List of Contents Preface Symbols Chapter
Title
1
Introduction; Dimensional Analysis; Similitude
2
Basic Thermodynamics, Fluid Mechanics: Definitions of Efficiency
,,
I
3
Two-dimensional Cascades
4
Axial-flow Turbines: Two-dimensional Theory
5
Axial-flow Compressors and Fans
6
Three-dimensional Flows in Axial Turbomachines
7
Centrifugal Pumps, Fans and Compressors
8
Radial Flow Gas Turbines
9
Hydraulic Turbines
\
I, i
I ::-
:.::j
,
,;
IJ )
Page
Preface
This manual is a supplement to the book Fluid Mechanics, Thermodynamics of
TurbomachinerY,5th edition and provides detailed solutions of the problems given at the end of each chapter of the parent book. The manual actually originated as
Worked Examples in Turbomachinery (Fluid Mechanics and Thermodynamics) and was written for the second edition of the above book. It now includes many
~
problems taken over a much broader field, so it has expanded considerably. The good news is that it is free!
•
In most engineering courses there is often only a limited amount of time available for problem solving and this is were the manual can be of great benefit. In courses on Turbomachinery it is vital for the student not only to understand the analytical development of a theory but can also apply the theory to the numerical solution of the problem. In my experience students often become better acquainted with the analytical processes of the theory after tussling for a while with a detailed calculation. The standard of the problems solved varies a lot. A large proportion of the problems are comparable in standard to those set in examinations for the Honours Degree in Mechanical Engineering (at a British University) but quite a few are considerably harder and would match the standard of questions set in some Masters degree examinations. Some of the problems are noticeably quite easy to do and are intended to lure the student on to the more difficult areas!
•
Theoretical derivations have been kept to a minimum in the manual unless some particular point required emphasis. On a point of clarification and to avoid needless repetition, I have used roman numerals e.g. egn (i), for referencing equations derived in this manual and arabic numerals, e.g. egn (2.1), when <;iting equations from the parent book. Similarly, the same logic applies to figures. S.L.Dixon
List of Symbols
•
I i
A
area
a
sonic velocity. position of maximum camber
b
passage width. maximum camber
C,
tangential force coefficient
C1 , CI'
lift and drag coefficients
C,
specific heat at constant pressure. pressure coefficient, pressure rise coefficient
C"
ideal pressure rise coefficient
C,
specific heat at constant volume
C" C,.
axial and tangential force coefficients
c
absolute velocity
C
spouting velocity
D
drag force. diameter
s
I
I
I
i
I
I
• I I I
equivalelll diffusion ratio
Dh
hydraulic mean diameter
E. e
energy. specifie energy
F,
centrifugal force in blade
f
acceleration. friction factor
g
gravitational acceleration
H
head. blade height effective head
H,
head loss fue to friction
H,;
gross head
H,
net positive suction head (NPSH)
h
specific enthalpy
/
rothalpy
illciJt:tIcc a"l:;Ic
L
K. k
constants
K~
nozzle velocity coefficient
L
lift force. length of diffuser wall
t
blade chord length. pipe length
M
Mach number
m
mass, molecular 'weight'
N
rotational speed. axial length of di ffuser
N,
specific speed (rev)
N"
power specific speed (rev)
N 5S
suction specific speed (rev)
n
number of stages, polytropic index
P
pressure
Pa
atmospheric pressure
P"
vapour pressure
Q
heat transfer. volume flow rate
•
•
q
dryness fraction
R
reaction. speci fie gas constant"
Re
Reynolds number
RH
reheat factor
RIJ
universal gas constant
r
radius
5
entropy, power ratio
s
blade pitch, specific entropy
T
temperature
[
time. thickness
U
blade speed. internal energy
u
specific internal energy
-
~-
v, v
volume, specific volume
W
work transfer
.j W
specific work transfer
w
relati ve velocity
X
axial force
x. y. z
cartesian coordinate directions
Y
tangential force, actual tangential blade load per unit span
Y:.J
ideal tangential blade load per unit span
YI{
tip clearance loss coefficient profile loss coefficient
•
Ys
net secondary loss coefficient
z
number of blades. Ainley blade loading parameter
a
absolute flow angle
{j
relative tlow angle
r
circulation
"(
ratio of specific heats
8
deviation angle
(
tluid deflection angle. cooling effect iveness
t;
enthalpy loss coefficient, total pressure loss coefficient
'7
efficiency
e
minimum opening at cascade exit
8
blade camber angle, wake momentum th ickness
I
I
I ;1 •
.::! ':::j
i
A
profile loss coefficient
- 3-
dynamic viscosity v
kinematic viscosity, blade stagger angle, velocity ratio
p
density
a
slip factor. solidity
a.
blade cavitation coefficient
a,.
Thoma' s coefficient, centifugal stress torque flow coefficient, velocity ratio
•
•
stage loading factor speed of rotation (rad/s) D,
specitic speed (rad)
D".
power specific speed (rad)
£2."
suction specific speed (rad)
(j)
vorticity
111
stagnation pressure loss coefficient
SUBSCRIPTS
av
average
c
compressor, critical
o
diffuser
-4 -
e
exJr
h
hydraulic, hub inlet, impeller
id
ideal
IS
isentropic
m
mean. meridional. mechanical. material
N
nozzle
n
normal component
0
stagnation property. overall
p
polytropic. constant pressure
R
reversible process. rotor
r
radial
rel
relative
I
s
isentropic. stall condition
I
ss
stage isentropic
~}
.-:;
.'.;.
ie I
I I
turbine. tip. transverse v
x, y,
e
velocity Z
cartesian coordinate components tangential
SUPERSCRIPT
•
time rate of change average
I
I
blade angle (as di~inct from flow angle)
•
nominal condition
-s-
Chapter 1 Introduction: Dimensional Analysis: Similitude
--
Rotor blades Outlet vanes
•
..
Flow-... __
-_.(a) Single stage axial flow Gefli\Jressor or pump
---1----'-
Flow direction
/
-
. _ . - _ . --1--
Oullel vanes
....-j----4-
(b) Mixed flow pump
Guide vanes
Outlet diffuser
•
-+\--
Vaneless diffuser
-........., Volute
~ Impeller (c) CentriflJ(J.a1 compressor or pump
I
II
I ,I
Draught tube.
(d) Francis turbine
Some Turbomachine Geometries
Runner blad
1
Chapter
I
Dimensional Analysis, Similitude 1.1 A fan operating at 1750 rev Imin at a volume flow rate of 4.25 m 3 Is develops a head of 153 mm measured on a water-filled U-tube manometer.
It is required to
build a larger. geometrically similar fan which will deliver the same head at the .same efficiency as the existing fan, but at a speed of 1440 rev/min.
Calculate the
volume flow rate of the larger fan.
Solution. For geometricaBy similar fans the dependent variables·gH.(the net energy
\. , I
transfer) and the efficiency
'7
are expressed in terms of two functional relationships
of the independent var iables, gH = f (Q,N,D,p,!J.) 1 "') = f (Q,N,D,p,l'-)
I
2
I
where Q is the volume flow rate, D is a characteristic diameter of a fan, N is the rotational speed,
J.L
the dynamic viscosity of the iluid and p the fluid density.
Using
either the formal procedure of dimensional analysis or the less formal but more direct process of dimensional elimination (see Q .1.4) with p, N, 0 as common
factors. the dimensionless groups are 2
eND ) !J. 2
pND ) I'-
The group pNI} II'- defines a flow Reynolds number Re based upon blade speed and fan diameter.
It is assumed for the purpose of chis problem that the effects of
changes in Re are small and can be ignored.
Thus. the performance character-
istics are reduced to (i)
!
(il)
i
I
I
~<.~~~_~~..-.,..
-
.
,
..~ ~:- ... " . :_-?-,'l'::~~': ."f~;'.;'("'-'-~ __ -: ,,:,- :..', . ,...~ .'_... " "~-'
- . . - .. '
,-
.
.:
'~
'.'~.
.. " " '
,"-".:
2
I
S.L. DIXON
I
I
For the two fans to have the same efficiency it follows from eqn. (ll) that the volume
flow coefficient Q/(NJ)3) must be the same.
Thus, the volume flow rate of the
second fan is (Iii)
Likewise, for the two fans to deliver the same head, then gH/(ND)2 must be the 3
same which follows from eqn. (I) and the fact that Q/(ND ) is fixed.
Hence, with
HI = H , then 2 NIDI =
N D 2 2
(Iv)
SubstitUting eqn. (Iv) into eqn. (iii)
•
Q
2
= Ql(N/N2) 2 = =
4.25(175011440)2 3 6.277 m Is
It will be observed that the' numerical value of the head developed is not actually used
in solving this problem. 1.2. An axial flow·fan 1.83 m diameter is designed to run at a speed of 1400 revI min with an average aXial air velocity of 12.2 m/s.
A quarter scale model has been
built to obtain a. check on the design and the rptational speed of the model fan is 4200 rev Imin.
Determine the axial air velocity of the model so that dynamical similarity
with the full-scale fan is preserved. neglected.
The effects of Reynolds number change may be
A sufficiently large pressure vessel becomes available in which the compl:,te model
•
can be placed and tested Wlder conditions of complete similarity.
The viscosity of
the air is independent of pressure and the temperature is maintained constant. what pressure must the model be tested?
At
r:--:-------------
. t? 2 . Solutlon. The volume flow rate Q ;; ryt-,rccD' where C 15 the axial velocity. 3 x x _ Thus, the volume flow coefficient Q/(ND ) can be replaced with c I(ND). For x axial velocity of dynamical similarity and ignoring changes in Reynolds number the the model is
_ \ Q=C,. A~-
C~J)"-
( OC '"E~ b_A..LA_J
r . T1 ""vv-..
tJ)
Dimensional Analysis
N D
c
xm
= c
~ xp N D
=
12.2 x 4200/(1400
3
'l(
4)
P P
= 9.15m/s
For complete similarity the Reynolds number of the model must equal that of the prototype.
Thus. Re
p
N
m m
D
m
2 m
(i)
I'm
I.
As the temperature remains constant fJ.
m
=
J.L
p
and. from the gas law, pdp.
Thus,
eqn. (i) becomes Pm Nm D: Pm
PpNpD: = pp(Np/Nm)(D/Dm )
1 x (1400/4200)4
2
2
5.33 atm. 1.3. A water turbine is to be designed to produce 27 MW when running at 93.7 revl
min under a head of 16.5 m.
A model turbine wit!' an output of 37.5 kW is to be
tested Wlder dynamically similar conditions with a head of 4.9 m.
model speed and scale ratio.
Calculate the
Asswning a model efficiency of 8810, estimate the
volume flow rate through the model. I
I I
I
It is estimated that the force on the thrust bearing of the full-size machine wl1l be 7.0 GN.
For what thrust must the model bearing be designed?
Solution. For geometrically similar hydraulic turbines the dependent variables are
the power output p. the efficiency., and the volume flow rate Q.
The independent
variables are the speed of rotation N, the characteristic diameter D, the useful head
H. the dynamic viscosity I' and the density p.
The functional dependences are
written as
P.,. Q .1
= f(p.N.D,gH.I')··
By the application of dimensional analysis the following non-dirnensional groups are
S.L. DIXON
4
formed using p, Nand D as the common ,dimensional factors to eliminate the
dimensions of the other variables (see Q.L4). p
-rs' p~D
7'
A remarkable feature of dimensional analysis is the ability to form a new nondimensional group from any other two such groups provided that the total number of groups in the functional relationship(s) remains the same.
Thus, by combining the
power coefficient with the headcoefficent to eliminate the diameter (which is not given in the problem) a new non-dirnensional group, the power specific speed is
formed, i. e. =
•
p~ p(gH)2.5
Taking the square root of the above expression, the power specific speed is,
2
N = NP1/ / (//l.(gH)%) sp
Assuming that changes in Reynolds number have negligible effect upon the periormance; the model (m) and prototype (P) will have the same N when sp operating under dynamically similar conditions. Thus,
Nplf2 mm
Nplf2
pP.
=
H "/4 m N
m
H "/4 p = N (P /P)
112
-
(H /H) "/4
5 3lf2 /4 = 93.7(27x 10 /37.5x 10) (4.9/16.5) ppm
mp
6
= 551. 2 rev/min
•
The scale ratio is determined from the head coefficient which is the same for model and prototype, Le. gH
P
(N D )2
(N D )2
:. D /D m p
(H
mm
P P 1 m
/H)I2 N / N
ppm
= 0.09264
1 = (4.9/16.5)1293.7/551.2
rI
Dimensional. Analysts
5
i.e. the prototype is 10.8 times bigger than the model. The turbine efficiency is defined as
'7 = P/(pgQH) hence, the volume tlow rate of the model turbine is
Qm = Pm/(pgHm
7 m)
3 3 = 37.5 x 10 /(10 x 9.81 x 4.9 x 0.88)
3
= 0.8865 m /s
The thrust force X is a new variable which can be related dimensionally to the other
·./i
known variables.
As force is the produce of pressure pgH and area which is 2
proportional to D , then a force coefficient can be defined as
, 2 X = X/(pgHD) For dynamical similarity this will·be the same for both model and prototype, hence
xm
= X (H /H )(0 /0)2 pmpmp 7 x 109 (4.9/16.5)(0.09264)2 = 17.84 MN
I
1.4. Derive the non-dimensional groups that are hormally used in the testing of gas
I ,
turbines and compressors.
A compressor has been deSigned for normal atmospheric conditions (101.3 kPa and 15°C).
In order to economise on the power required it is being tested with a
throttle in the entry duct to reduce the entry pressure.
The characteristic curve for
its normal design speed of 4000 rev/min is being obtained on a day when the ambient temperature is 20°C.
At what speed should the compressor be run?
At the point
on the characteristic curve at which the mass flow would normally be 58 kg/s the entry pressure is 55 kPa.
Calculate the actual rate of mass flow during the
I
Describe the relationship between geometry and specific speed for pumps ..
i
Solution. As the fluid density p can
\
1
ch~nge
te~t.
very appreciably across compressors
and gas turbines of large pressure ratio it is necessary to employ compressible fluid relations..
For a compressor of a given configuration and size represented by
:,!....,.-':.,,-- _ . '
.
, ' . -. .
-","'--.'
.-
I
.~. S.L. DIXON
6
a diameter D. operating at rotational speed N and mass flow rate rh and at specified
inlet stagnation conditions (PoI' T 01)' the dependent performance parameters are the outlet stagnation pressure p ~T.
temperature rise
~».
The
,the efficiency., and the overall stagnation 02 . Other dependent parameters may also be used (c.f. eqn.
depend~t variables can be expressed in the form of three unknown
( I.
functional relationships as PoZ'
'7 '
= f(N,D,m,pol,T o1 ''('f')
£lT o
where 1.1' is the dynamic viscosity and y = C Ie .
.
p
v
The dependent variables can be made dimensionless without difficulty by writing
•
where, for convenlence, Pol and T 01 have been replaced by Pol = P01/(RT01) and
1/2
= ('(RT01) . The most convenient and least formal method of finding the o1 remaining dimensionless groups is to take one variable of interest and reduce its
a
dimensions to zero by repeated multiplication with several other variables.
To do
this, thIee of the most easily measured variables P l' Nand 0 are selected. variables have the respective dirnensions.ML-3,
T-~ and L.
These
Considering in detail
the reduction of rh to a dimensionless group by repeated multiplication, Variable
Dimensions
MT- 1 L3T- 1
m m/Po l m/(Pol°3) 3
•
Eliminating
m/(p o1 ND )
T
M
-1
L
0
T
The same process is used to reduce the remaining dimensional variables a (= LT..1) -1 -I 01 and fL (= ML T ) resulting in
aND, y) 01
z
•
The group Pol ND If' is a Reynolds number Re based upon blade speed (d:ND) and compressor size.
The group NOla o l is a blade Mach number M.
::,~_~"<,.~/
":,-,,,. ,-"<,_"
_~~-_
._,,:~.-,_", :.~::.:_:~!;,,?",
.
-,
~ -,
-..
.
".
~', ~
',"
-..
0
',_','
The group
_ ':.,.
.
"
_,
,.,
,
11..,.tl.)
Dimensional Analysis
7
rh/(p 01 NOl) is not very convenient for compressor testing but may be easily trans-
formed into normal form as follows:rhRT
= P 01
ol 2 D (ND)
rhj(RT
Ol
)
2
Pol D /2
as the group (ao/(NDY lf2») =
(l/(MY 1
»)
is a combination of dimensionless
variables which have already appeared as separate independent groups it can be simply deleted from the above group.
I.
Thus, the final non-dlmensional form of the
compressor functional relationships is
tl T
P02
- , ;1 , -o
rh(RT = f(
T ol
Pol
01
Pol D
)1/2
Re, M, y)
2
(I)
For a given compressor of a given size and handling a specific gas it has become the custom in .practice_ to drop y, Rand D from the above set of dimensionless groups. The resulting relationships are then
t:;.T
mT
which are
1/2
01 P 01
o = f( 'y 01
Re)
(ia)
longer dimensionless.
DO
In the case of a turbine, the dependent variables are usually regarded as ~
m, 7
and
ii
.,~.
To so that the dimensional functional relationships are
"It It
11
\1
By a process of reasoning similar to that used for a compressor the variables .are
I!
ii
reduced to a smaller number of non-dimensional groups, i.e.
rh(RTo
//2
2
Pol D
.1 To '/.
T
= 01
f( NO- Pol a 'p' 01 02
Ii
m D}L'
y)
l !
I
r
.,'
~
~,
.
_
~ -<,~~.
~~
-l~
--
.
_"
.
":._";
"
-.
'
,-
-""~
S.L.. DIXON
8
J.K and ignoring any effects due to changes in Re, A . ~
Referring to Fig.
of a compressor is uniquely represented by one value of Nrr 1
~
rh T 1
the design point
and one value of
0
/p .
At the intersection of these two curves there is but one value for each 01 of the dependent variables, p 2/p 1" and AT rr l' Under normal atmospheric
o
00
00
"-
conditions, Pol: 101.3 kPa and Tal: 288 K, the compressor design speed N is 4000 rev/min and the mass now rate rh is 58 kg/s.
Design point conditions still
obtain for the new entry conditions Po;' : 55 kPa, T o{ : 293 K b( adjusting the SP~ed and maSs flow rate to maintain the design point values of rh T 01/2 /p 01 and Nrr 01
12
•
Thus,
•
h :
N' :
N(T' rr )l 01 01
:
4035 rev/min
4000 (293/288)
1/2
12
: (p o;';P 01)(T olrr 0 / rh : (55/101.3)(288/293/158
rho
:
31.22 kg/s
For a pump specific speed is defined, eqn. (1.8), by N
s
:
NQ1/2 /(gH)3/4
where N is the speed of rotation, Q the volwne flow rate and H the head rise.
For a
For
perf.
given speed N, high specific speed would be ob:ained in a pump of small head rise and large volume flow rate, e.g. an axial flow pwnp.
Conversely. a low specific speed
pump would be typified by a radial flow machine of relatively small flow rate and a high head rise. The
0
rall
• tel
press
defined
>.: ..-.>'.. ;-0
",~."\"-::
"
'~"." :.-
_ .
~. :.~,'
-
".,.
., .' , - '''''' . -
",';'
.-'.
',-.< : . . '~: ", ~"
.
.
,' .
I
I
Chapter 2 Basic Thermodynamics,
Fluid Mechanics ,
••
i
Flow direction
•
A
Control Volume of a Generalised Turbomachine
....-----~~~====== 9
Chapter
2
Thermodynamics 2.1.
For the adiabatic expansion of a perfect gas through a turbine, show that the
overall efficiency
7t where
e. =
r
"It
and small stage efficiency
,
."Ip are related by
= (1 - f. P )/(1 • £)
(1 - '1)1'; I
and r is the expansion pressure ratio. y is .the ratio of
specific heats.
e
An axial flow turbine has a small stage efficiency of 8610, an overall pressure ratio
of 4.5 to 1 and a mean value of '( equal to 1.333.
Calculate the overall turbine
efficiency. Solution~
The overall efficiency of a turbine is a.ssumed to mean the total to total
efficiency defined, eqn. (2.21), by 7)
It
=
(h
01
-h
02
)/(h
01
- h
a2s
) h
For a perfect gas, h = C T, so that p
I
"J t
= (T01 - T a2)/(T01 - T a2s)
!
=
(1 - T02{f01)/(1 - To2s/T01)
90'
." (i)
02 The overall total pressure ratio is
025
,e
s e=T/T= 02s 01
(ii)
Consider a small part of the expansion process as shown in the sk.etch.
This expansion is best imagined as a
small Stage with an enthalpy drop dh and corresponding
a
pressure drop dp. a defined as '" Ip
The small srage efficiency is
= dh0 /dh oS
(iii)
-=
S.L. DIxon
10
Now an elementary change in specifIc entropy can be related, using the laws of thermodynamics, to the elementary changes in other properties.
From eqn. (2.18)
Tds = dh - dp/p which is applicable to both reversible and irreversible processes on a pure substance. For a constant entropy process it follows that
dh
as
• dp /p 0
0
Substituting this result and the perfect gas relations, p /p = RT and dh = C dT <>
.'~
?l
Ip
•
<>
0
0
P
0
into eqn. (ill)
where C
p
= P C dT /dp
P
0
= yR/(y-I).
0
= (p /RT ) yR/(y-l) dT /dp
0
0
0
0
0
After rearranging the above equation
Integrating and putting in the limits for the overall process, T (p /p ) ,,!p(y-I)/y 02 01 02 01
rr -
=
r "}p(I-y)/y
=
~ '!p
(Iv)
SUbstItuting eqns. (ii) and (iv) into eqn. (i), the required relation is Obtained,
'!t
= (I - c?P)/(1 -
Withr =4.5, y = L333 and
"J p
~
)
=0.86,
E = r(l-y)/y = 1/4.5°.2498 = 1/1.456 = 0.6868.
e."lp = 0.6868°·86 = 0.7239
•
:. 'J t
= (1-0.7239)/(1-0.6868) = 88,16 per cent
2.2. Air is expanded in a multi-stage axial flow turbine, the pressure drop across each stage being very small.
Assuming that air behaves as a perfect-w:s with ratio
of specific heats y, derive pressure-tempe;-ature relationships for the following
,
processes: '.
(i) reversible adiahatic expansion; (ii) irreversible adiabatic expansion, with small stage efficiency 'Jp;
l
'.
Ii
j
I
-. Thermodynamics
11
(iii) reversible expansion in which the heat loss in each stage is a constant
fraction k of the enthalpy drop in that stage; (iv) reversible expansion in which the heat loss is proportional to the
absolute temperature T. Sketch the first three processes on aT. s diagram. If the entry temperature is 1100 K, and the pressure ratio across the turbine is 6 to
1,
caLcu~ate
the exhaust temperatures in each of these three cases.
Assume that y
.is 1.333. that Ip = 0.85, and that k = 0.1. Solution. (i) For a reversible adi2batic expansion the entropy does not change.
From eqn. (2.18l, with ds = 0,
•
Tds = dh - (I/p)dp = a dT = (I/p)dp = RTdp/p P dTIT = (R/C )dp/p = [(y-I)h dp/p .P :. dh
=C
1
Integrating this result bet:\veen limits, denoted by an initial srate 1 and a final state 2,
yields
(i)
(ii) It has already been shown in the solution of.Q.2.1 for an irreversible adiahatic expansion with small stage efficiency
'7 p'
that I
T IT 1 2
= (p /p ) ~p(y.l)h 1 2
It ~s a consequence of the !econd f..aw of 1hermodynamics that the entropy of a
•
substance (i. e. a system) undergoing an irreversible adiabatic process must
increase.
The magnitude of the entropy increase can be formulated from eqn. (2.18)
as follows:-
Tds = dh - dp/p = C dT - RTdp/p P :. ds ~ C dTrt' - Rdp/p p
Integrating and inserting limits, the entropy increase is
=
I
(ii) _
i
"---
i.e .
12
r
S.L. DIXON S2 - sl = C p -In(rlr l l
I
- R .In(Pip 1)
= R (In(P/P2) - (y/(y-l>] In(r1ff2>j
Substituting for Tiff2 from eqn. (ii), and simplifying s2 - sl = R(I-
"Jpl ~n(P/P2)
(iia)
.e .c.
(iii) From the !econd j-aw of thermodynamics, when an element of heat dQR is transfe::red reversibly
~
the surroundings ~ a unit mass of a substance at an
. absolute tempera.ture T, the specific entropy increases by an amount
ThUs, a reversible heat transfer from the substance to the sunoundings (dQR will cause the specific enttapy to decrease.
<
0)
In the reversible expansion thIough the
turbine with reversible heat loss the signs of the three elements ds) dh and dp in the
expression dQR
= Tds = dh
• dp/p are all negative.
WritingdQR
= kdh,
eqn. (2.1S)
gives
:. (1-k)C dT p
= =
= dh - dp(p dp(p = RTdp/p
dT/T
=
(dp(p)(y-l)/ [y(I-kl1
Tds
...
kdh
_
Integrating and inserting limits a.s before,
(iii) A.dditionally, it is easy tv determine the magnitude of the corresponding specific entropy change. as follows: ...
ds = kC dT/T P s2 - 51
=
-kC
p
tn(r1/T2)
= -kC p (y-I)/[Y(I-k)]ln(p/P2)
= .In(P/P2) kR/(1-k)
(iiial
(iv) The heat loss in each elementary stage is reversible and proportional to T. This condition is satisfied by
dQR = Tds
:.:.
=.
I
I
db - dp/p
_._----~~--========= 13
Thermodynamics
:. ds = C dTJT - R dP/p p (s2 - sl)/R = tn(p/P2) - [y/(y-l)]L.(TlJT 2)
After rearranging and exponentiating, (iv)
P/P2 = (TIJT2)y/(y-l). expC (s2 -SI)/R]
.-
The sketch shows the way the entropy changes as ~
the air is expanded from the initial srate 1 to the
T
final state 2 corresponding to the· first three processes.
•
The final temperatures (and
specific entropy changes) are easily determined
from the preceding equations. With T 1 = 1100 K, P/P2 = 6, "lp = 0.85, y = I. 333 and k = 0.1. From eqn. (i), T
2i
= 1100/6°·2498 =
703.1 K
2iii
From eqn. (ii) T
Zii
s
= 1I00/60.ZIZ3 = 751. 9 K
and the corresponding entropy increase is. from eqn. (iia)
1
From eqn. (iii)
•
T
Ziii
= 1100/6°·2776 = 669.0 K
and the corresponding entropy change is from eqn. (ilia) (sZiii-SI)/R
=
-(0.1/0.9)
Ln 6 =
-O.Z
2.3. A multi-stage high-pressure steam turbine is supplied with steam at a 0 stagnation pressure of 7 MFa abs .... and a stagnation temperature of 500 e. The
corresponding specific enthalpy is 3410 kJ/kg.
The steam exhausts from the
turbine at a stagnation pressure of 0.7 MFa abs .. the steam having been in a super-
-
-----~==-=-~======~
14
S.L • .DIXON
heated condition throughout the expansion.
It can be assumed that the steam behaves
like a perfect gas over the range of the expansion and that V = 1,3.
Given that the
turbine flow process has a small-stage efficiency of 0.82, determine
(i) the temperature and specific volume at the end of the expansion; (ti) the reheat factor.
The specific volume of superheated steam is represented by pv = 0.231(h -1943), 3 wherep is in kPa, v is in m /kg and h is in kJ/kg. Solution. (i) In the notation of Q.. 2.1 the actual temperature ratio across the turbine, from eqn. (2.37), is
•
1 T /T = (PO/P02)'?P(V- )!Y o 02 where
T
Ii
.,I
I,
= 0.82 x 0.3/1.3 = 0.1892 and Po/P02 = 10
"lp(v-I)/v o1
/T
02
The inlet stagnation temperature T 01 = 500 + 273 = 773 K, hence the outlet stagnation temperature is·
- T 02
= 773/1.5461 = 500 K
The specific volume v02 corresponding to stagnation conditions at outlet is
determined with the superheated steam relation pv = O. 231(b -1943) and the perfect gas law pv = RT.
Combining these two equations,
T 02/T01
••••
= 10°·1892 = 1,5461
= (h
02 -1943)/(ho1 -1943)
h 02 - 1943 = (bOl -1943)T02/Tol = (3410-1943)500/773 = 948.9
:.:
h • v
02 02
= 2891, 9 kJ/kg = 0.231 x 948.9/700
3 = 0.3131 m /kg (Ii) The reheat factor is defined, eqn. (2.39), as
I A
ThermodY,Damlc.
15
where the overall or total to total efficiency i.
[ =
1 • (p
/p )(y-l)/y] 02 01
(1 -1/1.5461)/(1 _1/100.2308)
= 0.5461 x
I. 7013/(1.5461 x 0.7013)
= 0.8568 :. R
H
= 0.8568/0.82 = 1.045
0 2.4. A 20 MW hack-pressure turbine receives stearn at 4 MFa and 300 C. •
exhausting from the last stage at 0.35 MFa.
The stage efficiency is 0.85, the
reheat factor 1. 04 and the external losses 2'1. of the ~
enthalpy drop.
Determine the rate of steam'.f1ow. At, the exit, from the first stage Dozzles the steam velocity is 244 mis, specific volume 68.6 dm 3 /kg. mean diameter 762 mm and steam exit angle 76 deg measured
from the axial direction.
Determine the nozzle exit height of this stage.
Solution. From. the definition. of reheat factor, ·eqn. (2.39), the turbine total to total efficiency can be immediately determined:T)
It
= "R
Ip H
= 0.85 x 1.04 = 0.884
Using the notation given in Q.2.1 the isentropic stagnation enthalpy drop, hal: h 02s ' can be determined using steam tables or, less accurately but more quickly. using a
•
Mollier diagram for steam. From steam tables,..at p 0 1 = 4 MFa (40 bar) and 0 0 T 01 = 300 C the initial steam condition i. superheated (about 50 C of superheat) with hal
=2963 kJ/kg and Sol =6.364 kJ/kg aC).
Inspection of the tables shows that at
P02 = 0.35 MFa (3.5 bar) the vapour saturation value of specific entroPy-sgo2> '01' This means the isentropic state point 02s is in the liqllid-vapour phase. dryness fraction q can be evaluated for point 025,
The
----------------S.L. DIXON
16
= (6.364 - 1. 727)/5.214 = 0.8893
Hence, the specific stagnation enthalpy at point 02s is
= h f02 + Q(hg02 - hf02 )
h 02s
= 584 + 0.8893 x 2148 = 2494 kJ/kg Thus, the isentropic stagnation enthalpy drop is
= 2963
hi· h 2 o
0 s
- 2494
= 469 kJ/kg
As the total to total efficiency is known the actual stagnation enthalpy drop can be found, i. e. h
01
- h
="It (h 0 I-h0 2S ) = 0.884x 469 = 414.6 kJ/kg
02
and this is the specific work done by the steam, /:) W.
The actual specific work
delivered at the output shaft is less than this because of the mechanical losses .
The
shaft power delivered; is
where 7Jm is the mechanical efficiency.
Thus, the rate of mass flow
m = W!('7m II W) 3 6 = 20 x 10 /(0.98 x 414.6 x 10 ) = 49.22kg/s
From the equation of continuity and assuming uniform flow at all radii,
m';.:
pAc
x
= pTrd
m
hccosa
Hence, the blade height is h =
mv/(rr dm c cos QI)
= 49.22 x 0.0686/(77" x 0.762 x 244 x .2419) = 0.0239Jlll"-
= 23.9 mm
"
I
turbine at a stagnation pressure of 1. 5 MFa and a stagnation temperatU!e of 350°C.
;
The steam leaves the last stage at a stagnation pressure of 7.0 kPa with a carres-
I
2.S. Stearn is supplied to the first ,~ta.ge of a five stage pressure-compoWlded steam
i
I
\
I t
Thermodynamics
ponding dryness fraction of 0.95.
17
By using a MoJller chart for steam and assuming
that the stagnation state paint locus is a straight line joining the initial and final
states, determine (i) the overall total-te>-total efficiency and total to static efficiency assuming the steam enters the condenser with a velocity of 200 mis, (Ii) the stagnation conditions between each stage assuming that each stage
does the same amount of work, (iii) the total to total efficiency of each stage,
(iv) the reheat factor based upon stagnation conditions. Solution.
•
~
In this problem it is more accurate to use steam tables to determine the
conditions as the final pressure is specified.
For the interstage
calculations where the pressures are not specified, the solutions required are greatly facilitated by the use of a Mollier chart for steam with only a small loss in accuracy.
0 (i) From the tables at p I ~ 1.5 MPa (15 b"ar) and T I ~ 350 C the stagnation enthalpy k a a hal ~ 3148 kJ/kg and the stagnation entropy Sol ~ 7.102 kJ/(kg 0C). Referring to the sketch of the Mollier diagram, the exhaust toral condition is the state point 06.
The exhaust stagnation enthalpy is h
06
~
h
f06
+ q(h - h ) (06 g06
= 163 + 0.95 x . ' = 2451.6 kJ/kg where h
and hgo are the liqUid and vapour saturation enthalpies at P06 r06 6 Hence, the actual speCific work done acroSs the whole turbine is,
•
4 W = hal -h
06
=7 kPa.
= 3148 - 2451.0
696.4 kJ/kg The corresponding isentropic stagnation enthalpy h 6 at the exhaust pressure p 6 o 55 ._ 0 is obtained by determining the dryness fraction q .
s
qs = (sol - s fo~)/(S go6 - s fa 6) = (7.102 - 0.559)/7.715 = 0.8481
h 06ss
~
h
f06
+q (h -h ) = 103 + 0.8481 x 2409 s goo foo
•
=.
':e-=-o
S.L. DIXON
18
= 2206 kJ/kg Hence, the overall isentropic stagnation enthalpy drop which is also the overall ideal specific work is fj W
max
=
h 1- h 6 0
0
55
=
3148 - 2206
=
942 kJfkg
The overall total to toral efflciency is, eqn. (2.21), 7tt =
-.
aWl L\Wmax = 696.4/942
= 73.94% The overall total to sratic efficiency is, eqn. (2.22), Jts = AW/(dWrnax + where it is assumed that c
2 12 6s
I
2"
= c 62/2 =:21
Jts = 696.4/(942
2 c 6s ) x 200
2
= 20 kJfkg
+ 20)
= 72.470 (ii) and (iii)
The specific work is divided equally between the flve stages so that the
specific work done per stage is A h
a
= Ll W/5 = 139.3 kJ/kg
A straight line is drawn on the Moiller chan between srate points 01 and 06 and the
intermediate stage srate points determined.
).
'\
The following table shows the
stagnation pressures, stagnation temperatures or dryness fractions, the actua.l and isentropic stagnation enthalpies for each stage.
The individual isentropic
stagnation enthalpy drops for each stage are shown ( Ll h ) and the srage toral to as total efficiencies \letermined from "'Itt = Llh/Ll has· It should be noted that the values of h
shown in the table are obrained from the os intersection of the isentrope at the beginning of a stage with the isobar .at the end of
I
!I
j
I I
that stage.
The reason the stage efficiencies increase as the flow proceeds through
the turbine is because of the reduced'"slope of the constant pressure lines at the lower
pressures (and lower temperatures) causing
~h
as
to reduce.
.~---~--~~-======~ Thermodynamics
•
State paint
Po (kPa)
01 02 03 04 05 06
1500 620 240 85 26 7
To
(oc.)
'Y
350 274 200 125 0.988 0.95
ho (kJ/kg)
3148 3008.7 2869.4 2730.1 2590.8 2451.5
19
Ahas
has (kJjkg)
225.0 212.7 199.4 193.1 185.8
2923 2796 2670
2537 2405
h
q=0'95
•
0655 5
(iv) The reheat factor in a turbine is defined for a finite number of stages ~y
:. R = (225 + 212.7 + 199.4 + 193.1 + 185.8)/942 H = 1.078
=
-
Stage effie. %
61.9 65.5 69.9 72.1 75.0
Chapter 3 Two-dimensional Cascades
.,
ie i
I
i
Screen
Settling
"Slngle-sl0ge fon
~'DJffuser
:'
-
ConlraCllon
. Cascade
section:
length
SuctIon slot \,
,
I >~. Screens TI \ 1 ,
Tcst ~ec
lion'
Layout of a conventional low-speed cascade wind tunnel. (adapted from Carter et al. (1950), see Ch. 3)
lip.
01
t rOverse"
.
Chapter 3 Two-dimensional Cascades
•
Screen
---
Drive
molor~
i
i
~
,.Slnqle-stage fon ./
,·D,ffuser
Setthng length
Contraction seo.;llon,'
Cascade Cascade
r
,
!
_ ;,~. Screer.s
I
"l"I I I
,
Tcst ~ec
lion'
Layout of a conventional low-speed cascade wind tunnel. (adapted from Carter et al. (1950), see Ch. 3)
lip" 01 . traverse ..
----=--==========:====c
-=:-' 20
Chapter 3 Two-dimensional Cascades 3.1. Experimental compressor cascade results suggest that the stalling lift coefficient of a cascade blade may be expressed as
where c 1 and c are the entry and exit velocities. Find the stalling inlet angle for a 2 compressor cascade of space-ehord ratio unity if the outlet air angle is 30 deg.
•
Sol ution.
The lift coefficient C L of a cascade blade is defined, eqn. (3 .16a), as the
force L per unit blade length acting in the direction normal to the average velocity
c
m
divided by the product of the average dynamic pressure and blade chord f. . i.e. C
1
L
= L/(zpc
2 m
0 1
= c Icos 11 and tan 11 = -2 (tan III + tan (1 ), For a compressor blade m x m m 2 cascade C L can be expressed, eqn. (3.18), _ in terms of the inlet flow angle a.1 , the
where c
outlet flow angle
11
2
the spacelchord ratio sll
and the drag coefficient CO' as (i)
cr,
It is assumed in this problem that Co = 0 and that ex is constant.
•
From the velocity triangles, (ii)
c,
Thus, with eqns. (i) and (ii) and the given expression for stalling,
= 2(sll..) cos
11
m
(tan III - tan 1l )(cos 1l1cos Ill)3 = 2.2 2 (iii)
With
sit
2 = 30 deg, only III remains unknown in eqn. (iii). Although it is possible to produce a polynomial equation in tan III from eqn. (iii) it is far less
=
= 1.0 and
11
L.
~------~=~-~======
Two-dimensional cascades
trouble to solve the equation numerically. values of
21
The procedure used is to select several
and calculate the numerical values of Q. etc. until a value of-a. m l 1 satisfies the equation, 0.
3 cos "m (tan "1 - tan "2)/cOS "I where tan "2
..~
3 cos "2
=0.5774,
tan tan
•
"I
"m 0 "m
cos a.
m LHS' eqn. (iv)
3 "2
= 1.6936
(iv)
= 0.6495. 47.5
49
50
1.0000
1. 0913
1.1504
1.1918
0.7887
0.8343
0.8639
0.8846
38.26
39.84
40.82
41.49
0.7852
0.7679
0.7567
0.7490
0.9387
1.280
1.5356
1.7320
45
"1°
= 1.l/cos
.[.J...S
By graphical interpolation the inlet flow stalling angle is "I = 49.81 deg. 3.2. Show, for a turbine cascade, using the angle notation of Fig. ..3-.!'r, thatthe lift
~
coefficient is
CL = 2(s/t )(tan "I + tan "2) cos "m + CD tan "m
;.
-:.,
I 1 2 where tan "m = 2" (tan "2 - tan "I) and CD = Drag/(2" p em :l. ).
A cascade of turbine nozzle vanes has a blade inlet angle angle ,,; of 65.5 deg, a chord length l T~e
~
= 0 deg, a blade outlet
of 45 mm and an axial chord b of 32 mm.
flow entering the blades is to have zero incidence and an estimate of the
deviation angle based upon similar cascades is that
outlet Mach number.
If the blade load ratio
"'f" T
S will
be about 1.5 deg: at low
defined by eqn . ..(XS1') is tobe 0.8S,
estimate a suitable space/chord ratio for the cascade.
I.
I
(3. S-s)
..
Determine the drag and lift coefficients for the cascade given that the profile loss coefficient
:1[. r
22
Solution.
S.L. DIXON
The figure shows part of a turbine blade cascade, the velocity triangle asswnlng c
.
~ I
is
x constant and the
S
force diagram.
, I
From the velocity triangle the mean flow direction a. is m defined by tan I
(1m = 2" (tan (12 tan (11) so that c
y
•
m
= c leas a. • x m
Referring to unit depth
(span) of blade, the lift force L acting on the blade is perpendicular to c
and the drag m force D acting on the blade is parallel to c
X L
o R
m
•
The resultant force R has components X and Y
in the axial and -'tangential' directions respectively. L = Y cos a
m
+ X sin
Q
Resolving forces,
m
D = X coS a. - Y sin a m m
(i) (ii)
With constant c the axial force acting on one blade is
x
x
= (p -p)s
I
2
(iii)
The tangential force acting on one blade is, from the momentum equation,
where p is a mean density through the cascade. With the 'incompressible' flow I 2 approximation (for simplicity), Po = p + '2 pc . then the total pressure loss acrosS the cascade is,
(v)
I
I ..,
.Ji
-'--
<
I .~
~~--=~===~
T
Two-dimensio~l
cascades
23
Substituting eqn. (v) mto eqn. (iii), 1
2 2 2 pCx (sec 02 - sec °1)s
+2'
X = sAPo
1
2
2
2
= SAPo
+ '2 pCx (tan 02 -tan 01)s
= sAp
+ p s cx tan °m (tan 02 + tan 01)
2
a
(vi)
After using eqns. (iv) and (vi) in eqns. (i) and (ii) it follows that
°
o
= sAp 0 cos . m
L = P s c 2 sec °
•
x
With the definitions C
L
=
1
(tan 02 + tan 01) + s 4 P sin ° m om 2
1
2d
L/('2 pC m l) and Co =O/('2pcm'~)'
C L = 2(s/e) cos om (tan 02 + tan 01) + CD tan om CD = 2(s/t) cos
tT
°m [Ap c 2) ] o/(p m
(vii) (viii)
. -------
_
The blade load ratio, eqn. ~), is A 2 't'T = 2(s/b) cos 02 (tan 02 + tan 01) At cascade exit the flow angle 02 is less than the blade outlet angle 0; by the amount of the deviation.
°2
= 0' - $
= 65.5 - 1. 5 = 64 deg
2
At cascade inlet the blade angle a; is zero, the .t1ow incidence is zero so that the flow angle
0.
1
= O.
Thus, with
0.
1
= 0, the space/chord ratio is
s/l = (b/l)(s/b) = (bIt) "f'T/ sin 202
'I
= (32/45)O.85/sin(2 x 64°) = 0.767 From the velocity triangles, C :;:;: c cos 0. = c cos a. • then c = c cos Q /cos 2 2 2 2 1 x m m m a. and tan a := -2 tan , Thus, <1 = 45.71°. Using this expression in eqn. m m m 2 (viii) the drag coefficient becomes '_
°
I
)
(3.sS)
24
S.L. DDWN
tip
=
(sl!') cos "m ( !..pc02 2 2 3
)(COS.,,)2 . cos
,,~
2
= (sl.e) A cos "m/cos "2 = 0.767 x 0.035 x cos 3 45.71 0 Icos 2 640 = 0.0476
From eqn. (vii) the lift coefficient can now be calculated 0
C L = 2x 0.767 xcos 45.71 x tan 64 0 +0.0476x tan 45.71 0 = 2.196+0.049 = ~
N.B. In a turbine cascade with "m> 0, the drag slightly Increases the lift which is
•
the converse of v.hat occurs in a compressor cascade .
3.3.
A compressor cascade is to be designed for the following conditions:
.
Nominal fluid outlet angle
"2
=
30 deg
Cascade camber angle
9
=
30 deg
Pitch/chord ratio
sl.e
=
1.0
Using Howell's curves and his formula for nominal deViation, determine the nominal incidence, the actual deviation for an incidence of +2.7 deg and the approximate lift coefficient at this incidence.
Solution.
-.
The nominal deviation angle. eqn. (3.39) is
II.! S" = m9(sll) where, from eqn. (3 AOa), the coefficient m is
•
m = 0.23(2altl+ "2"/500
Assuming a circular arc camber line, all
= 0.5. and
m = 0.23 + 30/500 = 0.29 :.~"
= 0.29x30x I =
8.7deg.
Referring to the notation giveI\.in the sketch, the blade angles are,
~.
a
Two-dimensional cascades
0.2,
=
o.{
=
25
0./ -s" = 30 - 8.7 = 21.3 deg 0.; + g = 21.3 + 30 = 51.3 deg
•
s
The nominal flow inlet angle can be obtained from the tangent difference approxi-
mation, eqn. (3.38), or less precisely from Fig.~, tan "'I" = tan
0./
3. 18
;... + 1.55/ [ 1+ 1.5(s/£)]
r
o = tan30 +1.55/2.5 = 1.197
. 0.I "
..
50.13degand E" = o.*-el· I 2
I
*I
1, = 20.13 deg
l; I, r ~
The nominal incidence is i" =
1 I , I 0.I " - 0.' I
= 50.13 - 51.3 = -1.17 deg _
For i =2.7deg, (i-i")/e" = (2.7+).(1)/20.13 =~. of relative deflection e/e" against relttive incidence of
cleo ;,
1.15.
Hence, the actual deflection
actual inlet flow angle is "'I = o.{ + i = 51.3 + 2.7 = 54 deg.
0.
angle is 2 = "I - e = 54 - 23.15 = 30.85 deg. for an incidence of 2.7 deg. is = "2 - "2' = 30.85 - 21.3 = 9.55 deg The approximate lift coefficient, eqn. (3.17), Is
I
From Howell's curve
(i~ i")/e",
e = 1.15 x
I
Fig.
~
the value
20.13 = 23.15 deg.
The
The actual Outlet flow
ThUS, the actual deviation angle
1
1 26
where It Is assumed that CD Is negligible. tanl1 m l1
m
The mean flow angle Is
I = 2"(tan11 +tan11 ) = 1 2
1 0 0 2"(tan54 +tan30.85)
= 0.9868
= 44.62 deg 0
C L = 2 x cos 44.62 (tan 54
0
-
tan 30.85
0
)
= 1.109
3.4.
A compressor cascade is built with blades of circular
ar~
camber line, a
space/chord ratio of 1.1 and blade angles of 48 and 21 deg at inlet and outlet.
•
Test
data taken from the cascade shows that at zero Incidence (i = 0) the deviation b =8.2 deg and the total pressure loss coefficient -w = A p /( 2" 1 pel) 2 =0.015. 0
positive Incidence over a limited range (0 ~ I ~ 6 ) the variation of both
S and
At W
for this particular cascade can be represented with sufficient accuracy by linear approximations, viZ.
dS di
d;;'
= 0.06,
dl
= 0.001
where i is in degrees. For a flow incidence of 5.0 deg determine
(I) the flow angles at inlet and outlet; (ii) the diffuser efficiency of the cascade;
(iii) the static pressure rise of air with a velocity 50 m/s normal to the
plane of the cascade.
•
3 Assume the denSity of air Is 1.2 kg/m • Solution. (i) At zero Incidence, i pressure loss coefficient
=0,
the deviation
w= wo = 0.015.
s = so + (d5/di)i
Ei = 5 =8.2 deg and the total o
At i = 5 deg,
= 8.2 + 0.06 x 5 = 8.5 deg
w = wo + (dw/di)i = 0.015 + 0.001 x 5 = 0.02. " The flow angles at 5 deg incidence are
-
1 j
S.L. DIXON
...
=--------~--=-~====--=-'_.==j
Two-dimensional cascades
"1 "2
,
~
"1
= 48 +5 = 53 deg 21 + 8.5 = 29.5 deg
+i
,
= "2
27
+.
~
(ti) The compressor cascade decelerates the flow oetween inlet and outlet and the
efficiency of the process, assuming incompressible flow, can be expressed by the diffuser efficiency, eqn.~.
70 As
•
1
~ (P2 -P )/[ 2"p(c
1
.
P2 - PI
2
~
""
2 -c 2 )
1 2 2 '-c ), = -.op o +2"p(c 1 2
then
- .1p / [1 - p(c'2 - c 2J ) o 2 1 2
=1-
1 2 2 2 .t.p i[ 2" P c (1 - cos ,,/cos "2) 1
J \i
2
2
1 - W/(1 - cos ,,/cos "2) 1 2 Llp/(2"pc 1 ) and c cos "I = c cos Q 1 2 2
= Cx are
used.
0],
Q
2
;I 2
0
2
0
= 1 - 0.02/(1 - cos 53 /cos 29.5 )
,
.•
2
P2 - PI =
"lo p(c 1
=
'70 p Cx
2
I
2 2 (tan "1 - tan "2)/2 2
2. 0
2 0 - tan 29.5 )/2
= 2.079 kPa
3.5. (a) A cascade of compressor blades is to be designed to give an outl~t.air angle "2 of 30 deg for an inlet air angle "1 of 50 deg measured from the normal to the
plane of the cascade.
The blades are to have a parabolic arc camber line with
all ~ 0.4 (1. e. the fractional diswce along the chord to the point of maximum
camber).
Determine the space/chord ratio and blade outlet angle if the cascade is
to operate at zero incidence and nominal conditions.
=
I
,II
2 2 2 2 -c 2 )/2 ~70 pCx (sec 0'1 - sec "2)/2
= 0.962 x 1.2 x 50 (tan ,3
.1
;11
I .
(iii) The static pressure rise is
.'::
I
I .
w,
and.
= 0.962
I I
':! I; ,
Substituting values for
"lo
1
)J
= P02 -Pol + 2" p(c 1
70 = 1
where
2 2 l -c 2
You may assume the linear
:
.----. S.L. DIXON
28
approximation for nominal deflection of HoweU's cascade correlation:
.. " = (16 - 0.2 0.2")(3 - sit) deg as well as the formula for nominal deviation:
s"
= [ 0.23 (
Y2a)2
o.z"]
+ 500
91f deg
(b) The spacelchord ratio is now changed to 0.8, but the blade angles remain as they
are in part (a) above.
Z.O deg.
Determine the lift coefficient when the incidence of the floW is
Assume that there is a linear relationship between
Ele~
and (i-i")/t"
over a limited region, viz. at (i-i')le' = 0.2, f./c.* = 1,15 and at i = i", ele" = 1.
In this region take CD = 0.02. •
Solution.
(a) As the cascade is designed to operate at the 'nominal' condition, then
the air angles given are also the nominal flow angles, i.e. a
Z
= n2*
= 30 deg.
a"
0.
1
=o.r" =50
deg and
ThUS,. the nominal deflection is =
at - 0z"
= (16 - O.Z 0Z")(3 - 511.) = ZO deg
:. 20 = (16 - O.Z x 30)(3 - 511.) = 10(3 - s/,l)
:. sll. = 1,0 The nominal deviation is
!J'* ~
[0. 23(Zall.,z + 0Z" 1500 J9(S/l/Iz
= [0.23xO.64+30/500]9 = 0.20729 As the incidence is zero the blade inlet angle (J1~
= at = SO
deg.
The nominal
deflection is used again to solve for the blade camber, i.e .
•
.: 9
= 20/(1
- 0.2072)
= 25.2 deg
Hence, the blade outlet angle is obtained from
a.;, = o{ - 9 = 50 - Z5. 2 = 24.8 deg
,
(b) The change to a smaller space/chord ratio will affect the nominal deviation and
=
.
- .:.. _--.-- "._-
-_.
-..
;
.;";:'%'~i . -':.:,;'~
Two-dimensional cascades
nominal flow outlet angles. ~"
29
The new nominal deviation Is
= (0.23 x 0.64 + 0. " /500) 25.2(0.8)
1/2
2
= 3.318 + 0.04508 0. "
I I
2
and the new nominal outlet angle is obtained from
= 0.2 + 0" = 24.8 +3.318 +0.045080.2" = 28.12/0.9549 = 29.45 deg
0. " 2 0. " 2
The new nominal deflection is
•
e"
= (16 - 0.2 x 29.45)(3 - 0.8) = 22.24 deg
Thus t the corresponding nominal inlet angle 1s
and the nominal incidence is obtained,
= o.r" - o.{ = 51.69
I"
- 50
= 1.69 deg
The linear. relationship betw'een,deflection and ;ncidence is in the form,
EI<." - 1 = k(i - i")/e:" which satisfies the initial condition, Le.
e
= £.'" when i = i*.
(i -i")/e:" = 0.2, the value of k is found to be 0.75.
With E:;t. = 1.15 at
Thus, at I = 2 deg, the actual
fluid deflection is e: = E" +0.75(I-i*) = 22.24+0.75(2 -1.69) = 22.47deg
The actual outlet angle is
0.2
=
0. - E 1
= 0.1'
= 50 + 2 -
+ I-€.
22.47
= 29.53 deg
The lift coefficient Is determined using eqn. (3.18),
I
\
tan 0.
m
..... -.
-
1
0
0
= I(tan 52 + tan 29.53 )
·~-----~~-~==========: S.L. DixoN
30
= ~(1.280+0.5665) = 0.9232 °m = 42.71 deg and cos °m = 0.7348 C
L
= 2 x 0.8 x 0.7348(1.280 - 0.5665) - 0.02 x 0.9232 = 0.820
2 3.6. (a) Show that the pressure rise coefficient Cp = t. p/(~ pc 1 ) of a compressor casca"de is related to the diffuser efficiency IDand the total pressure loss coefficient
.-
:I
by the follOWing expressions: C
•
2 2 2 sec (1) = 1 - (sec 02 + p = 7D(1 - sec .02/
where
aI' a
2
=
4P/(~P(c/ -C 2 »)
=
C.p/(I pCx )
=
floW' angles at cascade inlet and outlet.
r )/sec 2 01
2
1
2
(b) Determine a suitable maximum inlet flow angle of a compressor cascade having
a space/chord ratio 0.8 and O = 30 deg when the diffusion factor D is limited to 2 F 0.6. The definition of diffusion factor which should be used is the early Lieb1ein formula, COS U
DF
=
(
1 1 - cos 02 )
5
+
cos a
1 (l) --2
° - tan
(tan 1
<1 )
2 .
e (c) The stagnation pressure loss derived from flow measurements on
••
the above
cascade is 149 Pa when the inlet velocity c is 100 m/s at an air density p of 1 3 1. 2 kg/m . Determine the values of (i) pressure rise; (ii) diffuser efficiency;
(iii) drag and lift coefficients. Solution. (a) The loss in toral'pressure across a compressor cascade due to
irreversible processes is, for an incompressible flow,
-
N . B. This version of the diffusion factor was not quoted in the parent book but was obtained from the chapter "Experimental Flow in Two-dimensional Cascades" by S.Lieblein in "Aerodynamic Design of Axial-flow Compressors", NASA SP-36 , (1965).
.. Two-dlmens!l?nal cascades
=
1 2[1-(c Ic) 2J
-lIp+-pc 2 1
2
1
where tip:;: P2 -PI' is the static pressure rise across the cascade.
=
:;: c cos 0. 2 2
31
With c cos 1
Q.
1
ex = constant 1 2 2 2 = -tlp/(IPc ) + (1 - cos ,,/cos "2) 1
1 2 2 2 = 1 - tip/(IP c 1 ) - cos "l / cos "2 = 1-
! cos
2
2
2
"1 - cos ,,/cos "2
., 2 2 = 1 - ( , + sec "2)/sec "1
•
(i)
From the definition of diffuser efficiency
(ii) (b) For a compressor cascade of specified geometry the diffusion factor D
F increases rapidly with increasing inlet flow angle as the positive stall "point" is
approached.
With a
2
= 30 deg,
sit = 0.8 and D F = 0.6 substituted in the Lieblein
formula:-
0.6 = 1 - cos a/O. 866 + 0.4(sin a x
Putting
= cos
2
0.
1
, (I-x)
liz
=
1
- 0.5774 cos a ) 1
sin 01 and rearranging.
12
x(1/0.866+0.4xO.5774) = 0.4[1+(I.i/ ] 2 ..• (3.4 64x - 1)2 _- 1 _x
•
2 13x - 6. 928x + 1 = 1
:. x = cos a
1
= 6.928/13 = 0.5329
Thus, the maximum inlet flow angle (i.e. for positive stall) to give a diffusion
factor D
F
=0.6 is a
1
S7.8deg 0
= c cos a = 100 x cos 57.8 = 53.29 1 x 1 coefficient is immediately found, i. e.
(c) With
I
I
I
~
C
mis, the total pressure loss
32
S.L. DIXON
1 2 1 2 r = I!.PO/(IPC x ) = 149/(rx 1.2x 53.29) = 0.OS75
};:
':
Using eqn. (i),
+ sec 2 30 a)Isec 2 57. Sa
C = 1 - (0. OS75 P = 1 - (0.OS75
+ 1.3333) x 0.5329 2
4 1.
= 0.5965
The pressure rise is,
-;,;
:.-;
AP =
1
2
1
= rCppcl = r
P2 -PI
x 0.5965 x 1.2 x 10
4
= 3.579 kPa
!.
From eqn. (ii) the diffuser efficiency is, 2
= C/(1 - cos l1/cos
I
2 (1
) 2
= 0.5965/(1 - 0.5329210.S662) = 0.5965/0.6213
l
=
0.96
The drag coefficient is defined. eqns. (3. J 6b) and (3.17), as CD
=
1 2 D/(rpem
-e)
=
1 2 sLlp a cos"m I(Ipe m :()
= J (sl/. ) cos 3 "m where
tan"m
... "m .: CD
=
1 r(tan "1
1 a 0 + tan "2) = r(tan 57.S + tan 30 ) = 1.0S27
47.27deg
= 0.OS75 x O.S x cos 3 47.27 a =~
•
The lift coefficient is defined for a compressor cascade, eqn. (3.18), as
C L
44
I
!
= 2 (sll. )eos "m (tan "1 - tan 02) - CD tan om 7 SO 0 _ = 2xO.Sxeos47.27 o(lanS. -tan30)-0.0219x1.0S27 = 1.0972 - 0.0237 =~
Sho"
3.7 (a) A set of circular arc fan blades, camber 9
=8 deg., are to betested in a
cascade wind tunnel at a space/chord ratio, sll =1.5, with a stagger angle 5 = 68 deg. Using McKenzie's method of correlation and assuming optimum conditions at an axial velocity ratio of unity, obtain values for the air inlet and outlet angles.
(b) Assuming the values of the derived air angles are correct and that the cascade has an
effective liftJdrag ratio of 18, determine (i) the coefficient of lift ofthe blades (ii) the efficiency of the cascade (treating it as a diffuser).
•
Solution. (a) The mean flow angle a '" is determined with eqn (3.42) when 5 = 68 deg. tan am
:. a
~
m
tan 5 + 0.213 = 2.688
= 69.59 deg.
For a circular arc camber line With 9
•
a
•
1
~
9
= 8 deg., the blade angles are given by :-
.
5+'2 - 72 deg.
a; = 5-!!.2 - 64 deg. At the optimum efficiency condition, transposing eqn (3.44), we obtain the ideal static pressure rise coefficient, viz.
1s C.P' =0567 --• 9 l -04003 . 1
From eqn (3.43),
-Cz _ cosS - ( 1 - C )l2 e,. cos a 2 pi
-
0.7744
The deviation angle 0 is found using McKenzie's result, eqn (3.41),
j
,
o- (1.1 + 0.318 Xs Il)i - 4.098 deg.
oj
'~ -'
•
:. a 2
-
a~ + 0 - 64 + 4.098 - 68.1 deg.
cos ~ - Cz cos ~ - 0.373 = 0.2889 cj
a 1 = 73.21 deg.
(b) From eqn (3.18) the lift coefficient can be determined, Le.
• :. am - 70.98 deg.
2
Go - 2
x 1.5 x cos 70.98° (tan 73.21° - tan 68.1" ) - £..8 tan 70.98° . 1
ex. = 0.8085 .'. C
_
L
- 0.1611 CL
0.8085 _ Q.Q.2§ 1.161
The aerodynamic efficiency of the cascade can be calculated from eqn (3.26), 2CD
17 o - 1 - C sin 2a n L = 1-
2 - 08824 . 18 sin 70.98° .
'-- .
•
•
3
I
I
Chapter 4
I
I
I
Axial Flow Turbines
. \
. ,,
Cooled turbine rotor blade showing the cooling passages. . (courtesy of Rolls-Royce pic)
Chapter 4 Axial Flow Turbines 4.1.
"
Show, for an axial flow turbine stage, that the relative stagnation enthalpy
across the rotor row does not change.
Draw an enthalpy-entropy diagram for the
stage lalJelling all salient points.
Stage reaction for a turbine is defii1ed as the ratio of the static rotor to that in the stage.
•
enth~lpy drop
in the
Derive expressions for the reaction in terms of the flow
angles and draw velocity triangles for reactions of zero, 0.5 and 1.0 . Solution. c 1
It is assumed that the axial velocity through the stage is constant, Le.
=.cx 2 =c x 3 =cx
,that the absolute velocity at inlet to the stage c equals the . 1 absolute velocity'at outlet c and that the flow is adiabatic. Referring to the 3 velocity diagram, Fig. 4.1, and enthalpy-entropy diagram, Fig. 4.2, the specific
x
work done by the stage, which causes the specific stagnation enthalpy of the fluid to
decrease, eqn. (4.2), is (i)
As the nozzle flow is adiabatic and the nozzle does no work, then (Ii)
From the velocity triangles, using the Cosine Rule,
•
2 2 2 2 2 "2 = U + c - 2Uc cos(rrj2 -a ) = Lr + c - 2Uc 2 y2 2 2 2 2 2 2 w = +c - 2Uc cos(Trj2+a ) = +c + 2Uc 3 y3 3 3 3 3
tl
tl
(iii) (iv)
Subtracting eqn. (iii) from eqn. (iv) and re-arranging, U(c
y2
+ c
y3
I 2 2 2 2 ) = 2(C -c +w -w ) 2 3 3 2
" (V) 1 2
Combining eqns. (i), (ii) and ~ noting that h
f...
0
=h+ 2c ,
1 2 2 2 2 1 2 2 h 02 - h = (h2 -h ) + 2(c -c ) = 2(c -c +w -w2 ) 03 2 2 3 3 3 3
(vy
=
34
S. L. DIXON
.: h + 2
1
2
2"w2
1
2
= h + 2W3 3
(vi)
1 2 The relative stagnation enthalpy is deflned as h 1= h +-2 w and eqn. (vi) shows are that it is equal at inlet and outlet of the turbine rotor from which it is deduced that it
must be constant through the rotor.
The stage reaction, eqn. (4.17), is R = (h2 -h )/(hl -h3 ) 3
-.::~
.:.:-;
= (h2 -h 3)/(hOl -\3)
noting that c
I
I. I
= c for a "normal" stage. l 3 (i) and (h -h ) from eqn. (vi), 2 3 2 2 w - w 2 3 R= 2U(C +C ) y2 y3
After substituting for (hoI -\3) from eqn.
~
The nwnerator is facrorised as follows,
i As w + w = c + c ' eqn. (vii) reduces tCL y3 y2 y3 y2
\
R =
\ \
(W
y3
-W
y2
)/(2l!) =' (tan 13 -tan 13 ) c/(2l!) 3 2
(viii)
Alternatively, with w = U + c from the velocity diagram, y3 y3
'.
R = (U+c and, with
W y2
=c
y3
-W
y2
)/(2U =
I
+(tanI1 - tan I3 )C/(2U) 3 2
(bt)
y2 - U, R
= (2U + Cy3 - Cy2)/(2l!) = 1 + (tan 113 -tan 112)C/(2U)
(x)
The velocity and simplified MoUier diagrams for the three reactions R = 0, 0.5 and LO for arbitrary but constant values of Dow coefficient c /U and stage-leading
x
i
i
I \ i
factor (C
Y2
+c
y3
)/U ate shown below.
(i) R = 0, eqn. (viii) gives 13 =.. 13 , hence w = w and h = h • 3 2 2 3 2 3
._~.
---~----~==--=======
Axial flow turbines
35
/.. h
w, 5
U (ii) R = 0.5, eqn. (ix) gives u = ' \ ' hence c 3 = 3
f.,.
10
2
, c
2
= 10 and hI -h = h -h " 2 3 3 2
h
•
.,:(
1:i ,i 1
7
q~:l
U
(iii) R = 1.0, eqn. (x)
U
3
= u , hence c = c and h = hI" 2 2 2 3
(J
iI
r
I
h
I
,
, w,
_ w,
0,
c,
3
U
__5_ _
In a Parsons' reaction turbine the rotor blades are similar to the stator b ades but With the angles measured in the opposite direction.
The efflux angle relative to
each row of blades is 70 deg from the axial direction. the exit velocity of steam from the stator blades is 160 m/s. the blade speed is 152.5 m/s and the axial velocity is
constant.
Determine the specific work done by the steam per stage.
A turbine of 801" internal efficiency··consisrs of ten such stages as described above and receives stearn from the stop valve at 1.5 MPa and 300°C.
Determine. with the
aid of a Mollier chart, the condition of the steam at outlet from the last stage.
=
-r
~.2.
_ . ....-
r;rv. \' '3S'
_
J
(I'~
(a) An axial!flow turbine operating with an overall stagnation pressure of 8 to 1
has a polytropic efficiency of 0.85. Determine the total-ta-total efficiency ofthe turbine. (b) If the exhaust Mach number of the turbine is 0.3, determine the total-ta-static
efficiency. (c) If, in addition, the exhaust velocity of the turbine is 160 mis, determine the inlet total temperature.
Assume for the gas that Cp =1.175 kJ/(kg K) and R = 0.287 kJ/(kg K). Solution. (a) For an incremental pressure change dp the corresponding isentropic and
i. ,,
:
actual enthalpy changes for a perfect gas are, dp dp h (y -I) I dh" - pdp and dh -11 pdh,. -11 P pl(RT) - 11 p P Y
Rearranging the above we get, dh _ h
(y - 1) 11 y
dp , P
which, after integrating over a finite pressure ratio P3 I PI' becomes
With stagnation conditions under consideration, the total-ta-total efficiency is
Evaluating the con~~t needed, y -1/(1- RIC,) - 11(1- 0.287 11.175) - 1.323. y -1 Thus, -y- - 0.2441. Hence, with
11, - 0.85 and P, -
8, we get
1
ae. , t -/
11.#
_ 1- 0.6495 _ 0.8804 1- 06019
(b) We now employ the expression given in the solution Ex. 42 for evaluating the
total-to-static efficiency, viz.,
and a similar expression for the total-to-total efficiency, viz., 1Jtt -
11 r ( )<>-1)1 hol/ 1- POl
l
J
POI
Equating for M. in the last two equations and rearranging we obtain:
• , With P03 P3
=
lr1 + '!(y -1 )M: 1(,-I) ,then upon substituting the values M J J 2
1.323, we determine the value for P03 P3
•
:.
=
11..
0.8804{1- 0.6019) 1-0.601911.01454
=
=
= 0.3, y =
1.0609. Hence,
0.3505 0.4067
=0.8617
(c) The expressions for the Mach number M J = C J IOJ and the static enthalpy It ""3
=
C T.
p 3
=
' (yr R7;" _ 1) = (y a;_ 1) are comb'Ined to gIve
2
Substituting the values, e, - 160 mis, M, - 0.3 and y
= 1.323, then we obtain h J =
880.6 kJlkg. Also,
hm - ~ + 1c; = 880.6 + (160 2 (2) x 10 -, = 893.4 kJlkg.
By rearranging the equation for 11. used in part (b) above we get:
{ \,-11 r Pm J' hm -1-11 "11-1 hoI
L \ POI
I'" 1 - 0.8804 (1 - 0.6019) = 0.6495
J
:. h 01 = 893.41 0.6495 = 1375.5 kJlkg
•
::FrJl = hOlI C
p
= 1375.5/1.175 - 1170.6 K
4.3. The mean blade radii of the rotor of a miJted flow turbine are 0.3 m at inlet and A
0.1 m at outlet. The rotor rotates at 20,000 revlmin and the turbine is required to produce 430 kW. The flow velocity at nozzle exit is 700 mls and the flow direction is at 70 0 to the meridional plane.
Determine the absolute and relative flow angles and the absolute exit velocity if the gas flow is 1kgls and the velocity of the through flow is constant through the rotor.
Solution. The sketch Fig.(I) shows the components of the flow at inlet to and outlet from the rotor in the meridional plane. The through flow velocity is constant so that cm2
=
cm"
where
3
Cm2 -
"INsERT
~
COS
':P'
0~. (I) At N =20,000 revlmin, Q
4~c4ry
-,
~
az - 700 x co's 70° - 239·4 mis
=2094.4 rad/s, r z =0.3 so that U z = Q r z =628.3 mis;
and with r 3 =0.1 then U 3 'f.209.4 mis. ~.
F,L'V' (1-)
;;.
Considering the inlet velocity triangle, Fig. (2),
tanf3z _ ~ sinaz _ 700 x sin700-628.3 _ 0.12316 . cm2 239.4
•
:. f3 2
Wz -
-
7.02 deg,
cml lcos f3z - 239.4/cos 7.02 = 241.2 mis
IN~E.RTFt.d·b) --~
U sing Euler's turbine equation and referring to the velocity diagram at outlet from the
rotor, Fig. (3), we have:
.
W
DoW,--. -UZC62 +U,Ce, m
.
•• C6 , =
DoW, - U21 C6 2 U,
-
430,000- 628.3 x 700 x sin 70 209.4
8 =79. mis
i
Again, referring to the outlet velocity triangle, 4 1 I
f3 ,-tan
The absolute exit velocity
W, -
.JI ~ + Cg ,)j-tan .1(209.49+ 79.8) -50.37 dego \
em'
23 .4
e, - cm , I cosa, - 252.4 mis and the relative exit velocity
cm , I cosA
=375.3 mls
I, ,
~;j
.~:~
,
:. \
,1
i i i
ie I
i
I
~~~~I~I~~~liil~lrl~I~lr~i~I~~il~
~:i~~i~~-WW~~~~;l!~~,I~li~i~~#W~*~:~~~~~~j~i~W~I~!W~iI~~W~~ ~.~. l.•.s•.••.•. :~~ffin.:~~y.:I~~~~~~.:i~~~~~~i~l~i~~~ffi~~~~~~Bf~ . •:·:..· • •
i.•
.•
m.· ••
..<,..'. .. ;;: ;::;::;; ::"
~r~w~~ifuti~1~p¥Pi:8j/
::: ::~::::;: ;:: ::;;;;:: ,,;:
.
Using eqn (2.12e), viz.,
where
c JC 2+ c/ m -
z
is the meridional velocity.
Adding & subtracting .!.U2 to the rhs of the expression for I , we obtain 2
5
From the velocity triangle, c. - U - w., and so the above expression becomes
1- h +
'2I (w2 - uTO!) as w2 - w. 2 + c 2 m
As we assume that the rothalpy I is constant across the rotor from station 2 to station 3, (see the cautionary remarks under the heading Conservation of Rothalpy, Ch. 7), then
:. -"'VI,
•
Z) 1 (Z '2) 1 (2 2) -'21 (2 U2 -U, -'2 W z -w, +'2 c; -c,
The contributions of each term on the rhs of the above expression are:
1(628.32- 209.42) = 175.5 kJlkg 2"1 (Z U2 - U,2) = 2=
&( ~ - W;) &( 375.f - 241.i) = 41.35 kJ/kg =
•
(2
2) = -1 (2 700 - 252.42) = 213.2 kJlkg
-1 C2 - C,
2
2
--
6
The sum of the values of these terms equals 430 kJlkg in agreement with the given value of t:.. W,. It is seen that the contribution to the work output of the rotor due to the mixed flow effect is very significant (over 40 per cent of the total).
,
-
"1
• I
-,I
I I ~
7
S.L. DIXON
36
Solution. The velocity dJagram for the stage can be readily constructed from the data supplied and the specific work obtained from a scale draWing or. more accurately. by calculation.
It will be noticed that as the effllDC angle relative to
each blade row is equal. i.e. "2 the reaction
i~
50 per cent.
=133 =70 deg.
the velocity trJangles are similar and
The specific work per stage is
SolVing for the unknown swirl velocities using the usual sign convention
c
=
c 2 sin "2
150.4 m/s
y3
= =
-~
c
•
= 160
y2
:. 6.W
= = =
sin 70
0
w sin 13 - U 3 3
= c 2 sin "2 -
150.4 - 152.5
= -2.1 m/s
U
152.5(150.4 - 2.1)
22.62 kJlkg
This stage is rather lightly loaded and the stage loading factor is
'I'
= t,w/if =
(c
y2
+ c y3 )/U
=
148.3/152.5
= 0.9725
A turbine with ten similar stages to the one above-will prodllce a specific work of
226.2 kJlkg and this is equal to the change in stagnation enthalpy of the steam hoA -hoB between turbine inlet (A) and turbine exhaust (B). i.e. hoA - hOB = 226.2 kJ/kg
."
It is implied that the "internal" efficiency is the total to total efficiency, defined as
•
'tt
:. h
oA
- h
oSs
= (hoA
-\al/(hoA -h oss )
= 226.2/0.8 = 282.8 kJlkg
From steam tables or Mollier chart at PoA = 1.5 MFa (15 bar) and T h
oA
= 3039 kJlkg
:. hOB = 3039 - 226.2 = 2812.8 kJlkg
-
oA
0
= 300 C
r Axial flow turbines
37
A :. h
oBs
= 2756.2 kJ/kg
The less laborIous method of determining
h
the exhaust steam conditIon Is by plotting these specific enthalp·ies on a large scale
MoHler chart for steam!.
From such a
plot the exhaust steam lndition is
PoB = 420 kPa (4.2 bar);
T .~
I
oB
= InoC
l.e. the steam is stll! superheated at
!
s
exhaust.
it. ~~.
Values of pressure (kPa) measured at various stations of a zero""reaction gas
turbine stage, all at the mean blade height, are shown in the table given belowe
Stagnation pressure
Static pressure
Nozzle entry
414
Nozzle exit
207
Nozzle exit
400
Rotor exit
200
The mean blade speed is 291 mis, inlet stagnation temperature 1100 K, and the flow angle at nozzle exit is 70° measured from the axial direction.
Assuming the
magnitude and direction of the velocities at entry and exit of the stage are the same,
and the axial velocity remains constant. deternIi.ne the total to total efficiency of the stage.
"'~
\;,
j i
I
I 1
AssUJIE a perfect ll'lS with Cp = 1.148 kJ/(kg ·C) and y = 1.333.
Solution. The total to total efficiency of a turbine stage Is defined. in the usual notation, as
)'tt =
hoI - ho3 h -h 01
0355
With c = c this can be rewritten as 1 3
I
fj
r
1!
I
~------------------I
.-J
S.L. DIXON
38
h
In order to
with the perfect gas assumption.
determine the efficiency of the stage the velocity diagram must first be solved. The stage reaction is defined as R = (112 - h )/(II1 - h 3) 3 so that zero reaction means h equals h . The 3 2 1 w 2 is relative stagnation enthalpy h 1= h +-2 ore 2 I 2 constant in the rotor,then h2 +~W2 = h3+:2w3 and, therefore w2 = w3' The velocity at nozzle
•
35 3ss s
exit c must be determined to complete the velocity diagram.
2
= T
T 2
At nozzle exit,
(p Ip )(y-I)!"y 02 2 02
= 1100(207/400)0.2498 = 933.1 K 2
c2 :. c
2
= 2Cp(T02-T2) = 2xI148(1100-933.1) = 383,200 = 619.1 mls
o
c = c sin ~2 2 y2
= c cos ~2 = 619.1 cos 70 = 21l.8 2 0 = 61~.1 sin 70 = 58l.8 mls
= c y2 -
= 290.8 mls
Referring to the velocity diagram,
• wy2 ..
C
x
U = 581.8 - 291
An important point to note is that W = W (W = w ). y3 2 y2 3 c
•
25
y3
= wy3 -
U
= wy2 -
mls
Thus,
U = 290.8 - 291
= -0.2m/s
Le. the flow leaving the stage is very nearly axial in direction with a small angle of swirl ~3
=tan -I (-0.2/21l.8) =-0.05 deg.
Thus, with T
3
=T
2
= 933.1 K and T
o1
=T
Effectively c = c = C = 211.8 m/s. 3 i x 02
= 1100 K
2 T 1 -T3 = Tol-cl/(2Cp)-T3 2 = 1100,- 211.8 1(2 x 1148) - 933.1 0 = 147.4 C
-~-----------~-----
Axial flow turbines k
39
U=291 mls Using the isentropic relation benveen temperature and pressure
T :. T
3
• T
=
3ss
T
OI
(P/poi Y- 1)/Y =
= 933.1
3ss
:'''lrt
- 917.1.
1100(200/414)0.2498
=
917.1 K
= 16.00 C
= 1/ 1 +16/147.4
= 90.2%
~
In a certain axial flow turbine stage the axial velocity c is constant. The ~ x . absolute velocities entering and leaVing the stage are in the axial direction. If the flow coefficient c
x
IV
is 0.6 and the gas leaves the stator blades at 68.2 deg from the
axial direction, calculate:
Aw/if;
(i) the stage loading factor,
(il) the flow angles relative to the rotor blades;
(ill) the degree of reaction; (iv)" the total to total and total to static efficiencies.
The Soderberg loss correlation, eqn. (4.12) should be used. Solution. (i) The stage loading factor is
"1'= I1w/if = =
(ex/V) tan
U
cyz/U. as c
y3
z
= 0.6x tan 68.2 0 = 1.50 .1
(il) From the velocity diagram
tan ~3 :.
I
1
J
~3
=
U/c
x
=
= 59.04 deg
1/0.6
=
1.667
=0
S.L. DIXON
40 tan ~2 = tan a
2
- U/c = 2.5 -1.667 = 0.8a35 x
:. ~2 = 39.81 deg (ili) The stage reaction, eqn. (4.22a), is R = (tan
~3
-tan
~2)
cx /(2U)
= 0.3(1.667 -0.8335)
=
Q.:.32.
.~ " // '"X.t'
"",0
/
C : C
3
•
,
u (iv)
The total to total efiiciency of a normal stage (c = c ) is, 1 a
"'ltt
= (hal -h a3 )/(hal -ha3ss ) = (hI -h3 )/(hl- h3+ h3 -hass) = 1/ [I + (h3
-h3ss)~(h1
-hal]
Referring to Fig. 4.2, the enthalpy difference has -hass' equal to (h2 - h 2s )(r3 IT 2)' is usually simplified to h - h with only a small loss in accuracy in determining 2 2s efficiency.
•
The enthalpy differences h -h and h - h representing the effects of irreversible 2 2s a 3s flow in the nozzle and the rotor respectiveLy, can be expressed in terms of loss coefficients
?Nand JR
h2 -h 2s
J
-
I
2 7 2"c 2 ~N
1
2
h3 -has = 2"W3 SR Thus, the total to total efficiency becomes, eqn. (4.9a),
/
'.
~
..
Axial flow turbines
41
/. 7tt : [
1
(i)
+
The total to static efficiency, defined as '7ts
:
(hal -\3)/(hol -h3ss )
= 1/[1 + (h3 -h 3ss +t c /)/(hI-h3)] i~
1
2
used when the exhaust klnetic energy iC3 is wasted.
This efficiency is most
useful in the form,
(ti)
The enthalpy loss coefficients can be expressed, eqn. (4.12), in terms of the fluid deflection
~
(deg) of each blade row, that is,
I
+1.5(~/100lJ
= 0.04[1
where, for the nozzle, e:::: G = a + Q = 68.2 deg (i.e. Q : : 0) and, for the rotor 2 r 1 N row, ~: ~ :13 +13 =39.81 +59.04 =98.85 deg. Thus, .IN = 0.06791 and 2 3 R : 0.09863 after using the above equation.
J
From eqn. (i), with w : C sec 13 , c 2 =C sec "2 and hI -h x 3 x 3 3 2 2-1 = [1 + sec 133 + sec "2] tt (2 tan "2)/ II
IR
'7
=VCx tan "2
IN
2 2 = [1 + 0.09863/0.5144 + 0.06791/0.3714 ,-1 2 x 2.5/0.6
J
0.865]-1
= [ 1 + 8.334 '",
"/tt
:
From eqn. (ti), with c '7ts
90.6"'. /' I
=C x
= [1 + (0.865 +,1)/8. 334J -1 : 81,7%
-4-.5. h g:'s tUi'ei:nc seage develops 3.:36 MW lot a mass flow rate of 2; .2 kg/a.
,.~~.",:-,~:. ',~. ,
..
•
.
__ .. _.-
~:•.• ,\:.,.,.._~,"~', '~.<_~_'_ •
j
",.w
.',
~'.~
•• '
,
-
•
.
'~"..-'
.,'_l"....:,
1=a"
-"'~,--_:... - ~.~~- '~' ..
(q])
4.7. An axial flow gas turbine stage develops 3.36 MW at a mass flow rate of 27.2 ~
kg/so At the stage entry the stagnation pressUre and temperature are 772 kPa and 727 o C respectively. The static pressure at exit from the nozzle is 482 kPa and the
corresponding absolute flow direction is 72 0 to the axial direction. Assuming the axial velocity is constantacross the stage and the gas enters and leaves the stage without any absolute swirl velocity, determine,
(l) the nozzle exit velocity (2) the blade speed (3) the total-to-static efficiency (4) the stagereaetion.
•
The Soderberg correlation for estimating blade row losses should be used. For the gas assume that c;.
=1.148 kJ/(kg K) and R =0.287 kJ/(kg K).
Solution. (1) The velocity diagram for the stage is shown in Fig. (1) and the corresponding h-s diagram in Fig. (2). For the nozzle flow
•
_ 1.148
l:1~zN
.• {j
.!1
X
=
106 r1- (482) 41 _ 1.2753 x lOs Jlkg 772 j
l
1 + ~I C where eN is given by
1
:. {) h, _ "
:. C 2 =
5
fJI, _ 1.2753 X 10 1+ 11 0.0711 1 + 1/0.0711
_
8.465 Ilk g
488 mIs
(2) The specific work done is
~.w =~ _ m
As
C83 -
0 then ~W = UC82
:. U _
:1
I
-
~w 'i sin a 2
-.
).
6
3.36 X 10 27.2
UC2
_
123.5 kJlkg
=
sina 2
3
123.5 X 10 488 x sin 72 0
Also the axial velocity is c. - 'i cos a 2
-
(3) The fluid deflection in the rotor row is
_
266.1 mls
488 x cos 72 0 -_ 150.8 mls
f: R -
{32 + {33 where, from Fig. (1)
1
i
"
j
I
~
tan 132 _ C82 - U _ 488 x sin 72° - 266.1 -1.3128 Cx
150.8
i
I
I
2
U 266.1 tan fJ3 - - - c. 150.8 - 1.7648 0
Hence, fJ2 - 52.t, fJ3 - 60.46 and Sa _113.16
(;R - 0.0(1 + 1.5 x 1.1316
I._I.
where C'
_''3 R
''3, _
1 2 -1-\1. 2 3
2
]_
0
0.1168
f>h .tRot
1 (2 U2) - C + 2 x
1( 2+266.14 2) x 0.1168"" 5,466Jlkg f> hxRot - '2150.8
•
Thus,
1 '17" -1+(1 x 150.8 +8465+5466)/123.53 x 10 3 2
•
1 1+0.2048
(4) The stage reaction is most conveniently expressed by eqn (4.22c)
so that with a J =0,
3
R - 1-
{u tan ~ - 1
150.8 x tan72° 2 x 266.14
:.R =0.128
•
4
-------------------S',L. DIXON
44
absolute velocities at inlet and outlet are
~
equal.
A steam turbine stage of high hub/tip ratio is to receive steam at a stagnation 0 pressure and temperature of 1.5 MPa and 325 C respectively. It is designed for a blade speed of 200 m/s and the following blade geometry was selected:
Nozzles 0
inlet angle, deg
70.0
Outlet angle, deg
Rotor 48 56.25
~,
Space/chord ratio,
•
sll
0.42
Blade length/axial chord ratio, H(b
2.0
2.1
Max. thickness(blade chord
0.2
0.2
The deviation angle of the flow from the rotor row is known to be 3 deg on the evidence of cascade tests at the design condition.
In the absence of cascade data for
the nozzle row, the designer estimated the deviation" angle· from the "approximation
0.19
e sll
where G is the blade camber in degrees.
Assuming the incidence onto the
nozzles is zero, the incidence onto the rotor 1.04 deg and the axial velocity across the stage is constant, determine:
(i) the axial velocity; (il) the stage reaction and loading factor; (ili) the approximate total to total Stage efficiency on the basis of Soderberg's
loss correlation, assuming Reynolds number effects can be ignored; (iv) by means of a hrge steam chart (Mollier diagram) the stagnation temperature and pressure at stage exit.
Solution. The total to total efficiency of a turbine stage, applicable to the case c not
eq~al
1
to c ' is
3
7tt = (hol-h03)/(hol-ho3ss) = =
t!.W/(!lW+"losses")
J
1/ [1 + (h03 -h03ss )/ L\W
Employing the approximations h -h = h -h (i.e. this assumes c = C ) • 3 3ss 3 03 03ss 3ss and h3s - h = h - h ' then 3ss 2s 2
i
".:'
! I
Axial flow turbiDes
45
/..
I
Defining the enthalpy loss coefficients, eqns. (4.8a) and (4.8b),
for the nozzle (Le. stator) row, and
for the rotor row, the required expression for the efficiency is (i)
:.
wbere
I,
from the blade angles with suitable corrections for the incidence i and deviation
I
Ll.w = hoI -h03
= U(c
+ c y3 )'
y2
(i) The flow directions at inlet and exit of the nozzle and rotor blades are obtained
each blade row.
At nozzle exit the deviation is
bN
= 0.19 Q s/l
0.19x70xO.42 = 5.6deg Thus. the nozzle exit flow angle is 0.
2
=
0.;
-
bN
= 70 - 5.6 = 64.4 deg
For the rotor, the relative flow exit angle is
~
3
=~' 3
6R
= 56.25 - 3 = 53.25 deg
and the relative flow inlet angle is
~2 = ~;
+i
= 48
+ 1.04
From the velocity diagram, U = cx(tan C
x
= c
i
= U/(tan
0.
0.
2
2
= 49.04 deg
-tan
-tan
~2)'
~2)
= 200/(tan 64.4 -tan 49.04)
= 2l3.9m/s
J I
"'
i i I
,
I
therefore,
(il) The stage reaction is defined, eqn. (4. 22a), as
6
of
.>:-_~~.
::.:.: ..
l.--~---------------.~ 46
S.L. DIXON
R = (c /2U){ran 13 - tan 132) 3 x = (213.9/400)(ran 53.25 - tan 49.04) = (213. 9/400)(1.3392 - 1.1520) = 0.10
The srage loading factor is
"/1= e:,W/rJ = (Wy2 +WyS )/U = (c/U)(tan 13 2 + tan 133) = (1.152 + 1.3392)213.9/200 = 2; 664
•• - U = 200 mls (iii) The total to total efficiency of the stage for the case when c
is not equal to c1 3 The enthalpy loss coefficients for the nozzle and rotor
is given by eqn. (i) above.
are evaluated using the analytical simpliiication of Soderberg's loss correlation.
------
•
eqn. (4.12),
r
= 0.04 [1
+ 1.5{E/I00)2] and making suitable corrections for
blade aspect ratio in each case .
For the nozzle row, at the nominal aspect ratio H/b
= 3.0,
as the fiow defiection In the nozzle row, € N ="2 (i.e. "1
=0).
At aspect ratios other than the nominal. the enthalpy loss coefficient
,an be fOUnd, eqn. (4.13a), 1
r
+ Nl
= (1
+ 3N•)(0.993 + 0.021 b/H)
= (l.06488){0.993 + 0.021/2) = 1.06861
5Nl tor nozzles
47
Axial flow turbines
A.
:. rN1 1
= 0.06861
= 3~O,
For the rotor row, at H/b
I
I,
e R =~2 + ~3 =102.3
where the flow deflection In the rotor.
deg.
The correction
for the aspect ratio In the case of a rotor row. eqn. (4.13b). is = (1
+ 1 * )(0.975 + 0.075 b/H) R
= 1.1028(0.975 + 0.075/2.1) = 1.1146
ie !I
= 0.1146
The quantities in eqn. (i) are evaluated separately for convenience, i.e.
!
C
x
sec
<1
2
= 213.9 sec 64.4
= 0.06861 x 495 =
2
=
= 495.0 m/s
3 2 2 = 16.8xl0 m/s = 16.8kJ/kg
cxsec~3 = 213.9x sec 53.25* = 357.5m/s
= 0.1146 x 357.5
t!.W
a
,/,J
2
_
3
= 14.65 x 10 m
= 2.664 x 200
2
= [1
Is
2
= 14.65 kJ/kg
= 106.6 kJfkg
Using eqn. (i).
'7tt
2
+ (14.65 + 16.8)1(2 x 106.6)r
1
= 1.475-
1
= 87.15'70 (iv) At Pol
=1.5 MPa (15 bar).
Tal
=3250C.
hal = 3093.5 kJ/kg is obtained (tables).
the stagnation enthalpy at entry
Now
h01 - h0355 = (hal - h03 )/"?tt = I:J,W/'?tt = 106.6/0 • 8715 = 122.3 kJ/kg
i
I
j :j
i
:. ha 3ss = 3093.5 - 122.3 = 2971.2 kJ/kg From the Mollier chart. P03 = 0.9 MPa (9.0 bar). h
03
= 3093.5-106.6 = 2986.9kJ/kg
0 . T = 269 C .. 03
4.9 (a) A single stage axial-flow turbine· is to be designed for zero reaction without any absolute swirl at rotor exit. At nozzle inlet the stagnation pressure and temperature of the gas are 424 kPa and 1100 K The static pressure at the mean radius between the nozzle row and rotor entry is 217 kPa and the nozzle exit flow angle is 70°.
Sketch an appropriate Mollier diagram (or a T - s diagram) for this stage allowing for the effects of losses and sketch the corresponding velocity diagram. Hence. using Soderberg's correlation to calculate blade row losses. determine for the mean radius. (1) the nozzle exit velocity (2) the blade speed
•
(3) the total-ta-static efficiency.
(b) Verify for this turbine stage that the total-to-total efficiency is given by:
1
1
11..
= l A• -
(£11 '2)
where ¢' = cjU. Hence. determine the value of the total-ta-total efficiency. Assume for the gas that Cp = 1.15 kJ/(kg K) and y = 1.333.
•
Solution. (a) Fig (1) shows a sketch of the MoIlier diagram for this stage and Fig. (2) the corresponding velocity diagram. Now. considering the nozzle expansion.
}'-I
T" =1711
(h.)r = POI
033
1100 x ( -217)i33 =93156K 424 .
,', C 2
= 601.9 mls
(2) Considering the velocity diagram,
Cx
=c, cos
a'2
=205.86 mls and
c&
=Cx tan
~
=2U
.;
----.
i
1 ... U = -c 2 x tan
t>' •.,
= 282.8 mls
Fk.~\'&) (3) The total-te-static efficiency is detelTI1ined from the approximation, eqn (4. lOa),
From the velocity diagram,
It =tan
-1
(~) =53.95" and If =Aso that ~ =107.9
.,i
! I
i I
',1
=
2'1()2 2U =2U 2 = 159,952 m 2Is 2
0,
hence
,
=[
" 11t.
0.10985 1+
2
2
X
349.8 +0.0694 X 601.9 + 205.86 2 x 159952
•
2 ]- 1_
0.798 -
(b) The specific work done by the stage can be expressed of both 1], or 1],. i.e.
boW
• and as bo W = Uc OJ. = 2U'
1
1
11.,
= lit, -
and as ,A=
.
•
r
1 lit,
.',
(1:1' 2)
c 205.86 = = 0.7279 then U 282.8 •
-!.
1
=0.798 -
1~,
c
(205.86)' 2 x 282.8
0.8923
4.10 (a) Prove that the centrifugal stress at the root of an untapered blade attached to the drum of an axial-flow turbomachine is given by:
4
where
=density of blade material N =rotational speed of drum An =area of the flow annulus. (~
(b) The preliminary design of an axial-flow gas turbine stage with stagnation conditions at stage entry of POI = 400 kPa, TOl = 850 K, is to be based upon the following data applicable to the mean radius:
•
Row angle at nozzle exit, t; = 63.8 deg Reaction, R = 0.5 Row coefficient, cjUm = 0.6 Static pressure at stage exi t, p 3 = 200 kPa Estimated total-to-static efficiency, 11"
=0.85.
Assuming that the axial velocity is unchanged across the stage, determine
( 1)
the specific work done by the gas
(2)
the blade speed
(3)
the static temperature at stage exit
(c) The blade material has a density of 7,850 kg/m' and the maximum allowable stress in the rotor blade is 120 MFa. Taking into account only the centrifugal stress, assuming untapered blades and constant axial velocity at all radii, determine for a mean flow rate of 15 kg/s, (1)
the rotor speed (rev/min)
(2)
the mean diameter
(3)
the hub/tip radius ratio.
For the gas assume that Cp
=1050 J/(kg K) and R =2'07 J/(kg K).
Solution. (a) For an element of blade length d ("':; = -
.0 2 /1. r dr. hence
• 2
:n: N)2 A :n:p N A (; = p ( ~ = _m • as.Q = ", m 30 2:n: 1800'
:n N 130
(b) Fig. (1) shows the Mollier diagram for the stage and Fig. (2) the corresponding velocity diagram. For the overall pressure drop we find,
•
FJ.t'~) ~~)----->-:> 1',,,
( ):' = 11)°'=
=Ta,.l!l.. POI
tJ.h, = Cp(ToI
8 2
= 703.3 K where
Z:..!.1 = -R 3'
= 0.2733
Cp
- 1;,,)= 1050(850-703.3)= 154,000 = 154 kJlkg
As t/u = tJ. WI tJ.h, = 0.85 (given) • then
,'. tJ. W = 130.9 kJ/kg
v
(2) Considering the velocity diagram we can see that
bho
. u-- ( 2 ~ tan
..
~ - 1)
0.5
-
-
(
130.9 x 10' 2 x 0.6 x 2.0323 - 1)
0.5
= 301.6 mls
(3) The static temperature at stage exit can be determined from the enthalpy drop from point 01 to point 3:
r. I \ I
!
We need to determine the kinetic energy of the flow at exit,
C 93
:. :.~
I
Cx
= Cx tan
«, - U =
~81 x tan 63.8
tC3' = t (c: T c; ).
°-301.6 = 66.2 mls
= ~u = 0.6 x 301.6 = 180.96 mls
Upon substituting the derived values into the foregoing equation for T J' then
TJ
=707.6 K
I I
:j .,1
.,
'I I
(c) (1) From the equation proved in part (a) above, we obtain:
N' = 1800 ,~ .11pm A "
.
where A = A 3 = -
•
m
liE.,
.
mKI:
=:.:.::.::.J. =
p,cz]
151 x ?Z7 x7C17.9 , 5 = 0.08419 m 2 x 10 x 181
6 ,t,
(2)
•
N=
=:
1800 x 120 x 10 = 10,200 rev/min .11 x 7850 x 0.08419
D = 60 U " 60 x 301.6 = 0.5648 m m 11 N 11 x 10,200
. (3) The blade length IS H
=:
r, -
A,
r", " - -
211rm
5 =2110.08419 = 0.0474 m x 0.2824
!.i..=rm -HI2 "Dm-H =0.51735=0.845
r,
rm +Hl2
Dm+H
0.61225
--
4.11. The design of a single stage axial-flow turbine is to be based on constant axial
•
velocity with axial discharge from the rotor blades directly to the atmosphere. The following design values have been specified: Mass flow rate
16.0 kg/s
Initial stagnation temperature, TOI
llOOK
Initial stagnation pressure, Pol
230kN/m z
f~
7850kglm'
Density of blading"material,
Maximum allowable centrifugal stress at blade root, Nozzle profile loss coefficient, Yp Taper factor for blade stressing, K
=(POI - P02 )1(P02 - P2 )
1.7 x 108 N/m 1 0.06 0.75
In addition the following may be assumed:
Atmospheric pressure, P3
102 kPa
Ratio of specific heats, ]I
1.333
Specific heat at constant pressure, Cp
1150 J/(kg K)
In the design calculations values of the parameters at the mean radius are as follows:
Stage loading coefficient, If' = I:!. W/U 2 I
~i !
Row coefficient,
Isentropic velocity ratio, Uleo where
Co
=.J[2(h
O' -
1.2
0.35 0.61
h3SS )]
Detennine, (1) the velocity triangles at the mean radius (2) the required annul liS area (based on the density at the mean radius) (3) the maximum allowable rotational speed (4) the blade tip speed and the hub/tip radius ratio. Solution. (1) With the isentropic velocity Co defined above and the given data
Co = ~2(h01 -
h3 SJ ) =
1'-1 = 0.2498 and R = Cp ( '/.1-1) / J' = 1150 x 0.2498 = 287.3 J/kg I' .
-
,', Co
=
J2 x
1150 x 1100 x
(1 ~0.44348lU498) = 682 m1s
Hence, U = 0.61 x 682 = 416 mIs, II w = Uc IJJ. = 1.2 x U'
,', CO2
= 1.2U = 499.2 mIs, c. =
'fS U = 145.6 mis,
c, = 520 mls
From this derived data the velocity diagram for the stage may be constructed as shown in Fig.(l).
IN~r:.R\
Fla,.(')---7»
.
(2) The annulus area is determined from the equa!-i0n of continuity, m = rlA,.cx '
•
and, to find P2 we need to determine P 2 and T 2 •
(PO!) -1 From eqn (3.41), y" = PO! - P07. = P07. P07. - P2 1 _ P2 P07. ._2-1 P (
. 'P07. -
.', .& = 1 + P07.
2 2
C
- 2C T07. p
)*
520 2 )4.cxn 1150 x 1100 = 0.63604
•
..,..& = 1 + 0.06x(l- 0.63604) = 1.02184 P07.
P2 P07. 0.63604 ",P2=- - P O I = x230=143.16kPa P07. PO! 1.02184
'
hence
Y,,(I-.b..) P07.
, T, = T01. _..s.... = 1100 2Cp
R
520'/(2 x 1150) = 982.4 K
= p, = 143.16 x 1
m 16 , .'. f\, = A, = - - = = 0.2166m fic, 0.5073 x 145.6
(3) Eqn (4.30b) is used to determine the maximum speed of rotation
• I
!
~ = ~ \ 1_ (!h..)'l
2l
R,
r,J
which can be converted into a more useful form as follows.
t' = K!'i A" = :aN, KA" 11800
....;J;.
2:a
R,
as
n =:a N/30
1800 x t':, 1800 x 1.7 x 10 8 ,', N = = = 0.7638 x 108 :af1. KA" :a x 7850 x 0.75 x Q,2166
'
I
i . II
I
I,
... N = 8,740 revlmin (or 1J = 915.2 rad/s)
416 . U (4) The mean radius, r = ~ = - - = 0.4545 m m ~< 915.2
..A.. = 0.2166. = 0.07585 m · T he blade hetght, H = 2:nrm 2:11 x 0.4545 Hence, the tip radius, r, = r",
TH
/2 = 0.49243 m
.'. The blade tip speed, U, = .Q T, = 450.7 mls
From above,
•
•
=1- 2l~ , (.!&.)' r, PmKU
=1-
8
2x1.7x 10 =0.71566 7850 x 0.75 x 450.67'
Chapter 5 ,
i
-I
Axial Flow Compressors
i
Ii
Ie I
1
I
r- -J\...
(
•
!
_ -,~ .r ~ ~=====, I
\--
b
I
,
I
I 1 I ,i
Section of the compression system of the RB211 gas turbine engine. (courtesy of Rolls-Royce pIc).
48
Chapter 5 Axial Flow Compressors
L Note. In problems 5.1 to 5.4 assume that the gas constant R = 287 J/(kg °C) and that y = 1.4.
5.1. An axialllow compressor is required to deliver 50 kg/s of air at a stagnation I.. pressure of 500 kPa. At inlet to the first stage the stagnation pressure is 100 kPa and the stagnation temperatUre is 23°C.
are 0.436 m and 0.728 m.
•
The hub and tip diarneters at this location
At the mean radius, which is constant through all stages
of the compressor. the reaction is 0.50 and the absolute air angle at stator exit is
28.8 deg for all stages.
The speed of the rotor is 8000 rev/min.
Determine the
number of similar stages needed assuming that the pOlytropic efficiency is 0.89 and that the axial velocity at the mean radius is constant through the stages and equal to
1.05 times the average axial velocity. Solution. The number of stages is determined from the stagnation temperature rise • obtained from the specific work done equation and velocity diagram, a and the overall stagnation temperature ris~. through the compressor, TaB - T cA'
per stage
~T
' together with the oA The number of identical compressor stages, n, is
obtained from the overall stagnation pressure ratio. PoJiP polytropic efficiency,
'7p .
obtained to the nearest integer from n = (T
oB
•
- T
aA
)fAT
(i)
a
The specific work done by the rotor on the air, eqn. (S.l), is (j.W
=
h
02
-h
01
=
C 6T
P
a
= U(cy2 -cy1 )
(ii)
Referring to the mean radius velocity diagram and noticing the velocity triangles are
symmetrical for a reaction of 0.5 (i.e. c
y2
- c
y1
and, from eqn. (ii)
= U - 2 C tan a j
x
~2
= a ),
l
49
Axial flow compressors
f\ (iii)
The average axial velocity c is obtained from equation of continuity,
-
x
tit = pAc, x
the density being determined with the incompressible flow approximation
p = Poi = po/(RTal)'
Thus,
=
p
01
/(R T I) a
=
=
lOS /(287 x 296)
1.177 kg/m
3
ex = 4 riV[ Pol IT (d t / - dh/l] = 4 x 50/[". x 1.177(0.728
2
2 - 0.436 )}
= 159.1 m/s
The axial velocity at the mean radius is
•
c
x
= 1.05 x C x
= 167.1 m/s
The mean blade speed is U = 1TNd /60
m
=
= 'IT N(d
hl
+ d tl )/120
rrx8000(0.436+0.728)/120
=
243.8m/s
I
Mean radius velocity diagram
I
c,
U =243·8 m/s Polytropic efficiency for a small compressor stage is defined, eqn. (2.31), as '?l
/p
= dh.is /dh = vdp/C P dT = (y -1)TdP/(ypdT)
after using the perfect gas relations, pv = RT and C = yR/(y -I). P. .: T = constant x ph-I)/y 'lp
I
(iv)
As the stages are similar with ideiftical velocities, stagnation conditions can be used
I
:j
I
!
.. "
n
~
--r' S.L. DIXON
50
Thus, across the whole
In eqn. (iv).
compressor.
TopffoA
:.T oB -TeA
=
Ip )(y-1)/y"Jp oB oA 51/(3.5 x 0.89)
(p
h
= = 51/3.115 = 1.6764 = 0.6764 x 296 =
200. 2°C
From eqn. (iii).
•
t:.T
0'
=
243.8(243.8-2x167.1
OA
°
xtan28.8) 0
= 14.57 C
s
Using eqn. (i) n
=
200.2/14.57
=
13.74
The number of stages required is 14.
5.2.
Derive an expression for the degree of reaction of an axial compre.ssor stage
in terms of the flow angles relative to the-rotor and the flow coefficient.
Data obtained from eariy cascade tests suggested that the Umit of efficient working of an
axial~flow
compressor stage occurred when
(i) a relative Mach number of 0.7 onto the rotor is reached; (ii) the flow coefficient is 0.5; (iii) the relative 110w angle at rotor outlet is 30 deg measured from the axial •
direction;
(iv) the stage reaction is 50'7•• Find the limiting stagnation temperature rise which would be obtained ttl the first stage of an axial compressor working under the above conditions and compressing air Assume the axial velOcity is constant
at an inlet stagnation temperature of 289 K. across the stage.
Solution. The degree of reaction of an axial flow compressor stage is defined as the
to
-,
Axial flow compressors
SI
.
~
static enthaipy rise in the rotor divided by the static enthalpy rise in the stage, i.e, (i)
As the relative stagnation enthalpy is conStant in the rotor, then h Z • hI =
1
Z
Z
2" (wI' WZ )
Assuming a normal stage (Le. c == c ), then 1 3
Substituting into eqn, (i)
I.I
R = (W
2 2 ,w )/ [2U(c ,c )] y2 yl 1 2
= (W +WyZ)(W -W )/ [2U(C 'C )] yl yZ y2 yl yl where it is assumed that ex is constant across the stage.
(ii)
From the velocity
triangles for the compressor stage, c 2 == U - w 2 and c = U - w so that Y Y yl yl c - c =w • W ' Simplifying eqn. (li), yZ yl yl y2 (iii)
where the flow coefficient ~ = C /U,
x
The data given in the problem enables the velocity diagram shape to be drawn
c,
u W 2
'j
immediately.
!
Cy2
Tne lllagna"uJ,estSl{the.. velocity vectors must be calculated from the
information concerning maximum relative Mach number.
From the velocity
diagram the maximum relative velocity is wI and the corresponding relative Mach . '.
T !
52
S.L. DIXON
number (Iv) where the static temperature T
=T -C//(2C )' It Is most convenient to solve In OI p Writing WI = cx/cos PI and c = cx/cos "I
I
terms of the axial velocity cx'
i
= c /0.866. eqn. (IV) gives.
x
2 2 WI = Y RMrl T I 2 2 = yRM rl [TOl-c l /(2C p )] 2 2 Cx = YRMr/[T -C /(1.5 C )] ol x p
•
2 cos . ~I
(v)
Using the equation (iii), ~l can be determined as follows, tan ~1 = 2R/1" - tan ~2 = 2 - tan 30
...
~
~l
0
= 1.4227
= 54.9deg
Substituting values into eqn. (v),
2 C
x
= 1.4X287xO.49[289 -C = 1.882 x 10 4 - 0.0432
.
:. C
x
x
2 /(1.5xI005)]
x 0.5751
2
x2
C
= 134.3 m/s
The stagnation temperature rise in the stage
.a T o can now be immediately .
determined using the equation for the specific work, ~ W = C .1.T
•
p
o
= U(c
y2
-c
yl
) = U(W -W ) yl y2 2
x (tan~l-tan~)/1"
C
2
= Cx (tan ~I -tan ~2)/(I"Cp)
2
= 134.3 (tan 54.9
0
0
- tan 30 )/(0.5 x 1005)
0
= 30.35 C
5.3. Each stage of an axialfloloY compressor is of 0.5 reaction, has the same mean
/'
blade speed and the same flow outlet angle of 30 deg relative to the blades.
--
The
-~--------------,--------~
I I I
Axial flow compressors
53
~
mean now coefficient is constant for all stages at 0.5.
At entry to the first stage
the stagnation temperature is 278 K, the stagnation pressure 101.3 kPa, the static 2 pressure is 87.3 kPa and the flow area 0.372 m Using compressible flow analysis
determine the axial velocity and the mass flow rate. Determine also the shaft power needed to drive the compressor When there are 6
stages and the mechanical efficiency is 0.99 .
',i..3
Solution. It is tactitly assumed that the flow preceding the first Stage is deflected by
,:f~
inle: guide vanes to give an absolute flow angle 01 of 30 deg, the same as all the .- other
Stag~s.
The absolute inlet flow velocity c
1
is determined from t?e stagnation
enthalpy definition
•
hi
=
I
2
+ ICI
= 2C p (Tol -T I)
wherec =yR/h-I), and the isentropic temperature-pressure relationship, P (p /p )(y-I)/y Tlff ol = 1 01 (87.3/101.3)1/3.5
= 0.9584
= 2C Tol(I-TlfTol) p
= 2.325 x 10
c
i
2 x 1005 x 278(1 - 0.9584)
4
= 152.5 m/s
Thus, the axial velocity is
= 152.5 cos 30
0
= 132.1 m/s
Using the equation of continuity, the mass flow rate is
where
PI
3 = P/(RT I ) = 87.3 x 10/(287 x 0.9584 x 278) = 1.1417 kg/m
I
3
" m = 1.1417 x 0.372 x 132.1 = 56.1 kg/s
I
.
-.
S.L. DIXON
54
The specific work done on the gas per stage is !:J.W :
U(C
y2
-c
yl
)
:
= if(1 - 211 tan
U(U - 2 C tan (1 ) x 1
(1
1
)
as the velocity triangles are similar for a re.action of 0.5.
2 a :.6W : (2 x 132.1) (I - tan 30 )
= 29.5 kI/kg The shaft power needed to drive the compressor (including mechanical losses) is
Vic
•
=
n';"
6,Wh. . "/m
where n is the oum.ber of stages and ·'7m the mechanical efficiency.
W c
Thus,
3 = 6X56.lx29.5xI0 /0.99 = 10.03 MW
5.4. A sixteen-stage axial flow compressor is to have a pressure ratio of 6.3. Tests have shown that a
st~ge total
to total efficiency of 0.9 can be obtained for each
ot the first six stages and 0.89 for each of the remaining ten stages.
Assuming
constant work done in each stage and similar stages find the compressor overall total to cotal efficiency.
For a mass flow rate of 40 kg/s determine the power
required by the compressor.
Assume an bilet total temperature of 288 K.
Solution. The sketch shows the overall compressor process in the form of a Mollier diagram.
h
The overall stagnation pressure
ratio, PoB/P01' is the product of the pressure
•
ratio for the first six stages. PeA/Poi' and the pressure ratio for the remaining ten stages,
PoB/POA' Le, (i)
It is convenient to assume that the respective stage efficiencies of these two groups of
01
s
55
Axial flow compressors
.
~
stages is equal to the polytropic. (i.e. infinitesimally small stage) efficiency of each group.
1
Using eqn. (2.33), the stagnation pressure ratios in terms of the actuai
stagnation temperatures and polytropic efficiencies are
p
\ with,?
p
\
/p
oB
= (T
01
= 0.9 and p
"-=;
oA
y
/p
=1.4, = (T
oA
oA
oB
IT
) Y'Jp/(y-l)
=
(T
) Y"lp/(y-I)
=
(T
01
oA
IT )3.15
(ii)
01
and
IT
oA
oB
IT
)3.115
(iii)
oA
=
..:~~
with "l 0.59 and y = 1.4. P The work done in each stage is asswned to be constant so that the stignation
-.
temperarure rise for each stage tlT is constant (C is constant).
o
puc x =.Q.T
IT 0 I'
o
p
For convenience
The two temperature ratios can now be written as = 6 LlT IT I
o
0
+I
= I
+ 6x
ToBlT oA Substituting eqns. (ii) and (iii) into eqn. (i) and using the above temperature ratiOS, 6.3
= (I + 6x)3.15. [I + IOx/(I+ 6x)]3.115
= (1+6x)3.15. [(1+16X)/(I+6x)]3.115 = (I + 16x)3.115. (1+ 6x)0.035
The unknown x cannot be solved explicitly but can be determined qUite easily by a process of trial and error, Le.
i.
x
0.049
0.050
0.051
,
(I + 16x)3.1l5
6.0687
6.2398
6.4142
::
(I + 6x)0.035
1. 0091
I. 0092
1.0094
RHS
6.1237
6.2974
6.4744
.:.
By plotting these values the correct value of x :: 0; 05001 corresponding to a pressure
ratio of 6.3. Referring again to the MoHier diagram, the overall total to total efficiency of the compressor is
/
-,
56
S.L. DOWN
where T B- T 1 = 16 Ll T = 16x T 1 o 0 0 0 = /p )(y-1)!y _ 1] /(16x) ... ., tt oB 01 = [6.3 1/ 3 . 5 -1]/(16x 0.05002)
[(p
0.6919/0.8003 = 86.45% The power required by the compressor (e.xclu:ling mechanical losses) is given by, Wc = ;'Cp (ToB- T 01) = ;,C nt,T =;,C nxT 1 pop 0
•
40 x 1005 x 16 x 0.05002 x 288 = 9.266 MW
5.5. At a particular operating condition an axial flow compressor has a reaction of 0.6, a flow coefficient of 0.5 and a stage
loading~ defined as L1ho/t? of 0.35.
If
the flow exit angles for each blade row may be assumed to remain unchanged when the mass flow is throttled, determine the reaction of the stage and the stage loading when the air flow is reduced by
lo1~
at constant blade speed.
Sketch the velocity
triangles for the two conditions.
Comment upon the likely behaviour of the flow when further reductions in air mass flow are made. Solution. The velocity diagram shows the velocity vectors for the two
•
conditions (the broken lines denoting
the reduced mass flow condition).
It
is important to notice that it is the flow angle relatl'\'e to each blade row of the
stage which is assumed to remain the same.
Cascade test measurements
of compressor blade rows indicate that this asswnption is only approximately
u
----------------------~
Axial flow compressors
57
/I. correct for lightly loaded blades with a small space/chord ratio. The stage loading factor is V= Lih/U
2
=
6.W/J =
(C
y2 - Cy1 )/U =
y1 -
(W
= II (tan ~1 - tan ~2)
W
y2
)/U
(i)
The Stage reaction ratio is defined, eqn. (5.11), as (if)
Solving eqns. (i) and (if) for tan
~1
and tan
~2
tan ~l = (R + '1'/2) / II
ii
•
tan ~2 = (R - YJ/2) / II At the initial flow coefficient, II = 0.5, and with R = 0.6,,¥,= 0.35. the relative flow angles are
~1
l = tan- [(0.6+0.175)/0.5] = 57.17deg
~2
= tan
-I[ (0.6 - 0.175)/0.5]
= 40.36 deg
From the velocity triangles
-
..
tan "1 = 1/ II - tan ~I
(ifi)
tan "2 = 1/ II - tan ~2
(iv)
and, at the initial flow coefficient, the absolute flow angles are
"1 = tan -ll2 - 1.551 = 24.22 deg 1 "2 = tan- [2 - 0.851 = 49.0 deg The flow angles "1 and ~2 are asswned to be constant with variation of the flow coefficient.
Setting the stage loading factor and reaction in terms of thes~_fixed
angles, eqn. (i) can be rewritten
'f'= l-ll(tan"l.:tan~2) = I -1l(0.45+0.85) = I - 1.3
'.
II
-~--------:---~ S.L. DIXON
58 and eqn. (li) becomes R
= 0.5
-1l(tana -tan 13 )/2 2 1
after using eqns. (iii) and (iv).
= 0.5(1+0.411)
Thus, at the reduced flow rate, II = 0.9 x 0.5 = 0.45
and
R = 0.59,
¥'= 0.415
Further reduction of the flow rate would eventually lead to the condition of blade stall. probably in the rotor row as this is more heavily loaded than the stator
(i. e. R> 0.5), with a rapid increase in total pressure losses as stall is approached.
The assumption regarding constant flow angle relative to each
bl~de
row would also
fail to hold as stall implies flow separation off the blade suction surface.
5.6. The proposed design of a compressor rotor blade roW is for 59 blades with a
•
circular arC camber line.
At the mean radius of 0.254 m the blades are specified
with a camber of 30 deg, a stagger of 40 deg and a chord length of 30 mm. Determine, using Howell's correlation method, the nominal outlet angle, the nominal deviation and the nominal inlet angle.
The tangent difference'approximation,
proposed by Howell for nominal conditions (0 t:; a * ::; 40°). can be used: 2
tan at - tan a ' 2
= 1.55/(1
+ 1.5 s/l).
Determine the nominal lift coefficient given that the blade drag coefficient CD = 0.017. Using the data for relative deflection given in Fig. Jkt1 determine the flow outlet angle and lilt coefficient when the incidence i = 1.8
-.
/;g.
Assume that the drag
coefficient is unchanged from the previous value.
Solution. The nominal deviation S' can be determined directly from the blade geometry and space/chord ratio s/l using eqns. (3.39) and (3.40), viz.,
•
6' = m9(s/L//
2
where 9 = ai' - a 2' is the blade camber, m = 0.23 + a '/500, a{. ~2' ar~ the blade z inlet and outlet angles respectively and u • ::: a ' + &" is the nominal flow catlet angle. Z 2 For blades with a circular arc camber line the blade angles are
a; =
S + 9/2
= 40
+ 30/2 = 55 deg
3,
Axial flow compressors A. ~
/..
where
= s ~ g/~ = 40 ~
30/2
=
59
25 deg
Is the stagger angle.
The space/chord ratio at the mean radius is, sit. = 2-rrr/(Zt)
c,
= 2rrx 0.254/(59 X 0.03) =
)1.
,
a'
N.B. Stator blade
nototion
0.9017
Hence, uSing Howell '5 correlation, the nominal flow outlet angle is 11' = 11' + S" 2 2
'.
=
= 25 + (0.23 +11 '/500)30(0.9017//
25 + 6.552 + 0.05711 ' 2
..• 11 2" = 31. 55/0. 943
O·
= 11
=
2
2
" , 2 - 11 2
=
= 33.46deg 33.46 - 25
8.46 deg
Using Howell's. tangent difference approximation
tan Ill"
= tan 11 " + 1.55/(1 + 1.5 s/l)
2 = 0.6609 + 1.55/(1 + 1.5 x 0.9017) = 1. 320
Ill"
= 52.85 deg
Thus, the nominal incidence and nominal deflection angles are i"
= 11"-11' = 1 1
52.85-55
=
52.85 - 33.46
Q'
1
-11" 2
=
= -2.15deg = 19.39 deg
From eqn. (3.18), the lift coefficient at the nominal condition Is C
L
tan a. •
m
I 1
11 " m cos a .• m
=
2(s/'-) cos 11 ' (tan 11r" - tan 11 ') - CD tan 11 ' m 2 m
= (tan 11r" + tan 11 ' )/2
2
=
= (1.32+0.6609)/2 = 0,..9905
44.73 deg
= 0.7105
,.---------_._.
-
~~------~;l 2
S.L. DIXON
60
-1
'J
•
:. C
L
= 2xO.9017xO.7105xO.659 -0.017xO.9905 = 0.8444 - 0.0168 =
In Fig.
~
J
Is expret.ed as a function of the relative incidence (I-i")/£".
UsIng this curve the deflection
e
at
,/,'
1·25f------..,.~"
,·of-----..'
,
any arbitrary incidence i (within certain limits) can be found prOVided i* and t>ll are known.
At the
N.B. Not to scole
given incidence i :; 1. 8 deg.
(i-i")/o" = (1.8 +2.15)/19.39 = 0.2037
:.E/E" = 1.15 so that <'=01 -02 = 22.3 deg
•
o
04
Hence, the flow directions are
o{
= I. 8 + 55 = 56.8 deg
01
= i+
02
= 01 -E = 56.8 - 22.3 = 34.5 deg
and the deviation angle
s=
02 -02
= 34.5 - 25 = 9.5 deg (c.l. S" = 8.46 deg)
The lift coefficient at i = 1.8 deg can now be calculated with tan 01. = 1.5282,. tan 02 = 0.6873, tan .'. C
L
°m = 1.1077, °m
= 47.93 deg and cos
°m
= 0.6701.
= 2 x 0.901 x 0.6701 x 0.8409 - 0.017 x 1.1077 = 0.9974
5.7.
•
The preliminary design of an axial flow compressor is to be based upon a. II
.
simplified conSideration of the mean diameter conditions.
Suppose that the stage
characteristics of a repeating stage of such a design are as
follow~:
0
Stagnation temperature rise
25 C
Reaction ratio
0.6
Flow coefficient
0.5
Blade speed
275 m/s
The gas compressed is air with a specific heat at constant pressure of 1.005
kI/
i
-.-/""---------------------------~
I
Axial flow compressors
. 61
A
I i
(kg aC).
Assuming COnstaDt axial velocity acroSs the stage and equal absolute
velocities at inlet and outlet. determine the relative now angles for the rotor. Physical limitations for this compressor dictate that the space/chord ratio is unity
a t the mean diameter.
Using Howell' 5 correlation method, determine a suitable
camber at the mid-height of the rotor blades given that the incidence angle is zero. Use the tangent difference approximation
taD ~t
- taD ~2'
= 1.55/(1 + 1.5
for nominal conditions and the data of Fig.
sIt)
~
3. t
for finding the design deflection.
(Rint. Use several trial values of Q to complete the solution.) Solution. It is usually most convenient to solve the rotor relative flow angles
e
~1
and
13 in terms of the stage loading !actor .'P J the stage reaction Rand the flow coefficient 2 ~. The stage loading factor is defined, eqn. (5.l4a), as
-..p=
=
/::,.W/if' = C (r 3 -T 1)/U P a a 2 1005 x 25/275 = 0.3322
2 (i)
Referring to the velocity diagram, Fig. 5.2 (or see earlier solutions), and with /),W
= U(c y2 -c yl ) = U(wy1 -wy2) ' ''iJ= (wy1 -W )/U = ~(tan~l-tan~2) y2
(ia)
where
The reaction of a compressor stage, eqn. (5.11), is (ii)
Combining eqns. (ia) and (ii)
-e
tan~1
= (R+'If'/2)/~
tan~2
=
(R-'If'/2)/~
(0.6+0.3322/2)/0.5 = 1.532 = (0.6-0.3322/2)/0.5 = 0.8678
ThUS,
!
~I
I
= 56.87 deg,
~2
= 40.95 deg
At zero incidence the blade inlet angle~; = ~l = 56.87
I
/
'-
0
•
The blade camber
62
S.L. DIXON
' , =\3 -@ \i )= f.+S. where e = \3 - \3 = 15.92 deg is the fluid 9=\3 , -\3 2 1 1 2 1 21.. deflection and S is the aeviation angle. Thus. as Ii> O. then 9> €. = 15.92 deg. A suitable value for the camber angle 9 may be obtained by a process of trial and
error in which several values of 9 are selected and corresponding values of determined.
e. ar"e
Values of Eo can be found from Howell's curve of relative deflection
ele' against relative incidence
(i -
i')/e', Fig.
T
and this means that values of
e' and i' must be estimated for each value of 9 used. To simplify the calculation procedure values of the nominal deviation have been
estimated from Howell's expression S' = m 9 (slol) 112 with m = constant = 0.26
.-
•
rather than the more compllcated farm of m, eqn. (3.40a). used in the solution of the previous problem.
Noting that slL is unity•
E' = \3'-\3' = ~' - (\3 ' + S') = 9 - S' = 9-0.269 1 2 1 2
.
= 0.749 ~
,
= 13* 1
.. 2
-e-
,
= ~1 - 0.749 = 56.87 - 0.749
Using the tangent difference approximation, tan~t
=
tan~2'
+1.55/2.5 = tan(56.87 -0.749)+0.62.
A tabular form. of solution is desirable as follows:
0
9
20
22.5
25
5.2
5.85
6.5
36.87
34.37
31.37
42.07
40.22
38.37
0.9026
0.8457
0.7917
1.5226
1.4657
1.4117
56.7
55.7
54.69
··-13* 1 - 13' 1
-0.166
-1.174
-2.182"
£' = ~ , -
S'=0.269
,
•
~' = \3 - 9 2 1 ~'=\3' +S'
2 2 tan ~' 2 tan 13 • 1 ~
,
1
1
I
~ , 1 2 (i-i")/e"
14.63
15.48
16.32
0.0113
0.0758
0.1337
Elf,' (graph)
1.009
1. 060
1.105
63
Axial flow compressors {
8
By plotting values of 8 against
20
22.5
25
14.76
16.4
18.03
t (or by numerical interpolation), at E.
the camber is 8 = 21.76 deg
It is worth commenting that Howell recommended Q to lie in the range 1.2£'
<
8
<
1.8£'
which for C "" 15 deg gives
180 <: 8 < 27 0 Le. the value of 9 determined is satisfactory.
=15.92 deg
5.8. Air enters an axial flow compressor wi~ a stagnation pressure and temperature
...
of 100 kPa and 293 K, leaving at a stagnation pressure of 600 kPa. The hub and tip diameters at entry to the first stage are 0.3 m and 0.5 m. The flow Mach number after the inlet guide vanes is 0.7 at the mean diameter. At this diameter, which can be assumed constant for all the compressor stages, the reaction is.50 %, the axial velocity to mean blade speed ratio is 0.6 and the absolute flow angle is 30 deg. at the exit from all stators. The type of blading used for this compressor is designated "free-vortex", the axial velocity is constant for each stage.
Assuming isentropic flow through the inlet guide vanes and a small stage efficiency of
•
0.88, determine (1) the air velocity at exit from the IGVs at the mean radius (2) the air mass flow and rotational speed of the compressor (3) the specific work done in each stage (4) the overall
efficien~y
of the compressor
(5) the number of compressor stages required and the power needed to drive
the compressor.
•
(6) consider the implications of rounding the number of stages .down to an integer value if the pressure ratio must be maintained at 6 for the same values of blade speed and flow coefficient.
., Solution. (1) Fig. (1) shows the velocity diagram for a stage of the compressor. The velocity at exit from the IGV's, C l ' is determined from (and T 0/) to solve for T /.
hOI
= h,
+.! c~ 2
, using M /
•
.j
,
I = 2C pToJl-
.'. CI
T,) = .j2 x 1005 x 293 x (1
l
\
(2) To determine the air mass flow,
:.
-111.098) = 229.3
T OI
~ = ,.q~cz : the density and axial velocity must
be evaluated.
I
1; = Ten 11.098 = 266.85 K ,
,I
Po,
;
!
f? = .J!L = R1;
I
...1:...
~ ~-,
( 1;"
R1;
1 X 10 5 X
=
(
1
)3.5
--
1.098 287 X 266.85
= 0.9413 kg/m 3
I I
'-.
c z = c, cos
a;
= 198.5 mls
m
= c"" = n T then U = 198.5 = 330.8 mls If> ". m m 0.6
.Q = Um = 330.8 = 1654 rad/s Tm
0.2
.
.'. m = 23.47 kg/s
with A, = 0(0.5' - 0.3')/4 = 0.12566 m'
u
mls
.'. N = 30 n = 15,796 rev/min
n:
.', tlW = 33.614 kJ/kg • the specific work done.
•
(4) An expression for the overall efficiency of the compressor in tenns of the small stage efficiency and the overall pressure ratio was derived and given as eqn (2.36).
tAl
•
=~ 1 = PR
[PR
I'
-1
jr;,' -1
1
35 6 .
-1 = 0.6685 = 0.847
1
63.Oii -1
0.7892
--
(5) The number of stages N and the corresponding power P needed to drive the
compressor are:
N=
+(P uW l C:Z;l1
R
)I'-l I'
1 1005 x 293 [1.6685-1]= 5.856
-1 =
33,614
--
p = ~ I:J. W = 23.47 x 33.614 = 788.9 kW
,
j j
(6) Consider 6 stages to start with. As the pressure ratio, the blade speed and flow
coefficient are all specified, this increase will cause the effective stage loading to be reduced. i.e.
1 !
,
.,!
1 005 x 293
C T,
I:J. W = ~ (0.6685) = . N.
'.
I,
6
x 0.6685 = 32.8 kJ/kg
This implies that the flow geometry is changed such that
i
tt = tan -,
,.
f_1_(1 - I:J. ~)] = 30.26 deg. l2n,
Pm
If, on the other hand, the number of stages is made equal !O five, the stage loading will be increased so that the possibility of blade stall becomes greater. However, engine designers will usually consider this possibility together with the advantages of having a smaller, lighter engine with its reduced cost Eor 5 stages,
i
CT,
nw = ~ (0.6685) = 39.37 kJlkg N
It
= tan ~ f-1..(l- L\~)] =28.1 deg. l2¢..
Um
5.9. (a) The volume flow rate through an axial-flow fan fitted with inlet guide vanes is 2.5 m3 /s and the rotational speed of the rotor is 2604 rev/min. The rotor blade tip radius is 23 cm and the root radius is 10 cm. Given that the stage static pressure increase is 325 Pa and the blade element efficiency is 0.80, determine the angle of the
•
flow leaving the guide vanes at the tip, mean and root radii.
(b) A diffuser is fitted at exit to the fan having an area ratio of 2.5 and an effectiveness of 0.82. Determine the overall increase in static pressure and the air velocity at diffuser exit
:IT
Solution. (a) The angular velocity is .0 = - N = 272.7 radls, the flow area is 30 A
•
= TI (0.23 % - 0.1 % ) = 0.13477 m % and the axial velocity
ex = Q/ A = 2.5/0.13477
=18.55 mls . From eqn (5.37), blade efficiency is:
11, = P, - P', hence e = "
oUe yl r
yl
(P, - p,) 1J,pClr
=
325 = 1.2414 0.8x 1.2 x 272.7 r r
s
. = tan'I(C>,) = O.06692lr
q
Radial I>ositioll
II
'e I
Root
.'
Mean
,
33.79
Angle, «/ (deg)
,~
Tip'
,
~
22.08
~.
16.22
(b) From eqn (2.45 b)
I
!
1h = 2
3- p,
l'C'-C'
r
2
3
(e3 - p,)
=
2'
2)
1 2;'-;C2~1-1/AR
I I
,', (p, - P2) = ~ p C~ 1I (I-1I A~) = 0.6 X 18.552 x 0.82(1-112.5') = 142.2 Pa D
,'. Overall increase in static pressure, D.p = P3 - p, = 467.2 Pa
-
I
Velocity at diffuser exit, c3 = <; I AR = 18.55/2.5 = 7.42 mls
: "::
5.10. The rotational speed of a four-bladed axial-flow fan is 2900 rev/min. At the .j
I i
i
mean radius of 16.5 cm the rotor blades operate at CL = 0.8 with CD = 0.045. The inlet guide vanes produce a flow angle of 20· to the axial direction and the axial velocity through the stage is constant at 20 mls.
For the mean radius, determine
(1)
the rotor relative flow angles
(2) the stage efficiency
(3) the rotor static pressure increase (4) the size of the blade chord needed for this duty.
Solution.
•
(I) At the mean radius the blade speed U = n rm = 303.7 x 0.165 = SO.11 m1s :IT
where n =-N =303.7 rad/s. 30
\'NSER-I F)..~.(\)
~
From ilie velocity triangle, Fig. (1),
.
tan
A=
tan
Ar
U+c Y1
c.
C yt
= C < tan ~ = 20
= 2.8095 and so
U = - = 2.505 and so
c•
x
tan 20° = 7.279 mls
•
A = 70.79 deg.
A. = 68.24 deg.
• (2)
From eqn (5.40a),
where tan 'jJ =S.
CL
and tan
J~ = -21 (tan J? + tan Ji)
8
,'. p= 3.22 deg. and
, l' ..
{b
Pm =69.59 deg.
= ~[tan (69.59-3.22)-], 50.11 ~ 2
tan
20°] =0.8396
\
I
(3)
From eqns (5.37) and (5.38),
= 1.2 x 7.279 ( 0.8396 x 50.11 + ~ x 7.279) = 399.3 Pa
I I
(4)
.
1 , (I) sinV~
Eqn (5.36) gives p, - p, = - pc, CL . 2
hence,
-
S
'/7
-
p) '
cos mcosp
-p,)cos'/~ cos p = 399.3xO.1216 x 0.9984 =0.2756 {pc;C L sin(Jt. - p) 0.6 x 400 x 0.8 xO.9162
!.. = (p, S
2:ITr
As s = - - = 0.2592 then the blade chord 1 = 0.07144 m (7.144 cm) Z,.
:1
I
I! I
5.11. A diffuser is fitted to the axial fan in the previous problem which has an efficiency of 70 per cent and an area ratio of 2.4. AssU!lling that the flow at entry to
the diffuser is unifonn and axial in direction, and the"losses in the entry section and the guide vanes are negligible, detennine
(1) the static pressure rise and the pressure recovery factor of the diffuser (2) the loss in total pressure in the diffuser (3) the overall efficiency of the fan and diffuser.
Solution. (1) From eqn (2A5b) the diffuser efficiency is:
-.
• ,', (P3 - p,) = t pC~1b[I-I/ A~] = 0.6 x 400 x 0.7 x [1 - 112.5'] -
The pressure recovery factor of the diffuser is
•
c p
=
(.0, - p,) =
tpc;
(2) From eqn (2.50), we get
141.1
= 0.588
0.6x400
cp =Cp'"_ b.Po q,
--
141.1 Pa
.'. Dpo
=(Cp1 - Cp~: =(0.84 -0.588)240 = 60.48 Pa
(3) The overall efficiency of the fan and the diffuser is:
---._-
1p LiW
R< -~,L.Li P~ .!:J.w
where 2:.!:J. Po = sum of the total pressure losses =
where!:J. W
(1 - 1l)tlW t
=C/Slo =Ve =50.11 x 7.279 =364.8 m: / / Yl
".1}
=O.8396- 1.260.48 =0.7014 x 364.8
diffuser losses
Chapter 6 Three-dimensional Flows in Axial Turbomachines
• Moin
stream /
I
/ / / Boundary / layer / / / /
•
/
,
Pressure~~; force J
"
/
/
,, ,, LCenlrifugol force
(0 )
(a) Cross-channel flow
(b I
(b) Corner vortex.
Secondary Flow resulting from Boundary Layer on the Annulus Walls
• il
'_~_----------~------:--~ 64
Chapter 6 Three-dimensional Flows in Axial Turbomachines 6.1. Derive the radial equilibrium equation for an incompressible fluid flOWing with axisymmetric swirl through an annular duct. Air leaves the inlet guide vanes of an axial flow compressor in radial equilibrium
"
and with a free-vortex tangential velocity distribution.
The absolute static pressure
and static temperature at the hub, radius 0.3 m, are 94.5 kPa and 293 K respectively. At the casing, radius 0.4 m, "the absolute static pressure is 96:5 kPa.
•
Calculate the
flow angles at exit from the vanes at the hub and casing when the inlet absolute stagnation pressure is 101.3 kPa.
ible.
Assume the fluid to be inviscid and incompress-
(Take R =0.287 kJ/(kgOC) for air.)
. wo..,..,
~ t.-v..
Solution. A detailed derivation of the radial equilibrium equation)6ple:~
P"1'1O!t'
and so only a brief outline of the importaD.t equations (for an incompressible
flow) is given here.
I
= 0), rotating about an axis at r radius r with a tangential velocity component c g . the static pressure gradient is 2 For a fluid element which is in radial equilibrium (e
I
o
d = dr
Cg
(i)
= pr-
The total pressure Pain an incompressible fluid flow is I 2 I 2 2 Po = P+"2 Pc = P+"2P(cx +C g )
(ii)
is the axial component of the velocity c. Differentiating eqn. (ii) with x respect to r and combining the result with eqn. (i), the required form of the radiai
where c
equilibrium equation is found, viz.,
dc x c -X dr
I I .
c
+ -g- r d (c-r) - (...) 111 r
dr
The inlet guide vanes of the axial flow compressor deflect the incoming axial flow
away from the meridional plane imparting a free-vortex swirl to the flow.
\
~j
tree-vortex, rCg = K
,
I
--
= constant.
Substituting for Cg in eqn. (i),
For a
I=NT'I
4
Three-dimensional flows
65
(iv)
After integrating eqn. (iv) and putting limits at the hub and tip,
K2(_1 2 1_) 2
l(p _ p ) = p t h 2
r
r
h
(v)
t
The boundary values given are that at
r
= r
r
= r
.
= 0.4 m,
t
P = Pt = 96.5 kPit, and at
= 0.3 m, P = Ph = 94.5 kPa, T = T
h
= 293 K.
h
~
SolVing for p and K
3 3 P = Ph/(RTh) = 94.5 x 10 /(287 x293) = 1.124 kg/m 2 2 (Pt~Ph)' K = 2 2 P (l/r -I/r ) h t
2 (96.5-94.5)xI0 1.124 2 2 (1/0.3 -1/0,4 )
3 = 732.1
2 :. K .= rCg = 27.06m /s
Thus, for any radius r the magnitude of c g can be found. angles, C is needed and this can be found with eqn. (il),
To determine the flow
x
C
2
2
= 2(p -p)/p - (K/r) -
°
x
... Cx
= =
tan"t
=
C x Cgt C x
=
2
= 3965
= constant for all radii
62.97 m/s cgb.
"h
tan
.
3
= 2(101.3 - 94.5) x 10 /1.124 -732.1/0.3
27.06 62.97 x 0.3 27.06
= 62.97xO.4
= 1.432
= 1.074
Thus, the flow angles at the hub and tip are, respectively,
"h =
55.08°,
"t
= '47.05 0
6.2. A gas turbine stage has an .~itial absolute pressure of 350 ~a and a
•
I1~__------..;...-------~----~-I 1
I
S.L. DIXON
66
"
temperature of 56S"c with negllgible Initial velocity,
At the mean radius, 0.36 m,
conditions are as follows:
Nozzle exit flow angle
68 deg
Nozzle exit absolute static pressure
207 kPa
Stage reaction
0.2
Determine the flow coefficient and stage loading factor at the mean Fadius and the
reaction at the hub, radius 0.31 m, at the design speed of 8000 rev/min., given that the stage is to have a free vortex swirl at this speed. are absent.
You may assume that losses
Comment upon the results you obtain.
(Take C = 1.148 kJ/(kgOc) and y = 1.33) P
•
Solution.
Sufficient data are given to solve the mean radius velocity triangles from
which the flow coefficient and stage loading
fac~ors
are obtained.
Atr=r =0.36m. 02=68deg. P2 =207kPa. R=0.2, T I=T 2=838Kand m. 0 0 P 1 = P 2 = 350 kPa, assuming adiabatic frictionless nozzle flow. Since o 0 I 2 h 02 =h 2 +2'c 2 2 c = 2C (T02 -T 2) 2 p [1 - (p /p ly-I)/y) = 2C T P 02 2 02 = 2 x 1148 x 838 [1: (207/350)°·248] c
2
= 484.9 m/s
The mean blade speed is Urn = (211N/60)r = (211 x 8000/60)0.36' m = 301.6 m/s Hence, the mean flow coefficient is
= c /V = c cos "2/U x m 2 m = 484.9 x cos 68°/301. 6 = 0.6023 The stage reaction is defined as"
"
,,
J
= 23.51 x 10
4
Three-dimensional flows
67
(if c = c ) l 3
c
hol-ho3-h2+h3
1- R =
=
2 2
-c
2
=
hoI -\3 A t the mean radius
92
(i)
J
=
-C 93
C
-c 92 93 2U
C
3
2U (l-R ) m m
= C z sin 0. 2 = = -33.0 m/s
c 92 • c .. 93
= 2 x 301.6 x 0.8 = 482.6 m/s. 484.9 x sin 6B
o
= 449.6 m/s
The stage loading factor at the mean radius is
i,e
=
tJ,W/U 2
= (449.6 - 33)/301.6
m
= 1.381
From eqn. (i) above, the reaction
a z:: 68°
at any radius is
R = I - (c
92
-c
93
"
)/(2lJ)
where, for a free-vortex,
=K3/r and the 93 blade speed U = Jl.r. c 92
=K/r.
c
Substituting for c
92
R = I - k/(r /r
•
, c
m
93 2
and U Velocity triangles at mean radius
1
Soivlng for k with R = 0.2 at r = r 2 m R = I - 0.8/(r/r ) m The reaction at the hUb, r = r
R = I - 0.8/0.861 h = - 0.079
. --
2
h
= 0.31 m, is
,
-
s
S.L. DIXON
68
The negative reaction would imply that diffusion of the flow occurs iri the rotor row
(i.e. w
< w2) at the root.
a
For a turbine blade row, flow diffusion results in poor
efficiency caused by large toral pressure losses.
A poor flow distribution will
result and this can adversely affect the performance of any subsequent stages. Turbine designers always aim for a positive root reaction to avoid this problem. 6.3. Gas enters the nozzles of an axial flow turbine stage with uniform total
I-
pressure at a uniform velocity c in the axial direction and leaves the nozzles at a 1 cons;am: flow angle 0. to the axial direction. The absolute flow leaving the rotor c 3 2 is completely axial at all radi~.
:.
Using radial equilibrium theory and assuming no losses in total pressure show that
where U is the mean blade speed m c is the tangential velocity component at nozzle exit at the mean radius Sm2 r ;; r
m
(Note: The approximation c ;; c at r = r is used to derive the above expression.) m 3 1
Solution.
This problem and the one follOWing are both examples of the so-called
'direct problem' in which the flow angle is specified as some function of radius and
the velocity component distributions cx(r) and cg(r) are to be solved. In this question the essential idea to grasp is that as a result of the specific work
variation with radius, the total pressure at stage outlet P a is also non-uniform. o . The equation for specific work done by a turbine stage, eqn. (4.2), appli.id to an
•
incompressible, frictionless flow is (i)
where poj
J
2
=PI + 2" pel
and p03
J
2
=Pa + 2" pc a'
Noting that with no-swirl in the
flow at stage inlet and outlet, dpjdr = 0, and both p J and Pa have constant values. Thus, from eqn. (i), (il)
/
--
r
Three-dlnlenslonal nows
. 69
i
where k = (PI -P3)/P.
From eqn. (6.22)
c 92 = c 9m2
(r~ r sin ml
2 "2
(a slnlilar derivation of this equation is given In the solution of Q.6.4),
s (r:f 2
r = r
= m
(iii)
°2
Substituting eqn. (ili) into eqn. (il), 2 r cos °2 Um c 9m2 (r- ) . m
•
Noting that c = c at r = r , the constant k = i m 3
U~ c
9m2
•
Hence,
[ 1 - ( .E... )cos a 2] (c 2 - c 2)/2 = U c 3 1 m 9m2 r m
6.4. Gas leaves. an untwisted turbine nozzle at an angle a. to the axial direction and in radial-equilibrium.
Show that the variation in axial velocity from root to tip,
assuming total pressure is constant, is given by 5in 2 a c r = constant. :<
Determine the axial, velocity at a, radius of 0.6 m when the axial velocity is 100 mls
at a radius of 0.3 m.
The outlet angle a is 45 cteg.
Solution. In an incompressible flow the total pressure is Po = P +
I
pc
2
Diiferentiating this expression with respect to r and noting that p is assumed to be o constant
•
!.
dp P dr
+ cdcdr
=
0
For a swirling flow in radial equilibrium 2 1 ~ = c9 p dr r Combining equations (i) and (il),
--
• /
(I)
...._-_...
~------_
-r-'
S.L. DIXON
70
2
c
e + r
dc c dr
=
°
(iii)
which is another form of the radial equilibrium equation.
and c :::
C
x
sec
Noting that c = C tan
e
x
(l
substitution into eqn. (iii) gives.
Q,
\
C
2
x r
dc 2 x 2 tan a + ex dr sec a. = 0
After some simplification and re-arrangement to separate the variables, dc
x = --sin2 0dr .r x
-~
c
•
which, for constant flow angle a, can be immediately integrated to give sin 2 (l c r :. constant
x
The numerical part is easily solved by direct substitution. - 2 0.= 0sm .;:, C
x
At r = 0.6 m and with
= 100(0 . .>/0.6)°·5
= 70.7 m/s
6.5. The flow at the entrance and exit of an axial flow compressor rotor is in radial ~
,
equilibrium.
\
The distributions of the tangential components of absolute velocity
with radius are:
I i
c
e1
\,
C
Q2
;. ,
= ar - b/r. before the rotor, ::: ar + b/r, after the rotor,
where a and b are constants.
What is the variation of work done with radius?
Deduce expressions for the axial velocity distributions before and after the rotor, assuming incompressible flow theory and that the radial gradient of
s~gnation
pressure is zero. At the mean radius, r = 0.3 m, the stage loading coefficient, "l.p = /iW/u the reaction ratio is 0.5 and the __m ean axial velocity is 150 m/s. 7640 rev/min.
i /
\ I :
\
-.
is 0.3,
The rotor speed is
Determine the rotor flow inlet and outlet angles at a radius of 0.24m
given that the hub/tip ratio is 0.5.
i
2 t
Assume that at the mean radius the axial velocity
.. :,
Three-dimensional flows
remains unchanged (c
xt
71
= c"2 at r = 0.3 m).
(Note: t;.W is the specific work and U the blade tip speed.) t Solution. The specific work done on the gas, from eqn. (5.1), Is
Substituting for the blade speed U = n.r and the tangential velocity components ~W =
S1.r(2b/r) = 2bSl. = constant.
-Thus, the work done is constant with radius.
Note that for a uniform stagnation
pressure at rotor inlet and I:1W:= constant, the stagnation pressu~e after the rotor
will be constant provided that the flow Is either frictionless or that losses are distributed uniformly with radius. From the radial equilibrium equation, eqn. (6.8), with p = constant a d
2
dr (c" /2) +
cQ d dr (rc g) = 0
r
Considering the whirl distribution at rotor inlet, then
2 cx1 2 dr (-2-) + (a - b/r )2ar = 0 d
Integrating this equation
2 c"l -2-
J
2ab 2 I 2 2 ( - - -2a r)dr = 2ab-<.nr-a r +k/2 r I
=
Thus, the distributions in axial velocity at inlet and outlet of the rotor are,-
respectively, 2
c xl c
2
= k - 2a [ r
l
2
- (2b/a) inr]
2 2 2 x2 = k2 -2a (r +(2b/a)inr]
(i)
(Ii)
where k and k are arbitrary constants. 1 2 In order to determine the flow angles at any radius r it is necessary~first of all.. to
solve the four constants a, b. k and k . l 2
The stage loading factor. which is
\ S.L. DIXON
72
I
constant for all radii because
~W
f,W
'11= -2 = U t
is constant, is defined as
(c 92 - c 91 )t = U t
2b 2b =-tiT fu2 t t
= 0.3
t
2 :, b = 0.15 fu t
(iii)
The reaction ratio is defined, eqn. (5.11), as R
=
C
x
(tan ~1
= 1 - (c
At the mean radius r
•
m
91
+ tan
~2)/(2UJ
+ c 92)/(2UJ
the reaction ratio is 0.5, hence
(iv) after using the swirl equations given in the problem.
m
The angular velocity,
(iva)
.51.
= 21TN/60 = 2rrx 7640/60 = 800 rad/s.
The mean radius
is the arithmetic mean (other definitions of 'mean' are sometimes 1 r = 2" (r + r ), hence m t h
m
r
= flr m .
= S1./2
a
r
Hence, with U
t
= 2r
m /(1
I.e.
+ rh/r t )
= 2 x 0.3/(1
=
used~),
+ 0,5)
0.4 m
Thus, using eqns. (iva) and (iii), a = 400 and b = 19.2.
".
From eqns. (i) and (il), with c 1 = C 2 = 150 m/s at r = r = 0.3 m, x x m 2 2 k1 = C / + 2a [ r - 2 (b/a) .(.n r] . x 2 2 2 4 = 150 + 2 x 400 [0.3 - 0.0961.n 0.3J = 8.829 x 10 2 2 2 k = 150 + 2 x 400 [0.3 + 0.096 2
in 0.3]
= 1.431 x 10
Hence, atr=O.24m, C
x
/
= 8.829 x 10 = 2.601 x 10
/
--
4 4
2 4 - 32 x 10 [0.24 - 0.096ln 0.24J
4
Three-dimensional flows
73
=
• c 161.3 m/s ,. xl 2 4 4 2 = 1.431 x 10 - 32 x 10 [ O. 24 + 0.096 In 0.24] c x2 4 = 3.972 x 10 •c = 199.3 m/s " x2 From the swirl distribution equations, at r
c c
9l g2
==
0.24 m
= ar - b/r
= 400 x 0.24 - 19.2/0.24 = 16 m/s
= ar+b/r
= 400x 0.24+ 19.2/0.24 = 176m/s
For the axial compressor stage, the velocity triangles (Fig. 5.2) yield, at r = 0.24m
'.
tan ~1 = (Jtr - cgl)/cxl = (800" 0.24 - 16)/161.3 = 1.091 tan ~2
= (Sl.r - c 92 )/c,,2 = (800 x 0.24 -176)/199.3
= 0.0803
The relative flow angles at rotor inlet and outlet at r = 0.24 m are, respectively, ~l
=
47.5 deg.
~2
=
4.59 deg
6.6. An axia 1~IOW turbine" stage is to be designed for free-vortex conditions at exit from the nozzle row and- for zero swirl at exit fr.om the rotor.
The gas· entering the
stage has a stagnation temperature of 1000 K, the mass flow rate is 32 kg/5, the root
and tip diameters are 0.56 m and 0.76 m respectively. and the rotor speed is 8000 rev/min.
At the rotor tip the stage reaction is 5010 and the axial velocity is
constant at 183 m/s.
The velocity of the gas entering the stage is equal to that
leaVing. Determine:
•
(i) the maximum velocity leaving the nozzles; (il) the maximum absolute Mach number in the stage; (ili) the root section reaction;. (iv) the power output of the stage; (v) the stagnation and static temperatures· at stage exit.
(Take R = 0.287 kJ/(kgOC) and C =1.147 kJ/(kgOC) P Solution. A Simplifying assumption Which has been employed in deriving answers is that the axial velocity is constant both across the stage and radially.
/
--
S.L. DIXON
74
(i) At nozzle exit, as rC g2 = K = constant, the velocity will be largest at the hub, radius r = r =0.28 m. Now at the tip, radius r =r =0.38 m, the reaction R =0.5 t h (velocity triangles are symmetrical) and the leaVing absolute Velocity c = C = 183 3 x m/s. With Jl= 211 N/60 = IT x 8000130 = 837.8 radls, the blade tip speed
U
t
= ftr t = 318.4 m/s.
The tip section velocity diagram obtained from this data is
shown below,
• U. = 318-4 mls
It is seen from the tip section.velocity diagram that Cg2t ::; Ute Hence. c C r /r = 318.4 x 0.38/0.28 432.1 m/s. The maximum velocity at
Q2h = Q2t h nozzle exit is J
=
= (432.J:2 = 469.3
+ 183 2)
1/2
mls
(ii) The maximum absolute Mach number in the stage also occurs at nozzle exit at r
= r h and
is determined from
2
2
M2Max = c2h/(yRT2h)
•••••
2h
=Tal
2
2
1(2C ) = 1000 - 469.3 1(2 x 1147) 2h p and y = C I(C - R) = 1147/(1147 - 287) = 1.334 P p II :. M = 469.3/(1.334 x 287 x 904) 2 2Max
where T
- c
= 904.0 K
= 0.798
(iii) For a normal axial n'rbine .."tage (i.e. with c (4.22c), is
. --
l
= c ) the degree of reaction, eqn. 3
Three-dimensional flows
With c
e3
= a (I.e. axial exit flow), U =.Il.r and c
75
= K/r
e2
2 2 :. R = I - K/(2Jlr) = I - k/r 2 At the tip r = r , R =R =0.5, hence k=0.5 r • t t t is
R
h
Hence, the root section reaction
= 1 - 0.5 ( r / r / = 1 - 0.5 (0.38/0.28)2 = 0.079
.(iv) For a free-vortex turbine stage the specific work done is constant with radius. 2 At the tip radius c = 0, c = U and t;W = U ' Thus, the power developed by the e3 t t e2 stage is W t
=
m/iW
=
mcp (Tal -T03 )
= 32 x 318.4
2
= 3.244 MW
(v) The stagnation and, static temperatures at stage exit are
T
03
= T 1 - t;W/C
a
p
= T 1 a
2
2 ut /c p = 1000 - 318.4 /1147
= 1000 - 88.39 = 911,6 K T
3
= T
03
2 -c /2C = 3 p
9~1. 6
2
- 183 /(2 x 1147)
= 897.0 K 6.7. The rotor blades of an axial flow turbine stage are 100 mm long and are
"
designed to receive gas at an incidence of 3 deg from a nozzle row.
A free-vortex
whirl distribution is to be maintained between nozzle exit and rotor entry ~ exit the absolute velocity is 150 mls in the axial direction at all radii.
At rotor
The
deviation is 5 deg for the rotor blades and zero for the nozzle blades at all radii. At the hub, radius 200 mm, the conditions are as follows:
Nozzle outlet angle
70 deg
Rotor blade speed
180 mls
Gas speed at nozzle exit
450 m/s
Assuming that the axial velocity of the gas is constant across the stage, determine (i) the nozzle outlet angle at the tip;
I \
I
I'I '
\
I
I I
(ii) the rotor blade inlet angles at hub and tip;
I
(iii) the rotor blade outlet angles at hub and tip;
I,
, I ~
S.L. DIXON
J
I,
"
76
;
(iv) the degree of reaction at root and tip. Why is it essential to have a positive reaction in a turbine stage?
Solution. The gas flow angles are determined first from the velocity diagrams at the hub and the tip, then the blade angles are found by suitably correcting the flow angles
I
:;:!
Cor the incidence and deviation angles given.
'-.jJ.
shown below:-
The velocity diagram for the hub is
,
,
ex =70·
z
Un' 180 m/s
U
Sign convention for turbine rotor flow angles and blade angles with positive incidence, i = P2 - P ' 2 and positive deviation, S =
13; - 13
3
,
(i) The absolute tangential velocity at the hub is, c
g2h
= c 2h sin
Q
2h
= 450 sin 70·
= 422.9 mls For the free-vortex flow, r ... C
t
C
g2t
=r
h
c
g2h
g2t = c g2h rhlrt = 422.9 x 2/3 = 281.9 mls
tan "2t = cg2 lc x
= 281. 9/150
= 1.879
:. ale = a.2~ .;: 62 deg, as there is zero nozzle deviation. (ii) Referring to the velocity diagram, at the hUb,
Three-dimensional flows
77
tan 13 211 = (Cg21l-UIl)/cx = (422.9 -180)/150 = 1.619 :.13
= 58.3 deg.
211
TIle rotor blade inlet angle at the Ilub Is, A
=
,
~2h
A
~ 2h
- i = 58.3 - 3
= 55.3 deg.
-.
•
Similarly, at the rotor tip,
tan 13 2t = (c g2t - UJ/cx = (281.9 - 180 x 3/2)/150 = 0.0793 :. 13 :.
2t
= 4.54 deg
132~ = 4.54 - 3 = 1.54 deg
(iii) Again, from tile velocity diagram, at the Ilub,
311
= Uh/cx
3h
= 50.19 deg
tan 13 :. 13
... 13311'
= 13 + S 311
= 180/150
=
= 1.2
55.19 deg
Similarly, at tile tip tan 13
3t
= U/c = 1.5 x 180/150 = 1.8 x
133t = 60.95 deg . 13 ' " 3t = 65.95 deg (Iv) For a normal turbine stage (I.e. c = c ) with constant axial velocity across It, 1 3
reaction Is defined, eqn. (4.20), as
At the Ilub, R
Il
=
150(1.2 - 1.619)/360
-0.175
At the tip, Rt
/
"
= 150(1. 8 - 0.079)/540 = 0.478
J
-----~--------------~-----~]
,
S.L. DIXON
78
A positive degree of reaction in a turbine stage is necessary to avoid large total
pressure losses ca used by diffusion of the relative flow in the rotor. ~.
There is a
~
anomaly in the data given for this problem.
At nozzle exit
= C cos "Z = 450 cos 70 = 153.9 mls (at the hub) whereas the xZ z axial velocity is given as 150 m/s. This anomaly only slightly affects the numerical
the axial velocity c
answer5~
6.8. The rotor and stator of an isolated stage in an axial-flow turbomachine are to be represented by two actuator discs located at axial positions x = 0 and x = & ~
•
respectively.
The hub and tip diameters are constant and the hub/tip radius ratio
rhlrt is 0.5.
The rotor disc considered on its own has an axial.velocity of 100 mls
far upstream and 150 m/s downstream at a constant radius r
= 0.75
r tO . The stator
disc in isolation has an axial velocity of 150 mls far upstream and 100 mls far
= 0.75
downstream at radius r
rtO
Calculate and plot the axial velocity variation
between -0.51> x/r <; 0.6 at the given radius for each actuator disc in isolation and t
for the combined discs when
Sir
= 0.1,
t
0.Z5 and 1.0.
Solution. The variation in axial velocity c with axial distance x away from an x isolated actuator disc (located at x = a inside a cylindrical annulus) for a constant radial distance r is, eqn. (6.43), given to a first approximation by,
Cx = cx ""l -}(CX""l-cXOOZ)exp (rrx/(rt-rh»), Cx = cxcoZ +}(cX"'l -cxooZ) exp where
•
C
Xco
I' c
XeD
= 0.75
when Th/r
I
x
t
For the rotor, c"",l = 100 mis, c
= 0.5.
= 150 mls at
xooZ Substituting these values into the above equations
C
= 100+Z5 exp(Z1Tx/rt),
C
= 150 - Z5 exp (-z1I'x/rt),
x
C
(ii)
..
x
Values of
(i)
Z are the axial velocities far upstream and far downstream
respectively of the disc. rlrt
(-1I'x /(r t -rh»)
x~O , x ~a
x$
a
x
~
a
are shown in tabular form below and graphically in Fig. Q.6.8(a):-
• 79
Three-dimensional flows
x/r
t
c m/s x
-0.5
-0.3
-0.2 -0.1
101.1
103.8
107
113.4
0
0.1
0.2
0.3
0.5
125
136.7
143
146.2
148.9
The results for the isolated stator are not shown as they are merely the mirror
Image of those for the rotor with the origin" shifted according to the stator's location. When two actuator discs are in close proximity to one another, flow interference will occur, the magnitude of the effect being dependent upon the respective values of c and c
XCQ
2iar each disc.
The far upstream axial velocity at r/r = 0.75 is the "
t
same as the far downstream axial velocity and Is labelled c 2' Similarly, the far x downstream axial velocity of the stator disc at r/r = 0.75 -is the same as the far t
upstream axial velOCity of the rotor disc and is label1ed
C
XCQ
I'
The axial velocity
for the second disc in isolation is C
= c
c
=
x
for x
~S
forx)S
xoo2
_.!-(c 2
xCQ2
- c
xl:Ql
) exp[1T(x - S)/(r - r)]
(iii)
h
t
, and x
c
xO'l
+ .!-(c 2
xec2
Substituting values for c C
C
x
Xa:)
- c I'
xGOl C
xoo
) exp [-rr(x - S)/(r _r)l (iv) t
2 and rh/r
h~
t
= 150-25exp[2TT(x-S)/r ) , x,"S t
x = JOO+25 exp [-2TT(x-S)/r t
J'
x~S
The axial velocity variation for the two discs in combination appropriate to each
•
region, from eqns. (6.48), (6.49) and (6.50) are, Cx = JOO+25[exp(2rrx/r )-exp(21T(x-S)/r )] t t
x,,;oO
c
a ~x':S
x
1
In the problem, the second disc representing the stator
biade row Is located at x = S.
•
X""
= 150 -25 [exp(-21Tx/r)+exp(2TT(x-S)/r)] t
t
Cx = 100 + 25 [exp (-2TT(x -S)/r ) - exp(-21Tx/rJ] ,-t
J>
6 x
From the equations the variation in axial velocity for the three values of S/r = 0 1 t
0.25 and 1.0 have been calculated and are shown graphically.
"
--------------------------------------1
80
,I I
S.L. DIXON - - - - , - - - - , - - - 150r----r---r:====!='"-
u
.,j
.~
120
:-i
i.
110
i
02
1
I
I000~----=0~.2---..,0:l._4.,-----;:L,.---
(a) Variation in axial velocity with axial distance from an isolated actuator disc located at x = 0, for r/r = 0.75
e
------150
140
'"
"-
E
;; u
• 110
0"
o~oook---~0l.,;.2c:::----~0~_4;:=:::::==;0;l;.6;:.=--..,0,L.e;;---J.1.0=---J.1.2) x/r t
(b)
--
-
/
Variation in axial velocity with axial distance from twe actuator discs separated by a distance 5. one disc being located at the origin.
Chapter 7 Centrifugal Pumps, Fans and Compressors
•
• A centrifugal compressor
81
Chapter 7 Centrifugal Compressors and Pumps
_.:~. _,__ .
.. _. _ _
__ .
__ T Ju- -C..- ~ 7./
'__
-N~te. --In problems 7.1 37.5 assume that the gas constant R ; 287 J/(kgUC) and that --
f\
t
For problems 7l'':..~';. th.:- =-.tagnationpre:sure and stagnation temper~___ 7.
Y ; 1.4.
ature at compressor entry are assumed to be 101.3 kPa and 288 K respectively.
--~
,.
~-l1. ~ The air entering the impeller of a centrifugal compressor has an absolute axial velocity of 100
m/s.
At rotor exit the relative air angle measured from the radial.
:.
direction is 26° 36', the radial component of velocity is 120 m/s and the tip speed of
I
When the air flow rate is 2.5 kg/s and the mechanical efficiency is 95 %.
\,
radius ratio of the impeller eye i50.3. calculate a suitable inlet diameter assuming
,
the radial vanes is 500 m/s.
Determine the power required to drive the compressor
the inlet flow is incompressible.
If the
Determine the overall total pressure ratio of the
compressor when the total-to-toral efficiency is 80%. assuming the velocity at exit
from the diffuser is negligible. Solution. The specific work required to compress the air is /:iW U c g2 ' the. flow· being without swirl at impeller enn-y. 2 triangle at the impeller exit (see sketch) that
c
;
U - c tan 13 2 r2 2
;
500 - 120 x tan 26.6
;
440 mls
;
440 x 500
92
:. AW
;
It follows from the "velocity
0
.:' _ )0"
220 kJ/kg
~--------IU2
The theoretical power needed (i. e.
e 82
ignoring mechanical losses) is
W ; c ;
m A,W ;
2.5 x 220
550 kW
Hence, the actual power needed is ".
~
p
;
oW Ii c
m
=h03 -h01 =
;
550/0.95
7. Z. /;,7,,, ~
Z.
to 7.:; .
~
DU·
'-,
7.1. A cheap and simple radial-vaned centrifugal fan is required to provide a supply
of pressurised air to a furnace. The specification requires that the fan produce a total pressure rise equivalent to 7.5 cm of water at a volume flow rate of 0.2 m3 /s. The fan impeller is fabricated from 30 thin sheet metal vanes, the ratio of the passage width to circumferential pitch at impeller exit being specified as 0.5 and the ratio of the radial velocity to blade tip speed as 0.1.
Assuming that the overall isentropic efficiency ofthe fan is 0.75 and that the slip can be estimated from Stanitz's expression, eqn (7. 18a), determine
•
(1) the vane tip speed (2) the rotational speed and diameter of the impeller (3) the power required to drive the fan if the mechanical efficiency is 0.95 (4) the specific speed.
Fer air assume the density is 1.2 kglm 3 •
Solution. (1) The efficiency of the fan is defined as
•
!¥Jo _ !¥Jo
1'/ =
pUc 2 92
c U2
I~
where ..ll - 1- -
Z
pC! U22
. - 0.934 and L'1Po - P g 6h = lOt x 9.81 x 0.Q75 = 735.8 N/m 2 .
d P lO3 189 gI 3 • . . an p - RT - 287 x 293 - 1. k m. Upon substltutmg values, the vane tlp speed can be determined:
735.8 _ 27.9 mls 1.189 x 0.934 x 0.85
(2) As the volume flow rate is
Q - pAc, where A = 2it rb and c, - U 110 and b = it r 1Z, then
Q - 0.2(it r)2 p U1Z hence
•
D
2 ~
2
- ;;VD:2pU - it
Q =
0.2 x 30 _ 0,604 m 0,2 x 1.189 x 27,9
2U 1D = 2 x 27,9/0,604 - 92.4 rad/s :. N
= 882 revlmin
(3) The power required to drive the fan is obtained from:
p_ m.iW 11m
=
pQu U~ = 1.189 x 0,2 x 0.934 x 27.9 11ft 0.95
(4) The specific speed can be determined from eqn (1.8a),
. Q ..
s =
92.4 x 0.2 0.5 41.32 0 3 (735,8/1.189)0.75 - 124,1 - ,33 rad
2 _
182W
-~--------:-~--~ S .L. DIXON
82
= 578.9 kW
The equation of continuity, with e
m = PI Al c
= c ' is
l
xl
2
2
= trP (r sl • r hl )c I I
i
2 = rrP I c l r S1 [1. (rh/rSll] With the assumption that the now at entry is incompressible. the density PI can be determined from the stagnation pressure and temperature,
PI
= Pol
5
1.013 x 10 /(287 x 288)
= po/(RTol )
= 1.226 kg/m
3
2 . 2 ) :. r sl = m/{lTPl c l [1. (rh/r sl ) }
•
= 2.5/(lTx 1.226 x 100 x 0.91) = 0.7135 x 10
. d .. sl
·2
= 169 mm
(ecyn
The total to total efficiency of the compressor (eEtfl ... . 2g~ is
'c
=
h
·h
03ss 01 h -hal 03
=C
T p 01
r(p 03 IPol )(y-I)/y - IJ
(1+ ~ C T p ol
lAW
3 3.5
tJ.W)Y/(Y-li
=
j
f...
=(
I
+ 0.8x 220x 10 ) 1005 x 288
3 = 1.608 •5 = 5.273 ~
A centrifugal compressor has an impeHer tip speed of 366 m/s.
Determine
"
the absolute Mach number of the flow leaving the radial vanes of the impeller when the radial component of velocity at impeller exit is 30.5 mls and the slip factor is 0.90. 2 Given that the flow area at impeHer exit is 0.1 m and the total-to-total.efficiency of
,
i
the impeller is 90,%. determine the mass flow rate.
I
Solution. The absolute Mach number at impeller exit (eqn. .
-
M2 = czla = c/(yRT2) 2
1 !
--
1/2
~)
f...
is
7.
83
Centrifugal compressors
so that c and T 2 need to be determined. 2 exit
I
c
2
g2
+c
From the velocity triangle at impeller
2
2
(0.9x 366) +30.5
r2
2
5 = 1.094 x 10
7.10..
Thus,
M
2
=
-
[1.094 x 10'/(1.4 x 287 x 353.5)
]1/2
= 0.8778
The rate of mass flow is ~ = P2A2 c
where the density is P2 = P/(RT2)' The r2 static pressure P2 must be determined by first solVing for the impeller total pressure
ratio p 2/P I and then relating p to p 2 by means of the isentropic temperatureo a 2 0 pressure equation. Thus, the impeller total to total efficiency is
I
T
P02
...j
";:1
Pol
•
P2 P02
T
=
/p )(y-I)/y - I
(p
- T
025 01 = T -T 02 o1
-;0~2,-~01~
To/Tal - I 3.5
(1+ '7. (~-11 1
T
Tal
y/(y-I)
= ( To:)
_
408 - I))3.5 = 3.047 = ~1+ 0.9 x (288
353.5 3.5
= ("'408)
= 1/1.1543 . 5 = 1/1.651 l'OI'!y 186'12[5
3.047 x lOS = i81o' loP.' _
:. P2
1, 651
I..
3 = l8-h-4 x 10 x 0.1 x 30.5/(287 x 353.5) = ~kg/5
The
"
feye of a centrifugal compressor has a hub/tip radius ratio of 0.4,
I~b' et5
S' b'2. o'Z. a maxi-
----------------:------:--~ 84
S.L. DIXON
mwn relative flow Mach nwnber of 0.9 and an absolute flow which is uniform and
completely axial.
Determine the optimum speed of rotation for the condition of
maximum mass flow gIven that the mass flow rate Is 4.536 kg/so
Also, determine
the outside dIameter of the eye and the ratio of axial velocItyjblade speed at the eye tIp.
FIgure 7.4 may be used to assIst the calculations.
Solution. The compressible flow relation between mass flow rate, speed of rotation
and the flow parameters at the eye tIp is given by eqn. (7.11),
0.6598 x 10.8
when 'I
•
2 . 2
2
m~
=
M r1 SUI ~Sl cos ~SI
2 2 (I + 0.2 Mr1 cos
~SI)
4
=1.4, R =287 J/(kg °e), Pol =101.3 kPa and T 01 =288 K.
Now rather than
using the appropriate curve in Fig. 7.4 to determine the maximum mass flow condition, greater accuracy is obtained by differentiating the RHS of the above equation with respect of cos M
rl
~sl
and setting the result to zero.
= 0.9 and dIfferentiating etc., the following equatIon Is found,
(1
2
2
+ 0 .162 x )(1 - 3 x )
2
2
= I. 296x (I - x )
.42 .. 0.8Ix - 4.134x + 1 = 0 Solving this quadratic equation, the only valid rool: is
x
2
= 0.2546
:. cos ~Sl = jO.2546 = 0.5046 :. ~sl = 59.7 deg Hence,
•
Putting x
8 2 ~ = 0.729 x 10 x sin 59.7xcos59.7xO.84 l)
2 0.6598 x 4.536(1 + 0.162 x cos 59.7)4
6
= 6.547 x 10 (rad/s)
2
:. Jl. = 2559 rad/s = 24,430 rev/min From the equation of continuity, the rate of mass flow is
= cos
13
s1
'
85
Centrifugal compressors
2 where k = 1 - 0.4 "0.84 and PI (at the relatively high Mach number prevailing) is sought from compressible flow theory.
From the
inlet velocity triangle (see sketch), w
sl
Also M r1 :. M l
= c
sec 13
xl
w"
51
= ws/a and M = cx/a 1 1 1 = M
r1
cos 13
51
u"
= 0.9 x 0.5046
f3"
= 0.4541
Thus. using the compressible flow relation, i
1
~ I
2
= l+-(y-l)M 2 1
c xl
I
= M a
1 1
"
= 1+0.2xO.4541
Mj(YRT 1)
= 0.4541 (1.4 x 287 x 288/1.0412)
!
2
= 1.0412
Ii? -
= 151.4 m/s PI
= po1(T/To//(y-l) = P
o1
(l/1.0412)2.5 = p /1.106 o
Thus. r
mRToJ" 1.106
2 s1
= .1l'
=
:. d
s1
Pol kC x1
4.536" 287 " 288 x 1.106 5 'TT x 1.013 x 10 x 0.84 x 151.4
1.024 x 10
-2
= 202.4 mm
The ratio of axial velocity to blade tip speed at the eye is
••••
::~i -:~
cx l/Us 1 = cot ~5 1 = 0.5844 --
::~
,
7.. ~
i
I
I
~. An experimental centrifugal compressor is fitted with free-vortex_guide vanes in order to reduce the relative air speed at inlet to the impeller.
At the outer radius
of the eye, air leaving the guide-vanes has a velocity of 91.5 m/s at 20 deg to the
!
axial direction.
Determine thE! inlet relative Mach number, assuming frictionless
flow through the guide vanes, and the impeller total-to-total efficiency.
/
'.
86
S.L. DIXON
Other details of the compressor and' its operating conditions are:
Impeller entry tip diameter, 0.457 m Impeller exit tip dian1ter, 0.762 m Slip factor 0.9 Radial component of velocity at impeller exit, 53.4 m/s Rotational speed of impeller, 11, 000 rev/min Static pressure at impeller exit, 223 kPa (abs.) Solution. The inlet guide vanes reduce the inlet relative velocity WI by deflectiDg the absolute flow through an angle
°1,
produciDg a tangential velocity c~l in the same
direction as the blade motion (see sketch of velocity triangle at inlet). •
The required
relative Mach number, which will be greatest at the shrOUd, is defined as, M
lrel
= w/a l
where, from the inlet velocity triangle
2 WI
a
2 l
T
l
2 2 = (Usl - c gl ) + cxl '
= yRTl 1
2
= T ol
-ICI/Cp
= 288
-
i
2 x 91.5 /1005
= 283.8 K :. a
l
= (1.4 x
287 x 283.8)
1/2
= 337.7 m/s
•
c"
The inlet blade tip speed is U =Jlr = 11000 x (1T/30) x 0.457/2 sl sl = 263.2 m/s
c gl = c siD l
... -..
01
= 91.5 x siD 20
0
,I
Centrifugal compressors
87
= 247.3 m/s
. .. M lrel
= 247.3/337.7 = 0.7324
The impeller total to total effIciency is (see Sol. Q. 7 . 2)
From eqn. (.7-:1') the specific work is
f.t!.W
= Cp (T02- T ol)
2 o-U - U c 2 1 91 = C T p ol
= (0.762/0.457)263.
= 438.9 m/s 2
= (0.9 x 438.9 - 263.2 x 31.3)/(1005 x 288) = 0.5704
The impeller total pressure ratio is y/(y-l) P2
P02
Pol
P2
=-.NOWT
02
=T
1
2
+"2 c 2 /C p
2
2 2 2 :.TzlT = 1 - c /(2C T ) = 1 - (c + C )/(2C T ) 02 2 p 02 92 r2 p 02
I I !
c T
92 02
:.T2IT02
.,
= o-U
= 0.9 x 438.9 = 395 m/s
= 1.5704 x 288 =452.3 =1
- (395
(T
i
2
02
2
+ 53.4
K
2 )/(2 x 1005 x 452.3)
= 0.8252
IT)(P /p )(y-l)!y - 1 2 2 01
= (1/0.8252)(223/191.3)1/3.5 0.5704
.
- 1
= 0.9084
~. A centrifugal compressor has an impeller with 21 vanes, which are radial at
etit. a vaneless diffuser and no inlet guide vanes.
--
At inlet the stagnation pressure
_z------------------jJ S.L. DIXON
88
is 100 kPa abs. and the stagnation temperature is 300 K.
(i) Given that the mass flow rate is 2.3 kg/so the impeller tip speed is 500 m/s and the mechanical efficiency is 96%. determine the driving power on the shaft.
Use
eqn. (7.18a) for the slip factor. (11) Determine the total and static pressures at diffuser exit when the velocity
at that position is 100 m/s.
The total to total efficiency is 821••
(ill) The reaction. which may be defined as for an axial flow compressor by eqn.
(5.1Ob), is 0.5. the absolute flow speed at impeller entry is 150 m/s and the diffuser efficiency is 84%.
Determine the total and static pressures, absolute Mach number
and radial component of velocity at the impeller exit.
•
(iv) Determine the total-to-total efficiency for the impeller • (v) Estimate the inlet/outlet radius ratio for the diffuser assuming the conservation of angular momentwn.
(vi) Find a suitable rotational speed for the impeller given an impeller tip width of 6 mm. Solution. (i) For a radial vaned impeller, ~~ = 0 and the slip factor (eqn. 7.18a) is ""s =
j
-
0.631T/21 = 0.9057
The shaft power. P = = 226.4 kJ/kg
:. P
= 2.3 x 226.4/0.96 = 542.5 kW
(li) The total to total efficiency of the centrifugal compressor is
h
'7 c T
=
-h 03ss 01 = Cp Tol(Te:y/Tol - I) h03 - hal o-s U2 p
03ss
~=
(p:~)
[
(y -I)IY
.CT
P 01
2 1 + (y -1)7-c ""s u 2 /a
=
[
a
=
2
..E.!. y-I y/(y-I)
0/ 1
1+ 0.4xO.82x2.264xlO5J3.5 1.4 x 287 x 300
Centrifugal compressors
89
= 5.365
= 536.5 kPa
02
'h
3 = ( T: T
i i
)Y/(Y-I)
"
3
T =
03
y/(y-I)
[T ::2/(2C) 03 -
. y/(y-I)
= 1/
[I - c32/(2ho3)]
01
= (h03 - hoi) + hOI =
(226.1 I LeeS x 3SS) 16
3
s
527.9 k]/kgf. = 536.5[1-100 2/(2X52.79XI0 4)]3.5 = 518.9 kPa
(iii) At impeller exit the absolute Mach number M = c /a Where the speed of 2 2 2 sound a 2 = J{(y -1)h 2J · The enthalpy h and velocity c are evaluated as follows. 2 2 F:-om the definition of reaction R = (h2 - h )/(h3 - hi) = 0.5 and l 1 2 2 [ 22.64 + 2(2.25 I h03 - hoi + 2(c l - c 3 ) -I) ] x 10 4 h 3 - hi
=
=
232.7 k]/kg :. h 2 - hi
=
2 (h3 - hi) = 116.4 k]/kg
Now h 2
=
(hoi - tCll + (h2 - hi) = (301.5 - 11.25 + 116.4) x 103
=
406.7 k]/kg
•
2(h03 - h 2) since h = h 02 03
!
I I
I
I
=
2[(ho3- hOl)+(hol-h 2)]
=
2 [226.4 + 301.5 - 406.7] x 10 3 = 242.4 x 103
=
[242.4/(0.4 x 406.7)]
Thus,
M2
The diffuser efficiency is defined (eQI'
1/2
5 JOb)
= 1.221
2S
90
S.L. DIXON
(!j f.. . p3 P2
:. P2
=
h3 • - h 2 h - h2 3
P2
•
h - h2 3
=
[
= =
(I +0.84x 116.4/406.7] 3.5
1+
=
J
h [(p/p/y-I)/y - 1 2 h - h 3 2
'7ih3 - h2)/h2 r/(y-I)
518.9/2.126 T
P02
=
h 2(T3.(r2 - I)
= (T~)
=
=
2.126
244.1 kPa y/(y-I)
y/(y-I)
=[~t5 = 406.7
=(::)
.
2.492
= 608.2 kPa . p " 02
From the impeller tip velocity triangle. 2 2 2 2 2 c r2 = c - c 92 = c - (OS U ) 2 2 2 3 3 = 242.4 x 10 - (0.9057 x 500/ = 37.33 x 10
c
r2
193.2m/~.
~ <2 ,--
~.
-,
!\~-l2
,.c:..----c-:---:---'
I
2 U
I
C. 2
(iv) For the impeller. the total to total efficiency is
'?i
•
h - h 02. 01 = h02 - hal
h [(p /p )(y-I)/y _ 1] 01 02 01 = h03 - hal 1 3 5
= 301.5(6.082 / .
- 1)/226.4
= 0.899
(v) Assuming angular momentum is conserved, reg = constant. and
=
100 = 0.2208 0.9057 x 500
(Vi) The angular velocity of the impeller i.
f1.
= U/(2'TT r 2)'
The impeller tip
radius is found from the equation of continuity,m = P A c , where A = 2rrr b , Z Z r2 Z 2 2 and the equation of state. Thus,
/
'.
91
Centrifugal compressors
YP2
P2
·'-1
::;~
= 2.101 kg/m 3
P2
S1.
= U2 P2 b2 c r zlm = 500 x 2.101 x 0.006 x 193.2/2.3
2
2
= 529.5 rev/s = 31,770 rev/min
.i ,-.1
:::~
3 3.5 x 244.1 x 10 3 406.7 x 10
= RT = (y-l)h =
1·~1~
A centrifugal pump is used to raise water against a static head of 18.0 m. (\ The suction and delivery pipes, both 0.15 m diameter, have respectively, friction
..
head losses amounting to 2.25 and 7.5 times the dynamic head.
,i !
The impeller, which
rotates at 1450 rev/min, is 0.25 m diameter with 8 vanes, radius ratio 0.45, inclined
backwards at j3~ = 60 deg.
Ii
The axial width of the impeller is designed so as to give
constant radial velocity at all radii and is 20 mm at impeller exit.
Asswning an
hydraulic efficiency of 0.82 and an overall efficiency of 0.72, determine
i
(i) the volume flow rate; (ii) the slip factor using Busemann's method; (iii) the impeller vane inlet angle required for zero incidence angle;
(iv) the power required to drive the pump. Solution. (i) The acrual head H which is delivered by the pump is defined as the difference in head measured between the outlet and inlet flanges of the pump.
It is
equal to the static head H • defined as the difference in level between the two open s reservoirs, plus all external head losses. The external losses comprise the
friction losses in the suction and delivery pipes together with the kinetic energy leaVing the delivery pipe. H = H
Thus, 2
s
+ (2.25 + 7.5 + I) c /(2g)
where the average velocity in the pipes, both of diameter d, is
2
2
c = 4Q/(1Td) = 4Q/(1Tx 0.15) = 56.59Q m/s H = 18 + 10.75 x (S6.59Q//19.62 = 18 + 175SQ2 m Now the ideal head Hi = U cg/g and the hydraulic efficiency of the pump is defined 2 as
/
--
• •
7
7
S.L. DIXON
92
'" _ H _ ~ _ ~C fh - Hi - U2 c g2 /U) /""2 9" g2 2
::y
The Busemann slip factor O-B
~B~avoured
f..
in most pump design calcuIations, is
= CgiCg~ = (A - Bfl2 tan ~; )/(1 - fl2 tan ~; )
where A and B are constants determined by the geometry of a particular pump, ~; is the lmpeller vane outlet angle and fl = criU2'
2
The vane tip speed U is 2
U = ll'NDi60 = 1T x 1450 x 0.25/60 = 18.98 m/s 2
.-.
:. fl
2
= Q/(1TD 2 b 2 U2) = 4Q/crrx 0.02 x 18.98) ~'
•
= 3.351 Q
C8
2
C
C,2
T For this pump the radius ratio x /r 2
l
= 2.222 is sUfficiently large (Le.> ex:ponential
(211 cos~;; /2) = 1,481, see eqn. 7: 17c) for B to be assumed equal to Wllty and A can
0.9,.--------------="1
be fOWld by interpolating from the graph shown, which was obtained from Fig.
Thus, at
,
~2
s~stitution
•
A
= 60 deg and Z = 8 the value of
rf. = 0.818 with sufficient accuracy. 1\
~
After 0·8
=
gH A 2 U (O.818 -1.732 fl ) 2 2 2 0·7 H. U (0.818 - 3.351 x 1, 732Q)/g 2 2 2 18 + 17SSQ = 0.82 x 18.98 )(
= 'h
(0.818 - 5.804Q)/9.81
o· 6;;--:;-H--:---c~-:--L.--.-1--L---:'
o
After Slmplifying, . Q2
+ 0.0996Q _ 0.00378
= 0
2
4
6
8
Blade number,
10
12
Z
14
16
Centrifugal compressors
93
3 :. Q = 0.02932 m Is 3 = 29.32dm /s (li) With the volumetric flow rate solved the Busemann slip factor is easily obtained. Now li2 = 3.351 x 0.0293 = 0.09825
:'O"B = (0.818-0.09825x1.732)/(1-0.09825X1.732)
= 0.6478/0.8298 = 0.7807
(ill) The impeller vane inlet angle
Ie
cot
13{
I
=
\/U1
13; for zero flow incidence is obtained from
= (c r2 /U 2)(r / r I)
= li r 2/r 1 2
= 0.09825/0.45 = 0.2183
I
!
as crl = cr2 = c for this pump. r
13{
Thus,
= 77.68 deg
(iv) The power required is
W P
=
pQgH~
=
pQg(18~ 1755 Q2)/~ /..
= 29.32 x 9.81 x 19.51/0.72 = 7.794 kW
<
/
--
7.8.· A centrifugal pump delivers 50 dIn 3/S of water at an impeller speed of 1,450 rev/min. The impeller has 8 vanes inclined backwards to the direction of rotation with an angle at the tip of P/ = 6ff. The diameter ofthe impeller is twice the diameter of the shroud at inlet and the magnitude of the radial component of velocity at impeller exit is equal to that of the axial component ofvelocity at the inlet. The impeller entry is designed for the optimum flow condition to resist cavitation (see eqn (7.8», has a radius ratio of 0.35 and the blade shape corresponds to a well tested design giving a cavitation coefficient
G.
= 0.3.
Assuming that the hydraulic efficiency is 70 per revt and the mechanical efficiency is
•
90 per cent, determine (l) the diameter of the inlet
(2) the net positive suction head (3) the impeller slip factor using Wiesner's formula (4) the head developed by the pump (5) the power input. Also calculate values for slip factor using the equations of Stodola and Busemann, comparing the answers obtained with the result found from Wiesner's equation.
Solution. (1) At the optimum design condition to resist the_onset of cavitation, eqns
•
(7.8 a), (7.8 b) and (7.8 c) were derived, i.e.,
r
05
1
'" - l2(I+G.) J G.
(a)
(b)
02
3.42k
_
..
(e)
a. (1 +a. )11'
Cd whereiP-an d k - 1-
lJ.,
Using eqn (a), iP _
lJ.,
k
=
=
0 r"
Cx!
lJ.!
= 1C
l-l( r~!)\ r"
=(
(rr..!\)
2
h
11' . ) = 0.3397
03
2 x 1.3
N r" /30
=
10.717 m/s and cz1
-
0.3397 x 10.717 - 3.641 m/s
2
=
1-0.35 2 .0.8775
_ 0.0706 m
0.05 x 0.8775 x
3.641
I :. The inlet diameter d 1 - 14.12 em
(2) From eqn (b) above the net positive suction head at the optimum condition is:
.-.< --)
,
i (3) To determine the values ofthe various slip factors we use :
A. _
cr2
'1"2
U2
_
cz ' 2 xU"
•
3.641 2 x 10.72
=
0.17
dR'
an
"'2
=
600
The Wiesner slip factor, eqn (7.19 b), is:
a -1w
~
Zo.7 (1 -"2 tan tI~)
-1-
0.7072 _ 0.7664 4.287( 1- 0.17 x 1.732)
The Busemann slip factor is expressed as
a _ (A-B~tantl~) _ 0.825-0.17 x 1.732 _~ B
{1-t/>2 tan tl2'}
1-0.17x1.732
(with A =0.825 from Fig. 7.11 and B
•
Vi
= 1)
The Stodola slip factor is given ~ eqn (7.18 a) and is:
~ a, -1-
0.63Jf IZ
. - 0.7218
1- t/>2 tan P 2
(4) From the definition of pump efficiency, eqn (7.20), we have:
gH 7/H-
gH
U2C62 - aU22( I-t/>2 tan tl.) 2
hence H _ a7/ nU: (1-t/>2 tan p~)
_ 0.7664 x 0.7 x 21,433
g
9~1
(5) The shaft power input is found from:
w_pgQH _ 9.81 1) m1)H
x
3
10 x 0.05 x 17.73 _ 13.8 kW 0.9 x 0.7
2
(1- 0.7058) _ 17.73 m
7.9. (a) Write down the advantages and disadvantages of using free-vortex guide vanes upstream ofthe impeller of a high pressure ratio centrifugal compressor. What other sorts of guide vanes can be used and how do they compare with free-vortex vanes?
(b) The inlet of a centrifugal air compressor has a shroud diameter of 0.2 m and a hub
diameter of 0.105 In. Free-vortex guide vanes are fitted in the duct upstream of the impeller so that the flow on the shroud at the impeller inlet has a relative Mach number, Md ,, I,
:.
= 1.0 an absolute flow angle of a l =20°- and a relative flow angle fJI =
55°. At inlet the stagnation conditions are 288 K and Hf Pa.
Assuming frictionless flow into the inlet, determine (1) the rotational speed of the impeller (2) the air mass flow.
(c) At exit from the radially vaned impeller, the vanes have a radius of 0.16 m and a
design point slip factor of 0.9. Assuming an impeller efficiency of 0.9, determine
(1) the shaft power input (2) the impeller pressure ratio.
Solution. (a) See textbook for detailed explanation. (b) From eqn (7.12 a)
I
.
"M:, cos 2fJ,(tan a + tan fJI? kpo,a;, - (1+ t M;, cos 2fJt / cos a r
m Q2 H
I
I
l
2
and with the specified values of M" ~ 1.0, a l
l
=
20° and fJ, = 55°
Qo, -
01
•
.Jy RTo, - ./1.4 x 287 x 288 - 340.2 mis, Qo,
-
.
_ 05 -
(1+t xO.6103
340.2 _ 328 2 mI •
1.03966
From the velocity diagram, Fig. (1), U1 -
Q-
U, 1r" - 337.35/0.1
=
C,
M, _ M,. cos 11, - 0.61039 cos a,
s
cos al(tan a l + tan 11,)
- 337.35 mls
3373.5 radls
:. N _ 30 Q = 32,214 revlmin 7t
I N5tRT F~~. (\)
=r
Hence the mass flow rate is found:
•
,; _ 7t
3
kp01Qg, (rhs) _ 7t x 0.8775 x 1.2098 x ;40,2 x 0.45463 = 5.246 kg/s 3,373.5
Q2
(c) The specific work done on the air is:
0
where cel - £1 sin a l == 200.32 x sin 20 = 68.51 mls and
U2
-
0 Ii - 3373.5 x 0.16 - 539.76 nils
:. AW - 0.9 X 539.76
2
-
337.35 x 68.51- 239,095 Jlkg
. :. P - mAW - 1.254 MW
By combining eqns (7.21) and (7.22) an expression for determining the pressure
ratio of the impeller is obtained, viz.
.L
I.
I
I
. !
I ,! I
&'=(1+ POI
\
1JcAW \j'-1 =(1+ 0.9 X 239,095)"S -6997 CpA Tcn 1005 x 288 -'--
Chapter 8 Radial Flow Turbines .
•
f----l Hozzle blades - - -~:::
---2 Rotor
,'/
•
./ /
// '
-8/:~!~LL
Shro ud
3 '---V ~
4 Diftuser
"<'+----r-' - - ' -
Layout of a 90 deg. Inward Flow Radial Turbine
'-
'-
94
Chapter 8 Radial Flow Turbines 8.1. A small inward flow radial gas turbine, comprising a ring of nozzle blades I a radial vaned impeller and an axial diffuser, operates at its design point with a total to total efficiency of 0.90.
At turbine entry the stagnation pressure and
temperature of the gas is 400 kPaoand 1,140 K.
•
The flow leaVing the turbine is
diffused to a pressure of 100 kPa and has negligible final velocity.
Given that the
flow is just choked at nozzle exit, determine the impeller peripheral speed and the flow oudet angle from the nozzles.
o For the gas assume y = 1.333 and R = 287 J/(kg C). Solution. TA8 fig1::iI e below shows a meridional section of a 90 deg inward flow radial
7-
turbine and diffuser together with the design
point velocity triangles at rotor inlet and
I
rotor outlet.
At this condition the relative
i
velocity: w at rotor inlet is in the ,radial 2 direction and at rotor outlet the absolute
I
velocity c is in the axial direction (Le.
I
Nozzle
blades
a
c g2
=U2 and c g3 =0).
Thus, the specific
work, eqn. (8.4), is f:,W = hOI - h 03
=
U C g2 - U C g3 2 3
=
2 U 2
Referring to the simplified Mollier diagram
(,)
shown below, the total to total efficiency of the combined rurbine stage and diffuser is
7"
= (hOi - h04 )/(hol - h04SS ) = (hoi - h 03 )/(hol - h04SS ) 2 = U2 / [Cp T ol \1 - T04SSIT01)]
After transposing and substituting for the isentropic temperature T of the pressure ratio,
04ss
IT
01
in terms
Radial flow turbines
U2 = 2
'7tt Cp T 01
[1 - (P
04
95
lhJ
Ip 01 )('1- 1
= 0.9x 1149 x 1140 [1 _ (100/400)°.2498)
.0'
6 = 1.179 x 10 (1 - 0.7073)= 0.3451 x 10 6 Hence, the blade tip speed is
U = 587.4m/s. 2 At nozzle exit the absolute flow Mach number is
•
•• (i)
03ss 04ss 4 ss
In eqn. (i) the values of both
az and a are Z
unJatovm and another condition must be used to
3ss
3s
solve the flow angle a . Across the nozzle the 2 stagnation enthalpy remains constant. I.e.
s hoI :.C T p o1
= h oz = C T + p 2
1
Z
2" c 2
. 1 2 2 = C T + 2" U cosec a p Z 2 2
After rearranging,
1 2 2 = 1 - - U cosec a/(C T ) 2 2 p o1 1 2 2 1 - 2" ('I -1)(U/a ) cosec a Ol 2 C T = yRT0/('1 -1) = a 2/ (y-l) P 01 o1
T2/Tol
•
where Using
eqn.
(ii)
(i) (i11)
Combining eqns. (ii) and (iii) and rearranging sin a 2
= (U/a
• (.. ) 1/2 ) } ('I -1) + l/M 2 ol 2
(iv)
Substituting values,
= (1.333 x 287 x 1140)
11
2
-J!
Iiiiiiiiiiiii_--·iiiiiiII· ..- ._-. .......iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiz
S.L. DIXON
96
= 660.4m/s
:. sin a
2
1
= (587.4/660.4) (1 +0.1665)
/z
= 0.9607
Hence, the nozzle flow outlet angle is
8.2. The mass flow rate of gas through the turbine given in Problem 8.1 is 3.1 kg/s, the ratio of the impeller axialwidth/impeller tip radius OJir2) is 0.1 and the no"le Isentropic veloclty ratio (il2) is 0.96.
• 1
!
Assuming that the space' between nozzle exit
and impeller entry is negligible and Ignoring the effects of blade blockage, determine: (i) the static pressure and static temperature at nozzle exit;
(ii) the impeller tip diameter and rotational speed; (iii) the power transmitted assuming a mechanical efficiency of 93.5%. Solution. (i) The static temperature T 2 at nozzle exit can be obtained immediately
!:rom either eqn. (Ii) or eqn. (Iii) in the solution of Problem 8.1. T
2
USing eqn. (ill),
= T o1 U22/(M2aol sina2l
2
= 1140x 587.4 /(660.4 x 0.960i/
= 977.2 K The static pressure P2 at nozzle exit may be fOWld using the nozzle isentropic
velocity ratio, eqn. (8.17),
to determine the isentropic temperature ratio T
1
2
=2'c 2s '
= After
som~
h
ol -h 2
1
2s
IT
01
as follows:-
2
=2'c 2
(hoi - h 2)/(hol - h 2s )
= (Tal - T 2)/(Tol - T 2s)
_
rearranging,
. T2Srroi
= 1 - (I
-._T2rrot/~22
= 1-(1 - 977.2/1140)/0.96
= 0.8450
P2/Pol
=
(T
2s
rr
01
) y/(y-l)
= 0.845 4 . 003 = 0.5096
2
Radial flow turbines
97
= 0.5096 x 400 = 203.8 kPa
(il) Applying the equation of continuity at nozzle exit ,;,
= P2 A 2 cr2
where, from the design point velocity triangle,
•
and the flow area is
Hence, using the gas law,
mRT 2 tan 02
2 2
= 2
. r .. 2
= 0.03999 = 0.20 m
r
2 (b/r 2)
3.1 x 287 x 977.2 x tan 73.88 0 3 2'TTX 203.8 x 10 x 587.4 x 0.1
=
The impeller tip diameter is 40 em. The speed of rotation is
•
= 28,050 rev/min
(iii) The power transmitted taking into account the mechanical losses is
Wt
=
7m m /:;1'1
= 1m
.
2
m U2
=
0.935 x 3.1 x 582.4
2
I. 000 kW
8.3. A radial turbine is proposed as the gas expansion element of a nuclear powered Brayton cycle space pO\~er system.
The pressur~ and temperature
conditions through the stage at the design point are to be as follows:
=
Upstream of nozzles, Pol
= 699 kPa,
N:lzzle exit,
= 527.2kPa, T
p2
Tal 2
= 1,145 K; = 1,029K;
S.L. DIXON
98
P3
Rotor exit,
= 384.7 kPa,
T
3
=
914.5 K, T
03
=
924.7 K.
The ratio of rotor exit mean diameter to rotor inlet tip diameter is chosen as 0.49 and the required rotational speed as 24.000 rev/min.
Assuming the relative flow at
rotor inlet is radial and the absolute flow at rotor exit is axial, determine: (i)
,
tre
total to static efficiency of the turbine;
(ii) the rotor diameter;
•• 1
(iii) the implied enthalpy loss coefficients for the nozzles and rotor
~ow.
The gas employed in this cycle is a mixture of helium and xenon with a molecular
weight of 39.94 and a ratio of specific heats of 5/3.
The Universal gas constanr is,
R = 8.314 kJ/(kg-mol K). o Solution. (i) The tocal to static efficiency of a radial flow turbine is defined, eqn.
(8.6), as
7ts =
(1 - T
= (1 - T
o3IT 0 1)/(1 - T 3 ss /T aI)
IT )/ [I - (P
0301
ip
301
)(y-I)!y]
= (1 - 924.7/1145)/[1 - (384.7/699)°·4] = (I - 0.8076)/(1 - 0.7875) = 0.9054 (ii) At the design ccndition the relative flow at rotor inlet is radial and, from the
Velociry triangle (Fig. 8.3), c g2 = U . As the absolute flow at rotor exit is axial, 2 c g3 = 0, and the specific work /1W = U 2. . 2 2 ... U2 = hal - h03 = Cp(TOI - T 03 ) (i) The Universal gas constant R = Rm where m is the molecular 'weight' of the gas a
mixture.
Thus, R Cp
2
..'U2
= R/m = 8314/39.94 = 208.2 J/(kgOC) = yR/(y -I) = 2.5 x 208.2 = 520.51/(kg OC) = 520.5 (1I45
- 924.7)
=
11.47 x 10 4
Radial flow turbines
99
U = 338.6 m/s 2 Hence, the rotor tip diameter is 4 D = 60 U/(1I N) = 60 x 338.6/(rrx 2.4 x 10 ) 2 = 0.2694 m
(iii) The enthalpy loss coefficient for the nozzles, eqn. (8.16), is determined as
follows:-
•
= (h2 - h 2s)/(2"1 c 22) = (T2 - T 2s )/(Tol - T 2) = [T 2 - (T2sfTOl)Tol )/(TOI-T 2) = [T 2 -T ol (P/Pol)
(Y-l)!"Y}
/(Tol- T 2)
J
= [1029 - 1145 (527.2/699)°· 4 /(1145 - 1029)
= (1029 - 1022.8) /(1145 - 1029) = 0.05316
-'I'he enthalpy loss coefficient for the rotor is defined, eqn. (8.20), as
rR
1 2 = (h3 - h3s )/(2" w3 )
(ii)
The enthalpy loss h - h is determined using the specified static pressures and 3 3s temperatures) h3 - h3s
•
= C ( T - (T3sfT2)T2 ] 3 p
-_ Cp [ T 3 - T 2(P/P2) (Y-l)/Y] _
= 520.5 [914.5 - 1029(384.7/527.2)°·4) = 3.831 kJ/kg
From the rotor exit velocity triangle w
2 2 2 = c +U 3 3 3
where U 3
2 Substituting for U using eqn. (i), the relative kinetic energy at rotor exit is, 2 2 }w3 *p[T03-T3+}(r/r2)2(Tol-T03)]
=
f. ~
100
S.L. DIXON
~
520.5[924.7 - 914.5 +
~
19.07kJ/kg
tx
2 0.49 (1145 - 924.7)] The coo1i:
Substituting these values into eqn. (ii),
rR
~ 3.831/19.07 ~ 0.2009
where Tn
the specii
8.4. A film-cooled radial inflow turbine is to be used in a high performance open Brayton cycle gas turbine.
The rotor is made of a material able
to" withstand
ment
a
temperature of 1145 K at a tip speed of 600 m/s for short periods of operation. where
Cooling air is supplied by the compressor which operates at a stagnation pressure ratio of 4 to 1, with an adiabatic efficiency of 80%. When air is admitted to the compressor at a stagnation temperature of 2B8 K.
Assuming that the effectiveness
of the film cooling is 0.30 and the cooling air temperature at turbine entry is the same as that at compressor exit, determine the maximum permissible gas temperature at
The spec;
entry to the turbine.
radial tur
Take 'I
= 1.4 for
the air.
Take 'I
= 1.333
= 175.9 for the gas entering the turbine.
Assume
k
is considf
R = 287 J/(kg K) in both cases.
which flo\
Solution. The sketch shows a Brayton gas turbine cycle in the form of a Mollier diagram.
The compressor
purposes. h
cooling.
raises the stagnation pressure of the air from Pol to p02 at an adiabatic (total to total) efficiency
!. ::'l
"."1
I
"Ie of 0.80.
•0'
From the definitien of efficiency, eqn. (2.28e),
"-
\
7c = (h02s - ho1 )/(h02 - ho1 )
I
tk\l£l t£.F"i".
1
oressure
;
11-
= TOl(T02sIT01-1)/(T02-T01) i
1
~T01[(Po/Po/'I-l)/'i-l]/ (T02 -
T
ol )
L.-.----=-'--
Hence, the actual stagnation temperature at compresSor exit is,
T 02
1
= T ol [1+ {(po/Po/'I- )/'i -I}
1 3 5
= 288 [1 + (4 / .
- 1)/0.8
J=
\x
=:~I
/;;>cl
288 [1 + 0.486/0.8]
~s;__'c'[rr~:y)(lf;l) Q PI L
-
C:~1?i (fvo-(7WL
~
~- .=-----~--~--~===:==-._=-=Radial flow turbines
101
= 463.0 K The cooling effectiveness
E ,
eqn.Ja,....4-r), in the present notation is
I\. l;W/(2C )]/[T 3- T 2- ().W/(2C)]
€ = [T 3-T o m
p
0
p
0
where T
is the maximum permissible temperature of the rotor material and ljW is 2 the specific work done, assumed to be equal to u . Hence, after some rearrange .. 2 ment m
= (T m -
=
where
•
~T 0 2)/(1- Eo)
2
+ U,_ /(2Cp )
.
yR/(y-l)
= 4.003 x 287 = 1149 J/(kg
0
C)
2
= (1145 - 0.3 x 463)/0.7 + 600 /(2 x 1149) = 1594 K
,
The specific work done by the turbine, l;W = U - = 360 kJ/kg of gas admitted to the 2 t radial turbine. The specific wark required by the compressor, AW = C (T 2 -T 1) cpo 0 = 175.9 kJ/kg of air compressed (i.e. only about half of AW ). This is feasible if it t
is considered that the rate, of ma'55 flow through the compressor is about twice that
which flows through the turbine, the excess compressed air being used for other purposes. cooling.
Only about 10'70
or this diverted compressed air would be needed for
film
At outlet from the turbine the hot gases would still have a stagnation
pressure p 04 in excess of Pol'
Assuming the turbine has an adiabatic efficiency of
0.9 it is easy to show that p 2/ P 4 = 2.682 and, therefore, P /p I is 1.492. o 0 04 0
This
excess pressure ratio would be expanded through a second turbine stage to provide a
•
net power output . 8.5. The radial inflow turbine in Problem 8.3 is designed for a specific speed 0.55 (rad).
J'2
s
of
Determine,
(i) the volume flow rate at rotor exit and hence obtain the turbine P9wer outpy (li) the rotor exit hub and tip diameterj6
-
(iii) the nozzle exit flow angle and the rotor inlet passage width/diameter ratio, b/D2 · Solution.
(i) Specific speed is defined, c.f. egn. ~), as
/
S.L. DIXON
102
~
=
3/4 SJ.Q 31/2/Ah L> os
3 where Q is the volume flow rate at rotor exit, (m Is), fI, is the angular speed of 3 the rotor (rad/s) and t. h is the isentropic enthalpy drop between inlet stagnation . os conditions and exhaust static conditions, h 1 - h • This last definition conforms o 3 ss with the use of total to static efficiency in Problem 8.3. Thus, C>h
os
= ~c 2 = h 2 0
= C T
P 01
01
- h = C (T - T ) 3ss P 01 3ss
[1" (p
3
Ip
01
)(Y-I)/'{J .
= 520.5 x 1145 [1 - (384.7/699)°.4]
S2
= 126.6 kJ/kg
= 2Tr N/60 = 800TT = 2513 radls
Hence, from the specific speed definition, 112 = 3
Q
... Q
3
4 14 Llh 3/4 = (0.5512513) (12.66x 10 / = 1.469 os 3 = ~m Is
(S2s 1Jl)
~
The turbine power is
W t
=m
~W
·22 = P3Q" U2
= mU 2
= 384.7 x 1031(208.2 x 914.5) = 2.021 kg/m 3 :. m = P Q = 2.021 x 2.159 = 4.363 kg/s 3 3 2 ".W = 4.363x338.6 = 500kW
I
:.
t
(ii)
.'
where,
H c
2
3
:. c
3
=
D -D , 3h 3t
Om = (D
3t
+D
3h
)/2 = 0.49xD
2
= O,132m
= 2C (T03 - T ) = 2 x 520.5 (924.7 - 914.5) = 10620
p
= 103
3
mig
Hence. the rotor exit blade height is
, 2·15'8
'.-.
.
-=== === - .. :: .. ::'. ..'
.
.
103
Radial flow turbines
= 0.1011 m
:.03h
= 03m - H/2 = 0.132 - 0.0506 = 0.0814 m
:. 03t
= 03m + H/2 = 0.1826 m
The rotor exit hub/tip ratio 03h/03t = 0.4458 and the ratio 03/02 = 0.6i8. (iii) As hoI = h
02
c 22
•
' then -- 2C p'< I'"r" 01
-
4 T) - 2 x 520.5(1145 - 1029) = 12.08 x 10 2-
The nozzle exit velocity is C
2
= 34i.5 m/s
From the inlet velocity triangle sin 02 "
.
Q
2
= Uzlc = -'i.O
= 338.6/34i.5
2
= 0.9i439
o
~---
From the equation of continuity, m
= P2 A Z c
2
r2
= 7J' P (b2/ D2) D2 Uz cOt
2
. 2 = (mRT tan Q2)/(rr P D U ) 2 2 2 2
4.363 x 208.2 x 1029 x tan nO 2 3 tr x 52i.2 x 10 x 0.2694 x 338.6
•
= 0.0995
=
Q
Z
8.6. An inward flow radial gas turbine with a rotor diameter of 23.76 cm is designed to operate with a gas mass flow of 1.0 kgls at a rotational speed of 38,140 rev/min. At the design condition the inlet stagnation pressure and tempecarure are to be 300 kPa and 727 0 C. The turbine is to be "cold" tested in a laboratory where an air supply is available only at the stagnation conditions of 200 kPa and 102 0 C.
(al Assuming dynamically similar conditions between those of the laboratory and the projected design determine, for the "cold" test, the equivalent mass flow rate and the speed of rotation. Assume the gas properties are the same as for air.
(b) Using property tables for air, determine the Reynolds numbers for both the hot and cold running conditions. The Reynolds number is defined in this context as:
where llJ1 and flol are the stagnation density and stagnation viscosity of the air, N is the rotational speed (rev/s) and D is the rotor diameter.
Solution. (al From similarity considerations and using eqn (1.15)
•
(D ~;Po ).J RJ;l1
= constant
1
. .f
... mc = m I,
N
= - - x -200 = 1.089 his POll, 375 300
Ta,c
As IT = constant, then Nc = N
"Ta.
p.ooo
P
.:Jl11l..!:9J..<.
~ -2!£.. =38,140 Rl£75 - - = 23.356 rev/min
"
To
I h
1,000
(b) From gas tables, e.g. "Thermodynamic & Transport Properties ofFluids", Rogers & Mayhew (1995), values of the dynamic viscosity for air are obtained. At T 01 =1,000 K, j/. = 4.153 At Tal = 375 K,
X
10" kg/ms
f/. = 2.181 xlO" kg/ms
f.~IND' flol
3
= POIND' = 300 X10 x 635.7 x 0.238' = 9.063xlO' 3 s
287 X 10 x4.153 xl0-
RTQlJ()1
where N. = 635.7 revls and N, = 389.3 revls
•
3
Re = 200 x 10 x 389.3
,
X 0.238' = 1.879 x 10 ' 287 x 375x2.181xl0- s
8.7. For the radial flow turbine described in the previous question and operating at the prescribed "hot" design point condition, the gas leaves the exducer directly to the atmosphere at a pressure of 100 kPa and without swirl. The absolute flow angle at rotor inlet is 72 a to the radial direction. The relative velocity W J at the the mean radius of the exducer (which is one half of the rotor inlet radius r2 ) is twice the rotor inlet relative velocity w2• The nozzle enthalpy loss coefficient, ,= 0.06.
•
Assuming the gas has the properties of air with an average value of temperature range) and R = 287 J/kg K, determine
(1) the
to~- to-static efficiency of
the turbine
(2) the static temperature and pressure at the rotor inlet (3) the axial width of the passage at inlet to the rotor (4) the absolute velocity of the flow at exit from the exducer (5) the rotor enthalpy loss coefficient
y = 1.34 ( this
(6) the radii of the exducer exit given that the radius rntio at that location is
004.
Solution. (1) From eqn (8.11) the isentropic total-to-static enthalpy drop is related to the total-to static pressure rntio:
,
"oi
'.
vR iJ = ~ =1131.1 Jlkg K. Also .n =- N =3,994 radls P y -1 30
where C
,'. V,
=.n 'i =3,994 x 0.1188 =474.5 mls.
Fora"nominaldesign",DW" V~ =225,138m'/s'
r ... Co = 2 x 113.1x 10'll-
... 11,=
DW hOI -
11,,,
=
225,138 275,160
(1\0. '3) 1= 741.8 mls 2537
=0.8182
!
I
I
(2) From the velocity and h-s diagrams, Fig. (a), in Problem 8.1, we have
c, = V,I sin l'l', = 474.51 sin 72 0 = 498.9 mls
T. = T. ,
01
c' 2C
_...:.L = 1000 P
'
-
498.9' " 890K 2 x 1131.1 - -
It,. - h,., = '21 C,)' 'bN and so
hOI
1 '(1 + ~ 1') -h,., = '2c,
498.2' x 1.06 ]3.941 - [1 - 0.61433 -2 x l131.1x1,OOO -
•
.'. p, = 184.3 kPa
(3) From the equation of continuity m = I'lC"A. where A, = :n.D,b, the axial width of the passage b 2 can be den ved.
,.q
p -
=-' = RT"
184,3 X 10 3
m
,'.b,=
•
3
. = 0.7215 kg/m and c" = C, cos t; = 154 m/s 287 x 890
l'lc,,:n.D,
1
= ff
_ =O.01206m x 0.7215 x 154 x 0.2376
.., b 2 -12.06 mm
.
(4) At exit from the-exducer the continuity equation is written as m = f~ CX3~' Referring to the velocity diagranl, Fig, (a) in Problem 8.1, we obtain:
-U
C x3
= Jw~
u,
=.0 r3 = 3994 x 0.0594 = 237.24 m/s , and,
3'
where
W3
= 2 X w, = 307.9 m/s ,
tl =
(£A)
sin ~ c
= sin' ,(237.2)' 3m.9 = 50.4 deg.
3
... <; = cx3 = ,1308' - 237.24' = 196.26 mls
(5) The turbine total-ta-static efficiency is described in eqn (8.10) as:
i.
which can be re-arranged to give
I
All of the data needed are available in the earlier calculations. Substituting, we get:
_ (1- 0.8182) x2 x2.7516 x 10-' - (196.26 2 bR 3082
... bR
-I-
0.06 x 498.92 , )
=0.4915
(6) Using eqn (8.2 c)
T. (4745)'[ 1-0.10557-1- ( 05 ... J=I-0.17x--· . )') =0.8808 T, 585 1.2088
,', T J =784K and P = 3
m
1x10 5
287x784
=0.4445
1
A,=--= =0.01146m' RCX3 0.4445 x 196.26
With
r.Jr" r., =
•
= 0.4, the values of the hub and tip radii are evaluated, i.e.
2.636 em and
r"
= 6.59 em
8.8. One of the early space power systems built and tested for NASA was based on the Brayton cycle and incorporated an IFR turbine as the gas expander. Some of the data available concerning the turbine are as follows:
Total- to totaLpressure ratio (turbine inlet to turbine exit), POI fp OJ
PO/Pl = 1.613
Total-to-static pressure ratio, .
•
=1.560
Total temperature at turbine entry,
T OJ = 1083 K
Total pressure at inlet to ulrbine,
T o/=91 kPa
Shaft power output (measured on a dynamometer)
p." = 22.03 kW
11
Bearing and seal friction torque (a separate test),
D, = 15.29 em
Rotor diameter, Absolute flow angle at rotor inlet,
(1,
based on total-to-static pressure ratio)
0
r/r, =0.35
The hub to shroud 'radius ratio at rotor exit,
(CO
=72
a; = O·
AbsoJ ute flow angle at rotor exit,
Ratio of blade speed to jet speed,
=0.0794 Nm
I,r
=U,f =0.6958 Co
For reasons of crew safely, an inert gas argon (R = 208.2 J/(kg K), ratio of specific heats, ]I = 1.667) was uscd in the cycle. The turbine design scheme was based on the concept of optimum efficiency.
Detennine, for the design point,
,
I
(l) the rotor vane ti p speed
I i
(2) the static presfure and temperature at rolor exit
.~
(3) the gas exit veloci ty and mass f10w rate
j
I
.-.
•
(4) the shroud radius at rotor exit
(5) the relative flow angle at rotor inlet (6) the specific speed
N.B. The volume now ra'e to be used in the definition of the specific speed is based on the rotor exit conditions.
Solution. (1) For Argon
.J....: = 5/3 ]'-1
5/3-1
= 2.5 and C
=2.5 x R =520.5 J/kg
p
For the total-te-static exp'msion,
~
•
]-'
To, =(pC'! )--; = 1.613 0A =1.2108 T.,.,
r 3",.
II
p"
=
7.,
1.2lO8
Co = J2C.(To,
= 894.5 K
- r,,,) = ~2 x 520.5 x (1083 - 894.5)
= 443 mls
,'.
N
U~
=
Co = 0.6958 x 443
J.'
=308.24 mls
=60U, = 60 x 308.24 7JD~
7J x
'" 38,502 rev/min (and 0.1529 .
n = 4032 rad/s)
(2) We calculate the conditions at rotor exit as follows:
P, '"
.
P
91
01
(Po,/p,)
'" - -
1.613
=56.42 kPa
• where b.hg" = hOI
l(1- (Em) ~I] '"
520.5 x 1083 x (1-
POt}
1 1.56
04) = 91,856 J/kg
... &to = 0.88 x 91,856 '" 80.84 kJ/kg
... To> = 1083 - 80.84/05205'" 927.7 K
•
927.7 1.034 .
(
)04 '" 915.4 K
where
(3) The absolute exit velocity is given by:
C3
= C;d
=J2C (TO) -1;) '" .J2 x 520.5 x (927.7 - 915.4) = 113.6 m/s p
I !
I
The power developed by the rotor is equal to the measured shaft power plus the bearing and seal friction power, i.e.
I
p = pmrtU +
n
1:, = 22,030 + 0.0794 l x 4031.9 = 22,350 W J
,
•
P
22.35
,', m = == 0.2765 kg/s b.ho 80.84
Ie
(4) The gas density at exit is:
I,
·1
P3 _ 56,417 = 0"96k I 3 P3 -_ RT, - 208.2 .- g m x 9l5.4
I
I
~ =g
c...
= 0.9341 = 8.255 x 10-3 m' =:IT 1j',(1- )",)
113.2
I
I
(3
/:r 1 - [.
;<)
=O.05472m (or =2.15 inch)
This calculated size is in agreement with that in the original NASA report, i.e. 2.147 inch
I
I
(5) Fig. (1) shows the velocity triangle at inlet to the rotor. The flow thus represents
the optimum inlet condition rather than the nominal condition.
As LlW
c = "
= M. =Uc &
C f/.
tan a',
then c'"
=I1ho I U, =262.24 mls
= 262.2 = 85.2 m/s tan 72·
. j
(6) The specific speed is obtained from the expression:
.Q =.Q Q
•
•
j j j j j j j j j j j j j j j j j j j j j j j j j j J
Chapter 9
I
Hydraulic Turbines
I
i! '."'1
.;:~~
i . ,
i
I I
;
..
,Je }i :i
I I Computer-generated drawing of a Compact Kaplan turbine system for low head applications, (courtesy Sulzer Hydro Ltd, Zurich)
CIIAP"FER 9
9 .1. A generator is driven by a small, single jet Pelton turbine designed to have a power specific speed .fJ,p = 0.20. The effective head at nozzle inlet is 120 m and the nozzle velocity coefficient is 0.985. The runner rotates at 880 rev/min, the turbine overall efficiency is 88 per cent and the mechanical efficiency is 96 per cent.
•
If the blade speed to jet speed ratio,
I.'
=0.47, determine
(1) the shaft power output of the turbine
(2) the volume flow rate (3) the ratio of the wheel diameter to jet diameter.
Solution. (1) With eqn (9.1) the power specific speed is
where
•
n = - :n N = 92.15 rad/s. 30
nsp
=
n.JPTf (gH,)'
Hence,
P=
p[n~p (gH,t1' = 103[~(9.81.x120t15]' = 224,000 W 92.15 .
.·,P
=224kW
,~
(2) The overall eflJciency is given by
henceQ =
P
pg ~loH,
=
;OJ, =
P/ (pgQH,)
224
X
103
l~ x 9.81 x 0.88 x 120
= 0.2162 m 3/S
I
i
(3) The jet velocity is given by:
C,
= K~2gH. = 0.98s-J2 x 9.81 x 120 = 47.79 mis
The blade speed is: U =
I.'C,
== 0.47 x 47.79 == 22.46
mls
2U 2 x 22.46 04875 D=-= ==. m t:l 92.15
. J!Q =
:n: 4
As Q = -d'c, then d ==
I .j
I !
D
,'. -
d
==
--
4 x 0.2162 = 0.07589
:n:c,
;'TX 47.79
6.423
--
9.2. (a) Water is to be supplied to the Pelton wheel of a hydroelectric power plant by a pipe of uniform diameter, 400 m long, from a reservoir whose surface is 200 m vertically
.,
above the nozzles. The required volume flow of water to the Pelton wheel is 30 m3 /s.
?~i·
If the pipe skin friction loss is not to exceed 10% of the available head and f = 0.0075, determine the minimum pipe diameter.
.. -'
.. '
I
(b) You are required to select a suitable pipe diameter from the available range of stock sizes to satisfy the criteria given. The range of jiameters (m) available are: 1.6, 1.8,2.0,2.2,2,4,2.6, 2.8. For the diameter you have selected, determine
!
(l) the friction head loss in the pipe
i
) i
!
(2) the nozzle exit velocity assuming no friction losses occur in the nozzle and the water
f
leaves the nozzle at atmospheric pressure
3 (3) the total power developed by the turbine assuming that its efficiency is 75% (based upon the energy available at turbine inlet).
Solution. (a) Using eqn (9.6) with the pipe flow velocity, V = 4Q I(:n d
5
=
hence d
2flQ' ~z g(":"" h
4
•
5
,",d =
,", d
2 x 0.0075 x 400 x30 9.81 X(:Jl14Y x20
Z
)
f
2
=44.62
=2.138 m
(b) (1) Choosing d = 2,2 m from the available stock sizes, the pipe flow velocity is:
V = QI A =
•
4x30 = 7.892 m/s :nx2.22
2jlV2 2 x 0.0075 x 400 x 7.892 2 , h =-- = = 17.32m f gd 9.81 x 2.2
(2) The available head, h.
V '" .J2gh.
=200 - 17.32 =182.68 m, thecQrresponding velocity is: = 59.87 m/s
(3) The power developed is,
4
omgh. = 1JI-'Qgh. = 0.75 x30x 10
3
P = 11
x9.81 x 182.68 =40.3
X
lOS
.'. P =40.3 MW
9.3. A multi-jet Pelton turbine with a wheel 1.47 m diameter, operates under an effective head of 200 m at nozzle inlet and uses 4 m3/s of water. Tests have proved that the wheel efficiency is 88 per cent and the velocity coefficient of each nozzle is 0.99.
Assuming that the turbine operates at a blade speed to jet speed ratio of 0.47, determine
(1) the wheel rotational speed
(2) the power output and the power specific speed (3) the bucket friction coefficient given that the relative flow is deflected 165 0 (4) the required number of nozzles if the ratio of the jet diameter to mean diameter of the wheel is limited to a maximum value of 0.113.
Solution. (1) The jet velocity is;
VI = Cv.J2gH, = 0.99.J2 x 9.81 x 200 =62 mls :.,
.:::
and the blade speed is:
U =:.
n r = 00/2 = IN
j
= 29.147 mls
n = 2U / D = 2 x 29.147/1.47 = 39.66 radls N
=378.7 rev/min
1,~<7gQH,
(2) The power is P =
= 0.88 x 9810 x4x200 = 6,906,240
,'. P - 6.906 MW
0.5
'P
l(gHS
0-
=
p
:[[
.asp
= 39.66 x 6906°·5
15
(9.81 X 2(0)'15
=3295.6 =0252 ;mol 13,058
-'-
(3) From eqn (9.4), the runner efficiency is:
•
1'R I
= 21{\ - I.'V\ J' -kcos ,t/.'I ~-
k cos
jl=
I
1-
m
1
2 !.' (1-
I,')
= 1-
0.88 2 x 0.47 x 0.53
= -0.7664
-0.7664 . k= = 0.793 ,. -0.966--
•
d (4) As -l... D
=0.113
N=
then d j
4Q pd~Y;
=0.1661 m
=
4x4 P
2
x 0.1661 x62
=2.977 _
Thus, the choice is clearly N = 3
9.4. A four-jet Pelton turbine is supplied by a reservoir whose surface is at an elevation of 500 m above the nozzles of the turbine. The water flows through a single pipe 600 m long, 0.75 m diameter, with a friction coefficient,f
=0.0075. Each
6 nozzle provides ajet 75 mm diameter and then=le velocity coefficient K N = 0.98. The jets impinge on the buckets of the wheel at a radius of 0.65 m and are deflected (relative to the wheel) through an angle of 160 deg. Fluid friction within the buckets reduces the relative velocity by 15 per cent. The blade speed to jet speed ratio, t,· =
0.48 and the mechanical efficiency of the turbine is 98 per cent.
Calculate, using an iterative process, the loss of head in the pipeline and hence, determine for the turbine (I) the speed of rotation (2) the overall efficiency (based on the effective head)
•
(3) the power output (4) the percentage of the energy available at turbine inlet which is lost as kinetic energy at turbine exit.
Solution. (I) A simple iterative process is used to determine the frictional head loss.
Guess HE = 480 m
(i.e. hi' = 20 m)
c, = Kv .J2.gH e =0.98J2xO.98x480 = 95.1m1s
I
I.
j :1
With c,
=95.1 mis, then
V
=3.804 mls and h I = 1.223 x 33.804' = 17.7 m
Now, repeat the above calculation, this time with hi = 17.7 m. ,', He = 482.3 m
c, = 95.33 mis, V = 3.813 mls .., hi = 1.223 x 3.813' = 17.8 m This result shows that the calculation has converged sufficientlv.
7 (1)
The blade speed is U = I.' CI
.Q
=0.48 x 95.33 =45.76 mis
=U / r =45.76/0.65 =70.37 md/s
.'. N - 672.2 rev/min
(2) The overall efficiency is 110 = 1im 11. ;'!N
where 1/"
•
1JR
=2 I.' (1 - 1'7(1 - k cos
~!o =
•
11m
in =
=0.98
and
2 x 0.48 x 0.52 x 1.7987 =0.8979
0.8451
(3) The output power is:
Q
=K,~ =0.9604,
P = 1!opgQH£ where
=4<;A) =4 x 95.33 x ff x 0.075
2
/4 =1.6846 m 3 1s
... P = 0.8451 x 9,810 x 1.6846 x 482.2 = 6.735 x 10 6
The power output is 6.735 MW
(4) From the velOCIty diagram, Fig. (1),
W,
WI
=c
I -
= 0.85 X WI = 42.136 mls and
U
=95.33 - 45.76 =49.57 mls
c,
=~U' + w~ - 2Uw, cos a; =.J2094 + 1775.4 - 3623.7 =15.66 mls
Let R., = fraction of the energy available at turbine inlet which is lost as k.e. at exit, then
R
.,
.1 x ?453 ='1.' c, = " _. =0.0259
gH g
9.81 x 482.2
= 2.59 per cent
•
I
1
9.5. A Francis turbine operates at its maximul)l efficiency point at 110
=0.94,
corresponding to a power specific speed of 0.9 rad. The effective head across the turbine is 160 m and the speed required for electrical generation is 750 rev/min. The runner tip speed is 0.7 times the spouting velocity. the absolute flow angle at runner entry is 72 deg from. the radial direction and the absolute now at runner exit is without swirl.
I
Assuming there are no losses in the guide vanes and the mechanical efficiency is 100 per cent, deterrni ne,
':'~. -:r:
(1) the turbine power and the volume flow rate
."
(2) the runner diameter (3) the magnitude of the tangential component of the absolute velocity at runner· inlet (4) the axial length of the runner vanes at inlet
9. ..
Solution. (1) USIng eqn (9.1),
nSF = (nJPTj1 ',SI' gH
where
E)
",p
n =- :11 N = 78.54 radls 30
= 12.82 MW 6
andQ = __ P_ =
11PgHE
•
12.82 X 10 = 8.69 m 3/s 0.94>: 9810 x160
(2) The blade speed is V, = 0.7c o = 0.7.j2gH E = 39.22 mls
,'. D, = 2V, = 39.22 x2 = 0.9987
n
(3)
78.54
10 ... D ,-.m
P = pQI:!.W = pQV,C/lJ. as c~ = 0
6
•
P 12.82 '< 10 c = = 37.6 mls /lJ. - pQV, Ie? x 8.69 x 39.22
(4) As Q = 2:11r,b,c" and c" = c/I2l tan
... b,
a; = 37.61 tan n° = 12.22 mls
= QI (2:n r,c,,) =8.69/(2:11
x 0.5 x 12.22) = 0.226 m
9.6. The power specific speed of a 4 MW Francis turbine is 0.8, and the hydraulic efficiency can be assumed to be 90 per cent. The head of water supplied to the turbine is 100 m. The runner vanes are radial at inlet and their internal diameter are three-
10
quarters of the external diameter. The meridional velocities at runner inlet and outlet are equal to 25 and 30 per cent respectively of the spouting velocity. Determine, (1) the rotational speed and diametcr of the runner
(2) the flow angles at outlet from the guide vanes and at runner exit (3) the widths of the runner at inlet and at exit. Blade thickness effects can be neglected.
•
Solution. (1) From eqn (9.1) the rotational speed is:
=
0.8(9.81 x 100)'" ~
,,4,000
I
=69.44 rad/s
30 . N = -.0 = 663.2 rev/min :n
To determine the runner diameter we need to determine the blade speed, which is obtained as follows. With
C lJZ
= 0 then VI = COl
=
.J 11 gHE
= 29.71 m/s
D = 2V, = 2 x 29.71 = 0.8558 m I .0 69.44
(2) The spouting velocity is
Cd
Co
= .,J2gHE = 44.29 mls , thus
= em! = 0.25 x Co = 11.07m1s and cr2 = O.3xc o = 13.28m1s
II
111
. (c) = tan" ~ = tan '1(29.71) - - = 69.55 de!!.
j ~} =
11.07
C"
tan
.'(!!J..) -_0.75 x29.71 -- 592. dego c"
13.28
(3) At runner inlet the width is
=
2x
•
b2 =
Q 2:/1 r,c"
=
:/1X
4.53 = 0.1522 m 0.4279 x 11.07
4.53 I. = 0.1&92 m 20x 0.75 x 0.4279 x 13.28
9.7 .(a) Review, brief1y, the phenomenon of cavitation in hydraulic turbines and indicate the places where it is likely to occur. Describe the possible effects it can have upon turbine operation and the turbine's structural integrity. What strategies can be adopted to alleviate the onset of cavitation?
•
(b) A Francis turbine is to be designed to produce 27 MW at a shaft speed of 94 rev/min under an effective head of 27.8 m. Assuming that the optimum hydraulic efficiency is 92 per cent and the runner tip speed to jet sp~eed ratio is 0.69, detennine
(1) the power specific speed (2) the volume f10w rate (3) the impeller diameter and blade tip speed.
(c) A 1110 scale model is to be constructed in order to verify the performance targets of the prototype turbine anel to determine its cavitation limits. The head of water available for the model tesls is 5.0 m. When tested under dynamically similar conditions as the prototype. the net positive suction head H s of the model is 1.35 m. Determine for the model,
(1)
the speed and the volume now rate
(2) the power output, corrected using Moody's equation to allow for scale effects (assume a value for n
=0.2)
(3) the suction specific speed, .aSS
•
(d) The prototype turbine operates in water at 30 0 C when the barometric pressure is 95 kPa. Determine the necessary depth of submergence of that part of the turbine mostly likely to be prone to cavitation.
Solution. (a) Cavi lalion, which is local boiling, can occur in the low pressure conditions prevailing in the exit region of hydraulic turbines. Boiling commences when the static pressur~ of the water is reduced to the vapour pressure corresponding to the temperature of the water. It is found to occur on the suction surfaces of the turbine runner blades at or near outlet and can cause severe erosion of the blade surfaces. The best remedy requires an increase in the pressure at turbine exit, usually . I
..
by deeper submergence of the turbine. This remedy can entail some increase in civil engineering costs because of the bigger excavation needed and, because of the complete submergence of the turbine, increased problems of access for inspection or maintenance.
Under off-design operating conditions of the turbine another type of cavitation can occur in which a ropelike, twisting vortex of vapour is generated along the axis of the
13 diffuser from the turbine exit. This rotating vortex can cause a heavy vibration and lead to damage of the diffuser. It can be prevented or lessened by air injection along the axis through the tail spinner of the turbine.
:J1
(b) (1) With the given data and .0. = 30 N = 9.844 radls, the power specific speed is:
10' ~03 = 1617.5 = 1.459 'f'O..d (9.81 x 27.8Y 1108.3-
0. = 0..J'PT? = 9.844.J27 sp
•
•
H \4 ( g F.)
X
It is also worth noting that the specific speed, .0.s ' is
(2)
The volume now mte is,Q
(3)
Now c1 =
Co
P
27x10'
11PgHE
0.92 x 9810 x27.8
=-.:....-- =
= 107.6 m 3/s
= .j2gHE = 22.49 m/s. so the blade tip speed and diameter are
u" = 0.69 x 22.49 =15.52 mls
D, =2U,/,O = 3.153 m
(c) (1) For the model turbine, from considemtions of similarity, we have:
'.
gH
(ND)'
= constant
,'. N m
N = D (H )~ = 10 x (-5-)i = 4.241
,', ---!!!.
Np
-p
---!!!.
Dm H p
=4.241 x94 = 398.7 revlmin
27.8
14
-lL-3 =
constant
ND
.'. Qn = 0.456m 3/s
= N m)3( Dm)' = 4.241 Np Dp
II
Pm
As
(2)
P.
.. , Pm
3X(.l)5 = 0.7628 X10-3 10
=0.7628 X 10-3 X 27 X 10 6 =20.6 kW
Correcting this result to allow for the effects of scale (i.e. size!) using Moody's empirical formula, eqn (9.23), viz.
I
1- _ 11,p= _
(D)5
1 - l'fm
I\, Dp
.....:...t!l.
...
0.08 1-
(
= ,0.1
11m
)02
= 0.631 and so 11m
=0.873
Thus, the corrected power output for the model turbine is:
.
0.873 0.92
P = - - X 20.6 = 19.55 kW m
(3)
The suction specific speed of the model is:
j, j
D ss =
DJ2o.5
41. 75
(gHsF
= (9.81
X
0.4560.'
x1.35t"
= 4.06 (rad)
N.B. This result is in agreement with data quoted by Wislicenus (see eqn (1.12 b».
/5
(d) From the tables. (e.g. "ThemlOdynamic & Transport Properties of Fluids", by Rogers & Mayhew (1995», for water at 30° C, the vapour pressure p, = 4.242 kPa. From the definition of the suction specific speed, a value for the net positive suction head H s can be found:
i
1 ( - .0)3': Hs =Q3 = - 1 g 9,81 ss
,n
4 X
(9.8122)3 (105.3)3.. = 7.504 m 4.006
H s is the amount of head needed to avoid cavitation at turbine exit and, from eqn
•
(9.24), is equal to
(Po - p, )/(pg) - z . where z
is the elevation of the turbine exit
above the surface of the tailrace, as shown in Fig. (1). Thus,
z=
Pa - Pv
pg - H s "
3 (95-4.242) x 10 7504 -. "1.748 m 10' x 9.81
Clearly, z. is positi\'e (as dra,vn) when
(Po - p, )/(pg)
> H s . The height determined
here for z represents a limit which. if exceeded, would allow cavitation to occur.
•
9.8. The preliminary design of a turbine for a new hydro:electric power scheme has under consideration a vertical-shaft Francis turbine with a hydraulic power output of
200 MW under an effective head of 110 m. For this particlJlar design a specific speed,
.D, = 0.9 (rad), is selected for optimum efficiency. At runner inlet the ratio of the absolute velocity to the spouting velocity is 0.77, the absolute flow angle is 68 deg and the ratio of the blade speed to the spouting velocity is 0.6583. At runner outlet the absolute flow is to be without swirl.
Detennine:
16 (1) the hydraulic efficiency of the rotor
(2) the rotational speed and diameter of the rotor
(3) the volume no\\' rate of water (4) the axial length of the vanes at inlet.
I
Solution.
Data: P
=200 MW, HE =110 m
U2 1Co
=0.6583
.Qs = 0.9
I
,
(1) The hydraulic efficiency is determined from:
•
Uc
tlH = ~ and so U 2 and C /12 are required gH E
Co
=J2gH £ =46.456 mls
... U2
C2
= 0.77 x Co = 35.77 mls
,',
=0.6583 x 46.456 =30.58 m/s
C82
= c2 sin
172
=33.166 mls
.., ;" = 30.58 x33.166 = 0.94 HI
•
9.81 x 110
--
-
5
5
= ["~s ''.Ii 05(gH E )'4 = 0.9 x 0.949.81x 110 0.5 (
t:0; .., N
(2 x 105f
= 115.2 rev/min
and D2 = 2U2 /.Q = 2 x 30.58 /12.068 = 5.068 m
1
)'
= 12.068 rad/s
/7 1
(3) As .os =
.oQ'3
hence Q
.0 )' (gH f =197.2 m 3/S =(:g e 3
(gHer
Q=
Altematively,
P
1M'gHe
200 x 10·
= 0.94 x 981Ox110 = 197.2 m3 /s
(4) From the equation of continuity, Q = :lTD,b,c,,-, we get:
b,
•
where c"
Q
197.2
= :IT D, c" = :IT x 5.068 x 13.4 = 0.924 m
=c, cos a; =35.77 x 0.3746 =13.4.m/s
9.9. A Kaplan turbine designed with a shape/actor (power specific speed) of 3.0 (rad), a runner tip diameter of 4.4 m and a hub diameter of 2.0 m, operates with a net head of 20 m and a Shaft speed of 150 rev/min. The absolute flow at runner exit is axial. Assuming that the hydraulic efficiency is 90 % and the mechanical efficiency is
99 %, determine (1) the volume flow rate and shaft power output
•
I"~ (2) the relative flow angles at the runner inlet and outlet at the hub and at the tip. 'Ytl/IU.L . f,
_
Solution. (1) The power delivered to the rotor is given by:
(.0)'
P = P ~ (gHe}iS =19.67xl0·W
at.dlvrnA.M'-.
'8
Hence, the shaft power deli vered to external devices is:
p',>4t = 11m P = 0.99 x 19.67 x 106 = 19.47 MW
The volume flow rate is:
6 P 19.67 X 10 , Q= = = 111.4 m Is 11HPgHE 0.9 x 9810 x 20
6 P 19.67 X 10 (2) The specific work done is f!W = - = = 176.6 m'ls' pQ 10' x 111.4
The axial veloci ty and blade ti p speed are:
4Q
I
I'
ex - :n (D",. - D,) h
4x 111.4
- ff (4.4, -2 ,) = 9.234 m/s.. -
.1
u,
=
.or, = 15.078 x 2.2= 34.56 m1s
We can deteITI1ine the relative flow angles at the various radii from:
•
•
-
I
I I
I
i
l")
o
------
-~rccf FI~.
(,)
,,
Ie I
j I
iO-t Q 4- > 0
•
•
_FI G. (2.)
~ I,
.
,i' I
I
o/'2.~
(3'2
•
U
FI G-.'
•
-
72
0
I
I
J
I
•
2'5 03'55 '3$5.
5
/-
I
"oj
~-F
Q4·7
--~~~-r
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tc~
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•
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3ss
s
b:w
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-,
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s
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F1G-(I)
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