Budapest University of Technology and Economics Faculty of Mechanical Engineering
2002
Dr. Miklós Blahó
Selected Problems in Fluid Mechanics
1
Hydrostatics ................................. ................................................. ................................ ........................... ........... 3
2
Kinematics ................................... ................................................... ................................. ........................... .......... 8
3
Bernoulli Equation.............................. Equation............................................... ..................................10 .................10
4
Integral Momentum Equation................. Equation .................................. .............................. ............. 15
5
Hydraulics................ Hydraulics ................................. ................................. ................................. ............................ ........... 20
6
Compressible Flows....................................... Flows......................................................... ....................... ..... 24
RESULTS 1
Hydrostatics ................................. ................................................. ................................ ......................... ......... 27
2
Kinematics ................................... ................................................... ................................. ......................... ........ 29
3
Bernoulli Equation.............................. Equation............................................... ..................................31 .................31
4
Integral Momentum Equation................. Equation .................................. .............................. ............. 34
5
Hydraulics................ Hydraulics ................................. ................................. ................................. ............................ ........... 36
6
Compressible Flows....................................... Flows......................................................... ....................... ..... 39
Hydrostatics
1
Hydrostatics
1/5
upper tap? There is no flow in the pipe.
= ? [Pa ]
p A − p 0
The figure shows a vertical section of a gas pipe. At the lower tap there is an overpressure of 500 Pa. How big is the overpressure at the
= 287 J / kg K , g = 9.81 N / kg For all relevant problems R = 1/1
4
ρ air = 1.2 kg / m 3 ρ gas = 0.7 kg / m 3
1/6
= 10 5 N / m 2 ρ 0 = 1.2 kg / m 3
p 0 z=0 1/2
p1 − p 2
air R = 288 J / kgK
= ? [Pa ] a.) T0 = ? [K ] b.) p A
= ? [Pa ],
if the temperature is constant for 0 ≤ z < 2000m .
1/7
p A z
1/3
Section 1-2:
ρ12 = 1.3 kg / m 3
Section 3-4:
ρ 34 = 1.1 kg / m 3
p 4 − p1
1/4
p 0
≈ 10 5
1/8
Pa (for the calculation of
Outside (air): T1
=0
In chimney (smoke):
p1 − p 2
= ? [Pa ]
o
ρ)
C
p 2 T2
p = 10 5 N / m 2 =0 0 ρ 0 = 1.25 kg / m 3
zA
= ? [Pa ]
≈ 760 mmHg = 250 C o
= 0.5 ⋅ 10 5 N / m 2 air
= ? [m] if the temperature is constant for 0 ≤ z < z A .
The vehicle is filled with oil.
ρ oil = 950 kg / m 3 a = 3 m / s2 p A − p 0 = ? [Pa ]
Hydrostatics
1/9
5
The vehicle is filled with oil.
Hydrostatics
1/13
ρ oil = 950 kg / m 3 p A − p 0 = 0 Pa a = ? [m / s 2 ]
6
The pipe is filled with water. p 0 = 10 5 Pa How high angular velocity is needed to
= 0.8 ⋅ 10
5
a.)
reach p A
b.)
empty the A-B section and have pressure of
Pa ?
surface at standstill
0.8 ⋅ 10 5 Pa in it?
1/10
The tank wagon shown in the figure is taking a curve with a centripetal acceleration of a = 3 m / s 2 . The tank is filled with water. a.)
How high will climb the water surface on the A-B side?
b.)
1/14
when the vehicle is 1.6 m long?
1/11
Where are the both surfaces of the liquid situated if the pipe accelerates to the left with an acceleration of a =
1/12
Effect of gravity is negligible.
ρ = 800 kg / m 3 n = 6000 1 / min p A − p 0 = ?
How big force will affect the A-B side,
n = 1000 1 / min
g 2
1/15
Effect of gravity is negligible.
ω = 100 1 / s ρ water = 1000 kg / m 3 ρ oil = 800 kg / m 3 p A − p 0 = ? [Pa ]
?
1/16
ρ water = 1000 kg / m 3 p A − p 0 = ? [Pa ]
What area does an ice-floe have, which can carry a person weighing 736 N? The thickness of the ice-floe is 10 cm and its density is 900 kg/m3 ?
1/17
The rope is weightless.
ρ Cube = 2300 kg / m 3 ρ Water = 1000 kg / m 3 r Sphere = 300 mm G Sphere = 200 N a = ? m/s 2
Hydrostatics
1/18
7
A balloon is filled with hot air of 60°C. Its diameter is 10 m. The environmental temperature is 0°C . Pressure outside and inside the balloon is 10 Pa. The weight of the balloon material 5
2
Kinematics
is can be neglected. Determine the buoyant force! 2/1 1/19
p1 − p 2
= 20 N / m
qv
ρ liquid = 800 kg / m α = ? [°] if an error of ± 1mm at the reading of the fluid 3
column position causes
Pressure changes are negligible.
= 40 m 3 / s t 1 = 15 °C t 2 = 80 °C
2
± 2% relative error
of p1 − p 2 . v1 v2
1/20
= ? [m / s] = ? [m / s]
After having been filled the pipe both taps were closed. During the rotation the surface in the left pipe section sinks to the point B as shown in the figure.
2/2
Two dimensional flow: v
p 0
= 10 5 Pa
p saturated steam T
= 10
r
[(rot v)z ]A = ? [1 / s] = 2 ⋅ 10 4 Pa
= const
ω = ? [1 / s] 2/3
Axisymmetric flow.
v mean v max
2/4
=?
Unsteady, two dimensional flow. vy
=0
vx
= 5yt 2
Calculate the local and convective acceleration in point 'A' at t = 0.5 s .
Kinematics
2/5
9
Calculate the circulation along the dashed line. v =
3
Bernoulli Equation
2 r 2
Γ = ? [m 2 / s]
3/1
p t
= 3 ⋅105 Pa
= 105 Pa v = ? [m / s]
p 0
2/6
v1 = 20 m / s ρ = const. [a konv ]A = ? [m / s 2 ]
3/2
v = 10 m / s u = 4 m/s
ρ = 10 3 kg / m 3 p A − p 0 = ? [Pa ]
3/3
Friction losses are negligible.
ρ = 1.2 kg / m 3 v 2 = ? [m / s ]
3/4
Steady flow with
q V = 0.1 m 3 / min . h = ? [m]
Bernoulli Equation
3/5
11
p1
= 1.6 ⋅105 Pa
p 2
= 1.2 ⋅10 Pa = ? [m 3 / s]
qV
Bernoulli Equation
3/9
5
p 0
= 105 N / m 2
p A
=0 = 4 m/s = ? [m / s 2 ]
vA aA
3/6
a = 12 m s
2
3/10
= 10 Pa p t = 0.5 ⋅10 5 Pa q V = ? [m 3 / s] p 0
5
p 0 = 105 Pa p1
= 0.9 ⋅10 5 Pa
Friction losses are negligible. a.) How big is the starting acceleration ’a’ when opening the tap? b.) H = ? [m ] in case of steady flow?
3/7
ω = 25 1 / s w = ? [m / s]
3/11
(w: relative velocity)
3/8
w = 3 m / s ω = ? [1 / s] (w: relative velocity)
How big is the starting acceleration in point B when opening the tap?
3/12
How big is the starting acceleration at the end of the pipe?
= 2 ⋅10 4 N / m 2 (overpressure) v=0
p t
12
Bernoulli Equation
3/13
13
v = 1 m / s a = 1 m / s
Bernoulli Equation
3/18
14
Irrotational, horizontal, two-dimensional flow.
force is needed to push the
= 0.5 m = 0.8 m v0 = 5 m / s
piston?
a.) What kind of velocity distribution has
2
r 1
r 2
Friction is negligible. How big
developed in the arc? b.) p A − p B 3/14
u = 72 km / h
c.)
v = 4 m / s
ρ
Friction is negligible.
[
a.) q V = ? m / s 3
]
b.) How big power is needed to move the pipe?
3/15
ρalc = 800 kg / m 3 ρ air = 1.2 kg / m 3 v = ? [m / s]
3/16
The inner diameter of an orifice flowmeter is d = 200 mm . Flow coefficient Compressibility factor
ε = 1.
The measured difference pressure is
ρ = 1.3 kg / m 3 .
[
q V = ? m 3 / s
3/17
p A − p B
]
Width of the flow is 1 m. a.) Construct the velocity distribution diagram along the vertical line over the outlet. b.) Calculate the flow rate q V
[m / s]! 3
α = 0.7
∆ p = 600 N / m 2 .
v0 2
2
= ? [Pa ] r = f 2 ? (Draw a diagram!) r 1
Integral Momentum Equation
4
Integral Momentum Equation
4/5
v = 10 m / s
ρ Hg = 13600 kg / m 3 Friction and gravity are negligible. 4/1
Calculate the force acting on the cone!
Calculate the horizontal force acting on the conical part of the pipe! q V = 3.5 m 3 / min
4/6
A = 10 −4 m 2 v = 10 m / s
Friction losses are negligible.
Friction and gravity are negligible. Determine the weight of body ’G’ [N]!
4/2
v1 = 30 m / s
u = 13 m / s Friction losses are negligible. a)
v2
= ? [m / s]
b) Calculate the angle of deviation
4/7
G = 1 N v 0 = ? [m / s]
β [°] (angle
Friction is negligible.
between v1 and v 2 )! c) Determine the force acting on the blade! d) How is the kinetic energy of 1kg water changing, when passing the blade?
4/3
v = 10 m / s Friction and gravity are negligible. Calculate the force acting on the arc! 4/8
Two dimensional flow. v = 30 m / s a) F = ? [ N] b) A1 A 2
4/4
=?
v = 10 m / s u = 2 m / s Friction and gravity are negligible. Calculate the force acting on the moving conical body!
4/9
Two dimensional flow. Friction and gravity are negligible.
α = ? [°]
16
Integral Momentum Equation
4/10
Two dimensional flow.
17
Integral Momentum Equation
4/15
Friction losses are negligible.
v1 = 2 m / s
ρ 0 = 1.29 kg / m 3 t 0 = 0°C t 2 = 273°C
v = 10 m / s α = 15° G = ? [ N ]
Friction and density changes of the air because of pressure changes are negligible.
4/11
[
q V = ? m 3 / s
Friction losses are negligible.
]
The cylinder is balanced by the water jet. G h
= 10 N = ? [m]
4/16
v1 = 20 m / s
ρ = 1 kg / m 3 h = ? [m]
4/12
v = 10 m / s u = 6 m / s Friction is negligible.
4/17
There is no friction loss in the pipe. p1 − p 0
Calculate the power transmitted by the water jet to the
= ? [Pa ]
wheel!
4/13
v = 20 m / s u = 6 m / s Friction is negligible.
4/18
The flow rate through the lower and upper outlet is the same. The losses due to the rapid cross
Calculate the mean force acting on the wheel blades in
section change at the upper pipe must be
the direction x and y!
considered. h = ? [m]
4/14
v1 = 2 m / s
ρ1 = 1.2 kg / m 3 t 1 = 20°C t 1' = t 2 = 300°C Friction, gravity and density changes of the air because of pressure changes are negligible. p1 − p 2
= ? [Pa ]
4/19
Steady flow. h = ? [m]
18
Integral Momentum Equation
4/20
Determine the quotient of the flow rates with and without horizontal plate! qV
19
5
Hydraulics
without plate
qV
with plate
=? 5/1
The width of the gap is 100 m m (perpendicular to the paper plane). v = 0.5 m / s µ = 0.1 kg / ms F = ? [ N]
5/2
Friction loss in the confuser is negligible.
= 0.5 m / s ρ = 850 kg / m 3 ν = 10 −5 m 2 / s p1 − p 0 = ? [Pa ] v1
5/3
Friction loss of the transitional section is negligible.
= 10 m / s ρ = 1.2 kg / m 3 ν = 14 ⋅10 −6 m 2 / s v1
p1 − p 0 5/4
= ? [Pa ]
How do the Reynolds number and the pressure loss of a straight, smooth pipe depend on diameter in case of laminar and turbulent flow, if the flow rate is constant?
5/5
How does a straight, smooth pipe’s pressure loss depend on the flow rate in case of laminar and turbulent flow?
5/6
Oil flow rate of q V
= 2 ⋅10 −4 m 3 / s has to be transported through a 10 m long straight pipe
( ρ = 800 kg / m , ν = 10 −4 m 2 / s ). The available pressure difference is not more than 3
2 ⋅ 10 5 Pa . Determine the diameter D [mm] of the pipe!
Hydraulics
5/7
21
q V = 8000 m 3 / h
ρ = 1.2 kg / m λ = 0.025 ηD = 0.8 p1 − p 0
5/8
Hydraulics
5/11
ν water = 1.3 ⋅10 −6 m 2 / s
[
q V = ? m3 / s
3
]
= ? [Pa ]
5/12
q V = 1200 l / min
ρ Hg = 13.6 ⋅10 h = ? [m]
3
kg / m
Hydraulically smooth pipe walls.
3
ν water = 1.3 ⋅10 −6 m 2 / s qV = 5 l /s p1 − p 0
5/13
5/9
22
has to transport an oil flow rate of q V
= 0.05 ⋅10
Hydraulically smooth pipe walls.
ν water = 1.3 ⋅10 −6 m 2 / s q V = 180 l / min
The figure shows a part of a lubrication equipment, which −3
= ? [Pa ]
3
m / s . For
p1 − p 0
the calculation of the friction loss, it can be considered that
= ? [Pa ]
the pipe is straight.
ρ oil = 800 kg / m 3 ν oil = 10 −4 m 2 / s d = ? [mm] 5/10
The additional losses of the bends can be neglected. (It can be considered that the steel pipe is straight.)
ν water = 1.3 ⋅10 −6 m 2 / s q V = ? [m 3 / s ]
5/14
Steady flow, hydraulically smooth pipe.
ν water = 1.3 ⋅10 −6 m 2 / s v1 = 1 m / s a) H = ? [m] b) p1 − p 0
= ? [Pa ]
Hydraulics
5/15
23
What power is needed to drive the shaft of a glide bearing with 2880 1 / min , when the shaft is 60 mm wide, 100mm long and the gap between bearing and shaft is 0.2 mm?
6
Compressible Flows
( µ oil = 0.01 kg / ms ) How is it possible to decrease this power? 5/16
a) Determine the confuser’s output diameter d 2 , when the water jet is 12 m high!
6/1
T1
b) Calculate the flow rate
[
q V m3 / s
]
through
the
Isentropic change of state.
pipe! Friction losses of the
v 2 = ? [m / s]
bends, the confuser and friction effects between
= 1.5 bar , p 2 = 1 bar = 300 K c p = 1000 J / kg K κ = 1.4
p1
6/2
the water jet and the air
p1
= 1.3 ⋅ 10 5
Pa , p 2
= 10 5
Pa
= 273 K R = 287 J / kg K κ = 1.4
T1
are negligible.
Isentropic change of state. q m = ? [kg / s]
5/17
a) How wide pipe do we need to fulfill this task?
= 1.4 bar , p 2 = 1 bar = 20 °C κ = 1.4
b) Determine the maximal dike height where the transport is possible? (theoretical answer)
Isentropic change of state.
Water of q V = 18 m 3 / h flow rate has to be transported by the equipment shown in the
6/3
p1 t1
figure.
a)
t 2 static
= ? [°C]
b)
t 2 total
= ? [°C]
(temperature measured by the stagnation point thermometer)
6/4
p1
= 4 bar , p 2 = 1 bar
= 300 K R = 287 J / kg K κ = 1.4
T1
Isentropic change of state. q m = ? [kg / s ]
Compressible Flows
6/5
p1
25
= 4 bar , p 2 = 1 bar
6/11
Isentropic change of state.
b) Calculate th e thrust F [ N ] of the rocket
κ = 1.4 . the outflow needs to be isentropic?
= ? [mm]
engine!
What kind of formula can be used to calculate v 2 , if a)
b)
c)
p 2 p1 p 2 p1 p 2 p1
= 0.99 = 0.6 = 0.4
Isentropic change of state.
Air of temperature t = −40 °C flows at a velocity v = 180 m / s .
κ = 1.4 , R = 287 J / kg K .
Calculate the Mach number (Ma) !
6/8
Carbon-dioxide of the temperature t
= 20 °C flows
at a Mach number of Ma = 0.3 .
κ = 1.3 , R = 189 J / kg K . Calculate the velocity of the flow! [m / s]
6/9
6/10
c p = 1000 J / kgK ,
a) How wide should be the diameter d 2 , if
d min
6/7
R = 287 J / kg K ,
26
= 70 °C R = 287 J / kg K κ = 1.4 t1
6/6
Compressible Flows
A rocket flies in air of t c p
= 1000 J / kgK
tA
= ? [°C]
= −23 °C
An aircraft flies in air of t
at a velocity of u = 400 m / s .
= 0 °C at a velocity u = 200 m / s . The relative velocity w 2 in a κ = 1.4 . Calculate the Mach
definite point of the wing makes 250 m / s . R = 287 J / kg K , number in this point.
Results
RESULTS 1
Hydrostatics
1/1
p A − p 0
= 6200 N / m 2
1/2
p1 − p 2
= 12360 N / m
1/3
p 4 − p1
= 392 N / m 2
1/4
p1 − p 2
= 486 N / m 2
1/5
The overpressure at the upper
28
1/8
p A − p 0
1/9
a = 2.45 m / s 2
1/10
a)
h = 0.422 m
b)
F = 1400 N
1/11
= 7.23 ⋅ 10 3 N / m 2
The surface at the left side is situated at the left lower corner, the other surface in the right vertical section at a height of 100 mm.
2
1/12
Volumes are the same in standstill and rotation: R 2 πz 0
=
1 2
r 2 πz1
Points of equivalent potential: g ⋅ z1
r 2 ω 2
−
2
tap is 600 Pa.
= 0 ; r 2 =
2gz 1
ω2
After substitution: R 2z0
=
p A − p 0
1/13
2
a.) b.)
T0 dp dz p A
=
= 290 K
= −ρg = −
p
p 0
dp
∫ p
p 0
p ln A p 0 p A
ρ 0 R
=−
ρ0g p 0
=−
p 0
= 0.236 m
r 2 ω 2 2
+ const .
for the both known points (surfaces in the left and the right s ection), the angular velocity can be calculated.
ρ0g
a.) b.)
1/14
ω = 21.4 1 / s ω = 24.3 1/ s
Equation p = −ρ gz −
zA
= 0.788 ⋅ 10 5 N / m 2
h = 5650 m
g
After writing the equation
zA
ρ0g
z0
= R ω
R 2 ω 2 2 = −ρg z A − = 14300 N / m 2
p 0
z1 z 1
ω2
p = −ρ gz −
1/6
1/7
1 2gz 1
const. = p 0 p A − p 0
−ρ
=ρ
ω2 2
r 0
2
r 2 ω 2 2
+ const
written for the surface of the fluid:
ω2
2
[r
2 A
− r 02 ] = 19.7 ⋅ 10 5 N / m 2
Results
1/15
29
Results
r 2 ω 2
Apply the equation p = −ρ gz −
2
+ const at first for the oil-filled part and then for the
=
ω2
[ρ (0.1
2
oil
2
4
∂v y = 10 ∂x
water filled part of the pipe. It can be written then: p A − p 0
30
x2
x
1 4 2
(x
2
+y
− 34
+ y2 )
2x at po int A ⇒
2
∂v y ∂v x − = (50 + 100) ∂x ∂y
[(rot v)z ]A =
− 0.05 2 ) + ρ water (0.15 2 − 0.12 )] = 9.25 ⋅ 10 4 N / m 2
+ y2 − x
∂v x = −100 ∂y
0.1
0.1 = 47.5 1 / s
Solution with polar coordinates: 1/16
A = 7.5 m 2
dr
= 0.3 m
1/17
a
1/18
F = 1200 N
1/19
h= l=
20
= 2.55 mm
2.55 50
v mean
= 0.051 ⇒ α = 2.9°
1
15 r
=
15 0.1
= 47.5
1/ s
2 0
r 0
1
r
0
0
0
1
In general:
= 10 m / s ; v 2 = 6.9 m / s
2/1
v1
2/2
Solution with Cartesian coordinates:
2/4
vx
= c(− sin α ) = − v
vx
= −10
r x r x
1 r 7 r r r r d = 2 ∫ v max 1 − d = r r 0 r 0 r 0 0 0 0
9 1 r 1 r 2 7 2 v max − 2 v max = v max 1 − = v max ⇒ 2 r 0 9 r 0 9 9 0
1/ s
4
=
2r πv( r ) dr = ∫ 2 v r r r π ∫
=
r r 0
n
v = v max 1 −
∂v y = 10 ∂x
r
7 r r 0
2
r
10
elementary flow rate through the rings as follows:
Kinematics
= 10
+
The cross section has to be divided into rings o f elementary width 'dr'. Integrate the
2
vy
1 2 r
v = v max 1 −
± 1mm = 50 mm ± 0.02
ω = 81.8
r A
2/3
800 ⋅ 9.81
sin α =
1/20
[(rot c )z ]A = dc + c = 10
y r
= −10
; vy r y
x
= 10 2
y
r
r
4
x2
r
+y
2
+y
(x
x 2
2
+y
+ y2
2
− 34
)
2/5
Γ = ∫ v ds = −2.61 m 2 / s
2/6
r 1 πv1 2
= r 2 πv
v = v1r 1
2
2x at po int A : x, y = (0.1, 0) ⇒
v max
[a local ]ty==01.5 = 5 m / s 2
2
x 4
1
x
2
4
v mean
a convective = 0
x
y
= −10
= 10
+y −x 2
= v ⋅ cos α = v
⇒
∂v y = 50 ∂x
0.1
1 r 2
=
n n+2
v mean v max
=
7 9
= 0.778
Results
31
Results
∂v ∂v ∂r 2 ∆r = = v1r 1 2 − 3 ∂x ∂r ∂x r ∆x 2v 2 r 4 ∆r ∂v = − 15 1 a convective = v ∂x r ∆x 2 ⋅ 20 2 0.05 4 0.05 [a convective ]A = − = −132 m / s 2 5 0.075
(− r 1ω)2 2
3/1
=
+
ρ 2 ρ v = 19.8 m / s p A − p 0
3/3
ρ water ⋅ g ⋅ h =
3/9
1/ s
p 0
2
ρ
=
v 22
+g⋅h−
2
r 2
2
ω2
2
=
vA 2
A
+ g⋅h + ∫ 0
∂v
A
∂v ds ∂ t
⋅ l = a A ⋅ 3m
0
aA 3/10
100 4 − 1 ⇒ 50
v 2
= 24.1 m / s 2
a.) [a ]t =0
= 6.55 m / s
b.) H = 1.52 m
2
2
2
ω = 24
∫ ∂t d s = a
+ g⋅h
ρ
ω2
v 2 = 10.8 m / s
ρ = (v − u )2 = 1.8 ⋅ 10 4 Pa
3/2
2
0.8
A
v 2 p 0
r 1
the pipe.
Bernoulli Equation p t
−
Point 1 is situated on the water surface on an arbitrary radius r 1 , point 2 at the u pper end of
3/8
3
32
B
3/11
v = 7.4 m / s
∂v
∫ ∂t ds = a
B
A
10 5 + 5 = 7.5a B 20
[a B ]t =0 = 1.31 m / s 2
3/4
qV h=
3/5
qV
A 2g
= 0.141 m
= 0.793 m 3 / s
3/12
[a 2 ]t =0 = 7.94 m / s 2
3/13
F = 451 N
3/14
a)
The Bernoulli-Equation has to be written between the surface point (1) and the pipe’s
outlet point (2), in a co-ordinate system moving with the pipe. It means that v1 = 24 m / s . 3/6
p t
ρ
+ (g + a ) ⋅ h =
p 0
ρ
+
v2
From the Bernoulli-equation:
2
v2
q v = 0.00589 m / s 3
3/7
b)
= 2ω , so the term
= 0.116 m 3 / s
the power is necessary to lift the water and to increase its kinetic energy. The change
P = ρ ⋅ q V g ⋅ h +
∫ w × rot w ds is equal to
∫ 2 w × ω ds , the Coriolis force term. ( w – relative velocity) The Bernoulli equation can be written after simplifying the terms above:
qV
of the kinetic energy must be calculated with the absolute velocity ’v’.
Observing in an absolute co-ordinate system, the flow is irrotational (rot v = 0 ). In a coordinate system rotating with the pipe, rot w
= 23.4 m / s ⇒
3/15
v=
2 ⋅ ∆ p
ρ air
v2
2
− v1 2 = 8.85 kW .
2
= 36 m / s
Results
3/16
3/17
33
qV
= α⋅ε
d 2π
2∆ p
4
ρ
= 0.67 m / s
Results
4
Because the stream lines leaving the outlet are straight and parallel, there is only a hydrostatic pressure variation along the vertical axis. It follows that the outlet velocity is
3/18
4/2
After writing the Bernoulli equation for p oints situated upstream and downstream the blade we get the result:
r r 2
1
=
r 2
v2
K
a) in the arc v =
⇒ K =
− r 1 ∫r 1
, because rot v = 0 .
r K K dr = ln 2 Because of continuity: v mean r r 2 − r 1 r 1
v mean (r 2 − r 1 ) r ln 2 r 1
⇒ vA =
K r 2
= 3.2
= 4 m/s ,
vB
=
K r 1
= v0
ρ
=
2
(v
4/4
F = 109 N
4/5
F = 57 N
4/6
G = 14 N
4/7
The integral momentum equation written for a control surface including only the plate and the upper end of the jet:
= 6.4 m / s
G
ρ 2
= ρ ⋅ A ⋅ v2 = ρ ⋅ A0 ⋅ v0 ⋅ v
with v, the speed at the lower surface of the control surface. According to the Bernoulli equation: v=
c.) p A − p B v0
2
2
(n − 1)
3
... =
2
ln n
with n =
r 2 r 1
n +1 n
2
v0
2
− 2⋅g⋅h
v 0 = 4.55 m / s
2
v v = B − A = ... v 0 v 0
v1
F = 510 N , direction 45° from the horizontal plane (’Northeast’)
− v A 2 ) = 1.25 ⋅ 10 4 Pa
2 B
=
4/3
From the Bernoulli-equation: p A − p B
= 12100 N
Fx
= 3.15 m 3 / s .
b) v mean
Integral Momentum Equation
4/1
constant.
qV
34
4/8
Write the integral momentum equation for both directions x and y: a)
F = 636 N
b)
A1 / A 2
= 5.8
Solution with constructing the momentum rate vectors: (It has to be considered that
ρ ⋅ A 0 ⋅ v 2 = ρ ⋅ A1 ⋅ v 2 + ρ ⋅ A 2 ⋅ v 2
)
Results
4/9
35
α = arcsin
4/10
G = 52 N
4/11
h = 1 m
4/12
P
4/13
Fx
Results
a
36
v1
1− a
2
+
2 v2
= u ⋅ ρ ⋅ A ⋅ v ⋅ ( v − u ) = 302 W
= Fy = 280 N
p1 − p1'
= ρ1 ⋅ v1 ( v1' − v1 )
p1' − p 2
=
ρ2
(v 2 2 p1 − p 2 = 123 Pa
4/15
h = 0.8 m
4/19
h = 1 m qV
p1 − p 2
= ρ1 ⋅ v1 ( v 2 − v1 )
= 51 m
ρ1 2
+
p1
ρ
ρ =const,
=
=
2
2
− v1 )
dv
5/1
F = A⋅µ ⋅
= 7.5 N
5/2
p1 − p 0
= 72400 Pa
5/3
p1 − p 0
= 1500 Pa
5/4
Re =
dy
qv ⋅d d2π
∆ p lam =
ρ
const
=
d
ν qv
2 d
4
2
π2
16
pipe): 2
v2
v2
2
2
+
where ∆ p B− C
Hydraulics
v1
The Bernoulli-equation between point 1 and 2 (point 2 is situated at the outflow end of the
2
∆ p B−C ρ
/s
A 2 ( p1 − p 2 ) = ρ ⋅ A 2 v 2 ( v 2
v1
+g⋅h +
v 3 = 0 .
with plate
h = 6.5 mm 4/17
ρ
without plate
4 4/16
p 0
5
= (ρ1 − ρ 2 ) ⋅ g ⋅ h −
qV
+
2
− v1' 2 )
2
p1 − p 2
3
ρ
4/18
4/20
v 22
=
= v1 and
qV 4/14
p1
p 0
+ ρ⋅g⋅h ρ
2 d
4
because the area of cross section of the pipe is constant,
An other solution can be the Bernoulli equation between point 1 and 3 (point 3 is situated
64 d const d
=
2
L
0.316
π2
d
const
qv
16
= v1
on the water surface):
∆ p turb =
ρ
L
4
d
const d4
≈
const d5
ρ = (v 2 − v 3 )2 (Borda-Carnot-loss) 2
Results
5/5
37
∆ p lam =
ρ qv2
L
2
d
2 A
64 q vd
Results
= const ⋅ q V
38
Starting with
λ = 0.02 ,
v pipe
=
A⋅ν
ρ qv
∆ p turb =
5/6
2
L
0.316
2 A2 d
q vd
4
Re =
= const ⋅ q V 1.75
The Reynolds number is 189 which is less than 2300, so the flow is laminar.
5/8
h = 17 mm
5/9
g⋅h =
= 143 Pa
v
5/17
2
Considering laminar flow, the result will be d = 19.3 mm . Re = 33 < 2300 , so the flow is really laminar. 5/10
q v = 0.23 m / s
5/11
q v = 0.0817 m 3 / s
5/12
p1 − p 0
= 10900 Pa
5/13
p1 − p 0
= 28500 Pa
v2
=
2⋅g ⋅ h
= 15.3 m / s
d2
=
0.755 m / s
qv
= 1.47 ⋅10 −3 m 3 / s
15.3 m / s
⋅ 50 mm = 11 mm
a) At first the velocity without friction loss can be calculated: v ideal
=
2 ⋅ g ⋅ 3m
= 7.7m / s ,
18
1 + L λ d
2
= 3.2 ⋅10 4 ⇒ λ = 0.024
To reach h = 12 m , the necessary velocity at the confuser’s outlet must be:
λ = 64 / Re , we get
d = 13.4 mm .
p1 − p 0
1.3 ⋅ 10 −6
= 0.827 m / s
After the next iteration step, v pipe = 0.755 m / s , and the iteration can be finished.
A⋅ν
Considered that the flow will be laminar and using the formula
5/7
0.827 ⋅ .05
3m ⋅ 2 ⋅ 9.81m / s 2 200m 0.02 + 6 0.05m
m3 / s and A = 3600 7.7 m / s
= 6.5 ⋅10 −4 m 2
So the pipe diameter is in this case 29 mm. Because of friction losses, we need a pipe of larger diameter. We start the iteration with
3
v=
Re =
3m ⋅ 2 ⋅ 9.81m / s 14m 0.02 + 4 + 1 0.05m 2
2.36 ⋅ 0.052 1.3 ⋅10 −6
λ = 0.02
and d = 50 mm :
= 2.36 m / s ⇒ A = 21.2 ⋅10 −4 m 2 ⇒ d = 52 mm
= 9.45 ⋅10 4 ⇒ λ = 0.018
(At this Reynolds number we consider that the pipe is hydraulically smooth)
5/14
5/15
5/16
a)
H = 2 m
b)
p1 − p 0
In the next iteration step with
λ = 0.018 and d = 52 mm we get the new diameter of
51.2 mm . The iteration can be finished. b) If the dike is higher, the pressure in the pipe can reach the pressure of saturated steam. In
= 40000 Pa
this case, the water column is going to break. The lowest pressure appears after the valve, at
P = 77 W
the upper right point of the dike. From the equation
The power can be decreased by sinking the oil viscosity and by increasing the gap.
p min = p 0
The resultant height loss is h res g ⋅ h res
=
v
2
2
L λ + 2ζ d
= 15 m − 12 m = 3 m .
− ρ ⋅ g ⋅ h max −
ρ 2
h max can be calculated.
v 2 1 +
L1 + L 2 d
λ + ζ
Results
39
6
Compressible Flows
6/1
v 2 = 260 m / s
6/2
qm
6/3
a)
t 2 static
= −42°C
b)
t 2 total
= +20°C
6/4
= A 2ρ 2 v 2 = 10 −3 m 2 ⋅1.37 kg / m 3 ⋅ 200 = 0.274 kg / s
T*
=
T1
2
κ +1
= 0.833
a1
= κ ⋅ R ⋅ T1 = 346 m / s
a*
=
T* T1
a1
= 316 m / s = v*
1
T * κ−1 ρ* = ρ1 = 2.9 kg / m 3 T1 q m = v* ⋅ ρ* ⋅ A * = 0.018 kg / s 6/5
qm
= A 2 ⋅ ρ 2 ⋅ v 2 = 0.25 kg / s
A*
=
d min
6/6
qm v * ⋅ ρ*
= 2.34 ⋅10 −4 m 2
= d * = 17.3 mm 2
( p1 − p 2 )
a) v 2
=
b) v 2
=
κ−1 p κ 1 − 2 κ − 1 ρ1 p1
c) v 2
=
2κ p1 2 1− κ − 1 ρ1 κ + 1
ρ
2κ p1
6/7
Ma = 0.59
6/8
v = 80 m / s
Results
40
6/9
tA
= 56°C
6/10
T2
= 262 K ,
6/11
a)
d = 138 mm
b)
F = ρ2 ⋅ A2 ⋅ v2
Ma 2
= 0.77
2
= 9.8 ⋅103 N
