EXERCISE 13 - PERFORMANCES DGEN380
Performance calculation.
Topic.
We wish to estimate the isentropic thrust of the DGE380 turbojet engine from the data recorded in test.
For this it is necessary to determine the densities, temperatures and speeds of the gas leaving the secondary
and primary nozzles.
The station nomenclature for these two streams is as follows:
Primary flow (hot flow):
Station 5: primary nozzle inlet section – outlet turbine section.
Station 8: primary nozzle outlet section
Secondary flow (cold flow):
Station 13: secondary nozzle inlet section.
Station 18: secondary nozzle outlet section.
The primary thrust is given by the following formula: Fp = W8 .(V8 − V0 ) − K 8 . A8 .(P0 − Ps 8 )
The secondary thrust is given by the following formula:
Where
Fs = W18.(V18 − V0 ) − K18 . A18 .(P0 − Ps18 )
A18 and A8 are the outlet surface sections of the both nozzle.
A18 = 0,8286 and A8 = 1 are surface correction coefficient.
Constants.
Constant perfect gas:
R = 8,3144621 J.mol-1.K-1
Mean (Average) molar mass of dry air:
Specific constant of the air:
r=
Ma = 28,964 g.mol-1
R
= 287,058 J.kg-1.K-1
Ma
Specific heat of air at constant pressure: C p = 1004 ,64 J.kg-1.K-1
Adiabatic coefficient of dry air at 20 °C:
 = 1,4
Adiabatic coefficient of mixed gas (air + fuel) in section 5:
Specific constant of gas mixture (air + fuel):
  = 1,345
r  = 257,57 J.kg-1.K-1
Hypothesis: C p = Cte for the air alone and for gas mixture (air + fuel).
Recall the lesson.
r = C p − Cv and  =
Cp
Cv
so
Cp =
 .r
 −1
V2
Hypothesis: Isentropic flow in the nozzle: the total temperature T = TS +
is preserved in the nozzle.
2.C p
The mass flow is given by the following formula: W
= . A.V . The mass flow is preserved in the nozzle.

P  T   −1
=   adiabatic flow in the nozzle.
= r.T and
Perfect gas formula:
Ps  Ts 

P
Hypothesis: The thermodynamic evolutions in the nozzles are isentropic and there is no loss in its.
1 

T0   Ps  5.255 
z=
. 1−  
 For a given nozzle the previous formula allows to estimate the potential altitude
0.0065   P0 


as a function of the atmospheric conditions and the static pressure in the outlet section of the nozzle.
Work:
1.
In the inlet section (section 13) of the secondary flow, calculates:
•
The static temperature Ts13 ,
•
The speed
•
The density 13 .
Vs13
2. In the inlet section (section 5) of the primary flow, calculates:
•
the static temperature Ts 5 ,
•
The speed
•
The density 5
V5
3. In the outlet section (section 18) of the secondary flow, Calculates:
•
The total pressure P18 ,
•
The static temperature Ts18 ,
•
The speed V18
•
The density 18 .
4. In the outlet section (section 8) of the primary flow, Calculates:
•
The speed
V8
•
The density 8
•
The static temperature Ts8
•
The total pressure
P8
5. Calculates:
=
Wsec
,
Wprm
•
the by-pass ratio:
•
the thrust Fp of the primary flow,
•
the thrust
•
the total thrust FT (compare the result with measured thrust),
•
the following ratio:
Fs of the secondary flow,
F fan
FT
6. Calculates the specific consumption (in kg.h-1.daN-1)
7. Calculates the potential altitude expressed in meter and feet for the secondary nozzle.
Recall: 1 ft = 0.3048 m
8. Calculates the MACH number in the following cases:
•
The Mach number
M18 in the outlet section of the secondary nozzle,
•
The Mach number
M8 in the outlet section of the primary nozzle.
9. Calculates the ratio:
•
P0
P
In the secondary nozzle. Does the result confirm the MACH number calculated to the previous
question?
•
In the primary nozzle. Does the result confirm the MACH number calculated to the previous
question?
Pamb = 1.0275 bar
Tamb = 2.7 °C
Ftot = 264.6 daN
13
P13 = 1.2408 bar
Ps13 = 1.0965 bar
T13 = 293.3 K
W13 = 10.649 Kg/s
A13 = 56187 mm2
18
Ps18 = 0.7336 bar
A18 = 36698 mm2
P5 = 1.2498 bar
Ps5 = 1.1176 bar
T5 = 865.6 K
W5 = 1.609 Kg/s
A5 = 14389 mm2
Ps8 = 1.7336 bar
A8 = 9453 mm2
C = 0.0312 Kg/s
5
8