AEE 334 PROPULSION SYSTEMS I
Yusuf Özyörük - Oğuz Uzol
Department of Aerospace Engineering
Middle East Technical University
Spring 2023
NOTICE
1) These slides have been prepared for educational purposes
2) To be distributed to Middle East Technical University, Aerospace Engineering
students
3) Lecture material may contain pictures and illustrations from public domain web
pages or text book, if any
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
AEE 334 PROPULSION SYSTEMS I / 1
Content : Review of fluid mechanics
1
Conservation equations
1
2
3
4
5
6
7
8
2
Gas and thermodynamic laws
1
2
3
3
Fluid and continuum
Physical principles governing fluid flow
Control volume and surface
Conservation equations in integral form
Conservation equations in differential form
Conservation equations in alternative forms
Viscous stresses and Navier-Stokes eqs
Dimensionless numbers - Mach, Reynolds
Perfect gas law
Thermodynamic laws
Isentropic process
Ideal compressible gas flow
1
2
3
General 1-D energy conservation
Ideal 1-D energy conservation
Ideal gas flow relations
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AEE 334 PROPULSION SYSTEMS I / 2
2. Review of Fluid Mech: Fluid and continuum
Fluid is a substance that has no fixed shape and yields easily to applied force.
Fluid particle is collection of molecules which obeys continuum principle.
Continuum is the situation in which the mean free path is much shorter than the characteristic length of the
physical problem.
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2. Review of Fluid Mech: Phys. principles governing flow
Mean free path is the distance a molecule travels between two consequtive collisions.
System is a fixed identity matter.
Fluid motion is governed by following facts :
Matter cannot be created nor destroyed – time rate of change of mass of a system is zero :
dmsys
=0
dt
Newton’s law of motion – time rate of change of linear momentum of a system equals the net force on it:
X
⃗sys
dP
⃗on sys
=
F
dt
First law of thermodynamics – time rate of change of energy state of a system equals the sum of net rates of
work on and heat transfer to it:
dEsys
= Ẇon sys + Q̇to sys
dt
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2. Review of Fluid Mech: Control volume and surface
Control volume (C.V., V) is a defined region in space whose boundaries allow passage of matter.
Control surface (C.S., S) is the surface of a control volume, which may deform or be stationary.
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
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2. Review of Fluid Mech: Leibniz’ rule
Suppose f (x) is an integrable function. Then,
I(x) =
Z
x
c is a constant. Then,
f (ξ) dξ,
c
Similarly, if
I(x) =
Z
c
c is a constant. Then,
f (ξ) dξ,
x
dI(x)
= f (x)
dx
dI(x)
= −f (x)
dx
Now, consider a two-variable function g(x, t), and its definite integral over one variable
I(t) =
Z
b
g(ξ, t) dξ
a
Then, derivative of I(t) with respect to the other variable, t in this case, is
dI(t)
d
=
dt
dt
Z
b
g(ξ, t) dξ =
a
Z
b
a
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
∂
g(ξ, t) dξ
∂t
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2. Review of Fluid Mech: Leibniz’ rule (cont’d)
Now, consider a case in which both the integrand and the limits are functions of t (we often have
time-dependent phenomena),
I t, a(t), b(t) =
Z
b(t)
g(ξ, t) dξ
a(t)
Then, total derivative of integral I using the chain rule,
∂I
∂I da
∂I db
DI
=
+
+
Dt
∂t
∂a dt
∂b dt
Using previous results, we get
DI
=
Dt
⃗ (ξ)
Here, U
Z
b(t)
a(t)
is surface velocity
db
da
∂g(ξ, t)
dξ + g b(t), t
− g a(t), t
∂t
dt
dt}
|
{z
R
=
S
⃗ (ξ)·n̂(ξ) dS
g(ξ,t)U
S
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2. Review of Fluid Mech: Reynolds Transport Theorem
Extension all these for a function F to 3-D converts the boundary terms in Leibniz integral to surface integrals:
D
Dt
Z
V(t)
F dV =
Z
V(t)
∂F
dV +
∂t
Z
S(t)
⃗ S · n̂ dS
FU
Meaning of this equation is as follows:
Rate of change of total amount of F inside V (left-hand side) = Rate of change due to changes of F within V
(first integral on right-hand side) + Rate of change due to surface motion (second integral on right-hand side)
⃗ S is the velocity of the control surface. If medium is moving (i.e. flow exists!), U
⃗ S is defined
Velocity U
⃗
independently of the fluid velocity U
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2. Review of Fluid Mech: Reynolds Transport Theorem (cont’d)
Now, consider we would like to always have the same fluid particles within the control volume. Then we
must move the control surface with the same velocity as the fluid. That is,
⃗S = U
⃗
U
This corresponds to the Lagrangian frame approach
Then, Leibniz integral rule produces
D
Dt
Z
F dV =
V(t)
Z
V(t)
Z
∂F
dV +
∂t
(Note D/Dt used for moving boundaries !)
⃗ · n̂ dS
FU
S(t)
This is known as the Reynolds Transport Theorem
Using Gauss’ theorem, surface integral may be converted to volume integral. Then,
D
Dt
Z
F dV =
V(t)
Z
V(t)
∂F
⃗ ) dV
+ ∇ · (F U
∂t
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2. Review of Fluid Mech: Reynolds Transport Theorem (cont’d)
If both fluid and surface are in motion with different velocities, then we may write, conservation equations for
fluid motion in most general integral forms:
Mass
dmsys
=0
dt
Linear momentum
X ⃗sys
dP
⃗
=
F
on sys
dt
D
Dt
∴
D
Dt
∴
Z
Z
ρ dV +
Z
⃗ (U
⃗ −U
⃗ S )·n̂ dS = −
ρU
⃗ dV +
ρU
V(t)
S(t)
V(t)
Z
S(t)
⃗ −U
⃗ S ) · n̂ dS = 0
ρ(U
Z
pn̂ dS +
S(t)
Z
⃗τ dS +
S(t)
Z
ρ⃗b dV
V(t)
Energy, ei internal, eZ
k kinetic
Z
Z
D
⃗
⃗
⃗ dS
ρ(ei + ek ) dV +
ρ(ei + ek )(U − US ) · n̂ dS = −
pn̂ · U
Dt V(t)
S(t)
S(t)
+
Z
S(t)
⃗ dS +
⃗τ · U
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
Z
V(t)
⃗ dV −
ρ⃗b · U
Z
S(t)
q⃗′′ · n̂ dS +
Z
q̇ dV
V(t)
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2. Review of Fluid Mech: Conserv. eqs. in integral form
⃗ S = 0. Then,
When control volume has fixed boundaries, U
Mass
d
dt
Momentum
d
dt
Z
⃗ dV +
ρU
V
Z
S
Energy, ei internal, ek kinetic
d
dt
Z
ρ(ei + ek ) dV +
V
Z
V
ρ dV +
Z
S
⃗ · n̂ dS = 0
ρU
Z
⃗ (U
⃗ · n̂) dS = −
ρU
Z
S
⃗ · n̂) dS = −
ρ(ei + ek )(U
+
Z
V
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
pn̂ dS +
S
Z
⃗τ dS +
S
Z
S
Z
⃗ · n̂ dS +
pU
⃗ dV −
ρ⃗b · U
ρ⃗b dV
V
Z
S
Z
S
q⃗′′ · n̂ dS +
⃗ dS
⃗τ · U
Z
q̇ dV
V
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2. Review of Fluid Mech: Conserv. eqs. in integral form (cont’d)
Exercise
Write down the x-component of the momentum eq. in integral form.
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2. Review of Fluid Mech: Conserv. eqs. in integral form (cont’d)
Exercise
Consider steady flow of an incompressible fluid through a pipe of dimater D oriented horizontally. If the fluid
velocity profile does not change between two sections which are separated by a distance L, but a pressure drop
of ∆p is observed, what can you say about the average friction coefficient between these two sections?
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2. Review of Fluid Mech: Fundamental math (cont’d)
Vector operations
Gradient :
∂
∂
∂
∇ ≡ ê1
+ ê2
+ ê3
=
∂x1
∂x2
∂x3
∂
∂
∂
,
,
∂x1 ∂x2 ∂x3
short notation
=
z}|{
∂
∂xi
Gradient of scalar function F → vector (first-order tensor) :
∇F ≡
∂F ∂F ∂F
,
,
∂x1 ∂x2 ∂x3
=
∂F
∂F
∂F
∂F
ê1 +
ê2 +
ê3 =
∂x1
∂x2
∂x3
∂xi
⃗ → second-order tensor :
Gradient of a vector function F
⃗ ≡
∇F
∂
∂
∂
,
,
∂x1 ∂x2 ∂x3
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
F1 , F 2 , F 3
 ∂F1
∂x1
∂Fi
2
=
=  ∂F
∂x1
∂xj
∂F3
∂x1
∂F1
∂x2
∂F2
∂x2
∂F3
∂x2
∂F1
∂x3
∂F2
∂x3
∂F3
∂x3


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2. Review of Fluid Mech: Fundamental math (cont’d)
Vector operations
Second-order tensor :
 ∂F1
∂Fi
=
∂xj
∂x1
∂F2
∂x1
∂F3
∂x1
∂F1
∂x2
∂F2
∂x2
∂F3
∂x2
∂F1
∂x3
∂F2
∂x3
∂F3
∂x3
 ∂F1

≡
ê ê
∂x1 1 1
∂F2
ê ê
∂x1 2 1
∂F3
ê ê
∂x1 3 1
∂F1
ê ê
∂x2 1 2
∂F2
ê ê
∂x2 2 2
∂F3
ê ê
∂x2 3 2
∂F1
ê ê
∂x3 1 3
∂F2
ê ê
∂x3 2 3
∂F3
ê ê
∂x3 3 3


Tensor function G in general, Gij
Gij
"
G11 ê1 ê1
¯ = G ê ê
≡ Ḡ
21 2 1
G31 ê3 ê1
G12 ê1 ê2
G22 ê2 ê2
G32 ê3 ê2
G13 ê1 ê3
G23 ê2 ê3
G33 ê3 ê3
#
⃗ → scalar :
Divergence of a vector, ∇ · F
⃗ ≡ ∂Fi = ∂F1 + ∂F2 + ∂F2 = trace(∇F
⃗)
∇·F
∂xi
∂x1
∂x2
∂x3
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2. Review of Fluid Mech: Fundamental math (cont’d)
Vector/tensor operations
¯ → vector :
Divergence of a second-order tensor, div(Ḡ)
¯ ≡ Gij = ∇ · Ḡ
¯T =
div(Ḡ)
∂xj
 ∂G11
=
∂x1
∂G21
∂x1
∂G31
∂x1
∂
∂
∂
,
,
∂x1 ∂x2 ∂x3
+
+
+
∂G12
∂x2
∂G22
∂x2
∂G32
∂x2
+
+
+
∂G13
∂x3
∂G23
∂x3
∂G33
∂x3
"G11
G12
G13
G21
G22
G23
G31
G32
G33
#


¯ T = Ḡ,
¯ and hence,
In mechanics/fluid mechanics we have symmetric tensors. Then, Ḡ
¯ T ) = ∇ · Ḡ
¯
div(Ḡ
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2. Review of Fluid Mech: Conserv. eqs in differential form
Mass
∂ρ
⃗) = 0
+ ∇ · (ρU
∂t
∂ρ
∂ρuj
+
=0
∂t
∂xj
Momentum
⃗
∂ρU
⃗U
⃗ ) = ρ⃗b − ∇p + ∇ · τ̄¯
+ ∇ · (ρU
∂t
∂ρui
∂ρui uj
∂p
∂τij
+
= ρbi −
+
∂t
∂xj
∂xi
∂xj
Energy, e = ei + ek (internal+kinetic)
∂ρe
⃗ )] = ρ⃗b · U
⃗ + ∇ · (τ̄¯ · U
⃗ ) + ∇ · (k∇T ) + q̇
+ ∇ · [(ρe + p)U
∂t
∂[(ρe + p)uj ]
∂(ui τij )
∂ρe
∂
∂kT
+
= ρbi ui +
+
+ q̇
∂t
∂xj
∂xj
∂xj ∂xj
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2. Review of Fluid Mech: Conserv. eqs in diff. form (cont’d)
Exercise
Write down the components of the momentum equation in cartesian coordinates.
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2. Review of Fluid Mech: Conserv. eqs in alternative forms
⃗ and subtracted from the momentum equation, it
When mass conservation equation is multiplied with U
becomes
momentum (non-conservative form)
ρ
∂
⃗ ·∇
+U
∂t
| {z }
D
material derivative= Dt
∂
∂
ρ
+ uj
∂t
∂xj
⃗ = ρ⃗b − ∇p + ∇ · τ̄¯
U
ui = ρbi −
∂p
∂τij
+
∂xi
∂xj
Left-hand side : total acceleration (local+convective) times mass per unit volume
Righ-hand side : body force + pressure force + viscous force, all per unit volume
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2. Review of Fluid Mech: Conserv. eqs in alternative forms (cont’d)
Second law of thermodynamics combined with first law with application to fluid flow, neglecting of body
forces (for gas flows), gives
entropy
ρT
ρT
∂s ⃗
+ U · ∇s
∂t
∂s
∂s
+ uj
∂t
∂xj
⃗
= −∇ · q⃗′′ + q̇ + τ̄¯ : ∇U
=−
∂qj′′
∂ui
+ q̇ + τij
∂xj
∂xj
Infer from equation : Entropy increases along a stream line (material derivative is non-zero) if heat transfer, or
heat generation or viscous forces, or any combinations of them exist!
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2. Review of Fluid Mech: Viscous stresses and N-S eqs.
Origin of viscous forces
Molecules enter and leave a fluid particle volume continuously
Collisions (interactions) of high and low energetic molecules cause transfer of momentum
Momentum transfer imbalance occurs when velocity gradient exists, as layers of fluid will have changing levels
of molecular activity
Hence a relative force results which is called viscous force
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
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2. Review of Fluid Mech: Navier-Stokes (N-S) eqs.
Navier-Stokes
When following form of τ̄¯ is used, the set of conservation equations is known as the Navier-Stokes equations.
¯∇ · U
⃗ , or
⃗ + (∇U
⃗ )T − 2 µI¯
τ̄¯ = µ ∇U
3
τij
∂ui
∂uj
=µ
+
∂xj
∂xi
−
2
∂uj
µδij
3
∂xj
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AEE 334 PROPULSION SYSTEMS I / 22
2. Review of Fluid Mech: Navier-Stokes (N-S) eqs. (cont’d)
Exercise : incompressible 2-D, steady flow between infinite parallel plates
When flow is incompressible, density is constant, and mass conservation simplifies to
∂u
∂v
+
=0
∂x
∂y
Having plates infinitely large, we may assume ∂/∂x = 0 for any quantity q.
Then, ∂v/∂y = 0 in mass conservation, also.
This result dictates v = 0 everywhere, as boundary conditions (BC) at lower and upper plates require v = 0
there.
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2. Review of Fluid Mech: Navier-Stokes eqs. (cont’d)
Assume µ=cons, and show that the x and
y-momentum eqs are
0=−
∂p
∂2u
+µ 2
∂x
∂y
0 = −ρg −
∂p
∂y
Note : for incompressible flows
ρ
⃗
DU
⃗
= −ρ⃗b − ∇p + µ∇2 U
Dt
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2. Review of Fluid Mech: Dimensionless numbers - Mach and Reynolds
Viscous flow (originally momentum) equations are called Navier-Stokes equations
Ratio of velocity to sound speed is called Mach number, M =
When M < 0.3, flow is usually considered incompressible
M < 1 → subsonic flow
M ≈ 1 → transonic flow
M > 1 → supersonic flow
V
c0
Ratio of intertia forces to visoucs forces is called Reynolds number, Re =
ρV Lc
µ
When Re is significantly high, inertia forces dominate and viscous forces may be dropped (flowfield dominated
by inviscid process, but viscous stresses at wall still exist!)
Equations without viscous terms are called Euler equations
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2. Review of Fluid Mech: Perfect gas law
For perfect gases pressure p, density ρ, and tempersture T are related to each other by
p = ρRT
R is gas constant. For air R = 287 J/kgK
Enthalpy is composed of internal energy of a system plus the work needed for it to establish its volume at its
pressure state
Internal energy is total energy contained within a system but it does not contain kinetic nor potential energy
as a whole.
For calorically perfect gases, internal energy ei , and enthalpy h are given as
ei = Cv T, and h = Cp T
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
dei
dh
= Cv , and
= Cp
dT
dT
R = Cp − Cv
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2. Review of Fluid Mech: Thermodynamic laws
First law of thermodynamics
The first law states that the energy of a system may be changed by heat transfer to/from and work on/by its
surrounding :
∆E = Qto sys + Won sys
in differential form, per unit mass
de = dqto sys + dwon sys ,
dE
= Q̇to sys + Ẇon sys ,
dt
de
= q̇to sys + ẇon sys ,
dt
in terms of time rates
in terms of time rates per unit mass
When flow exists,
2
Energy is, e = ei + ek = ei + U2 , where ei is internal energy, U 2 /2 is kinetic energy, all per unit mass. Work
due to body forces is considered as part of the right-hand side work in above eq.
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2. Review of Fluid Mech: Thermodynamic laws (cont’d)
Second law of thermodynamics
Entropy of a system is a thermodynamic quantity representing the unavailability of a system’s thermal energy
for conversion into mechanical work
Second law of thermodynamics states that total entropy S of a system can never decrease over time
Second law is expressed as, dS ≥
δQ
T
All real processes are irreversible processes in thermodynamic sense, and entropy increases
Following are examples to real processes, and cause entropy increase
diffusion of momentum (fluid-fluid, fluid-wall friction)
shocks, unrestrained expansion
diffusion of mass (mass transfer)
diffusion of heat
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2. Review of Fluid Mech: Isentropic process
We can write for a reversible process,
dw = pdv,
dq = T ds.
The first law may then be expressed as
de = dqto sys − dwby sys = T ds − pdv
It is often more useful to work with the enthalpy,
h = e + pv = e + p/ρ,
∴
dh = de + pdv + vdp
Using the first law for de,
dh = T ds − pdv + pdv + vdp = T ds + vdp = T ds +
dp
ρ
From the first law eq. we may also write
T ds = Cv dT + pdv,
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
or
ds = Cv
dT
p
+ dv
T
T
AEE 334 PROPULSION SYSTEMS I / 29
2. Review of Fluid Mech: Isentropic process
Using the perfect gas relation p = ρRT , and integrating between two states yields
∆s = s2 − s1 = Cv ln
T2
ρ1
+ R ln
T1
ρ2
For isentropic processes involving perfect gases, ∆s = 0 and this yields,
p
= constant ;
ργ
Cp
Cv
γ=
This means between two states 1 and 2, we have the following relations
p2
=
p1
T2
T1
γ
γ−1
ρ2
=
ρ1
,
T2
T1
1
γ−1
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2. Review of Fluid Mech: 1-D energy eq.
Control volume bounded by streamlines and rotating blades
Control surfaces
⃗ · n̂ = 0 on them.
”As ” overlap streamlines ∴ U
”A
”
overlap
rotating
blades
and shaft ∴ fluid
b
R
R particles move together with them, and
⃗ −U
⃗ S ) · n̂ dS = 0, always; BUT,
⃗ dS ̸= 0 !
ρ(U
(−pn̂ + ⃗
τ) · U
Ab
”Ai ” inlet boundary ∴
R
”Ao ” outlet boundary ∴
R
Ab
R
Ai
⃗ · n̂ dS < 0 on it.
ρU
Ao
⃗ · n̂ dS > 0 on it.
ρU
What is the value of
p n̂ dS ?
As
It is not practical to take form a C.V. bounded by streamlines that curve in the nearfield ...
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Review of Fluid Mech: 1-D energy conservation
Steady conservation of energy in integral form
Z
S
⃗ r · n̂) dS = −
ρe(U
Z
S
⃗ dS +
pn̂ · U
Z
S
⃗ dS +
⃗τ · U
Z
V
⃗ dV −
ρ⃗b · U
Z
S
q⃗′′ · n̂ dS +
Z
q̇ dV
V
⃗r = U
⃗ −U
⃗ S , which is the relative fluid velocity wrt S, as some part of S is moving (blades)
where U
⃗ r · ⃗n = 0 always, by definition (solid surfaces are impermeable).
Note : on impermeable surfaces, U
Any surface integral may be split into
Hence, energy eq. reduces to
+
Z
So
Z
Si
R
S
=
⃗ · n̂ dS +
ρ(ei + ek )U
⃗ dS +
(−pn̂ + ⃗τ ) · U
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
Z
Sb
R
Si
Z
So
+
R
Ss
+
R
Sb
+
R
So
⃗ · n̂ dS =
ρ(ei + ek )U
⃗ dS −
(−pn̂ + ⃗τ ) · U
Z
S
Z
Si
⃗ dS
(−pn̂ + ⃗τ ) · U
q⃗′′ · n̂ dS +
Z
V
⃗ dV +
ρ⃗b · U
Z
q̇ dV
V
AEE 334 PROPULSION SYSTEMS I / 32
Review of Fluid Mech: 1-D energy conservation (cont’d)
For gas flows, gravitational effects (body force) may be neglected.
Doing so, and shifting the pressure terms of inlet and outlet boundaries from right-hand side to left-hand side :
=
Z
|
Si
Z
Si
⃗ · n̂ dS +
ρ(ei + ek + p/ρ) U
⃗ dS +
⃗τ · U
{z
Z
So
⃗ dS
⃗τ · U
}
Ẇshear
+
|
Z
Sb
Z
So
⃗ · n̂ dS
ρ(ei + ek + p/ρ) U
⃗ dS
(−pn̂ + ⃗τ ) · U
{z
}
Ẇout,shaft
−
|
Z
S
q⃗′′ · n̂ dS +
{z
Z
Q̇out
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
V
q̇ dV
}
AEE 334 PROPULSION SYSTEMS I / 33
2. Review of Fluid Mech: General 1-D energy conservation
One dimensional flow relations are very useful in analysis.
Consider a stream tube (like stream of air captured into en engine).
Steady-state conservation laws for mass and energy applied to stream tube between two stations denoted inlet
(”in”) and outlet (”out”) yield, with usage of averaged properties at ”in” and ”out” stations,
ei +
1 2 p
U + + gz
2
ρ
+
in
Ẇin,shaft
Q̇in
+
=
ṁ
ṁ
ei +
1 2 p
U + + gz
2
ρ
+
out
Ẇshear
ṁ
Here we have included the potential energy (work) due to gravity for completeness, and Q̇in and Ẇin,shaft are
heat transfer and shaft work rates into content of stream tube between the inlet (in) and outlet (out), and
Ẇshear is the shear (friction) work rate
When no heat transfer, no work input and no frictional forces exist, the energy equation simply becomes
1
p
ei + U 2 + + gz
2
ρ
in
=
1
p
ei + U 2 + + gz
2
ρ
out
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
AEE 334 PROPULSION SYSTEMS I / 34
2. Review of Fluid Mech: Ideal 1-D energy conservation
Noting enhtalpy h = ei + p/ρ, energy equation may also be written as, for gas flows,
1
h + U2
2
=
in
1
h + U2
2
= constant = h0
out
Here h is static enthalpy, and h0 is called total enthalpy per unit mass.
It is clear that for an adiabatic and reversible process (isentropic), total enthalpy remains constant !
This is an extremely useful result. It is often used, especially for internal flows
For example, for quick determination of flow properties at the fan face in the intake of a turbofan engine, this
equation may be employed
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
AEE 334 PROPULSION SYSTEMS I / 35
Review of Fluid Mech: Ideal gas flow relations
Note, for perfect gases
h = Cp T =
γR
c2
T =
γ−1
γ−1
Then, show that 1-D ideal energy equation becomes
T0
γ−1 2
=1+
M
T
2
And using isentropic relations
p0
=
p
ρ0
=
ρ
γ−1 2
1+
M
2
γ−1 2
1+
M
2
γ
γ−1
1
γ−1
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
AEE 334 PROPULSION SYSTEMS I / 36
Review of Fluid Mech: Ideal gas flow relations (cont’d)
Exercise
Find an expression for the relationship of the Mach numbers of two sections along a stream tube in terms of the
ratio of areas of these sections.
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
AEE 334 PROPULSION SYSTEMS I / 37
Review of Fluid Mech: Ideal gas flow relations (cont’d)
Exercise
Find an expression for the mass flow rate through a given cross-sectional area of a duct, given the upstream
stagnation pressure and temperature.
Y. Özyörük - O. Uzol / Rev. Fluid Mech.
AEE 334 PROPULSION SYSTEMS I / 38