MMME2028 – Aerospace
Propulsion
Energy balances of open systems
Steady flow energy equation
Learning objectives
• Understand mass and energy balances of open systems
• Understand the steady flow energy equation and how it can be
simplified
• Apply SFEE to simple engineering scenarios
Open systems
In open systems, mass can cross a system boundary.
As such, we need to consider the motion of matter – matter will
have kinetic energy and potential energy.
The way that we evaluate open systems must take into account
these additional aspects.
Open systems
Closed System
Open System
Processes
Fluid undergoing a process is
retained within the system
State
State is uniform throughout
State varies with location through
the system, but varies during a the system. State is constant at a
process
fixed location
Relationships
between
properties
Fluid undergoes a process as it
passes through the system
Same for closed and open systems
Open systems
In its broadest sense, the boundaries (and hence volume) of
open systems can vary with time.
We can simplify the evaluation of open systems by assuming
fixed boundaries and steady flow conditions:
For steady flow through an open system with fixed boundaries,
the quantities of matter and energy within the system
boundaries are constant and do not change with time
Net flows of matter and energy into the system must equal zero.
Conservation of mass
To maintain a constant mass within the control volume:
෍ 𝑚ሶ = ෍ 𝑚ሶ
𝑖𝑛𝑙𝑒𝑡
𝑜𝑢𝑡𝑙𝑒𝑡
So 𝑚ሶ 𝑎 + 𝑚ሶ𝑏
= 𝑚ሶ 𝑐
a
c
We calculate mass flow rates based
on fluid velocity and inlet/outlet area:
b
𝑚ሶ = 𝜌𝐴𝐶
where ρ is fluid density, A is cross-sectional area of inlet/outlet,
and C is fluid velocity
Conservation of mass – Group exercise
a
c
b
Gas is transported in the piping system shown,
where the pipes a, b, and c have internal
diameters of 20cm, 25cm, and 40cm,
respectively.
The mass flowrate of gas passing through the
inlet pipe a is 2 kg/s. At the outlet, the gas
density is 11 kg/m3 and flows with a velocity of
3m/s.
If the density at inlet b is 10 kg/m3, what is the
gas velocity at this inlet?
Conservation of mass – Group exercise
𝑚ሶ 𝑐 = 𝜌𝑐 𝐴𝑐 𝐶𝑐
a
c
b
𝜋 × 0.402
𝑚ሶ 𝑐 = 11 ×
× 3.0 = 4.147 𝑘𝑔/𝑠
4
𝑚ሶ 𝑏 = 𝑚ሶ 𝑐 − 𝑚ሶ 𝑎 = 4.147 − 2.0 = 2.147 𝑘𝑔/𝑠
𝑚ሶ 𝑏
𝐶𝑏 =
=
𝜌𝑏 𝐴𝑏
2.147
= 4.374𝑚/𝑠
2
𝜋 × 0.25
10 ×
4
Steady flow energy equation (SFEE)
In steady flow problems, the energy contained within the system
boundaries does not increase or decrease with time.
So the energy input must be equal to the energy leaving the
system within a time interval.
The steady flow energy equation (SFEE) is used to represent
the conservation of energy in steady flow open systems for
reversible and irreversible processes
Steady flow energy equation (SFEE)
The steady flow energy equation (SFEE) relates to open systems
where matter is moving (flow process). This can be used in a
wide range of thermodynamics and fluid mechanics problems
Steady flow energy equation (SFEE)
Work transfer
Kinetic energy,
Flow work
Heat transfer
Kinetic energy,
Flow work
Potential energy
Specific flow work
This is work done on and by matter flowing across the system
boundaries:
𝜕𝑥2
𝜕𝑥1
𝑝2
𝑝1
Matter entering the system does work on the fluid ahead of it; matter
exiting the system does work on fluid outside of the system.
Specific flow work:
𝑝1 𝑣1 − 𝑝2 𝑣2
Steady flow energy equation (SFEE)
If all the energy inflows and outflows are balanced, then:
Heat
transfer
+
Work
transfer
+
Flow work
=
Matter
energy,
out
−
Matter
energy,
in
Since we know that mass flow rate into and out of the system are
equal and constant, we can express energy transfers per unit of
mass passing through the system:
𝑄ሶ
𝑞=
𝑚ሶ
𝑊ሶ
𝑤=
𝑚ሶ
Steady flow energy equation (SFEE)
We can then write the SFEE as:
𝐶22
𝐶12
𝑞 + 𝑤 + 𝑝1 𝑣1 − 𝑝2 𝑣2 = 𝑢2 +
+ 𝑔𝑧2 − 𝑢1 +
+ 𝑔𝑧1
2
2
The SFEE considers changes in:
Specific internal energy
𝑢2 − 𝑢1
Specific kinetic energy
𝐶22
𝐶12
−
2
2
Specific potential energy
𝑔𝑧2 − 𝑔𝑧1
Steady flow energy equation (SFEE)
We can rearrange the SFEE as:
𝐶22
𝐶12
𝑞 + 𝑤 = 𝑝2 𝑣2 − 𝑝1 𝑣1 + 𝑢2 +
+ 𝑔𝑧2 − 𝑢1 +
+ 𝑔𝑧1
2
2
𝐶22
𝐶12
𝑞 + 𝑤 = 𝑢2 + 𝑝2 𝑣2 +
+ 𝑔𝑧2 − 𝑢1 + 𝑝1 𝑣1 +
+ 𝑔𝑧1
2
2
𝐶22 𝐶12
𝑞 + 𝑤 = ℎ2 − ℎ1 +
−
+ 𝑔𝑧2 − 𝑔𝑧1
2
2
If ΔKE and ΔPE are negligible:
𝒒 + 𝒘 = 𝒉𝟐 − 𝒉𝟏
Group exercise - SFEE
Calculate the volumetric flow rate of water (L/minute) through a
hot water radiator when the rate of heat output from the radiator
to the room is 2.3 kW.
The water inlet temperature is 75°C and the water outlet
temperature is 66°C. Assume water has a specific heat capacity
of 4.18 kJ/kg-K. The pipe diameter is the same at inlet and outlet.
Group exercise - SFEE
0 – no shaft work
0 – pipe diameter same
𝑄ሶ + 𝑊ሶ = 𝑚ሶ ℎ2 − ℎ1 +
𝐶22
𝐶12
−
2
2
0 – assume same elevation
+ 𝑔𝑧2 − 𝑔𝑧1
𝑄ሶ = 𝑚ሶ ℎ2 − ℎ1 = 𝑚ሶ 𝑐𝑝 ∆𝑇
𝑄ሶ
𝑚ሶ =
𝑐𝑝 ∆𝑇
=
𝑉ሶ =
2.3 𝑘𝐽ൗ𝑠
4.18
𝑘𝐽
𝑚ሶ
ൗ𝑘𝑔−𝐾 ×(75−66℃)
= 0.0611 𝑘𝑔/𝑠
0.0611𝑘𝑔/𝑠 60𝑠
×𝑡 =
×
= 3.6 𝐿/𝑚𝑖𝑛
𝜌𝑤𝑎𝑡𝑒𝑟
1 𝑘𝑔/𝐿
𝑚𝑖𝑛
Reversible heat transfer and work in
open systems
In differential form, the SFEE becomes
d𝑞 + d𝑤 = d𝑢 + d(𝑝𝑣) + d 𝐾𝐸 + d(𝑃𝐸)
Reversible heat transfer (1st Law):
d𝑞 = d𝑢 + 𝑝d𝑣
Simplifying SFEE with this definition of dq gives:
d𝑤 = 𝑣d𝑝 + d 𝐾𝐸 + d(𝑃𝐸)
Reversible heat transfer and work in
open systems
Reversible work:
2
𝑤 = න 𝑣 d𝑝 + ∆𝐾𝐸 + ∆𝑃𝐸
1 2
𝑊ሶ = 𝑚ሶ න 𝑣 d𝑝 + ∆𝐾𝐸 + ∆𝑃𝐸
1
In some cases, ∆𝐾𝐸 and ∆𝑃𝐸 are negligible and work can be
calculated as
2
𝑤 = න 𝑣 d𝑝
1
Perfect gas: Work in open and closed
systems
Closed system
Specific work in a
reversible process
2
𝑤 = − න 𝑝 d𝑣
1
Open system
(negligible ΔKE, ΔPE)
2
𝑤 = න 𝑣 d𝑝
1
Isochoric
0
𝑤 = 𝑣(𝑝2 − 𝑝1 )
Isobaric
𝑤 = 𝑝(𝑣2 − 𝑣1 )
0
Isothermal
𝑣2
𝑤 = −𝑅𝑇 𝑙𝑛
𝑣1
𝑝2
𝑤 = 𝑅𝑇 𝑙𝑛
𝑝1
Polytropic
𝑝2 𝑣2 − 𝑝1 𝑣1
𝑤=
𝑛−1
𝑛
𝑤=
𝑝2 𝑣2 − 𝑝1 𝑣1
𝑛−1
Perfect gas: reversible work in open
systems – polytropic process
2
𝑤 = න 𝑣 d𝑝
1
Substitute value for v with respect to p, accounting for polytropic
process:
𝑝𝑣 𝑛 = 𝑐𝑜𝑛𝑠𝑡
𝑐𝑜𝑛𝑠𝑡 ∗
𝑣 = 1/𝑛
𝑝
2
d𝑝
𝑤 = 𝑐𝑜𝑛𝑠𝑡 ∗ න 1/𝑛
1 𝑝
𝑛
𝑤=
𝑝2 𝑣2 − 𝑝1 𝑣1
𝑛−1
Group exercise
Calculate the rate of work input needed for an isentropic air
compressor operating with an inlet pressure of 1 bar and an
outlet pressure of 3 bar. The mass flow rate is 5 kg/s and the inlet
temperature is 300K. Assume 𝛾 = 1.4 and R=287 J/kg-K.
1
Work
2
Group exercise
Calculate the rate of work input needed for an isentropic air
compressor operating with an inlet pressure of 1 bar and an
outlet pressure of 3 bar. The mass flow rate is 5 kg/s and the inlet
temperature is 300K. Assume 𝛾 = 1.4 and R=287 J/kg-K.
𝛾
𝑤=
𝛾−1
For an isentropic process:
𝑝2 𝑣2 − 𝑝1 𝑣1
𝛾
𝑤=
R 𝑇2 − 𝑇1
𝛾−1
1
Work
𝑇2 = 𝑇1
2
𝑝2
𝑝1
𝛾−1
𝛾
3
= 300𝐾
1
0.4
1.4
= 410𝐾
1.4
𝑤=
× 287 × 410 − 300𝐾 = 111𝑘𝐽/𝑘𝑔
0.4
5𝑘𝑔 111𝑘𝐽
ሶ
𝑊 = 𝑚ሶ 𝑤 =
×
= 555 𝑘𝑊
𝑠
𝑘𝑔
Summary
• Open systems – mass and energy balances
• Steady flow energy equation – derivation and simplification
• Application of SFEE to simple engineering problems