Lewis R.I. Turbomachinery Performance Analysis

54

Principles of performance analysis for axial turbines

Pl h o

P2

J

/p3

y3

Pol

Po2

c12/2

Aho loss /o3s~~ ~ c22/2

l c32/2

"3

s s

(a) (b)

Fig. 3.4 Condition line for a gas turbine stage: (a)

h-s

condition line; (b) ho-s condition line

3.2 Total-to-total efficiency I~T T

So far we have used total-to-total efficiency r/Tr without formal definition. Let us

now correct this by consideration of the stage thermodynamics. The condition line

1-2-3 recording the thermodynamic changes of state through the blade row can be

plotted on an

h-s

diagram as illustrated in Fig. 3.4(a). Alternatively we may derive

the stagnation enthalpies hol = hi + ~/2 etc. and plot the

ho-s

condition curve

ol-o2-o3, Fig. 3.4(b). From this we may then more easily visualise the definition

of total-to-total efficiency r/Tr:

Stagnation enthalpy drop

rrrr = Ideal stagnation enthalpy drop

ho 1

_

ho 3

(3.16)

hol - ho3s

hol- ho3s represents the maximum possible stagnation enthalpy drop for isentropic

flow through an ideal loss-free turbine having the same overall pressure ratio

Pl[P3.

Thus we can identify the lost stagnation enthalpy (Aho)loss of the real turbine stage

through

(Aho)loss = ho3- ho3s

(3.17)

Also, since Aho = hol- ho3, Eqn (3.16) for r/Tr may be expressed as

Aho 1

= (3.18)

r/TT = Aho + (Aho)loss 1 + (Aho)loss

Aho

3.3 50% reaction stages

55

Now our previous definitions of loss coefficient, Eqns (3.6), involved stagnation

pressure losses Apo s and Apo R rather than stagnation enthalpy loss. To accommodate

this we make use of the well-known thermodynamic relationship (Rogers and

Mayhew, 1992):

1

dh = Tds +-dp (3.19)

P

If we apply this along the reversible path 1-3s of the perfect turbine for which ds = 0,

we obtain

dh= 1 dp _

P

(for the isentropic line 1-3s)

If (Aho)loss is small we may thus make the assumption that

1

(Aho)loss

=.

(Apo)loss

P

where

(mpo)los s

represents the total loss of stagnation pressure for the stage, rrrr then

approximates to

1

T/T T = (Apo)los s (3.20)

1+

pAho

Since (Apo)loss = Apos + ApoR we may introduce the dimensionless groups from Eqns

(3.6) and (3.18) to obtain finally

1

r/Tr = 1 + 2-~[ (U) ~$2+ (~)~R3 ] (3.21)

To proceed further we need to derive relationships for

c2/U

and

w3/U

in terms of

~b, @ and R by reference to the velocity triangles. We will first tackle the easier case

of 50% reaction stages (R = 0.5), Sections 3.3 and 3.4. Following this we will go on

to deal with stages of arbitrary reaction, Section 3.5.

3.3 50% reaction stages

From the definition of stage reaction, Eqn (3.4), for 50% reaction we have

h2- h3 0.5

hi-h3

from which we obtain

hi - h2

= h2- h3 (3.22)

56

Principles of performance analysis for axial turbines

k.. Z~Co = Z~Wo

I-

o2 w3_ Ol

C 1 -- C3 Cx

v

02 __. w03

Fig. 3.5 Velocity triangles for a 50% reaction turbine stage

The specific enthalpy drop across stator and rotor are thus equal. From this we may

deduce that the velocity triangles are symmetrical as shown in Fig. 3.5. To be certain

of this we will prove it as follows. Since the stagnation pressure remains constant

through the stator,

hi_h2= 1 2

~(c2 c 2) (3.23)

Similarly, relative to the rotor (see Section 7.2), the relative stagnation enthalpy

remains constant, resulting in

h2_ h3 = 1 2

~(w 3 w 2)

(3.24)

Equating these and introducing c 2 = c 2

+ c21

etc.,

c~- Cl ~ - w3 ~ - w~

hence

cb- ~1 - w~03- wb

and

(C02- C01)(C02 W col ) -" (w03- Wo2)(w03 4- w02 )

but from the velocity triangles we observe that

C02 4- Co1 = W03 Jr- W02

3.3 50% reaction stages

57

The previous equation then reduces to

c02- col =

w03- w02

Solving these last two equations we have finally

C01 -- W02

and c02

=

w03

(3.25)

Thus for 50% reaction the velocity triangles are symmetrical as drawn in Fig. 3.5.

Example 3.1

Problem

Given the design input data of the turbine shown in Fig. 3.1, calculate ~b and q~

assuming identical blade geometry at the mean radius for all three stages.

Solution

The calculation for ~b is as follows.

r m = l(r h + rt) =

0.375

m

U

= rmO, =

0.375 x 27r x 6000/60 = 235.62 m s -1

From the equation of state

p = pl/RT1

= 2.5 x 105/(287.0 x 1200)

-3

= 0.7259 kg m

From the continuity equation

Therefore

and

th = pCx zr(r 2 - r 2)

Cx

= 35.0/(0.7259

x 7r(0.452 - 0.32)) =

136.42 m s

4~ = cx/U

= 0.578 98

-1

Next we calculate ~.

(Aho)total =

hol

-- ho3 = hi - h3

since

C 1 = C 3

= cp(T 1 - T3) =

"yR (TI_ T3 )

7-1

To find T3,

hol

- ho3

hi

--

h3 T1- T3

r/Tr = hom- ho3s ~ hi - h3---~ = T1 - T3s

58 Principles of performance analysis for axial turbines

t_. r

~W2/U ~~. ~Cl/U

U/U = 1.0

I - ~ I

Fig. 3.6 Dimensionless velocity triangles for a 50% reaction turbine stage

Therefore

T 3 = T 1 - r/Tr(T 1 - T3s )

= 1200{1 - 0.92(1

- (1.02/2.5)~

= 950.53 K

Substituting this into the equation for

(Aho)total

above,

(Aho)tota 1 --

Hence for one stage

1.4

x

287.0

(1200 - 950.53) = 250 589 J kg -1

0.4

~ho = 89 • 250 589 = 83 529.7 J kg-1

= Aho/U 2 = 1.504 58

3.3.1 Dimensionless velocity triangles

Let us now consider the factors which determine the general shape of the velocity

triangles. By inspection of Fig. 3.5 we see that the overall shape is governed entirely

by the three velocities Cx, Aco and U. Now we can show that Cx and Aco are uniquely

related to the flow coefficient ~b and work coefficient ~ respectively. Thus from Eqn

(3.7),

Cx = ~U (3.26)

3.3 50% reaction stages

59

Also from the Euler turbine equation (1.10)

Aho

=

U(Col + co2)= UAc 0

where

Col

and c02 are the absolute values as shown in Fig. 3.5 and

Aco

is thus the

change in peripheral w~hirl through either the stator or rotor. By definition

Aho = OU 2, Eqn (3.8), so that finally

Aco = ~U

(3.27)

The obvious strategy to follow at this point is to make the velocity triangles

dimensionless by dividing all velocities by the blade speed U. The outcome of this

is portrayed in Fig. 3.6 from which important and interesting conclusions may be

drawn. We observe that the dimensionless velocity triangles and therefore the general

blade shapes required to achieve them are totally determined by the stage duty

coefficients ~b and ~. The dimensionless blade speed is of course unity. It follows

also that all other angles and velocities may be expressed explicitly as functions of

~b and ~. These may be summarised as follows.

a l = f12 = arc

tan ~b

(0+1)

if2 = f13 = arc

tan 24~

c_ _w3 j

_ _ q~2+ (~b+l)

2

C...~_ 1 = C_.~. 3 __ = t~ 2 + ~(~-- 1) 2

U U

The fluid deflection angles for stator and rotor thus become

(3.28)

8 S = 8 R -- arc

tan (a I

+ Or2)

- arc tan q~2_ 1

1 44~2

(3.29)

It is quite clear therefore that the designer's choice of overall duty requirements and

therefore of duty (th, ~), Fig. 3.2, will have a crucial effect upon the velocity triangles

and thus upon blade aerodynamics. We will now go on to study some of these effects

in relation to published data from experimental tests.

Example 3.2

Problem (a)

Calculate the angles and velocities identified in Fig. 3.5 for the stages considered in

Example 3.1 and comment.

60

Principles of performance analysis for axial turbines

Solution

From solution of Example 3.1, the duty coefficients were ~b = 0.578 98, qJ = 1.504 58.

Introducing these into Eqns (3.28) and (3.29) we have

or1 = f12- 23.45~

or2 = f13 -- 76.98~

e S = e R = 88.73 ~

c2 = w___2 = 1.3797

U U

Cl = c3 = w__g2 = 0.631 56

U U U

Modest interstage swirl

Fairly high efflux angles from stator and

rotor

Fairly high fluid deflections required of

stator and rotor

Stator and rotor exit velocities fairly high

compared with blade speed

Interstage velocities kept well below blade

speed

Problem (b)

Calculate exit Mach number from the stator.

Solution

M 2

is given by

a2 V'yRT2

where a2 = speed of sound, and T2 = temperature at exit from stator.

To find T2 we can note the following. For identical stages with the same velocity

levels

Cx

and U, the temperature drop per stage will be one-third of the overall

temperature drop, namely

ATstag e =

1(1200 - 950.53) = 83.157 deg C per stage

Since the stages are 50% reaction, half of this occurs through stator and half through

rotor. Thus

T2 = 1200-1 x 83.157 = 1158.42 K

Hence from the equation for ME above,

1.3797 x 235.62

ME = V'l.4 x 287.0 x 1158.42 = 0.476 49

3.4 An experimental correlation based on model tests

All large manufacturers of gas and steam turbines do of course undertake extensive

test work on models or prototypes leading to experimental correlations which act as

a vital long term data base for future design and development. Such a correlation

was published by S. F. Smith (1965) based upon 70 model turbine tests related to

3.4 An experimental correlation based on model tests

61

Fig. 3.7

Model turbine instrumentation (Smith, 1965) (by courtesy of the Von Karman Institute)

the post-war aircraft propulsion gas turbines of Rolls-Royce such as the Avon, Dart,

Spey, Conway and others. The special model four-stage turbine test facility,

illustrated in Fig. 3.7, had blade rows 15 to 20 inches (38 to 51 cm) in diameter and

was supplied with air at 100 ~ to 200 ~ Such low temperatures, as compared with

1000 ~ or more in the prototype, not only simplified the instrumentation needed

for accurate assessment but also permitted easy matching of prototype Mach numbers

at the smaller model scale. An additional advantage was the small power output of

800 to 2000 bhp as compared with, for example, the 70000 bhp produced by the

Conway. Matching of Reynolds number between model and prototype is another

requirement to be met, but Smith reported little influence of blade chord Reynolds

number upon efficiency for the bigger engines for which Ret tended to be in excess

of 105 .

The important outcome of this test series was the turbine efficiency correlation

shown in Fig. 3.8. Each turbine was tested over a range of pressure ratios to determine

its maximum efficiency point, for which ~b and q~ were then calculated. Thus each

point plotted on Fig. 3.8 represents one particular test model turbine at its best duty

point (~b, qJ), and its efficiency is entered adjacent to that point. A pattern emerges

from the data immediately, from which contours of constant efficiency can be

constructed revealing very distinct trends. The following comments are worthy of

note:

(1) All data points represent best efficiencies likely to be achievable. For

example, a turbine designed to operate at ~b = 0.9, q~ = 1.2 might achieve at

best 92% efficiency.

(2) The best turbine design duty point to aim at, if efficiency is all that matters,

is in the region of ~b = 0.6, q~ = 1.0. However, it should be observed that this

lies on the edge of the available test data in this correlation.

(3) The majority of test points, being related to a practical series of engines,

were aimed at higher ~b and qJ values. A designer may thus accept some

reduction of efficiency in order to increase mass flow (hence engine thrust)

and stage loading (hence less blade rows/weight/cost).

One important point which should be mentioned is that efficiencies were evaluated

on the assumption of zero tip clearance. Losses due to flow through the clearance

62

Principles of performance analysis for axial turbines

CURVES GIVE STAGE EFFICIENCY)

AT ZERO TIP LEAKAGE

30-- r

9 ! ,

~--~-(~~ I

.

t

STAGE LOADING AH ~ ../~,.a :

' ... ~ I,,.o~

",~ \ },

.

119.11 N.~

9 3 -4 'S "6 -7 -8 '9

I0

I'l

1"2 I 3

STAGE FLOW FACTOR~Va/U

Fig.

3.8 Efficiency correlation for turbine stage (Smith, 1965) (by courtesy of the Von Karman

Institute)

gap between blade end and annulus wall can be enormous compared with other

losses. They can be eliminated from the correlation by repeating tests with increasing

tip clearance and then extrapolating the results to simulate zero tip clearance.

Alternatively, formulae analogous to those developed in Section 8.4.5 for fans can

be used to estimate the tip leakage losses. These may then be subtracted from the

measured stage losses before working out the total-to-total efficiency.

3.4.1 Theoretical analysis by S. F. Smith

S. F. Smith developed a most interesting theoretical analysis to explain the shape

of these efficiency contours, as follows. He argued that the losses in any blade row

will be proportional to the dynamic head or kinetic energy in the row. He proposed

that the average kinetic energy in a blade row could be represented by 89 2 + ~),

Fig. 3.6. For his 50% reaction stages Smith therefore defined the coefficient

fs = Aho/(~ + ~) as the ratio of the shaft work output to the sum of the mean kinetic

energies within stator + rotor. He argued that we should expect a high efficiency when

this ratio of work output to kinetic eneragy is high and to confirm this action he

proceeded to plot contours of fs = Aho/(C~l+ c 2) onto a (4', ~') chart.

It is indeed striking to compare the r/Tr contours of Fig. 3.8 with the fs contours

of Fig. 3.9 which show a remarkable similarity, confirming Smith's hypothesis. By

reference to the dimensionless velocity triangles, Fig. 3.6, we can derive an analytical

expression for fs as follows:

Aho

Aho/U 2

fs = 4 + c2 = (Cl/U) 2 + (c2/U) 2 (3.30)

2q,

44,2 + q,2 + 1

3.4 An experimental correlation based on model tests

63

3.0

~'opt =. ~/4~b2 +

1

2.5-L

0.6 ~ 0.5

lk ~. 0.7

2.0-

1.5

0.8

fs=O.

1.0

0.5

O. 0 4------,-----,-

0.0 0.2

0.4 0.6 0.8 q~ 1.0 1.2

A

4~b2 + 1/,2 +

1

Fig.

3.9 Contours of blade row kinetic energy coefficients, after Smith (1965)

From this expression we can estimate the optimum work coefficient for a given flow

coefficient and hence obtain a theoretical curve of the form qJ = f(~b) which defines

the optimum desirable design duty coefficients. Thus writing

4fs

,gg,

2(4q~ 2-

I//2 4- 1) _ 0

(4q~2 + ~2 + 1)2 --

the optimum duty coefficient locus becomes

~top t =

N/'4q~2 + 1

(3.31)

This is shown on Fig. 3.9 and to some extent follows the trend of optimum efficiency

turbines of the Rolls-Royce correlation, Fig. 3.8. We will return to this later (see

Eqn (3.35)).

3.4.2 Theoretical performance analysis by R. I. Lewis (1978)

A more rational and explicit approach which links directly with the generalised

dimensional analysis of Sections 3.1 and 3.2 has been developed by the present author

(Lewis, 1978). Let us pick up the argument from Eqn (3.21). For 50% reaction stages

Lewis R.I. Turbomachinery Performance Analysis

Подождите немного. Документ загружается.