MEC E 430
Lecture Notes
MEC E 430
Lecture Notes
Fall 2023
Table of Contents
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MEC E 430
Lecture Notes
Fall 2023
Lecture 1
Example. Consider a 90◦ reducing elbow as shown in Fig. 1.1. Assuming steady state and a
fluid density of 999 mkg3 , what is the outlet flow speed? Also, what horizontal force must be
applied by the bolts at location 1 to keep the piping in place?
Figure 1.1: Schematic
Begin by defining a fixed control volume as shown in Fig. 1.2.
Figure 1.2: Schematic
We assume steady state and incompressible flow. Then the mass conservation equation becomes,
X
X
A1 v1
m
ṁin =
ṁout ⇐⇒ A1 v1 = A2 v2 =⇒ v2 =
= 16
A2
s
The momentum equation for a fixed control volume is,
X
X
ṁout~vout −
ṁin~vin = F~
|
{z
}
~ =m~a
Equivalent to Newton’s 2nd law: F
In the next lecture, we will find Fbolt using the momentum equation.
2
MEC E 430
Lecture Notes
Fall 2023
Lecture 2
Let us finish the example from the previous lecture.
Example. Recall the schematic of the CV as shown in Fig. 2.1.
Figure 2.1: Schematic
The momentum equation for a fixed control volume is,
X
X
ṁout~vout −
ṁin~vin = F~
|
{z
}
~ =m~a
Equivalent to Newton’s 2nd law: F
Here F~ represents the total/net force on the CV. We convert the momentum equation into two
scalar equations. Begin with the horizontal component, ~i direction.
X
−ṁin vin,x =
Fx (since piping is vertical at location 2) =⇒ vout,x = 0
External forces acting on the CV include:
• the force applied by the bolts, Fbolt
• the surface force associated with pressure i.e. P1 A1 . Note that we are not focused on
P2 A2 since it points in the vertical direction.
Thus we have,
−ṁin vin,x =
X
Fx = Fbolt + p1 A1 =⇒ Fbolt = −p1 A1 − ρA1 v1 2
Note that we use Pgage since Patm cancels on each side. Evaluating gives Fbolt = −3.39 kN. The
negative sign makes sense since flow is going in the +ve ~i direction.
Consider the following,
X
X
ṁin =
ṁout
X
X
ṁout~vout −
ṁin~vin = F~
They are said to be a special form of the integral form of the governing equations (mass and
momentum conservation equations for a control volume). That is, we consider average quantities
e.g. an average velocity at inlet/outlet. A further handy equation that considers average quantities
3
MEC E 430
Lecture Notes
Fall 2023
is Bernoulli’s equation.
p1
v1 2
p2
v2 2
+
+ z1 =
+
+ z2 + hf + hs − hq
ρg
2g
ρg
2g
where terms,
hf ≡ head loss due to friction between points 1 and 2
hs ≡ change of head due to shaft work done by the fluid
hq ≡ change of head due to heat that is added to the fluid
Example. What is the minimum diameter D1 at section 1 [Fig. 2.2] that just avoids cavitation?
Figure 2.2: Schematic
Begin by listing assumptions.
• The tank is broad such that the free surface descends very very very slow implying that
v3 = 0. More apparently, it helps us make a steady flow assumption (sufficiently slow
draining).
• The horizontal length of piping is short meaning any friction losses are small and therefore
negligible.
L v2
hf = f
(frictional losses)
D 2g
• The expansion that appears just downstream of point 1 is smooth and therefore not
associated with any additional losses.
hf 0 = kexp
v1 2
2g
(frictional losses at the expansion)
Begin by applying Bernoulli’s equation between point 2 and 3 .
p3
v3 2
p2
v2 2
+
+ z3 =
+
+ z2
ρg
2g
ρg
2g
At the surface of the tank and at the outlet, both faces are subject to Patm so they cancel.
Ap3
A
ρg
A
+
v3 2
p2
v2 2
+ z3 = AA +
+ z2
2g
ρgA
2g
4
MEC E 430
Lecture Notes
Fall 2023
Let us define the datum as the centerline of the pipe meaning z2 = 0.
Ap3
A
ρg
A
+
v3 2
p2
v2 2
+ z3 = AA +
+Z
zZ2
2g
ρgA
2g
Since we are assuming the tank is big, we have v3 = 0.
Ap3
A
ρg
A
vA 3 2
p2
v2 2
+ A + z3 = AA +
+Z
zZ2
2gA
ρgA
2g
Rearranging now,
v2 =
p
m
2gz3 = 9.90
s
Next, apply Bernoulli’s equation at points 1 and 2 .
p1
v1 2
p2
v2 2
+
+Z
zZ1 =
+
+Z
zZ2
ρg
2g
ρg
2g
The pressure at point 1 is pvap while point 2 is subject to atmospheric pressure. Rearranging
and solving for v1 gives,
s s
p2
v2 2
p1
2∆p1→2
m
v1 = 2g
+
−
=
+ v2 2 = 17.118
ρg
2g
ρg
ρ
s
Finally, we recall mass conservation,
r
A1 v1 = A2 v2 =⇒ D1 = D2
v2
= 0.114 m
v1
Note we round up our diameter so that velocity v1 will be smaller and therefore p1 will be
bigger by Bernoulli’s equation.
5
MEC E 430
Lecture Notes
Fall 2023
Lecture 3
Integral equations are the “bread and butter” of MEC E 331, but they have their limitations.
Consider, for example, an airfoil as shown in Fig. 3.1.
Figure 3.1: Schematic
To obtain an estimate for the drag force requires knowing the pressure and shear force all along
the airfoil surface. Our usual control volume equations can’t give us this information because
they consider average quantities only. Instead, we need to consider the differential form of the
dynamical equations.
Example. Consider blasting the water from a firehose against a horizontal wall as shown in
Fig. 3.2.
Figure 3.2: Schematic
The flow field is given by ~v = hkx, −kyi,
k, y > 0.
1. Is the flow incompressible?
2. Sketch the streamlines of the flow.
By the continuity equation, we have
∇ · ~v = 0 ⇐⇒
∂u ∂v
+
= 0 ⇐⇒ k − k = 0
∂x ∂y
Therefore the flow is incompressible.
From MEC E 331, streamlines (which equal streakliones and pathlines for a steady flow) are
given by
dx
dy
dx
dy
dx dy
=
⇐⇒
=
⇐⇒
+
=0
u
v
kx
−ky
kx ky
Integrating above yields,
ln x + ln y = C 0 ⇐⇒ xy = C =⇒ y(x) =
C
x
y is a family of curves i.e. streamlines are hyperbolas. Different hyperbolas are recovered by
choosing different values for C.
6
MEC E 430
Lecture Notes
Fall 2023
Example. A 1D radial flow in the r − θ plane, describing, for example, oil flowing into a well,
is given by
vr = f (r), vθ = vz = 0
What is the functional form of f if the fluid has constant density?
By the mass continuity equation,
∇ · ~v = 0 ⇐⇒
1 ∂
1 ∂vθ ∂vz
rur +
+
=0
r ∂r
r ∂θ
∂z
Recalling vθ = vz = 0,
1 ∂
1 ∂vθ ∂v
z
rur +
+ =0
r ∂r
r ∂θ
∂z
Then,
1 ∂
C
rur ⇐⇒ rur = C ⇐⇒ vr =
r ∂r
r
Note that C ∈ R because vr is a function of r only. Also note that fluid velocity decays as we
move further and further from the well. Conversely, vr ↑ as r ↓.
Example. In a square channel [Fig. 3.3], the velocity of a compressible gas is given by
x ~
~v = vo 1 +
i
L
Figure 3.3: Schematic
Find the acceleration of a particle moving with the flow.
The acceleration of a fluid particle is given by
D~v
,
Dt
Hence we have,
Dv
Dw
=
Dt
Dt
Du
∂u
∂u
∂u
∂u
=
+u
+v
+w
Dt
∂t
∂x
∂y
∂z
Again recalling v = w = 0, then
Du
∂u
∂u
∂u
∂u
=
+u
+ v + w
Dt
∂t
∂x ∂y
∂z
7
MEC E 430
Lecture Notes
Fall 2023
Substituting the definition of u,
Du
x uo
uo
x
= uo 1 +
=
1+
Dt
L L
L
L
As an extension, we can determine that the flow is compressible through the aid of the mass
continuty equation. Recall for incompressible flow,
∇ · ~v = 0 =
∂u ∂v ∂w
+ + =0
∂x ∂y ∂z
But u 6= 0 =⇒ ∂x u 6= 0. Therefore ∇ · ~v 6= 0 implying the flow is compressible.
Example. As shown in Fig. 3.4, the upper plate translates and lower plate is stationary. What
is the fluid velocity distribution?
Figure 3.4: Schematic
Assumptions:
• Steady, 2D flow (i.e. v = 0),
• Parallel flow (i.e. w = 0)
• Incompressible, laminar flow.
From the mass continuity equation for incompressible flow,
∇ · ~v = 0 =
∂u ∂v ∂w
∂u
+ + = 0 =⇒
=0
∂x ∂y ∂z
∂x
This means that u = u(z). Now consider Navier Stokes equation in the x direction.
!
∂u
∂u
∂u
∂u
1 ∂p µ ∂ 2 u ∂ 2 u ∂ 2 u
+u
+v
+w
=−
+
+
+
∂t
∂x
∂y
∂z
ρ ∂x ρ ∂x2 ∂y 2 ∂z 2
Since the flow is steady,
∂u
∂u
∂u
∂u
1 ∂p µ ∂ 2 u ∂ 2 u ∂ 2 u
+u
+v
+w
=−
+
+
+
∂x
∂y
∂z
ρ ∂x ρ ∂x2 ∂y 2 ∂z 2
∂t
!
Since the flow is 2D and parallel,
∂u
∂u
∂u
∂u
1 ∂p µ ∂ 2 u ∂ 2 u ∂ 2 u
+u
+ v + w
=−
+
+ +
∂x ∂y
∂z
ρ ∂x ρ ∂x2 ∂y 2 ∂z 2
∂t
8
!
MEC E 430
Lecture Notes
Fall 2023
Because the flow is driven by a translating boundary,
∂u
∂u
∂u
∂u
1 ∂p µ ∂ 2 u ∂ 2 u ∂ 2 u
+u
+ v + w
=−
+
+ +
∂x ∂y
∂z
ρ ∂x ρ ∂x2 ∂y 2 ∂z 2
∂t
!
Finally, by the mass continuity equation,
∂u
∂u
∂u
∂u
1 ∂p µ ∂ 2 u
∂ 2 u ∂ 2 u
+ +
+u
+ v + w
=−
+
∂x ∂y
∂z
ρ ∂x ρ ∂x2 ∂y 2 ∂z 2
∂t
!
Therefore, the Navier Stokes equation in the x direction simplifies to
∂ 2u
= 0 ⇐⇒ u(z) = Az + B
∂z 2
Above is a linear profile. To determine A and B, apply boundary conditions (no slip). That is,
u(z = 0) = 0 =⇒ B = 0
u(z = H) = V =⇒ B =
Hence,
V
H
V
z
H
The shear stress at the wall can also be determined.
u(z) =
τ =µ
du
µV
=
dz
H
This means the shear stress is constant. Additionally, z momentum equation can also be
applied. One obtains,
1 ∂p
0=−
− g =⇒ p = po + ρg(H − z)
ρ ∂z
Here po is the atmospheric pressure.
In the next lecture, we will consider how the results change when a 2nd fluid is added to the
channel.
9
MEC E 430
Lecture Notes
Fall 2023
Lecture 4
Example. Simple Couette flow involving two fluids is shown in Fig. 4.1.
Figure 4.1: Schematic
The flow is not driven by horizontal pressure gradients, that is
d2 ui
= 0,
dz 2
i = 1, 2
We need to define the boundary conditions of the problem.
1. u1 (z = 0) = 0
(no slip)
2. u2 (z = h1 + h2 ) = V
(no slip)
3. u1 (z = h1 ) = u2 (z = h1 ) (equality of velocities at interface)
4. τ1 (z = h1 ) = τ2 (z = h1 )
(equality of velocities at interface)
From our previous lecture, we determined the general solution for Couette flow.
(
u1 = C1 z + C2 ,
u(z) = Az + B =⇒
u2 = C3 z + C4
From the boundary condition u1 (z = 0) = 0,
C2 = 0 =⇒ u1 = C1 z
From the boundary condition u1 (z = h1 ) = u2 (z = h1 ),
C3 (h1 + h2 ) + C4 = V
From the boundary condition u1 (z = h1 ) = u2 (z = h1 ),
C1 h1 = C3 h1 + C4
From the boundary condition τ1 (z = h1 ) = τ2 (z = h1 ),
µ1 C 1 = µ2 C 3
Therefore, solving the above system of equations yields
u2 V
C1 =
,
µ2 h1 + u1 h2
µ1 V
C3 =
,
µ2 h1 + µ1 h2
10
C4 = V
µ2 h1 − µ1 h1
µ2 h1 + µ1 h2
MEC E 430
Lecture Notes
Fall 2023
With this information, u1 (z), u2 (z) and the associated shear stresses can be determined
explicitly. As before, note that shear stress is independent of position z.
Question, what can we say about pressure inside of the channel?
Consider the pressure in z direction for fluid 1,
dP1
= −ρ1 g
dz
As well, by defining pb as the pressure at the channel’s base (z = 0),
p1 = pb − ρ1 gz
For the fluid 2,
dP2
= −ρ2 g
dz
As well,
p2 = −ρ2 g(z − h1 ) + p2 (z = h1 )
= −ρ2 g(z − h1 ) + p1 (z = h1 )
= −ρ2 g(z − h1 ) + pb − ρ1 gh1
= pb − (ρ1 − ρ2 )gh1 − ρ2 gz
Let us shift our focus to laminar flow in an annular pipe.
Example. Laminar flow in an annular pipe. is shown in Fig. 4.2.
Figure 4.2: Schematic
Assumptions:
1. Steady flows
2. Parallel and laminar flows
3. Incompressible flows
4. Constant pressure gradient ∂x p
By the mass continuity equation,
∇ · ~v = 0 ⇐⇒
1 ∂
1 ∂vθ ∂vx
rur +
+
=0
r ∂r
r ∂θ
∂x
11
MEC E 430
Lecture Notes
Since the flow is parallel,
1 ∂v
1 ∂ ∂vx
θ
ru
+
=0
r +
r ∂r
r ∂θ
∂x
This means u = u(r). Now writing Navier Stokes equations in the x direction,
!
∂u
∂u uθ ∂u
∂u
1 ∂p µ 1 ∂
∂u
1 ∂ 2u ∂ 2u
+ ur
+
+u
=−
+
r
+ 2 2 + 2
∂t
∂r
r ∂θ
∂x
ρ ∂x ρ r ∂r ∂r
r ∂θ
∂x
Noting steady state, ur = uθ = 0 and the mass continuity equation, one obtains
dp
µ d
du
=
r
dx
r dr dr
We will complete this example in the next lecture.
12
Fall 2023
MEC E 430
Lecture Notes
Fall 2023
Lecture 5
Example. Recall where we left off in the previous lecture.
Figure 5.1: Schematic
Mass continuity implies that u(r). Navier-Stokes equations in the x direction,
dp
µ d
du
=
r
dx
r dr dr
Integrating gives,
u=
1 2 dP
r
+ C1 ln r + C2
4µ dx
By boundary conditions, at r = a, b, velocity u = 0
=⇒ 0 =
1 2 dp
a
+ C1 ln a + C2
4µ dx
=⇒ 0 =
1 2 dp
b
+ C1 ln b + C2
4µ dx
Solving for C1 and C2 yields,
b2 − a2 dp
C1 = −
,
4µ ln ab dx
"
#
1 dp b2 − a2
ln a − a2
C2 =
4µ dx ln b/a
Rearranging for u(r),
"
#
2
2
1 dp 2
b
−
a
r
ln
u=
r − a2 −
4µ dx
a
ln b/a
Now we can evaluating the shear stress,
τrx = µ
du
r dp 1 dp
b 2 − a2
=
−
−
dr
2 dx 4 dx r ln b/a
When τrx = 0, then velocity is a maximum. Define r? as the radius r when the maximum
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MEC E 430
Lecture Notes
Fall 2023
occurs. By solving, one obtains,
s
r? =
b 2 − a2
2 ln a/b
Suppose that a → 0, then r? → 0.
Let us now turn our focus towards frictional losses in laminar and turbulent flows.
Example. Heat exchangers often consist of many triangular passages. An example is shown
in Fig. 5.2 with L = 60 cm and an isosceles triangle cross-section of side length a = 2 cm and
included angle β = 80◦ . If the average velocity is v = 2 ms and the fluid is SAE oil at 20 ◦ C
kg
(fluid density is 870 mkg3 ; dynamic viscosity is 0.104 m·s
), estimate the pressure drop. Assume
that the product of friction factor and Reynolds number is f × Re = 52.9.
Figure 5.2: Schematic
First, we need to calculate Reynolds number. What length do we consider? We use the
hydraulic diameter Dh = 4A
. Denote the cross sectional area as A and the wetted perimeter
P
as P . Considering the geometry,
Dh = 1.2 cm
With Dh , we can calculate Re.
ρvDh
= 201
ν
This value of Re implies that we are in a laminar flow regime therefore using f × Re = 52.9 is
applicable. So the friction factor is,
ReDh =
f=
52.9
= 0.263
Re
The pressure gradient can therefore be calculated as,
∆P =
f L ρv 2
= 23 kPa
Dh 2
The pressure drop is quite large for a short pipe. This is because a sticky SAE oil is passing
through. Additionally, the geometry (triangular prism) is sharp.
This last example considered a case of laminar flow so that Dh was the only characteristic length
scale of interest. In turbulent flows by contrast, things are more complicated. In turbulent flows,
both Dh and Deff is required. Therefore, when solving for friction factor f , we use Deff . When
14
MEC E 430
Lecture Notes
Fall 2023
solving for ∆p or hf , use Dh .
ReDeff =
ρvDeff
,
µ
where Deff =
64
Dh
c
Example. A ventilation duct measuring 1 ft×1 ft in cross-section is to deliver 3600 cfm of air
at 68 ◦ F. What pressure drop, in inches of water, can be expected in a horizontal run of 200 ft?
Take e = 0.0002 ft as the wall roughness.
Begin by considering the hydraulic diameter.
Dh =
4A
= 1 ft
P
The mean velocity is v = Q/A = 60 fts . If the flow were laminar and given that a = b, C = 56.91
then the effective diameter is given by,
Deff =
ReDeff =
64
Dh = 1.1246 ft
C
ρvDeff
= 4.13 × 105 2200
µ
Based on Re, the flow is turbulent. Therefore, we can use the Moody chart to calculate the
friction factor.
e
relative roughness =
= 1.79 × 10−4
Deff
=⇒ f = f e/Deff , Re = 0.015
Finally, the head loss can be determined.
hf =
f L v2
= 2.4 in of water
Dh 2g
15
MEC E 430
Lecture Notes
Fall 2023
Lecture 6
Example. Air at 20 ◦ C flows through a smooth duct of diameter 20 cm at an average velocity
of 5 ms as shown in Fig. 6.1. It then flows into a smooth square duct of side length a. Find the
square duct size a for which the pressure drop per unit meter will be the same as that for the
circular duct.
Figure 6.1: Schematic
Begin by considering the pressure drop ∆p given by
f l ρv 2
∆p
f ρv 2
∆p =
=⇒
=
Dh 2
L
2Dh
Since we don’t know length L of each duct, we consider ∆P
, which must be the same between
L
the cylindrical and square duct (as per the problem statement). For example, friction factor f
changes with the geometry so we can’t cancel the term. However, density ρ can be cancelled
since both ducts have the same working fluid. We remark that velocity v should be different
for both ducts because both ducts have different cross sectional areas. Consider the volumetric
flow rate,
2
Q
Q
2
Q = vA =⇒ v =
=⇒ v =
A
A
Here, cross sectional area A = π4 D2 for a cylinder duct and A = a2 for the square duct.
2
vcylinder
=
16Q2
,
π 2 D4
2
vsquare
=
Q2
a4
We also need to consider the hydraulic diameters Dh which is D for the cylindrical duct and a
for the square duct. Combining these relations,
f ·16Q2
π2 D4
D
=
f ·Q2
a4
a
=⇒
16 fcylinder
fsquare
=
2
5
π
D
a5
The question becomes, how do we find f values? Start with the cylinder for which D = Deff =
20 cm.
ρvD
Re =
= 6.6 × 104
µ
The above result indicates turbulent flow. Referring to the Moody chart assuming a smooth
pipe meaning e = 0, fcylinder = 0.001961.
What can be done about the square duct case for which we don’t even know the value of a?
We will need to adopt an iterative solution technique.
(i) Assume a = D = 20 cm.
(ii) Solve for vsquare recalling volumetric flow rate is conserved.
16
MEC E 430
Lecture Notes
Fall 2023
2
πD2
π D
2
=⇒
vcylinder = a vsquare =⇒ vsquare =
vcylinder
4
4 a
(iii) Compute Deff
Deff =
64
Dh
56.91
(iv) Compute Reynolds number.
Re =
ρvsquare Deff
µ
(v) Use Moody chart with Re and e = 0 to solve for fsquare .
(vi) Check if the pressure drop per unit length to see if both sides are equal. If the expressions
aren’t equal, re-iterate to find a new value of a.
16 fcylinder
fsquare
=
2
5
π
D
a5
Running this scheme, a = 18.2 cm < D = 20 cm. But note that a2 > π4 D2 .
Question: We went to some effort to distinguish the laminar flow case from the turbulent flow
case. Would the results to the last example be all that different if we were to dispense with Deff
and use Dh everywhere. No, not really.
Question: Is it always the case that laminar and turbulent flows can be resolved using similar
methods. No.
Previewing this discussion, module no. 9 studied a case of a large scale open channel flow. In
this circumstance, there is a very significant difference between the laminar and turbulent flow
solutions. That is, if a laminar flow solution technique was applied, a profoundly incorrect solution
would be derived. After conquering this material, the question will be become how do we actually
model these turbulent flows? In module no. 10, we will cover Reynolds decomposition, and take
a parameter such as velocity v = v̄ + v 0 .
Z
1 T
v̄ =
v dt, v̄ 0 = 0
T 0
17
MEC E 430
Lecture Notes
Fall 2023
Lecture 7
In module no. 10, we were introduced to the idea of Reynolds decomposition.
Z
1 T
0
v = v̄ + v , v̄ =
v dt
T 0
Is there a way to have v 0 = 0? Therefore, we need to derive the Reynolds average Navier Stokes
equations (RANS).
Z
1 T
RANS =
NS dt
T 0
But performing these integrations yields within RANS equations terms like v 0 v 0 6= 0.
In module no. 11, we argue that the above problematic terms can be related to an eddy viscosity
µT . RANS equations then read as follows:
ρ
D~v
= ρ~g − ∇p + µ∇2~v + µT ∇~v
Dt
In turn, µT is a function of the flow and not of the fluid. This means µT is resolved using a k − ε
model where k ≡ turbulent kinetic energy and ε ≡ homogeneous dissipation rate i.e. the rate at
which turbulence is dissipated (note that dissipation ultimately occurs at the smallest of length
scales where the dynamic viscosity becomes important).
Some of the challenges associated with using a CFD approach include:
• For complicated flows, there is no particular reason why the turbulence terms can be rolled
into an eddy viscosity term.
• The transport equations for the k − ε are challenging to solve and lack sound theoretical
justification.
• Coupled PDEs contain a number of empirical constants. While this gives us an opportunity
to tune the equations to a particular flow, there is a certain lack of generality, meaning when
it comes to an unknown flow, we are in the dark i.e. investigation of new flows is a challenge.
The moral of the story is to use good engineering judgment when interpreting the results of a
CFD simulation.
We will now focus on laminar boundary layer equations such as Fig. 7.1.
Figure 7.1: Schematic
18
MEC E 430
Lecture Notes
Fall 2023
Typically BL thickness is defined based on BL velocity as compared to free stream velocity U .
Often δ corresponds to locus of points where u = 0.99U . Begin by applying mass continuity.
∇ · ~v = 0 ⇐⇒
∂u ∂v
+
=0
∂x ∂y
Define the length of the flat plate as L units long. Hence L is the distance in x over which velocity
U changes.
∂u
U
∼
∂x
L
In a likewise fashion,
∂v
v
∼
∂y
δ
It is helpful to think of δ as a “characteristic” vertical length scale i.e. δ might represents an
average on terminal BL thickness.
Combining both terms,
U
v
Uδ
∼
=⇒ v ∼
L
δ
L
We have defined a characteristic vertical velocity scale. But since δ/L 1, we have
vU
Keeping these separation of scales in mind, it is helpful to go ahead and non-dimensionalize our
variables.
x
y
u
vL
p − p∞
?
?
?
?
?
x = , y = , u = , v =
, p =
L
δ
U
Uδ
ρU 2
Here p∞ is the pressure measured outside of the BL. We will use these non-dimensionalized variables when “doing battle” with the Navier Stokes equations. This will allow us to cancel nonimportant terms in the equations.
19
MEC E 430
Lecture Notes
Fall 2023
Lecture 8
We will continue to discuss the non-dimensionalization of the NS equations for a laminar boundary
layer as shown in Fig. 8.1. Consider the dimensional form of the NS equation in the y direction
for a 2D case.
Figure 8.1: Schematic
∂v
∂v
∂v
1 ∂p µ ∂ 2 v ∂ 2 v
+u
+v
=−
+
+
∂t
∂x
∂y
ρ ∂y ρ ∂x2 ∂y 2
!
We need to determine which terms are important and which aren’t. With this information, we
can simplify the NS equations.


u
?
2
?

∂v
∂ 
 v U δ  = U δ u? ∂v
= |{z}
u? U
∂x
∂ |{z}
x? L | {z
L }
L
∂x?
u
x
v
∂p
∂v
Using a similar approach, we can replace terms such as v ∂y
and − ρ1 ∂y
. After this manipulation,
2 ?
2
L ∂p
1 ∂ 2v?
1 L ∂ 2v?
u
+v
=−
+
+
∂x?
∂y ?
δ ∂y ? Re ∂x? 2 Re δ ∂y ? 2
? ∂v
?
? ∂v
?
Here, we assumed steady flow so the term with time derivative is canceled. Additionally, we take
Re 1. To cancel terms, we consider the magnitude of each term. As an example,
Re 1 =⇒
1
1
Re
So with this logic, we have term 4 is insignificant when compared to term 1 and 2 . Similarly,
we expect term 3 to be greater than term 5 .
How about term 3 to term 1 and 2 . Recall,
2
L
δ/L 1 =⇒
1
δ
Therefore we conclude that term 3 is significantly larger than term 1 and 2 .
20
MEC E 430
Lecture Notes
Finally, our PDE simplifies to
Fall 2023
∂p?
∂p
= 0 ⇐⇒
=0
?
∂y
∂y
The pressure inside of the BL is in fact imposed by the outer flow and they must be the same. So
p = p(x).
Let us apply the same approach to NS in the x-direction.
∂u
∂u
∂u
1 ∂p µ ∂ 2 u ∂ 2 u
+u
+v
=−
+
+
∂t
∂x
∂y
ρ ∂x ρ ∂x2 ∂y 2
!
Using the non-dimensionalized parameters,
∂u?
u? ?
∂x
+
∂u?
v? ?
∂y
2
∂p?
1 ∂ 2 u?
1 L ∂ 2 u?
=− ? +
+
∂x
Re ∂x? 2 Re δ ∂y ? 2
Again noting that Re 1, we conclude term 4 is small and can be canceled. We are also
tempted to cancel term 5 because of the 1/Re coefficient. However it should not be canceled
2
because of the magnitude of L/δ which is unknown. Additionally, term 5 is the only term
remaining containing viscosity µ. On physical ground, viscosity should be retained since it is an
important factor to include. Suppose,
r
r
2
1 L
L √
UL
L
νL
∼ 1 =⇒
∼ Re =
, δ∼q
=
Re δ
δ
ν
U
UL
ν
This suggests that the BL↑ in thickness as we ↑ ν and L and ↓ U .
We then have to solve an equation with four terms, none of which can be eliminated:
2
?
?
∂p?
1 L ∂ 2 u?
? ∂u
? ∂u
u
+v
=− ? +
∂x?
∂y ?
∂x
Re δ ∂y ? 2
21
MEC E 430
Lecture Notes
Fall 2023
Lecture 9
Recall from the previous lecture:
L
δ∼q
r
=
UL
ν
νL
=⇒ δx =
U
r
νx x
U
where x is the distance measured from the leading edge. δ is the thickness of the boundary layer
and L is the length of the flat plate. Additionally, the Navier Stokes equation in the x direction
(non-dimensional) reads:
∂u?
u? ?
∂x
+
∂u?
v? ?
∂y
2
∂p?
1 L ∂ 2 u?
=− ? +
∂x
Re δ ∂y ? 2
Let’s re-express the last result using dimensional rather than non-dimensional variables.
u
∂u
∂u
1 dp
∂ 2u
+v
=−
+ ν 2,
∂x
∂y
ρ dx
∂y
ν=
µ
ρ
Can the pressure gradient term be re-expressed? Indeed, this term can be re-written in terms of
the external flow velocity using Bernoulli equation while ignoring gravitational effects i.e. change
in elevation. Along a streamline from 1 to 2 , Bernoulli’s equation reads:
p1 1 2 p2 1 2
+ u1 =
+ u2
ρ
2
ρ
2
More generally,
p 1 2
1 dp
dU
+ U = constant =⇒ −
=U
ρ 2
ρ dx
dx
Therefore the x momentum equation can be re-written as:
u
∂u
∂u
dU
∂ 2u
+v
=U
+ ν 2,
∂x
∂y
dx
∂y
ν=
µ
ρ
Clearly, if U is not a function of x (e.g. flow over a flat plate), then du
= 0 which simplifies the x
dx
momentum equation further.
∂u
∂u
∂ 2u
µ
u
+v
= ν 2, ν =
∂x
∂y
∂y
ρ
..............................................................................................
Let us summarize the results over the past few lectures. The BL equations for flow over a flat
plate are as follows:
p = p(x) is a function of x and not y. If the free stream velocity is constant, then p = constant =
p∞ . This leads to
∂u
∂u
∂ 2u
u
+v
=ν 2
∂x
∂y
∂y
∂u ∂v
+
=0
∂x ∂y
We will need 4 boundary conditions to solve the PDEs.
22
MEC E 430
Lecture Notes
Fall 2023
• u = 0 when y = 0 i.e. no slip
• u → U as y → ∞ i.e. relaxation to the far field condition
• u = U for all y when x = 0
• v = 0 when y = 0 i.e. no penetration or no normal flow at the solid boundary
Note that with L → ∞, there is no inherent length scale to the problem. Contrast this with
Couette flow where there is a length scale “built in”. I.e. the depth of the channel is problem. Or
consider Poiseuille flows which are pressure driven. The flow can’t be reconciled without making
reference to a radius r or diameter d. Indeed, the BL flow over a flat plate problem enjoys the
property of self-similarity i.e. the flow appears the same no matter how we zoom in or zoom
out. With this revaluation of self-similarity, Blasius in 1908 proposed a similarity variable η that
combines x and y into non-dimensional independent variable where,
r
U
η=y
νx
It can be verified that the units work out.
s
m
m
s
m2 /s · m
=1
Blasius then summarized correctly that the solution for u can be written in terms of η i.e. u/U =
f (η). This is “good news” however the bad news is twofold i.e. (i) lots of tedious algebra that is
needed in order to simplify the governing equations and (ii) the Blasius solution involves solving
a non-linear ODE for which there is not a known analytical solution.
1 00
f f + f 000 = 0
2
The silver lining however is that one can applying a numerical approach (shooting method).
23
MEC E 430
Lecture Notes
Fall 2023
Lecture 10
Recall from the previous lecture:
1 00
f f + f 000 = 0
2
The silver lining however is that one can applying a numerical approach (shooting method). The
solution is shown in Fig. 10.1.
Figure 10.1: Schematic
q
U
Recall that η = y νx
. Often times, we define the boundary layer thickness based on the 99%
criteria i.e. based on the position where u = 0.99U . As the figure suggests, this locus of points
is defined by a particular value of η = 4.91. Recalling the definition of η and re-arranging with
y = δ i.e. the BL thickness,
r
r
U
νx
δ
4.91
=⇒ 4.91 = δ
⇐⇒ δ = 4.91
⇐⇒
=
νx
U
x
Rex
This aligns with the previous notion determined in previous lectures
r
νx
δ∼
U
Example. Air at 20 ◦ C and 1 atm flow at 20 ms past a flat plate as shown in Fig. 10.2. A pitot
stagnation tube, placed at a vertical distance of 2 mm from the wall, develops a manometer
head of h = 16 mm of Meriam red oil whose density is 825.35 mkg3 . Estimate the downstream
position of the pitot tube relative to the leading edge of the plate. Assume laminar (but high
Re) flow.
24
MEC E 430
Lecture Notes
Fall 2023
Figure 10.2: Schematic
The pressure difference associated with the manometer is given by
∆pmano = (ρoil − ρair )gh = 129 Pa
Applying Bernoulli’s equation, the velocity in front of the manometer is given by
s
2∆pmano
m
u=
= 14.7
ρair
s
Since the free stream velocity U is known, we can calculate their ratio
u
= 0.735
U
Referring to the Blasius BL solution shown in Fig. 10.1, η = 2.42.
r
2
2
U
U y
U y
η=y
⇐⇒ x =
=
= 0.911 m
νx
ν η
µ/ρ η
Example. A sharp flat plate with L = 50 cm and width b = 3 m is parallel to a stream of
velocity 1.0 ms . Find the drag on one side of the plate and the BL thickness, δ, at the trailing
edge assuming (a) air, and (b) water. Assume 20 ◦ C and 1 atm as the temperature and pressure,
respectively.
kg
m2
ρair = 1.2 3 , νair = 1.5 × 10−5
m
s
kg
m2
ρwater = 998 3 , νwater = 1.005 × 10−6
m
s
(a) Let us begin by calculating Reynolds number of air
Re =
UL
= 33333 < 500000 =⇒ laminar
ν
When considering the terminal BL thickness δL ,
δL
4.91
=√
=⇒ δL = 1.34 cm
L
ReL
25
MEC E 430
Lecture Notes
Fall 2023
To determine the drag for width b of the plate and surface shear stress τwall ,
Z L
Fdrag = b
τwall dx
0
Recall,
τwall = µ
du
dy
y=0
We need to determine the gradient,
du
dy
Here
∂ u/U
∂η
y=0
∂ U · u/U
p
=
∂η νx
U
r
=
y=0
U 3 ∂ u/U
νx ∂η
η=0
can be evaluated from the slope of Blasius solution. One obtains
η=0
r
τwall = 0.332µ
U3
ρU 2
= 0.332 √
νx
Rex
Therefore, τwall can be integrated
√
Fdrag = 0.774bρ νLU 3
Returning to the example,
Fdrag,air = 0.774bρair
p
νair LU 3 = 0.006 55 N
(b) We now consider water.
Re =
UL
= 497512 < 500000 =⇒ laminar flow just...
ν
δL
4.91
=√
=⇒ δL = 0.348 cm
L
ReL
p
Fdrag,water = 0.774bρwater νwater LU 3 = 1.41 N
Summary: Comparing the answers, we see that the BL thickness in water is thinner. Since the
fluid density is also larger, the drag is larger (consider a jet flying in water vs air). Additionally,
the drag force can alternatively be calculated in a different way i.e. where we first consider a
drag coefficient.
26
MEC E 430
Lecture Notes
Fall 2023
Lecture 11
Regarding the previous lecture, in water, the BL is thinner, the fluid density is larger and the drag
force is larger, too. The drag force can be evaluated in a more circuitous way whereby we first
calculate the non-dimensional drag coefficient, CD , whose definition is
CD =
drag force
1
ρU 2 · bL
2
Note similarity to the definition of skin friction coefficient
τwall
1
ρU 2
2
Cf,x =
What is the relationship between CD and Cf,x ? By the definition for plate width b,
drag = D = b
L
Z
τwall dx
0
r
τwall = 0.332µ
U3
,
νx
(laminar BL flow)
Hence,
r
L
U3
0.332µ
dx
νx
0
r
Z
U 3 L −1/2
= 0.332µb
x
dx
ν 0
r
L
U 3 1/2
= 0.664µb
x
ν
0
r
3
U L
= 0.664µb
√ ν
= 0.664ρb νU 3 L
Z
D=b
Recalling the connection between drag D and CD ,
1
D
D = CD ρU 2 · bl =⇒ CD = 1 2
2
ρU bL
2
√
r
0.664ρb νU 3 L
ν
1.328
=⇒ CD =
= 1.328
=√
1
2
UL
ρU bL
ReL
2
To infer the relationship between CD and Cf,x , we need to go back to the beginning.
Z
L
D=b
τwall dx
0
1
=⇒ CD ρU 2 · bL = b
2
27
Z
0
L
1
Cf,x · ρU 2 dx
2
MEC E 430
Lecture Notes
1
=⇒ CD =
L
If CD =
1.328
√
ReL
Z
Fall 2023
L
Cf,x dx
0
then it can be shown that
0.664
Cf,x = √
Rex
The above results apply for a laminar BL flow. If the BL is turbulent,
Cf,x =
0.027
,
Rex 1/7
CD =
0.031
ReL 1/7
Example. A hydrofoil 1.2 ft long and 6 ft wide is placed in a seawater flow of 40 fts with ρ =
2
1.99 slug
and ν = 0.000 011 fts . Estimate the BL thickness at the end of the plate. Estimate
ft3
the friction drag for turbulent smooth wall flow.
The problem statement assumes a turbulent BL. Is this a valid assumption?
ReL =
UL
= 4.36 × 106 > 5 · 105
ν
This means we have a turbulent BL everywhere. We can then evaluate a BL thickness from
δL =
0.16L
= 0.0216 ft
Re1/7
To calculate the drag force, begin with the evaluation of the drag coefficient CD .
CD =
0.031
ReL
1/7
= 0.00349
We have drag on both sides of the hydrofoil,
1
D = 2 × ρU 2 bL = ρU 2 bL = 80 lbf
2
Example. A very thin sheet of fiberboard weighs 90 N and lies on a flat rooftop as shown
kg
below. Assume ambient air with ρ = 1.2 mkg3 and µ = 1.8 × 10−5 m·s
. If the coefficient of
solid friction between the fiberboard and the roof is 0.12, what wind velocity, U , will generate
sufficient friction to dislodge the board? (Hint: Assume a turbulent boundary layer with
Cf,x = 0.027/Rex 1/7 .
28
MEC E 430
Lecture Notes
Fall 2023
Figure 11.1: Schematic
Previously in considering the connection between drag and wall shear stress,
Z L
D=b
τwall dx
0
Z 5
1
=b
Cf,x · ρU 2 dx
2
Z2 5
0.027 1 2
=b
· ρU dx
1/7 2
2 Rex
Z 5
1 2
0.027
= ρU b
1/7 dx
2
ρU x
2
µ
Z 5
0.027 6/7 13/7 1/7
=
ρ U
µ b
x−1/7 dx
2
2
5
0.027 6/7 13/7 1/7 7 6/7
=
ρ U
µ b· x
2
6
2
In order to dislodge fiberboard, the above drag force must exceed the friction force of 0.12 ·
weight.
5
0.027 6/7 13/7 1/7 7 6/7
0.12 · 90 N =
ρ U
µ b· x
2
6
2
Solving for velocity,
m
s
Indeed with the value of velocity, it can be verified the boundary layer is turbulent.
U = 33
29
MEC E 430
Lecture Notes
Fall 2023
Lecture 12
Example. A three-bladed helicopter rotor rotates at 200 rpm as shown in Fig. 12.1. If each
blade is 4 m long by 0.4 m wide, estimate the torque needed to overcome friction on the blades
kg
assuming they act as smooth flat plates. Take ρ = 1.2 mkg3 and µ = 1.8 × 10−5 m·s
.
Figure 12.1: Schematic
The problem complexity increases since the “free stream” velocity is variable. We must think
about the boundary layers of the cross sections running along r. First, we need to consider
areas in which the flow is laminar and turbulent. Indeed (conjecture), there exists a laminar
flow regime up to radius rtr for which ReL < ReL,tr = 5 × 105 . Afterwards, the flow becomes
turbulent.
ρ(rtr ω)L
µReL,tr
ReL,tr =
=⇒ rtr =
= 0.897 13 m
µ
ρωL
This means the flow regime is laminar from 0 ≤ r < rtr . On the other hand, the flow regime is
turbulent from rtr < r ≤ 4. Now, let’s determine the torque. Since the velocity is variable the
force is variable as well. Consider an infinitesimal element from the blade with length dr and
a width of L = 0.4 m with velocity U . Then the force of the element is:
dF = CD
ρv 2
L · dr × 2
2
Note that the dF element is multiplied by 2 since there is drag on both sides of the blade. Not
we can calculate the differential torque using
dM = r dF
However recall that the blade has two flow regimes. For laminar regime,
1.328
CD = √
ReL
30
MEC E 430
Lecture Notes
For turbulent regime,
CD =
Therefore (noting ReL =
ρ(rω)L
µ
Fall 2023
0.031
ReL 1/7
and v = rω),
Z
rtr
Mlaminar =
r dF
0
Z
rtr
rCD ρv 2 L dr
=
Z0 rtr
1.328
r·√
· ρv 2 L dr
Re
L
0
Z rtr
1/2 1/2 1/2 3/2
= 1.328ρ µ L ω
r5/2 dr
=
0
= 0.0726 N · m
Similarly,
Z
4
Mturbulent =
r dF
r
Z tr4
=
rCD ρv 2 L dr
rtr
4
Z
0.031
· ρv 2 L dr
1/7
Re
rtr
L
Z
6/7 1/7 6/7 13/7
= 0.031ρ µ L ω
=
r·
rtr
r20/7 dr
0
= 53.4 N · m
Thus the total torque is,
Mtotal = 3 × (Mlaminar + Mturbulent ) = 160 N · m
Comments:
• In this example, the drag force (corresponding the torque) are much larger over the turbulent portion of the blade. Why? The velocity and moment arm is larger. Additionally,
the area of the turbulent regime is greater. Furthermore, the skin friction associated with
the turbulent BL is larger.
• The final answer in this example must be regarded as conservative (Fig. 12.2). To be
completely rigorous, we should separately account for laminar and turbulent drag at all
distances beyond rtr .
31
MEC E 430
Lecture Notes
Figure 12.2: Schematic
32
Fall 2023
MEC E 430
Lecture Notes
Fall 2023
Lecture 13
Up until now, we’ve implicity assumed that the BL is defined by the locus of points where u =
0.99U . From the Blasius solution for laminar, high Re BLs, we know that
√
δ99
4.91
=√
=⇒ δ99 ∼ x
x
Rex
Meanwhile, for turbulent BL, is is found from semi-empirical evidence that
δ99
0.16
=
=⇒ δ99 ∼ x6/7 > x1/2
x
Re1/7
x
It is important to remark that the 99% criterion is arbitrary. Therefore, can we use a different
criteria i.e. 97.5% etc. A less arbitrary approach is to define a displacement thickness δ ∗ as the
vertical distance that streamlines outside of the boundary layer are displaced from the wall by the
boundary layer. This is depicted in Fig. 13.1.
Figure 13.1: Schematic
It is important to note that δ ∗ is a function of x starting from the leading trail. Additionally, δ99
cannot cross the streamlines. To calculate δ ∗ , we first consider the amount of mass that is flowing
through transects situated up-stream versus down-stream of the leading edge. This is depicted in
Fig. 13.2.
Figure 13.2: Schematic
Therefore, there is a nice geometric relation between these three parameters.
ho + δ ∗ = h
Additionally, the following assumptions are important.
• No flow through plate surface
33
MEC E 430
Lecture Notes
Fall 2023
• No flow across the streamline
Applying these considerations, the mass flow rate measured below the streamline at point 1 and
2 must be equal.
Z ho
Z h
ρu dy =
ρu dy
0
0
The density is constant and on the LHS, velocity u = U (free stream velocity) is constant. On the
other hand, u on the RHS is variable (BL).
Z
ρU ho = ρ
h
u dy
0
Applying the geometric relation,
∗
Z
h
u
dy
0 U
Z h
u
∗
δ =h−
dy
0 U
Z h
u
=
1−
dy
U
0
∗
ho = h − δ =⇒ h − δ =
Finally, we recognize that the value of h is irrelevant so long as we are outside of the boundary
layer since all streamlines outside of the BL are displaced by the same vertical distance.
Z ∞
u
∗
δ =
1−
dy
U
0
In a mathematical sense, this is how we define the displacement thickness. For example, if the BL
flow was laminar, then we could evaluate δ ∗ based on Blasius BL solution.
r
µx
1.721x
δ∗
1.721
∗
δ = 1.721
= √
=⇒
=√
ρU
x
Rex
Rex
Equivalently, for turbulent BLs,
δ∗
0.02
=
x
Re1/7
x
One final measure related to the BL thickness is the momentum thickness, which we define as
Z ∞ u
u
θ=
1−
dy
U
U
0
The momentum thickness is important because it is related to the drag D on a flat plate as follow.
Consider again a control volume CV surrounding a BL as shown in Fig. 13.3.
34
MEC E 430
Lecture Notes
Fall 2023
Figure 13.3: Schematic
Let’s apply the x-component of Newton’s 2nd law for the CV defined by the red dashed lines.
X
X
X
Fx =
ṁout vout −
ṁin vin
Note that
P
Fx = −D where D is the drag force that is exerted by the plate on the fluid.
Z
2
−D = −ρU bho + ρb
| {z }
|
inflow at 1
h
u2 dy
{z
}
0
outflow at 2
Here, b represents the width into the page. As well, it is important to note that at 1 , u = U
which is constant however at 2 , u is variable in the BL. Recall that,
ho + δ ∗ = h
Substituting this relation,
Z
∗
2
h
D = ρU (h − δ ) − ρb
u2 dy
0
Z h
2
= ρU b h −
0
u
1−
U
dy
Z
− ρb
0
h 2
u
u
dy − ρU 2 b
dy
2
0 U
0 U
Z h u
u
2
= ρbU
1−
dy
U
0 U
Z ∞ u
u
2
= ρbU
1−
dy
U
U
0
|
{z
}
= ρU 2 b
Z
h
!
Z
θ
2
= ρbU θ
35
h
u2 dy
MEC E 430
Lecture Notes
Fall 2023
Thus, the drag exerted on a plate can be calculated if the momentum thickness is known. We
already have expressions for drag force D. Let’s use these to derive expressions for the momentum
thickness θ. It can be shown that,
laminar =⇒
turbulent =⇒
36
θ
0.664
=
x
Re1/2
x
θ
0.016
=
x
Re1/7
x
MEC E 430
Lecture Notes
Fall 2023
Lecture 14
Example. A sharp flat plate with L = 50 cm and width b = 3 m is parallel to a stream of
velocity 1.0 ms . Find the terminal displacement thickness, d∗ , and momentum thickness, θ.
2
Assume the fluid flowing over the plate is water with ρ = 998 mkg3 and ν = 1.005 × 10−6 ms .
From our previous calculations, ReL = 497512 while δ99 = 0.348 cm. The Re value indicates
we are close to the transition point. Additionally the one-sided drag D = 1.41 N. Now,
δ∗
1.721
=√
x
Rex
To determine the terminal boundary layer thickness, x = L (trailing edge of the plate),
1.721L
δ∗ = √
= 0.122 cm < δ99
ReL
Also, let us consider the momentum thickness θ for x = L,
θ=
0.664L
1/2
ReL
= 0.047 cm < δ ∗ < δ99
Recall, drag force D
D = ρbU 2 θ = 1.41 N
Indeed, either way, the drag force result is the same (using CD or θ approach).
Example. The ideal flow velocity along a wall varies with a distance along the wall as U (x) =
1 − x2 as shown in Fig. 14.1. Close to the wall, a laminar BL develops. (i) Estimate the drag
on this wall over a distance x. (ii) Estimate the BL separation point and compare your answer
to the measured value of xseparation = 0.271 m. Assume that lengths are measured in m through
out.
Figure 14.1: Schematic
(i) To estimate the drag (using Twaites’ equation to calculate θ),
D = ρbU 2 θ
37
MEC E 430
Lecture Notes
Fall 2023
The momentum thickness is given by
Z x
5
0.45µ
2
θ =
U (τ ) dτ
6
ρU (x) 0
Z x
0.45µ
2 5
=
1
−
τ
dτ
ρ(1 − x2 )6 0
0.45µ
5 3
10 7 5 9
1 11
5
=
x − x + 2x − x + x − x
3
7
9
11
ρ(1 − x2 )6
Substituting this result into the the drag equation,
1/2
D = (0.45ρµ)
1/2
b
5 3
10 7 5 9
1 11
5
x − x + 2x − x + x − x
1 − x2
3
7
9
11
(ii) Recall that separation occurs when
θ2 dU
= −0.09
ν dx
1
0.45µ
5 3
10 7 5 9
1 11
5
×
x − x + 2x − x + x − x
× (−2x) = −0.09
ν ρ(1 − x2 )6
3
7
9
11
0.45
5 3
10 7 5 9
1 11
5
x − x + 2x − x + x − x
× (−2x) = −0.09
3
7
9
11
(1 − x2 )6
The cancellation of kinematic viscosity meas the separation does not depend on density or
viscosity. Solving the above equation numerically (i.e. MATLAB’s roots or fsolve)
xseparation = 0.268 m
Comparing against 0.271 m, there is good agreement is observed.
38
MEC E 430
Lecture Notes
Fall 2023
Lecture 15
Example. Consider the left-to-right flow over a solid cylinder of radius R as shown in Fig. 15.1.
Figure 15.1: Schematic
Use ideal flow theory, it can be shown that the flow just exterior to the BL is given by
s
U = 2U∞ sin φ = 2U∞ sin
R
(i) Derive an expression for θ/R. (ii) What is the pressure gradient as measured along the
cylinder surface? (iii) Does the flow separate from the cylinder?
(i) Applying Twaites equation,
0.45µ
θ =
ρU 6 (s)
2
Z sh
U s0
i5
ds0
0
where s0 is a dummy variable. It follows that
0.45µ
θ =
·R
ρ[2U∞ sin φ]6
2
φ
Z
2U∞ sin φ0
5
dφ0
0
where φ0 is another dummy variable (derived from φ0 = s0 · R). Exploiting some cancellations,
0.45µR
θ =
2ρU∞ sin6 φ
2
φ
Z
sin5 φ0 dφ0
0
We are interested in θ/R so divide the expression by R2
θ2
0.45µ
=
2
R
2RρU∞ sin6 φ
We can remark that
Z
φ
sin5 φ0 dφ0
0
µ
1
=
2RρU∞
Re
Z
φ
θ2
0.45
=
sin5 φ0 dφ0
6
R2
Re sin φ 0
39
MEC E 430
Lecture Notes
Fall 2023
Performing the integration,
θ2
0.45
=
2
R
Re
0.45
=
Re
φ
1
5
5
1
0
0
0
·
− cos φ +
cos 3φ −
cos 5φ
8
48
80
sin6 φ
0
1
5
5
1
·
0.53̄ − cos(φ) +
cos(3φ) −
cos(5φ)
8
48
80
sin6 φ
Interestingly, applying L’Hopital’s rule suggests that
lim
0.53̄ − 58 cos(φ) +
φ→0
5
48
cos(3φ) −
sin6 φ
1
80
cos(5φ)
6= 0
We conclude that θ is not zero at the leading edge! Why is this? It is because the leading edge
represents a forward stagnation point where all the velocities go to zero.
(ii) To find the pressure distribution, we recall that the pressure inside of the BL must match
the pressure that we measured on the outside. By Bernoulli’s equation,
1 2
1
p∞ + ρU∞
= p + ρU 2
2
2
Take the derivative with respect to s,
=⇒ 0 =
dp
+ ρU dU s
ds
Recalling U = 2U∞ sin φ where φ = s/R,
=⇒
dp
2U∞
= −p · 2U∞ sin φ ·
cos φ
| {z } | R {z }
ds
U
Simplifying,
dU
ds
2
dp
2ρU∞
=−
sin(2φ)
ds
R
It can be noted that dp
> 0 when π/2 ≤ φ ≤ π since sin(2φ) < 0 in this domain. Such an
ds
“adverse” pressure gradient may cause flow to separate.
(iii) Flow will separate when
Recall that,
θ2 dU
= −0.09
ν ds
2
θ
0.45
1
5
5
1
=
·
0.53̄ − cos(φ) +
cos(3φ) −
cos(5φ)
R
Re sin6 φ
8
48
80
40
MEC E 430
Lecture Notes
Fall 2023
θ2
0.225R2
5
5
1
2U∞
dU s =
0.53̄ − cos(φ) +
cos(3φ) −
cos(5φ) ·
cos φ
6
ν
8
48
80
R
RU∞ sin φ
We note the cancellation of R meaning the radius of the cylinder doesn’t affect flow separation.
θ2 dU
0.45 cos φ
5
5
1
=
0.53̄ − cos(φ) +
cos(3φ) −
cos(5φ) ≡ f (φ)
ν ds
8
48
80
sin6 φ
Is there any
value of φ such that f (φ) = −0.09? We check critical values of φ = π/2, π leading
to f π/2 = 0 and f (π) = −∞. It follows that f (φ) is a continuous function of its argument
thus their must exist an in-between value of φ (i.e. −π/2 < φ < π) for which f (φ) = −0.09.
Hence the flow will separate.
41
MEC E 430
Lecture Notes
Fall 2023
Lecture 16
What if we were to apply Twaite’s equation to a very special case of, for example, uniform flow
over a flat plate?
Z
0.45µ x 5 0
2
θ =
U dx
ρU 6 0
If U is independent of x,
θ2 =
0.45µU 5
0.45µx
θ2
0.45µ
0.45
x
=
=⇒
=
=
6
2
ρU
ρU
x
ρU x
Rex
Hence,
θ
0.671
=√
x
Rex
Why is the coefficient 0.671 not 0.664? Twaite’s equations, though bench-marked against the
Blasius boundary layer solution, also incorporates laboratory date (through best fit) and so don’t
reproduce Blasius result with 100% accuracy.
..............................................................................................
Compressible flows
The incompressible assumption is a good approximation for liquids but not always for gases.
Compressiblity effects arise when Ma = av ≥ 0.3 where v is the fluid speed and a is the speed of
sound. Another way of expressing the above inequality is to state
∆T
≥ 0.1
T
where ∆T is the temperature change in the flow and T is the absolute temperature (Kelvin).
Examples include:
• Aerodynamics
• Gas pipelines
• Gas compressors and turbines
The bad news → variable ρ introduces a suite of new complications both experimentally and analytically. The good news → most compressible flows can be modeled as inviscid i.e. µ = ν = 0
which simplifies matters.
The hierarchy of compressible flows are as follows:
• Ma < 0.3, incompressible flow
• 0.3 ≤ Ma < 0.8, subsonic flow (most commercial aircrafts)
• 0.8 ≤ Ma < 1.2, transonic flow
• 1.2 ≤ Ma < 3, supersonic flow (military aircraft)
• Ma ≥ 3, hypersonic flow
42
MEC E 430
Lecture Notes
Fall 2023
..............................................................................................
Review of relevant thermodynamics
Most compressible flows can be assumed to satisfy the ideal gas law.
p = ρRT
J
where R = RMu where Ru = 8314.3 kmol·K
is the universal gas constant and M is the molar mass in
g
units of mol . Commonly for air,
J
R = 287
kg · K
Typically, one also assumes that cp and cv are the specific heat capacities measured at a constant
pressure and volume. Generally, cp and cv are constant so long as the temperature changes are
moderate, around 100 to 200 Kelvin.
43
MEC E 430
Lecture Notes
Fall 2023
Lecture 17
Last lecture, we introduced cv and cp . Why are these needed in a fluid mechanics? We are
interested in changes in energy. For internal energy for example,
Z T2
û2 − û1 = cv (T2 − T1 ) =
cv T 0 dT 0
T1
where ûi is the internal energy per unit mass. Likewise for changes in enthalpy,
Z T2
h2 − h1 = cp (T2 − T1 ) =
cp T 0 dT 0
T1
where again h is defined on a per unit mass basis. The relationship between internal energy and
enthalpy is given by,
p
h = û +
ρ
Applying the ideal gas law p = ρRT ,
h = û + RT
For a differential enthalpy dh,
dh = dû + R dT
dh
dû
=⇒
=
+ R ⇐⇒ cp = cv + R
dT
dT
By convention, the specific heat ratio γ√= k = cp /cv . The value of k is important because it
lets use calculate the sound speed a = kRT (ideal gas). This contrasts the more formal (and
general) definition of a given by
s
a=
dp
dρ
This expression applies for gases, liquids, and solids.
Entropy changes can be important and are measured as follows.
T ds = dh −
dp
ρ
where s is the entropy per unit mass. The above relation unifies the first and second laws of
thermodynamics. Again applying the ideal gas law,
T ds = dh −
=⇒ ds = cp
dp
P
RT
dT
dp
−R
T
p
Integrating yields,
T2
p2
s2 − s1 = cp ln
− R ln
T1
p1
44
MEC E 430
Lecture Notes
Fall 2023
If ∆s = 0 such that the flow is isentropic,
T2
p2
0 = cp ln
− R ln
T1
p1
Recall that R = cp − cv and cp /cv = k,
R = cp
1
1−
k
= cp
k−1
k
Substituting this result,
Rk
T2
p2
ln
= R ln
k−1
T1
p1
k/(k−1)
T2
p2
=
T1
p1
Is there a possibility that k/(k − 1) → ∞ ⇐⇒ k → 1? Recall that k = cp /cv . If k = 1, then
cp = cv but recall that cp = cv + R which is a contradiction. Therefore such scenario is impossible.
Applying ideal gas law again,
k
ρ2
p2
=
ρ1
p1
These two main results are important relations for isentropic ideal gas. With these results, the
goal is to find a relationship between velocity and density.
Consider that temperature T and pressure p must obey the isentropic ideal gas. relationship.
T
=
To
P
Po
(k−1)/k
where To and Po are arbitrary references.
ln T − ln To =
k−1
(ln p − ln p0 )
k
ln T − ln To =
R
(ln p − ln p0 )
cp
Differentiate vertically with respect to z,
1 dT
R 1 dP
=
T dz
cp p |{z}
dz
−ρg
Applying the ideal gas law again,
1 dT
R 1
=
· ρg
T dz
cp ρRT
45
MEC E 430
Lecture Notes
Fall 2023
−g
cp
|{z}
1 dT
=
T dz
adiabatic temperature gradient
g/cp is known as the “lapse rate” which specifies the largest rate at which temperature can decrease
without causing a convective overturn in the atmosphere. For standard atmospheric conditions,
◦C
g/cp ≈ 10 km
. When
dT
=0
dz
the atmosphere is called highly stable.
Example. Consider a bullet in air (from the reference frame of the bullet) traveling such that
Ma = 3. What is the stagnation temperature at the stagnation point?
To
Ma2 (k − 1)
=1+
T
2
where To represents the stagnation temperature. Solving to To noting kair = 1.4 yields
To = 2.8T = 840 K
In Module 21, we will consider isentropic flow in a variable area duct with a noteworthy result i.e.
linking A, A∗ , Ma particularly a parabolic curve of A/A∗ vs Ma.
46
MEC E 430
Lecture Notes
Fall 2023
Lecture 18
Last lecture, we emphasized the importance of this figure from Module 21 (Fig. 18.1).
Figure 18.1: Schematic
From the above diagram, we find a variety of behavior that is associated with isentropic flow in
converging or diverging channels.
• Subsonic nozzle (accelerates flow)
• Subsonic diffuser (decelerate flow)
• Supersonic nozzle (accelerates flow)
Figure 18.2: Schematic
Why does this happen? By mass balance, we have
ρvA = constant = C1 =⇒ A =
47
C1
ρv
MEC E 430
Lecture Notes
Fall 2023
This means that if A increases, then ρv must go down and vice-versa.
In incompressible flow,
rhov versus x curve is “scaled” by constant ρ. That is, v ∝
1
.
A
Figure 18.3: Schematic
However in compressible flows, ρ is not constant but rather changes as a function of Ma.
"
#
ρ
1
=
ρ0
1 + 12 (k − 1)Ma2
Figure 18.4: Schematic
In diverging ducts, ρ decreases so rapidly that v must increase so that mass balance is preserved.
That is, increasing v “counteracts” decreasing ρ to maintain the ρv versus x curve.
Figure 18.5: Schematic
Alternative explanation: Consider mass balance,
ρvA = ṁ = constant
Apply the logarithm to both sides,
ln ρ + ln v + ln A = ln ṁ
48
MEC E 430
Lecture Notes
Differentiate,
Fall 2023
dρ dv dA
+
+
=0
ρ
v
A
However, by differential momentum conservation,
dp
+ v dv = 0,
ρ
(Bernoulli equation without elevation term)
=⇒ dp = −ρv dv
Also by definition,
a2 =
dp
dp
=⇒ dρ = 2
dρ
a
Substituting,
dρ = −
ρv dv
a2
Substituting again,
1
ρv dv
dv dA
− 2
+
+
=0
ρ
a
v
A
!
dv
v2
dA
=⇒
1− 2 =−
v
a
A
dA
dv
dA
dv
1 − Ma2 = −
⇐⇒
Ma2 − 1 =
v
A
v
A
dv
1
dA
=
2
v
Ma − 1 A
The second last expression shows that dA = 0 when Ma = 1. Flow can be sonic only when the
duct is neither increasing or decreasing in area with x. On the other hand, the last expression
shows that dv/v exhibits a qualitative change in behavior with dA/A where flow is subsonic (Ma
< 1) versus a case of supersonic flow (Ma > 1).
If we want to accelerate a flow from a subsonic state to a supersonic state, then we must apply a
converging-diverging duct or nozzle/channel (“Laval nozzle”).
49
MEC E 430
Lecture Notes
Fall 2023
Lecture 19
Recap from the previous lecture focusing on a converging-diverging (Laval) nozzle as shown in
Fig. 19.1.
Figure 19.1: Schematic
Example. Air flows isentropically at 0.5 kgs through a choked converging-diverging nozzle with
subsonic flow upstream and supersonic flow downstream of the throat as shown in Fig. 19.2.
At the inlet, the pressure is 680 kPa, the temperature is 295 kPa and the area is 6.5 cm2 . If the
exit area is 13 cm2 , calculate, (i) the stagnation temperature and pressure, and, (ii) the exit
Mach number (assume the flow remains supersonic).
Figure 19.2: Schematic
(i) We know T1 and P1 so if we knew the Ma number, we could obtain T0 and P0 from the
isentropic flow relations. That is,
"
#k/(k−1)
P
1
=
,
P0
1 + 12 (k − 1)Ma2
T
1
=
1
T0
1 + 2 (k − 1)Ma2
To evaluate Ma, we realize that ṁ = ρ1 v1 A1 . We can rearrange (since other quantities are
known),
ṁ
v1 =
ρ1 A1
50
MEC E 430
Lecture Notes
Fall 2023
We can find ρ1 by applying the ideal gas law, that is
ρ1 =
and combining both relations,
v1 =
P1
RT1
ṁRT1
P1 A1
It follows that Ma1 is
v1
ṁRT1
1
ṁ
Ma1 =
=
·√
=
a1
P1 A1
P1 A 1
kRT1
r
RT1
= 0.278
k
At this point, we have two paths, (a) return to the isentropic flow relations, rearrange, and
solve for T0 and P0 , and (b) use the isentropic flow tables (IFT) which are more convenient.
Reading the table, we have
Ma1 = 0.278 ≈ 0.28,
P
= 0.9470,
P0
T
= 0.9846
T0
Solving,
P1
T1
= 718 kPa, T0 =
= 300 K
0.9470
0.9846
(ii) We look to solve for Ma2 . We are tempted to use,
P0 =
"
A
1 1
=
∗
A
Ma
+ 12 (k − 1)Ma2
1 + 12 (k − 1)
#(k+1)/2(k+1)
however the analytical solution is complicated (tedious). So what other approach can we take?
We will try to use the ITF tables. If we can find A2 /A∗ , we can find Ma2 . Let’s try to determine
A∗ . Since the flow is choked,
√
P0 A∗
ṁ RT0
∗
ṁ = ṁmax = 0.6847 √
=⇒ A =
= 2.98 cm2
0.6847P0
RT0
Thus,
Reading the table,
A2
= 4.36
A∗
0.12 ≤ Ma ≤ 0.14
This result is unexpected since the flow is supersonic. Thus we look further down the table
(i.e. A/A∗ rebounds). Reading the table and interpolating,
Ma2 ∈ [3.02, 3.04] =⇒ Ma2 = 3.03
51
MEC E 430
Lecture Notes
Fall 2023
Lecture 20
Example. Air exits from a large reservoir maintained at 20 ◦ C and 500 kPa absolute into a
receiver (Fig. 20.1) maintained at (i) 300 kPa absolute, or, (ii) 200 kPa absolute. Estimate the
mass flux if the exit area is uniform and equal to 10 cm2 .
Figure 20.1: Schematic
Since the reservoir is large, velocity is approximately zero. From the ITF tables, it follows for
Ma = 0 that p/p0 = T /T0 = 1. Hence p0 = 500 kPa and T0 = 300 kPa.
In the uniform connecting pipe, air is flowing at approximately the speed of sound, that is
Ma = 1. From the IFT tables,
ITF
Ma = 1 ==⇒
p
= 0.5283 ⇐⇒ p = 0.5283p0 = 264.2 kPa
p0
• If receiver pressure pr > p = 264.2 kPa, we do not choke the flow. That is, the flow
remains subsonic everywhere.
• If receiver pressure pr < p = 264.2 kPa, we choke the flow. That is, the flow transitions
from subsonic to sonic.
1. pr = 300 kPa > p = 264.2 kPa (unchoked flow)
pr
ITF
= 0.6 ==⇒ Ma = 0.8864
p0
where pr = 300 kPa. Note that Ma is measured at the pipe exit/inlet to receiver (and
consequently p = pr and T = Tr ). It follows that
ṁ = ρAv
p
=
A(Ma · a)
RT
√
p
=
A Ma · kRT
RT
r
k
= pAMa
RT
52
MEC E 430
Lecture Notes
We look to write ṁ in terms of Ma2 and k.
v
u
p
k
u
ṁ = p0 ·
AMat p0
R T ·
0
Fall 2023
T
T0
k+1
2(1−k)
k
1
2
= p0
Ma · A 1 + (k − 1)Ma
RT0
2
kg
= 1.167
s
r
An alternative approach is to use ITF tables.
ITF
Ma = 0.8864 ==⇒
Now at the pipe exit,
Tr
= 0.864 =⇒ Tr = 253 K
T0
p
m
kRTr = 282
s
v=
By ideal gas law,
ρ=
pr
kg
= 4.13 3
RTe
m
Then the mass flow rate is
ṁ = ρAv = 1.165
kg
s
2. pr = 250 kPa < p = 264.2 kPa (choked flow) As before, we find pressure and temperature
at the exit of the pipe i.e. inlet of the receiver.
ITF
Ma = 1 ==⇒
Tr
= 0.8333,
T0
pr
= 0.5283
p0
Then we find that
Tr = 244.2 K,
Now at the pipe exit,
v=
By ideal gas law,
pr = 264.15 kPa
p
m
kRTr = 313.24
s
ρ=
pr
kg
= 3.769 3
RTe
m
Then the mass flow rate is
ṁ = ρAv = 1.180
kg
s
...........................................................................................
The results of choked flow aligns with the maximum mass flow rate.
p0 A∗
kg
ṁmax = 0.6847 √
= 1.180
s
RT0
In (ii) there is an imbalance between exit pressure and receiver pressure on the other hand
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MEC E 430
Lecture Notes
Fall 2023
(Fig. 20.2). Fluid must be continually withdrawn from receiver. We might also expect some
kind of more interesting flow behavior inside of the receiver (expansion waves or oblique shocks
possibly).
Figure 20.2: Schematic
54
MEC E 430
Lecture Notes
Fall 2023
Lecture 21
Example. A pitot tube and a thermocouple (Fig. 21.1) give the following measurements for an
air flow: p0 = 180 kPa, p = 157 kPa and T0 = 1250 K. Estimate the velocity of the air stream
if the flow is supersonic and there is a normal shock in front of the instrument.
Figure 21.1: Schematic
We remark that entropy is only generated at the shock face itself. When we look specifically at
the supersonic or subsonic regions, they are entirely isentropic. However at the singularity, we
cannot apply isentropic relations. We begin by observing that the measurements are obtained
on the downstream side i.e. subsonic region. It follows that,
p2
= 0.8722
p02
The isentropic flow relation gives
k
"
# k−1
p2
1
=
=⇒ Ma2 = 0.446
p02
1 + 12 (k − 1)Ma2
Alternatively, we can use the ITF tables. Both results for Ma2 indicate subsonic flow as
expected. Furthermore, it is important to note that stagnation pressure changes across normal
shock face therefore we should distinguish between p01 and p02 . To connect both sides of the
shock, we use the normal shock relations.
Ma2 2 =
(k − 1)Ma1 2 + 2
2kMa1 2 − (k − 1)
Solving for Ma1 is tedious so we refer to normal shock tables.
NST
Ma2 = 0.448 ===⇒ Ma1 = 3.64 = √
It follows that
v1 = Ma1
v1
kRT1
p
kRT1
To find T1 , we recall that T0 (T01 = T02 = 1250 K) is constant across the shock. Then by
55
MEC E 430
Lecture Notes
isentropic relations,
Fall 2023
T1
1
=
=⇒ T1 = 343 K
1
T0
1 + 2 (k − 1)Ma1 2
The same result could be obtained by using ITF tables. Finally, the velocity upstream is
calculated as
m
v1 = 1354
s
Example. A shock wave from an explosion propagates through still air with p = 100 kPa and
T = 300 K. If the absolute pressure just inside the shock is 700 kPa, estimate the shock speed
and √
the flow velocity just inside the shock. Where necessary, estimate the speed of sound from
a = kRT .
In our last example, we had a stationary shock. This example is rather different... that is, we
have a moving shock. We can transform the propagating shock into a stationary shock by a
simple change of reference frame.
The static pressure ratio is as follows,
p2
NST
= 7 ===⇒ Ma1 = 2.48,
p1
Ma2 = 0.5149,
T2
= 2.118
T1
Solving for T2 gives
T2 = 2.118T1 = 635.4 K
With T1 , we can find v1 . Then,
v1 = Ma1 · a1 = Ma1 ·
p
m
kRT1 = 861
s
v1 represents the shock speed in the stationary frame of reference. Following a similar approach,
p
m
v2 = a2 Ma2 = Ma2 · kRT2 = 260
s
Then the flow velocity inside of the shock is v1 − v2 or 601 ms .
56
MEC E 430
Lecture Notes
Fall 2023
Lecture 22
Example. A supply tank at 500 kPa and 400 K feeds air to a converging-diverging nozzle whose
throat area is 9 cm2 . The exit area is 46 cm2 . With specific reference to the appearance and
location of any normal shocks, specify the flow conditions in the nozzle if the pressure outside
the exit plane is (i) 400 kPa, (ii) 119.5 kPa, and, (iii) 9 kPa. Find the mass flux for each of (i),
(ii) and (iii).
Begin by drawing a schematic as shown in Fig. 22.1.
Figure 22.1: Schematic
Let’s begin with analysis of design conditions.
Aexit
ITF
= 5.1 ==⇒ Maexit = 3.2
A∗
Note that there is another value of Ma based on Aexit /A∗ but we note that for the diverging
portion, we must have Ma > 1 (supersonic branch of solutions). Additionally from the ITF,
p
= 0.0202 =⇒ p = 0.0202p0 = 10.1 kPa
p0
In case (iii), the ambient pressure i.e. receiver pressure is 9 kPa which is less than 10.1 kPa.
Therefore, the direction of the inequality precludes the appearance of pressure increasing
shocks. Rather, we expect within the receiver to see expansion waves. This will cause a further
drop of pressure i.e. from 10.1 kPa to 9 kPa.
Now suppose that there is a normal shock sitting at the nozzle exit as shown in Fig. 22.2.
57
MEC E 430
Lecture Notes
Fall 2023
Figure 22.2: Schematic
Conditions at 1 are exactly what we determined for design conditions i.e. Ma1 = 3.2 and
p1 = 10.1 kPa. Across the shock,
p2
1 =
2kMa1 2 − (k − 1) =⇒ p2 = 119.5 kPa
p1
k+1
This pressure matches exactly exit condition (ii) of receiver pressure.
categorized by a shock at the exit plane.
Therefore (ii) is
In the case (i) i.e. 400 kPa, let’s consider the scenario where the flow is sonic at the throat but
then reverts to a subsonic state on the downstream side. We have,
Aexit
p
ITF
= 5.1 ==⇒ Maexit = 0.114 =⇒
= 0.990 =⇒ p = 495 kPa
∗
A
p0
Note here we use the branch of solutions for Mach number in the subsonic branch since we
consider the scenario of flow reverting to a subsonic state on the downstream side. Therefore
we find that,
pamb = 400 kPa < p = 495 kPa
This suggests that flow scenario (i) has a pressure smaller than a subsonic→sonic→subsonic
flow but larger than a case where a shock appears at the exit. It follows that a normal shock
must appear between the throat and exit plane as shown in Fig. 22.3.
Figure 22.3: Schematic
This above physics is best illustrated in the plot of pressure as a function of position (Fig. 22.4).
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MEC E 430
Lecture Notes
Fall 2023
Figure 22.4: Schematic
Because cases (i) - (iii) all entail sonic flow at the throat, the mass flow rate must be the same
in each case i.e.
p0 A∗
kg
ṁ = ṁmax = 0.6847 √
= 0.909
s
RT0
evaluated at p0 = 500 kPa, T0 = 400 K, and A∗ = 9 cm2 .
59