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Turbomachinery Lecture Notes KTH Course MJ2429/MJ2241

Turbomachinery for Compressible Fluids

DRAFT version Turbines

Damian M. Vogt KTH Heat and Power Technology

Turbomachinery Lecture Notes 1 2007-10-08

Compressible Aerodynamics

Damian Vogt CourseMJ2429

Nomenclature

Symbol Denotation Unit a Speed of sound m/s c Flow speed m/s cp Specific heat at constant

pressure J/kgK

cv Specific heat at constant volume

J/kgK

h Enthalpy J/kg M Mach number - p Pressure Pa R Gas constant J/kgK s Entropy J/kgK T Temperature K v Specific volume m3/kg γ Ratio of specific heats

(also referred to as “isentropic exponent”

-

ρ Density kg/m3

Subscripts

0 Total s Isentropic

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Compressible Aerodynamics The theory of compressible aerodynamics deals with variable density flow. In real life every fluid is to some extent compressible. For many fluids, e.g. for liquids, the compressibility is however so low that it usually does not need to be accounted for. In case if incompressible flow we can use Bernoulli’s equation as given by

20 2

cpp ρ+= Eq. 1

This equation can also be used for gas flows at low speeds, as the change in density can then be neglected. At higher flow speeds it is however necessary to take into account the compressibility of the gas. A measure of significance is the Mach number that relates the flow speed to the speed of sound as follows

acM = Eq. 2

The speed of sound tells us at which speed information can be transmitted in a gaseous medium. To derive an expression for the speed of sound recall that the air around us consists of molecules that are continuously undergoing random motion at a certain velocity referred to as “molecular speed”. From statistical thermodynamics the molecular speed can be determined to πRT8 . Imagine a sound source that creates a pressure disturbance at a certain point. From that point sound waves will propagate equally in all directions similar to the waves that occur when throwing a stone into water. This sound wave is nothing else than a wave front that travels through the air leaving a “track of changed properties” behind as sketched in figure 1.

a

T

ρ

p

a+da

T+dT

ρ+dρ

p+dp

Figure 1. Change of properties across sound wave front Note that the sound wave is regarded as a weak wave and that the changes in properties are therefore infinitesimal. To derive an expression for the speed of sound we set up the continuity equation as follows

( )( ) daddaadadaada ρρρρρρρ +++=++= Eq. 3

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After dropping the second order terms, i.e. products of infinitesimal quantities, and simplifying, we obtain

daad ρρ −= Eq. 4 or

ρρddaa −= Eq. 5

To substitute da we set up the momentum equation as follows

( ) ( )( )22 daaddppap ++++=+ ρρρ Eq. 6 Resolving yields

22222 22 dadadadaddaadaadppap ρρρρρρρ +++++++=+ Eq. 7 Again we drop second order terms and simplify such that

ρρ dadaadp 220 ++= Eq. 8 From which we can obtain an expression for as follows da

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−=

ρρ

adadpda

2

2

Eq. 9

Substituting Eq. 9 into Eq. 5 we obtain

a

addp

adadp

da

22

22 +

=⎟⎟⎠

⎞⎜⎜⎝

⎛ += ρ

ρρ

ρρ

Eq. 10

, which can be simplified to

ρddpa =2 Eq. 11

At this position we recognize that the speed of sound square is proportional to the derivation of pressure with respect to density. Further above it has been stated that a sound wave is a weak wave and that the changes in properties are therefore infinitesimal small. Considering that there is no heat addition (i.e. the process is adiabatic) and that irreversible effects can be neglected the process of sound wave propagation must be isentropic and therefore

s

pa ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=ρ

2 Eq. 12

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For a perfect gas the relation between pressure and density is given by

.constp=γρ

Eq. 13

Deriving Eq. 13 with respect to density we obtain

ργ

ρpp

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

Eq. 14

And finally

ργpa =2 Eq. 15

Keeping in mind that RTp=

ρ it yields that

RTa γ= Eq. 16

From this expression for the speed of sound the following can be recognized:

• The speed of sound is only dependent on gas properties and the temperature • The expression for the speed of sound is similar to the one for the molecular velocity;

indeed the speed of sound amounts to around 75% of the molecular velocity • The higher the temperature the higher the speed of sound; this is due to increased

activity of the gas molecules From the expression of total enthalpy it has previously been shown that there is a relation between flow velocity and static enthalpy in a fluid flow as follows

2

2

0chh += Eq. 17

Below it is demonstrated that this expression is valid with the consideration of compressible flow and that it leads to expressions for the Mach number in dependency of total-to-static temperature and total-to-static pressure. As Eq. 17 can be rewritten as Tch p=

2

2

0cTcTc pp += Eq. 18

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Dividing by T and leads to pc

pcc

TT

21

20 += Eq. 19

The specific heat at constant pressure can be expressed by the gas constant and the ratio of specific heats as follows

1−=

γγRc p Eq. 20

Substituting Eq. 20 into Eq. 19 leads to

( )RT

cTT

γγ

2)11

20 −

+= Eq. 21

wherein the speed of sound can be identified in the denominator, thus

( )2

20

2)11

ac

TT −

+=γ

Eq. 22

Expressing 2

22

acM = then leads to

20

211 M

TT −

+=γ

Eq. 23

To obtain a relation between the total-to-static pressure ratio and the Mach number we apply the isentropic relation as follows

γγ 1

00

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pp

TT

Eq. 24

leading to

120

211

−

⎥⎦⎤

⎢⎣⎡ −

+=γγ

γ Mpp

Eq. 25

Resolving for M we can also write

21

1

0 11

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

−γ

γ

γ pp

M Eq. 26

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Ernst Mach

Source: wikipedia

Ernst Mach (1838-1916) was an Austrian physicist and philosopher and is the namesake for the "Mach number". Mach was born in Chrlice (now part of Brno), Czech Republic. He was educated at home until the age of 14, then went briefly to gymnasium before entering the University of Vienna at 17. There he studied mathematics, physics and philosophy, and received a doctorate in physics in 1860. His early work was focused on Doppler effect in optics and acoustics. In 1864 he took a job as professor of mathematics in Graz, in 1866 he was also appointed as a professor of physics. During that period Mach became interested also in physiology of sensory perception. In 1867 Mach took the chair of a professor of experimental physics at Charles University, Prague. Mach returned to the University of Vienna as professor of inductive philosophy in 1895, but he suffered a stroke two years later and retired from active research in 1901, when he was appointed to the Austrian parliament. He continued to lecture and publish in retirement. Mach died on 19 February 1916 in Haar, Germany.

Daniel Bernoulli

Source: wikipedia

Daniel Bernoulli (1700-1782) was a Dutch-born mathematician who spent much of his life in Basel, Switzerland. He worked with Leonhard Euler on the equations bearing their names. He went to St. Petersburg in 1724 as professor of mathematics, but did not like it there, and a temporary illness in 1733 gave him an excuse for leaving. He returned to the University of Basel, where he successively held the chairs of medicine, metaphysics and natural philosophy until his death. His father wanted him to do this because he believed that there was no money in mathematics. Undoubtedly the most important work which Daniel Bernoulli did while in St Petersburg was his work on hydrodynamics. Even the term itself is based on the title of the work which he produced called Hydrodynamica. This work contains for the first time the correct analysis of water flowing from a hole in a container. This was based on the principle of conservation of energy which he had studied with his father in 1720. Daniel also discussed pumps and other machines to raise water.

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http://upload.wikimedia.org/wikipedia/commons/5/57/Ernst-Mach-1900.jpg


Review of Basic Laws of Turbomachinery

Damian Vogt Course MJ2429

Nomenclature

Symbol Denotation Unit A Surface area m2

E Internal energy J F Force N I Rothalpy J/kg M Moment Nm Q Heat energy J W Mechanical energy J c Absolute velocity m/s g Gravitational constant m/s2

h Enthalpy J/kg m& Mass flow rate kg/s p Pressure Pa r Radius m t Time s u Tangential velocity m/s w Relative velocity m/s z Height coordinate m ρ Density kg/m3

ω Rotational speed rad/s Subscripts

0 Total 1 Inlet stator 2 Outlet stator (inlet rotor) 3 Outlet rotor c Total in absolute frame of reference n Normal R Radial component w Total in relative frame of reference X Axial component θ Tangential component

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Turbomachine System Discretization

30

20

10

0

10

20

30

−30

−20

−10

0

10

20

c2

c1

w2

u

u

w3 u

1

2

1 2 3 stator rotor

ω

Turbine shown

stator rotor

Stage velocity triangles

Stage denotations

1 stator inlet 2 rotor inlet 3 rotor outlet

Schematic representation

1 inlet 2 outlet

c3

A turbomachine consists of one or several stages

Q&

W&

The thermodynamic and kinetic properties in a stage are defined by velocities

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Conservation of Mass Sum of mass flow rates over all system boundaries equals to change in mass in control volume

∑ ∂∂

=i t

mm& Eq. 1

For steady process mass in control volume constant over time 0=∂∂

tm

thus

Eq. 2 ∑ =i

m 0&

Mass flow rate through boundary Acm n ⋅⋅= ρ& Eq. 3 Conservation of mass for control volume featuring one in- and one outflow 22,211,1 AcAc nn ρρ = Eq. 4 Note: The indexes “1” and “2” refer to inlet and outlet of the control volume respectively as depicted below

1 2

Control volume e.g.: stage

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Conservation of Energy First law of thermodynamics applied to closed process, i.e. system taken through a complete cycle

∫ =− 0)( dWdQ Eq. 5

Change in internal energy during change in state from one point to another in the cycle Eq. 6 dWdQdE −= For a steady flow process the conservation of energy per unit time is regarded, i.e. conservation of power Eq. 7 WQdzgdhmEd &&&& −=⋅+⋅= )( 0

Where denotes the change in total enthalpy and the term 0dh dzg ⋅ change in specific potential energy. Apart from hydraulic machines the latter can be neglected. Furthermore the process can be assumed as adiabatic leading to the conservation of energy for a steady turbomachine process being written as Eq. 8 )( 0201 hhmW −⋅= &&

Note:

• work producing machines (turbines) 00201 >⇒> Whh &

• work absorbing machines (compressors, pumps) 00201 <⇒< Whh &

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Conservation of Momentum According to Newton’s second law of motion the sum of all forces acing on a body equals the change in momentum

∑ ∂∂

=t

mcF x

x)(

Eq. 9

For a steady flow process the change in momentum is exclusively due to the change in flow velocity Eq. 10 ∑ −⋅= )( 21 xxx ccmF &

Note:

• From the perspective of the fluid the forces are acting as pressure forces ( ). ApF ⋅=• A change in velocity indicates a change in pressure remember Bernoulli’s equation for

incompressible fluids: .21 2

0 constcpp =⋅+= ρ

• From the perspective of the turbomachine the pressure forces on the fluid are yielding a

resultant reaction force (actio=reactio). In a turbomachine the moment of momentum rather than the momentum itself is of interest

∑ ∑ ∂∂

=⋅=t

rmcFrM z

)( θθ Eq. 11

For a steady flow process the change in moment of momentum is due to the change in tangential flow velocity Eq. 12 )( 2211 θθ crcrmM z∑ −⋅= &

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Euler’s Turbine Equation At this point the conservation of energy and the conservation of moment of momentum shall be combined. The mechanical work per unit time ( power) equals the product of moment and rotational speed Eq. 13 ω⋅= zMW& Thus the conservation of energy can be related to the conservation of momentum as follows ωθθ ⋅−⋅=−⋅ )()( 22110201 crcrmhhm && Eq. 14 Substituting ω⋅r by the tangential speed u and eliminating yields m& 22110201 θθ cucuhh −=− Eq. 15 The above equation is referred to as Euler’s turbine equation. Note:

• A change in total enthalpy is equivalent to a change in tangential flow speed and/or tangential engine speed

• For engines with little change in mean radius 21 uu ≈ (e.g. axial turbines, axial compressors, fans) the change in total enthalpy is entirely due to change in tangential flow speed θcuh Δ⋅≈Δ 0 blades are bowed

• For engines with large change in mean radius (e.g. radial engines) the change in enthalpy is to a large degree due to the change in radius θcuh ⋅Δ≈Δ 0 centrifugal effect, possibility for larger change in enthalpy

Leonhard Euler

Leonhard Euler (1707-1783) was arguably the greatest mathematician of the eighteenth century and one of the most prolific of all time; his publication list of 886 papers and books fill about 90 volumes. Remarkably, much of this output dates from the the last two decades of his life, when he was totally blind.

Though born and educated in Basel, Switzerland, Euler spent most of his career in St. Petersburg and Berlin. He joined the St. Petersburg Academy of Sciences in 1727. In 1741 he went to Berlin at the invitation of Frederick the Great, but he and Frederick never got on well and in 1766 he returned to St. Petersburg, where he remained until his death. Euler's prolific output caused a tremendous problem of backlog: the St. Petersburg Academy continued publishing his work posthumously for more than 30 years. Euler married twice and had 13 children, though all but five of them died young.

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Rothalpy Reformulating Euler’s turbine equation leads to a fundamental aspect of turbomachine thermodynamics referred to as rothalpy Icuhcuh =−=− 22021101 θθ Eq. 16 The rothalpy is a function that remains constant throughout a rotating machine. Note: in a non-rotating machine the total enthalpy is constant (non-rotating in the above equation)

0=u

The general notation of rothalpy is

θcuchI ⋅−+= 2

21

Eq. 17

This expression can be reformulated by expressing the velocities in the relative frame of reference as follows

c

w cx

u

Cθ

uwcwuc +=→=− θθθθ Eq. 18 Eq. 19 222222 2 uuwwcccc xx +++=+= θθθ

Substituting and in the expression of rothalpy 2c θc

[ ] ( )22

1221 2

222222 uwchuuwuuwwchI xx −++=−−++++= θθθθ Eq. 20

With and the reformulated expression of the rothalpy yields xx wc = 222

θwww x +=

22

21

21 uwhI −+= Eq. 21

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Efficiencies


Nomenclature

Symbol Denotation Unit c Flow speed m/s cp Specific heat at constant

pressure J/kgK

cv Specific heat at constant volume

J/kgK

h Enthalpy J/kg n Polytropic exponent - p Pressure Pa RH Reheat factor - s Entropy J/kgK T Temperature K v Specific volume m3/kg γ Ratio of specific heats

(also referred to as “isentropic exponent”

-

η Efficiency - Subscripts

0 Total 1 Start change of state 2 End change of state p Polytropic s Isentropic tt Total-to-total ts Total-to-static

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Efficiencies In the turbomachinery context a large number of efficiencies are defined such as thermodynamic or mechanical efficiency. In the sections below the focus is put on the thermodynamic efficiencies. For a given change of state of a fluid the efficiency is defined as the ratio between actual change in energy to ideal change in energy in case of expansion or the inverse in case of compression

• Expansion

energy in change idealenergy in change actual

=η Eq. 1

• Compression

energy in change actualenergy in change ideal

=η Eq. 2

The symbol for efficiencies is the Greek letter η (say “eta”). For adiabatic processes the efficiency lies between 0 and 1. Isentropic Efficiency Depending on which process is taken as ideal process efficiencies are referred to as isentropic or polytropic efficiencies. In case of an isentropic efficiency the ideal process is represented by an isentropic change of state from start to end pressure, i.e. the same pressures as for the real process. This is illustrated in figure 1 for an expansion process by means of an enthalpy-entropy diagram (h-s diagram).

h [kJ/kg]

s [kJ/kgK]

h01

h02

h02s

Ideal process

Real process

Δh0

Δh0s

p01

p02

Figure 1. Expansion process

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In the above depicted process the changes in total energy are referred to, which is expressed by indexing the efficiency by “tt”, i.e. “total-to-total”. Recall that the total energy is defined as follows:

2

2

0chh += Eq. 3

The total-to-total isentropic efficiency (expansion) is thus given by

sstt hh

hhhh


0201

0201

0

0−−

=ΔΔ

==η Eq. 4

In case of a compression process the situation is as follows:

h [kJ/kg]

s [kJ/kgK]

h02s

h02

h01

Ideal process

Real process Δh0s

Δh0

p02

p01

Figure 2. Compression process

Total-to-total isentropic efficiency (compression)

0102

0102

0

0hhhh

hh

energy in change actualenergy in change ideal ss

tt −−

=ΔΔ

==η Eq. 5

Note:

• For adiabatic real processes the entropy must always increase during the change of state • Due to this increase in entropy the real change in energy is smaller than the ideal during

expansion. In other words, you get out less energy from the real process than you could have from an ideal one

• For the compression process the increase in entropy signifies that you need to put in more energy to compress a fluid than you would have in an ideal process

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• Therefore the efficiency is always smaller or equal to unity • The only way to reduce entropy would be to cool a process. However in such case we do

no longer look into adiabatic processes In certain cases the kinetic energy that is contained in the fluid (i.e. the amount of energy that is due to the motion) can not be used at the end of a process. An example for such a process is the last stage of a energy producing turbine where the kinetic energy in the exhaust gases is not contributing to the total energy produced. In such case a so-called total-to-static isentropic efficiency is used, identified by indexing the efficiency by “ts”, i.e. “total-to-static”. An expansion line is drawn in figure 3. Note that it is necessary to include total and static states in this case.

h [kJ/kg]

s [kJ/kgK]

h01

h2

h2s

Ideal process

Real process

h01-h02

h1

h02h01-h2s

2

21c

2

22c

p01

p1

p2

p02

Figure 3. Expansion process; total-to-static efficiency

The total-to-total isentropic efficiency (expansion) is thus given by

2

22

0

0

201

0201

ch

hhhhh


ss

ts

+Δ

Δ=

−−

==η Eq. 6

By reformulating the above expression a relation between total-to-total and total-to-static efficiency can be obtained as follows

1

0

22

1

0

22

0

0

1

0

22

0

21

22

−−

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

Δ+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

Δ+

ΔΔ

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

Δ

+Δ=

hc

hc

hh

h

ch

tt

ssts η

η Eq. 7

This relation shows that for values of the total-to-static efficiency is always smaller than the total-to-total efficiency.

02 >c

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Calculating with Isentropic Efficiencies Next the focus is drawn towards the calculation of efficiencies and states. For perfect gases with constant specific heat the enthalpies are only a function of temperature as follows pc

Tch p ⋅= Eq. 8 Furthermore the gas law for a perfect gas relates temperatures and pressures for an isentropic process as given below

.1

constTp =

−γγ

Eq. 9 , where γ is non-dimensional and stands for the ratio of specific heats

v

p

c

c=γ Eq. 10

The two states 1 and 2s at the same entropy are thus related by

γγ 1

02

01

02

01

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

ss pp

TT

Eq. 11

By expressing by sT02

γγ 1

01

020102

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pp

TT ss Eq. 12

the isentropic enthalpy difference can be written as sh 0201→Δ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=Δ=Δ

−

→→γ

γ 1

01

0201020102010201 1)(

pp

cpTTTcpTcph ssss Eq. 13

To obtain the real change in enthalpy the efficiency must be accounted for as shown above yielding

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−=Δ

−

→γ

γ

η

1

01

02010201 1

pp

cpTh stt Eq. 14

Note that the above equation represents a rather common problem; very often the inlet state to a gas turbine is given by (p,T), e.g. from conditions after a combustion chamber. Furthermore the exit pressure of the turbine might be set. As approximation it can also be assumed that . By knowing (or assuming) the efficiency the real change in enthalpy can thus easily be calculaed.

0202 pp s ≈

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Polytropic Efficiency As for the isentropic efficiency the polytropic efficiency relates a real process to an ideal one. The main difference however is that the ideal process in this case is not taken as the single isentropic change of state but rather the flow work, which is defined as follows

∫= vdpy Eq. 15

Thus the definition of polytropic efficiency is given by

yh


p0Δ

==η Eq. 16

The flow work is not easily visualized in the h-s diagram. It can be understood as infinite number of infinitesimal small isentropic changes of state that follow the real expansion line like a saw tooth curve, see figure 4. This consideration also leads to the polytropic efficiency sometimes being referred to as “small-stage efficiency”.

Δh0s,1

h [kJ/kg]

s [kJ/kgK]

h01

h02

Ideal processes

Real

Δh0

p01

p02

Δh0s,2

Δh0s,N

h02s

Figure 4. Illustration of flow work

Note that the sum of all these infinitesimal isentropic changes is greater than the single isentropic change from 1 to 2s.

∑ Δ>Δ sis hh 0,0 Eq. 17 This is due to the fact that the isobars are spread apart with increasing entropy, which in turn is due to the slope of the isobars being proportional to the temperature as follows

Tsh

constp=⎟

⎠⎞

⎜⎝⎛

∂∂

= . Eq. 18

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The growing spreading of the isobars is an indication for increased energy content of the fluid at the same pressure due to increased entropy (hint: see the entropy as measure for disorder; higher temperature leads to greater disorder). By approximating the flow work by the aforementioned infinitesimal isentropic changes a phenomenon known as “reheat” gets apparent; due to the reheating phenomenon part of the heat generated due to losses (i.e. efficiency smaller than unity) is fed back to the fluid as energy and can be used during the process. Following this consideration a reheat factor is defined as follows

10

,0>

Δ

Δ= ∑

s

isH h

hR Eq. 19

For an expansion the isentropic and polytropic efficiencies can now be related by this reheat factor following

ttHsHis

p RhRh

hh ηη 1

0

0

,0

0 =Δ

Δ=

ΔΔ

=∑

Eq. 20

As the reheat factor is larger than unity the polytropic efficiency is smaller than the isentropic efficiency. In case of a compression the polytropic efficiency yields from

ttHsHis

p RhhR

h

hηη =

ΔΔ

=Δ

Δ= ∑

0

0

0

,0 Eq. 21

, which leads to the polytropic efficiency being greater than the isentropic efficiency. From that point of view it is apparent that the polytropic efficiency reflects a different aspect of a change of state of a perfect gas as it takes into account the effect of reheating. By knowing the polytropic efficiency it is possible to apply the gas law as introduced further above to polytropic changes by reformulating

.1

constTp nn

=−

Eq. 22 The coefficient “n” is thereby referred to as polytropic coefficient and is related to the isentropic exponent γ as follows

• Expansion process

γγ

η11 −

=−

pnn Eq. 23

• Compression process

γγ

η111 −

=−

pnn Eq. 24

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Two states 1 and 2, which do not need to be at the same entropy, are now in case of an expansion process related by

γγ

η1

2

1

2

1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

p

pp

TT Eq. 25

Going back to the expansion sketched in figure 1 a relation between the isentropic and the polytropic efficiency and thus the reheat factor can be derived. The total-to-total isentropic efficiency was given by

sstt h

hhh

0201

0201

0

0

→

→ΔΔ

=ΔΔ

=η Eq. 26

It has also been shown that the isentropic change in enthalpy could directly be determined from the gas law for a perfect gas by

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=Δ

−

→γ

γ 1

01

020102010201 1)(

ppcpTTTcph s

ss Eq. 27

By applying the polytropic relation the actual change in enthalpy can be obtained directly from

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=Δ

−

→γ

γη

1

01

020102010201 1)(

p

ppcpTTTcph Eq. 28

Note the presence of the polytropic efficiency in the exponent reflecting the polytropic coefficient. By substituting these expressions into Eq. 26 and under assumption of 0202 pp s ≈ we obtain

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−−γ

γγ

γη

η

1

01

02

1

01

02 11pp

pp p

tt Eq. 29

For small pressure ratios 10201 ≈pp the polytropic and isentropic efficiencies therefore differ very little. With increasing pressure ratio this difference also increases. The reheat factor in case of an expansion process is then obtained from Eq. 20 as follows

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−==

−−γ

γγ

γη

ηηη

1

01

02

1

01

02 111pp

ppR

p

pp

ttH Eq. 30

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Turbine and Compressor Lines in h-s Diagrams


Nomenclature

Symbol Denotation Unit c Absolute flow speed m/s

Enthalpy J/kg h Rothalpy J/kg I Pressure Pa p Entropy J/kgK s Temperature K T Relative flow speed m/s w

Subscripts

Isentropic s Total 0 Inlet stator (T), inlet rotor (C) 1 Outlet stator (T), outlet rotor (C) 2 Outlet rotor (T), outlet stator (C) 3

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Axial Turbine Reaction Turbine, R=0.5

h [kJ/kg]

s [kJ/kgK]

Δh0

2

23w

p1

p02

p01

p3

p03

2

21c 2

22c

2

22w 2

22u

p2

2

23u

I

h1

h2

h03

h3

1

01

2

02

3

03

3s

2

23c

h01=h02

03s

Note:

• Enthalpy drop in stator equal to enthalpy drop in rotor; therefore 321 ppp >>• Static conditions at stator outlet equal to rotor inlet • Polytropic expansion line connects static conditions • Total enthalpy constant in stator, rothalpy constant in rotor

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Action Turbine, R=0

h [kJ/kg]

s [kJ/kgK]

Δh0

2

22w

p1

p02

p01

p03

2

21c

2

22c

2

23w

2

22u

p2=p3

2

23u

I

h01=h02

h1

h03

h2=h3

1

01 02

2=3

03

3s

2

23c

03s

Note:

32 pp ≈• Enthalpy drop in rotor equal to zero; therefore (in reality a small pressure drop is necessary)

• Static conditions at stator outlet equal to rotor inlet • Polytropic expansion line connects static conditions • Total enthalpy constant in stator, rothalpy constant in rotor

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Axial Compressor Degree of reaction R=0.5

h [kJ/kg]

s [kJ/kgK]

Δh0

2

22w

p3

p03

p02

p1

2

21c

2

22c

2

21w

2

22u

p2

2

21u

I

h02=h03

h1

h2

h01

h3s

1

01

2

02

3

03

3s

2

23c

p01

h3

03s

Note:

• Enthalpy increase in rotor equal to enthalpy increase in stator; therefore 321 ppp <<• Static conditions at stator outlet equal to rotor inlet • Polytropic expansion line connects static conditions • Total enthalpy constant in stator, rothalpy constant in rotor

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Turbine Design Parameters


Nomenclature

Symbol Denotation Unit c Absolute velocity m/s h Enthalpy J/kg m& Mass flow rate kg/s r Radius m u Tangential velocity m/s w Relative velocity m/s α Absolute flow angle deg β Relative flow angle deg ω Rotational speed rad/s C Normalized absolute

velocity ucC = -

W Normalized relative velocity uwW =

-

U Normalized tangential velocity 1== uuU

-

Subscripts

0 Total 1 Inlet stator 2 Outlet stator (inlet rotor) 3 Outlet rotor n Normal r Radial component x Axial component θ Tangential component

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Turbine Stage Denotations and Conventions

1 2 stator

3 rotor

Reference radius

Stage denotations


ω

30

20

10

0

10

20

30

−30

−20

−10

0

10

20

c1

c2

w2

u

u

w3

c3

rotor stator Stage velocity triangles u

Velocity triangles denotations and conventions

Relative flow angle

Axial direction

Origin

Circumferential speed

Absolute velocity

Relative velocity

u

wθ

cθ

β α

w

c

Absolute flow angle

Relative circumferential velocity component

Absolute circumferential velocity component

Axial velocity component (absolute and relative)

cx=wx

θ neg α, β neg

x pos

α, β pos

θ pos

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Stage Velocity Triangles The stage velocity triangles are commonly employed as graphical measure to represent averaged kinetics of the flow at a reference radial position throughout the stage. Usually one of the following reference radial positions is used:

• Mean radius

2sh

mrr

r+

= Eq. 1

• Euler radius ( radius that splits the annular cross section in half)

2

22sh

Err

r+

= Eq. 2

The absolute frame of reference is bound to the stator and is therefore non-rotating. The relative frame of reference is bound to the rotor and rotates with the circumferential speed of the rotor u at the reference radius obtained from ω⋅= refru Eq. 3 The relation between the velocities in the absolute frame of reference (denoted “absolute velocities”) and the ones in the relative frame of reference (respectively denoted “relative velocities”) is the following Eq. 4 xx cw = Eq. 5 ucw −= θθ where cx and cθ are the axial and circumferential components of the respective velocity as follows Eq. 6 222

θccc x +=

Eq. 7 222θwww x +=

The flow angles are defined as

xccθα =tan Eq. 8

xwwθβ =tan Eq. 9

Note:

• The relative velocity is the velocity that an observer sees while sitting on the rotor • The rotor blades thus see the relative flow velocities • The direction of the absolute flow velocity at stator outlet corresponds approximately

to the stator blade metal angle at the trailing edge • The direction of the relative flow velocity at rotor outlet corresponds approximately to

the rotor blade metal angle at the trailing edge

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First Design Parameter: Degree of Reaction The degree of reaction relates the change in enthalpy effectuated in the rotor to the change in enthalpy of the stage as follows

stage

rotorhh

RΔΔ

= Eq. 10

, which can be rewritten as

3221

32hhhh

hhhh

hhh

Rrotorstator

rotor

stage

rotor−+−

−=

Δ+ΔΔ

=ΔΔ

= Eq. 11

The change in enthalpies in stator and rotor respectively are related to the velocities as follows

In the stator the stagnation enthalpy 2

2

0chh += is constant, thus

22

22

2

21

1c

hc

h +=+ Eq. 12

leading to

( ) Eq. 13 21

2221 2

1 cchh −=−

In the rotor the rothalpy 22

22 uwhI −+= is constant, thus

2222

23

23

3

22

22

2uw

huw

h −+=−+ Eq. 14

leading to

( 22

23

22

2332 2

1 uuwwhh +−−=− ) Eq. 15

Substituting these expressions into the equation of stage reaction above leads to the following general expression

22

23

22

23

21

22

22

23

22

23

uuwwcc

uuwwR

+−−+−

+−−= Eq. 16

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For a normal repetition stage with the following restrictions 31 cc rr

= Eq. 17 constccc xxx === 3,2,1, Eq. 18 Eq. 19 32 uu = the expression of the degree of reaction can further be simplified. Firstly it can be noted that the circumferential speed u cancels out. Secondly the velocities shall be written in terms of their components as and respectively. This yields the following expression

222θccc x += 222

θwww x +=

22,

23,

21,

22,

22,

23,

θθθθ

θθ

wwcc

wwR

−+−

−= Eq. 20

The relative velocity components in the denominator shall be expressed by the absolute velocity components as leading to ucw −= θθ

2

2,22,

23,

23,

21,

22,

22,

23,

22 uuccuucccc

wwR

−+−+−+−

−=

θθθθθθ

θθ Eq. 21

After canceling out elements the expression can be rewritten as

( )3,2,

22,

23,

2 θθ

θθ

ccuww

R−⋅

−= Eq. 22

At this stage the enumerator shall be expressed as ( ) ( )2,3,2,3,

22,

23, θθθθθθ wwwwww +⋅−=− . The

absolute velocity components in the denominator shall be expressed in terms of relative velocities as leading to uwc += θθ

( ) ( )

( )33,22,

2,3,2,3,2 uwuwu

wwwwR

−−+⋅

+⋅−−=

θθ

θθθθ Eq. 23

Both the circumferential speeds in the denominator and the relative components

cancel out finally yielding 2,3, θθ ww −

( 2,3,21

θθ wwu

R +−= ) Eq. 24

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At this position a more intimate analysis of the degree of reaction is appropriate. For this purpose the relative circumferential velocity component at position 2 shall be expressed in the absolute frame of reference as yielding ucw −= θθ

( ) ( 2,3,2,3, 21

21

21

θθθθ cwu

ucwu

R +−=−+−= ) Eq. 25

By expressing the circumferential velocity components in terms of flow angles as xccθα =tan the

following expression is obtained for the degree of reaction

( 23 tantan22

1 αβ +−=uc

R x ) Eq. 26

In the above equation the degree of reaction is expressed in terms of axial velocity component, circumferential speed and stator and rotor outflow angles respectively, which are approximately equal to blade metal angles at trailing edge. According to the convention of velocity components depicted above flow angle β3 is negative whilst flow angle α2 is positive. This leads to the following observations

• An increase in flow angle α2 leads to a decrease in degree of reaction ( ), i.e. the contribution of enthalpy change in the stator to the total change in enthalpy in the stage gets larger

↓↑⇒ R2α

• An increase in flow angle β3 leads to an increase in degree of reaction ( ), i.e. the contribution of enthalpy change in the rotor to the total change in enthalpy in the stage gets larger

↑↑⇒ R3β

• For turbine stages the degree reaction usually lies in the range [0…1]

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Second Design Parameter: Loading Factor The loading factor relates the change in total enthalpy effectuated in the stage to the rotational speed as follows

20

u

hΔ=ψ Eq. 27

Under application of Euler’s turbine equation the change in enthalpy can be expressed as

leading to 3,32,20 θθ cucuh −=Δ

2

3,32,2

u

cucu θθψ−

= Eq. 28

For a normal repetition stage with the following restrictions 31 cc rr

= Eq. 29 constccc xxx === 3,2,1, Eq. 30 Eq. 31 32 uu = the expression of the loading factor can further be simplified to

ucc 3,2, θθψ

−= Eq. 32

Expressing the absolute flow velocities in the relative frame of reference as the loading factor can be expressed as

uwc += θθ

uww 3,2, θθψ

−= Eq. 33

An equivalent expression can be obtained by substituting the relative velocity component at position 2 in the absolute frame of reference as uwc += θθ yielding

uwc 3,2,1 θθψ

−+−= Eq. 34

, which also can be expressed in terms of flow angles α2 and β3 as follows

( 32 tantan1 βαψ −+−=ucx ) Eq. 35

According to the convention of velocity components depicted above flow angle β3 is negative whilst flow angle α2 is positive. This leads to the following observations:

• Increase in flow angles α2 and β3 lead to increase in loading factor ( ) ↑↑⇒ ψβα 32 ,• To obtain a loading factor smaller than one 3tan β must be greater than 2tanα

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Third Design Parameter: Flow Coefficient The flow coefficient relate the axial velocity component to the circumferential speed as follows

ucx=φ Eq. 36

The only observation to make for this coefficient is that the higher the axial velocity in the stage the higher the flow coefficient. As can be recognized below the flow coefficient stretches the velocity triangles in the axial direction.

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The Normalized Velocity Triangle At this position the normalized velocity triangle shall be introduced. The normalization consists therein that all velocity components are depicted with reference to the outlet circumferential velocity u3. The normalized velocity components are denoted by the respective capital letters and yield from

3ucC = Eq. 37

3uwW = Eq. 38

3uuU = Eq. 39

The special case of a normal repetition stage shall be regarded here for the sake of simplicity. The applied principle is however valid for all types of turbine stages. Conveniently the velocity triangle is drawn with a common origin for stator and rotor outlet. As a normal repetition stage with the condition constccc xxx === 3,2,1, is considered the height of the

triangle corresponds to φ==uc

C xx , i.e. the flow coefficient.

W3

C3 W2

C2

U=1 U=1

22,3, θθ WW +−

R

ψ

Φ

Note:

• The height of the velocity triangle corresponds to the flow coefficient Φ • The loading coefficient corresponds to the circumferential distance between C2 and

C3. In the case of a repetition stage this equals to the circumferential distance between W2 and W3.

• The degree of reaction equals to the distance between axial and half the midpoint between W2 and W3.

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Special Cases The special cases are here analyzed for the case of normal repetition stage. Similar analysis can be performed in a general manner for other types of stages. Degree of Reaction equal to Zero (R=0; Action Turbine) The expression of the degree of reaction yields the following

( ) 3,2,2,3,210 θθθθ wwwwu

R −=⇒+−== Eq. 40

Substituting this expression into the equation of loading coefficient yields

( 1222,

2,3,2, −= )⋅=⇒

−= θ

θθθ ψψ cuu

wuww

Eq. 41

Velocity triangle

W3

C3 W2

C2

U=1 U=1

22,3, θθ WW +−

ψ

Φ

Note:

• As and normal stage it follows that 3,2, θθ ww −= 32 ww = and consequently . The change in enthalpy in an action stage is thus entirely due to change

in enthalpy in the stator. 0=Δ rotorh

• The forces acting on the rotor are action forces, as the fluid is not accelerated through the rotor. This leads to the denotation of “action stage”

• The rotor only effectuates deflection of the fluid but not expansion as 32 ββ −= • As the fluid is not expanded throughout the rotor the pressure up- and downstream of

the rotor is (practically) unchanged. In reality a minimum pressure drop is necessary due to losses to drive the fluid, thus 23 pp ≈

• As a consequence there is little axial force on the rotor in an action turbine

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Degree of Reaction equal to one half (R=0.5; Reaction Turbine) The expression of the degree of reaction yields the following

( ) 3,2,2,3,21

21

θθθθ wuwwwu

R −=+⇒+−== Eq. 42

, which is equivalent to Substituting this expression into the equation of loading coefficient yields

3,2, θθ wc −=

12

12 2,2,3,2, −=+

⋅=⇒

−=

uc

uw

uww θθθθ ψψ Eq. 43

Velocity triangle

W3

C3 W2

C2

U=1 U=1

22,3, θθ WW +−

ψ

Φ

R Note:

• As and normal stage it follows that 3,2, θθ wc −= 32 wc = and with the assumption of repetition stage ( ) consequently 31 cc = statorrotor hh Δ=Δ . The change in enthalpy in a reaction stage is thus equally split on stator and rotor.

• The forces acting on the rotor are partially action and partially reaction forces, as the fluid is accelerated through the rotor. This leads to the denotation of “reaction stage”

• Both stator and rotor effectuate expansion of the fluid and thus and 12 pp < 23 pp < • As a consequence there is a considerable axial force on the rotor in a reaction

turbine. In most cases this force is too large to be submitted to an axial bearing and thus must be compensated for. Possible compensations are appropriate arrangement of components such as to cancel out axial forces or application of a thrust compensation devices (e.g. piston), see further below

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Zero Exit Swirl (cΘ,3=0) With the degree of reaction writes to uwuwc −=⇒+== 3,3,3, 0 θθθ

( )u

www

uR

221

21 2,

2,3,θ

θθ −=+−= Eq. 44

The loading coefficient yields from

12,3,2, +=−

=uw

uww θθθψ Eq. 45

Combining the two expressions leads after reformulation to a relationship between degree of reaction and loading coefficient as follows )1(2 R−⋅=ψ Eq. 46 Velocity triangle

Φ

W3

C3

W2

C2

U=1 U=1

R 2

2,3, θθ WW +−

ψ

Note:

• The flow exists the stage purely axial, i.e. there is no swirl at stage exit • For a zero exit swirl stage the degree of reaction and the loading factor are

dependent • Example for a zero exit swirl stage are for example last stages of jet engines thrust

maximized when jet flow purely axial

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Turbine Flow in Various Frames of Reference


Stator Exit Rotor Exit

0 20 40 60 80 100 120

Stator Rotor Stator Rotor

0 20 40 60 80 100 120

w

u

cw

c u

u

u

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Axial Turbine – Effect of Degree of Reaction


Source: Traupel, W., 1977 “Thermische Turbomaschinen” Vol. 1, 3rd Ed. ISBN 3-540-07939-4

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Turbine Design Aspects


Disk Rotor

• Used when pressure difference small across rotor ( action stage) • Sealing (labyrinth sealing) across stator on small diameter less leakage • Little to no axial thrust often minimized by holes in rotor disks

Drum Rotor

• Used when pressure difference present across rotor ( reaction stage) • Considerable axial thrust, which must be balance • Possible thrust balancing thrust piston


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Sealing Aspects

• Commonly labyrinth sealing are used (touch-free) due to high contact speeds as other types of sealing (e.g. O-ring, rubber lip) would fail

• In labyrinth sealing a minimum leakage flow is accepted note: the higher the leakage flow the lower the stage efficiency

• Both shrouded and shroudless turbines are used (the sketch below depicts a shrouded turbine)

• Shrouded turbines have the general advantage of less leakage however this has to be bought by increased mechanical load on the rotor blades (large mass on large diameter high centrifugal load high stresses at blade root risk for failure)

• To reduce leakage flow in labyrinth sealing a increased back-pressure might be applied at one or more axial positions throughout the sealing

High pressure

Low pressure (e.g. ambient)

Medium pressure


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Action Turbines


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Partial admission Possibility for partial admission as pressure difference small across rotor (note: partial admission would not work for reaction stage as pressure difference across rotor could not be maintained)

Partial admission in steam turbine


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Curtis stage

• Used to maximize power and for control stage • Little pressure difference across rotor 1, stator 2 and rotor 2 • Main pressure (and enthalpy) drop across first stator


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Reaction Turbines Standard reaction turbine (drum rotor) Note: thrust balancing piston (to the right)

Reaction turbine with control stage (action stage partial admission)


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Two-flow turbine Axial forced balanced by two identical turbine flows with opposite orientation on common axle

Large steam power turbo group

Low-pressure steam turbine


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Turbine blade geometry

• After determining the velocity triangles the shape of the blade profiles can be determined

• The blade geometry yields the flow passage (e.g. to be determined by tangent circles) and by this the flow velocity distribution around the blade

• Although in- and outflow cross sections are identical the change in cross section throughout the passage might differ task of the designer to find optimum nowadays assessed by means of CFD (computational fluid dynamics)

Example of CFD prediction of flow around blade profile

• Colors mark pressure (red high – green medium – blue low) • Ribbons to visualize secondary flow structure • CFD results are usually employed to determine efficiency (rather than empirical

correlations)


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Three-Dimensional Effects

• For long blades (e.g. last stage blades) the effects of three-dimensionality cannot be neglected

• Commonly a radial balance between stator and rotor is aimed at, e.g. constant swirl constcr =⋅ θ

• This yields a twisted blade shape


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Design of Multistage Turbine


Nomenclature

Symbol Denotation Unit c Absolute velocity m/s cp Specific heat J/kgK h Enthalpy J/kg m& Mass flow rate kg/s p Pressure Pa r Radius m u Tangential velocity m/s v Specific volume m3/kg w Relative velocity m/s z Number of stages - R Degree of reaction - R Gas constant J/kg/K T Temperature K W& Power J/s Y Radius ratio hs rrY = - Φ Flow coefficient - Ψ Loading coefficient - Ω Cross section m2 α Absolute flow angle deg β Relative flow angle deg γ Isentropic coefficient - ε Turning deg ρ Density m3/kg η Efficiency - ζ Loss coefficient - ω Rotational speed rad/s

Subscripts

0 Total 1 Inlet stator 2 Outlet stator (inlet rotor) 3 Outlet rotor n Normal s Shroud (tip) h Hub r Radial component x Axial component α Engine inlet ω Engine outlet θ Tangential component


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System Discretization

30

20

10

0

10

20

30

−30

−20

−10

0

10

20

c2

c1 w2

u

u

w3 u

1 2 3 stator rotor

ω

stator rotor Stage velocity triangles

Stage denotations


c3

Reference radius

α

ω

Schematic representation α inlet ω outlet

It is assumed that the turbine consists of one or several stages

W&


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Problem Statement The task is given to perform a preliminary design of a multistage turbine. It is thereby assumed that the following parameters are known

• Inlet pressure and temperature • Outlet pressure • Required power

Furthermore the following limitations shall be made:

• Repetition and normal stages shall be valid. • Zero exit swirl

The expressions of the design parameters are given below for this special case

( )u

www

uR

221

21 2,

2,3,θ

θθ −=+−= Eq. 1

12,3,2, +=−

=u

wu

ww θθθψ Eq. 2

u

cx=φ Eq. 3

Note that the following relationship is valid )1(2 R−⋅=ψ Eq. 4 Step 1: Determination of approximate flow parameters (isentropic expansion) As the geometry and design parameters of the turbine are not known a priori in this first step isentropic expansion shall be assumed. The efficiency is then calculated based on this first assumption and one or several iterations are performed thereafter. Determination of inlet specific volume

α

αα p

RTv = Eq. 5

Determination of outlet specific volume

γ

ω

ααω

1

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pp

vv Eq. 6

Determination of change in enthalpy

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=Δ

−γ

γ

α

ωααγ

γ1

11 p

pvphs Eq. 7


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Determination of mass flow rate

As 2

22

0ωα cc

hh ss−

+Δ=Δ and due to the assumption of normal repetition stages ωα cc = it

follows that ss hh Δ=Δ 0 leading to

sh

WmΔ

=& Eq. 8

Step 2: Determination of inlet and outlet annular geometry Assume: u=const, implying that r=const The radius ratio Y is dependent on flow coefficient and mean radius and yields from the conservation of mass ucnm ⋅⋅⋅Ω=⋅⋅Ω= φρρ& Eq. 9

with h

srr

Y = and 2

shm

rrr

+= it follows that

12

+=

Yr

r mh and

12

+=

YYr

r ms leading to

( )114 222

+−

=−=ΩYYrrr mhs ππ Eq. 10

Substituting this expression into the conservation of mass above it follows that

ωφρπφρπ ⋅⋅⋅+−

=⋅⋅⋅+−

=114

114 32

YYru

YYrm mm& Eq. 11

By expressing Ar

m

m=

⋅⋅⋅ ωφρπ 34

& the radius ratio is obtained from

11

−+

=AAY Eq. 12

Choose • Rotational speed • Flow coefficient phi

Determine Y in appropriate range

Note:

• Too short blades give bad efficiency • Too long blades shall be avoided due to high mechanical loads


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Step 3: Determination of number of stages The total change in enthalpy between inlet and outlet is distributed over an appropriate number of stages as follows

∑=

=Δz

iiiuh

1

20 ψ Eq. 13

where z denotes the number of stages. Above the assumption of a normal repetition stage at constant radius has been made, thus Ψ=Ψi and uui = . In this first step the isentropic enthalpy change is regarded leading to 2uzhs ⋅⋅=Δ ψ Eq. 14 The number of stages yields from the following expression

22ωψ m

s

r

hz

Δ= Eq. 15

Choose

• Loading coefficient Note:

• For the present special case of normal repetition stage at zero exit swirl the loading coefficient depends from the degree of reaction as )1(2 R−⋅=ψ .

• The choice of the loading coefficient has to be made such that the number of stages yields an integer number. In case the number of stages is non-integer it has to be rounded to the next higher integer


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Step 4: Determination of stage efficiency Determination of loss coefficients and total-to-total stage efficiency as follows

( )

1

31

23

22

2

3

21

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

++=

hh

wcTT

RN

tt

ζζη Eq. 16

Firstly it is assumed that 32 TT ≈ . For a normal repetition stage with zero exit swirl the stage efficiency then yields from

( ) ( )( )1

222 1211

−

⎥⎦

⎤⎢⎣

⎡++++= φζψφζ

ψη RNtt Eq. 17

In a first approximation the loss coefficients can be determined from Soderberg’s correlation as follows

2

*100

06.004.0 ⎟⎠⎞

⎜⎝⎛+=

εζ Eq. 18

where ε denotes the turning, which for stator and rotor respectively yields from

⎟⎟⎠

⎞⎜⎜⎝

⎛==−= −

φψ

αααε 1221 tanN Eq. 19

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=−= −−

φφψ

ββε 1tan1tan 1132R Eq. 20


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Step 5: First iteration of flow parameters (polytropic expansion) With the knowledge of the approximate total-to-total efficiency the polytropic expansion between inlet and outlet can now be determined yielding the flow parameters at these stations more accurately. The change of enthalpy is now given by

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=Δ

−γ

γ

α

ωααγ

γη

1

11 p

pvph tt Eq. 21

The outlet temperature yields from

pchTT Δ

−= αω Eq. 22

, which allows us to determine the outlet specific volume by

ω

ωω p

RTv = Eq. 23

Finally the mass flow rate needs to be updated by the updated enthalpy difference as

h

WmΔ

=& Eq. 24

Step 6: Finalization of first iteration The finalization of the first iteration comprises the steps of determination of number of stages and loading factor. From the values obtained an updated polytropic efficiency can be determined, which could be used in another iteration step. This iteration process should be carried out until convergence is obtained. Criteria for obtaining convergence should be established by the user. Convergence could for example be measured by relating a number of parameters from two subsequent iteration steps. Usually one or two iterations will do if the change of state is close to the isentropic one.


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Example: Design of Multistage Turbine in Excel sheet The above equations have been implemented in an Excel sheet, which is made available with the present document. The purpose of the Excel sheet is to recognize the effects of design choices on the resulting engine as well as on engine costs. Parameters that are open to choose are the following:

• Thermodynamic parameters: inlet pressure and temperature, outlet pressure and poser

• Engine parameters: engine rotational speed and mean radius • Design parameters: flow coefficient and loading factor (note: as the assumptions of

normal repetition stages and zero exit swirl has been made the degree of reaction is directly related to the loading coefficient)

To recognize the impact of certain parameter choices on costs a simplified cost analysis has been included. The cost analysis covers the factors of engine purchasing, engine service and engine fuel costs. Output is provided in numerical and graphical format. Have fun!

isentropic 1st iteration --> polytropic Cost analysis Parameter Unit α ω α ω p [Pa] 2,00E+06 5,00E+05 2,00E+06 5,00E+05 base costs - 3,00E+06 T [K] 800,00 538,36 800,00 565,01 cost per stage - 3,00E+05 v [m3/kg] 0,11 0,31 0,11 0,32 fuel cost per l 2 P [J/s] 3,00E+06 3,00E+06 fuel heating value J/l 4,00E+07 m dot [kg/s] 11,41 11,41 12,71 12,71 Δh [kJ/kg] 2,63E+05 2,36E+05 gamma [-] 1,40 cp [J/kgK] 1004,5 R [J/kgK] 287 phi [-] 0,50 0,50 n [rpm] 3,00E+04 3,00E+04 3,00E+04 3,00E+04 fixed costs om [rad/s] 3141,59 3141,59 3141,59 3141,59 machine costs 3,62E+06 rm [m] 0,080 0,080 0,080 0,080 A [-] 0,13 0,35 0,14 0,41 runtime # years 10 Y [-] 1,30 2,07 1,34 2,38 rh [m] 0,070 0,052 0,068 0,047 service (per stage & year) 2,00E+05 rs [m] 0,090 0,108 0,092 0,113 service (period) 4,16E+06 s [m] 0,021 0,056 0,023 0,065 u [m/s] 251,33 251,33 fuel run time [h] per year 7000 psi [-] 2,00 1,80 total energy per year 8,37E+13 z [-] 2,08 2,08 total fuel costs per year 4,18E+06 R 0,00 2 fuel (period) 4,18E+07 stat rot eps [rad] 75,96 126,87 74,45 121,31 subtotal (var, period) 4,60E+07 zeta [-] 0,07 0,14 0,07 0,13 η tt [-] 0,898 0,904 total costs (period) 4,96E+07

Legend of parameters to choose thermodynamic parameters fluid properties engine parameters design parameters operating parameters


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Losses


Nomenclature

Symbol Denotation Unit c Flow speed m/s h Enthalpy J/kg p Pressure Pa s Entropy J/kgK T Temperature K η Efficiency - ζ Loss coefficient - Φ Flow coefficient - ψ Loading factor -

Subscripts

0 Total 1 Inlet stator 2 Outlet stator (inlet rotor) 3 Outlet rotor x Axial component θ Tangential component N Nozzle (stator) R Rotor s Isentropic tt Total-to-total ts Total-to-static


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Losses Losses are an indication for the irreversibility of a process. Considering an adiabatic machine, losses therefore lead to increase in entropy. Previously efficiencies have been defined as the ratio between actual change in energy to ideal change in energy (expansion) as follows


=η Eq. 1

The difference between ideal and actual energy is referred to as loss and is apparent in an enthalpy-entropy diagram as depicted in figure 1.

Figure 1. Expansion process

A common way to express losses is to use a normalized loss coefficient that relates the loss to the exit kinetic energy as follows

ζ2200 2

1 chh s =Δ−Δ Eq. 2

, where ζ stands for the normalized loss coefficient. The relation above represents a very simple and straightforward estimation of losses and thereby also for the efficiency of a turbomachine component. It is therefore often used during preliminary engine design. To obtain more accurate predictions of losses in components it is necessary to resolve the three-dimensional flow field in detail, which nowadays is commonly done by computational fluid dynamics (CFD).

h [kJ/kg]

s [kJ/kgK]

h1

h2 h2s

Ideal process

Real process

Δh Δhs

p1

p2

loss


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Losses in a Turbine Stage Rather than expressing the overall loss of a turbine stage by reference to only the stage outlet velocity the losses of the stator and the rotor are expressed separately and summed up. The reason for doing so is to be able to relate losses to geometrical properties such as the deviation of the flow in each blade row. Consider a turbine stage as sketched in figure 2 and the corresponding expansion as included in figure 3.

Figure 2. Sketch of turbine stage

Figure 3. Expansion in turbine stage

h [kJ/kg]

s [kJ/kgK]

h1

h3

Stator ideal

Δh0

p1

p3

h3ss

Rotor ideal

3s 3ss

3

2

2s

1

h2s h2

30

20

10

0

10

20

30

−30

−20

−10

0

10

20

c2

c1 w2

u

u

w3 u

ω

stator rotor Stage velocity triangles

Stage denotations 1 stator inlet 2 rotor inlet 3 rotor outlet

c3

Reference radius

1 2 3 stator rotor


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The losses in the stator are apparent between points 2s and 2, the ones in the rotor between 3s and 3. For the sake of simplicity we shall consider a repetition stage where 31 cc = . For that reason the static rather than the total conditions are indicated in the above figure. Furthermore to perform an intimate analysis of the present expansion the following states are defined:

• 1 stage inlet • 2 stator outlet • 2s stator outlet after isentropic expansion in stator • 3 rotor outlet • 3s rotor outlet after isentropic expansion in rotor only • 3ss rotor outlet after isentropic expansion in stator and rotor

The total-to-total isentropic efficiency is defined as follows

sssssstt hhhhhh

hhhhhh

333331

31

31

31−+−+−

−=

−−

=η Eq. 3

Note the presence of the losses in the denominator. The losses in the stator and the rotor respectively are expressed by the respective outflow velocity such that

• Stator loss

Ns chh ζ2222 2

1=− (see note 1) Eq. 4

• Rotor loss

Rsss whh ζ2333 2

1=− Eq. 5

Note the following:

• The stator loss is referred to the stator outflow velocity, i.e. c2 • The rotor loss is related to the rotor outflow velocity. As the rotor moves we have to

consider the relative velocity w3 Next the efficiency shall be expressed in terms of these loss coefficients. Whereas the expression for the rotor loss can readily be included in Eq. 3 this is not the case for the stator loss as

ssss hhhh 3322 −≠− Eq. 6 The reason for this is the spreading of the constant pressure lines in the enthalpy-entropy diagram with increasing temperature, which is know as reheating phenomenon. It is known that the slope of the constant pressure lines is proportional to the temperature in a given point such as

Tsh

constp=⎟

⎠⎞

⎜⎝⎛

∂∂

= . Eq. 7

, leading to

1 The index ”N” in the loss coefficient stands for ”nozzle”, which is a synonym for stator


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sTh Δ=Δ Eq. 8

Applied on the present case this implies the following

( )ss ssThh 22222 −=− Eq. 9

( )sssss ssThh 23333 −=− Eq. 10 As

ssss ssss 3322 −=− Eq. 11 , we obtain

( )ssss hhTT

hh 222

333 −=− Eq. 12

At this position the total-to-total efficiency can be expressed in terms outlet velocities and loss coefficients to

( )

1

31

23

22

2

3

23

22

2

331

312

1

21

21

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

++=

+⋅+−

−=

hh

wcTT

wcTThh

hh RN

RNtt

ζζ

ζζη Eq. 13

For preliminary design computations (i.e. a first shot during the design process) it is often accurate enough to set 123 ≈TT as the temperatures throughout the stage are not know at this stage. To obtain the total-to-static efficiency we have previously seen that

1

0

23

21

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

Δ+=

hc

ttts η

η Eq. 14

, which for the present case (recall: repetition stage) leads to

( )

1

31

23

23

22

2

3

21

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

+++=

hh

cwcTT

RN

ts

ζζη Eq. 15


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Loss Coefficients As pointed out above the purpose of using normalized loss coefficients is to have a simple and straightforward tool to estimate losses during preliminary engine design. One of the tasks of a turbomachine engine designer is to select potential stage geometries at an early stage in the design process and pursue these further using more advanced (as well as more time consuming) CFD tools. The question that opens up here is how these loss coefficients are related to a turbine or compressor geometry. Considering the various types of losses as listed below, this task is rather complex.

Type of loss Due to Profile losses Deviation of the flow as well as wall

frictionDischarge losses Uneven velocity distribution at the

outlet of a blade row Leakage losses leakage across sealings (note that it

is necessary to have a certain gap between stationary and rotating parts)

Disk wheel friction losses Friction between stationary and rotating parts

Ventilation losses Ventilation phenomena at sector ends (e.g. during partial admission)

Table 1. Types of losses One possibility to determine the impact of geometrical parameters such as blade profile shape, blade length and width, surface roughness and sealing type just to name a few is to build a large number of machines and actually measure the losses. The dependency of the loss coefficients from these parameters can then be determined experimentally. Traupel (1977) has presented one of the most comprehensive experimental studies on losses in turbomachines and is still nowadays regarded somewhat as a standard in loss prediction during preliminary design. The overall loss coefficients are thereby composed of coefficients from various loss sources that are obtained from diagrams. To simplify this problem Soderberg (1949) has previously presented a method for estimating the losses of a “good design”. Rather than taking into account a large number of parameters it is assumed that the regarded turbine stage is close to an optimum design, for example that the solidity of the blade row (blade pitch over blade chord) and the aspect ratio (blade length over blade chord) are within certain limits. Soderberg’s correlation states that under assumption of these restrictions the loss coefficient can be expressed in terms of flow deviation as follows

2

10006.004.0 ⎟

⎠⎞

⎜⎝⎛+=

εζ Eq. 16

, where ε is the deviation of the flow in deg. Soderberg went even further to account for the effects of designs that differed from the optimum, in other words, give an estimate for how much the loss coefficient changes if the design lies outside the imposed limits. A few of these expressions are contained in Dixon (1998).


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Dependency of the Efficiency from Design Parameters At this stage the focus is drawn towards how the design parameters affect the loss coefficients and the efficiency. During preliminary engine design it is important to keep in mind that any decision that is taken affects the engine in some way and might have good or bad consequences of the final performance. To name an example, it might be possible to reduce stage count of a multi-stage engine, and thereby manufacturing and service costs, by increasing the stage loading factor, however this might lead to a reduction in efficiency. As most of the decisions that an engine designer will face represent a compromise between conflicting aspects, it is essential to understand the contexts. For the sake of simplicity, consider a normal repetition turbine stage. The design parameters are given as

• Degree of reaction

( )2321

θθ wwu

R +−= Eq. 17

• Loading factor

uww 3,2, θθψ

−= Eq. 18

• Flow coefficient

ucx=φ Eq. 19

Below the two special cases of reaction and action turbines are treated in detail.


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Reaction Turbine (R=0.5) Velocity triangle

Note the following:

• 3,2, θθ wc −= and 32 wc = due to the assumption of normal repetition stage • Identical stator and rotor geometries, thus RN ζζ =

By assuming 123 ≈TT the expression for the total-to-total efficiency simplifies to

1

31

231

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+=

hhw R

ttζ

η Eq. 20

Due to the assumption of repetition stage ( 31 cc = ) it follows that

310 hhhh −=Δ=Δ Eq. 21 and

231 uhh ψ=− Eq. 22

Substituting Eq. 22 into Eq. 20 we obtain

1

2

231

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

uw R

ttψ

ζη Eq. 23

Next the velocity w3 shall be expressed in terms of design parameters. For this purpose it is split into its axial and circumferential components 2

32

32

3 θwww x += yielding

W3

C3 W2

C2

U=1 U=1

Φ

ψ

22,3, θθ WW +−

R


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22

23 φ=

u

wx ; 2232

23

23

23

2

23 tan φβθθ ⋅=⋅=

u

c

c

w

u

w x

x Eq. 24

The expression of the efficiency now writes to

( )1

23

2tan11

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡++= β

ψζφ

η Rtt Eq. 25

Furthermore the flow angle 3β can be expressed from design parameters as

φψ

β2

1tan 3+

−= Eq. 26

, which finally leads to the following expression for the total-to-total efficiency

122

21

11

−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ +++=

φψ

ψζφ

η Rtt Eq. 27

Next, to obtain the loss coefficient using the Soderberg’s correlation the flow deviation in the rotor (note: identical to the stator) must be determined. The flow deviation yields from

32 ββε −= Eq. 28 Whereas an expression for 3β has already been given above the flow angle 2β is obtained from

φψ

β2

1tan 2−

= Eq. 29

leading to the following expression for the flow deviation in the rotor

⎟⎟⎠

⎞⎜⎜⎝

⎛ ++⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

φψ

φψ

ε2

1arctan2

1arctan Eq. 30

By implementing this expression into the Soderberg’s correlation (Eq. 16) and finally into Eq. 27 an expression is obtained that directly relates efficiency to the loading factor and the flow coefficient. The result can be displayed graphically as included in figure 4 and indicates the interdependence of the design parameters for obtaining an optimum design.


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0 0.5 1 1.50

0.5

1

1.5

2

2.5

3

φ, −

ψReaction stage R=0.5

0.94

0.94

0.92

0.92

0.92

0.9

0.9

0.9

0.9

0.90.88

0.88

0.88

0.88

0.88

0.88

0.88

0.86

0.86

0.86

0.86

0.86

0.86

0.840.84

0.84

0.84

0.84

0.82

0.82

0.82

0.8

0.8

0.8

Figure 4. Loss diagram; reaction turbine


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Action Turbine (R=0) Velocity triangle

Note the following:

• 3,2, θθ ww −= and 32 ww = due to the assumption of normal repetition stage • Different stator and rotor geometries

By assuming 123 ≈TT the expression for the total-to-total efficiency simplifies to

( )

1

2

23

22

1

31

23

22

21

21

−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ++=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+

+=uwc

hhwc RNRN

ttψ

ζζζζη Eq. 31

Similar to the case of the reaction turbine above the velocities can be expressed in terms of design parameters as follows

22

2

22

2

22

2

22 1

2⎟⎠⎞

⎜⎝⎛ ++=+=ψφθ

uc

uc

uc x Eq. 32

2

22

23

2

23

2

23

2⎟⎠⎞

⎜⎝⎛+=+=ψφθ

uw

uw

uw x Eq. 33

, which finally leads to the following expression for the total-to-total efficiency

12

22

222

1211

−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛++⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛ +++=

ψφζψφζψ

η RNtt Eq. 34

W3

C3 W2

C2

U=1 U=1

Φ

ψ

22,3, θθ WW +−


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The flow deviation in stator and rotor respectively can be determined from design parameters in a similar way than demonstrated above. Again the result can be displayed graphically as included in figure 5 such as to indicate the interdependence of the design parameters for obtaining an optimum design.

0 0.5 1 1.50

0.5

1

1.5

2

2.5

3

φ, −

ψ

Action stage R=0

0.92

0.92

0.92

0.9 0.9

0.9

0.9

0.9

0.880.88

0.88

0.88

0.88

0.88

0.86

0.860.86

0.86

0.86

0.86

0.84

0.84

0.840.84

0.84

0.82

0.820.82

0.82

0.80.8

0.8

0.8

Figure 5. Loss diagram; action turbine


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References Dixon, S.L., 1998 "Fluid Mechanics and Thermodynamics of Turbomachinery" Fourth edition, Butterworth-Heinemann, Woburn, MA, USA, 1998 ISBN 0-7506-7059-2 Traupel, W., 1977 “Thermische Turbomaschinen” Vol. 1, 3rd Ed. ISBN 3-540-07939-4 Soderberg, C.R., 1949 Unpublished Note, Gas Turbine Laboratory, Massachusetts Institute of Technology

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