Turbomachinery Aero-Thermodynamics - crans.orgperso.crans.org/epalle/M2/Turbomachines/Cours1.pdf · You said Turbomachines? Turbomachines are machines that transfer energy between

Turbomachinery Aero-ThermodynamicsIntroduction – Thermodynamics

Alexis. Giauque1

1Laboratoire de Mecanique des Fluides et AcoustiqueEcole Centrale de Lyon

Ecole Centrale Paris, January-February 2015

Alexis Giauque (LMFA/ECL) Turbomachinery Aero-Thermodynamics I Ecole Centrale Paris 1 / 70

Table of Contents

1. IntroductionSome historyTurbomachinery now and in the near future

2. Compressible flows: A refresher crash courseIsentropic flow relations

3. Dimensionless quantities and similitude lawsDimensionless numbersSimilitude laws

4. ThermodynamicsEnergiesEffective workKinetic energy / Work of internal forcesInternal energy / mechanical dissipationEntropy / Gibbs equationSummary


You said Turbomachines?

Turbomachines are machines that transfer energy between a rotor and afluid.If the energy is transferred from the fluid the turbomachine, it is a turbine.If the energy is transferred to the fluid, it is a compressor.Turbomachines

have been here for long

are almost everywhere

are key ingredients in projects that will address climate change andressource scarcity issues


Let’s have a closer look to turbomachinery components

Figure : Schematic views of axial and centrifugal rotors


How it all began...

-120 — The first turbomachinery : The aeropile (Hero of Alexandria)


How it all began...

1500 – Chimney Jack (Leonardo Da Vinci)


How it all began...

1629 – First centrifugal impeller (Papin)

1791 – The first concept of gas turbine cycle (John Barber)


How it all began...

1827 – The first underwater hydraulic turbine (Fourneyron)


How it all began...

1883–1897 The first modern steam turbines (De Laval, Rateau,Parsons, Curtis)

1897 – Demonstration of first modern steam engine boat (Parsons)


How it all began...

1905 – First self-sustained gas turbine cycle (Societe Anonyme desTurbomoteurs - Paris)


How it all began...

1939 – First 4 MW utility power generation gas turbine (NeuchatelSwitzerland) – Thermal efficiency 17.4%


How it all began...

1934 – First turbojet engine (von Ohain - Germany)


How it all began...

1939 – First turbojet airplane (Heinkel-178)


How it all began...

1947 – First Mach 1 flight (Charles ”Chuck” Yeager with the X-1)


What are they used for right now: Propulsion inaeronautics - Civil Applications

(a) Pratt & Whitney 4156. Fan diameter: 2.4m.Equips A310-300, A300-600, B747-400, B767-200,MD-11


What are they used for right now: Propulsion inaeronautics - Military Applications

(b) GE F404. Turbofan with post-combustion


What are they used for right now: Electricity Production -Thermal

Figure : Gas Turbine for electricity production (43 MW with a thermal efficiencyof 33%)


What are they used for right now: Electricity Production -Nuclear

Figure : Steam Turbine used in nuclear facilities for electricity productionAlexis Giauque (LMFA/ECL) Turbomachinery Aero-Thermodynamics I Ecole Centrale Paris 18 / 70

What are they used for right now: Electricity Production -Hydraulic

(a) Pelton Turbine (b) Francis Turbine (c) Kaplan Turbine

Figure : Hydraulic Turbines


What are they used for right now: Electricity Production -Wind

Figure : Wind Turbine


And now what are the stakes and technologies?

Sustainable progressPropulsion – Hybrid plane



Sustainable progressPropulsion – Contra-Rotative Open Rotors (CRORs)


http://www.aeronewstv.com/fr/industrie/motoristes-aeronautiques/1373-quand-snecma-developpe-le-moteur-du-futur.html


Sustainable progressElectricity production – Break Efficiency Barriers



Share progressElectricity production – Improve on existing technologies


Future technologies were hard to predict in the 20thcentury...

A few statements from eminent scientists and engineers(source: Cyrus B. Meher-Homji, ASME)

”The energy produced by the breaking down of atoms is a very poorkind of thing. Anyone who expects a source of power from thetransformation of these atoms is talking moonshine.” –ErnestRutherford, circa 1930.

”As far as sinking a ship with a bomb is concerned, it just can’t bedone.” –Rear Admiral Clark Woodward, 1939, US Navy.

”That is the biggest fool thing we have ever done?. The atomicbomb will never go off, and I speak as an expert in explosives.”–Admiral William Leahy, US Navy, to President Truman, 1945.

”Space travel is utter bilge.” –Sir Richard van der Riet Wooley,Astronomer Royal, 1956.


And it goes on...

A few statements from eminent scientists and engineers

”Cellular phones will absolutely not replace local wire systems.” –Marty Cooper, Director of research at Motorola (1981)

”I predict the Internet in 1996 (will) catastrophically collapse.” –Robert Metcalfe co-inventor of Internet (1995)

”The subscription model of buying music is bankrupt.” – Steve Jobs(2003)

”There’s no chance that the iPhone is going to get any significantmarket share.” – Steve Ballmer, Microsoft CEO (April 2007)

”In five years I don’t think there’ll be a reason to have a tabletanymore.” – Thorsten Heins, BlackBerry CEO (2013)


Table of Contents






Some physical phenomena in which compressibility cannotbe ignored


Isentropic flow relations

Let’s consider the nozzle below in which the fluid is accelerated.

In this nozzle we will consider that the compressible fluid undergoes anreversible adiabatic or isentropic transformation.



Adiabatic transformation : – no heat is exchanged between the fluid andthe nozzleIsentropic transformation : – the entropy is constant during thetransformation1

Since in this nozzle, there is also no work (no moving parts) exchangedwith the fluid, the following relations hold:

1 h0 = CpT0. The total enthalpy per unit mass is constant along theflow2

2 ∆s = Cv ln[PP1

(ρ1ρ

)γ]= 0 regardless of the reference state ”1”.

1The term entropy actually refers in statistical mechanics to the volume of the phasespace. To know more about entropy have a look at this video

2All quantities are considered per unit massAlexis Giauque (LMFA/ECL) Turbomachinery Aero-Thermodynamics I Ecole Centrale Paris 30 / 70

http://ocw.mit.edu/resources/res-tll-004-stem-concept-videos-fall-2013/videos/governing-rules/entropy/


Obtaining

T0 = T

(1 +

γ − 1

2M2

)



T0 = T

(1 +

γ − 1

2M2

)This equation provides the relation existing between the total (stagnation)temperature and the static (actual) temperature.Providing there is no heat or work exchange with the fluid, it is a fonctionof the Mach number only.



Obtaining

Isentropic Flow relations

(T0

T

)=

(1 +

γ − 1

2M2

)(P0

P

)=

(1 +

γ − 1

2M2

) γγ−1

(ρ0

ρ

)=

(1 +

γ − 1

2M2

) 1γ−1


Critical section and mass flow rate

Obtaining

m = P0T−1/20

(γr

)1/2M

(1 +

γ − 1

2M2

)− γ+12(γ−1)

A


Critical section and mass flow rate

Obtaining

m = P0T−1/20

(γr

)1/2M

(1 +

γ − 1

2M2

)− γ+12(γ−1)

A︸︷︷︸(1+ γ−1

2 )− γ+1

2(γ−1) A?

A? is fixed at the design stage (geometry) and the fluid (usually air) is also

fixed so that the quantity m√T0P0

enables to compare turbomachinesregardless of the external conditions (P0, ρ0, T0).

Standard mass flow rate

The standard mass flow rate, largely used in turbomachinery, is defined as

mst = m√T0P0

Pst0√T0st


Table of Contents






Defining the parameters...

The most important job of an engineer/scientist is to define theparameters upon which the system he/she studies depends.The following list can be proposed

ρ0 (kg.m-3), fluid density,

µ (kg.s-1.m-1), fluid viscosity,

U (m.s-1), reference velocity,

D (m), reference dimension,

Q (m3.s-1), volume flow rate,

∆p0 (kg.m-1.s-2), change in total pressure,

P (kg.m2.s-3), power.


Using Vashy-Buckingham Theorem

There are 7 parameters and 3 dimensions [L,T,M], Vashy-Buckinghamtheorem tells us that there are 7-3=4 dimensionless numbers describingthe system.It also tells us that the initial relation f (ρ0, µ,U,D,Q, (∆p0),P) = 0 canbe recast as f (Π1,Π2,Π3,Π4) = 0 in which the Πs are defined as:

Π1 = ρa10 ∗ µa2 ∗ U ∗ Da4 ∗ Qa5 ∗ (∆p0)a6 ∗ Pa7

Π2 = ρb10 ∗ µb2 ∗ Ub3 ∗ Db4 ∗ Q ∗ (∆p0)b6 ∗ Pb7

Π3 = ρc10 ∗ µc2 ∗ Uc3 ∗ Dc4 ∗ Qc5 ∗ (∆p0) ∗ Pc7

Π4 = ρd10 ∗ µd2 ∗ Ud3 ∗ Dd4 ∗ Qd5 ∗ (∆p0)d6 ∗ P



Obtaining

Π1 = ρ0 ∗ µ−1 ∗ U ∗ DΠ2 = U−1 ∗ D−2 ∗ QΠ3 = ρ−1

0 ∗ U−2 ∗ (∆p0)

Π4 = ρ−10 ∗ U−3 ∗ D−2 ∗ P



Let’s define those dimensionless numbers

the Reynolds Number (Re = ρUDµ ). It assesses the nature of the

flow (laminar/turbulent)

the flow coefficient (φ = QUD2 ). It provides a comparison of the

output velocity with the reference velocity.

the load coefficient (Ψ = ∆p0

ρ0U2 ). It compares the change in pressureto the available dynamic pressure.

the power coefficient (P = PρU3D2 ). This coefficient is a

dimensionless form of the power output.


Other numbers/coefficients can be important

the Mach number (Ma = Uc ). It assesses the importance of

compressibility effects

the specific speed, Ns = NQ0.5

(∆h0)3/4 . This coefficient is a normalization

of the rotation speed.

the specific diameter, Ds = D(∆h0)1/4

Q0.5 . This is a normalization ofthe reference dimension of the turbomachine.

The efficiency, η (We will come to that later)

The degree of reaction, Λ (We will come to that later)


Specifics for Hydraulic turbines

Hydraulic turbines work with water which is practically incompressible sothat the previous list can be revisited. The following coefficients aredenoted the ”Rateau” coefficients.

the flow coefficient, δ = QND3 ,

the manometric coefficient, µ = gHm

N2D2 , where Hm is the waterheight,

the power coefficient, τ = PρN3D5 ,

the specific rotation speed, Ns = NQ0.5

(gHm)3/4 ,

the torque coefficient, γ = CρgHmD3 ,

the opening coefficient, Φ = Q(gHm)0.5D2 .


Why are these numbers important

A good reason to consider dimensionless numbers

Dimensionless numbers enable to categorize turbomachines (for ex: pistonvs axial vs centrifugal compressors)


Why are these numbers important

Let’s consider a compressor you need to design. You know the volume flowrate (Qv ) and how much work you can afford (∆h0). The following chartlets you decide depending on the size (D) and the rotation speed (N)which technology should be used.


Similitude laws

What you can do with dimensionless numbers and similitude laws

Determine the most important parameters of your system.Limit experimental cost by ’a priori’ limiting the number of variables takeninto accountGuide the design of representative prototypes for the system (for examplea smaller one)


Similitude laws

The whole idea of similitude laws is to analyze a simpler system than thereal one but in which most of the dimensionless numbers are kept constant.

It is important to remember that if all dimensionless numbers are keptconstant, the physical problem is the same for the prototype and the realmachine.


Unfortunately, it is quite difficult so that many scenarios are possibledepending on the problem at hand.Similitude laws can be:

Geometrical. In this case dimensions in different directions are allscaled by the same factor.

Kinematic. The flow coefficient is kept constant. Velocity angles arealso conserved.

Dynamic. The load coefficient is conserved. The ratio of forcesapplied to the blades are the same as for the real machine.

Energetic. The power coefficient is kept constant. The energy ratioare conserved.


Table of Contents






System of interest

We will consider the following system


Total energy

The total energy e0 is composed of the internal energy e and the kinetic

energyV 2

2. Following the first principle of thermodynamics, we have:

De

Dt+

D(V 2/2)

Dt=

Dq

Dt︸︷︷︸heat exchange

+Dwe

Dt︸︷︷︸work of external forces

Total energy balance

De0

Dt=

Dq


+Dwe



External forces power

External forces that apply to a volume of fluid Dm are of two types:

The volume forces denoted ~f . When applied to Dm, their powerwrites Pev =

∫Dmρ~f .~Vdv

The surface forces due to the stress tensor and denoted ¯σ.~n. Theirpower can be written as:

Pes =

∫∂Dm

¯σ.~VdS

where ¯σ = −p¯I + ¯τ . We therefore have:

Pes =

∫Dm

div(¯τ ~V )dv −∫Dm

pdiv(~V )dv︸︷︷︸Compressibility

−∫Dm

~V . ~grad(p)dv︸︷︷︸Transport


External forces work

So that

Dwe

Dt= ~f .~V +

1

ρdiv(¯τ ~V ) − p

ρdiv(~V ) −

~V

ρ. ~grad(p)

Obtaining

Dwe

Dt= ~f .~V +

1

ρdiv(¯τ ~V ) − D(p/ρ)

Dt+

1

ρ

∂p

∂t


Total enthalpy balance

Total enthalpy balance

Dh0

Dt=

Dq


+Dwu

Dt︸︷︷︸effective work


Effective work

Obtaining

Dwu

Dt= ~f .~V +

1

ρdiv(¯τ ~V ) +

1

ρ

∂p

∂t


Effective power

Let’s consider the effective power applied to the fluid in the following

system

Pu =

∫Dm

ρDwu

Dtdv

Replacing the effective work by our previous findings we have

Pu =

∫Dm

ρDwu

Dtdv =

∫Dm

ρ~f .~Vdv +

∫Dm

div(¯τ ~V )dv +

∫Dm

∂p

∂tdv


Effective power

The following hypothesis can be made that simplify the previousexpression

The flow has reached a steady state

The velocity is zero on the solid boundaries

The viscous stress is negligible at the inlet and outlet

Using all these assumptions, one can state that:

Pu ≈∫Dm

ρ~f .~Vdv

showing that the effective power is indeed equal to the power exchangedbetween the flow and the machine.


Link between mechanical effective powers

Obtaining

Pu − Pe =

∫∂D1 U ∂D2

p~V .~nds︸︷︷︸Flowtransferpower

This term represents the power necessary to impose a given flow ratebetween the inlet to the outlet. It is called the transfer power.


Link between effective power and effective work

the effective power writes

Pu =

∫Dm

ρDwu

Dtdv

Pu =

∫Dm

∂ρwu

∂tdv +

∫∂Dm

ρwu~V .~nds

Assuming the system is in a steady state, the effective work is constant atthe inlet and outlet, and since the velocity is zero on solid boundaries, wehave

Pu =

∫∂D1 U ∂D2

ρwu~V .~nds

Pu = −mwu1 + mwu2 = m∆wu


Kinetic energy / work of internal forces

Obtaining from ρDV 2/2Dt

Dwi

Dt=

DV 2/2

Dt− Dwe

DtDwi

Dt= −1

ρ¯σ : ¯D


Kinetic energy

Let’s pause a moment on the expression of the conservation of kineticenergy and use the fact that ¯σ = ¯τ − p¯I

DV 2/2

Dt= ~f .~V +

1

ρ~div(¯τ.~V ) − 1

ρ¯τ : ¯D −

~V

ρ. ~grad(p)

~f .~V represents the work done by the volume forces (the machine)1ρ~div(¯τ.~V ) represents the work done by the viscous forces

−1ρ

¯τ : ¯D is the dissipation of kinetic energy due to the viscosity

~Vρ .

~grad(p) represents the work done by the pressure force (transport)


Internal energy / mechanical dissipation

By subtracting the conservation equation of kinetic energy to the one forthe total energy, one obtains the conservation equation for the internalenergy which writes:

De

Dt=

Dq


− Dwi

Dt︸︷︷︸work of internal forces

TRICKYDe

Dt=

Dq


+1

ρ¯τ : ¯D︸︷︷︸

mechanical dissipation

− pD(1/ρ)

Dt︸︷︷︸compression work

mechanical/viscous dissipation

We see that the mechanical dissipation decreases kinetic energy (slowsdown the fluid) and increases the internal energy (heats up the flow). Thisterm therefore does not appear in the balance of total energy.


Internal enthalpy

To obtain the internal enthalpy conservation equation (useful for open

systems), we add D(p/ρ)Dt on both sides of the previous equation

Dh

Dt=

Dq


+1

ρ¯τ : ¯D︸︷︷︸


+1

ρ

Dp

Dt


Entropy / Gibbs equation

Entropy has two definitions that refer to the same quantity.

Following Boltzmann, it is a measure of the number of micro-statesall corresponding to the same macro-state. Sb = kblnΩ

Following Gibbs, the entropy is a state function that can be computedknowing the thermodynamic state of the system.

They both refer to the same quantity but it has only been definitely provenin 1965.

Second principle of thermodynamics

The second principle of thermodynamics states that entropy of a closedsystem can only grow.

dSclosed system >= 0


Entropy / Gibbs equation

The previous expression can be applied to the universe (which is supposedto be a closed system). The universe can be split between the system ofinterest and its surrounding. In this case we have

dSsystem + dSsurrounding >= 0

So that if the entropy of the system decreases, the entropy of thesurrounding must have increased by a larger quantity.Let’s now come back to Clausius/Gibbs definition of entropy. It is definedas follows:

Gibbs Equation

Tds = dh − 1

ρdp


Physical interpretation

From the previous equation one can write that:

TDs

Dt=

De

Dt+ p

D1/ρ

Dt

Using the conservation equation for the internal energy, one can write that

TDs

Dt=

Dq


+1

ρ¯τ : ¯D︸︷︷︸


The entropy variation is therefore due to the entropy creation due to

the heat exchange of the fluid with its surrounding (It can beradiation, convection or conduction),

the mechanical dissipation


Physical interpretation

Obtaining

TDs

Dt=

2µ

ρ

∑i

∑j

D2ij +

2λ

ρ

∑j

∂Vj

∂xj

2

≥ 0

which shows that the entropy can only grow in a fluid that flows withoutexchanging heat with its surrounding.


Stagnation entropy

Showing that∆s = ∆s0

so that

T0Ds0

Dt=

Dh0

Dt− 1

ρ0

Dp0

Dt

Stagnation entropy is equal to static entropy

Because one goes from the static to the stagnation state by an isentropicdeceleration, stagnation entropy and static entropy are equal.


Thermodynamics – Summary – energy/enthalpy

De0

Dt=

Dq


+Dwe


Dwe

Dt= ~f .~V +

1

ρdiv(¯τ ~V ) − p

ρdiv(~V ) −

~V

ρ. ~grad(p)

Dh0

Dt=

Dq


+Dwu

Dt︸︷︷︸effective work

Dwu

Dt= ~f .~V +

1

ρdiv(¯τ ~V ) +

1

ρ

∂p

∂t

Pu − Pe =

∫Dρ

(Dwu

Dt− Dwe

Dt

)dv =

∫∂D1∪∂D2

p~V .~nds︸︷︷︸Flow transfer power


Thermodynamics – Summary – works

DV 2/2

Dt=

Dwt

Dt︸︷︷︸work of all forces

Dwt

Dt= ~f .~V +

1

ρ~div(¯τ.~V ) − 1

ρ¯τ : ¯D −

~V

ρ. ~grad(p)

Dwt

Dt︸︷︷︸work of all forces

=Dwe


+Dwi

Dt︸︷︷︸work of internal forces

Dwi

Dt= −1

ρ¯σ : ¯D


Thermodynamics – Summary – entropy

TDs

Dt=

Dh

Dt− 1

ρ

Dp

Dt

T0Ds

Dt=

Dh0

Dt− 1

ρ0

Dp0

Dt

TDs

Dt=

Dq


+1

ρ¯τ : ¯D︸︷︷︸



Documents

Turbomachinery Aero-Thermodynamics - crans.orgperso.crans.org/epalle/M2/Turbomachines/Cours1.pdf · You said Turbomachines? Turbomachines are machines that transfer energy between