal the total
amount of mass leaving it.

KEITH VAUGH

For steady-flow processes, we are
interested in the amount of mass
flowing per unit time, that is, the mass
flow rate

&
in
&
∑m = ∑m
out
( s)
kg Multiple inlets
and exits

& &
m1 = m2 → ρV1 A1 = ρV2 A2 Single
stream

Many engineering devices such as
nozzles, diffusers, turbines,
compressors, and pumps involve a
single stream (only one inlet and one Conservation of mass principle for a two
outlet). inlet - one outlet steady flow system

KEITH VAUGH

Special case: Incompressible Flow
The conservation of mass relations can be simplified even further when the
fluid is incompressible, which is usually the case for liquids.

& &
∑V = ∑V
in out
( s)
m3

& &
V1 = V2 → V1 A1 = V2 A2

There is no such thing as a “conservation of
volume” principle.

However, for steady flow of liquids, the volume
flow rates, as well as the mass flow rates, remain
constant since liquids are essentially
During a steady-flow process, incompressible substances.
volume flow rates are not
necessarily conserved
although mass flow rates are. KEITH VAUGH

FLOW WORK & THE ENERGY
OF A FLOWING FLUID
Flow work, or flow energy
The work (or energy) required to push the mass into or out of the
control volume. This work is necessary for maintaining a
continuous flow through a control volume.
}

In the absence of acceleration, the force applied on a fluid by a piston is equal to the force applied
on the piston by the fluid.
KEITH VAUGH

F = PA
W flow = FL = PAL = PV ( kJ )

W flow = Pv ( kg)
kJ

Schematic for ﬂow work

KEITH VAUGH

TOTAL ENERGY OF A
FLOWING FLUID

e = u + ke + pe = u +
V2
2
+ gz ( kg)
kJ
}
θ = Pv + e = Pv + ( u + ke + pe)

KEITH VAUGH

TOTAL ENERGY OF A
FLOWING FLUID

e = u + ke + pe = u +
V2
2
+ gz ( kg)
kJ
}
θ = Pv + e = Pv + ( u + ke + pe)

But the combination Pv + u has been previously deﬁned as the
enthalpy h

KEITH VAUGH

TOTAL ENERGY OF A
FLOWING FLUID

e = u + ke + pe = u +
V2
2
+ gz ( kg)
kJ
}
θ = Pv + e = Pv + ( u + ke + pe)

But the combination Pv + u has been previously defined as the
enthalpy h
The flow energy is automatically
V2
θ = h + ke + pe = h +
2
+ gz ( kg)
kJ taken care of by enthalpy. In
fact, this is the main reason for
defining the property enthalpy.

KEITH VAUGH

The total energy consists of three parts for a non-flowing fluid and four parts for a
flowing fluid.

KEITH VAUGH

ENERGY TRANSPORT BY
MASS
Amount of energy transport

Emass
⎛
= mθ = m ⎜ h +
⎝
V2
2
⎞
+ gz ⎟
⎠
( kJ )
}
Rate of energy transport

& ⎛ V2 ⎞
Emass & &
= mθ = m ⎜ h + + gz ⎟ ( kW )
⎝ 2 ⎠

The product ṁiθi is the energy
transported into control volume by
mass per unit time
KEITH VAUGH

When the kinetic and potential
energies of a ﬂuid stream are
negligible

Emass = mh
& &
Emass = mh

The product ṁiθi is the energy When the properties of the mass at
transported into control volume by each inlet or exit change with time
mass per unit time.
as well as over the cross section

⎛ Vi 2 ⎞
Ein,mass = ∫ θ i δ mi = ∫ ⎜ hi + + gzi ⎟ δ mi
mi mi ⎝ 2 ⎠

KEITH VAUGH

ENERGY TRANSPORT BY
MASS
Many engineering systems
such as power plants operate
under steady conditions.
}

KEITH VAUGH

Under steady-flow conditions, the mass Under steady-flow conditions, the fluid
and energy contents of a control volume properties at an inlet or exit remain
remain constant. constant (do not change with time).

KEITH VAUGH

Mass and energy balances for a steady-ﬂow process

& &
∑m = ∑m
in out

& &
m1 = m2 Mass
ρ1V1 A1 = ρ2V2 A2 balance

A water heater in steady operation.

KEITH VAUGH


& &
∑m = ∑m
in out

& &
m1 = m2 Mass

Energy balances
Z 0
dEsystem
& &
Ein − Eout = =0
14 2 4 3 dt 3
Rate of net energy transfer 14 2 4
by heat, work and mass Rate of change in internal,
kinetic, potential, etc... energies

KEITH VAUGH


& &
∑m = ∑m
in out

& &
m1 = m2 Mass

Energy balances
Z 0
dEsystem
& &
Ein − Eout = =0
14 2 4 3 dt 3
&
Ein = &
Eout ( kW )
{ {
Rate of net energy transfer Rate of net energy out,
by heat, work and mass by heat, work and mass

KEITH VAUGH


& &
∑m = ∑m
in out

& &
m1 = m2 Mass

Energy balances
Z 0
dEsystem
& &
Ein − Eout = =0
14 2 4 3 dt 3
&
Ein = &
Eout ( kW )
{ {
Rate of net energy transfer Rate of net energy out,
by heat, work and mass by heat, work and mass

⎛
& + Win + ∑ m h +
& V2 ⎞ & ⎛
& + ∑m h + V2 ⎞
Qin &⎜ + gz ⎟ = Qout + Wout &⎜ + gz ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
1 4 44 2 4 4 4
in
3 1 4 44 2 4 4 4
out
3
for each inlet for each exit

KEITH VAUGH

Energy balance relations with sign conventions (i.e., heat input and work
output are positive)

⎛
&− W = + ∑ m h +
& V2 ⎞ ⎛ V2 ⎞
Q &⎜ &
+ gz ⎟ − ∑ m ⎜ h + + gz ⎟
⎝ 2 ⎠ in ⎝ 2 ⎠
1 4 44 2 4 4 4
out
3 1 4 44 2 4 4 4 3
for each exit for each inlet

⎡
&− W = m h2 − h1 +
& & V22 − V12 ⎤
Q ⎢ + g ( z2 − z1 ) ⎥
⎣ 2 ⎦

V22 − V12
q − w = h2 − h1 + + g ( z2 − z1 )
2

q − w = h2 − h1
Q&
q=
W&
W&
w=
&
m
Refer to pages, 212 - 214 KEITH VAUGH

STEADY FLOW
ENGINEERING DEVICES
Many engineering devices operate essentially under the same
conditions for long periods of time. The components of a steam
power plant (turbines, compressors, heat exchangers, and
}
pumps), for example, operate nonstop for months before the
system is shut down for maintenance. Therefore, these devices
can be conveniently analysed as steady-ﬂow devices.

A modern land-based gas turbine
used for electric power
production. This is a General
Electric LM5000 turbine. It has a
length of 6.2 m, it weighs 12.5
tons, and produces 55.2 MW at
3600 rpm with steam injection.

KEITH VAUGH

Nozzles and Diffusers
Commonly utilised in jet engines, rockets,
spacecraft, and even garden hoses.

A nozzle is a device that increases the
velocity of a fluid at the expense of pressure.

A diffuser is a device that increases the
pressure of a fluid by slowing it down.

The cross-sectional area of a nozzle decreases
in the flow direction for subsonic flows and
increases for supersonic flows. The reverse is
Nozzles and diffusers are shaped so
true for diffusers.
that they cause large changes in fluid
velocities and thus kinetic energies.
& &
Ein = Eout

⎛ V12 ⎞ ⎛ V22 ⎞ & &
& &
m ⎜ h1 + ⎟ = m ⎜ h2 + ⎟ Since Q ≅ 0, W ≅ 0, and Δpe ≅ 0
⎝ 2 ⎠ ⎝ 2 ⎠
Refer to Example 6-4 & 6-5, page 216/7 KEITH VAUGH

Turbines and compressors
Turbine drives the electric generator in power
plants. As the fluid passes through the turbine,
work is done against the blades, which are
attached to the shaft. The shaft rotates, and the
turbine produces work.

Compressors, as well as pumps and fans,
increase the pressure of a fluid. Work is
supplied to these devices from an external
source through a rotating shaft. Energy balance for the compressor
in this figure:
A fan increases the pressure of a gas slightly
and is mainly used to mobilise a gas. & &
Ein = Eout
& & & &
Win + mh1 = Qout + mh2
A compressor is capable of compressing the gas
to very high pressures. Since
Δke = Δpe ≅ 0
Pumps work very much like compressors
Refer to Example 6-6 & 6-7, page 218/9
except that they handle liquids instead of gases. KEITH VAUGH

Throttling valves
Throttling valves are any kind of flow-
restricting devices that cause a significant
pressure drop in the fluid.

The pressure drop in the fluid is often
accompanied by a large drop in
temperature and for that reason throttling The temperature of an ideal gas does not
change during a throttling (h = constant)
devices are commonly used in refrigeration
process since h = h(T).
and air-conditioning applications.

h2 ≅ h1 Energy
balance
u1 + P1v1 = u2 + P2 v2

Internal energy + Flow Energy = Constant

Refer to Example 6-8, page 221 During a throttling process, the enthalpy of a
fluid remains constant. But internal and flow
energies may be converted to each other.
KEITH VAUGH

Mixing chambers
In engineering applications, the section 60°C

where the mixing process takes place is
commonly referred to as a mixing
chamber.
140 kPa

10°C 43°C

Energy balance for the adiabatic
mixing chamber in the ﬁgure

& &
Ein = Eout
& & &
m1h1 + m2 h2 = m3h3
The T-elbow of an ordinary Since
shower serves as the mixing & &
Q ≅ 0, W ≅ 0, Δke ≅ 0 and Δpe ≅ 0
chamber for the hot and the
cold-water streams.

Refer to Example 6-9, page 223 KEITH VAUGH

Heat exchangers
Heat exchangers are devices where two
moving ﬂuid streams exchange heat without
mixing. Heat exchangers are widely used in
various industries and come in a variety of
designs.

A heat exchanger can be as simple
as two concentric pipes

The heat transfer
associated with a heat
exchanger may be zero
or nonzero depending
on how the control
volume is selected.
KEITH VAUGH

Mass and energy balances for the
adiabatic heat exchanger in the
ﬁgure are:

& &
Ein = Eout
& & &
m1 = m2 = mW
& & &
m3 = m4 = m R
& & & &
m1h1 + m3h3 = m2 h2 + m4 h4


Pipe and duct flow
The transport of liquids or gases in pipes
and ducts is of great importance in many
engineering applications. Flow through a
pipe or a duct usually satisfies the steady-
flow conditions.

Heat losses from a hot fluid flowing through Pipe or duct flow may involve more
an uninsulated pipe or duct to the cooler than one form of work at the same time.
environment may be very significant.

KEITH VAUGH

Energy balance for the pipe ﬂow
shown in the ﬁgure is

& &
Ein = Eout
& & & &
We, in + m1h1 = Qout + m2 h2
& & &
We, in − Qout = m1c p (T2 − T1 )


ENERGY ANALYSIS OF
UNSTEADY FLOW PROCESSES
Many processes of interest, however, involve
changes within the control volume with time. Such
processes are called unsteady-flow, or transient-flow,
processes.
}
Most unsteady-flow processes can be represented
reasonably well by the uniform-flow process.

Uniform-flow process
The fluid flow at any inlet or exit is uniform and
steady, and thus the fluid properties do not change
with time or position over the cross section of an
inlet or exit. If they do, they are averaged and
treated as constants for the entire process.

KEITH VAUGH

Charging of a rigid tank from a supply
line is an unsteady-ﬂow process since
it involves changes within the control
volume.

KEITH VAUGH

Charging of a rigid tank from a supply The shape and size of a control volume
line is an unsteady-ﬂow process since may change during an unsteady-ﬂow
it involves changes within the control process.
volume.

KEITH VAUGH

Mass balance
& & &
min − mout = Δmsystem

&
Δmsystem = m final − minitial

{
i = inlet
e = exit
mi − me = ( m2 − m1 )CV 1 = initial state
2 = ﬁnal state

KEITH VAUGH

Mass balance
& & &

&

{
i = inlet
e = exit
2 = ﬁnal state

Energy balance
Ein − Eout = Esystem
14 2 4 3 {
Net energy transfer Change in internal,
by heat, work and mass kinetic, potential, etc... energies

KEITH VAUGH

Mass balance
& & &

&

{
i = inlet
e = exit
2 = ﬁnal state

Energy balance
14 2 4 3 {

⎛ ⎞ ⎛ ⎞
⎜ Qin + Win + ∑ mθ ⎟ − ⎜ Qout + Wout + ∑ mθ ⎟ = ( m2 e2 − m1e1 )system
⎝ ⎠ ⎝ ⎠
in out

KEITH VAUGH

Mass balance
& & &

&

{
i = inlet
e = exit
2 = ﬁnal state

Energy balance
14 2 4 3 {

⎛ ⎞ ⎛ ⎞
⎜ Qin + Win + ∑ mθ ⎟ − ⎜ Qout + Wout + ∑ mθ ⎟ = ( m2 e2 − m1e1 )system
⎝ ⎠ ⎝ ⎠
in out

θ = h + ke + pe
e = u + ke + pe

KEITH VAUGH

The energy equation of a uniform-ﬂow system
reduces to that of a closed system when all
the inlets and exits are closed.

KEITH VAUGH

The energy equation of a uniform-ﬂow system A uniform ﬂow system may involve
reduces to that of a closed system when all electrical, shaft and boundary work all
the inlets and exits are closed. at once

KEITH VAUGH

Conservation of mass
 Mass and volume flow rates
 Mass balance for a steady flow process
 Mass balance for incompressible flow
Flow work and the energy of a flowing fluid
 Energy transport by mass
Energy analysis of steady flow systems
Steady flow engineering devices
 Nozzles and diffusers
 Turbines and compressors
 Throttling valves
 Mixing chambers and heat exchangers
 Pipe and duct flow
Energy analysis of unsteady flow processes

KEITH VAUGH

SSL8 Mass & Energy Analysis of Control Systems

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