Theory of turbomachinery Chapter 1 · The continuity of flow equation – mass conservation First...

Theory of turbomachinery

Chapter 1

Department of Energy Sciences / Division of Thermal Power Eng. / JK

Introduction: Basic Principles

Take your choice of those that can best aid your action. (Shakespeare, Coriolanus)


Introduction

Turbomachinery describes machines that transfer energy between a rotor and a fluid, including both turbines and compressors (source: Wiki).

Devices in which energy is transferred, either to, or from, a continuously flowing fluid by the dynamic action of one or more moving blade rows (Dixon)

Definition

The words rotor and continuous separate turbomachines from e.g. reciprocating (piston) engines


Introduction

Close to all electric power is produced by turbomachines

They consume large parts of energy used in many industrial processes

They are integral parts of gas turbines used in e.g. aircraft engines and as (shaft-) power supply in oil and gas industry (for pumps and compressors) as well as propulsion of ships

Why a course on turbomachines?


Samples

Windpower Steam turbines Hydropower Turbochargers of cars and trucks Vacuum cleaners Pumps Dental drills

http://en.wikipedia.org/wiki/Image:Axial_compressor.gif

http://en.wikipedia.org/wiki/Image:Axial_compressor.gif


Samples


Introduction

Energy may flow to the fluid (increasing velocity and/or pressure) or from the fluid producing shaft power

Flowpath: Axial or radial machines (mixed flow)

Changes in density, compressible or incompressible analyses.

Impulse or reaction machines: Does the pressure change in the rotor, or in a set of nozzles before the rotor?

Classifications


Samples

FIG. 1.1. Fig 1.1 Examples of turbobomachines


Axial-flow turbines

FIG. 4.1. Large low pressure steam turbine (Siemens)


Axial-flow turbines

FIG. 4.2. Turbine module of a modern turbofan jet engine (RR)


Axial-flow Turbines: 2-D theory

FIG. 4.3. Turbine stage velocity diagrams.

Note direction of α2


Coordinates

Fig 1.2: The coordinate system

22rxm ccc +=

”streamwise” coordinate


Coordinates (2)

Fig 1.2: The coordinate system


Velocity triangles

Fig 1.3: Velocity triangles for an axial compressor stage

angle flow Relative :angle flow Absolute :

velocityRelative : velocity Absolute :

speed Blade:

βαwcU

1 2 3


Fundamental laws

The continuity of flow equation – mass conservation

First law of thermodynamics and the steady flow energy equation

– The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed.

The momentum equation, F = m ∙ a The second law of thermodynamics

– Total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process


System vs. Control Volume

System: A collection of matter of fixed identity– Always the same atoms or fluid particles– A specific, identifiable quantity of matter

Control Volume (CV): A volume in space through which fluid may flow

– A geometric entity independent of mass


Equation of continuityThe mass flow through a surface element dA:d d dnm c t Aρ=

θcosc cn =

So that

Or, if A1 and A2 are flow areas at stations 1 and 2 along a passage:

Actmm nd

ddd ρ==

AcAcAcm nnn ρρρ === 222111

Fig 1.4: Flow across an element of area


The first law

( ) 0dd =−∫ WQ

( )∫ −=−2

112 dd WQEE

WQE ddd −=

The first law of Thermodynamics: For a system that completes a cycle during which heat is supplied and work is done:

If a change is done from state 1 to state 2, energy differences must be represented by changes in internal energy:

or


The first law

Energy is transferred from fluid to the blades of the machine, positive work is at the rate xW

Mass flow, , enters at 1 and exits at 2m

Heat transfer, , is positive from surrounding to machineQ


( ) ( ) ( )[ ]1221

2212 2 zzgcchhmWQ x −+−+−=−

( )0102 hhmWQ x −=−

0=Q 0>xWFor an adiabatic work producing machine (turbine):

Neglecting potential energy and using total (stagnation) enthalpy:

The steady state energy equation becomes (Reynold’s transport theorem)

( )0201 hhmWW tx −==

And for adiabatic work absorbing machines (compressors): 0<xW

( )0102 hhmWW xc −=−=

The first law


Newton's second law: The sum of all forces acting on a mass, m, equals the time rate of momentum change:

( )xx mct

Fdd

=Σ

Here, only x-component of force and velocity is considered. For steady state the equation reduces to:

( )12 xxx ccmF −=Σ

If shear forces (viscosity) are neglected, the Euler’s equation for one-dimensional flow can be obtained:

0ddd1=++ zgccp

ρ

The momentum equation


( ) 02

d112

21

22

2

1

=−+−

+∫ zzgccpρ

Integrating Euler’s equation in the stream direction yields Bernoulli’s equation:

Control volume in a streaming fluid.

The momentum equation


Bernoulli’s equationFor an incompressible fluid (constant density) using total or stagnation pressure:

( ) ( ) 01120101 =−+− zzgpp

ρ

220 cpp ρ+=

Using the Head, defined as reduces Bernoulli’s eq. to:

( )gpzH ρ0+=012 =− HH

For an compressible fluid, changes in potential are often negligible:

02

d1 21

22

2

1

=−

+∫ccp

ρ

For small pressure changes (or isentropic processes) :

00102 ppp ==


Units

( ) ( ) ( )[ ]1221

2212 2 zzgcchhmWQ x −+−+−=−

( ) WsJ

kgJ

skg:12 ==− hhm

( ) WsJ

sNm

sm

skg:

2 2

221

22 ===

− ccm

1 Newton is the force needed to accelerate 1 kilogram of mass at the rate of 1 meter per second squared

1 Joule is the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of its motion through a distance of one meter

( ) WsJ

sNmm

sm

skg: 212 ===− zzgm


For one-dimensional steady flow, entering at radius r1 with tangential velocity cθ 1 and leaving at r2 with cθ 2 :

Moment of momentumFor a system of mass m, the sum of external forces acting on the system about the axis A-A is equal to the time rate of change of angular momentum:

( )θτ rct

mA dd

=

Where r is the distance of the mass center from the axis of rotation and cθ is the tangential velocity component.

( )1122 θθτ crcrmA −=

Multiplication with the angular velocity Ω = U/r, where U is the blade speed, yields :

( )1122 θθτ cUcUmA −=Ω


Euler’s pump and turbine equations (1.6)

The work done on the fluid per unit mass (specific work) becomes:

01122 >−=Ω

==∆ θθτ cUcU

mmWW Ac

c

02211 >−==∆ θθ cUcUmWW t

t

Control volume for a generalized turbomachine.


Example: radial pump


Rotor inlet: Relative velocity tangent to blade

Rotor exit: tangential velocity induced

Velocity triangles at in- and outlet

[ ] 2211122 0 θθθθτ cUccUcU

mmWW Ac ===−=

Ω==∆


Combining the first law of thermodynamics and Euler’s pump equation (from Newton’s second law):

01021122 hhcUcUmWW cc −=−==∆ θθ

2 21 1 1 1 2 2 2 22 2h c U c h c U c Iθ θ+ − = + − =

Rearranging and using the definition of stagnation enthalpy, allows the definition of the rothalpy, I:

Where does not change from entrance to exit.2 2I h c Ucθ= + −

Rothalpy


The second law of ThermodynamicsClausius Inequality: For a system passing through a cycle involving heat exchange,

where dQ is an element of heat transferred to the system at an absolute temperature T.

If the entire process is reversible, dQ = dQR, equality holds true:

0d≤∫ T

Q

0d=∫ T

QR

From this, the entropy is defined. For a finite change of state:

∫=−2

112

dTQSS R or

TQsmS Rddd ==

m being the mass of the system


EntropyFor steady one-dimensional flow in which the fluid goes from state 1 to state 2:

For adiabatic processes, dQ = 0 and this becomes: 12 ss ≥

For a system undergoing a reversible process , dQ = dQR = m T dsand dW = dWR = m p dυ, the first law becomes:

dE = dQ - dW = m T ds - m p dυ or with u = E / m

T ds = du - p dυ

Further, with h = u + pυ, dh = du + p dυ + υ dp:

T ds = dh - υ dp

( )12

2

1

d ssmTQ

−≤∫


Definitions of efficiencyConsider a turbine: The overall efficiency can be defined as

Mechanical energy available at coupling of output shaft in unit time

Maximum energy difference possible for the fluid in unit time η0 =

If mechanical losses in bearings etc. are not the aim of the analyses, the isentropic or hydraulic efficiency is suitable:

Mechanical energy supplied to the rotor in unit time

Maximum energy difference possible for the fluid in unit time ηt =

The Mechanical efficiency now becomes η0 / ηt


EfficiencyFrom the steady flow energy equation,

and the second law of thermodynamics,

dQ can be eliminated to obtain:

[ ]zgchmWQ x d2dddd 2 ++=−

( )phmsmTQ dddd υ−=≤

[ ]zgcpmWx d2ddd 2 ++−≤ υ

For a turbine (positive work) this integrates to:

( ) ( )1222

21

2

2

2d zzgccpmWx −+−+≤ ∫υ


Efficiency

Once more applying T ds = dh - υ dp = 0 for the reversible adiabatic process:

and hence the maximum work from state 1 to state 2 is:

where the subscript s denotes an isentropic change from state 1 to state 2

gzcpgH ++= 22ρ

In the incompressible case, neglecting friction losses:

[ ]zgchmWx d2ddd 2max, ++−=

[ ] ( ) ( )[ ]120201

1

2

2max, d2dd zzghhmzgchmW sx −+−=++= ∫

[ ]21max, HHgmWx −=

where


The picture can't be displayed.

Enthalpy-entropy diagrams for turbines and compressors.

Efficiency


Neglecting potential energy terms, the actual turbine rotor specific work becomes:

And, similarly, the ideal turbine rotor specific work becomes:

where the subscript s denotes an isentropic change from state 1 to state 2

( ) 222

21210201 cchhhhmWW xx −+−=−==∆

( ) 222

21210201max,max, sssxx cchhhhmWW −+−=−==∆

Efficiency


If the kinetic energy can be made useful, we define the total-to-total efficiency as

Which, if the difference between inlet and outlet kinetic energies is small, reduces to

( ) ( )sxxtt hhhhWW 02010201max, −−=∆∆= η

( ) ( )stt hhhh 2121 −−=η

If the exhaust kinetic energy is wasted, it is useful to define the total-to-static efficiency as

( ) ( )sts hhhh 2010201 −−=η

Since, here the ideal work is obtained between points 01 and 2s

Efficiency


Efficiencies of compressors are obtained from similar considerations:

( ) ( )01020102 hhhh sc −−=η

Minimum work input

Actual work input to rotorηc =

Which, if the difference between inlet and outlet kinetic energies is small, reduces to

( ) ( )1212 hhhh sc −−=η

Efficiency


Small stage or polytropic efficiencyIf a compressor is considered to be composed of a large number of small stages, where the process goes from states 1 - x - y -…. - 2, we can define a small stage efficiency as

( ) ( ) ( ) ( ) ...11min =−−=−−== xyxysxxsp hhhhhhhhWW δδη

If all small stages have the same efficiency, then

However, since the constant pressure curves diverge:

WWp δδη ΣΣ= min

( ) ( ) ( )121 ... hhhhhhW xyx −=+−+−=Σδ

and thus

( ) ( )[ ] ( )121 ..... hhhhhh xysxsp −+−+−=η

( ) ( )1212 hhhh sc −−=η

( ) ( ) ( )121 ..... hhhhhh sxysxs −>+−+− and cp ηη >


If T ds = dh - υ dp,

then for constant pressure:

(dh / ds)p = T

or

At equal values of T:

(dh / ds)p = constant

For a perfect gas, h = Cp T,

(dh / ds)p = constant

for equal h

Small stage or polytropic efficiency


Assume two identical stages with:

Expansion

consthconst 0 =∆= ,η

0

0,s

h consth

η∆

= =∆ p01

p03

p02

h

02

0302,s

03,s03,ss

01

01 03 0013

01 03 01 03

2

,ss ,ss

h h Δhh h h h

η− ⋅

= =− −

( ) ( )

( )02 03, 02, 03,

001 03,

013

2s s ss

ss

h h h h

hh hη

η η

− > −

⋅∆− <

>

s

Define the total turbine efficiency as:

By observation from the figure:

Usage of polytropic and isentropic efficiencies?


Small stage efficiency for a perfect gas

Tp

hhis

p dCd

dd

p

υη ==

For the isentropic process

T ds = dh - υ dp = 0

and with h = Cp T,

the polytropic efficiency becomes:

Now compression!



Tp

pTR

p dd

Cp

=η

pp

TT

p

d1dγηγ −

=

And with Cp = γ R / (γ − 1):

( ) p

pp

TT

γηγ 1

1

2

1

2

−

=

With constant γ and efficiency, this integrates to

Substituting υ = RT / p into Tp

hhis

p dCd

dd

p

υη ==



For the ideal compression, ηp = 1, and the temperature ratio becomes:

( ) ( )

−

−

=

−−

111

1

2

1

1

2p

pp

pp

c

ηγγγγ

η

Which is also obtainable from pγ = constant and pυ = RT. If this is substituted into the isentropic efficiency of compression for a perfect gas,

( ) γγ 1

1

2

1

2

−

=

pp

TT

a relation between the isentropic and polytropic efficiencies is obtained:

( ) ( )1212 TTTT sc −−=η





For a turbine, similar analyses results in

( ) ( )

−

−=

−− γγγγη

η1

1

2

1

1

2 11pp

pp p

t

and

( ) γγη 1

1

2

1

2

−

=

p

pp

TT

Thus, for a turbine, the isentropic efficiency exceeds the polytropic (or small stage) efficiency.






Reheat factor

For e.g. steam turbines

Hps

is

isst R

hhh

hhh

hhhh ηη =

−Σ∆

Σ∆−

=−−

=21

21

21

21

i.e. the ratio of the sum of small isentropic enthalpy changes to the overall isentropic enthalpy change.

Thus:

( ) ( )[ ] ( )sysxxsH hhhhhhR 211 ... −+−+−=


Reheat factor

Mollier diagram showing expansion process through a turbine split up into a number of small stages.


The inherent Unsteadiness

a: Pressure tap at *b: Static pressure measurements vs time

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Theory of turbomachinery Chapter 1 · The continuity of flow equation – mass conservation First...

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