Solutions gas turbine theory - MTF171
Exam. 2003-06-03
Problem 1:
Assume polytropic efficiencies of 88% on compressor and turbine. Since the turbine operates uncooled
a realistic value on a conventionally cast turbine is 1200 K.
ISA standard => t1 = 288.15, p1 = 101.325 kPa. Compressor exit station:
T2  P2 
 
T1  P1 
1  1
c ,  
t2 = 566.01, p2 = 810.6 kPa
 p 
P03  P02 1  b 
p02 

pb
 2  3% (industria l gas turbine)
p02
Assume a pressure loss value of 3%, which gives p3 = 786.3 kPa, t3 = 1200 K. The turbine exit station
obtained from:
t ,
T03  P03 


T04  P04 
 1

t4 = 764.8 K, p4 = p1= ambient pressure. The power output is
wt  wtc  217.5 kW
Power requirement per engine is 17.90 MW which gives a required mass flow per engine of about
m = 82.28 kg/s.
Problem 2:
m T03
P03

m T01 P01 P02
P01 P02 P03
T03
T01
Interpolation in compressor map at the n=150*0.922=138.9 rps rotational speed line, and calculating
the corrected compressor mass flow gives (pressure in bars):
m
64.5000
70.4250
75.6000
m*sqrt(T01)/P01
1080.6
1179.8
1266.5
rc
4.4750
4.0800
3.2900
etac
0.8125
0.8462
0.8000
The x-function can be used to determine the corrected turbine massflow (pressure in bars and M=1):
m RT0
  1 2 
  M 1 
M 
P0 A
2



( 1)
2 ( 1)
 ( , M )
which directly yields the turbine inlet temperature at the three selected off-design points.
m*sqrt(T03)/P03
516.56
516.56
516.56
t03 (K)
1318.7
919.5
518.8
The power output from the load turbine is then determined by (neglecting mechanical efficiency):
W  mc pg T03  T04   c pa T02  T01 
where the temperature drops are obtained from:
t ,
T03  P03 


T04  P04 
 1

using the relation:
P03 P02 P03
P

 Neglect pressure loss   02
P04 P01 P02
P01
and
 1



T01   P02  


T02  T01 

1

c   P01 



the powers and temperature drops are:
tdeltc
189.5258
168.3531
145.9854
tdeltt
366.5463
242.1127
118.3135
power (MW)
14.856
7.659
negative
2*power (MW)
29.712
15.318
negative
Inserting the result in the propeller map (double the gas turbine powers) shows that the first point is
very close to the propeller power at that speed. Thus the match point is approximately that of the first
selected point which is close to surge. This is in agreement with the experiences of the russians. A
bleed valve in the rear stages, variable geometry or modifications to the turbine area (increasing), may
be used to overcome this difficulty. The engine is likely to have been started with any or all of these
relieving schemes. The problem would most likely be solved if these schemes were used also at such
high rotational speed. This would, however, reduce the performance of the engine.
Problem 3:
Cooling reduces efficiency but increases specific power output.
Title:
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To maximize the output power one should minimize the power requirement by the compressors.
wc  c p T2  T1   c p T4  T3   c p T2  T1   c p T4  T1 

T
 T

wc
  2  1   4  1
c p T1  T1
  T1

t4 is obtained from the isentropic relations
T2  P2 
  
T1  P1 
T4  P4 
 
T3  P3 
rc  rc1rc2
 1

 1

 1
 rc1   
 rc2
 1


rc2
 1


rc
 1



c

which gives (T3=T1)
c

wc
   1    1
c pT1



 c


Differenti ate with respect to beta

1- c
2
0
rc1  rc2  rc