3 Fluid kinematics

Transcription

3 Fluid kinematics

1
3 Fluid kinematics
3.1 A circular cylinder is moving in motionless environment without friction at constant velocity. Draw the the streamlines and the pathlines for some special fluid particles.
3.2 Determine for the velocity field
u = u0 cos(ω t) v = −v0 sin(ω t)
with u0 / ω = v0 / ω = 1 m
a) the streamlines for ω t = 0 , π / 2 , π / 4 ,
b) the pathlines,
c) the pathline for the particle, that is at t = 0 s in x = 0 m, y = 1 m!
2
4 Basic equations for fluids
vektors, tensors, operators
1.) Scalars (Tensors of rank 0)
e.g. pressure p, density ρ, temperature T
−→ real number (+ physical unit)
2.) Vectors (Tensors of rank 
1) 
x
 
~

~
~
e.g. Position vector ~r = 
 y  = x · i + y · j + z · k,
z
 
u
 
~

~
~
Velocity vector ~v =  v 
= u·i+v·j+w·k
w
−→ n dimensions
3.) Dyads (Tensors of rank 2)
e.g. Stress tensor τ , (~v ~v )
−→ n dimensions (n × n Matrix)
~ =
Nabla - Operator ∇
∂
∂
∂
,
,
∂x ∂y ∂z
~ p =
Gradient grad p = ∇
~ · ~v =
Divergence div ~v = ∇
~ × ~v =
Rotation ∇
~i
∂
∂x
u
~j
∂
∂y
v
!
∂p ∂p ∂p
,
,
∂x ∂y ∂z
!
∂
∂
∂
,
,
∂x ∂y ∂z
~k
∂
∂z
w


u
 
∂u
∂v
∂w

·
 v  = ∂x + ∂y + ∂z
w
∂w
∂v 
−
∂z 
 ∂y

=
!










∂w 
∂u

−

∂z
∂x 

∂u 
∂v
−
∂x
∂y
3
Partial derivatives
Total differential of a function f = f (x, y, z)
df =
∂f
∂f
∂f
· dx +
· dy +
· dz
∂x
∂y
∂z
The total differential describes the increas of a function, e. g. for the velocity ~v = ~v (t, x, y, z)
d~v =
=⇒
=⇒
∂~v
∂~v
∂~v
∂~v
· dt +
· dx +
· dy +
· dz
∂t
∂x
∂y
∂z
∂~v
∂~v dx
∂~v dy
∂~v dz
d~v
=
+
·
+
·
+
·
dt
∂t
∂x dt
∂y dt
∂z dt
d~v
∂~v
∂~v
∂~v
∂~v
∂~v
+ v·
+ w·
+ (~v · ∇) · ~v
=
=
+ u·
dt
∂x
∂y
∂z
∂t
|{z}
| ∂t
{z }
|
{z
}
substantial
local
convective acceleration
4
4.1
a) Proof the following identities
~ × ∇p
~ = 0
1)
∇
~ ×∇
~ 2~v = ∇
~ 2 (∇
~ × ~v )
2)
∇
2
~ v = ∇
~ ~v − ~v × (∇
~ × ~v )
3)
(~v · ∇)~
2
b) Formulate the conservation law of mass and momentum for a three-dimensional, incompressible and unsteady flow in vector notation.
4.2
a) A piston is moving in a tube of infinite length and with constant cross section A with the
velocity vpiston (t). The density of the fluid is constant.
A
vKolben( t )
x
Kolben
Determine the substantial acceleration in the tube.
b) A fluid of constant density flows into a diffusor with the constant velocity v = v0 . The cross
section of the diffusor is A(x). Determine the substantial acceleration of the fluid along the
axis x.
5
4.3 The following continuity equation is formulated in cartesian coordinates.
dρ
∂v
∂w
∂u
+ ρ·
+
+
dt
∂x
∂y
∂z



!
= 0 (Eq. 1)

u
vx
 
 



with v =  v  =  vy 

w
vz
Transform the equation into
a) cylindrical coordinates
Hint: x = r · cos φ
y = r · sin φ
z = z
√
r = x2 + y 2 y
φ = arctan
x
z = z
b) sperical coordinates
Hint: x = r · sin Θ · cos φ
y = r · sin Θ · sin φ
z = r · cos Θ
√
x2 + y 2√ + z 2
!
x2 + y 2
Θ = arctan
z
y
φ = arctan
x
r = +
4.4 An incompressible fluid with the viscosity η is flowing laminar and steady between two
parallel plates
The flow is radial from inside to outside. The determining differential equations in cylindrical
coordinates are
1 ∂(ρ vΘ )
∂(ρ vz )
1 ∂(ρ r vr )
+
+
= 0
r
∂r
r
∂Θ!
∂z
!
2
∂vr
∂p
vΘ ∂vr
vΘ
∂vr
∂ 1 ∂(r vr )
ρ vr
= −
+
−
+ vz
+ η
∂r
r ∂Θ
r
∂z
∂r
∂r r
∂r
2 ∂vΘ
∂ 2 vr
1 ∂ 2 vr
−
+
+ 2
r ∂Θ2
r 2 ∂Θ
∂z 2
Simplify the equations for the flow problem described.
!
6
4.5 The Navier-Stokes equations for rotationally symmetric, unsteady, incompressible flows in
cylindrical coordinates read:
1 ∂(r · vr )
∂vz
+
= 0
r
∂r
∂z
∂vr
∂vr
∂vr
+ vr
+ vz
∂t
∂r
∂z
ρ
ρ
∂vz
∂vz
∂vz
+ vr
+ vz
∂t
∂r
∂z
∂T
dp
1 ∂
r
+λ
dt
r ∂r
∂r
!
∂2 T
+
∂z 2
"
∂p
∂
= −
+ η
∂r
∂r
!
#
1 ∂
(r vr )
r ∂r
"
∂p
∂vz
1 ∂
= −
r
+ η
∂z
r ∂r
∂r
∂T
∂T
∂T
+ vr
+ vz
∂t
∂r
∂z
ρ cp
"
!


∂ vr
+ 2η
 ∂r
!2
+
vr
r
!
2
!
!
∂ 2 vr
+
∂z 2
∂ 2 vz
+
∂z 2
#
#
=
+
∂ vz
∂z
!2 

∂ vr
∂ vz
+η
+

∂r
∂z
Consider a steady, laminar, fully developed, incompressible duct flow with temporal and spatial
variable temperature distribution.
Simplify the equations for the case described.
4.6 The Navier-Stokes equations for unsteady, incompressible flows in graviational field read:
∇ · ~v = 0
ρ
d~v
= − ∇p + η∇2~v + ρ~g
dt
Formulate the equations for a steady, frictionless, twodimensional flow in a cartesian coordinate
system (x, y).
!2
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5 Hydrostatics
5.1 A cube swims in two fluids that are arranged in layers.
ρ1 = 850 kg/m3
ρ2 = 1000 kg/m3
ρK = 900 kg/m3
a = 0, 1 m
Determine the height h!
5.2 A container is filled with a fluid of the density ρ. The drain of the container, filled up to a
height h, is closed with a hollow hemisphere ( radius R, weight G ).
Given: h, ρ, R, G, g
Determine the necessary force F to open the drain.
Hint: Volume of a sphere: Vk =
4
π R3
3
5.3 A buoy with the mass mB , which is open on the lower side, is fixed with a cable at the
bottom of a sea, and sticks out with a third of its height if the cable is not strained (Sketch
1). With increasing water level the buoy is drawn under the water surface (Sketch 2) and sinks
when the immersion depth is H.
Determine H!
Given: mB , pa , ρW , h, g, ρL << ρW
8
rL , pa
g
rL , pa
2h
3
H
rW
h
rW
1
2
Hint: Assume, that the temperature inside of the buoy is constant and the enclosed air can be
treated as a perfect gas. Neglect the weight of the cable.
5.4 A ship with vertical side walls has a weight of G0 Its draught is h0 and it displaces the
volume τ0 in sea water. At the entrance of an estuary the weight is decreased by ∆G in order
to avoid the ship runs aground. Now the draught is h1 and the displaced volume is τ1 . The
density of sea water is ρM , the density of the rivers water is ρF .
G0 = 1, 1 · 109 N ∆G = 108 N h0 = 11 m h1 = 10, 5 m
ρM = 1, 025 · 103 kg/m3 ρF = 103 kg/m3 g = 10 m/s2
Determine
a) the volume τ0 ,
b) the deck area A,
c) the difference τ2 − τ1 between the displaced volumes in fresh water and sea water,
d) the draught h2 in fresh water!
9
5.5 The sketched weir of length L separates two basins of tdifferent depth.
Determine the force of water onto the weir.
Given: ρ, g, L, a
5.6 A boiler with a small hole at the top is filled with water and is screwed on a plate.
R=1m
ρW = 103 kg/m3
g = 10 m/s2
Determine the force on the screws by neglecting the boilers weight.
5.7 A vehicle filled with water is moving under constant acceleration ax in x-direction.
1. Determine the pressure gradient ∂p/∂x by using the balance of forces at the plotted volume
element of length dx and of an area dA.
2. Determine the vehicle length l.
Given: h, ρW , g, ax
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5.8 A liquid rotates in an open cylindrical jar with a constant angular speed which is such
that the liquid just reaches the upper border. Under quiescent condition the liquid fills the jar
up to a height of h0 .
D = 0, 5 m
h0 = 0, 7 m
H=1m
ρ = 103 kg/m3
pa = 105 N/m2
g = 10 m/s2
Determine
a) the height h and the angular speed ω,
b) the pressure distribution along the side wall and the bottom!
Hint:
∂p
= ρω 2 r
∂r
∂p
= −ρg
∂z
dp =
∂p
∂p
dr + dz
∂r
∂z
5.9 A spherical, open, and rigid gas balloon is designed for a ceiling of H = 10km in isothermal
atmosphere (temperature T0 ).
H = 104 m,
RL = 287
Nm
,
kg K
g = 10
m
,
s2
T0 = 287 K
11
What is the ceiling h, if a hole is in the envelope (see sketch)?
5.10 A wheather balloon with mass m and initial volume V0 ascends in an isothermal atmosphere. Its envelope is loose up to the achievement of the maximal volume V1 .
p0 = 105 N/m2 ρ0 = 1, 27 kg/m3
R = 287 Nm/kgK g = 10 m/s2
m = 2, 5 kg
V0 = 2, 8 m3
V1 = 10 m3
a) What is the necessary force to hold down the balloon before launch?
b) In what altitude the balloon reaches its amximum volume V1 ?
c) What ceiling reaches the balloon?
12
6 Continuity and Bernoulli’s equation
6.1 A large lake is connected via overflow to a reservoir and via pipe system with a lower lake
(see sketch).
Determine the total volume flux V̇ that flows into the lower lake when the overflow is opened.
Hint: Neglect the altitude difference in the horizontal pipes and neglect friction!
Given: ρ, g, h, A
pa
g
Reservoir
h
A/2
A
6.2 In order to determine the velocity in a duct flow the pressure difference ∆p is measured.
Through strong obstruction the pressure difference deviates from the dynamic pressure of the
undisturbed incoming flow.
Outline the distribution of v∞ /
s
2∆p
d
in terms of
for frictionless flow.
ρ
D
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6.3 An incompressible fluid of density ρ is flowing stationarily from a large tank through a
well rounded exit of radius R into the surrounding.
hB
g
r
R
H
r(z)
z
Given: H, hB , R, ρ, g
Determine the radius r(z) of the jet as a function of the altitude z.
6.4 Water flows from a large pressurized tank into the open air. The pressure difference ∆p is
measured between the cross sections A1 and A2 .
A1 = 0, 3 m2 ,
A2 = 0, 1 m2 ,
A3 = 0, 2 m2 ,
h = 1 m,
3
3
5
2
ρ = 10 kg/m ,
pa = 10 N/m ,
∆p = 0, 64 · 105 N/m2 ,
g = 10 m/s2
14
Compute the
a) velocities v1 , v2 , v3 ,
b) pressures p1 , p2 , p3 and the pressure p above the water surface!
6.5 Two large basins located one upon the other are connected with a duct.
A = 1 m2 ,
Ad = 0, 1 m2 ,
h = 5 m,
H = 80 m,
5
2
3
3
pa = 10 N/m ,
ρ = 10 kg/m ,
g = 10 m/s2
a) Determine the volume rate!
b) Outline the distribution of static pressure in the duct!
c) At what exit cross section bubbles are produced, when the vapour pressure is
pD = 0, 025 · 105 N/m2 ?
6.6 The flap at the exit of the water pipe (constant width B) of a large container is opened
abruptly. The appearing flow is without any losses.
Given: H, h1 , h2 , g, L ;
L >> h1
Determine
a) the differential equation for the exit velocity v3
b) - the local acceleration
- the convective acceleration
- the substantial acceleration
at x =
L
when the exit velocity reaches half of its asymptotic final value!
2
Hint: The computation of v(t) is not necessary for solving this problem.
15
6.7 The exit of a large container is build as diffusor. The flap at the end of the diffusor is
opened at t = 0 abruptly.
Given: L, pa , p2 ≈ pa , A2 , h
Determine
dv2
a) the acceleration
in point ”2” immediately after the opening,
dt
b) the position x in the diffusor, where the the pressure has its maximum, when v2 reaches
half of its asymptotic final value.
16
6.8 Water flows from a large reservoir into a lower lying basin whose drain orifice is decreased
to one third of its origin size abruptly.
Determine the time interval in that the water surface quadrupels from its origin altitude h.
2
A = 0, 03 m ,
Hint :
Z
2
AB = 1 m ,
dh
dt
2
h = 5 m,
g = 10 m/s ,
!2
<< (vA )2
h
√
√ i
dx
√ = 2 (a − x) − a ln(a − x) + C
a− x
6.9 A spring well fountain is feeded from a large container. When the pump is switched on a
pressure difference of ∆p is created. After a time of 7. 6 seconds the velocity at the end of the
duct reaches 99,9 % of its asymptotic value (v(t→∞) ).
2
g
H
pa
0
h
1
L
kW
)p
T99,9 = 7.6 s, L = 10 m, g = 10 m/s2
At ”1” the fountain is a free jet.
a) Determine the altitude h of the fountain for t → ∞.
b) At what time t1/2 it reaches 50 % of this altitude?
17
Hint: Neglect the friction losses!.
Z
a2
1
1
a+x
dx =
ln
2
−x
2a
a−x
6.10 A piston pump surges water with a vapour pressure pD from a reservoir. The piston has
the frequency ω and the stroke ξ0 . The pump has a constant cross-section A and the pistons
position is in the altitude h above the water surface. The length of the inlet is L.
Given: pa , L, ξ0 , h, A, ρ, pD , g
a) Determine the nondimensional pressure pK / (ρL2 ω 2 ) = f (ω) at the piston head in terms
of the piston frequency.
b) Determine the critical pump frequency for ξ0 << L when the vapour pressure pD is just
reached at the piston head.
c) Determine the average displaced volume per time V̇ .
6.11 Water flows under the influence of gravity from a large reservoir through a convoluted
duct into open air. At t = 0 the convoluted duct is accelerated by an engine up to the angular
speed ωr .
a) At what time ∆T the flow reaches 50 % of the velocity which is at t → ∞ in the vertical
part of the duct?
b) Determine the pressure at point 1 for t → ∞ in the duct that rotates with ω = ωr .
Given: h, l, R, A, ωr , ρW , pa , ~g
18
rW
h
g
l
A
R
A
1
Hint:
Z
Motor
1
a+x
dx
=
ln
+ C
a2 − x2
2a
a−x
;
∂p
= ρ ω2 r
∂r
19
7 Momentum and momentum of momentum equation
7.1 A pitot tube in a gas pipeline is connected via two u-tubes with a pressure hole (see sketch).
Determine the force F on the fitting of the pitot tube. Assume an incompresible fluid! Neglect
the friction in the pipeline!
Given: ρHq ≫ρGas , g, ∆h1 , ∆h2 , D
7.2 Determine the pressure difference ∆p = p2 − p1 in the plotted bifurcation by neglecting
the friction.
1
Given: v1 , v2 , A3 = A, α, ρ = konst.
4
20
7.3 Water is flowing through a bifurcation into open air stationarily. The pressure in the
incoming tube is ∆p higher than in the surrounding.
A1 = 0, 2 m2
A2 = 0, 03 m2
∆p = 104 N/m2 ρ = 103 kg/m3
A3 = 0, 07 m2
α2 = 30
α3 = 20
Determine
a) the velocities v1 , v2 , v3 ,
b) the force Fs in section 1,
c) the angle α3 , when Fsy vanishes.
7.4 A flat plate of constant thickness with the mass m and the length l is hung at a hinge and
is passed by a planar water jet of height h and width B with the velocity v. The flow in the
jet is without any losses.
a) Determine the force F1 , at the lower end of the plate, that is necessary to fix the plate
in a vertical position.
b) Determine the necessary forceF2 , when a deflector blade is mounted on the plate.
c) Compute the rotation angle θ for the steady state if the plate is swinging undisturbedly..
Given: m, a, l, h, B, v, ρ, g, pa
Hint: The volume forces are neglectable.
21
7.5 A rocket is moving at constant velocity. The passing air is displaced radially. Inside the
jet the velocity is vA , outside v1 .
Given: v1 , vA , ρ1 , ρA , AR
Determine
a) the displaced air mass,
b) the thrust and the engine power.
7.6 A propeller is passed with constant velocity v1 . At a certain distance downstream of the
propeller the velocity inside the jet is v2 , and v1 outside the jet.
A′ = 7, 06 m2
v1 = 5 m/s
v2 = 8 m/s
Determine
a) the velocity v ′ in the propeller plane,
b) the efficiency.
ρ = 103 kg/m3
22
7.7 The shell of a propeller is passed with constant velocity. The inlet is well rounded.
A = 1 m2
v1 = 10 m/s
p1 = 1, 345 · 105 N/m2
p1′ = 105 N/m2
a) Outline the distribution of the static pressure along the axis
and determine
b) the mass flux,
c) the thrust,
d) the power that the propeller emits to the flow.
7.8 Two fans sucking air from the surrounding differ in their inlets
Given: ρ, A, ∆p
ρ = 103 kg/m3
23
Compute
a) the volume flux,
b) the power of the fans,
c) the force on the fitting.
7.9 An air cushion vehicle of weight G, width B and length L is hovering above the ground in
steady state.
Compute by neglecting the friction and the compressibility
a) the pressure p1 and the volume flux V̇ that is flowing through the hover craft,
b) the power that is emitted from the fans to the flow and the power losses throught the
accumulation of vortices.
Given: ∆p, ρ, G, A,
BL = 20 A,
h(B + L) = A
Hint: Neglect the differences between the geodetic altitudes.
24
7.10 A ball of weight G is passed frictionless by a fluid jet (e. g. air jet) and is hovering from
the forces of the jet. Tne jet is flowing with the velocity v1 under the influence of the angle α1
(see sketch).
Determine
a) the downstream velocity v2
b) the angle α2 of the downstream jet,
c) the mass flux ṁ, to hang the ball in balance.
Given: v1 , G, α1 .
Hinweis: Neglect the force of gravity of the jet!
7.11 A sprinkler with three arms is supplied by a large tank and rotates with the angular
velocity ω = const.. The angle between the outflowing jets and the circumferential direction is
α.
H = 10 m
R = 0, 5 m
h=1m
A = 0, 5·10−4 m2
A1 = 1, 5.10−4 m2
pa = 105 N/m2
ρ = 103 kg/m3
g = 10 m/s2 ,
ω = 15 s−1
Determine
a) the relative exit velocity,
b) the torque and the volume flux,
c) the pressure p1 ,
d) the maximum torque.
α = 30
25
7.12 A atnk with the weight G is fixed in a rotatable bearing in D. Its drain-pipe has a 90◦ bend. The center of gravity of the system has the distance h to point D. What is the angle α
between the pipe-axis and the vertical axis, if the water flows without friction?
Given: G, l, h, A, ρ
Hint: The tank is such large that the water surface is not moving.
26
9 laminar viscous flows
9.1 An oil film of constant thickness and width is flowing on an inclined plane.
m
δ = 3 · 10−3 m B = 1 m α = 30o ρ = 800 kg/m3 η = 30 · 10−3 Ns/m2 g = 10 2
s
Calculate the volume flux.
9.2 A viscous oil film is flowing on an inclined plane (angle α) under the influence of gravity.The wall temperature of the plane is T0 and the temperature at the surface of the oil is Tδ .
The temperature distribution of the oil is linearly in terms of y and constant in x-direction.
The thickness of the film is δ and is constant. The kinematic vviscosity ν is constant and the
density in the important temperature range is Temperaturbereich mit
− TT
ρ = ρ0 · e
0
The density is independent of the pressure
Given: g, δ, α, ν, ρ0 = ρ(y = 0), T0 = T (y = 0), Tδ = T (y = δ)
27
a) Compute the shear stress τ (y).
b) Compute the velocity u(y).
∂p
c) Show that
= 0 by using the momentum equation in y-direction.
∂x
9.3 A car is in a wind tunnel with the velocity u∞ . To simulate the relative motion between
the vehicle and the roadway a bend-conveyor with the velocity u∞ is mounted unde the fixed
car (see sketch). Between the lower side of the vehicle and the bend-conveyor a gap flow is
formed. The flow shall be analyzed for
1. for the non-moving bend-conveyor (uB = 0)
2. for the moving bend-conveyor (uB = u∞ )
and
Assume, the flow is fully devloped in both cases. The following relation is valid:
d2 u
dp
=η 2
dx
dy
mit
dp
= konst < 0
dx
a) Compute and sketch the velocity profiles u(y) in the gap for the non-moving and the
moving bend-conveyor.
b) Compute the friction per unit width, that the flow affects on the lower side of the car over
the length l for the non-moving and the moving conveyor.
Given:
dp
, h, l, η, u∞
dx
9.4 A Newtonian fluid is flowing between two horizontal plates. The upper one is moving at a
velocity uw . The lower oneis standing still. The pressure is decreasing linearly in x-direction.
Given: H, uw , ρ, η, dp/dx
Determine for a fully developed laminar flow
a) the velocity distribution,
28
b) the relation between the shear stresses at y = 0 and y = H,
c) the volume flux for a width of B,
d) the maximum velocity for uw = 0,
e) the momentum flux for uw = 0,
f) the wall shear stress in non-dimensional form for uw = 0.
g) Outline the distribution of the velocity and the shear stress for uw > 0, uw = 0, and uw < 0.
9.5 An incompressible fluid is flowing through a pipe with radius R. The volume flux is V̇ .
The shear stresses are depicted with the model of Ostwald-de-Waele
τ = −ηOdW
Given: V̇ , L, R, ηOdW
du dr du
dr
The pressure decreases within a length of L with ∆p.
u
R
r
.
V
L
)p
Compute the pressure decrease ∆p for a fully developed pipe flow.
Hinteis: The equation of motion for a fully developed flow
r
dp d(τ r)
+
=0
dx
dr
29
9.6 The velocity distribution of a laminar pipe flow can be described in the inlet with the
following approximation
u
=
um
f
2
y
y
y
2 −
f( ) =
δ
δ

δ
 1
um
y
y
δ
2δ
1
1−
+
3R 6



Given: um , R, ρ, η
δ
R
!2
0 ≤ y ≤ δ(x)
δ(x) ≤ y ≤ R
R
*(x)
x
Le
Determine in the inlet cross section, at the end of the inlet section and for δ/R = 0, 5
a) the momentum flux,
b) the wall shear stress.
9.7 The velocity u1 in the inlet cross-section (1) of a circular pipe with radius R is constant.
In the cross-section (2) the velocity" of a fully #developed laminar pipe flow can be written as
2
r
.
parabola of the form u2 (r) = u2max 1 −
R
Given: ρ, R, u1
30
Determine
a) the pressure loss (p1 − p2 ),
ρ
b) the coefficient of the pressure loss ζE = (p1 − p2 )/ ū2
2
Hint: Neglect the wall friction.
9.8 A Bingham fluid is flowing between two infinite parallel plates under the influence of
gravity.
Given: b, ρ, η, τ0 , g, dp/dz = 0
Assume a fully developed flow and determine
a) the distance a,
b) the velocity distribution.
9.9 A Bingham fluid is in a basin. A vertical bend-conveyor, moving with the velocity uB is
used to transport the fluid in another basin.
Given: uB , δ, g, ̺, η, τ0 = ̺g
δ
2





du
du
für
<0
dy
dy
τ =
du
du


für
>0
 −τ0 − η
dy
dy
+τ0 − η
31
a) Determine and sketch the distribution of the shear stress in section I.
b) Determine and sketch the distribution of the velocity in section I.
c) What is the minmum velocity uB,min for conveying mass?
Hint: The flow at the conveyor is fully developed and laminar. The thickness δ of the fluid is
assumed to be constant.
9.10 A Newtonian fluid is moving between two coaxial cylinders.
Given: R, a, η, dp/dx
Bestimmen Sie für eine ausgebildete laminare Strömung
a) the velocity distribution. Outline the result,
b) the relation of the shear stresses for r = a and r = R,
c) the average velocity.
32
9.11 A Couette viscosimeter consists of two concentrical cylinders with the length L. The
interstice is filled with a Newtonian fluid. The oute cylinder rotates with the angular velocity ω,
and the inner one is standing still. At the inner cylinder the hinge momentum Mz is measured.
Ra = 0, 11 m
Ri = 0, 1 m
L = 0, 1 m
ω = 10 s−1
Mz = 7, 246 · 10−3 Nm
Determine
a) the velocity distribution,
b) the dynamic viscosity of the fluid.
Hint: The differential equations for the velocity- and the shear stress distribution are:
"
d 1 d
(rv)
dr r dr
#
= 0
d
τ = −ηr
dr
v
r
33
10 Turbulent pipe flows
10.1 Proof the following rules:
a) f = f
b) f + g = f + g
c) f · g = f · g
d)
e)
∂f
∂f
=
∂s
∂s
R
s
f · ds =
R
s
f · ds.
10.2 The velocity profile in a fully developed flow in a pipe with a smooth surface can be
approximated with the potential law:
v̄
v̄max
Re
n
1 · 105
6 · 105
1.2 · 106
2 · 106
7
8
9
10
=
1 −
r
R
1
n
, mit n = n(Re).
a) Use the continuity equation to compute the relation between the average velocity v̄m and
v̄m
the maximum velocity v̄max , i. e.
= f (n).
v̄max
r
b) At what position
is v̄(r/R) = v̄m ?
R
c) How can the results of a) and b) be used, if the volume flux shall be measured?
10.3 The velocity distribution of a turbulent pipe flow can be approximated with the following
34
law: ū / ūmax = (y / R)1/7 .
Determine
a) the ratio of the velocities ūm / ūmax ,
b) the ratio of the momentum fluxes I˙ / ρ ū2m π R2 .
10.4 Water is flowing through a hydrauliccaly smooth pipe.
D = 0, 1 m
Re = 105
ρ = 103 kg/m3
η = 10−3 Ns/m2
Determine
a) the wall shear stress,
b) the ratio of the velocities ūm / ūmax ,
y u∗
y u∗
c) the velocity for
= 5 and for
= 50,
ν
ν
y u∗
d) the mixing length for
= 100.
ν
10.5 Water is pumped through a pipe (roughness ks ) from h1 to h2 .
V̇ = 0, 63 m3 /s
ρ = 103 kg/m3
h1 = 10 m h2 = 20 m L = 20 km D = 1 m
ν = 10−6 m2 /s pa = 105 N/m2 g = 10 m/s2
ks = 2 mm
35
a) Sketch the distribution of the static pressure along the pipe axis.
Determine
b) the pressure at the pump inlet,
c) the pressure at the pump exit,
d) the net power of the pump.
10.6 The pressure decrease ∆p along L is measured in a fully developed pipe flow with the
volume flux V̇ .
V̇ = 0, 393 m3 /s L = 100 m
η = 5 · 10−3 Ns/m2
D = 0, 5 m
∆p = 12820 N/m2
Determine
a) the skin-friction coefficient,
b) the equivalent roughness of the pipe,
c) the wall shear stress and the force of the support.
d) What is the pressure decrease, if the pipe is smooth?
ρ = 900 kg/m3
36
10.7 Air is pumped with a fan through a rough pipe with a well rounded inlet.
L = 200 m
ks = 1 mm
Determine the ratio of the net power for D = 0, 1 m and D = 0, 2 m for a constant volume
flux. Assume a very large Reynolds number.
10.8 Water is pumped through a channel with a quadratic cross-section. At a certain position
the water is pumped through a bundle of 100 pipes with the length L.
V̇ = 0, 01 m3 /s
L = 0, 5 m
a = 0, 1 m
D = 0, 01 m
ρ = 103 kg/m3
η = 10−3 Ns/m2
Compute the length of the channel that produces the same pressure loss as the pipe bundle.
Hint: The pipes are hydraulically smooth.
37
11 Similarity theory
a) dimensional analysis
Frequently used quantities and their dimensions
Quantity
Dimension
*
*
*
*
meter, m
kilogram, kg
second, s
Kelvin K [1K] = [o C] + 273, 16
kg m
Newton N,
s2
m/s
m/s2
kg/m3
Pascal, P a = N/m2
kg/s
m3 /s
Joule, J = Nm
Watt, W = Nm/s
kg
Ns/m2 =
m·s
m2 /s
m2
J
= 2
kg K
s ·K
m2
J
= 2
kg K
s ·K
Length L
Mass M
Time t
Temperature T
Force F
Velocity u, v, w, ~u
Acceleration a, b,~a
Density ρ
Pressure, Stress p, τ, σ
Massflux ṁ
Volumeflux V̇
Momentum, work, energy M, W, E
Power P
dyn. viscosity µ, η
kin. viscosity ν
spec. heat capacity cp , cr
spec. gas constant R
* Basic dimensions
38
11.1 The wake of a long cylinder with the diameter D is anlyzed experimentally in a windtunnel. On certain circumastances a periodic vortex configuration is generated, the Kármán
vortex street. The dimensionless parameters of the problem shall be determined. How many
variations of parameters are necessary in this investigation to measure the frequency of the
vortex street?
11.2 A liquid is flowing steadily through a hydraulical smooth pipe. The flow is laminar and
fully developed.
a) Deduce from the Ansatz V̇ =
sional analysis!
∆p
L
α
η β D γ the Hagen-Poisseuille law by using the dimen-
b) Show, that the skin-friction coefficient in a pipe is inversely proportional to the Reynolds
number!
11.3 An upward directed flow develops along a vertical plate with the temperature TW . At a
large distance away from plate the temperature is T∞ . The temperature distribution can be
written with the equation
y
T − T∞
,
, g, ρ, η = 0
F
TW − T∞ x1/4
T − T∞
Deduce, using the dimensional analysis, an expression for the temperature ratio
in
TW − T∞
terms of the nondimensional coordinate
µ=f
y
x1/4
, g, ρ, η !
11.4 What is the ratio of drag for spheres with different diameter and with the same Reynolds
numbers, if one is flown against with air and the other with water and if the drag coefficient
is only a function of the Reynolds number?
ρL
= 0, 125 · 10−2
ρW
ηL
= 1, 875 · 10−2
ηW
39
11.5 The necessary power of a car with quadratic surface A to overcome the air drag shall be
analyzed experimentally in a windtunnel test. The surface of the model must not exceed Am
from technical reasons.
A = 4 m2
Am = 0, 6 m2
v = 30 m/s
a) Choose the wind velocity for the model tests?
′
b) Determine the power of the car, if the drag force FW
= 810N is measured at the largest
possible model!
11.6 An axial fan (diameter D, number of revolutions n) shall be designed for air. In a model
test with water (reduced scale 1:4) the increase of total pressure ∆p′0 is measured.
V̇ = 30 m3 /s
D=1m
n = 12, 5 s−1
ρ′ = 103 kg/m3
ρ = 1, 25 kg/m3
η ′ = 10−3 Ns/m2
η = 1, 875 · 10−5 Ns/m2
∆p′0 = 0, 3 · 105 N/m2
Determine
a) the volume flux and the number of revolutions in the model test,
b) the increase of the total pressure for the fan,
c) the power and the torque for the model and for the fan.
11.7 A pontoon swims at the bank of a river. A test with a model and the reduced scale 1:16
shall be carried out.
L = 3, 6 m
B = 1, 2 m
′
FW
=4N
H = 2, 7 m
h′ = 2, 5 cm
v = 3 m/s
ρ = 103 kg/m3
40
Determine
a) the flow velocity in the experiment,
′
,
b) the resulting force on the pontoon, if the model force is FW
c) the drag coefficient of the pontoon,
d) the height of the wave h upstream of the pontoon, if the wave height in the model is h′ !
11.8 The following Couette flow can be described with a partial differential equation
∂p
∂2u
=η 2
∂x
∂y
a) Determine the dimensionless parameters of this problem using the method of differential
equations.
b) How many parameters are achieved with the Π-theorem?
11.9 Deduce from the momentum equation in x-direction
∂u
∂u
∂u
ρ
+u
+v
∂t
∂x
∂y
!
∂p
∂2u ∂2u
=−
+η
+
∂x
∂x2 ∂y 2
!
the dimensionless parameters!
11.10 In a gas flow the heat transfer is determined from the viscous
and from heat
#
" effects
kgm
, the dynamic visconduction. The influencing quantities are the heat conductivity λ 3
sK
"
#
kg
cosity η
and the reference values for the temperature, the velocity, and the length. The
ms
physical relationship can be described with the energy equation
∂u
∂2T
λ 2 +η
∂y
∂y
!2
= 0.
41
Deduce the dimensionless parameters of the problem
a) with the method differential equations
b) with the Π-theorem.
c) Expand the resulting parameter with the specific heat capacity cp and formulate the new
coefficient as a product of three different parameters.
Hint: The material quantities are constant. The fourth basic dimension is the temperature.
11.11 The equilibrium of forces in a fully developed laminar pipe flow is described with the
differential equation
!
∂p
du
η d
−
r
= 0.
+ ρg +
∂x
r dr
dr
Determine the parameters of the problem
a) with the method of differential equations
b) with the Π-theorem.
Interpret the relationship between the solutions of a) and b)
11.12 The hydrodynamic attributes of a motor ship shall be analyzed with a model in a water
cahnnel.
a) Determine the dimensionless parameters of the problem with the method of differential
equations using the momentum equation in z-direction, that is decisive for the wave motion.
ρ
Use only the given quantities.
Given: l, u∞ , η, ρ, g
dw
∂p
= − − ρg + η∇2 w
dt
∂z
42
b) Compute the velocity u′∞ and the kinematic viscosity ν ′ of the model fluid such that the
flows are similar.
Given: u∞ , ν, l/l′ = 10
c) Compute the power of the motor ship at the velocity u∞ .
Given: l/l′ = H/H ′ = 10, u∞ , u′∞ , ρ, ρ′ , Drag force in the experiment F′
11.13 The energy equation for steady, compressible flows with constant material quantities is
∂
u2
ρu
cp T +
∂x
2
!
∂
u2
+ ρv
cp T +
∂y
2
!
∂2u
∂u
= uη 2 + η
∂y
∂y
!2
+λ
∂2T
∂y 2
Determine with the method of differential equations
a) the dimensionsless form of the differential equation,
b) the dimensionless parameters of the problem.
c) Determine the isentropic coefficient γ , if the equation is independent of the Mach-number
u∞
M∞ =
.
c∞
Hint: c =
√
γRT
cp =
γR
γ−1
43
11.14 The steady flow of a water jet colliding with a deflector blade shall be analyzed experimentally.
a) Determine with the method of differential equations from the x-momentum equation
∂2u ∂2u
+
∂x2 ∂y 2
∂u
∂u
1 ∂p η
u
+v
=−
+
∂x
∂y
ρ ∂x ρ
!
the dimensionless parameters of the problem. Use only the given quantities as reference quantities and define the reference velocity by using the pressure difference p0 − pa .
b) Interpret the solutions of a).
Given: D, pa , p0 , ηW , ρW
11.15 The laminar boundary layer flow on a flat plate, neglecting the viscous heat, can be
described with the continuity, the momentum, and the energy equation in the following form:
∂u ∂v
+
=0
∂x ∂y
∂u
∂u
ρ u
+v
∂x
∂y
ρcp
!
∂T
∂T
+v
u
∂x
∂y
=η
!
∂2u
∂y 2
=λ
∂2T
∂y 2
a) Determine the dimensionless parameters of the problem.
44
b) Reformulate the resulting parameters by using well known parameters of fluid mechanics.
Assuming constant material quantities the flow field is independent of the temperature field.
Both distributions can be computed separately.
c) Specify the assumptions to determine the temperature distribution in the boundary layer
directly from the velocity distribution.
Hint for c): Compare the differential equations and assume that the velocity distribution ~v (x, y)
is already known.
45
14 Potential flows
Complex Potential
F(z)
Potential
φ(x, y)
Streamfunction
ψ(x, y)
(u∞ − iv∞ )z
Parallel flow
u∞ x + v∞ y
u∞ y − v∞ x
E
ln z
2π
Source E > 0, Sink E < 0
E
E q 2
ln r =
ln x + y 2
2π
2π
E
E
y
ϕ=
arctan
2π
2π
x
Γ
i ln z
2π
Vortex, Γ < 0 clockwise
Γ > 0 counterclockwise
Γ
y
arctan
2π
x
m
z
Dipole
mx
x2 + y 2
−
E
ln z
2π
Parallel Flow+Source/Sink
u∞ z +
R2
)
z
Parallel Flow + Dipole =
Cylinder Flow
u∞ (z +
R2
Γ
)−
i ln z
z
2π
Cylinder Flow + Vortex
u∞ (z +
Parallel Flow + Vortex
u∞ x +
u∞ x(1 +
u∞ x(1 +
E
ln r
2π
R2
)
x2 + y 2
−
Γ q 2
ln x + y 2
2π
−
my
x2 + y 2
u∞ y +
u∞ y(1 −
E
ϕ
2π
R2
)
x2 + y 2
R2
R2
Γ
Γ
u
y(1
−
)
+
ϕ
)−
ln r
∞
2
2
2
2
x +y
2π
x +y
2π
u∞ x +
Γ
ϕ
2π
u∞ y −
Γ
ln r
2π
46
velocity
u
velocity
v
u∞
v∞
E
x
2
2π x + y 2
E
y
2
2π x + y 2
velocity
√
c = u2 + v 2
c∞ =
Streamlines
ψ = const
q
2
u2∞ + v∞
E
2πr
y
−
Γ
y
2
2π x + y 2
m
y 2 − x2
(x2 + y 2)2
u∞ +
x
E
2π x2 + y 2
Γ
x
2
2π x + y 2
−m
−
2xy
(x2 + y 2 )2
Γ
2πr
m
r2
E
y
2π x2 + y 2
on the cylinder:
2u∞ sin2 ϕ
−2u∞ sin ϕ cos ϕ
2u∞ | sin ϕ|
on the cylinder:
2u∞ sin2 ϕ
−
Γ
sin ϕ
2πR
u∞ −
y
Γ
2
2π x + y 2
+2u∞ sin ϕ cos ϕ
+
+
Γ
cos ϕ
2πR
x
Γ
2
2π x + y 2
2u∞ | sin ϕ|
−
Γ
2πR
x
47
14.1 The twodimensional flow around the sketched body shall be described with the potential
theory by superimposing a parallel flow, a source, and a s ink with a distance a from the center,
respectively.
Determine
a) the positions xs of the stagnation points.
b) What is the equation to determine the contour of the body?
14.2 Proof, if the stream function and the potential exist for the following velocity fields!
a) u = x2 y,
v = y2x
b) u = x,
v=y
c) u = y,
v = −x
d) u = y,
v=x
Compute the stream, function and the potential
14.3 The stream function is given ψ = ψ1 + ψ2
Γ q
with ψ1 = − ln (x − a)2 + y 2
2π
2Γ q
ψ2 = − ln (x + a)2 + y 2
2π
Given: a, Γ > 0
Determine
a) the coordinates of the stagnation point,
b) the pressure coefficient on the x-axis cp (x, y = 0) such, that cp = 0 in the origin of the
48
coordinate system.
14.4 The complex stream function is given
F (z) =
E
2 u∞ 3
√ z2 +
ln(z)
3 L
2π
Given: L, u∞
Determine
a) the potential φ(r, θ) and the streamfunction ψ(r, θ).
b) the components of the velocity vr , vθ .
c) the constant E such that a stagnation point is at (x = −L, y = 0) ,
d) the equation that describes the contour rk (θ).
Hints:
z = x′ + iy ′ = reiθ
vr =
1 ∂ψ
∂φ
=
,
∂r
r ∂θ
vθ =
1 ∂φ
∂ψ
=−
r ∂θ
∂r
U
14.5 A planar flow is described by the stream function ψ = ( )xy. The pressure in xref =
L
0, yref = 1 m is pref = 105 N/m2 .
U = 2 m/s
L=1m
ρ = 103 kg/m3
49
a) Proof, if the flow has a potential!
Determine
b) the stagnation points, the pressure coefficient, and the lines of constant total velocity
c) the velocity and the pressure at x1 = 2m, y1 = 2m,
d) the coordinates of a particle at t = 0.5s, if it passes at t = −0 the point x1 , y1 ,
e) the pressure difference between these two points.
f) Sketch the stream lines.
14.6 A plane half-body with the width 2h is flown against with the velocity u∞ .
Gegeben: u∞ , p∞ , h
Determine:
a) the stagnation point and the velocity at x = xs , y = h,
b) the contour of the half-body,
c) the pressure distribution on the contour,
d) the isobars,
ρ
e) the curve on that the pressure is u2∞ larger than the pressure p∞ at infinity,
4
f) the lines of constant velocity,
u∞
,
g) the area in which the velocity v is larger than
2
h) the curve on which the streamlines have an inclination of 45o ,
i) the maximum deceleration for a particle on the x-axis between x = −∞ and the stagnation
point!
50
14.7 The following stream function is given
ψ = u∞ y(1 −
R2
) .
x2 + y 2
a) Sketch the stream lines for x2 + y 2 ≥ R2 .
Determine
b) the pressure distribution on the contour ψ = 0,
c) the time, a particle needs to come from point x = −3R, y = 0 to point x = −2R, y = 0.
14.8 A bridge pylon with a circular cross-section is flown against with the velocity u∞ . Far
away from the pylon the water depth is h∞ .
u∞ = 1 m/s
h∞ = 6 m
R=2m
ρ = 103 kg/m3
Determine
a) the water depth at the pylon wall as a function of θ,
b) the water depth in the stagnation points,
c) the smallest water depth over the ground.
Hint: Assume a twodimensional flow.
g = 10 m/s2
51
14.9 A wind energy facility shall be positioned on a hill. Assume, that the wind flow can be
described with a potential flow to determine the energy. The flow over the hill is given by the
stream function
u∞ H 2
F (z) = u∞ z + 0.8
,
z
assuming that the countour of the hill is a streamline itself.
a) Compute the contour of the hill fk (x, y) = 0 in cartesian coordinates,
b) Compute the power that can be extracted from the flow between the contour and the contour
line in a distance H
I. far away from the hill (x → −∞) and
II. on top of the hill (x = 0).
c) Sketch the potential theoretical velocity profile on the hill (x = 0) as well as a realistic
profile for viscous flow.
Given: u∞ , ρ, H
14.10 A rotating cylinder of length L is flown against with the velocityu∞ normal to its axis.
At the surface, the rotational part of the velocity is vt .
a) Determine the circulation.
b) Discuss the flow field for vt = u∞ .
c) Compute the force on the cylinder.
52
14.11 The twodimensional flow at a 90o -corner can be described with the complex stream
function
2
F (z) = Az 3 , A = konst.
Given: |~v|(r=l,
θ=0)
= u1
Determine:
a) the constant A
b) the pressure distribution cp along the wall. Sketch it!
c) Sketch carefully the lines of constant pressure.
d) What is the equation r = r(θ) for the streamlines?
e) Sketch the flow field.
53
14.12 The stream function ψ(x, y) for the flow of an incompressible fluid through the sketched
plane nozzle is given.
y
ψ(x, y) =
u∞ L
h(x)
1
Given: u∞ , L, B, h1 = L, h2 = L
3
a) Determine the upper and the lower contour h(x) such that the flow can be described with
the potential theory,
b) Compute the velocity distribution u(x, y) and v(x, y).
c) Determine the volume flux for a nozzle with the width B.
54
15 Laminar boundary layers
15.1 A flat plate is flown against parallel to the surface with water. Determine for a laminar
boundary layer the momentum thickness as an integral of the wall shear stress
−
Z
0
x
τ (x′ , y = 0) ′
dx !
ρu2∞
15.2 A flat plate (length L, width B) is flown against parallel to the surface with air.
u∞ = 10 m/s
L = 0.5 m
B=1m
ρ = 1.25 kg/m3
ν = 1.5 · 10−5 m2 /s
a) Sketch the velocity profiles u(y) for different x!
b) Specify the bouindary conditions for the boundary equations!
c) Sketch the distribution of the shear stress τ (y) at a position τ (y)!
d) Determine the boundary layer thickness at the end of the plate and the drag force!
15.3 From exercise 15.2 the drag of a flate plate, wetted on both sides, of length x and width
B can be determined with
W =
Z
x
0
τw (x)Bdx = ρ
Z
0
δ(x)
u(u∞ − u)Bdy.
Using this equation and with the approximation of the velocity profile
2
y
y
u(x, y)
= a0 + a1 + a2
u∞
δ
δ
the boundary
s layer thickness δ(x) is to be determined and compared with the Blasius solution
νx
δ(x) = 5.2
.
u∞
Given: ν, u∞
55
15.4 The velocity profile of a laminar incompressible boundar layer with constant viscosity η
can be described with a polynomial:
2
u(x, y)
y
y
+ a2 (x)
= a0 + a1 (x)
ua (x)
δ
δ
+ a3 (x)
3
y
δ
The outer velocity ua (x) is given by the following approach:
ua (x) = ua1 − C · (x − x1 )2 .
ua1 is the outer velocity at x1 and C is a positive constant. The boundary layer thickness at
x2 is δ(x2 ).
Given: ρ, η, x1 , ua1 , δ(x2 ), C, mit: C > 0
Determine:
a) the pressure gradient∂p/∂x in the flow as a function of x.
b) the coefficient a0 and the coefficients a1 (x), a2 (x), a3 (x).
15.5 In the boundary layer of a flat plate the velocity profiles and the pressure distribution
2
p(x)
x
are measured. Thre pressure distribution on the surface is described with
= 1−k
,
p0
l
with k = const < 1 and the velocity profiles are presented with
y
u(x, y)
=
ua (x)
δ0
with a constant boundary layer thickness δ0 .
1
2
56
Determine the wall shear stress τw using the Kármán integral equation
dδ2
1 dua
τ (y = 0)
=0
+
(2δ2 + δ1 ) +
dx
ua dx
ρu2a
Given: p0 , k, δ0 , l
Boundary layer equation (x-momentum):
u
∂u
1 ∂p η ∂ 2 u
∂u
+v
=−
+
∂x
∂y
ρ ∂x ρ ∂y 2
15.6 The fluid along a flat plate is sucked off through the porous wall with the velocity va .
ua
Element
dx
d
u( y )
r,h
y
x
va
Determine the suction velocity va by using a balance for the sketched element under the
condition that the boundary layer thickness δ is independent of the length x. Assume that the
tangential component of the velocity increases linearly.
Given : ρ, η, δ
57
15.7 The velocity profile in the laminar boundary layer of a flat plate (length L) is described
by a polynomial of fourth order
2
u
y
y
+ a2
= a0 + a1
ua
δ
δ
+ a3
3
y
δ
+ a4
4
y
δ
.
a) Determine the coefficient of the polynomial!
b) Proof the following relationships:
δ1
= 3/10
δ
δ2
= 37/315
δ
q
δ
= 5.84/ Rex
x
q
cw = 1.371/ ReL
15.8 In the stagnation point of a flat plate that is flown against normally to the outer flow
ua (x) is accelerated in such a way that a constant boundary layer thickness δ0 is generated.
The velocity profile is assumed to be linear as a first approximation.
y
u(x, y)
= a0 + a1
ua (x)
δ0
Determine:
a) the constants a0 , a1
b) the distribution of the outer velocity ua (x) using the von Kármán integral equation.
c) the tangential force that is applied between x = 0 and x = L on the plate with the width
B.
Given: δ0 , L, η, ρ, B
von Kármán integral equation
1 dua
τw
dδ2
+
(2δ2 + δ1 ) = 2
dx
ua dx
ρua
58
u a (x)
d0
y
x
x=0
x=L
15.9 The velocity profile of a laminar boundary layer in a flow along a flat plate (length l) is
approximated with
A) a polynomial of third order
3
3 y
1 y
u
=
−
ua
2 δ
2 δ
and
B) a sinusoidal approach
π y
u
.
= sin
ua
2 δ
a) Determine δ1 , δ2 , δ and cw !
b) Compute for
ua = 1 m/s
L = 0.5 m
B=1m
ρ = 103 kg/m3
ν = 10−6 m2 /s
the boundary layer thickness at the end of the plate and the drag force.
59
16 Turbulent boundary laters
16.1 A flat plate is flown against parallel to the surface with air.
ν = 1, 5 · 10−5 m2 /s
u∞ = 45 m/s
Determine
a) the transition point for Recrit = 5 · 105 ,
b) the velocity in the point x = 0, 1 m, y = 2 · 10−4 m using the Blasius solution. What
is the coordinate y with the same velocity at x = 0, 15 m?
Sketch
c) the distribution of the boundary layer thickness δ(x) and a velocity profile for x < xcrit
and x > xcrit .
d) the wall shear stress as a function of x for dp/dx < 0, dp/dx = 0 and dp/dx > 0.
16.2 Representing the Reynolds averaged Navier-Stokes equations the averaged form of the
x-momentum equation of a twodimensional, unsteady, incompressible flow is to be determined.
∂u ∂u2 ∂(uv)
ρ
+
+
∂t
∂x
∂y
!
∂p
∂2u ∂2u
= fx −
+η
+
∂x
∂x2 ∂y 2
!
a) Proof
∂u ∂u2 ∂(uv)
+
+
ρ
∂t
∂x
∂y
!
!
∂u
∂u
∂u
=ρ
.
+u
+v
∂t
∂x
∂y
b) Determine the averaged form odf the x-momentum equation in the time intervall [0,
T ].
16.3 Two planar rectanguular plates have the same lengths L1 and L2 . Plate no. 1 is flown
against parallel to L1 and die plate no. 2 is flown against parallel to L2 with the velocity u∞ .
L1 = 1 m
L2 = 0, 5 m
ν = 10−6 m2 /s
a) Determine the ratio of viscous forces for u∞ = 0, 4m/s; 0, 8m/s; 1, 6m/s.
b) What is the velocity for plate 2, assuming that the velocity for plate 1 is u∞ = 0, 196 m/s
and the drag coefficients are the same?
Hint: cw =
0, 074
1/5
ReL
−
1700
for 5 · 105 < ReL < 107
ReL
60
17 Boundary layer separation
17.1 The lower border of a divergent channel is formed by a flat plate. At x = xa the flow
separates. The velocity profile is described with a polynomial of third order:
u(x, y)
y
y
= a0 (x) + a1 (x)
+ xa2
ua (x)
δ(x)
δ(x)
!2
y
+ a3 (x)
δ(x)
!3
Given: xa
Determine the velocity profile
u(xa , y/δ(xa ))
at the separation point.
ua (xa )
Sketch 3 velocity profiles for x < xa ; x = xa ; x > xa .
17.2 Assume that the flow on a circular cylinder separatesat α = 120o, the pressure up to the
separation is determined from the potential flow, and the pressure in the separation region is
constant. bestimmt und der Druck im Totwasser konstant ist.
Determine thedrag coefficient of the cylinder by neglecting the viscous drag.
61
17.3 A cylinder with radius R is flown against normal to its axis with the velocity u∞ . The
velocity of the frictionless outer flow at the cylinder wall is
x
ua (x) = 2u∞ sin
R
Assume
δ(x) = 5
s
ηx
ρua (x)
for the boundary layer thickness.
a) Determine the local velocity profiles in the boundary layer for the (x, y) coordinate
system with the following approach
2
y
y
u
+ a2 (x)
= a0 (x) + a1 (x)
ua (x)
δ
δ
+ a3 (x)
3
y
δ
b) Determine the angle where the separation occurs (τw = 0)!
Hints:
- boundary layer equation
∂u
∂u
dp
∂2u
ρu
+ ρv
=− +η 2
∂x
∂y
dx
∂y
- Approximation for ϕ ≈ π/2:
sin ϕ ≈ 1
cos ϕ ≈ π/2 − ϕ
17.4 A sphere and a cylinder made from the same material are falling at constant velocity
through air. The axis of the cylinder is normal to the falling direction. For 0 < Re ≤ 0, 5 the
drag coefficient for a sphere cw = 24/Re and the cylinnder cw = 8π/[Re(2 − lnRe)] are given.
ρ = 800 kg/m3
ρL = 1, 25 kg/m3
νL = 15 · 10−6 m2 /s
Determine
a) the maximum diameters for these laws being valid,
b) the corresponding sink velocity.
g = 10 m/s2
62
17.5 A sphere is falling steady with the constant velocity v1 through non-moving air. A downward blast increases the velocity up to v2 .
D = 0, 35 m
G = 4, 06 N
v1 = 13 m/s
3
−6
ρL = 1, 25 kg/m
νL = 15 · 10 m2 /s
v2 = 18 m/s
a) What is the drag coefficient before the blast?
b) What is the final steady velocity of the sphere after the decay of the blast?
17.6 A sphere with the diameter D and the density ρK is shot vertically with the initial velocity
v0 upwards through non-moving air.
D = 0, 1 m
v0 = 30 m/s
g = 10 m/s2
ρH = 750 kg/m3
ρM = 7, 5 · 103 kg/m3
ρL = 1, 25 kg/m3
Determine generally with the assumption of a constant drag coefficient
a) the ceiling,
b) the ascend time,
c) the velocity when the sphere hits the ground,
d) the descend time.
e) Determine these values for a wooden sphere with the density ρH and for a metallic sphere
with the density ρM , if cw = 0, 4 and cw = 0.
63
19 compressible flow
64
65
19.1 A plane flies over an observer horizontally
H = 577 m
v = 680 m/s
T = 287 K
R = 287 Nm/kgK
γ = 1.4
a) What is the Machnumber of the plane?
b) What is the distance covered by the plane before it can be heard by the observer?
c) When was the sound created?
19.2 One jet passes another at a distance b.
vA = 510 m/s vB = 680 m/s b = 170 m
T = 287 K
R = 287 Nm/(kgK)
γ = 1.4
How mauch later the pilot of the passed jet can hear the sound of the faster plane?
19.3 Air is flowing isentropically (γ = 1.4) from a large and frictionless supported container
through a well rounded nozzle into the open air.
a) Determine the dimensionless thrust Fs /p0 AD for the pressure ratios pa /p0 = 1; 0.6; 0.2; 0 !
b) What are these values for an incompressible fluid?
19.4 An airplane flies at supersonic speed. A shock is generated, that is normal in front of the
airplane nose.
u1 = 680 m/s
T1 = 287 K
R = 287 Nm/(kgK)
Determine the temperature change across the shock.
γ = 1.4
66
19.5 A turbine engine sucks air from the atmosphere. Immediately before the compressor the
pressure is p1 .
p0 = 105 N/m2 T0 = 287 K p1 = 0.74 · 105 N/m2
R = 287 Nm/(kgK) γ = 1.4
A = 9 · 103 mm2
Compute the mass flux flowing through the engine!
19.6 Air is flowing from a large reservoir through a well rounded nozzle into the open air. At
the exit cross section A1 a normal shock develops.
A1 = 0.018 m2
T0 = 287 K
A∗ = 0.01 m2
pa = 105
N
m2
R = 287
a) Compute the mass flux
b) Sketch the distribution of the static pressure along the nozzle axis.
J
kg K
67
Hints:
p2
2γ
=1+
(M12 − 1)
p1
γ+1
3,0
A
A*
2,0
1,0
0
0
0,5
1,0
1,5
2,0
M
2,5
19.7 A rocket is equipped with a Laval nozzle. At liftoff (1) the jet has the velocity ME1 =
2 with a jet temperature TE1 . The surrounding pressure is pa . At the altitude H (2) the
1
surrounding pressure is only pa . Under the condition that p0 , T0 and AE are constant during
4
the flight, the smallest cross section A∗ has to be adapted in such a way that no shocks develop
in the jet. Determine the ratio between the smallest cross sections A∗2 /A∗1.
γ = 1.4
ME1 = 2
γR
cp =
;
γ−1
T
=
T0
p
p0
! γ−1
γ
;
68
A∗
M
="
# 1 γ+1
A
γ − 1 2 2 γ−1
2
1+
M
γ+1
2
19.8 Air flows through a Laval nozzle. At positio ’1’ a normal shock is located.
ṁ = 200 kg/s T0 = 300 K p0 = 2.2 · 105 N/m2 pa = 105 N/m2
p01
= 1.39 (ratio of total pressure across the shock) R = 287 Nm/(kg K)
p01′
γ = 1.4
Determine at the exit of the nozzle (’2’)
a) the Mach number M2 ,
b) the velocity u2,
c) the stagnation density ρ0 2 ,
d) the exit cross section A2 .
Hint:
T0
p0
=
p
T
γ
γ−1
19.9 The velocity V1 in front of a straight shock is given by the normal component un1 =
400m/s and the tangential component ut1 = 300m/s. The static temperature changes to
T2 = 1.2 · T1
69
Determine with γ = 1.4 and R = 287 Nm/(kgK):
a) the Mach number M1 and the statictTemperature T1 in front of the shock,
b) the Mach number M2 and the deflection angle β behind the shock,
c) for M1 = const. the velocity components un1 and ut1 in such a way that M2 = 1. Compute
the deflection angle β.
19.10 An oblique shock in a supersonic flow at M1 = 2.2 hits a flat wall at an angle of 40o . At
the impinging point the wall is kinked outwars with 6o .
γ = 1.4
M1 = 2.2
Determine: M2 ,
R = 287 Nm/(kgK)
M3 ,
p2
,
p1
p3
,
p2
T2
,
T1
T3
T1
19.11 Atmospheric air (T = 280 K, p = 1 bar) is sucked through a supersonic wind tunnel
into an evacuated boiler (see sketch). The laval nozzle is designed for ME in the test section.
p = 1 bar
ME = 2.3
T = 280 K γ = 1.4 R = 287 Nm/(kgK)
AH = 0.1 m2 VK = 1000 m3
During the test the reservoir temperature in the boiler is TK = 280 K = const.. At the time
70
t = 0 the boiler pressure is pk (t = 0) = 0.08 bar.
a) Determine the available testing time ∆t. (undisturbed flow in the test section, ME = 2.3.)
b) Determine for the boiler pressure pK = 0.16 bar the following quantities:
Shock angle σ, deflection angle β, p0 2′ , M2 , T2 , T02 and the velocity V2 behind the shock.
c) A wedge 2 · βK = 40o is mounted in the test section.
What is the maximum angleof attack ǫ without the development of a detached shock? What
is the pressure difference pu − pob and the Mach numbers Mob and Mu for this case?
19.12 The sketched diffusor with three shocks decelerates the flow from M∞ to M3 . The conditions of the incoming flow, and the normal velocity component un in front of the last shock
are known.
71
M∞ = 3
β1 = 15o
T∞ = 270 K p∞ = 1 bar
β2 = 10o un = 649 m/s
γ = 1.4
R = 287 J/(kg K)
a) Compute the Mach numbers M1 and M2 as well as the shock angles σ1 and σ2 .
b) Determine for point ’3’ the Mach number M3 , the contour angle β3 , the shock angle σ3 and
the static pressure p3 .
19.13 The interaction of two oblique shocks of different intensities is sketched. The flow behind
the shocks is splitted from a discontinuity line. The ratio of the total pressures in the areas ’1’
and ’2’ and the Mach number in ’1’ are known.
γ = 1.4
R = 287 Jkg −1K −1
p01 /p02 = 1.2
M1 = 1.6
a) Calculate the Mach number M2 in area ’2’.
~2 |/|V
~1| and the density ratio ρ2 /ρ1 .
b) Compute the ratio of the velocities |V
72
3 Fluid Kinematics
3.1
Unsteady flow for the resting observer, steady flow for the moving observer.
3.2
v
v0
dy
=
= −
tan(ωt)
dx
u
u0
a)
Integration :
y =
−
v0
tan(ωt) x + c
u0
Straight lines with slope 0, -1, -∞
b)
x(t) =
Z
u dt + c1 =
u0
sin(ωt) + c1
ω
y(t) =
Z
v dt + c2 =
v0
cos(ωt) + c2
ω
=⇒
Circles with radius 1 m
c) Circle around the origin
ω
u0
2
(x − c1 )2 +
ω
v0
2
(y − c2 )2 = 1
73
4 Basic equations for fluids 4.1
a)
 ∂   ∂p 
∂p 
 ∂x 
 ∂x   ∂x 


 




 


 ∂p 
 ∂   ∂p 
~
~
~






∇ × ∇p = ∇ × 
 = 
×

 ∂y 
 ∂y   ∂y 

1)
=
!


∂
−
∂z
∂
∂y
∂p
∂z
∂p
∂y
=
∂2p
∂2p
−
∂y ∂z
∂z ∂y
∂p
∂z
!


,
,


∂
∂z






∂
∂p
∂z
∂z
!
!
∂ ∂p
∂p
−
∂x
∂x ∂z
∂2p
∂2p
−
∂z ∂x
∂x ∂z
,
,
∂
∂x
∂p
∂y
∂2p
∂2p
−
∂x ∂y
∂y ∂x
!
∂
−
∂y
∂p
∂x
!T
= 0
,
If the second partial derivatives of p exist the orser of the differentiation is irrelevant.
2)
3)
analogous to 1)
analogous to 1)
~ · ~v = 0
∇
b)
2
∂~v
~ ~v
+ ∇
ρ
∂t
2
!
~ × ~v
− ~v × ∇
!
~ p − η∇
~ × ∇
~ × ~v
= ρ · ~g − ∇
4.2
∂v
∂v
dv
=
+ v·
(onedimensional)
a) in general (substantial acceleration):
dt
∂t
∂x
From the continuity equation A · v = const follows v ·
∂v
= 0
∂x
∂v
∂vpiston
=
6= 0
∂t
∂t
=⇒
dv
∂vpiston
=
(local acceleration)
dt
∂t
b) in general (substantial acceleration):
∂v
∂v
dv
=
+ v·
dt
∂t
∂x
Constant inflow velocity v = v0
∂v
= 0
∂t
=⇒
Continuity :
A(x) · v(x) = konst
!!T
74
=⇒
=⇒
=⇒
∂A(x)
∂v(x)
· v(x) +
· A(x) = 0
∂x
∂x
∂A(x) v(x)
∂v(x)
= −
·
6= 0
∂x
∂x
A(x)
dv
∂v(x)
∂A(x) v(x)
= v(x) ·
= −
·
· v(x)
dt
∂x
∂x
A(x)
(only convective acceleration)
4.3
∂
∂r ∂
∂φ ∂
∂z ∂
=
· +
+
·
∂x
∂x ∂r ∂x ∂φ ∂x ∂z
a)
∂r
∂x
∂φ
∂x
∂z
∂x
=
=
=
− 1
x
1 2
x
=
x + y 2 2 · 2x = √ 2
= cos φ
2
2
r
x + y
y
y
y
sin φ
−2
= − 2
−
= − 2 = −
2 · x
2
y
x + y
r
r
1 +
x
0
∂
=⇒
∂x
∂
analogous
∂y
∂
∂z
vx
=
vy
=
vz
=
=
=
=
∂
(cos φ) ·
+
∂r
∂
(sin φ) ·
+
∂r
∂
∂z
!
− sin φ ∂
r
∂φ
!
cos φ ∂
r
∂φ
dx
dr
dφ
=
· cos φ − r · sin φ
= vr · cos φ − sin φ · vφ
dt
dt
dt
dy
dr
dφ
=
· sin φ + r · cos φ
= vr · sin φ + cos φ · vφ
dt
dt
dt
dz
dz
=
dt
dt
=⇒
mit vφ =
∂vx
∂vx ∂r
∂vx ∂φ
∂vx ∂z
∂u
=
=
·
+
+
·
∂x
∂x
∂r ∂x
∂φ ∂x
∂z ∂x
r · dφ
dt
75
∂r
∂φ
− sin φ
∂z
= cos φ
=
= 0
∂x
∂x
r
∂x
∂vr
∂φ
∂vx
=
· cos φ − sin φ
∂r
∂r
∂r
∂vx
∂vφ
= − vr · sin φ − cos φ · vφ − sin φ
∂φ
∂φ
∂vx
= 0
∂z
∂w
∂v
and
are computed analogous.
∂y
∂z
Introducing in eq. 1) results in
1 ∂
∂
1 ∂
∂ρ
+ ρ·
(r · vr ) +
vφ +
vz
∂t
r ∂r
r ∂φ
∂z
!
= 0
b) analogous to a)
vx = vr · sin Θ · cos φ − vφ · sin φ + vΘ · cos Θ · cos φ
vy = vr · sin Θ · sin φ − vφ · cos φ + vΘ · cos Θ · sin φ
vz = vr · cos Θ − vΘ · sin Θ
∂ρ
1
1
1 ∂ 2
∂
∂vΘ
=⇒
r · vr +
+ ρ· 2
(sin Θ vΘ ) +
∂t
r ∂r
r · sin Θ ∂Θ
r · sin Θ ∂φ
4.4
incompressible, steady flow in r-direction:
=⇒
∂
= 0 , vΘ = 0 , vz = 0 , ρ = const.
∂t
!
= 0
76
Symmetrical problem =⇒
∂(r vr )
= 0
∂r
=⇒ Conti:
Momentum eq.:
4.5
steady flow
ρ vr

 ∂
∂p
∂ vr

= −
+ η
 ∂r
∂r
∂r
invompressible
1


r


∂(r vr ) 
∂ 2 vr 



 +
2

∂z 
| ∂r
{z }
= 0
∂ vr
∂ vz
=
= 0
∂z
∂z
ρ = konst
∂(r vr )
= 0 =⇒ vr = 0
∂r
momentum eq.:
∂p
= 0
∂r
∂vz
η ∂
∂p
r
+
−
∂z
r ∂r
∂r
energy eq.:
4.6
steady:

∂ vr
∂ vz
=
= 0
∂t
∂t
fully developed
Conti.:
∂
= 0
∂Θ
ρ cp
∂T
∂T
+ vz
∂t
∂z
!
!
= 0
∂p
∂T
1 ∂
= vz
r
+ λ
∂z
r ∂r
∂r
!
∂
= 0, frictionless: η ∇2 ~v = 0
∂t
=⇒ Navier-Stokes eqns for incompressible, steady, frictionless flows:
∂v
∂u
+
= 0
∂x
∂y
∂u
∂u
ρ u
+ v
∂x
∂y
!
= −
∂p
+ ρ gx
∂x
∂v
∂v
+ v
ρ u
∂x
∂y
!
= −
∂p
+ ρ gy
∂y
∂2 T
+
∂z 2
!
∂ vz
+ η
∂r
!2
77
5 Hydrostatics
5.1
Lift = Weight
FA = G
FA = ρ1 h a2 g + ρ2 (a − h) a2 g
G = ρK a3 g
ρ2 − ρK
=⇒ h =
a = 6, 67 · 10−2 m
ρ2 − ρ1
5.2
X
F = 0
F − G + Fp = 0
F = G − Fp
Fp is the resulting pressure force. Fp can be determined with the principle of Archimedes taking
into account that the body is not fully wet.
Fp = VHK · ρ g − ρ g h · AHK
VHK =
ˆ Volume of the half sphere
Fp =
AHK =
ˆ Base area of the halph sphere
1 4
π R3 ρ g − ρ g h π R2
2 3
2
R − h
Fp = ρ g π R
3
2 2
R − h
=⇒ F = G − ρ g π R
3
2
5.3
Force equilibrium( τ : volume of the buoy,τH : encosed volume of air )
1) at the surface : g mB
2) indepth H :
g mB
1
=
τ g ρw
3
= τH g ρw





=⇒ τH =
1
τ
3
78
enclosed air mass is constant:
=⇒
2
1
τ ρL0 = τH · ρLH =
τ · ρLH =⇒ 2 ρL0 = ρLH
3
3
1
1
h
pa + ρw g H +
pa +
ρw g h
3
3
=⇒ 2
=
RL TL
RL TL
1
pa
+
h
=⇒ H =
ρw g
3
5.4
a)
G0
= 1, 07 · 105 m3
ρM g
∆G = ρM A(h0 − h1 ) g
τ0 =
b)
=⇒ A = 1, 95 · 104 m2
c)
G0 −
G0 − ∆G
=⇒ τ2 − τ1 =
ρM · g
d)
τ2 −
=⇒ h2 = 10, 625 m
∆G = ρM
! τ1 g = ρF τ2 g
ρM
− 1 = 2, 44 · 103 m3
ρF
τ1 = (h2 − h1 ) A
5.5
1
F1 =
Z
d F1 =
Z
coordinate transformation : with S =
p (z1 )·L·ds
z1
cos α
;
ds =
d z1
cos α
F1x = F1 · cos α
=⇒ F1x =
Z2a
0
F1z = − F1 · sin α
Z2a
d z1
cos α p(z1 ) · L
ρ g z1 L d z1 = 2 ρ g a2 L
=
cos α
0
79
F1z = −
Z2a
0
d z1
sin α p(z1 ) · L
= −
cos α
Z2a
0
tan α · ρ g z1 L d z1 = −
with tan α =
2
3
F3x = +
2
3
F2z = 2ρ g a2 L
F2x = 0 ;
Z4a
p(z1 )·L dz1 = +
Z4a
ρ g z1 L dz1 = 6 ρ g a2 L ; F3z = 0
2a
2a
4 , 5 , 6
1
ρ · g · a2 · L ; F6z = 0
2
=⇒ F45z =
F456x = −
Fx =
25
ρ g 5a
· 5a · L = −
ρ g a2 · L
2
2
X
Fix = −
9
ρ g a2 L
2
X
Fiz = +
7
ρ g a2 L
6
i
Fz =
i
Fges =
q
4
ρ g a2 L
3
Fx2 + Fz2 = 4.65 ρ g a2 L
80
5.6
Screw force FS = ?
The origin of the force is the different pressure distributions pi (z) , pa (z). The force FS depends
on the shape of the boiler.
the resulting pressure force Fp res can be
With GGefäß = 0 =⇒ Fp res − FS = 0
determined.
pressure distributions
EHE (inside) pi (z) = pa + ρW · g(R − z)
outside:
d pi
d pL
= ρL g << ρW g = −
dz
dz
due to ρL << ρW pressure gradient can be neglected outside.
=⇒
pa (z) = pa
Resulting pressure force from the integration of all forces:
z
d Fa
R
d Fi
.
d Fi sin Q
d Fa . sin Q
dA
Q
dFi = pi · dA
81
dFa = pa · dA
z - component: dFpz = (dFi − dFa ) · sin Θ = (pi (z) − pa ) dA sin Θ
surface element dA
dA = R · dΘ · r · dϕ
mit
r = R · cos Θ
dA = R2 · cos ΘdΘ dϕ
=⇒
EHE: pi − pa = ρW · g(R − z) = ρW · g(R − R · sin Θ)
Fpz =
R
dFpz =
2π
R π/2
R
0
0
ρW · gR(1 − sin Θ) R2 cos Θ · sin Θ dΘ dϕ
substitution:
sin Θ = η
Fpz = 2πρW · g R
=⇒
Fpz =
3
Z1
0
(η − η 2 ) dη
π
ρW · g R3 = FS
3
alternative solution
Assume a completely submerged body
FS = 0, pi (z) unchanged
the difference is the weight of the hatched water
=⇒
FS = GW
=
=
=
ρW · g (Vcylinder −
1
ρW · g (π R3 −
2
3 π
ρW · g · R ·
3
Vhalfsphere )
4
· R3 π)
3
V =
ˆ volume
82
5.7
1.
X
F = 0 = p dA −
!
∂p
dx
p +
∂x
2. =⇒ p = − ρW ax · x + c1 and
=⇒ p∗ = pa + ρgh = pa + ρax l
g
=⇒ l =
h
ax
5.8
boundary condition: r = 0, z = h :
dA − ρW ax dx dA =⇒
p = − ρW g · z + c2
p = pa
∂p
= − ρW ax
∂x
83
p − pa = ρ g(h − z) +
1
ρω 2 r 2
2
a) Surface: p = pa ; z0 (r) is the surface coordinate
ω 2 r2
2g
r = R : z0 = H
H −h
=⇒ ω 2 = 2g
R2
=⇒
=⇒
z0 (r) = h +
z0 (r) = h + (H − h)
r2
R2
Volume of water:
2
π R h0 =
ZR
z0 (r) 2π r dr = π R2 h +
0
=⇒
=⇒
b)
1
π R2 (H − h)
2
h = 2h0 − H = 0, 4 m
ω =
s
4g
(H − h0 ) = 13, 9 s−1
R2
r = R : p = pa + ρ g (H − z)
z = 0:
p = pa + ρ g h +
ρ 2 2
ω r
2
from hint: dp = ρ ω 2 r dr − ρ g dz
=⇒
Zp
pa
dp =
Zr
r=0
2
ρ ω r dr −
Zz
z=h
ρ g dz
84
ρ ω 2 r2
− ρ g (z − h)
2
=⇒ p − pa =
1
ρ ω 2 r2
2
=⇒ p − pa = ρ g (h − z) +
5.9
z = H:
z = h:
0 = FA − FG − FN ; FA = ρL
(H)
0 = FA − FG − FN ; FA = ρL
(h)
=⇒
ρL
(H)
ρL
(H)
ρL
(H)
− ρG
gH
,
= ρ0 exp −
RL T0
open balloon: pL = pG
FN =
RL
1 −
RG
= ρL
ρL
(h)
g
(h)
RL
1 −
RG
1
2
(H)
τB
, FG = ρG
2
gτB =
ρL
(h)
gτB
(h)
− ρG
gh
= ρ0 exp −
RL T0
=⇒ ρL RL T0 = ρG RG T0
h = H −
5.10
(H)
gτB , FG = ρG
!
gH
=⇒ ρ0 exp −
RL T0
RL T0
ln 2 = 4290 m
g
τB
2
1
g τB
2
(h)
(T = const)
=⇒ ρG = ρL
g
RL
RG
1
gh
= ρ0 exp −
2
RL T0
!
85
a)
FA = FS + FN
and FS = FA − FN
=⇒ FS = ρ0 · V0 · g − m · g = (ρ0 · V0 − m) · g
(mit m = mGas + mstructure )
b)
General equation für V (z) :
V =
mg
ρg (z)
with perfect gas assumption
=⇒
pressure equalization:
=⇒
V (z) =
pg (z)
Rg · Tg
mg · Rg · Tg
pg (z)
pg = pa (z) = ρa (z) · Ra · Ta
V (z) =
mg · Rg · Tg
ρa (z) · Ra · Ta
ρa (z) = ?
,,Scale Height Relation”:
temperatureequalization
=⇒
=⇒
−g
·z
pa (z) = p0 · e Ra · T0
ρa (z) =
pa (z)
Ra · T0
−g
· ρ0 · z
ρa (z) · Ra · T0 = ρ0 · Ra · T0 · e p0
=⇒
For z = 0
ρg (z) =
g · ρ0
·z
mg · Rg
· e p0
V =
ρ0 · Ra
V = V0
=⇒
mg · Rg
V0 =
ρ0 · Ra
=⇒
g · ρ0
·z
V (z) = V0 · e p0
Tg = Ta
86
for z = z1 follows:
g · ρ0
· z1
V 1 = V 0 · e p0
p0
V1
=⇒ z1 =
· ln
g · ρ0
V0
c)
−→ For z ≥ z1 the balloon is rigid.
−→ gas mass and volume is contant =⇒ no pressure equalization.
Ceiling is reached when Fa (z2 ) = G
−g · ρ0
·z
Fa (z2 ) = ρa (z2 ) · V1 · g = ρ0 · V1 · g · e p0
= m·g
=⇒
z2 = ln
ρ0 · V1
p0
·
m
g · ρ0
87
6 Continuity and Bernoulli’s equation
6.1
a) Volume flux V̇ = v5 · A
1
2
h
4
5
H
z
3
ρ 2
v
2 3
ρ 2
v
Bernoulli 2 −→ 4 : pa + ρ g (H + h) = p4 +
2 4
with p3 = p4 =⇒ v3 = v4
=⇒ no mixing losses:
ρ 2
v
2 5
√
√
with p5 = pa + ρ g H =⇒ v5 = 2 g h =⇒ V̇ = 2 g h A
6.2
Bernoulli :
p0 = p∞ +
ρ
ρ 2
v∞ = p2 + v22
2
2
=⇒ ∆p = p0 − p2 =
ρ 2
v
2 2
v∞ π D 2 = v2 π (D 2 − d2 )
Conti :
=⇒ v2 =
s
2 ∆p
D2
= v∞ 2
ρ
D − d2
v∞
s
= 1 −
2∆p
ρ
=⇒
1
vh
2D p
w r
d
D
1
d
D
!2
88
6.3
1
hB
r
R
H
r(z)
2
z
Bernoulli 1 −→ 2 : pa + ρ g H = pa + ρ g z +
=⇒ v(z) =
ρ 2
v (z)
2
q
2gH − 2gz =
q
2 g(H − z)
π R2 vR = π r 2 (z) v(z)
conti :
=⇒ r(z) = R
r(z) = R
r(z) = R
v
u
u v(z
t
s
vR
v(z)
= H − hB )
v(z)
v
u
4u
t 2g(H
− H + hB )
2g(H − z)
s
r(z) = R 4
hB
H −z
6.4
a) Bernoulli :
p1 +
conti :
v1 A1 = v2 A2 = v3 A3
=⇒ v2 =
s
2 ∆p
= 12 m/s
ρ [1 − (A2 /A1 )2 ]
=⇒ v1 = 4 m/s
b) Bernoulli :
ρ 2
ρ
v1 = p2 + v22
2
2
p2 +
v3 = 6 m/s
ρ
ρ 2
v2 = p3 + v32
2
2
89
p3 = pa = 105 N/m2 (Outflow to the surrounding)
=⇒ p2 = 0, 46 · 105 N/m2
=⇒ p1 = 1, 1 · 105 N/m2
Bernoulli :
p + ρ g h = pa +
p = 1, 08 · 105 N/m2
6.5
v
2
0
3
h
1
g
pa
r
A
H
z
4
s
5
Ad
a) Bernoulli 0 −→ 5 :
pa + ρ g H = p5 − ρ g s +
ρ 2
v
2 5
p5 = pa + ρ g s
=⇒ V̇ = Ad
b)
q
2 g H = 4 m3 /s
ρ 2
v
2 3
90
c) bubbles occur, if p2 = p3 = pD .
v5∗ A∗d = v ∗ A
conti :
Bernoulli :
pa = pD + ρ g h +
=⇒
A∗d
= A
s
ρ ∗2
v
2
pa − pD
h
−
= 0, 244 m2
ρgH
H
6.6 Bernoulli 0 −→ 3 :
Z3
ρ 2
v +
pa + ρ g H = p3 +
2 3
Z3
0
=
ρ
∂v
ds ≃ 0 (h1 << L)
∂t
=
Z2
ρ
∂v
ds ,
∂t
I23
=
=
dv2
ρ
dt
ρ
Z3
2
Z2
1
v = v2
h2
h
Z2
,
+
,
p3 = pa ,
v2 = v3
Z3
2
1
=
1
I12
+
Z1
0
I12
Z1
0
0
I01
∂v
ds
∂t
ρ
h = h1 +
h2 − h1
x
L
dv2 h2 L
h2
dv2
h2
L
dx = ρ
ln
= ρ
h2 − h1
dt h2 − h1
h1
dt
x
h1 +
L
∂v
dv2
ds = ρ L
∂t
dt
introduce in Bernoulli:
dv3
ρ 2
v3 + ρ
(L + L)
2
dt
!
v32
dv3
1
gH −
=
dt
2
L + L
pa + ρ g H = pa +
t→∞:
gH −
q
1 2
v3e = 0 =⇒ v3e = 2 g H
2
local acceleration:
∂v
dv3 h2
=
, bl
∂t
dt h
convective acceleration:
bl =
bk = v
(v3 = 21 v3e , x= L
)
2
h2 dh
∂v
= − v32 32
∂x
h dx
,
=
1
2 h2
3
gH
4 h1 + h2
L+L
dh
h2 − h1
=
dx
L
91
g H h22 (h1 − h2 )
L
(h1 + h2 )3
bk(v3 = 1
v3e ,
bs =
dv
∂v
∂v
=
+ v
= bl + bk
dt
∂t
∂x
2
x= L
)
2
= 4
substantial acceleration:
h2
g H h22 (h1 − h2 )
3 gH
+ 4
2 L + L h1 + h2
L
(h1 + h2 )3
bs =
6.7
a) Bernoulli 0 −→ 2 :
ρ 2
v + ρ
pa + ρ g h = p2 +
2 2
(1)
conti:
ZL
2/3 L
∂v
dx
∂t
v · A = v2 · A2
=⇒
ρ
ZL
2/3 L
∂v
∂v2
dx =
ρ
∂t
∂t
Introduce in (1):
∂v2
∂t
b) Bernoulli x −→ 2 :
t=0
ZL
2/3 L
dv2
=
dt
A2
∂v2
dx =
ρ
A
∂t
=
t=0
ZL
2/3 L
L2
∂v2 L
dx =
ρ
2
x
∂t 2
2gh
2 g h − v22
=
L
L
ρ 2
ρ 2
p +
v = p2 +
v + ρ
2
2 2
ZL
x
∂v
dx
∂t
92
conti:
A2
· v2
A
v =
ρ
p = p2 + v22
2
=⇒
1 −
A2
A
L4
ρ 2
v2 1 − 4
=⇒ p = p2 +
2
x
Extremum for p(x) from
2 !
!
ZL L2
ρ 2
+
2 g h − v2
dx
2
L
x
x
+ ρ 2gh −
v22
L
−1 +
x
dp
1 q
2 g h:
= 0 bei v2 =
dx
2
!
ρ 1
4 L4
=⇒ 0 =
· 2gh
2 4
x5
L4
3 L
=⇒ 0 = 5 −
x
2 x2
d2 p ρgh
=
2
d x x=xm
L2
1
L
2gh · − 2
4
x
+ ρ 2gh −
=⇒
s
3 2
L ≈ 0, 87 L
3
x =
3
1
−5
+ ·2·
2
(2/3)
2
2/3
!
ρgh
27
=
<0
−
2
L
4
s
3 2
·L
=⇒ Maximum bei x =
3
6.8
conti :
AB
Torricelli :
q
v1 =
2 g h1
√
introducing
= √
3 AB
2
2g A
A
dh
= v1 A − vA ·
dt
3
,
vA =
q
2 g h(t)
3
h1 −
q
h(t)
!
q
q
3 AB dh
= 3 h1 − h(t)
2 g A dt
4h1
3 AB Z
dh
q
=⇒ T = √
√
2g A
h
h(t)
−
3
1
h1
q
dh
<< vA
dt
− 3
q
h1 ln
q
3
h1 −
q
h(t)
4h1
h1
= 108 s
93
6.9
a) steady Bernoulli 0 −→ 1
for t −→ ∞
pa + ρ g h = pa + ρ g h +
=⇒ vmax =
s
ρ 2
v
− ∆p
2 max
2∆p
ρ
unsteady Bernoulli 0 −→ 1
Z
ρ 2
∂v
pa + ρ g h = pa + ρ g h +
v(t) − ∆p +
ds
ρ
2
∂t
L
dv
ρ 2
v(t) + ρ L
2
dt
2L
2L
ρL
dv = 2
dv
=⇒ dt =
ρ 2 dv = 2 ∆p
2
vmax − v(t)
2
∆p − v
− v
2
ρ
=⇒ ∆p =
vmax + v v=.999 vmax
L
1 + .999
ln
ln
=
T =
vmax
vmax − v v=0
vmax
1 − .999
L
10 m
vmax =
· 7.6 =
· 7.6 = 10 m/s
T
7.6 s
Integration :
Bernoulli 1 −→ 2
L
for t −→ ∞
ρ 2
v
= ρgH
2 max
b)
vH/2
=⇒
H =
vmax
L
ln
= √ =⇒ tH/2 =
vmax
2
2
vmax
= 5m
2g
vmax
vmax
!
1
1+ √
2!
= 1, 763 sec
1
1− √
2
6.10
a)
Bernoulli 0 −→ K :
pa = pK
conti :
=⇒ ρ
L+ξ
Z
0
ρ 2
v + ρgh + ρ
+
2 K
v · A = vK · A
dvK
∂v
ds = ρ
(L + ξ)
∂t
dt
L+ξ
Z
0
∂v
ds
∂t
94
vK =
dξ
2
= − ξ0 ω sin (ωt); vK
= ξ02 ω 2 sin2 (ωt)
dt
dvK
d2 ξ
=
= − ξ0 ω 2 cos (ωt)
dt
dt2
=⇒ pK = pa −
=⇒
ρ 2 2
ξ ω sin2 (ωt) − ρ g h + ρ ξ0 ω 2 cos (ωt) (L + ξ0 cos (ωt))
2 0
pa − ρ g h
1 ξ02
ξ0
ξ02
pK
2
=
−
sin
(ωt)
+
cos
(ωt)
+
cos2 (ωt)
2
2
2
2
2
2
ρL ω
ρL ω
2 L
L
L
pa − ρ g h
ξ0
pK
=
+
=⇒
2
2
2
2
ρL ω
ρL ω
L
ξ0
b) Minimum of
L
"
"
1 ξ0
cos (ωt) −
(1 − 3 cos2 (ωt))
2 L
#
1 ξ0
π
cos (ωt) −
(1 − 3 cos2 (ωt)) at t = für ξ0 << L!
2 L
ω
2
=⇒ pD = pa − ρ g h − ρ ξ0 (L − ξ0 ) ωK
=⇒ ωK =
s
pa − ρgh − pD
ρ ξ0 (L − ξ0 )
c)
∆T =
2π
ω
1
V̇ =
∆T
Zω
Period of piston movement
2π
π
ω
2π
ω
A
vK (t) A dt =
2π
Zω
vK dt
π
ω
2π
ω
V̇ =
A
2π
Zω
(−ξ0 ω sin (ωt)) dt =
π
ω
=⇒ V̇ =
ω
A ξ0
π
2π
ω
cos ωt ω
A ξ0 ω
2π
ω ωπ
#
95
6.11
a) Bernoulli from surface −→ pipe end :
1) ω
=
0 : pa + ρ g (h + l) = pa +
=⇒ v (t < 0) =
2) ω
=
q
ρ 2
v
2
2 g (h + l)
ρ 2
ρ 2 2
v −
ω R
2
2 r
q
=⇒ v (t → ∞) = 2 g (h + l) + ω 2 R2
ωr : pa + ρ g (h + l) = pa +
unsteady Bernoulli:
pa + ρ g (h + l) = pa +
=⇒
ρ 2
ρ 2 2
dv
v −
ωr R + ρ(l + R)
2
2
dt
dv
2g(h + l) + ωr2 R2 − v 2
a2 − v 2
=
=
dt
2(l + R)
2(l + R)
1
v(t→∞)
2
Integration :
Z
∆T =
v(t<0)
1
2(l + R) dv
a + v 2 v(t→∞)
2(l + R)
=
ln
a2 − v 2
2a
a − v v(t<0)
with v(t → ∞) = a
=⇒ ∆T =
3 ωr2 R2
l+R
ln
a
(a + v(t < 0))2
b)
∂p
1
= ρ ω 2 r =⇒ p(r) =
ρ ωr2 r 2 + c
∂r
2
boundary condition :
r = R =⇒ p = pa =⇒ c = pa −
ρ ωr2
(R2 − r 2 )
2
ρ ωr2 2
R
point 1 : r = 0 =⇒ p1 = pa −
2
=⇒ p = pa −
1
ρ ωr2 R2
2
96
7 Momentum and momentum of momentum equation
7.1
Force equilibrium in the u-tubes:
p1 + ρHg g∆h1 = p01 = p2 + ρHg g∆h2
=⇒ p1 − p2 = ρHg g(∆h2 − ∆h1 )
Momentum equation in x-direction:
0 = (p1 − p2 )AD − F
=⇒ F = ρHg g(∆h2 − ∆h1 )
7.2
Momentum:
conti:
πD 2
4
dIx
= −ρv12 A + ρv22 A − v3 cos αρv3 A3 = (p1 − p2 )A
dt
v1 A + v3 A3 = v2 A
=⇒
1
v1 + v3 = v2
4
=⇒
v3 = 4(v2 − v1 )
mit
1
A3 = A
4
=⇒
F = (p1 − p2 )AD
97
=⇒
p2 − p1 = ρ (v12 − v22 + 4(v2 − v1 )2 cos α)
7.3
a)
ρ
ρ
ρ
p1 + v12 = pa + v22 = pa + v32
2
2
2
∆p = p1 − pa
v1 A1 = v2 A2 + v3 A3
v1 =
v
u
u 2∆p
u
u ρ t
v2 = v3 =
1
A1
A2 + A3
2
= 2, 58 m/s
−1
A1
v1 = 5, 16 m/s
A2 + A3
b) Momentum in x-direction:
−ρv12 A1 + ρv22 A2 cos α2 + ρv32 A3 cos α3 = (p1 − pa )A1 + Fsx
=⇒
Fsx = −866, 4 N
Momentum in y-direction:
ρv22 A2 sin α2 − ρv32 A3 sin α3 = Fsy
=⇒
Fsy = −238, 4 N
c)
A2 sin α2 − A3 sin α3∗ = 0
=⇒
α3∗ = 12, 37◦
98
7.4
a)
X
~ =0
M
R
Momentum:
x-direction:
~va ⊥~v ;
a
=⇒ F1 = Fw1
l
=⇒ aFw1 − lF1 = 0
ρ~v (~v · ~n)dA = ΣF
P
F = −Fw1 ;
p = pa
a
=⇒ −ρv 2 hB = −Fw1 =⇒ F1 = ρ v 2 hB
l
b)
Momentum in x-direction:
Bernoulli:
−ρv 2 hB − 2ρv22 h2 B = −Fw2
ρ
ρ
1 −→ 2 p1 + v12 = p2 + v22
2
2
p1 = p2 = pa ;
v1 = v
=⇒ v2 = v
99
conti:
vh = 2vh2
1
=⇒ h2 = h
2
=⇒ Fw2 = 2ρv 2 hB = 2Fw1
X
~ = 0 =⇒ F2 = a Fw2 =⇒ F2 = 2ρ a v 2 hB = 2F1
M
l
l
c)
X
~ = 0 =⇒ − l mg sin θ + a Fw3 = 0 =⇒ Fw3 = 1 l mg sin θ cos θ
M
2
cos θ
2a
Momentum in z-direction:
ρv cos θvhB = Fw3
=⇒ ρv 2 cos θhB =
1l
mg sin θ cos θ
2a
2a hB 2
v
l mg
!
2a ρhBv 2
=⇒ θ = arcsin
l mg
=⇒ sin θ = ρ
100
7.5
a) Conti:
ρ1 v1 A∞ = ρ1 v1 (A∞ − AR ) + ∆ṁ
=⇒
∆ṁ = ρ1 v1 AR
b) Momentum in x-direction
Z
−ρ1 v12 A∞ + ρ1 v12 (A∞ − AR ) + ρA vA2 AR +
Für A∞ /AR ≫ 1 : vx = v1
Z
AM
ρ1 vx vr dA = v1
Z
AM
AM
ρ1 vx vr dA = Fs
ρ1 vr dA = v1 ∆ṁ
=⇒
Fs = ρA vA2 AR
=⇒
P = Fs v1 = ρA vA2 v1 AR
7.6
a)
Bernoulli1 −→ 1′ :
Bernoulli2′ −→ 2 :
ρ
ρ
pa + v12 = p1′ + v ′2
2
2
ρ
ρ
p2′ + v ′2 = pa + v22
2
2
Konti :
v1 A1 = v ′ A′ = v2 A2
Momentum in x − direction for the small surface :
Momentum in x − direction for the large control surface
0 = (p1′ − p2′ )A′ + Fs
−ρv12 A∞ + ρv22 A2 + ρv12 (A∞ − A2 ) − ∆ṁv1 = Fs
(see Exercise 7.5)
101
conti:
=⇒
ρv1 A∞ + ∆ṁ = ρv2 A2 + ρv1 (A∞ − A2 )
v1 + v2
v′ =
= 6, 5 m/s
2
b)
η=
7.7
a)
v1
Fs v1
= ′ = 0, 769
′
Fs v
v
102
b) Bernoulli 1 −→ 1’:
ρ
ρ
p1 + v12 = p1′ + v12′
2
2
s
2
=⇒ v1′ = v2 =
(p1 − p1′ ) + v12
ρ
ṁ = ρAv1′ = 13 · 103 kg/s
c) Momentum in x-direction:
−ρv12 A∞ − ∆ṁv1 + ρv22 A + ρv12 (A∞ − A) = Fs
Konti:
(see Exercise 7.5)
ρv1 A∞ + ∆ṁ = ρv1 (A∞ − A) + ρv2 A
=⇒
Fs = ρv2 (v2 − v1 )A = 0, 39 · 105 N
d)
P = V̇ (p02 − p01′ ) = V̇ (p1 − p1′ ) = 448, 5 kW
7.8
well rounded inlet
a) Bernoulli ∞ −→ 1 :
ρ
pa = p1 + v12
2
∆p = p2 − p1 = pa − p1
s
2∆p
A
=⇒ V̇ = vA =
ρ
103
b)
P = V̇ (p02 − p01 ) =
s
2∆p
∆pA
ρ
c) Momentum in x-direction:
ρv 2 A = Fs = 2∆pA
sharp edged inlet
a) Momentum in x-direction for the dotted control surface:
ρv 2 A = (pa − p1 )A
∆p = p2 − p1 = pa − p1
s
∆p
=⇒ V̇ = vA =
A
ρ
b)
P = V̇ (p02 − p01 ) =
s
∆p
∆pA
ρ
c) Momentum in x-direction for the dashed line
ρv 2 A = Fs = ∆pA
7.9
a)
Bernoulli ∞ =⇒ 1 :
ρ
pa = v12 + p1 − ∆p
2
104
dIy
= −ρv12 2A = (p1 − pa )BL − G
dt
=⇒ ρv12 =
1
(G − (p1 − pa )BL)
2A
1
(G − (p1 − pa )BL) + p1 − ∆p
4A
BL
G
BL
= p1 1 −
− ∆p +
=⇒ pa 1 −
4A
4A
4A
G
∆p − 4A
1 G
=⇒ p1 = pa +
= pa +
− ∆p
4 4A
1 − BL
4A
=⇒ pa =
ρ
=⇒ v12 = pa − p1 + ∆p
2
=⇒ v1 =
s
s
V̇ = v1 2A = 2A
b)
Power:
pressure loss:
conti:
ρ
ρ
∆pv = po1 − po2 = p1 + v12 − pa − v22
2
2
=⇒ v1 = v2
1G
1
5∆p −
2ρ
4A
P = ∆pV̇
2Av1 = 2h(B + L)v2
1
1G
5∆p −
2ρ
4A
1 G
− ∆p
=⇒ ∆pv = p1 − pa =
4 4A
105
Pv = ∆pv V̇ =
V̇
4
G
− ∆p
4A
7.10
In frictionless flow the Bernoulli equation is valid. If the distance from the ball is large enough
a constant flow velocity is assumed. In a constant velocity field the pressure gradient across
the flow is zero.
a) Bernoulli ∞ −→ 1 −→ 2 −→ ∞ :
pa + ρ
v2
v2
v2
v12
= p1 + ρ 1 = p2 + ρ 2 = pa + ρ 2
2
2
2
2
=⇒ v1 = v2
b)
−ṁv1 cos α1 + ṁv2 cos α2 =
X
Fpx + Fstx = 0
p = pa
=⇒ v1 = v2
=⇒ α1 = α2 ,
with the absolute values α1,2 .
c)
−ṁv1 sin α1 + (ṁv2 (− sin α1 )) = −G
106
=⇒
ṁ =
G
2v1 sin α1
7.11
a)
~va = vabsolut
~vr = vrelativ
ω
~ · R = vFahrzeug
vr Bernoulli with acceleration term
~va = ~vr + ~ω R
Z s2
ρ
w 2 R2
ρ
ρ(~b · d~s) = pa + vr2 − ρ gH +
0 → 2 : pa = pa + vr2 −
2
2
2
s0
q
m
=⇒ vr = 2gH + w 2R2 = 16
s
b) Steady flow in a moving coordinate system.
=⇒
Z
KF
~ = Σ ~r × F~
(~r × ~v)ρ~v · ~ndA = ΣM
~ =
=⇒ M
Z
KF
KV
(~r × ~va ) ρv~r · ~ndA
|~r × ~va | = |R(ωR − vr cos α)|
torsional moment for 3 arms:
=⇒ M = 3ρvr AR (ωR − vr cos α) = −6.8 Nm
!
107
~ operates in the direction of ~ω . F~ and ~v are in opposite direction.
The torsional moment M
V̇ = 3vr A = 2.4 · 10−3
m3
s
c) Bernoulli 0 −→ 1:
ρ
pa + ρg(h + H) = p1 + v12
2
V̇
v1 =
A1
2
ρ V̇
N
=⇒ p1 = pa + ρg(h + H) −
= 0.82 · 105 2
2
2 A1
m
d)
dM
=0
dω
=⇒ maximum for ω = 0
q
q
=⇒
vr =
√
2gH
=⇒ M = 3ρ 2gHAR − 2gH cos α = −6ρgHAR cos α = 13 Nm
7.12
Momentum of momentum:
R
KF
~
ρ(~r × ~v)~v · ~ndA = ΣM
=⇒ ρlv12 A = Ghg sin α
Bernoulli 0 −→ 1 :
ρ
ρgl cos α = v12
2
108
=⇒ v1 =
q
2gl cos α
=⇒ Aρl2gl cos α = Ghg sin α
=⇒ Gh tan α = 2 ρ l2 A
!
Aρl2
=⇒ α = arctan 2
Gh
109
9 Laminar viscous flow
9.1
R
Volume flux :
Momentum in x−direction :
boundary conditions :
=⇒
=⇒
V̇ = B 0δ u(y)dy
dτ
= ρg sin α
dy
du
τ = −η
dy
y =0: u=0
y =δ: τ =0
!
y2
ρg sin α
δy −
u(y) =
η
2
ρgB sin α 3
V̇ =
δ = 1, 2 · 10−3
3η
9.2
a)
−
∂τ
+ ρg sin α = 0
∂y
m3 /s
110
with
− TT
ρ = ρ0 e
0
y Tδ − T0
T
=
+1
T0
δ T0
and
and the constant k = −
Tδ − T0
T0
follows
y
∂τ
eδk
= ρ0
g sin α
∂y
e
Integration:
τ=
Boundary condition:
τ (y = δ) = 0
=⇒
ρ0 gδ sin α y k
e δ + C(x)
ke
=⇒
τ (y) =
C=
−ρ0 δg sin α
ke
ek
ρ0 δg sin α y k
e δ − ek
ke
b)
τ = −η
∂u
∂u
ρ0 gδ sin α y k
e δ − ek
= −ρν
=
∂y
∂y
ke
y −gδ sin α ∂u
1 − ek(1− δ )
=
∂y
kν
!
y
δ
−gδ sin α
y + ek(1− δ ) + C(x)
u =
kν
k
Integration :
Boundary condition:
u(y = 0) = 0
=⇒
=⇒
C=
gδ 2 sin α k
e
k2ν
y
gδ sin α
δ
u(y) = −
y + (ek(1− δ ) − ek )
kν
k
!
c)
Momentum in y-direction: −
Integration:
Boundary condition:
∂p
= ρg cos α
∂y
p = f (y) + C(x)
p(y = δ) = pa
=⇒
=⇒
C = pa + f (y) 6= f (x)
∂p
= 0
∂x
111
9.3
a)
PDE :
dp
d2 u
= η 2
dx
dy
1. Integration :
1 dp
du
=
y + C1
dy
η dx
2. Integration :
u=
1 dp 2
y + C1 y + C2
2η dx
1) non-moving bend
Boundary conditions
y=0:
u = 0
−→
C2 = 0
y=h:
u = 0
−→
C1 = −
=⇒
u(y) =
h2
2η
h
2η
dp y 2 y
( ) −
dx h
h
dp
dx
2) moving bend
Boundary conditions:
y=0:
u = u∞
−→
C2 = u ∞
!
h dp u∞
+
y = h : u = 0 −→ C1 = −
2η dx
h
2
h
y
dp y 2 y
=⇒ u(y) =
( ) −
+ u∞ (1 − )
2η dx h
h
h
112
b)
Friction per unit width at the bottom:
1) non-moving bend
F
du = lη
B
dy y = h
(
h
dp
h−
dx
2η
=⇒
lh dp
F
=
B
2 dx
1
F
= ηl
B
η
dp
dx
)
2) moving bend
=⇒
=
h dp u∞
1 dp
ηl
h−
−
η dx
2η dx
h
F
B
=
lh dp
l
− η u∞
2 dx
h
=⇒
9.4
a)
(
F
B
|F |2)
>
|F |1)
)
113
dp dτ
+
= 0
dx dy
du
dy
1 dp
d2 u
=
dy 2
η dx
τ = −η
=⇒
Boundarycondition :
y= 0: u=0
y = H : u = uw
1 dp 2
H
u(y) =
2η dx
Integration :
"
y
H
2
b)
1 dp 2
H
τ (y = H)
2η dx
=
1 dp 2
τ (y = 0)
H
uw −
2η dx
uw +
c)
V̇ = um BH = B
Z
H
0
u(y)dy =
!
uw
dp H 2
BH
−
2
dx 12η
d)
umax = −
dp H 2
dx 8η
e)
dIx
=B
dt
Z
H
0
6
ρu(y)2dy = ρu2m BH
5
f)
τw = η
um =
∂u
∂y
2
umax
3
using u from a)
see script p. 140
and dimensionless:
g)
τw
12
ρ 2 = Re
u
2 m
#
y
y
+ uw
−
H
H
114
9.5
a) Computation of τ (r)
dp
∆p
d(rτ )
∆p
= const. =
=⇒
= r
dx
L
dr
L
∆p
dr
=⇒
d(rτ ) = r
L
r 2 ∆p
rτ = c +
2L
c r∆p
+
=⇒
τ =
r
2L
fully developed :
Integration :
boundary condition:
r −→ 0 =⇒ τ −→ 0 =⇒ c = 0
=⇒ τ =
r ∆p
2 L
Computation of u(y)
τ = −ηOdW
=⇒
for a given r is
du
<0
dr
du dr du dr s
du
r∆p
=
dr
2L
du
r∆p
= −
dr
2ηL
s
r∆p
∆p √
du
rdr
=−
=⇒ du = −
=⇒
dr
2ηOdW L
2ηOdW L
=⇒
u = c−
s
∆p
2 3
r2
2ηOdW L 3
115
boundary condition:
u(r = R) = 0
=⇒
1
u(r) =
3
=⇒
2
c=
3
s
s
∆p
3
2ηOdW L
R2
3
2∆p 3
R2 − r2
ηOdW L
Computation of V̇
V̇ =
Z
R
0
2π
u(r)2πrdr =
3
s
2∆p
ηOdW l
=⇒
=⇒
Z
R
0
3
2
R r−r
π
=
7
V̇
∆p =
s
5
2
2π
dr =
3
s

2∆p 7
R2
ηOdW L
7V̇
πR
7
2
!2
ηOdW L
2
9.6
a)
r = R−y
Z 1
u
dIx Z R 2
2
2
˙
I=
=
ρu 2πrdr = 2ρum πR
dt
um
0
o
δ = 0:
=⇒
δ = R:
2
u
=1
um
I˙ = ρu2m πR2
"
u
r
=2 1−
um
R
2 #
7
7

2∆p  R 2
2R 2 
−
ηOdW l
2
7
r
r
d
R
R
116
I˙ = 1, 33ρu2mπR2
=⇒
R
δ =
:
2
=⇒ I˙ =
2ρu2m πR2
(Z
0,5
0
24
17
2




r
96 r
1−
17 R
R
r
96 r
1−
u
17 R
R
=
24
um 


17
r
r
+
d
R
R
Z
1
0,5
R
≤r≤R
2
R
0≤r≤
2
2
r
r
d
R R
)
= 1, 196ρu2mπR2
b)
τw = ηum
2/δ
1
2δ
+
1−
3R 6
δ
R
!2
δ −→ 0 : τw −→ ∞
δ = R : τw = 4
δ=
η · um
R
R
η · um
: τw = 5, 65
2
R
9.7
−ρu21 πR2 +
=⇒
=⇒
=⇒
−
ρu21 πR2
+
Z
0
R
ρu22 (r)2πrdr = (p1 − p2 )πR2
ρ2πu22max
Z
0
R
"
r
1−
R
2 #2
rdr = (p1 − p2 )πR2
1
− ρu21 πR2 + ρπu22max R2 = (p1 − p2 )πR2
3
1 2
p1 − p2 =
u
ρ − ρu21
3 2max
117
Continuity:
ūπR
2
=⇒
u1
=⇒
p1 − p2
b)
2
= u1 πR =
Z
R
0
u2 (r)2πrdr =
1
= ū = u2max
2
1
1
ρ4ū2 − ρū2 = ρū2
=
3
3
2
p1 − p2
=
ζE = ρ
3
· ū2
2
9.8
a)
For y ≤ a the fluid behaves like a rigid body
equilibrium of forces :
ρga∆z = τ0 ∆z
τ0
=⇒ a =
ρg
b)
a ≤ y ≤ b:
Momentum in z−direction :
dτ
= ρg
dy
τ = −η
dw
+ τ0
dy
u2max 2
πR
2
118
Boundary conditions :
y =a:
y =b:
τ = τ0
w=0
ρg
[(b − a)2 − (y − a)2 ]
2η
ρg
w(y) = (b − a)2
2η
Integration : w(y) =
0≤y≤a:
9.9
a)
Force equilibrium:
dτ
dy) dxB = 0
dy
dτ
= −ρg
=⇒
dy
τ dxB − ρgdxdyB − (τ +
Integration:
Boundary condition:
τ = −ρgy + C1
y = δ : τ = 0 =⇒
τ = ρg(δ − y)
b)
du
<0
dy
=⇒
τ = τ0 − η
du
dy
119
τ0 − η
=⇒
du
dy
du
dy
=⇒
Integration :
Boundary condition:
y=0
=⇒
Remark:
δ
2
=
τ0 − ρg(δ − y)
for |τ | > |τ0 |
η
τ0 − ρg(δ − y)
dy
η
τ0
ρg
y2
y − (δy − ) + C2
η
η
2
=
du
=
u(y)
=
:
ρg(δ − y)
u = uB
u(y) = uB +
ρg 2
(y − δy) for |τ | > |τ0 |
2η
from problem formulation
c)
Assumption:
m2
with [V̇ ] =
s
V̇ = 0
with
u
Z
δ
2
Z
δ
V̇
=
!
= uB −
δ
2
0
Z
δ
2
u(y)dy =
Z
!
0
δ
2
V̇
=0
B
oder:
u(y)dy +
Z
δ
δ
2
!
δ
dy = 0
u
2
ρg 2
δ
8η
!
ρg
δ ρg
1
1
u(y)dy =
uB + (y 2 − δy) dy = uB + δ 3 ( − )
2η
2 2η 24 8
0
0
!
Z δ
Z δ
ρg
δ ρg 1
δ
u( )dy = δ uB − δ 2 dy = uB − δ 3
δ
2
8η
2
η 16
2
2
ρg 3 5
ρg 5
=⇒ V̇ = uB δ − δ
= 0 =⇒ uB,min = δ 2
η 48
η 48
120
9.10
a)
dp 1 d(τ r)
+
=0
dx r dr
τ = −η
=⇒
Boundary conditions :
=⇒
du
dr
!
du
1 dp
1 d
r
−
=0
r dr
dr
η dx
r=a :
u=0
r=R :
u=0
"
1 dp 2
ln(r/R)
R2 − r 2
u(r) = −
(R − a2 )
−
2
2
4η dx
R −a
ln(a/R)
b)
a
2
2
2
R 2a ln R + R − a
τ (r = a)
=
τ (r = R)
a 2R2 ln a + R2 − a2
R
#
121
c)
um
Z R
V̇
1
=
u(r)2πrdr =
=
π(R2 − a2 )
π(R2 − a2 ) a
"
1 dp 2
a
= −
R 1+
8η dx
R
2
1 − (a/R)2
+
ln(a/R)
9.11
a)
"
#
d 1 d
(r · v) = 0
dr r dr
1 d
(rv) = c1
r dr
1. Integration:
=⇒
d(rv) = c1 rdr
1
rv = c1 r 2 + c2
2
2. Integration:
1)
r = Ri :
v=0
2)
r = Ra :
v = ωRa
1. BC:
2. BC:
1
c1
0 = c1 Ri2 + c2 =⇒ c2 = − Ri2
2
2
1
ωRa2 = c1 Ra2 + c2
2
=⇒
1
c1
c1 Ra2 − Ri2
2
2
2ωRa2
=
Ra2 − Ri2
ωR2 R2
= − 2 a i2
Ra − Ri
ωRa2 =
=⇒
c1
=⇒
c2
#
122
1 2 2ωRa2
ωRa2 Ri2
r 2
−
2 Ra − Ri2 Ra2 − Ri2
ω(r 2 − Ri2 )Ra2
=⇒ rv =
Ra2 − Ri2
ωRa2 r 2 − Ri2
=⇒ v =
r Ra2 − Ri2
=⇒
rv =
b)
=⇒
d v
τ = −ηr
dr r
τ
η =
d v
−r
dr r
Mz = −τ (r = Ri )2πRi2 L
"
#
Ns
Ri 2
Mz
1−
= 10−2 2
=⇒ η =
2
4πωRi L
Ra
m
123
10 Turbulent pipe flows
10.1
The first is trivial and no proof is necessary. The time average of a quantity m in the time
intervall [0, T ] is defined as
1
m̄ =
·
T
ZT
m · dt .
0
Using this for the averaging of f + g results in:
b)
1
·
f + g =
T
ZT
0
ZT
1
(f + g) · dt =
·
T
0
1
f · dt +
·
T
0
g · dt = f¯ + ḡ .
T
T
1 Z
1 Z ¯
¯
g · dt = f¯ · ḡ .
f · g · dt = f · ·
·
f ·g =
T
T
c)
0
d)
e)
ZT
Z
s
∂f
1
=
·
∂s
T
ZT
1
·
f · ds =
T
ZT
Z
s

0

∂f
∂ 1
· dt =
·
∂s
∂s T
0
0
1
 ·
T


ZT
0
Z
s
f
1
·
f · ds·dt =
T
· dt ·
ds =
Z
s
f · dt =
0



ZT
Z
s


ZT
0
∂f
.
∂s

f · dt· ds =
f · ds
10.2
The ratio between the average and the maximum velocity is
Z1
1
v̄m
2n2
ξ(1 − ξ) n dξ =
= 2
v̄max
(n + 1)(2n + 1)
0
mit ξ =
r
.
R
The integral is solved using partial integration. The average velocity is at a distance
124
rm
= 1 −
R
v̄m
v̄max
n
see table.
Re
n
v̄m /v̄max
rm /R
1 · 105
6 · 105
1.2 · 106
2 · 106
7
8
9
10
0.8166
0.8366
0.8526
0.8658
0.7577
0.76
0.762
0.7633
Measuring v̄(r) at a distance R − rm from the wall, and with the known v̄max the average
velocity can be determined, and the volume flux V̇ = vm π R2 can be computed.
10.3
a)
r = R−y
V̇
ūm
=
= 2
2
ūmax
π R ūmax
b)
RR
I˙
=
ρ ū2m π R2
ρ ū2 2π r dr
0
ρ
ū2m
π
R2
Z1 0
r
1 −
R
1/7
ūmax
= 2
ū
2 Z1 λ =
0, 316
4√Re
r
d
R
1 −
0
r
R
r
R
2/7
=
r
d
R
10.4
a) 2300 ≤ Re ≤ 105 :
ūm =
λ =
b)
8 τw
ρ ū2m
η
Re
ρD
=⇒ τw =
λ ρ ū2m
= 2, 22 N/m2
8
ūmax
ūm
=
− 4, 07
u∗
u∗
λ =
u∗
8 τw
=
8
ρ ū2m
ūm
2
49
60
r
R
=
50
49
125
=⇒
ūm
1
q
= 0, 84
=
ūmax
1 + 4, 07 λ/8
yu∗
ū
= 5 =
ν
u∗
c)
with
ū = 5
s
u∗ =
s
τw
ρ
(viscous sublayer) =⇒ ū = 5 u∗
and
τw =
λ ρ ū2m
8
follows
λ
ūm = 0, 236 m/s
(for y = 0, 11 mm , λ from diagram)
8
yu∗
= 50
(log. velocity distribution)
ν
ū
yu∗
+ 5, 5
= 2, 5 ln
u∗
ν
=⇒
d)
ū = 0, 720 m/s
l = 0, 4 y = 0, 4
(for y = 1, 1 mm)
ν
yu∗
q
= 0, 85 mm
ν
λ/8 ūm
10.5
a)
b) Bernoulli with losses:
pa + ρ g h1 = p1 +
L
1 + λ
2D
ρ 2
ū
2 m
126
V̇ = ūm
Re =
λ = 0, 024
π D2
4
ūm D
= 8 · 105
ν
ks
= 0, 002
D
(from Moody − diagram)
p1 = 1, 22 · 105 N/m2
=⇒
c) Bernoulli with losses:
p2 = pa + ρ g h2 + λ
d)
L ρ 2
ūm = 3, 77·105 N/m2
2D 2
P = V̇ (p2 − p1 ) = 160, 5 kW
10.6
a)
∆p = λ
L ρ 2
ū
D 2 m
V̇ = ūm
=⇒
b)
λ =
Re =
π D2
4
π 2 ∆p D 5
= 0, 0356
8 ρ L V̇ 2
ρ ūm D
= 1, 8 · 105
η
127
ks
= 0, 0083 (from Moody diagram)
D
=⇒ ks = 4, 2 mm
c)
momentum equation for the inner control surface:
∆p
π D2
− τW π D L = 0
4
=⇒
τW = ∆p
D
= 16 N/m2
4L
momentum equation for the outer control surface:
F = − ∆p
d)
π D2
= − 2517 N
4
λ = 0, 016
=⇒
10.7
P1
P2
(from diagram)
∆p = 5, 8 · 103 N/m2
L
1 + λ1
D1
= L
1 + λ2
D2
V̇ = ūm
π D2
4
D
1
√ = 2, 0 log
ks
λ
=⇒
P1
P2
ρ 2
ū
2 m1
ρ 2
ū
2 m2
+ 1, 14
L
4
D1 D 2
=
= 39, 2
L
D1
1 + λ2
D2
1 + λ1
128
10.8
P B: pipe bundle ;
C: Channel
λP B
L ρ 2
LK ρ 2
ūmP B = λC
ū
D 2
dh 2 mC
dh = a
ReC =
ReP B =
ρ a V̇
= 105
η a2
ρD
η
V̇
= 1, 27 · 104
π D2
100
4
LK = 13, 57 m
λC = 0, 018
λP B = 0, 030
129
11 Similarity theory
11.1 The frequency f depends on the following quantities:
incoming velocity u∞
[u∞ ] =
˙
density ρ
[ρ]
=
˙
kin. viscosity ν
[ν]
=
˙
diameter of the cylinder D [D]
=
˙
=⇒ f = F (u∞ , ρ, ν, D)
=
˙
[f ]
m
s
kg
m3
m2
s
m
1
s
5 variables exist with 3 different dimension (s, m, kg) =⇒ from Π-theorem the problem has
5 − 3 = 2 dimensionless parameters.
The quantities D, ρ, u∞ are chosen as reference quantities. They include all basic dimensions
and are linearly independent
Determining the dimensionless parameters:
π1 = f · D α · uβ∞ · ργ
L [m] : 0 = 0 + 1 · α + 1 · β + (−3) · γ
3γ = α + β
t [s] : 0 = −1 + 0 · α + (−1) · β + 0 · γ
β = −1
M [kg] : 0 = 0 + 0 · α + 0 · β + 1 · γ
γ=0
=⇒ α = 1;
β = −1;
=⇒ π1 = f ·
D
u∞
π2 = ν · D α · uβ∞ · ργ
γ=0
130
[m] : 0 = 2 + 1 · α + 1 · β + (−3) · γ
3γ = 2 + α + β
[s] : 0 = −1 + 0 · α + (−1) · β + 0 · γ
β = −1
[kg] : 0 = 0 + 0 · α + 0 · β + 1 · γ
γ=0
=⇒ α = −1;
=⇒
β = −1;
π2 = ν ·
γ=0
1
D · u∞
The dimensionless parametersusually can be expressed in terms of wellknown parameters or
combination of them.
π1 =
f ·D
= Sr
u∞
Strouhalnumber
π2 =
ν
1
=
D · u∞
Re
Reynoldsnumber.
The functional relationship for this problem is
Sr = F (Re)
Only one variation of parametyers is necessary.
Hint: It is possible to choose the dynamic viscosity ν instead of the kinematic viscosity η. It is
η
not necessary to use the density ρ, since ν = is the relevant quantity. In this case 4 variables
ρ
and 2 basic dimensions are existing. In both cases the result is the same.
11.2
a)
L3 t−1 = (M L−2 t−2 )α (M L−1 t−1 )β Lγ
α=1
=⇒
β = −1
V̇ ∼
∆p D
Lη
γ=4
4
131
b)
D ∆p
L ρ u2
2 m
λ=
V̇ η
um η
∆p D
∼ 3 ∼
L
D
D
(with V̇ ∼ um · D 2 )
λ∼
11.3
µ = f(
y
x1/4
1
Re
, g, η, ρ); [g] =
Ansatz :
µ=
m
kg
kg
; [η] =
; [ρ] = 3
2
s
ms
m
y
x1/4
· g α · η β · ργ
=⇒ M : 0 = 0 + β + γ
3
L : 0 = + α − β − 3γ
4
t : 0 = 0 − 2α − β
1
1
1
=⇒ α = ;
β=− ;
γ=
4
2
2
!1/4
y g x3 ρ2
y
1/4
−1/2
1/2
·ρ =
=⇒ µ = 1/4 · g · η
x
x
η2

!1/4 
y g x3 ρ2
T − T∞

= F̃ 
=⇒
TW − T∞
x
η2
11.4
π
cwL DL2
FwL
4
=
π 2
FwW
D
cwW
4 W
ρL 2
ρW
u∞L
Re2L
cwL ηL2
2
2
=
ρW 2
2 ρL
u∞W
Re2W
cwW ηW
2
2
ReL = ReW : cwL = cwW
FwL
= 0, 281
FwW
132
11.5 a) Re = Re : v =
′
′
s
A
v as small as possible in order to use incompressible equations:
A′
A′ = Am :
v ′ = 77, 46 m/s
ρ
cw v 2 A
Fw ′
′
P = ′ Fw v = ′ ρ2 ′
Fw v
2
Fw
cw v Am
2
b)
′
′
Re = Re :
cw = cw
′
P = Fw v = 24, 3 kW
11.6
′
a)
Re = Re :
with
′
′
D2 π
4
V̇ = v
′
D v ρ
Dvρ
=
η
η′
′
′
V̇ ρ 4
V̇ ρ 4
= ′ ′
Dηπ
D η π
=⇒
′
=⇒
′
n D
nD
=
v
v′
′
Sr = Sr :
′
η ρD
V̇ = V̇ ·
= 0, 5 m3 /s
η ρ′ D
′
′
′
′
v D
v D 2 D3
=⇒ n = n · · ′ = n ·
·
v D
v D2 D′3
′
′
V̇
D
=n·
·
D′
V̇
3
= 13, 3
′
b)
c)
′
Eu = Eu :
ρ V̇ 2 D
ρ v2
′
∆p0 = ′ ′ 2 ·∆p0 = ′ ′ 2 ·
ρ v
D
ρ V̇
P = V̇ ∆p0 = 15, 82 kW
′
′
′
P = V̇ ∆p0 = 15 kW
!4
1
s
′
·∆p0 = 527, 34
N
m2
133
M=
P
= 201 Nm
2πn
′
P
= 179 Nm
M =
2 π n′
′
11.7
′
a)
′
F rL = F rL :
v = v·
s
′
L
= 0, 75 m/s
L
′
b)
cW = cW
(Drag coefficient with respect to the cross section perpendicular to the incoming velocity)
FW
ρ 2
v BH
′
FW = 1, 64·104 N
= ρ2 ′
′
′
v2 B H
2
c)
cW = 1, 12
′
v
v
′
√
= q ′ : h = 16 h = 0, 4 m
gh
gh
d)
11.8
a)
The reference values are chosen in such a way that the dimensionless quantities are of the order
of magnitude O(1).
ū =
u
uref
p̄ =
p
∆p
x̄ =
x
L
ȳ =
Introducing into PDE:
∂ 2 (uref · ū)
∂(∆p · p̄)
= η̄ · ηref
∂(L · x̄)
∂(ȳ · h)2
y
h
η̄ =
η
ηref
134
with η̄ = 1:
∆p ∂ p̄
ηref · uref
·
=
L ∂ x̄
h
=⇒
∂ ū
ηref · uref ∂ 2 ū
∂(ȳ h)
·
=
· 2
∂ ȳ
h2
∂ ȳ
∂
L · ηref · uref ∂ 2 ū
∂ p̄
· 2
=
∂ x̄
h2 · ∆p
∂ ȳ
|
{z
}
1stparameter
2 terms =⇒ 1 parameter
b)
f (L, ηref , uref , h, ∆p) = 0
5 variables and 3 basic dimensions =⇒ 2 parameters
The Π-theorem says something about the maximum number of dimensionless parameters. The
method of differential equations gives more informations about the actual problem and the
terms can be neglected. The resulting number of parameters can be smaller.
11.9
ū =
u
v1
v̄ =
v
v1
p̄ =
p
∆p1
ρ̄ =
∂ ū
∂ ū
∂ ū
ρ̄ Sr
+ ū
+ v̄
∂ t̄
∂ x̄
∂ ȳ
Sr =
L1
v1 t1
!
ρ
ρ1
η̄ =
η
η1
x̄ =
∂ p̄
η̄
= − Eu
+
∂ x̄
Re
Eu =
∆p1
ρ1 v12
Re =
x
L1
ȳ =
∂ 2 ū
∂ 2 ū
+
∂ x̄2
∂ ȳ 2
y
L1
!
ρ1 v1 L1
η1
Comment: usually, the election of the refernce values depends on the actual problem.
11.10
a)
u2R
λ TR ∂ 2 T̄
+
η
l2 ∂ ȳ 2
l2
∂ ū
∂ ȳ
K∗ =
b)
!2
= 0
η u2R
λ TR
:
K ∗ = λα η β TRγ uδR lε
choose β = 1
λ TR
l2
t̄ =
t
t1
135
kg
m
s
K
:
:
:
:
0=α+1
0=α−1+δ+ε
0 = −3α − 1 − δ
0 = −α + γ
α = γ = −1
δ = 2
ε = 0
K∗ =
K∗ =
c)
η u2R
λp TR
η cp u2R
η cp u2R
=
(γ − 1)
λ cp TR
λ γRT
cp =
γR
γ − 1
K ∗ = P r · Ma2 (γ − 1))
11.11
η d
∂p
+ ρg +
−
∂x
r dr
du
r
dr
!
= 0
∆p d p̄
d ū
η
η̄ d
−
r̄
+ ρ g ρ̄ ḡ + 2 u∞
L d x̄
R
r̄ d r̄
d r̄
!
η̄ d
d ū
ρ g R2
∆p R2 d p̄
ρ̄ ḡ +
r̄
+
−
L η u∞ d x̄
η u∞
r̄ d r̄
d r̄
|
{z
K1∗
}
| {z }
K2∗
Einflußgrößen: ∆p, ρ, um , η, l, R, g
= 0
!
= 0
136
K = " ∆p1 # " ρ β # uγm D δ
kg
kg
m
[m]
2
3
s m
m
s
kg
m
1
-1
+β
−3β
s
2
kg
m
η [kg/s m]
1
-1
s
-1
+γ
+δ
−γ
+β
−3β
+γ
+δ
−γ
β = −1
γ = − 2, δ = 1 − 3 + 2 = 0
∆p
= Eu
K1 =
ρ u2m
β = − 1, γ =
δ = 1−3+1 =
η
=
K2 =
ρ um D
−1
−1
1
Re
l [m]
kg
m
+β
−3β
1
s
1
s
-2
+β
−3β
+γ
+δ
−γ
K1∗ = Eu · Re ·
K2∗ =
+δ
−γ
g [m/s2 ]
kg
m
+γ
β = 0, γ = 0
δ = 1
l
K3 =
D
β = 0, γ =
δ = −1+2
gD
K4 = 2 =
um
−2
= 1
1
Fr2
R
L
Re
Fr2
11.12
a)
ρ
dw
∂p
= −
− ρ g + η∇2 w
dt
∂z
Reference values: l, u∞ , constants: η, ρ, g
=⇒
ρu2∞ ∂ p̄
u∞
ρ u2∞ dw̄
= −
− ρ g + η 2 ∇2 w̄
l
dt̄
l ∂ z̄
l
=⇒
∂ p̄
1
1
dw̄
= −
−
+
∇2 w̄
2
dt̄
∂ z̄
Fr
Re
with F r 2 =
u2∞
u∞ l
and Re =
gl
η/ρ
·
1
ρ u2∞
l
137
′
b)
u2∞
u∞2
′
F r = F r −→
=
′ =⇒ u∞ = u∞
gl
gl
′
′
l′
u∞
= √
l
10
′
′
′
s
u l
ν
u l
u∞ l
′
√
= ∞′ =⇒ ν = ν ∞
=
Re = Re −→
ν
ν
u∞ l
10 10
′
FW
cW = ρ
u2 A
2 ∞
c)
with A = l·H
F
′
F
′
cW = cW =⇒ ρ
= ′
ρ ′2 ′
u2∞ A
u A
2
2 ∞
2
2
′ ρ u∞ A
′ ρ u∞
=⇒ F = F ′ ′2 ′ = 100 F ′ ′ 2
ρ u∞ A
ρ u∞
P = F · u∞ = 100 F
′
ρ u3∞
ρ′ u′∞2
11.13
a)
cp ρ u
∂T
∂T
+ ρv
∂x
∂y
!


u2
u2
∂
∂ 



+ ρ u 2 + ρ v 2 

∂x
∂y 
∂u
∂2 u
+
η
= ηu
∂ y2
∂y
!2
+ λ
∂2 T
∂ y2
reference values: ρ∞ , u∞ , T∞ , L, η∞ , cp
=⇒ ū =
∞,
λ∞
u
v
T
ρ
x
; v̄ =
; T̄ =
; ρ̄ =
; x̄ = ;
u∞
u∞
T∞
ρ∞
L
ȳ =
y
cp
λ
η
; c¯p =
; λ̄ =
; η̄ =
L
η∞
cp∞
λ∞
introduce:
=⇒ a1 c¯p ρ̄ ū
∂ T̄
∂ T̄
+ ρ̄ v̄
∂ x̄
∂ ȳ
!


ū2
ū2
∂
∂ 



+ a2 ρ̄ ū 2 + ρ̄ v̄ 2 

∂ x̄
∂ ȳ 
∂ ū
∂ 2 ū
+ a4 η̄
= a3 η̄ ū
2
∂ ȳ
∂ ȳ
!2
+ a5 λ̄
∂ 2 T̄
∂ ȳ 2
138
with
a1 =
cp∞ ρ∞ u∞ T∞
;
L
a2 =
ρ∞ u3∞
;
L
a3 = a4 =
η∞ u2∞
;
L2
a5 =
λ∞ T∞
L2
dimensionsless form via division of the equation with e. g. a1 :
b) =⇒ dimensionless parameters
(i) K1 =
ρ∞ u3∞
u2 γ R
a2
L
2
=
= ∞ 2 (γ−1) = (γ−1) M∞
a1
L
γ R c∞
(ii) K2 =
1
a4
η∞ u2∞
η∞
u2∞
L
a3
2
=
=
=
=
(γ−1)
(γ−1) M∞
2
2
a1
a1
L
u∞ ρ∞ L
c∞
Re
(iii) K3 =
λ∞ T∞
η∞
1 1
λ∞
L
a5
=
=
=
2
a1
L
u∞ ρ∞ L η∞ cp∞
Re P r
c)
K1 = K2 = 0 =⇒ (γ−1) = 0 =⇒ γ = 1
11.14
ρref = ρw ;
uref =
s
p0 − pa
ρw
ηref = ηw ;
m
s
∆pref = p0 − pa ;
from Bernoulli
√
D ρw
D
[s]
=
= √
p0 − pa
uref
tref
Dimensionless parameters:
ρ̄ =
lref = D
ρ
η
x
y
p
; η̄ =
; x̄ =
;
; ȳ =
; p̄ =
ρw
ηw
D
D
p0 − pa
ū = s
u
v
t
√
; v̄ = s
; t̄ =
D ρw
p0 − pa
p0 − pa
√
p0 − pa
ρw
ρw
introduced in PDE:
∂ ū p0 − pa
·
+
∂ t̄
ρw D
!
p0 − pa
∂ ū
∂ ū
·
ū
+ v̄
∂ x̄
∂ ȳ
ρw D
139
1 ∂ p̄ p0 − pa
ηw
= −
·
+
·
ρ̄ ∂ x̄
ρw D
ρw
s
∂ 2 ū
∂ 2 ū
+
∂ x̄2
∂ ȳ 2
p0 − pa 1 η̄
·
ρw
D 2 ρ̄
∂ 2 ū
∂ 2 ū
+
∂ x̄2
∂ ȳ 2
∂ ū
∂ ū
1 ∂ p̄
η̄
∂ ū
+ ū
+ v̄
= −
+ K·
=⇒
∂ t̄
∂ x̄
∂ ȳ
ρ̄ ∂ x̄
ρ̄
ηw
·
with K =
ρw
b)
uref =
s
s
p0 − pa
ρw
!
1
· p − p
0
a
ρw D
!
ρw
1
p0 − pa D
=⇒ K =
1
1
ηw
=
ρw uref D
Re
Friction forces
Inertia forces
K =
11.15
a)
conti:
∂u ∂v
+
= 0
∂x ∂y
∂u
∂u
+ v
momentum: ρ u
∂x
∂y
qnergy: ρ cp
!
∂T
∂T
u
+ v
∂x
∂y
= η
!
∂2u
∂y 2
= λ
∂2T
∂y 2
dimensionsless parameters:
ū =
u
v
ρ
x
y
; v̄ =
; ρ̄ =
; x̄ = ; ȳ = ;
u∞
u∞
ρ∞
L
L
η̄ =
b)
η
cp
λ
T
; c¯p =
; λ̄ =
; T̄ =
η∞
cp ∞
T∞
λ∞
u∞
conti:
L
momentum: ρ∞
∂ ū
∂v̄
+
∂ x̄
∂ ȳ
!
= 0
=⇒ no parameter
∂ ū
∂ ū
u∞2
ρ̄ ū
+ v̄
L
∂ x̄
∂ ȳ
∂ ū
∂ ū
+ v̄
ρ̄ ū
∂ x̄
∂ ȳ
!
∂ 2 ū
= η̄ ¯2
∂y
|
K1 =
!
= η∞ η̄
η∞ u∞ L
L2 ρ∞ u2∞
!
{z
}
η∞
1
=
L ρ∞ u∞
Re
K1
u∞ ∂ 2 ū
L2 ∂ y¯2
140
ρ∞ cp∞ u∞ T∞
∂ T̄
∂ T̄
energy:
ρ̄ c¯p ū
+ v̄
L
∂ x̄
∂ ȳ
∂ T̄
∂ T̄
ρ̄ c¯p ū
+ v̄
∂ x̄
∂ ȳ
!
=
λ∞ T∞ ∂ 2 T̄
λ̄
L2
∂ ȳ 2
∂ 2 T̄
λ∞ T∞ L
λ̄
=
L2 ρ∞ cp∞ u∞ T∞
∂ ȳ 2
{z
|
K2 =
!
K2
!
}
η∞
λ∞
1
1
=
·
L ρ∞ cp∞ u∞ η∞
P r Re
c) dimensionsless PDE.:
conti:
∂v̄
∂ ū
+
= 0
∂ x̄
∂ ȳ
∂ ū
∂ ū
momentum: ρ̄ ū
+ v̄
∂ x̄
∂ ȳ
∂ T̄
∂ T̄
energy: ρ̄ c¯p ū
+ v̄
∂ x̄
∂ ȳ
!
!
1
=
η̄
Re
!
1
1
∂ 2 T̄
=
·
λ̄
P r Re
∂ ȳ 2
konstante fluid properties: ρ̄ = c¯p = λ̄ = η̄ = 1
Comparison between momentum and energy equation:
By replacing T with u and proposing
Pr = 1
the energy and the momentum equation are the same
i.e.
∂ 2 ū
∂ y¯2
η∞ cp∞
= 1
λ∞
!
141
14 Potential flows
14.1
a) Parallel flow x-direction: F1 (z) = u∞ z
Source in x = −a: F2 (z) =
Sink in x = +a : F3 (z) =
E
ln((x + a) + iy)
2π
−E
ln((x − a) + iy)
2π
Superposition: F (z) = u∞ z +
E
ln((z + a) − ln(z − a))
2π
E
dF
= u∞ +
u − iv =
dz
2π
w̄ =
E
2π
= u∞ +
=
−2a
2
z − a2
u∞ −
1
1
−
z+a z−a
with b = x2 − y 2 − a2
Ea
b − 2ixy
π (b − 2ixy)(b + 2ixy)
=
u∞ −
Ea x2 − y 2 − a2 − 2ixy
π (x2 − y 2 − a2 )2 + 4(xy)2
u=
u∞ −
x2 − y 2 − a2
Ea
π (x2 − y 2 − a2 )2 + 4(xy)2
v=
−
s
for u = 0, v = 0 : y = 0, xs = ±
2xy
Ea
π (x2 − y 2 − a2 )2 + 4(xy)2
Ea
+ a2
πu∞
b) The contour is a streamline ψ = const crossing the stagnation points.
ψ = u∞ y +
for y = 0, x = ±xs
E
y
E
y
arctan
−
arctan
2π
x + a 2π
x−a
(from table)
follows ψ = 0
=⇒ Equation for the contour
u∞ y +
y
E
y
E
arctan
−
arctan
=0
2π
x + a 2π
x−a
142
Comment: This equation additionally describes the stagnation streamlines on the x-axis.
14.2
a)
b)
c)
d)
▽ · ~v
4xy
2
0
0
| ▽ ×~v |
y 2 − x2
0
-2
0
The streamfunction exists for c) and d), the potential exists for b) and d).
Determination of the stramfunction:
c)
y2
udy + f (x) =
+ f (x)
2
∂ψ
= −f ′ (x) = −x
v= −
∂x
1 2
ψ=
(x + y 2 ) + c
2
ψ=
R
d)
1
ψ = (y 2 − x2 ) + c
2
Determination of the potential:
b)
x2
+ f (y)
udx + f (y) =
2
∂φ
v=
= f ′ (y) = y
∂y
1 2
(x + y 2) + c
φ=
2
φ=
R
d)
φ = xy + c
143
14.3
a) Condition for the stagnation point: u = 0, v = 0
∂ψ
Γ
u=
=−
∂y
2π
y
2y
+
2
2
(x − a) + y
(x + a)2 + y 2
2
1
Γ
+
u=− y
2
2
2π
(x − a) + y
(x + a)2 + y 2
Γ
∂ψ
=
v=−
∂x
2π
for y = 0 :
Γ
v=
2π
stagnation point:
a
xs = ,
3
alternatively:
vθ =
Γ
2Γ
Γ
: vθ1 = vθ2 →
=
→ r2 = 2r1
2πr
2πr1
2πr2
2
a
2a = r1 + r2 → 2a = 3r1 → r1 = a → xs =
3
3
b)
u2 + v 2
cp (x, y = 0) = 1 − 2
2
u(0,0) + v(0,0)
=⇒ cp (x, y = 0) = 1 −
v(0,0) =
v2
2
v(0,0)
mit u = 0 for y = 0.
,
Γ 1
(see ⋆) d.h. cp(0,0) = 0
2π a
= 0 wenn y = 0
x−a
2(x + a)
+
(x − a)2 + y 2 (x + a)2 + y 2
Γ 3x − a
2
1
=
+
x−a x+a
2π x2 − a2
v = 0, if 3x − a = 0 →
!
!
⋆)
ys = 0
!
144
3x − a 2
 x2 − a2 

with ⋆) cp (x, y = 0) = 1 − 

1 
a

3xa − a2
=⇒ cp (x, y = 0) = 1 −
x2 − a2
!2
for x 6= a, −a
14.4
a)
F (z) =
=
E
2 u∞ 3
√ z2 +
ln z
3 L
2π
2 u∞ 3 i 3 θ
E
√ r2e 2 +
(ln r + ln eiθ )
3 L
2π
= Re(r, θ) + iIm(r, θ) = φ(r, θ) + iψ(r, θ)
=⇒ φ(r, θ) =
E
2 u∞ 3
3
√ r 2 cos θ +
ln r
3 L
2
2π
=⇒ ψ(r, θ) =
b)
3
2 u∞ 3
E
√ r 2 sin θ +
θ
3 L
2
2π
r
∂φ
r
3
E
vr =
= u∞
cos θ +
∂r
L
2
2πr
r
r
3
1 ∂φ
= −u∞
sin θ
vθ =
r ∂θ
L
2
c) stagnation point at
(x = −L, y = 0) →
2
r = L, θ = π
3
=⇒ vθ = 0 = −u∞
s
L
sin π
L | {z }
=0
145
=⇒ vr = 0 = u∞
=⇒ u∞ =
s
L
E
cos
π} +
|
{z
L
2πL
= −1
E
→ E = 2πu∞L
2πL
d)
3
2
2
L2
E
E 2
π} +
ψ r = L, θ = π = u∞ √ sin
π=
|
{z
3
3
2π 3
3
L
=0
2
2
!
=⇒ ψ r = L, θ = π = πu∞ L = ψk (r, θ)
3
3
3
3
2
2
r2
=⇒ πu∞ L = u∞ 1 sin θ + u∞ Lθ
3
3
2
L2
π − 23 θ
=⇒ rk (θ) = L
sin 32 θ
!2
3
14.5
a) | ▽ ×~v | = 0 : Potential exists.
b)
u=
U
x,
L
U
v=− y
L
stagnation points:
u=v=0:
x=y=0
pressure coefficient:
x2 + y 2
p − pref
u2 + v 2
=
1
−
cp = ρ
=1− 2
2
2
2
uref + vref
x2ref + yref
~vref
2
Isotach:
~v 2 = u2 + v 2 =
2
2
x +y =
2
(x2 + y 2 )
~vL
U
!2
U
L
146
Circles around the origin with radius
|~v |L
U
c)
u1 = 4 m/s,
v1 = −4 m/s,
|~v1 | = 5.66 m/s
ρ 2
= 0.86 · 105 N/m2
p1 = pref + cp1 ~vref
2
d)
t=
ψ = const :
e)
f)
Z
L x2
dx
= ln
U x1
x1 u
x2 = 5.44 m
x2
x1 y1 = x2 y2 ,
y2 = 0.74 m
ρ 2
p1 − p2 = (cp1 − cp2 ) ~vref
= 0.442 · 105 N/m2
2
147
14.6
a)
#
"
E
h
y
ψ = u∞ y +
θ + c = u∞ y + arctan
2π
π
x
u = u∞
"
h x
1+
π x2 + y 2
v = u∞
stagnation point: u = v = 0 :
+c
#
h y
π x2 + y 2
h
xs = − , ys = 0
π
u(xs , h) = u∞
π2
1 + π2
v(xs , h) = u∞
π
1 + π2
b) Contour: Streamline crossing the stagnation point
rk =
h π−θ
h θ′
=
π sin θ
π sin θ
with
θ′ = π − θ
148
c)
u2 + v 2
h 2x + πh
cp = 1 −
=− 2
u2∞
π x + y2
cpk
sin (2θ′ )
sin θ′
=
−
θ′
θ′
!2
d)
"
cp = const :
h
x+
πcp
#2
h
+ y = (1 − cp )
πcp
2
!
√
h 1 − cp
h
, 0 with radius
Circles around −
πcp
πcp
e)
1
cp = :
2
f)
2h
x+
π
√
!2
h
+y =2
π
2
!2
u2 + v 2
=k
u∞

h 2
kh 2



π 
x −
 + y2 =  π 

 k2 − 1 
k2 − 1 

Circles around
!
h
kh
, 0 with radius
2
π(k − 1)
π(k 2 − 1)
g)
v = u∞
u∞
h y
>
2
2
πx +y
2
!2
149
h
x + y−
π
2
h)
tan α =
h
x+
2π
!2
!2
<
h
π
!2
v
=1
u
h
+ y−
2π
!2
1
=
2
h
π
!2
i) Acceleration on the x-axis:
∂u
h
du
=u
= −u∞
b=
dt
∂x
π
1
h 1
+
2
x
π x3
!
150
db
=0:
dx
xmax = −
bmax = −
4 π 2
u
27 h ∞
ψ=0:
y=0
3h
2π
14.7
a)
x2 + y 2 = R2
r=
√ 2
x + y2 → ∞ :
ψ → u∞ y(Parallel flow)
Streamfunction describes the flow around a cylinder.
b)
cp = 1 −
vr = u∞
"
vθ = −u∞
vr2 + vθ2
u2∞
R
1−
r
"
2 #
R
1+
r
cos θ
2 #
sin θ
r = R : cp = 1 − 4 sin2 θ
151
c)
△t =
Z
−2R
−3R
u(x, 0) = u∞
△t =
14.8
a)
R x−R
1
x + ln
u∞
2 x+R
dx
u(x, 0)
R2
1− 2
x
−2R
=
−3R
!
R
(1 + 0.5 ln 1.5)
u∞
ρ
ρ
ρgh∞ + u2∞ = ρgh(θ) + ~v 2
2
2
r = R : ~v 2 = vθ2 = 4u2∞ sin2 θ
h(θ) − h∞ =
b) Stagnation points:
u2∞
(1 − 4 sin2 θ)
2g
θ = 0 and θ = π
h = h∞ +
u2∞
= 6.05 m
2g
c)
θmin =
hmin = h∞ −
π 3π
,
2 2
3u2∞
= 5.85 m
2g
14.9
a)
u∞ H 2
u∞ H 2
= u∞ (x + iy) + 0.8 2
(x − iy)
F (z) = u∞ z + 0.8
z
x + y2
152
=⇒ ψ = u∞ y − 0.8
u∞ H 2
y
x2 + y 2
ψk = ψ(x = 0, y = H) = u∞ H − 0.8
u∞ H 2
H = 0.2Hu∞
H2
Contour equation:
0.2Hu∞ = u∞ y − 0.8
=⇒
u∞ H 2
y
x2 + y 2
y
H2 y
− 0.8 2
= 0.2
H
x + y2 H
b)
ρ
P
= ~v 2
2
V̇
u=
=⇒
ρ
V̇
P = (u2 + v 2 )
2
B
(based on the width)
∂ψ
(x2 + y 2 ) − 2y 2
x2 − y 2
2
= u∞ − 0, 8H 2u∞
=
u
−
0.8H
u
∞
∞
∂y
(x2 + y 2 )2
(x2 + y 2 )2
v=−
−2x
xy
∂ψ
= −0.8H 2 u∞ y 2
= 1.6u∞ H 2 2
2
2
∂x
(x + y )
(x + y 2 )2
I. far before the mountain (x → −∞)
u = u∞ , v = 0 mit
V̇
ρ
= Hu∞ → PI = Hu3∞
B
2
II. on the top (x = 0, y)
H2
u = u∞ + 0.8u∞ 2 ;
y
ρ
P =
2
Z
2H
ρ
P = u3∞
2
Z
2H
H
H
ρ
u (y)dy =
2
3
Z
2H
H
v=0
(u∞ + 0.8u∞
H2 3
) dy
y2
4
6
H2
2H
3H
(1 + 3 · 0.8 2 + 3 · 0.8 4 + 0.8 6 )dy
y
y
y
153
ρ
H2
H4
H6
= u3∞ y − 3 · 0.8
− 3 · 0.82 3 − 0.83 5
2
y
3y
5y
ρ
P = 2.86 Hu3∞
2
c) Parallel flow + dipole =
ˆ cylinder flow
14.10
a)
"
2 #
−
"
2 #
cos θ
R
ψ = u∞ r sin θ 1 −
r
vr = u∞
vθ = −u∞
r=R:
R
1−
r
Γ
ln r
2π
R
Γ
1 + ( )2 sin θ +
r
2πr
vθ |Wirbel = vt =
Γ
2πR
Γ = 2πRvt
b) Flow field vt = u∞ :
ψ = u∞
"
R2
r sin θ 1 − 2
r
!
− R ln r
#
!2H
H
154
vr = u∞
vθ = u∞
"
"
#
R2
1 − 2 cos θ
r
!
R2
R
− 1 + 2 sin θ
r
r
#
contour:
ψ = ψK = ψ(r = R, θ = 0) = u∞ (−R ln R)
⇒ u∞
"
R2
r sin 1 − 2
r
!
r
− R ln
R
2 stagnation points on the contour (r = R):
π 5π
θs = , ; no free stagnation points
6 6
c)
dFx = −pLR cos θdθ
dFy = −pLR sin θdθ
#
= 0 ()
155
ρ
p = p ∞ + cp u ∞ 2
2
r=R:
Fx = −LR
Z
"
2π
(
Z
2π
0
Fy = −LR
0
cp = 1 −
vt
− 2 sin θ
u∞
vt
ρ 2
u∞ 1 −
− 2 sin θ
2
u∞
(
"
2 #
ρ 2
vt
u∞ 1 −
− 2 sin θ
2
u∞
2
)
+ p∞ cos θdθ = 0
2 #
)
+ p∞ sin θdθ
= −2πρLRvt u∞ = −ρu∞ ΓL
14.11
a)
2
3
iθ
F = Az = A re
2
3
= Ar
2
3
2
2
cos θ + i sin θ = φ + iψ
3
3
2
2
ψ = Ar 3 sin θ
3
vr =
1
2
2
1 ∂ψ
= Ar − 3 cos θ
r ∂θ
3
3
vθ = −
1
2
∂ψ
2
= − Ar − 3 sin θ
∂r
3
3
1
2
3 1
θ = 0, r = l : vr = Al− 3 = u1 −→ A = u1 l 3
3
2
b)
− 2
r
~v 2
cp = 1 − 2 = 1 −
u1
l
3
156
c)
Lines p = const.:
r = const. → cp = const. → p = const.
Circles around the origin
d)
Streamlines:
2
2
ψ = Ar 3 sin θ = const.
3
=⇒
2
3
rs =
ψ
2
A sin θ
3
=⇒
rs (θ) =




3
2
ψ
2
A sin θ
3



157
e)
y
Symmetrieachse
Q
| v |= u1
x
l
14.12
a)
Condition:
ω=0→
∂2ψ ∂2ψ
+ 2 =0
∂x2
∂y
∂ 2 ψ(x, y)
∂
∂2ψ
∂
yu∞L
=
0
→
=
2
2
∂y
∂x
∂x
∂x
∂
h′ (x)
−yu∞ L 2
∂x
h (x)
integrate 2 times: −
!
d
=0→
dx
1
h(x)
h′ (x)
h2 (x)
!
!!
=0
=0
1
= C1 x + C2
h(x)
B.C.:
x = L, h = L
x = 3L,
=⇒
=⇒
1
h= L
3
C2 = 0,
−
=⇒
1
= C1 L + C2
L
−
3
= 3C1 L + C2
L
1
L2
=⇒
∂ψ
;
∂x
ψ=
C1 = −
h(x) =
b)
u=
∂ψ
;
∂y
v=−
u∞
xy
L
L2
x
158
x
u = u∞ ;
L
v = −u∞
y
L
c)
V̇ = ψ(y=h) − ψ(y=−h) |x=L
V̇ = (h1 + h1 )Bu∞
V̇ = 2U∞ LB
159
15 Laminar boundary layers
15.1
−ρu2∞ δ + ρ
Z
0
δ
△ṁ = ρ
Z
0
δ
Z
u2 dy + △ṁu∞ =
Z
0
δ
0
τ (x′ , y = 0)dx′
(u∞ − u)dy
u
u
1−
dy = δ2 = −
u∞
u∞
Z
x
0
15.2
a) ReL = 3.33 · 105 : laminar boundary layer
b)
y=0:u=v=0
y → ∞ : u → u∞
c) from boundary layer equation :
x
τ (x′ , y = 0) ′
dx
ρu2∞
160
∂τ
= 0 for y = 0 and y = δ
∂y
d) from Blasius solution:
5L
δ(x = L) = √
ReL
0.664
cf = √
Rex
δ(x = L) = 4.33 mm
Fw = 2
Z
0
L
Bτw dx = ρu2∞ B
Z
0
L
cf dx = 0.144 N
15.3
Determination of the coefficients a0 , a1 , a2 by using the boundary conditions:
B.C.1: no slip u(x, y = 0) = 0
B.C.2: boundary layer edge u(x, y = δ) = u∞
∂p
∂ 2 u(x, y = 0)
=
=0
R.B 3: at the wall (from x-momentum) η
2
∂y
∂x
y
If additional boundary conditions are necessary, a steady transition at = 1 ac be assumed,
δ
i.e. ∂ n u = 0 with n ≥ 1
∂y n y=δ
161
=⇒ from B.C.1 follows a0 = 0
from B.C.2 follows a1 + a2 = 1
from B.C.3 follows
∂2u
1 ∂ 2 (u/u∞)
1
=
u
= 2u∞ 2 a2 = 0
∞ 2
2
2
∂y
δ ∂(y/δ)
δ
=⇒ a2 = 0,
=⇒
y
u(x, y)
=
u∞
δ
a1 = 1
linear distribution.
Permutation of the order of boundary conditions, e,g,:
B.C.1 u(x, y = 0) = 0
B.C.2 u(x, y = δ) = u∞
∂u B.C.3
=0
∂y y=δ
=⇒ a0 = 0; a1 = 2; a2 = −1
parabolic distribution
Since the approximation of the velocity profiles is not an exact solution of the boundary layer
equations, the boundary conditions are not satisfiable in all cases. Hence, it has to be paid
attention, that the physical boundary conditionsa re satisfied first.
Computation of the boundary layer thickness δ(x):
u∞ ∂u/u∞ u∞
∂u =η
=η
τw = η
∂y y=0
δ ∂y/δ y=0
δ
y
y
u∞ − u∞
u(u∞ − u) = u∞
δ
δ
with the equation
B
=⇒
Z
Z
x
0
x
0
Z
τw (x)dx = Bρ
u∞
η
dx = ρu2∞
δ(x)
0
Z
0
1
δ(x)
=
u2∞
2 !
y
y
−
δ
δ
u(u∞ − u)dy
2 !
y
y
−
δ
δ
δ
d
y
δ
162
ηu∞
Z
x
0
dx
= ρu2∞ δ(x)
δ(x)
2
1 y
2 δ
= ρu2∞ δ(x)
3 !1
1 y
−
3 δ
0
1
6
Differentiating gives
1ρ
dδ(x)
1
=
u∞
δ(x)
6η
dx
=⇒ dx =
Integration:
x=
1ρ
u∞ δ(x)dδ(x)
6η
1 ρ
u∞ δ 2 (x)
12 η
=⇒ δ(x) =
=⇒ δ(x) =
√
Compare with Blasius solution: δ(x) = 5.2
s
s
12νx
u∞
s
νx
νx
12
≈ 3, 5
u∞
u∞
s
νx
u∞
15.4
a)
∂p
in the flow:
∂x
frictionless outer flow: ρua
=⇒
∂p
∂ua
=−
∂x
∂x
∂p
∂ua
= −ρua
= −ρ(ua1 − C(x − x1 )2 ) · (−2C(x − x1 ))
∂x
∂x
∂p
= +2ρC(x − x1 )(ua1 − C(x − x1 )2 )
∂x
∂p =0
∂x x
1
∂p ∂x x2
= 2ρC(x2 − x1 )(ua1 − C(x2 − x1 )2 )
163
b) 4 B.C:
I
y
=0
δ
II
u=0
y
=1
δ
H.B. =⇒ a0 = 0
u = ua
III
y
=0
δ
∂p
∂2u
=η 2
∂x
∂y
IV
y
=1
δ
∂u
=0
∂y
at the wall
u =
a1
y
δ
II:
1 = a1 + a2 + a3
IV:
0 = a1 + 2a2 + 3a3
III:
2a2
∂p
= η ua 2
∂x
δ
=⇒
=⇒
a3 = −
a1 =
+ a3
3 !
y
δ
!
1
y
+ 2a2 2 + 6a3 3
δ
δ
a2 =
ua
ua
ua
1 δ 2 ∂p
2 η ua ∂x
δ 2 2 ρ C (x − x1 )(ua1 − C (x − x1 )2 )
a2 (x) =
2η
(ua1 − C · (x − x1 )2 )
a2 (x) =
II / IV:
y
δ
1
y
y2
a1
+ 2a2 2 + 3a3 3
δ
δ
δ
∂u
=
∂y
∂2u
=
∂y 2
+ a2
2
1
1
−
a2
2
2
1
3
−
a2
2
2
δ2 ρ C
(x − x1 )
η
164
therefore:
3
1 δ2 ρ C
−
(x − x1 )
2
2
η
a1 (x) =
1
1 δ2 ρ C
a3 (x) = −
−
(x − x1 )
2
2
η
15.5
a)
1 dua
τ (y = 0)
dδ2
=0
+
(2δ2 + δ1 ) +
dx
ua dx
ρu2a
•
δ1
=
δ0
Z
0
1
u
y
1−
d
ua
δ
=
Z
y
1−
δ0
1
0
=⇒ δ1 =
•
•
δ2 Z 1 u
u
y
1−
d
=
δ0
ua
δ
0 ua
=
Z
1/2
1 !
!
y
−
δ
=⇒ δ2 =
1
δ0
6
0
y
d
δ
y
d
δ
dδ2
= 0, since δ0 = const.
dx
into the above equation: ρ ua
dua
(2δ2 + δ1 ) = τw
dx
from the x-momentum equation for y = δ0 :
ρua
dp
dua
=−
dx
dx
dp
(2δ2 + δ1 )
dx
2 !
dp
2x
d
x
1−k
= −p0 k 2
= p0
dx
dx
l
l
=⇒ τw = −
with
=⇒ τw = p0 k
15.6
conti.:
∂u ∂v
= 0 → v = va = konst
+
∂x ∂y
|{z}
=0
=
3 !1
y 2 y
−
δ 3 δ
2
0
1
δ0
3
y
δ
1
2
4
δ0 x
2x 2
δ0 = p0 k 2
2
l 3
3
l
=
3/2
2 y
3 δ
2 !1
1 y
−
2 δ
0
165
momentum equation:
dI~
=
dt
momentum balance x-direction:
Z
ρ ~u (~u · ~n) dA =
X
F~KV = F~R
−ρ va ua dx B − ρ u2(y) dy B + ρ u2 (y) dy B + ρ va 0 dx B = τy=0 dx B
−ρ va ua dx B = τy=0 dx B
with
τy=0 = −η
du
ua
= −η
dy
δ
=⇒
ρva =
va
η
ρ
=⇒
va =
ua
u( y )
y
t
x
va
dx
15.7
a) boundary conditions:
y
u
v
=0:
= 0,
=0
δ
ua
ua
u
y
=1:
=1
δ
ua
from boundary layer equation
∂u
∂u
+v
ρ u
∂x
∂y
y
=0:
δ
y
=1:
δ
!
=η
u=v=0:
∂u
∂u
=
=0:
∂x
∂y
∂2u
:
∂y 2
∂ 2 (u/ua)
=0
∂(y/δ)2
∂ 2 (u/ua)
=0
∂(y/δ)2
η
ρδ
166
frictionless outer flow:
y
=1:
δ
∂(u/ua )
=0
∂(y/δ)
τ∼
3
u
y
d
ua
δ
u
y
y
−2
=2
ua
δ
δ
+
4
y
δ
b)
δ1
=
δ
Z
Z
1
δ2
=
δ
0
1
0
1−
=
u
u
y
1−
d
ua
ua
δ
3
10
=
37
315
dδ2 τ (y = 0)
=0
+
dx
ρu2a
Integration:
δ
5.84
=√
x
Rex
ηua
ηua d(u/ua) = −2
τ (y = 0) = −
δ d(y/δ) y/δ = 0
δ
2
cw =
L
Z
0
L
2
τw
dx = −
2
ρua
L
Z
0
L
1.371
τ (y = 0)
dx = √
2
ρua
ReL
15.8
a)
y
u(x, y)
= a0 + a1
ua (x)
δ0
B.C. y = 0 =⇒ u = 0 =⇒ a0 = 0
y = δ0 =⇒ u = ua (x) =⇒ a1 = 1
=⇒
y
u(x, y)
=
ua (x)
δ0
b)
1 dua
τw
dδ2
+
(2δ2 + δ1 ) = 2
dx
ua dx
ρua
167
displacement thickness
δ1
=
δ0
Z
1
Z
1
δ2
momentum thickness
=
δ0
1−
0
0
τw = η
u
1−
ua
∂u ∂y u
y
d
ua
δ0
u
y
d
ua
δ0
= 1/2
=
1
6= f (x) =⇒
6
u
∂
ua
ua
=η
δ ∂ y
y=0
δ
dua
dx
=⇒
=η
y=0
6 η 1
dua
=
dx
5 ρ δ02
=⇒ ua (x) =
6 η x
+C
5 ρ δ02
B.C.: x = 0 =⇒ ua (x) = 0 =⇒ C = 0
=⇒ ua (x) =
6 η x
5 ρ δ02
c)
F =
Z
L
0
=⇒ F =
τw (x)Bdx,
6 η2 B
5 ρ δ03
Z
0
L
τw = η
xdx =
15.9
a) Solution see 15.7
A)
δ1
3
=
δ
8
39
δ2
=
δ
280
4.641
δ
=√
x
Rex
1.293
cw = √
ReL
ua (x)
δ0
3 η 2 BL2
5 ρ δ03
ua
δ
dδ2
=0
dx
168
B)
δ1
2
= 1 − = 0.363
δ
π
2 1
δ2
= − = 0.137
δ
π 2
s
2π 2
δ
4.795
4−π
= √
=√
x
Rex
Rex
q
2 2 − π/2
1.310
cw = √
=√
ReL
ReL
b)
A)
δ(x = L) = 3.28 mm
ρ
Fw = cw u2a 2LB = 0.91 N
2
B)
δ(x = L) = 3.39 mm
Fw = 0.93 N
169
16 Turbulent boundary layers
16.1
a)
xkrit =
νRecrit
= 0, 167 m
u∞
b)
yq
Rex = 1, 095
ξ =
x
u
= 0, 36 :
u = 16, 2 m/s
u∞
(see Diagram ξ = f (u/u∞), script chapter 15.4)
x = 0, 15 m
:
Rex = 1, 095
:
yq
x
c)
d)
Rex = 4, 5 · 105
y = 2, 45 · 10−4 m
170
16.2
a)
with the product rule
∂u ∂u2 ∂(uv)
ρ
+
+
∂t
∂x
∂y
!
∂u
∂u
∂v
∂u
=ρ
+ 2u
+u
+v
∂t
∂x
∂y
∂y
!
∂u
∂u ∂v
∂u
∂u
ρ
+ ρu
+u
+v
+
∂t
∂x
∂y
∂x ∂y
!
!
with ∂u/∂x + ∂v/∂y = 0 (conti.) follows
∂u ∂u2 ∂(uv)
ρ
+
+
∂t
∂x
∂y
!
!
∂u
∂u
∂u
=ρ
.
+u
+v
∂t
∂x
∂y
b)
′
u=u+u,
′
v =v+v ,
p=p+p
′
∂(ū + u′ ) ∂(ū + u′ )2 ∂((ū + u′ )(v̄ + v ′ ))
ρ
+
+
∂t
∂x
∂y
∂ 2 (ū + u′ ) ∂ 2 (ū + u′ )
∂(p̄ + p′ )
+η
+
fx −
∂x
∂x2
∂y 2
!
=
!
The computational rules 2 and 4 result with f x = fx
(u + u′ ) ∂(u + u′ )2 ∂((u + u′ )(v + v ′ ))
ρ
+
+
∂t
∂x
∂y
!
=
∂(p + p′ )
∂ 2 (u + u′ ) ∂ 2 (u + u′ )
fx −
+η
+
∂x
∂x2
∂y 2
!
∂u ∂u′ ∂(u + u′ )2 ∂((u + u′ )(v + v ′ ))
+
+
+
ρ
∂t
∂t
∂x
∂y
!
=
=
171
∂p ∂p′
∂ 2 u ∂ 2 u′ ∂ 2 u ∂ 2 u′
fx −
−
+η
+
+ 2+
∂x
∂x
∂x2
∂x2
∂y
∂y 2
!
with u′ = 0 and p′ = 0 follows
∂u ∂(u + u′ )2 ∂((u + u′ )(v + v ′ ))
ρ
+
+
∂t
∂x
∂y
!
= fx −
∂p
∂2u ∂2u
+ η 2 + 2.
∂x
∂x
∂y
Regard
∂(u + u′ )2
∂x
und
∂(u + u′ )(v + v ′ )
∂y
follows
∂(ū + u′ )2
∂(u¯2 + 2ūu′ + u′ 2 )
∂u2 ∂(2ūu′ ) ∂u′ 2
∂u2 ∂u′ 2
=
=
+
+
=
+
∂x
∂x
∂x
∂x
∂x
∂x
∂x
and
∂(u + u′ )(v + v ′ )
∂(uv + uv ′ + u′ v + u′ v ′ )
=
=
∂y
∂y
∂ ūv̄ ∂u′ v ′
∂(ūv̄) ∂(uv ′ ) ∂(u′ v) ∂(u′ v ′ )
+
+
+
=
+
.
∂y
∂y
∂y
∂y
∂y
∂y
put into the Reynolds averaged x-momentum equation
∂u ∂ ū2 ∂u′ 2 ∂(ūv̄) ∂(u′ v ′ )
ρ
+
+
+
+
∂t
∂x
∂x
∂y
∂y
!
∂p
∂2u ∂2u
= fx −
+η
+
∂x
∂x2 ∂y 2
!
algebraic transformation
∂u ∂ ū2 ∂(ūv̄)
ρ
+
+
∂t
∂x
∂y
16.3
a)
!
!
∂p
∂u′ 2 ∂(u′ v ′ )
∂2u ∂2u
= fx −
+η
+
+
ρ
−
−
∂x
∂x2 ∂y 2
∂x
∂y
cw1
Fw1
=
Fw2
cw2
!
172
u∞
Re1
103 cw1
Re2
103 cw2
2, 10
2, 76
3, 19
2 · 105
4 · 105
8 · 105
2, 97
2, 10
2, 76
0, 4m/s 4 · 105
0, 8m/s 8 · 105
1, 6m/s 1, 6 · 106
b) Re1 = 1, 96 · 105
cw1 = cw2 = 3, 0 · 10−3
1)
Re2 = Re1 :
u∞2 = 0, 392 m/s
2)
Re2 ≈ 1, 3 · 106 :
u∞2 = 2, 6 m/s
3)
Re3 ≈ 9 · 106 :
u∞2 = 18 m/s
(see diagram, chapter 16.2)
Fw1
Fw2
0, 707
1, 313
1, 156
173
17 Boundary layer separation
17.1 The coefficients a0 (x), a1 (x), a2 · x, a3 (x) of the polynomial are determined first. At
dp
least 4 boundary conditions are necessary. Since
is unknown the momentum equation at
dx
the wall gives no further information. The condition for separation gives an additional boundary
condition
∂u/ua = 0.
Separation occurs at
∂y/δ y = 0
for x = xa
y = 0 =⇒ u = 0
=⇒ a0 (x) = 0
y = δ =⇒ u = ua (x)
=⇒ 1 = a1 (x) + a2 x + a3 (x)
∂u/ua y = δ =⇒
= 0 =⇒ 0 = a1 (x) + 2a2 x + 3a3 (x)
∂y/δ y = δ
y = 0 =⇒
∂u/ua = 0 =⇒ 0 = a1 (xa )
∂y/δ y = 0
At x = xa
a2 xa + a3 (xa ) = 1 ;
=⇒ a2 =
3
xa
2a2 xa + 3a3 (xa ) = 0
a3 (xa ) = −2
y
u(xa , y/δ(xa))
=3
=⇒
ua (xa )
δ(xa )
!2
y
−2
δ(xa )
!3
174
17.2
cw =
2
Rπ
0 dFw (α)
ρ 2
u DL
2 ∞
dFw (α) = dF (α) cos(α) = p(α)L
D
cos(α)dα
2
ρ
p(α) = cp (α) u2∞ + p∞
2
2
ρ
0 ≤ α ≤ π : p(α) = (1 − 4 sin2 α) u2∞ + p∞
3
2
2
2
π ≤ α ≤ π : p(α) = 1 − 4 sin2
π
3
3
cw =
√
ρ
ρ 2
u∞ + p∞ = −2 u2∞ + p∞
2
2
3
17.3
a)
Determination of the coefficients ai with the following boundary conditions:
(I)
(II)
(III)
(IV)
u(y = 0) = 0
u(y = δ) = ua
∂ 2 u dp
=η
momentum at the wall: y → 0 : u → 0, v → 0,
dx
∂y 2 y = 0
τ (y = δ) = 0
dpa
dua (x)
u2
2x
dp
=
= −ρua (x)
= −2ρ ∞ sin
dx
dx
dx
R
R
175
Using the ansatz:
u ∂
dp
ηua
ua
ρu2a
u2∞
x
x
ua =
=η 2 2a
=
2a
=
−4ρ
sin cos
2
2
2
2
y dx
δ
δ
25x
R
R
R
∂
y
δ
=0
δ
2
=⇒ a2 = −
25 x
x
cot
2 R
R
Following the boundary conditions
a0 = 0
a1 + a2 + a3 = 1
a1 + 2a2 + 3a3 = 0
1
a3 = − (1 + a2 )
2
3 1
a1 =
− a2
2 2
=⇒
b)
Separation: τw = 0
τw = η
∂u ∂y u !
ua = 0
y ∂
y
δ
δ
∂
!
=0
y=0
=⇒
a1 = 0
x
cos
25 x
RA = 3 ;
−
2 R sin x
R A
x
assumption:
R
A
π
≈
2
=⇒
(≈
x
R
A
π
x
−
2
R
A
6
=−
25
π
assumption is justifiabe)
2
ΘA = 98o
=⇒
x
R
A
= 1.71 ,
176
17.4
a)
equilibrium of forces:
Fw = G (buoyancy can be neglected)
sphere:
cw
2
3
ρL 2 πDK
πDK
v
=ρ
g
2
4
6
v = Re
DK =
νL
DK
v
u
2
3u
t18Re ρL νL
ρ g
DKmax = 6, 81 · 10−2 mm
Re = 0, 5 :
cylinder (Lnge L):
cw
ρL 2
πDZ2
v DZ L = ρ
Lg
2
4
DZ =
v
u
3u
t
16 Re ρL νL2
2 − lnRe ρ g
DZmax = 4, 71 · 10−2 mm
Re = 0, 5 :
b)
vK = 0, 110 m/s
vZ = 0, 159 m/s
17.5
a)
ρL 2 πD 2
v
(buoyancy neglected)
2 1 4
= 0, 4
(Re1 = 3, 03 · 105 )
G = Fw1 = cw1
cw1
177
b)
Re2 =
from diagram:
v2 D
= 4, 2 · 105
νL
cw2 = 0, 1
Fw2 = cw2
ρL 2 πD 2
v
= 1, 95N < G
2 2 4
acceleration onto v3 , so that G = Fw3 ist:
G = cw3
from diagram:
ρL 2 πD 2
v
2 3 4
cw3 = 0, 1
v3 = 26, 0 m/s
17.6
a)
ρK
ρL πD 2
πD 3
πD 3 dv
= −cw v 2
− ρK
g
6 dt
2
4
6
steady sink velocity:
vs2 =
−
1
g
1Z 0
H=−
g v0
4 ρk Dg
3 ρL cw
dv
dz
2 = dt =
v
v
1+
vs
"
vs2
vdv
v0
ln 1 +
2 =
v
2g
vs
1+
vs
2 #
b)
1
TH = −
g
Z
0
v0
dv
v0
vs
arctan
2 =
v
g
vs
1+
vs
c)
ρK
ρL πD 2
πD 3
πD 3 dv
= −cw v 2
+ ρK
g
6 dt
2
4
6
dv
1
dz
2 = dt =
v
g
v
1−
vS
178
1
H=
g
Z
"
VB
vdv
vB
vs2
2 = − ln 1 −
v
2g
vS
1−
vS
vS
vB = s
2
vS
1+
v0
Z
dv
vS vS + vB
ln
2 =
v
2g vS − vB
1−
vS
0
2 #
d)
1
TB =
g
vB
0
e)
H[m]
TH [s]
vB [m/s]
TB [s]
cw = 0, 4
wooden sphere metal sphere
37, 2
44, 0
2, 64
2, 96
24, 9
29, 3
2, 81
2, 98
cw = 0
45
3
30
3
179
19 Compressible flows
19.1
a)
M=√
v
= 2.0
γRT
b)
L=
H
tan α
1
α = arcsin
Ma
L = 1001 m
c)
S
a
H
cos α =
S
∆t = 1.96 s before passing the observer
∆t =
19.2
∆t =
b
(vB − vA ) tan α
MB = 2;
α = 30o
∆t = 1.73 s
180
19.3
a)
momentum:
ρe ve2 AD = (pa − pe )AD + Fs
energy:
ve2 = 2cp (T0 − Te )
cp =
γR
γ−1
!γ−1
p
p0
Fs
Te
pa pe
2γ ρ
1−
−
=
+
p0 AD
γ − 1 ρ0
T0
p0 p0
T
=
T0
ρ
ρ0
2γ
Fs
=
p0 AD
γ−1
pe
p0
!1
=
γ
−
! γ−1
γ
γ + 1 pe pa
−
γ − 1 p0 p0
subcritical: pe = pa
supercritical: pe = p∗ = 0.528 · p0
b)
momentum:
Fs = ρve2 AD
Bernoulli:
ve2 =
2(p0 − pa )
ρ
Fs
pa
=2 1−
p0 AD
p0
!
181
Fs
p0 AD
pa
p0
a) compr.
0
0.66
1.07
1.27
1
0.6
0.2
0
19.4
Energy:
b) incompr.
0
0.8
1.6
2
u2
u21
= cp T2 + 2
2
2
2
u1
(γ + 1)(M1 )
=
u2
2 + (γ − 1)M12
cp T1 +
cp =
T2 − T1 =
γR
γ−1
(see script)
(γ − 1)(u21 − u22 )
= 197.9 K
2γR
19.5
ṁ = ρ1 v1 A =
s
γ
p1 M1 A
RT1
momentum:
ρ1 v12 A = (p0 − p1 )A
M1 =
energy:
v
u
u1
t
γ
p0
−1
p1
cp T0 = cp T1 +
!
v12
2
T0
γ−1 2
M1
1+
2
ṁ = 1.41 kg/s
T1 =
182
19.6
s
q
ρ1 p0
T1
u
√ 1
γRT0
ṁ = ρ1 u1 A1 =
A1
ρ0 RT0 γRT1
T0
M1 > 1, M2 < 1; p2 = pa
0 → 1 isentropic flow
A∗
1
=
(fron diagram)
→ M1 = 2,
A1
1.8
p2
p1 =
= 2.22 · 104 N/m2
2γ
2
1+
(M − 1)
γ+1 1
T0
T1 =
γ −1 2
1+
M
2 γ 1
γ
γ−1
γ − 1 2 γ−1
T0
= 1.74 · 105 N/m2
= p1 1 +
M1
p0 = p1
T1
2


ṁ = 


1

γ − 1 2
1+
M1
2
from hint:


γ+1
2(γ−1)
M1 √
p0 √
γA1 = 4.43 kg/s
RT0
b)
19.7
A∗2
ME2
=
∗
A1
ME1

γ−1 2
2
1+
ME1 
+1
2


2
γ−1 2 
1+
ME2
γ+1
2

γ


with ME1 = 2
Energy equation:
u2
= cp T0
cp T +
2
with
cp =
γR
γ−1
1γ+1
2γ −1
∗)
183
=⇒
=⇒
T+
1 u 2 c2
= T0
2 c2 cp
1
T0 = 1 − M 2 (γ − 1) T =⇒ T0 = (1 + 0.2M 2 )T
2
T0
= (1 + 0.2ME2 1 ) = 1.8
=⇒
TE1
ground
:
altitude H
:
pa
TE1
=
p0
T0
γ
γ−1
1
γ
pa 4 = TE2 γ−1
p0
T0
1 γ−1
=⇒ TE2 = ( ) γ TE1 = 0.673
4
with ∗∗) ME2 =
into ∗):
1
γRT
T + M2
(γ − 1) = T0
2
γR
=⇒
v
u
u
t5









γ
γ
TE1 =
0.673
T0 =⇒
1.8
!
A∗2
= 0.44
A∗1
M2 =
u2
.
a2
h0 = const : T0 = T01 = T02
Energy:
cp T0 = cp T2 +
c2 =
u22
2
q
γRT2
γ
R
cp =
γ−1
M2 =
T0
=
T2
s
2
T0
−1
γ − 1 T2
p0 2
p2
! γ−1
γ
= 1.14
p0 1′
p0
p0
pa = p2
p0 2 = p0 1′ =
M2 < 1 :
γ
T γ−1
T γ−1
T0γ−1
= E1 = E2

1
p0
pa


pa



4


T0
− 1 = 2.89
TE2
19.8
a)
∗ ∗)
M2 = 0.837
T0
= 2.67
TE2
184
b)
q
u2 = M2 γRT2
T2 =
T2
· T0 = 263.12 K
T0
=⇒ u2 = 272.2 m/s
c)
ρ0 2 =
p0 2
= 1.838 kg/m3
RT02
d)
ṁ = ρ2 u2 A2
ρ2 =
A2 =
p2
= 1.324 kg/m3
R · T2
ṁ
= 0.555 m2
ρ2 u2
19.9
a)
T2
= 1.2
T1
T2
= f (M1 sin σ)
T1
Diagram or formula: M1 sin δ = 1.31
un1
σ = arctan
= 53.13o
ut1
=⇒ M1 = 1.64
v
u
| ~v1 | u
u2 + u2t1
M1 =
= t n1
c1
γRT1
=⇒ T1 =
u2n1 + u2t1
= 229 K
γRM12
b)
momentum equation tangential to the shock + continuity:
ut2 = ut1 = 300 m/s
un1
= f (M1 , sin σ)
un2
diagram or formula:
un1
= 1.55
un2
185
=⇒ un2 =
un2
un1 = 2.58 m/s
un1
q
u2n2 + u2t2
M2 = s
= 1.2
T2
γR T1
T1
tan(β + ε) =
tan(ε) =
ut
un2
ut
un1
ut
ut
− arctan
= 12o
un2
un1
or β from diagram β = f (M1 , σ)
=⇒ β = arctan
c)
M1 = const = 1.64
M2 = 1
fro diagram β = f (M1 , σ):
β = 15o;
σ = 61o
q
un1 = M1 sin σ γRT1 = 440 m/s
M1 =
=⇒ ut1 =
q
u2n1 + u2t1
c1
q
M12 c21 − u2n1 = 234 m/s
19.10
σ = 40o M1 = 2.2
Formula or diagramm β = f (M1 , σ) :
M22 sin2 (σ12 − β12 ) =
=⇒
M1 sin σ12 = 1.41
β12 = 14o
(γ − 1)M12 sin2 σ12 + 2
=⇒ M2 = 1.65
2γM12 sin2 σ12 − (γ − 1)
186
Formula or diagram
p1
= f (M1 , σ12 ) =⇒
p2
p2
= 2.17
p1
T2
= f (M1 , σ12 ) =⇒
T1
T2
= 1.26
T1
∆β = 14o − 6o = 8o = β23
M2 = 1.65
from diagram σ = f (β23 , M2 ), schwacher Stoß:
σ23 = 46o
using formula M3 = f (M2 , σ23 , β23 ):
from diagramm
p3
= f (M2 , σ23 ) =⇒
p2
M3 = 1.37
p3
= 1.53
p2
T3
= f (M2 , σ23 ) =⇒
T2
T3
= 1.13
T2
19.11
The suction leads to an increasung pressure pk (t) in the boiler. The flow is undisturbed until
a shock is located in the exit cross-section AE of the measuring chamber.
ME = 2.3
p0
pE Ausl
γ −1 2
ME
= 1+
2
with p0 = 1 bar, follows pE Ausl = 0.08 bar = pk (t = 0)
γ
γ−1
= 12.5
187
Determining of the boiler pressure pk (∆t) at the end of the measuring time:
relations across the normal shock:
pk (∆t)
2γ
(M 2 − 1) = 6.0
=1+
pE Ausl
γ+1 E
=⇒ pk (∆t) = 0.48 bar
s
T∗q
ρ∗
T0
ṁ = ρ A c = ρ0 A∗ γR
ρ0
T0
∗
=
∗ ∗
2
γ+1
!
1
γ−1
s
p0 ∗
T∗q
T0
A γR
RT0
T0
A∗ = AH
=⇒ ṁ = 24.2 kg/s
pk (t = 0) = ρk (t = 0)RTK
pk (∆t) = ρk (∆t)RTK
∆m = ṁ∆t = Vk [ρ(∆t) − ρ(t = 0)]
=⇒ ∆t =
Vk 1
[pk (∆t) − pk (t = 0) ] = 20.6 s
ṁ RTK
b)
pk
pE Ausl
from diagram
pk
pE Ausl
ME sin σ = 1.36
= f (ME sin σ):
with
ME = 2.3 follows σ = 36.2o
β from diagram β = f (M, σ) :
from diagram
=2
β = 12o
p01
= f (ME · sin σ) :
p02
p01
= 1.03
p02
T2
= 1.22
T1
with p01 = 1 bar follows p02 = 0.97 bar
T1 =
1
T = 136 K
γ −1 2 0
1+
ME
2
=⇒ T2 = 166 K
T1
T0 =
T0
188
(γ − 1)M12 sin2 σ + 2
2γM12 sin2 σ − (γ − 1)
M22 sin2 (σ − β) =
=⇒
M2 = 1.83
T02
γ−1 2
=1+
M2 =⇒ T02 = 277.2 K
T2
2
V2 = c2 M2 =
q
γRT2 M2 = 473 m/s
c)
βab = βk − ε = 20o − ε
βu = βk + ε = 20o + ε
βk + ε ≤ βmax (ME = 2.3) = 27.5o =⇒ ε ≤ 7.5o
βob = 12.5o ME = 2.3
from diagram σ = f (β, M) :
σob = 37o
=⇒
ME sin σob = 1.384
βu = 27.5o , ME = 2.3
from diagram σ = f (β, M) :
σu = 62o
=⇒
pu − pob =
pu
pE Ausl
and
pob
pE Ausl
from diagram
ME sin σu = 2.03
pu
pE Ausl
−
pob
pE Ausl
!
p2
= f (M1 sin σ):
p1
=⇒ pu − pob = 0.22 bar
pE Ausl
189
using M22 sin2 (σ − β) =
19.12
a)
M∞ = 3;
(γ − 1)M12 sin2 σ + 2
follows Mob = 1.85 and Mu = 0.94
2γM12 sin2 σ − (γ − 1)
β1 = 15o →
with M12 sin2 (σ1 − β1 ) =
β12 = β1 − β2 = 5o →
σ1 = 33o
diagram σ1 = f (β1 , M∞ ):
2
(γ − 1)M∞
sin2 σ1 + 2
follows M1 = 2.2
2 sin2 σ − (γ − 1)
2γM∞
1
diagram σ2 = f (β12 , M1 ) follows σ2 = 31o
using M2 = f (β12 , σ2 , M1 ) follows M2 = 2.02
b)
∗
c =
s
γR
M2∗ =
T ∗ T0
T∞ = 503.2 m/s
T0 T∞
un
= 1.29
c∗
v
u
u
u
u
t (γ
γ+1
2
− 1) + 2
M2
un
M2 sin σ3 =
c
=⇒ M2∗ sin σ3 =
=⇒ σ3 = 52o →
= 1.64
un
c∗
Diagramm β = f (σ3 , M2 ) : β = 20o
β3 = β − β2 = 10o
using M3 = f (β, δ3, M2 ) follows M3 = 1.27
M∞ sin σ1 = 1.63 =⇒ diagram
p1
p1
= f (M∞ sin σ1 ) :
= 2.9
p∞
p∞
M1 sin σ2 = 1.13 =⇒ diagram
p2
p2
= f (M1 sin σ2 ) :
= 1.4
p1
p1
M2 sin σ3 = 1.59 =⇒ diagram
p3
p3
= f (M2 sin σ3 ) :
= 2.7
p2
p2
p3 =
p3 p2 p1
·
·
· p∞ = 10.96 bar
p2 p1 p∞
190
19.13
a)
M1 = 1.6
p1
γ−1 2
= 1+
M1
p01
2
−
γ
γ−1
= 0.235
p2 = p1
p2
p01 p1
=
= 0.282
p02
p02 p01
γ −1 2
p2
= 1+
M2
p02
2
−
γ
γ−1
=⇒ M2 = 1.48
b)
M2 · c2
M2
|V~2 |
=
=
~
M1 · c1
M1
| V1 |
s
T2
M2
=
T1
M1
s
T2 T02 T01
T02 T01 T1
T01 = T02
γ−1 2
T2
= 1+
M2
T02
2
analogous
=⇒
−1
= 0.69
T1
= 0.66
T01
~2 |
|V
= 0.95
~1 |
|V
p2 /(RT2 )
T1
ρ2
=
=
= 0.96
ρ1
p1 /(RT1 )
T2

3 Fluid kinematics

Transcription

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