![Solutions Gas Turbine Theory 4 PDF [PDF]](https://pdfs.asia/img/200x200/solutions-gas-turbine-theory-4-pdf.jpg)
5 0 13 MB
Lecturer's Solutions Manual
GAS TURBINE THEORY 4TH EDITION
by
H Cohen, G F C Rogers and H I H Saravanamutoo
�
© Addison Wesley Longman 1 9 9 6 Lecturers adopting the main text are permitted to photocopy the pack as required.
Preface
·
Since
for
the
introduction
solutions
processing
properly
All
systems
printed
problems
involved.
done
is
to
It
calculations
is
This
then
students
be
very
I
will
glad
am
C ar l e t o n
gas
the
received.
now
and
made
I
am
turbine
by
digital
an
perhaps
Second
Edition
The
it
glad
to
do
calculations
computer
originally
historical
done
on
a
1972
advent
convenient
u n d e r s t an d i n g
of
in
of
to
requests
modem
p r e p ar e
word
these
in
a
so.
c arr i e d
and
of
out
the
the
rule
in
i n d u s t ry
p u rp o s e
of
engineering
s i g n i fi c an c e
slide
many
and
that
these
principies
all
many
are
of
were
these
former
questions.
m an u a l
are
has
provide
were
examination
been
format
s i g n i fi c an t
universally
who
have
of
will
be
available
p e rm i tt e d
tackle
to
h e ar
indebted
to
the
of
to
U n i v e r s i ty ,
to
photocopy
problems
any
Mr
for
instructors
the
befare
corrections
P.M.
his
adopting
material,
looking
but
a t
the
it
the
is
main
text,
hoped
manual.
I
that
will
needed.
Reddy,
formerly
invaluable
a
graduate
as s i s t a n c e
in
student
p r e p ar i n g
at
this
manual.
H.I.H.
SARA VAN AM U T T O O
Ottawa,
Gas Turbine Theory
F e b r u ary
1 9 9 6
4th edition
ISBN: O 582 23632 O
©
ADDISON WESLEY LONGMAN
1996
�
�
-·
•
Problem 2 . T
4
5
LD A D
ir,
( -, - 1 )
To2 - T. = :� ( ( �:)
'
-
1]
288 [ 1 1 f.! - 1] = 345.598 K
= 0.82
Compressor and turbine work required per unit rnass flow is :
Wtc
=
C,,a(To2 - Ta)
=
Cp,(Tos - To4)
'7m.
1.005
Ta3 - Ta4
X
=
345.598
=
x 1.147
308.992 K
0_98
�
Ta4 = 1 1 5 0 - 309 = 8 4 1 K 1�'
"""
Po.)
To3 - To• = 'IT To3 [1 - ( p
l
03) 1
3 08 . 9 9 2 = 0 . 8 7
X
1150
Pog 3
Po
=
) ¡]
/ l
[1 - (
Po4
4.382
Pa4
Pa3 = 1 1 . 0 - 0 . 4 = 1 0 . 6 bar
Pa4
=
2.418 b a r ,
P o s = r, 1
Ta4 - Tas = 0.89 x 8 4 1 [1 - (
)
t]
= 148.254 K
2.418
Gas Turbine Theory
4th edition e ADDISON WESLEY LONGMAN
ISBN: O 582 23632 O
1996 �
lspecific power output ·:
WN
= 1 . 1 4 7 x 0.98 x 1 4 8 . 2 5 4 = 1 6 6 . 64 k W s/Kg
. H c n c e mass flow required
20
X
l a3
=
=
1 2 0 . 0 1 9 Kg/Sec
1 66 _ 6 4
To2
=
+
288
To3 - To2
=
3 45 . 6 = 633.6 K
1 1 5 0 - 633.6 = 5 1 6 . 4 K
Theoretical f = 0 . 0 1 4 1 5 ( from
Fig.
2.15 )
Actual f = 0 . 0 1 4 1 5 / 0 . 9 9 = 0.01429
i
. S . F . C . = 3:0/ = 3600 x 0 . 0 1 4 2 9 = 0.308 kg/kW-h N
Gas Turbine Theory
1 66 . 6 4
4th edition
ISBN: O 582 23632 O
©
ADDISON WESLEY LONGMAN
1996
ij
•
I
Problem 2.8
T3 = T or T
and let e
+
sr
= (P Pi2 ) � =
Compressor
=
work
(Ps)7 p_.
T1 Q C p - ( c - 1)
'le
= ( Q - éi.Q)Cp Ts
Turbine work
= ( Q - e:.. Q )
Heat supplieci
� But T3 - Ts
=
then heat supplied
. W ith
=
Ts - T�
.
e,
'11
(1-
D
when T3 = T
+ sr
(Ts - Ts)
Ts '11
(1 -
= ( Q - éi. Q ) e, (Q - e:.. Q ) ( T
�)
Ts n« ( 1 - � )
+ e:..T) f'Jt
(1 -
i) -
(T3
=
T + éi. T )
Q 1i ( c - 1)
coolmg : ,., =
'7c
( Q - e:.. Q ) ( T •
+ e:..T)
f'J t ( l -
i)
T
1
T 'lt ( 1 - -) - .!..1. ( c - 1) Without cooling : '1 =
e
'7c
T 'lt ( 1 -
l)
These efficiencies are equal w hen:
( Q T 1 / '1 c ) ( c - 1 )
��
�
(Q - e:.. Q ) ( T
_
+ e:.. T ) 'l t ( l - �) -
Gas Turbine Theory
(T¡/r¡c)(c - 1 ) T 'l t ( l -
l)
n
4th edition
ISBN: 0 582 23632 ()
© ADDISON WESLEY LONGMAN 1996
l::!:i
i
_·.!_
1
+
( 1 - �)(T
6T) - T
6Q or
T
1 - - = - - -
Q
T+6T
6Q
6T/T
Q =
+
.
ó.T /T
and mdependent of r7c & '7c:
1
e = 6i.J' = 1 . 66 8
for the given plant :
1 e - 1
=
& ( 1 - -) e
0.668
=
0.401
W i t h cooling :
_ l _
T1/'1c:(c - 1)
11
i
-
(1·-
�)(1
= l _
+ �)
t)
T '1 t ( l -
= l _ 0.516
(288/0.87)(0.668)
11 0.95
X
1.25
X
1000
X
0.9
0.401
X
11 = 48.40%
Without cooling:
1J
=
l _
1 00 0
( a)
=
(288/0.87)(0.668) X
0.9
X
38.72%
0.401
T h e percentage increase in efficiency :
9.68
=
� = 2 5 . 0 0 ;1 0
38.72 L)\Vithout c o o l i n g the specific power output
1
i v N = C p [ T 11 t ( l - � ) - T ( c - 1 ) ] e '7 c 28�
=
Cp [ 1 00 0
X
0 . 90
X
0.40 -
X
0.668]
0.8,
=
1 3 8 . 8 6 8 Cp C h u / l b
\Vith c o o l i n g specific work o u t p u t :
ivN
=
cp [ ( 1 - il Q ) ( 1 + ó.T ) T 11 t ( l - �) - Ti ( e - 1 ) ] T
Q
= Cp [ ( 0 . 9 5
1'
X
1.25
X
1 00 0
e
X
0.9
X
1J c
288 0.401) . 0.87
X
0.668]
= c p ( 4 2 8 . 5 6 8 - 2 2 1 . 1 3 1 ) = 2 0 7 . 4 3 7 cp Ch u/lb
Gas Turbine Theory
4th edition
ISBN: 0 582 23632 0
�
© ADDISON WESLEY LONGMAN 1996
E::::j
i}(b) The percentage in�rease in speci.6.c work output
68.569
=
=
49.3%
138.868
Plant with no heat e x ch an g e r :
Gain in specific work output would be unchanged.
ficiency
requires
would
be
a much
much
greater
less,
because
increase
a rise
of heat
in
input
cycle
in
the
Gain in ef
temperature
combustion
chamber than is the case when a heat exchanger is fitted.
In other
11,words, if no heat exchanger is fitted, the higher temperature of the
exhaust gases is wasted.
N.B.:
[ 1 - 1 / ( l - �)(1 % increase in
+
�>]
efficiency: =
[(T/T1)(r¡ 0 r¡,/c) -
1]
% increase in specific work output:
[ ( 1 - AQ/Q)(l
[ 1 - (T1/T)
Gas Turbine Theory
+
AT/T) - 1]
x ( e/ r¡ 0 r¡,)]
4th edition
ISBN: O 582 23632 O
© ADDJSON WESLEY LoNGMAN 1996
�
o
•
�
illiproblem 2 . 9
8LEE..!> ¡.
·(
1
1
f
4 288
�oz -
r,
=
[ 3 . 8 /�
= 157.3 K
- 1]
0.85
= 3 . 8 0 - 0 . 1 2 = 3 . 6 8 bar
?03
?03
= 3 . 68
r. 1
T03
-
T04 = 1050 x 0.88 [ 1 - ( 3 . � 8 ) ']
= 256.87 K
N et work output
_
iv - 1J m (Load)
[(
m -
) ( mCpa(Toz - Ta) l me Cpg To3 - To4) - _....;.._ ...;_
n-« m ,00 = 0 . 9 8
[
(m - 1 . 5 )
X
1.147
X
X
( c o m. p . r o t o r )
1 . 00 5
X
15i.3]
256.87 0_99
2 00 = 2 8 8 . i 3 m - 4 3 3 . 1 - 1 5 6 . 5 m
m = 4.i88 kg/sec
(a)
W i t h no bleed fíow:
N e t work output
_ _ [ ... " _ 1.005 X 1 5 7 . 3 ] - 0.98 x 4 . , 8 8 1 . 1 4 , x ... 5 6 . 8 , - ----0.99 = 6 3 3 . 1 1 k\V
(b)
T h e power output with no b l e e d = 6 3 3 . l l k\V.
m, Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
e ADDISON
WESLEY LONGMAN
1996
�
_.
•
�roblem 2.10
lk.1
t_----1_
�
H.E
�------
7
s
R
F======�HP�t==========�T
LP�
,__ �-----2 '
3
n - 1
1
= -
For compression : n .
4
(1 -
=
1)
1)
=
'l oo t
n
0.66
=
= 0.4518 0.88
1
'7ooc
1
n -
For e x p a n s i ó n :
(1 -
0.88
X
X
1.66
0.66
=
0.3498
1.66
1 1
To2 - To1 = Ta1 [ ( P Poi o2)
":
- 1
=
To• - To3 = 300[2°-•sis - 1]
Q
500
To6 - Tos = C = m
.·.
=
T06
=
[
o 4518 l 2 · - 1 = 114 K
(given)
3
10
X
=
· · . 180
p
310
110.3 K
and To5 = 700 K
To4 = 4 1 0 . 3 K
�
]
X
535.2 K
5.19
1235.2 K
Po3 = 2 x 1 4 . 0 - 0.34 = 27.66 bar
Po4
=
2 x 2 7 . 66
=
5 5 . 3 2 bar
Pos = 5 5 . 3 2 - (0.27
Po1
=
14.0
+
0.34
+
+
1 . 0 3 ) = 54.02 bar
0.27
=
1 4 . 6 1 bar
P o s = 3.697 Po1 3498]
1
Tos - To1 = 1 2 3 5 . 2 [1 -
(
) o.
= 453.3 K
3.697
iTo1
= i81.82 K
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
© ADDISON WESLEY LoNGMAN 1996
�
ltower output
=
=
mcp [2.T061 -
= 180 X 5 J 9 x 229.0
ÁT034 - Á T 0 1 2 ]
2 1 3 99 6 kW or 2 1 3 . 9 96 MW
. Therrnal e ffi ciency
%
2 1 3 . 996
=
= 0.4279 or 42.8 500
H . E . effectiveness
=
Tos - To.
To1 - To.e
=
700
410·3
=
-
O. 7798 or 78%
7 8 1 . 8 - 410.3
n �
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
©
ADDISON WESLEY LONGMAN
1996
�
•
C3
=
=
336.87
=
p
pCr
1.596
2.755
2.41
1 99 . 5
1.140
1.581
2.78
2.43
177.5
1.140
1.581
2.78
2.43
176
7 2 . 4 m/s,
T3
=
399.1 K
= 3 3 6 . 8 7 m/s
� a = J 1 . 4 x 0.287 x 399.1 x
M
P,
To/T, Po/P,
10
3
= 400.44 m/s
0.841
4 00 . 4 4 ,
72.4 tan B = . 329
=
0.22
f3 = 1 2 ° 2 4
a
• Gas Turbine Theory
4th edition
ISBN· O 582 23632 O
© ADDJSON
WESLEY LONGMAN
1996
�
�, •
fim) Problem 4.5
T O?:/
-
.
1/
------- �,
�
� lll'
288
.6.To13 =
[
o 286 l 4 · - 1
= 175.1 K
0_80 pqU:2 2
.6.To13 = X
Cp
. ' .
U2
10
U:2
_
2
-
= 175.1
3
1.005
X
175.1
X
l . 0 4 x 0 . 90
103
_
..
s
-l.8,6x10
m/s
= 433
433
= 0 . 68 9 m
d = íT
To3
=
200
X
+
= 175.1
To2
. M At tip =l.O
2 8 8 = 4 63 . 1 K 2
'T1
= 'l
a= �
C2
v' l . 4
= a =
463 . 1
.l 2 = T 0 2 x
0.287
X
393.7
+
385.9
X
1
X
. =385.9K
1.2 103
m/s
��'7'
C2w = 0 . 90
X
U2
=
0 . 90
433
X
c?r
= 393.72 - 389.72
C2r
= 55.97
=
=
389.7
m/s
3133.6
m/s
W i t h a 50% loss in impeller
, 11 1 = 0 . 9 0
1
.6.T 012 = 0 . 90 x 1 7 5 . 1 = 1 5 7 . 5 9 K 5 Paz
� =
[
1
+
157.5913·
ro1
Gas Turbine Theory
= 4 . 60 7 288
.
4th edition
ISBN: O 58?. 23632 O
©
ADDISO� WESLEY LONGMAN
1996
�
Po2 = 1
-
p;
Po.,
('
=
+ 1) �
=
4.607 bar
3·5
=
= 1.2
-·
1.893
2 _ 4.607 _
p. 2
_
-
1 00
1 . 8 9 3 - 2 . 4 3 4 bar X
P'l - 0.287
m
4.60'T
X
2.434 385.9
X
= P Ac2r
. .
=
A .
2·198
14.0
=
.1138 1r
x
2
= 0.1138 m 2.198
h =
3
kg/m
X
55.97
_
x 100 - 5.257 cm 0_689
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
© ADDISON WESLEY LONGMAN 1996
�
t
Problem 5 . 1
• l o 8 · I
I S o
200
D
mOAC10 = mCr,AT
At m e a n 'radius,
1 . 00 5 .'.
ACw
X
=
20
10
X
3
=
X
m/s
108.1
200
0_93 2 00 - 1 0 8 . 1 Cow
=
=
45.95
m/s
22 45.95
tan a0
=
=
0.307
150 ,
200 - 45.95
=
tan a1 =
1 . 02 7
ª1
= 45º44
150
With free vortex, Cwr
=
constant i , e .
Cwu
=
constant
2So
Tip Cow
X
.'.
2 5 0 = 46.0
X
Cow = 3 6 . 8
Gas Turbine Theory
200
m/s
4th edition
JSRN: O 582 23632 O
e ADDISON
WESLEY LONGMAN
1996
�
-A C
=
.ó.Cwu
constan] . ' .
u
_
1 08 . 1 x 200 _ 86
w -
-
_:/ . 4 m s
250
º • ao = 1 3 46
36.8
= 0.245,
tan ao = 150 ·
.
250 - 36.8
• 0
tan
= 1.421,
=
a1
= 54 53
a1
150 250 - (36.8 tan
+
86.4)
, 0
= 0.845,
=
a2
= 40 1 4
a2
150
+
36.8 tan
86.4
,
=
=
a3
0.821,
a3
=
39°26
150 \ 4 4
.·\
·-
\
\ '
\./ -
-¡
- 1 4 4
I S O
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
©
ADDISON WESLEY LONGMAN
1996
�
..,
I
•
Problem 5 . 3
With no inlet guide vanes the inlet velocity is axial.
(a)
e
V
140 x 140
_ "88 _
T
-
1
-
X
2
1.005
X
B ,,
_ 27
- -
1 03
a =
.JilIT =
.'.
V = 0.95 x 3 3 4 . 4 = 3 1 7 . 7 m/s
v
2
= u
2
V l . 4 x 0 . 2 8 i x 2 7 8 . 2 5 x 1 03
2
+ c
. ' .
U
K
.-5
u
.- .
=
2
=
2
311.1
2 -
=
140
=
3 3 4 . 4 m/s
81333.3
2 8 5 . 2 m/s 285.2
x D x N
1í
=
285.2
=
D
.'.
= 1í
�
.'.
(b)
Po
- -
P
3·5
(T º) -
-
T
-
-
0 . 90 8 m
100
= 0.454 m or 4 5 . 4 0 cm
Tip radius 3·5
,w
X
288
(
)
- 1 1"8 . ..
2 7 8 . 25
1.01 P =
= 0 . 8 9 5 bar 1.1�8
P
=
0.895
o
Dhub
A =
·-
I
9-3
X
- I
= 0.60
:íT
(
= 1.l2l
1 00
X
11 3 -
k CJ' / m 3 o
11 • ..;.�
•
0.908 = 0 . 5 4 5 m
X
.,
")
0.908· - 0.545·
.,
= 0 . 4 1 4 3 m"
4 m
=
1 . 1 2 1 x 0 . 4 1 4 3 x 140
=
6 5 . 0 2 or m
=
65 k g / s
3·5
Po2 ( e)
- =
[ 1
+
0.89 x 2 0 ]
= 1.2335
Poi
288
D Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
©
ADDISON WESLEY LONGMAN
19%
�
.
D ( el)
At the root we have axial inlet velocity .
Free vortex gives con
stan t work at all radii and constant axial velocity. --' - · ·
1 7 1 ·
U h = 0 . 60
1
m/s
285.2 = 1 7 1 . 1
X
1
=
mU �Cwn l 03
mCp�T 3
_
Cp�T x 10
� e,,, -
3 _
1.005 x 20 x 10 171.1
171.1 a1
=
=
. . .
1.2221
X
s
0.93 I
o
a1
/ .3 m
-
un tan
126
_
-
=
50
44
140 171.1-126.3 tan
a2
=
,
=
0.320,
a2
=
0
17
46
140
At tip
C)
�Cw
=
126.3
X
0 . 60
=
75.78
285.2 tan
a1
=
. . .
= 2 . 03 7 1
m/s º a1
=
63
, 50
140 285.2 - 7 5 . 7 8 tan
o-
•
Power input
,
=
=
1.4958,
a2
=
0
56
14
140
= 65 x 1.005 x 20 = 1 3 0 6 . 5 k\.V
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
© ADDISON WESLEY LONGMAN 1996
�
!ffl'Problem 5 . 4
n-l
-
n
-
To2
•
O 325
� 0.88 -
·
= ( 4.0)o.32s = 1.569
To1
�Ton= 2 8 8 [ 1 . 5 6 9 - 1] = 163.9 K
1 63 . 9
=
N um b er of stages
=
6.556
i.e. 7 stages
25 6. T
1 63 . 9
=
f
=
2 3 . 4 K/stage
1
stages
# o
288
Tout
(DF'or first stage
:
- T¡n
=
1.0812
Ti«
=
=
23 . 4 _ 08 - l. -12 288
325
(R)º·
288
+
R¡irJt
• '.
F or last stage
Tout
+
= 1.271
:
1 63 . 9
=
451.9 K
4 5 1 . 9 - 23.4 = 428.5 K;
Tout
= 1.0546
r: =
1.0546
325
(R)º·
Rza.,t = 1 . 1 7 8
!' I
/
'
·
1
/• . . •
,
/
>--�
_,,,,/_/ I
I G S
.N-c = 1022 - 0 . 0 1 x 1 1 8 = 1 0 0 9 . s K 2 p
Poi
8.0
8.0
P« =
l, .
= 4
(T
01
/T�)
(1200/1009.5)
Gas Turbine Theory
4
= 4.008 bar 1.995
4th edition
ISBN: O 582 23632 O
©
ADOISON WESLEY LONGMAN
1996
�
49
e, 2
p
4.008
=
0.287
X X
1 00
k
•
1 02 2
=
1.366
m
/ g
3
m
36
=
C a :: = C2A3
= 348 m/s
(agrees with
given 346 m/s)
1 . 3 6 6 x 0.0756
At the root, for free vortex design, we ha ve:
rm = 0.2037 = l . l 7 r;
0.1741
( C w 2 ) ,.
=
(C,a::)m
rm r;
X
=
537.6
Ca:: is constant at 346 m/s.
X
1.17
=
629 m/s
Hence:
D C2,. = V629
2
+ 346
2
= 7 1 7 . 9 m/s
2
718
T2,.
=
=
1200 -
975.3 K
2294 320
U,.
= - =
273.5 m/s
1.17
V:z,.
=
J ,...... 3 4 _ 6 _ _ + _ ( _ 6 2 9 2 7 3 .5 ) 2 2
=
a2,.
=
V,RT2r = J 1 . 3 3 3 X 0 . 2 8 7 X 975.3 X 10
496.1 m/s
3
=
6 1 0 . 8 6 m/s
496.1 ( Mu2 ) ,. =
= 0.812 6 10.8 6
�
�ill1v'
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
© ADDISON WESLRY I.Ol'GMAN 1996
�
•
�
�_,,roblem 7 . 4
\J
1
1
V
0:2
=
constant
=
Cwzr
(1)
(2)
constant
Ca-:. = Cw2 cot
From ( 2 ) , since
c.zr =
constant
c.2
and
0:2
= ( c.2)m
( r;) 2
Also U = Um. (
.!.._) rm
2
u - = tan
Now, �
0:2
tan f32
-
Caz 2
tan f3-:. = tan
0:2
-
Um
(.!.._)
(Ca2)m
rm/
2
tan a: :: = ( t a n a: -:. ) m. = tan 5 8 º 2 3 ' = 1 . 6 2 4
Um.
(
)
,
=
tan 5 8 ° 2 3
, - tan 2 0 º 2 9
=
1 . 6 2 4 - 0.37 4
=
1.25
Ca: m
( � . B . F l o w c o e ffi c i e n t ( C 11 / U ) m = 0 . 8 far this mean diameter design)
2
H e n c e tan f32 = 1 . 6 2 4 - 1 . 2 5 ( � ) rm
Gas Turbine Theory
2
4th edition
ISB?\T: O 582 ?3632 O
e ADDISON
WESLEY LONGMAN
!996 �
.::n •
1
-·
h>
/h
(�)2
(�);
tan
root
1.164
1.357
0.709
35°20'
tip
0.877
0.769
o
Oº
/J2
W e therefore ha ve un twisted nozzles.
Cw2r== constant where x = sin
202
= sin
258°231
= 0.726 With
o 2 =constant, we also have C 02 r==constant.
�ence
C02
= (C
0
2)m (
r�):
This with
U = Um.
yields
( �)
rm
2
r
tan
/J2
. tan 02 -
(-
):+1 ( M )
rm
e a2
2
m
1
root
1 . 3 00
0.662
33°30
tip
0.797
0.056
3º12
1
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
e ADDISON
WESLEY LONGMAN
1996
�
•
�oblem 7 . 5
Assuming isen tropic expansion,
A v'2G; m =
X
R
For t h e g i v e n inlet conditions m is a m á x im u m when
(Ti
2
- �T
1 -
) T ( J - ..., ) / ( -r - 1 ) - 1 + 1 T./.. 1
w
01
1
3
1
3TJ
=O i.e. when:
= O
3
1 ?
T3 =
(To1 - .ó.Tw)
._
1 + 1
lence,
writing K
= A�
X
�(��-l) 01
( T o 1 - !);.T
� [ ( ) ., 1
w
But
CP R
=
2
) ., : 1
(
) ., : 1
2
-
1 + 1
1 + 1
(
2
\ l J
1 + 1 1
1
, - 1
Gas Turbine Theory
4th edition
ISBN: O 582 ?.3632 O
©
ADDISON WESLEY LoNG"AN
1996
�
•
, AP01
m
ma:
A ,r,
(,r, �01
----
r,h/'l-1)
- u�
)
l.±! ( 2
..,-i
111
v'Tcñ
AP01
2
i ( R
)
¡
) :r�i (l _
+
�T"')
1
2 :;-=-i ( j
-
1)
l.±!
( ; + l ) .., - 1
R ; - 1
01
m ma :z:
I
-
1
4!! rJ¡+
T01
h-l)
r. b + l h - 1 ) 01
mma:z: v'Tcñ AP01
Hence
1- ( ( 2 ) (l _ � T ., ) ] � ;+
R
maximum mass
1
flow will vary
9varies with Poi/ Pos-
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
To1
with
N/v'Tcñ
because
�Tw
•
�roblem 8 . 1
Note :
In the problems for Chapter 8 , we are dealing with stagnation conditions at
all times and the su.ffix
'O'
has dropped throughout.
( o \o"'f' p
33-4
Flow compatibility is expressed by :
m� =
my"T;
P3
X
P3
P1
X
P1
{Ti
V ;¡;
and:
P3
��
p;
6.903
�
5.0
34.515
32.9
4.i
32.444
33.8
4.5
31.064
34.3
�
Frorn curves, equilibrium poin t is a t :
p.,
p:
=
4.835,
my'Tí = 3 3 4 P1
Gas Turbine Theory
.
n, =
0.795
'
4th edition
ISBN: O 582 23632 O
e
ADDISON WESLEY l,ONGMA.:
1996
�
1
.
Ps
- = 0. 95
X
4.835 = 4.59
P1
and from turbine characteristic graph : 'lt = 0.8497
33.4 x 1.01
m =
.
K = l . 98 8
Mo'o
/ g
s
v288 1
= mCpgAT3,c - -mCpaAT12
N et power output
'lm
1
Poutput
D
= 1 . 98 8
X
.
- 1 . 98 8
1.147
X
X
1 . 00 5
0.8497
X
288
X
[
1 1 0 0 [ 1 - (-1-) 4.59
1
( 4.835) D°
-
1
¡]
l
0.795
Poutput
= 6 7 5 . 1 8 9 - 4 1 1 . 6 2 0 = 263.989 Kw
Net power output
Gas Turbine Theory
ISBN: O
·"S2
=
264 Kw
4th edition
23632 O
©
ADDISON WESLEY LONGMAN
1996
�
.
�
Problem 8 . 2
•
¡
2 - 2 7
?3 r = -
and
f'J t
=
0.85
?-4
1
?3/?4
(?3/ ?4) r
ó.T3,4/T3
T,4/T3
vT•7T3
m,/T;,/ ?3
my'T¡/P•
2.00
1.189
0.135
0.865
0.930
88.2
164.05
2.25
1.2247
0.156
0.844
0.918
90.2
186.31
., �o .... . o
1.257
0.174
0.826
0.909
90.2
205.0
D
mvl'i
= 188
?4
when
?3 - = 2.270
(graphical solution)
?4
.
m� = 9 0 . 2
?3
�T3 T
4
= 0.85
[
Gas Turbine Theory
1 -
(
1
)
il =
0.1575
2.27
3
4th edition
ISBN: O 582 2�632 O
� ADDISON WESLEY LONGMAN
1996
�
'rom compatibility oí ffow :
{f;
-·
;!;¡
1
= 90.2
Y Ti
e:
m
1
(A) P1
compatibility of w o r k :
sr sr T3 Cpgf'Jm - = -T¡
T3
6.T12
T1
e;
0.1575
T1
1.147
T1
T3 0.1762T1
. . . . . . . . . (B)
1]
P1
'7e
P1
0.98 T3
1.005
(�) 0.286
( mv'T1} �
X
=
�[(P2)0.2ss -
=
X
=
r¡
R- ( =
tlfi¡'l
(!
�
(A)
*
(
0.1 �62 ) {
*))
(B)
ffiflS
'"�.2
220
2.13
4.54
0.82
1.602
0.734
4.17
5.0
236
1.91
3.65
0.83
1.584
0.704
3.99
4.8
244
1.77
3.15
0.82
1 . 5 66
0.690
3.92
P2 Pi
hence
=
5.105
and
T3 Ti = 4 . 06 ,
from
graphical
solution
T3 = 1 1 7 0 K
Gas Turbine Theory
4th edition
ISBN: O 582 23632 O
©
ADDISON WESLEY LONGMAN
1996
�
� 1 ¡h ,c
'rro
•
bl em 8 . 3
At outlet of gas generator turbine
fj¡
m..;r; = my'Tj° x Ps x
r,
and
·
P3
�:
=
Y
P,.
1 - ,¡, [ 1 -
"f;
c ;P.fl 3
_ m .¡r¡ _ 4 T= my'Ti x _ P 3 { l _ 'lt [i _ (
P3
P4 TJ t
=
1
)
il } !
P3/ r,
P,..
0.85
1 · --------------------------------
l
(�) t
( -
�
1
)
i
{ 1
P37P,
-
'1t [ 1
-
(
p 3�
p.)
il }
t
�
�
�
1.3
1.0678
0.064
0.972
20.0
25.27
1.92
1.5
1.1067
0.097
0.958
44.0
63.22
1.664
1.8
1.1583
0.137
0.940
62.0
104.90
1.387
T h e value of P4/ Pa in the table is found from Pi
�n �a
P4 P3 P: = - - -?3 P-:.
r,
P4
= --
X
0.96
X
2.60
P3
',,, 3 3. t .
=
r, 2 . 4 9 6 - P3
º "'
t
I
e le
H e n c e curves - from which power t u r b i n e pressure ratio is r = 1 . 5 5
1
Gas Turbine Theory
•
4th edition
ISBN: O 582 7�632 O
© A•JDISON �ESLEY LONGMAN 1996
�
flnd gas generator turbine pressure ratio r = 1 . 6 1
Use work compatibility equation to find T3
T1 must be given .
Compressor efficiency from compressor characteristic once oper
ating point found from "''(;1
= "''(;3 x �
X
¡Ti;
where T3 unknown, so trial and
error method required.
Gas Turbine Theory I S P. N :
O 582 'B612
i
1
'
j
1
4th edition O
.
.
1
j
'
© ADDISON WESLEY LONGMAN 1996
n
�
�Problem 8 . 4
•
�H}1
For gas generator turbine
and rJ t is constant.
mi{i
Thus :
=
« �)
frorn gas generator turbine characteristics.
Since power turbine is choking at all conditions considered, the gas generator turbine Dis operating at a fixed pressure ratio and hence fixed value of � :;; • .
At 95 % of speed, work compatibility yields
( a)
At 100 % s p e e d we have sirnilarly :
1
� '&
T �
3'4
=
CpaTl [ 4 . 6 3 . l>
- 1)
C
pa = _;..
rJ m. C p g 0 . 8 5 9
But:
_
rJ m C p g 0 . 8 5 9
(�T34)
1075
T1 0 . 5 4 6 5
9 5 % .s p e e d
( il T 3 4 ) T3 1 00 % .s pe e d
Therefore ;
T3 = 1 0 í 5 x
º·
0·863
5455
0.486
0.859
(b)
95
= 1214.5 K.
x
% mechanical speed at 273 K .
Gas Turbine Theory
4th edition
ISBN: O 58?. 23632 O
0
Jri:
�·95vm
(% design)
= 97.57 %
From the operating line � = 439,
273
285
6.T =
[4.29°·
-
R
= 4.29 and '7 e = 0.862
1 ) = 1 6 3 . 6 K.
0_862 439 m
=
X
0.76
y'273
= 20.19 Kg/s.
273 Power = 20.19 x 1.005 x 1 6 3 . 5 = 3 3 1 8 Kw.
4 3 3
- z�
-
I
o c,
1
Gas Turbine Theory
:
4th edition
ISBN: O 582 23632 O
©
ADDIS01' WESLEY LONGMAN
1996 �
1
�
•
i ili! P r o b l e m
Work
At
8 . 5
compatibility
design
point
yields
:
:
1
[ 4 :n- - 1 ] l
[ 1 - ( t ) Ll l
1 � = me - m b
mb
=
= 0.6622
1 - 0.6622
=
=
0.3378
mb vT i / Pi
m c vf 1 / P 1
m,
Therefore
at
design
point
:
m b vfí
P1
= 0 . 3 3 7 8 x 2 2 . 8 = 7 . 70
(me - mb)vT3
P3.jl - �
and
JTj
is
constant.
m, - mb (m, - mb)J
-
2
2
?3
JI -
1/r
(P3)d
Ji -
l / r 1 - rd.jl - 1/r�
rJI -
r y' l - 1 / r
1/r2
4Jl - 1 / 1 6
Jr� -
1
JI5
. . . . . . . . . . . . . ( 1 )
G as Turbine Theory
4th edition
ISBN: O 582 23632 O
