AE4660_Module_2_Review_of_Fundamentals-Lecture-2023.pdf

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AE 4660 Aerospace Propulsion I
Fall 2023
Instructor: Dr. William W. Liou
1

and may lead to Student Conduct charges and other enforcement actions.
Module 2
Review of Fundamentals
Chapters 1, 2 and 3
2

and may lead to Student Conduct charges and other enforcement actions. 3
Objective
Review the fundamentals of thermodynamics and fluid dynamics for air-breathing
propulsion studies. (ME2320, AE3610, AE 3710)

Enthalpy (h, H)
Internal Energy (u, U)
Kinetic Energy (ke, KE)
Potential Energy (pe, PE)

Mass Conservation Equation (2.3.1)
• Time rate of change of mass in the control volume must equal the net mass flow rate
out of the control volume.
• X=1 and source terms = 0 in eq. (2.11) and Table 2.1.
• For steady, one-dimensional (1-D) flow:
• In addition, one inlet and outlet (Fig. 2.1):

One-Dimensional Flow (1D)
• The flow is normal to the boundary as it enters or exit the control volume. Or, 𝛼 =
0𝑜
𝑜𝑟 180𝑜
.
• Over each inlet/exit, all intensive properties are uniform with position.
• Sometime, still call 1D, even if the velocity is not normal to the boundary
• (Ex. 2.1)

Momentum Equations
(ME 2320, AE 3610, AE 3710):
In (2.9) and Table 2.1, set x=𝑉 and source term equal to the total force:
ෝ
𝐧1
ෝ
𝐧2
Thrust Equation#1: Assume steady state 1D flow, the first term vanishes. Let’s
use the thrust stand as an example :
Thrust is in the x-direction, so

Mass Conservation Equation
Combine (2) and (3),
Therefore, the thrust equation for this engine is (p1 and p2 are now gauge pressures),
Put the 𝑔𝑐 in (DIY).
About 𝒈𝒄 …….
1 lbf = (1 slug) (1 ft/𝑠2) = (32.174 lbm) (1 ft/𝑠2)
1 lbf = (1 lbm) (32.174 ft/𝑠2)
On page 2,

Example:
Solution:
What is the inlet size (diameter)?

Homework Assignment #1 (Elearning):
A turbojet engine has four components, i.e. compressor, diffuser, turbine, and
nozzle, with the following conditions,
Compressor Inlet Airflow Rate 75 kg/s
Compressor Inlet velocity small (~0 m/s)
Compressor
Exit Velocity 125 m/s
Exit Pressure 650 kPa
Exit area 0.12 m2
Diffuser
Exit Velocity 75 m/s
Exit Pressure 700 kPa
Exit area 0.14 m2
Combustor
….
Find (1) the engine thrust, and the thrust produced by each of the 4 components
of the engine and (2) graph the distribution of the gas velocity, gas pressure, and
the thrust of the components along the gas path.

Homework Assignment #1:
Example Solution:
Use thrust equation #1, compressor thrust
𝑇𝑐𝑜𝑚𝑝 = 75
𝑘𝑔
𝑠
× 125 − 0
𝑚
𝑠
+ 0.12 𝑚2 × 650,000 𝑃𝑎 = 87,375 N

Thrust Equation#2 (1.5)
Properties at station 1, the engine inlet, in thrust equation #1 is generally not
known. To calculate a/c engine thrust using the flight speed and the ambient
air properties (aka the free stream conditions), a control volume as in Fig. 4.5
and
for steady 1-D flow, the momentum equation becomes, or thrust equation#2,
Or,

Notes:
(1) The pressure force is not zero only when the exhaust gas is not fully
expanded, or the exhaust jet flow speed is supersonic.
(2) The air was assumed inviscid here. The actual, or installed thrust T, can be
much less.
(3) Using fuel/air ratio f,
It becomes
ሶ
𝑚0[ 1 + 𝑓 𝑉9 − 𝑉0]

(4) Gross Thrust : stationary engine, or 𝑉0 = 0
(5) Ram Drag: Loss in thrust due to aircraft velocity 𝑉0

Zeroth Law of Thermodynamics
If two systems are separately in thermal equilibrium with a third
system, they are in thermal equilibrium with each other.
Note:
Temperature is the common property of the systems that are in
thermal equilibrium.

First Law of Thermodynamics
Conservation of Energy (e)
.

The way we use the energy equation here… (2.3.3)
• For steady, 1-D flow with one inlet and one outlet,
•
• ሶ
𝑊 = ሶ
𝑊𝑜𝑢𝑡 − ሶ
𝑊𝑖𝑛
ሶ
𝑄 = ሶ
𝑄𝑖𝑛 − ሶ
𝑄𝑜𝑢𝑡

Q
▪ Adiabatic process: a thermodynamics process that does not
involve heat transfer
▪ A system that does not interact in any way with its surroundings
is called an isolated thermodynamics system.
▪ Transitory phenomena. Not a property of the system.

▪ Transitory phenomena. Not a property of the system.
W

Second Law of Thermodynamics
The second law postulates the impossibility of certain
thermodynamic processes through a system property named
entropy (s,S).
• For an isolated system, the entropy can not decrease.
• Entropy can be produced by, for instance, friction and heat conduction.
• The entropy is related to other properties of the system (or state variables) by
the Gibbs equations:

Reversible and Irreversible Process
If both heat is added and work is done in an infinitely slow and quasi-
static manner, which cause no motion and no non-uniformity in the
system, during a change of the state of a thermodynamic system, the
thermodynamic process is a reversible process.
• All real processes are irreversible.
• Depending on circumstances, a real process may be approximated by a reversible
process.
• 𝑇 𝑑𝑠 = 𝑑′
𝑞 for reversible process
• A reversible and adiabatic process is a constant entropy process (s=constant) and
is named an isentropic process where there is no energy loss.
• Isentropic process are often applied in ideal engine performance analysis (Ch. 5).

The Perfect Gas (2.4)
• A perfect gas is defined as a substance with the following equation of state:
Where
P: pressure
𝜌: density
T: temperature
𝑅: gas constant
Speed of Sound :
Mach Number :

Calorically Perfect Gas
• CPG is a perfect gas with constant specific heats.
• For isentropic process (𝑠2 = 𝑠1),
In fact,
1
1

Example: Calorically Perfect Gas

Nozzles and Diffusers
• A device of varying cross-sectional area through which the flow accelerate
(nozzles) or decelerate (diffusers)
• Steady state (
𝜕𝑚𝑐𝑣
𝜕𝑡
=
𝜕𝐸𝑐𝑣
𝜕𝑡
= 0)
• ሶ
𝑊 = 0, ሶ
𝑄 = 0
• Negligible potential energy change (∆𝑝𝑒 = 0)
0 = ሶ
𝑚 ℎ2 − ℎ1 +
𝑉2
2
− 𝑉1
2
2𝑔𝑐
ሶ
𝑚𝑖𝑛 = ሶ
𝑚𝑜𝑢𝑡 = ሶ
𝑚
Applications of the governing equations

Example: Nozzle

• A device in which power ( ሶ
𝑊) is developed and transferred out of (in the case
of a turbine) or into (in the case of a compressor) the fluid flow system.
• Steady state Steady state (
𝜕𝑚𝑐𝑣
𝜕𝑡
=
𝜕𝐸𝑐𝑣
𝜕𝑡
= 0)
• ሶ
𝑄 = 0
• Negligible potential energy change (∆𝑝𝑒 = 0)
ሶ
𝑊𝑖𝑛 − ሶ
𝑊𝑜𝑢𝑡 = ሶ
𝑚 ℎ2 − ℎ1 +
𝑉2
2
− 𝑉1
2
2𝑔𝑐
ሶ
𝑚𝑖𝑛 = ሶ
𝑚𝑜𝑢𝑡 = ሶ
𝑚
Compressor and Turbine
Applications of the governing equations

Example: Single spool compressor and Turbine
For an equal mass flow rate through the compressor and the turbine of a single
shaft engine of 185 lbm/s (f<<1), determine the compressor power and the
turbine exit temperature for the following conditions (𝑐𝑝𝑔𝑐 = 6000 𝑓𝑡2
/(𝑠2
∙
𝑜
𝑅)):
• Compressor:
𝑇2 = 740 𝑜
𝑅, 𝑇3 = 1230 𝑜
𝑅, 𝑉2 = 𝑉3
• Turbine:
𝑇4 = 2170 𝑜
𝑅, 𝑉4 = 𝑉5
ሶ
𝑊𝑡,𝑜𝑢𝑡 = ሶ
𝑊𝑐,𝑖𝑛
ሶ
𝑊𝑖𝑛
Turbine ሶ
𝑊𝑜𝑢𝑡
ሶ
𝑚𝑐𝑝 𝑇4 − 𝑇5 = ሶ
𝑚𝑐𝑝 𝑇3 − 𝑇2
Solution:
Compressor

(AE 3710)
Stagnation Enthalpy
In the energy equation, for ሶ
𝑊 = 0, ሶ
𝑄 = 0, and 𝑝𝑒 = 0, for instance, for inlets,
Therefore, define the stagnation enthalpy ℎ𝑡 as
𝑇𝑡: stagnation temperature (or total temperature), which relates to T through ke.
Stagnation Properties (3.2)
0 = ℎ2 − ℎ1 +
𝑉2
2
− 𝑉1
2
2𝑔𝑐

So,
Since,
Example - What does 𝑻𝒕 matter ?
Airplane flying at M=3.3 at 25, 000 ft,
𝑇𝑡 = 1366 𝑜
𝑅 = 906 𝑜
𝐹
Note the melting temperature of aluminum is about 1220 F. The heat (high temperature) added to
the aircraft surface is aerodynamic heating at high speeds.

Aerodynamic Heating ?

Stagnation (Total) Pressure, 𝑷𝒕
The pressure obtained when the gas is in motion is brought to rest isentropically,
Relations for total pressure (and total density) with Mach number can be
obtained using isentropic equations.
For example,

For an adiabatic (Q=0) and no-shaft-work (W=0) flow, such as the inlet and
nozzle flow in engine,
𝑇𝑡2 = 𝑇𝑡1
And,
Therefore,
• Since all real processes are nonisentropic, there will always be loss of energy and
therefore loss of efficiencies.
• Equation (3.5b) and (3.6) show that the energy loss can be reflected in the decrease of
total pressure.
• Total pressure loss is used extensively as a way to quantify performance of turbine
engine components.
≥ 0

END

Aerodynamic Heating ?

AE4660_Module_2_Review_of_Fundamentals-Lecture-2023.pdf

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