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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)
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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?
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Samples
Windpower Steam turbines Hydropower Turbochargers of cars and trucks Vacuum cleaners Pumps Dental drills
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Samples
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Samples
FIG. 1.1. Fig 1.1 Examples of turbobomachines
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Axial-flow turbines
FIG. 4.1. Large low pressure steam turbine (Siemens)
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Axial-flow turbines
FIG. 4.2. Turbine module of a modern turbofan jet engine (RR)
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Axial-flow Turbines: 2-D theory
FIG. 4.3. Turbine stage velocity diagrams.
Note direction of α2
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Coordinates
Fig 1.2: The coordinate system
22rxm ccc +=
”streamwise” coordinate
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Coordinates (2)
Fig 1.2: The coordinate system
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
( ) ( ) ( )[ ]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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
( ) 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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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 ==
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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.
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Example: radial pump
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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==∆
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
−≤∫
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
The picture can't be displayed.
Enthalpy-entropy diagrams for turbines and compressors.
Efficiency
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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 >
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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?
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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!
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Small stage efficiency for a perfect gas
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 ==
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Small stage efficiency for a perfect gas
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
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Small stage efficiency for a perfect gas
The picture can't be displayed.
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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.
Small stage efficiency for a perfect gas
Department of Energy Sciences / Division of Thermal Power Eng. / JK
The picture can't be displayed.
Small stage efficiency for a perfect gas
Department of Energy Sciences / Division of Thermal Power Eng. / JK
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 ... −+−+−=
Department of Energy Sciences / Division of Thermal Power Eng. / JK
Reheat factor
Mollier diagram showing expansion process through a turbine split up into a number of small stages.
Department of Energy Sciences / Division of Thermal Power Eng. / JK
The inherent Unsteadiness
a: Pressure tap at *b: Static pressure measurements vs time
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