424521 /2018 Pages 65-68 Fluid & Particulate Systems Fluid...

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Fluid and Particulate systems 424521 /2018

DYNAMIC FLOWS /COMPRESSIBLE FLOWS /MOMENTUM BALANCES

Ron ZevenhovenÅA Thermal and Flow Engineering

[email protected]

4

Pages 65-68added 3.4.2018

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4.1 Dynamic, time-dependentflow

Åbo Akademi University - Värme- och Strömningsteknik Biskopsgatan 8, FI-20500 Åbo / Turku Finland

RoNz 2april 2018

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Dynamic Flow

t

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Two processes

tseconds minutes

Fastprocess

Slowprocess

april 2018 RoNz 4Åbo Akademi University - Värme- och Strömningsteknik Biskopsgatan 8, FI-20500 Åbo / Turku Finland

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Slow process

minutes

Fastprocess

Slowprocess

seconds

Calculate the pseudo-stationary state at several points during the process.Integrate the result to achieve the value of the searched variable.

4.1

Lowering of level in storage vessel;Lowers tube inlet pressure

2

1 )()(:Xproperty of change 12

X

X Xf

dxttXf

dt

dX


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Fast process

minutes

Fastprocess

Slowprocess

seconds

The state is changing within the control volume; time derivates of energyneed to be accounted for.

Acceleration of the water mass in the of tube: F = mꞏa = Δp/L


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Time derivates of energy dt

zgmd

dt

umd

dt

dU

dt

wmd 221

Potential energy:

Internal energy:

Kinetic energy:

Neglected at non-compressible flowwhen assuming that the liquid massin a pipe is constant.

Partial energy balance for internal energy

dt

dwwAl

4.27

Useenthalpy

hinstead

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Time-variant pressure balance for fast processes

dt

dwlpppwwzzg loss21

22

212

121

4.3, 4.4april 2018 RoNz 8Åbo Akademi University - Värme- och Strömningsteknik

Biskopsgatan 8, FI-20500 Åbo / Turku Finland

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4.2 Example: water hammer


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Water hammer

What happens if the valve is suddenly closed?

Theoretically: Deceleration of an incompressible fluid and a non-expandablepipe would cause infinite high pressure before the valve.

Practically: The kinetic energy of the liquid in motion will compress the fluid andexpand the pipe (a slightly bigger perimeter). This results in a high pressureshock wave bouncing back and forth, which is experienced as a loud bangresembling a hammering noise. Friction will, however, decay the wave fast.

l


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Compressibility of a liquid

pc

c

p

v

dp

dV

V

1

For water at 20ºC

For ideal gases (isentropic compression)

-110 Pa108.4

4.5, 4.6

Compressibility

11

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Pressure rise by water hammer

Ed

wp 0

0wdE

velocity of the liquid before the valve closes m/sDensity kg/m3

Compressibility Pa-1

pipe diameter mcoefficient of elasticity Papipe wall thickness m

Ed

c1

speed of sound in the liquid in tube m/s

cwp 0

speed of sound in free flow m/s

1c

Ekinetic → Wliquid,kompression + Wpipe,expansion → Einternal

Transformation of energy

For examplesteel pipeE = 2.1×1011 Pa at wo = 10 m/s: Δp ~ 10 MPa

12

(Unit term √ .. = s/m)

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Water hammer period

What will happen if the valve is suddenly closed?

Length l


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Water hammer period

Consider water flowing to the right as a spring. At t = 0, the water is hittingthe instantaneously closed valve.

c (velocity of the phenomena)

The water flow at the valve will stop and water is compressed, at thevelocity of sound c in water. At t = ½ꞏl /c, half of the water has stoppedflowing and is compressed (kinetic energy → flow energy = enthalpy). The rest is still moving to the right.

t=0

t=½l/c

l

cFLOW

0

FLOW

0


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Water hammer period

c

t=l/c

All the water has stopped flowing at t = l /c and the pressure in thepipe is very high. The water is beginning to flow backwards into thetank, since the pressure is lower there.

c

t=1½l/c

Half of the water in the pipe is flowing back into the tank and pressureis reduced (flow energy (enthalpy) → kinetic energy). The rest has still a high pressure, which will soon will be released.

FLOW

0

FLOW

0


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Water hammer period

t=2l/c

c

All the water is moving to the left and the high pressure isreleased at t = 2l /c.

c

t=2½l/c

At t = 2½ l /c, half of the water has stopped flowing to the left and is“stretched”, which means that the pressure will become low.

FLOW

0

FLOW

0


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Water hammer period

c

t=3l/c

The flow backwards has stopped completely and the pressurein the pipe is very low at t = 3l /c. Water will start to flow forwards intothe pipe again.

c

t=3½l/c

Half of the pipe has water flowing to the right, while the other halfhas a low pressure causing the water to flow at t = 3½l /c.

FLOW

0

FLOW

0


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Water hammer period

t=4l/c

c

All the water is flowing towards the valve at t = 4l/c. The situation isequal to t = 0.

The phenomenon will decayfast because of friction.

http://www.queensu.ca/cloe/projects/public/waterhammer/Simulation.html dead link

FLOW

0


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Critical closing time

Water hammer occurs in a pipe if a valve is closed faster than

c

lt

2critical

4.7

More risk for water hammer with long tubes!!

http://wn.com/1-d_water_hammer video # 14 (checked April 2018)

See for a simulation of water hammer:


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Water hammer arrester

Hydraulic ram

http://schou.dk/animation/

Hydropower application


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Pressure surges during valve closure

Fluid volume = AL; Fluid mass = ρ ∙ A ∙ L ;

Gradual closure of valve:

Decelaration = u/t; t= time for reducing velocity to 0.

Inertial force required = F = m ∙ a = ρ∙A ∙ L ∙ u/t

pressure rise Δp = F/A = ρ ∙ L ∙ u/t

april 2018 Åbo Akademi University - Värme- och Strömningsteknik Biskopsgatan 8, FI-20500 Åbo / Turku Finland

RoNz 21

Pipe diameter DCross section A

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Fluid volume = A∙L; Fluid mass = ρ∙A∙L ;

Sudden closure of valve, with compressibility κ:

a shock wave with velocity c = 1/√(κꞏρ)

The change in volume ~ change in pressure

Average pressure rise is ½Δp; storing compressionenergy ½Δp∙ΔV, which is equal to kinetic energychange ½m∙u2 pressure rise Δp = cꞏuꞏρ


RoNz 22

Pipe diameter DCross section A

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Δp may cause pipe swelling.An elastic wall (elasticitymodulus E) will lower the pressure rise. Kinetic energywill be absorbed by the wall.

With wall thickness δ the stress is σ = ΔpD/2δ;givingstrain energy S = σ2/2E×material volume = Δp2DAL/2δE

Wall strain energy + fluid strain (compression) energy is equal to the kinetic energy lost by the flow which is ½mu2 = ½ρALu2. Fluid strain energy ½ΔpΔV = ½Δp2ALκ

Δp2DAL/2δE + ½ΔpΔV= ½ρALu2


RoNz 23

Wall (material volume = πDδL

ED

up

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4.3 Excercises 4


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Exercises 4 marked** will not be discussed; answers will be distributed

4.1 Two containers, A and B, with cylindrical cross sections have the diameters dA = 4 m and dB = 3 m. They are joined together with a horizontal pipe (l = 20 m, d = 75 mm, wall roughness k = 0.5 mm, Σζ´ = 4.5). The water level zA in container A is 6.5 m over the pipe, while the water level in container B has the same level (zB = 0m) as the pipe at the time t0. Calculate the time needed for transporting 28 m3 water (10ºC) from A to B, under the conditions that only the mentioned pipe is utilized and the storing capacity of B is big enough.

4.2 Derive the expression for the time derivate of kinetic energy.

4.3 Derive the expression for the time-variant pressure balance for fast processes.

** 4.4 Water (ρ = 998 kg/m3, kinematic viscosity ν =10-6 m2/s) is flowing through a horizontal pipe (l = 100 m, d =100 mm, k = 0.2 mm) with the velocity w = 3.0 m/s. The static pressure before the inlet is p1 = 220 kPa and the static pressure after outlet is p2 = 100 kPa. A valve near the outlet is instantly closed at the time t0 and the ζvalve value changes then from 1 to 50. Calculate and illustrate the velocity w during the first 4 seconds after the valve was closed. Calculate the pressure in front of the valve at the time of closing.


RoNz 25

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Exercises 4 4.5 A volume of 1 m3 water at 20ºC and 100 kPa is compressed by 0.0010%.

a) What is the pressure at the end of the compression?b) How much theoretical compression work is needed per kg of water?c) How much does the specific enthalpy of the water increase?

**4.6 To what value will the pressure rise in a 400 m long pipe (d=100 mm) where water at 20ºC is flowing with a velocity of 5 m/s if a valve in the far en is closed immediately? Assume that all kinetic energy of the moving water is used for compression of the liquid.

4.7 a) What is the speed of sound in water at 20ºC in a 400 m long steel pipe (d =100 mm, δ = 5 mm, E = 2.1·1011 Pa)?

b) What is the critical closing time for a valve in the far end?c) What pressure increase can be expected if the valve is closed too fast and the water is flowing with a velocity of 5 m/s?


RoNz 26

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4.4 Compressible flow


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Compressible Flow

static pressure velocity

static pressure velocityNon-compressible flow

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Compressible fluid

p is highp is low

specific volume v is low(density high)

specific volume v is high(density low)

Concerns gases. Liquids are usually approximated as non-compressible.


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Gas flow in a pipe calculations

The normal pressure balance can be used if Δp = p1 – p2 < 20%. Use then w and ρ pertaining at pmean; gas behaves like liquid

Calculate as a connection in series. Constant values of w and ρ can beassumed for each section. Temperature changes due to heat transfer canalso be accounted for.

1

2

static pressure velocity

p decreases → v increases → V increases → velocity w increases.


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Calculations

dlw

ddp d

2

2

3 Consider the case in which the temperature is constant and the changes in the potential and kinetic energies between the in- and outlet can be neglected. The decrease in static pressure –dp for a thin slice dl of the pipe (duct) is equal to the loss of pressure due to fluid friction.

The velocity w of a flowing gas can anywhere in a pipe be expressed as a function of the pressure p and the state at the inlet p1, w1 (alternatively thestate at outlet p2, w2), using the ideal gas law, for molar mass M kg/kmol:.

TR

pM

V

nMTRnVp

The density ρ of a flowing gas can likewise anywhere in a pipe beexpressed as a function of the pressure p and the state at the inlet p1, ρ1

(alternatively the state at the outlet p2, ρ2), according to the ideal gas law.

31

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Calculations

loss,1p

dlw

ddp d

2

23 continued…

Integration of and expressing w=f(w1,p1,p),

and ρ = f(ρ1,p1,p), the pressure at the outlet p2 can be calculated if the state of the inlet “1” is known according to

where

)2( loss,1112 pppp

is the fictive pressure loss if the state at the

inlet would pertain for the whole pipe.

32

*see pages 65-68

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Calculations

loss,2p

dlw

ddp d

2

2

3 continued…

and expressing

w = f(w2, p2, p) and ρ = f(ρ2, p2, p), the pressure at the inlet p1 can be calculated if the state of the outlet “2” is known according to

where is the fictive pressure loss if the state at the

outlet would pertain for the whole pipe.

Alternatively, integration of

)2( loss,2221 pppp

33

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Mass flow always constant

5.1 34

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Temperature constant?

2211 VpVp

/122

/111

pTpT

122

111

VTVT

An isentropic expansion of an ideal gas would decrease the temperature in an isolated system.

An isentropic compression of an ideal gas would increase the temperature in an isolated system.

Isentropic changes from a beginning state “1” to an end state “2” can be described by

v

p

c

c

“1”

“2”

5.2 35

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Temperature constant?

The expansion of a flowing ideal gas in a pipe is not isentropic since the pressure loss due to fluid friction involves an entropy generation.

Temperature decrease due to expansion

Temperature increase due to fluid friction→ temperature remains constant

Changes in potential and kinetic energies are neglected.

The temperature increase for a real gas, e.g. air, can also be neglected.

5.3, 5.4 april 2018 RoNz 36Åbo Akademi University - Värme- och Strömningsteknik


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4.5 Excercises 5


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Exercises 5 marked** will not be discussed; answers will be distributed,

5.1 a) Derive the expressions for

b) Can it be assumed that the values of Re and , for a compressible gas, are constant although the pressure decreases?c) In what cases can the change in kinetic and potential energies, between the in-and outlet of the control volume, be neglected?

5.2 Explain qualitatively what temperature and pressure changes occur after the valve has been opened.


RoNz 38

)2( loss,1112 pppp )2( loss,2221 pppp

pA,1 = 2ꞏpB,1

A B

TA,1=TB,1

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Exercises 5 5.3 What is special with the enthalpy of a flowing gas in a pipe?

**5.4 Air with a pressure of 600 kPa and a temperature of 200ºC is transported through a pipe (l=500 m, d=50 mm, roughness k=0.05 mm). The pressure at the outlet is 100 kPa.a) Calculate the mass flow rate under the assumption that the process is isothermal at 200ºC.b) Calculate the mass flow rate in a case where the surrounding air at 0ºC has a cooling influence; the heat transfer coefficient related to the inner surface of the pipe is assumed to be 3.0 W/(m2 K). Calculate the pressure distribution along the pipe.Pressure losses due to inlet, outlet and valves can be neglected.


RoNz 39

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4.6 Momentum balances, the ejector


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Applied Momentum BalancesF

luid

& P

artic

ulat

eS

yste

ms

4245

21 /

201

0Fl

uid

& P

artic

ulat

eSy

stem

sÅ

A 4

2452

1 /

201

8

The Ejector

The drive medium a is supplied at high pressure and flow through the nozzle at high speed into the mixing space. The pump medium b is sucked into the mixing space due to the high momentum flux provided by the drive medium a. The drive medium a and pump medium b are mixed (s) in the mixing space and the pressure increases in the flow direction. A further increase in the pressure is achieved in the diffuser, where kinetic energy is partly transformed into flow energy.

a

b

smixing space diffusernozzle


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The Ejector

a

b


“3”“2”“1”

“0”

“0”

If the drive medium a is steam, which condensates in the mixing space, the outletpressure p3 might reach values higher than the inlet pressure pa0 of the steam.This kind of ejector is called injector, which can be used as feed pump for steam boilers.

This is an easy and inexpensive pump to manufacture, and the operation is safesince it does not contain any moving parts. It has, however a low efficiency andit might be a negative aspect that the drive medium a is mixed together with thepump medium b.


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The Ejector

Energy and momentum balances can be utilized as tools for picturing theworking boundaries and the efficiency of the ejector when put into practice.

Following ideal conditions are assumed:- densities are constant for the drive medium ρa and the pump medium ρb

- no losses due to fluid friction in the mixing space- kinetic energy and momentum ξ coefficients are assumed to be 1- ideal inflow; no losses when flow energy is transformed to kinetic energy

at boundary “1”- constant diffuser efficiency (80%)


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The Ejector

a

b


“2”“1”

“0”

“0”“3”

7.145

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The Ejector

a

10aa1

2

pp

w

b

10bb1

2

pp

w

b1a1s2 1 www

s2

bb1aa1s

1

w

ww

2s2s

2b1b

2a1a12 1 wwwpp

2s2s23 4.0 wpp

b00a1

b03s2e ppw

ppw

a

7.2

The following equations can be utilized for calculations of ejectors.

46

)/m(m 22

cs

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spacemixing

nozzle

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A

A

A

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The Ejector (Injector)

If the drive or pump medium (or both) are gaseous (steam), changes in the specific volume (density) have to be accounted for.

In such case it might be appropriate to utilize energy and entropy balances for rough estimates of the ejector performance.

7.3april 2018 RoNz 47Åbo Akademi University - Värme- och Strömningsteknik


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4.7 Momentum balances, propellors and wind turbines

see also ÅA course New Energy Technologies 424517and REN21 annual reports on renewable energy for windenergy status


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REN21

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49/64

Wind energy – Betz’ Law

Theoretically (Betz’ Law) a wind turbine can exploit 60% (16/27) of incoming windpower.

In practice it is ~40%, due to (exergy) losses, mainly as heat, for example friction betweenrotor shaft and bearings, cooling fluids in gear box and generator.

Pic

s: h

ttp://

guid

edto

ur.w

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ower

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/en/

stat

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zpro

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Power input P > 0 (boats, airplanes)

w1 undisturbedinput velocity w2 velocity at a

distance of ~ Dp

after the propeller

wp velocity throughthe propeller

“1”

“2”

“1” “2”

Sweep area ofthe propeller: 4

2p

p

DA

Dp

Input power PInput axle force F


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Power output P < 0 (wind power plants)w1 undisturbedinput velocity

w2 velocity at a distance of ~ Dp

after the propeller“1”

“2”

Output power-POutput axle force-F


“1” “2”

: mass flow if fluid through the propeller


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Energy and momentum balancesThe simplified theory is based on the assumption that the propeller power is used for changing the kinetic energy (without losses) of the mass flow going through the sweep area of the propeller.

Idealizations: -The propeller is considered as a pump (η =100%) in a pipe with the diameter

Dp and the length 0- The kinetic energy coefficients ξ are assumed to be 1- Neglecting that the propeller jet is spinning around its axis (= power loss)- Neglecting the influence of the number of propeller wings- The pressure is constant along the balance boundaries. However, there is a

pressure difference Δp on both sides of the propeller distributed over thewhole sweep area

7.4april 2018 RoNz 52Åbo Akademi University - Värme- och Strömningsteknik


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Velocity wp through the propeller

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Betz’ Law (1926)

P is power extracted from the wind

Power of undisturbed flow through area Ap

Efficiency of power exraction from wind

with maximum 16/27 at w2/w1=1/3


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p2122

21airp21

22

21

air A)w½(www½ρA)w(www4

ρP

wAΔpwFAw2

ρP p

31

air0

)1(12

1

1

221

22

0 w

w

w

w

P

P

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Propeller pitch s (sv: stigning)

nsw p n: rotational speed

7.5


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Recent developments

See S Langton „Fitting a pitch“ Mechanical Engineering (ASME) December 2009, p. 38-42


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Savonius turbines


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@ Vaasan yliopisto

SOURCES:http://vaasalaisia.info/savonius-tuulivoimala-saa-lisaa-tehoa/ (dr thesis S Marmutova, May 2016)https://www.researchgate.net/publication/228957621_Increase_in_the_Savonius_rotors_efficiency_via_a_parametric_investigationhttp://vaasalaisia.info/wp-content/uploads/2016/05/savonius_tuuliv-oimala_vaasan_yliopisto.jpg

Efficiency etc. follow from Betz law assessment, power :with fluid density ρ, rotor height h, radius r and fluid flow velocity v. In practice, constant 16/27 0.36Angular frequency ω ≈ v/r (depending on turbine).For example, r = 0.5 m, h = 1 m, v = 10 m/s, ω ≈ 20 rad/s ~ 190 rpm, gives P ~ 180W

http://www.savonius.net/project-background.html

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Rotor sails, rotor ships


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Principle of operation: Magnus effect

”Norsepower”: Fuel savings for shipping with one rotoron a 9700 DWT were 2.6%, assuming potential of ~20% with several rotors.

SOURCES:https://lucasbioblog.weebly.com/the-magnus-effect.htmlhttps://www.norsepower.com/https://www.engineering.com/ElectronicsDesign/ElectronicsDesignArticles/

ArticleID/10259/Shipping-Company-Takes-Rotor-Sail-for-a-Spin.aspx

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4.8 Excercises 7


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Exercises 7 marked** will not be discussed; answers will be distributed

7.1 Derive equations for evaluating the ejector pump in practice.

7.2 An ejector pump is used for emptying a flooded basement floor. The drive medium consists of tap water with a pressure pa0 of 900 kPa. A suitable inlet pressure pb0 of the pump medium would be 80 kPa, and the outlet pressure p3 is expected to be 130 kPa. The tap water flow is 1 l/s. Design the ejector (calculate Acs and Acs,a) that maximizes the mass flow rate of the pump medium.


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a

b


“2”“1”

“0”

“0”“3”

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Exercises 7 **7.3 A steam injector is consuming 0.105 kg/s saturated drive medium steam

with a pressure of 420 kPa for compression of 0.120 kg/s wet steam (containing 2% water) with a temperature of 15ºC to the pressure of 4.0 kPa.a) Calculate the temperature of the out coming steam at 4.0 kPab) Calculate the efficiency of the injector by comparing the turbine power, which the drive medium steam (0.105 kg/s) would theoretically give off at an isentropic expansion to 4.0 kPa, with the compressor power, which would be needed for isentropic compression of 0.120 kg/s steam to 4.0 kPa.

7.4 Derive momentum balances fora)


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w1 undisturbedinput velocity w2 velocity at a

distance of ~ Dpafter the propeller

“1”

“2”

Input power PInput axle force F

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Exercises 7 7.4 Derive momentum balances for

a)

b)

c) What is the relationship between the power (input or output) P and the pressure difference Δp on both sides of the propeller?


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w1 undisturbedinput velocity

w2 velocity at a distance of ~ Dpafter the propeller

“1”

“2”

Output power -POutput axle force -F


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Exercises 7

**7.5 A wind motor has a propeller diameter of 6 m.a) How much axle power can theoretically be achieved if the wind flow (0 ºC) is w1 8 m/s and the wind velocity w2 (ca 6 m after the propeller) is 2 m/s.b) What will the rotational speed be if the propeller pitch is 15 m.c) Is w2 = 2 m/s optimal? What w2 would be optimal for reaching maximum power output?d) Show that the maximum power output is 16/27 (=59%) of the kinetic energy that an air flow is possessing, that undisturbed with the velocity w1 is flowing through a ring with the propeller diameter Dp.


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Further reading DWIA09 “Proof of Betz’ Law” Danish Wind Industry Association

http://guidedtour.windpower.org/en/stat/betzpro.htm

KJ05: D. Kaminski, M. Jensen ”Introduction to Thermal and Fluids Engineering”, Wiley (2005) Chapter 4, chapter 10

L09: S Langton „Fitting a pitch“ Mechanical Engineering (ASME) December 2009, p. 38-42 (as pdf at course material www)

SWE = magazine Sun & Wind Energy http://www.sunwindenergy.com T06: S.R. Turns ”Thermal – Fluid Sciences”, Cambridge Univ. Press (2006) Z13: R. Zevenhoven ”Introduction to process engineering / Processteknikens

grunder” (PTG), # 6: Fluid mechanics: fluid statics, fluid dynamics. Course material ÅA (version 2013) available on line http://users.abo.fi/rzevenho/PTG%20Aug2013.pdf

ÖS96: G. Öhman, H. Saxén ”Värmeteknikens grunder”, Course compendium ÅA (1996)

Ö96: G. Öhman ”Teknisk strömningslära” Course compendium ÅA (1996) sections 3, 4, 7 (p. 8-14, 22-26)

CEWR10: C.T. Crowe, D.F. Elger, B.C. Williams, J.A. Roberson ”Engineering Fluid Mechanics”, 9th ed., Wiley (2010)

REN21: Renewables 2017 Global Status Report (GSR 2017) http://www.ren21.net/wp-content/uploads/2017/06/17-8399_GSR_2017_Full_Report_0621_Opt.pdf (302 pages)http://www.ren21.net/status-of-renewables/global-status-report/ (links to Figures etc.)

april 2018 Åbo Akademi University - Värme- och StrömningsteknikBiskopsgatan 8, FI-20500 Åbo / Turku Finland

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Add: compressible flow pressure drop


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424521 /2018 Pages 65-68 Fluid & Particulate Systems Fluid...

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