# Sammanfattning hydraulik

```Sammanfattning hydraulik
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Bernoullis ekvation
R&ouml;relsem&auml;ngdsekvationen
Energiekvation applikationer
R&ouml;rstr&ouml;mning
Friktionskoefficient, Moody&acute;s diagram
Pumpsystem
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BERNOULLI’S EQUATION
p


V
2
2g
 z  H  const .
Bernoulli’s equation is a useful
relationship between pressure, p,
velocity, V, and geometric height,
z, above a reference plane
(datum).
H: energy head (m)
z: elevation head above datum
(m)
V: velocity (m/s)
g: gravity acceleration (m/s2)
p: pressure (Pa)
γ: weight density for the flowing
fluid (N/m3)
Quantity
Name
Measure of
H
Energiniv&aring;
Total energi
P/γ
Tyckh&ouml;jd
“tryckenergi”
Z
Geometrisk
h&ouml;jd
L&auml;gesenergi
V2/(2g)
Hastighetsh&ouml;jd
R&ouml;relseenergi
p

 z  piezometri c head or
H .G.L  Hydraulic Grade Line
=&gt;Tryckniv&aring;
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γ=w=ρg
Grundl&auml;ggande ekvationer
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Pitotr&ouml;r (manometri, peizometer)
&Aring;ngtryck, kavitation
Energi- och trycklinjer
Str&ouml;mning med energif&ouml;rlust
Bernoullis ekvation (K.E.; B.E.; E.E.)
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Momentum Equation (RRM ekv.)
 FX    Q ( VX ,OUT  VX , IN ) ( x  direction)
 FY    Q ( VY ,OUT  VY , IN ) ( y  direction)
 FZ    Q ( VZ ,OUT  VZ , IN )
F:
:
Q:
VOUT:
VIN:
:
( z  direction)
Sum of all external forces acting on the control
volume (like the streamtube).
Density of fluid
Flowrate
Velocity out of the control volume
Velocity in to the control volume
Correction coefficient for momentum (non-uniform
velocity)
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Methodology Using The Momentum Equation For
A Fluid Flow Problem
1. Define an appropriate control volume (Draw picture)
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2. Define coordinate axes (For F and V ; find out +
and -)
3. Determine all forces (magnitude and direction)
acting on the control volume (Draw picture)
4. Determine flowrate, outflow and inflow velocities to
the control volume (if not given, use continuity
equation and Energy / Bernoulli equation)
5. Solve momentum equation:
ΣF = ρQ(V2 - V1) = Total forces acting on CV
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R&ouml;rstr&ouml;mning I--IV
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Energif&ouml;rluster vid r&ouml;rstr&ouml;mning
Lokala energif&ouml;rluster
Trereservoirproblemet
Allm&auml;nfriktionslag
Turbulent r&ouml;rstr&ouml;mning
Friktionskoefficient, Moody&acute;s diagram
Icke-cirkul&auml;ra r&ouml;r
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(total energi)
(tryckniv&aring;)
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Bell-mouth = Trattformig
Pipe systems – branched pipe systems
J
Solution
3 Possible flow situations:
1) From reservoir 1 and 2 to reservoir 3
2) From reservoir 1 to reservoir 2 and 3
3) From reservoir 1 to 3 (Q2 = 0)
For the situation as shown:
Energy equation 
HJ = PJ/w + zJ + V2J/2g
hf1 + hlocal,1 = z1 – HJ
hf2 + hlocal,2 = z2 – HJ
hf3 + hlocal,3 = HJ – z3
Continuity equation  Q3 = Q1 + Q2
As HJ (HJ is total head at J) is initially
unknown, a method of solution is
as follows:
1) Guess HJ
2) Calculate Q1, Q2, and Q3
3) If Q1 + Q2 = Q3, then the solution is
correct
4) If Q1 + Q2 ≠ Q3, then return to 1).
Ett vattenfl&ouml;de p&aring; 60 l/s str&ouml;mmar genom r&ouml;rledningen i Figur 2.5
a) Om vattnet stiger 3.0 m ovan r&ouml;rets centrum i Pitot-r&ouml;ret vid B,
hur h&ouml;gt stiger vattnet i piezometern vid A?
a) Hur stort &auml;r det statiska trycket (i Pa) vid r&ouml;rcentrum i B?
F&ouml;rsumma alla energif&ouml;rluster.
+ 3.0 m
Figur 2.5
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Vad &auml;r skillnad mellan A o B?
p
V 
 w  z  2 g   H  konst.

A
2
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I figuren visas en Venturimeter (fl&ouml;desm&auml;tare). Vilket vattenfl&ouml;de g&aring;r
genom r&ouml;rledningen f&ouml;r det aktuella differentialmanometerutslaget i
ledningen? Relativa densiteten f&ouml;r kvicksilver &auml;r SHg = 13.56.
F&ouml;rsumma eventuella f&ouml;rluster mellan 1 och 2.
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2.5 En vattenstr&aring;le avb&ouml;rdas med ett konstant fl&ouml;de av 35 l/s fr&aring;n
den &ouml;vre tanken, se figuren nedan. Om jetdiametern vid sektion 1
&auml;r 10 cm, vilka krafter m&auml;ts av v&aring;garna vid A och B. Anta att en
tom tank v&auml;ger 135 kg och att tankens ytarea &auml;r 0.4 m2. H = 3 m
och h = 0.3 m.
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1
z  V t   g t2
z0
2
V  V  g t
z
z0
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(Based on Cast Parable)
I37: This corrugated ramp is used as an energy dissipator in a twodimensional open channel flow. For a flowrate of 5.4 m3/(sm)
calculate the head lost, the power dissipated, and the horizontal
component of force exerted by the water on the ramp.
2
X
1
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The hydraulic characteristics for the pipe
system, Hsyst, is obtained from the energy
equation 
L Q2
H
 z  h
 z  f
syst
losses
D 2 gA 2
(local losses neglected in this case)
Hsyst states how much energy that is
needed to transport 1 kg of water
from A to B
Hp states how much energy the
pump can provide to the water
When the pump is introduced in the
pipe system the flowrate and pump
head will adjust so that Hsyst = Hp
Hsyst: Systemkurvan
best&aring;r av skillnad i niv&aring; + friktionsf&ouml;rluster
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PARALLEL PUMPING
To deal with cases where you have pumps operating in
parallel you can consider them as being replaced by a fictive
equivalent pump with a pump curve obtained by horizontal
addition of the single pumps pump curves
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PUMPS IN SERIES
To deal with cases where you have pumps operating in series
you can consider them as being replaced by a fictive
equivalent pump with a pump curve obtained by vertical
addition of the single pumps pump curves
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