PROYECTO TURBOPROP 1

Axel Oswaldo Malagon Olea

September 2024

1
Part I
Provided Data
Data name Value
SHP 300 HP
ALTITUDE MSL
Cruising Speed Mach 0.4
High speed rotor RPM 10000-20000 RPM
Low speed Rotor RPM 2500-15000 RPM
Compressor Pressure Ratio 3.9
Maximun Temperature 1300 K
Intake Isentropic Efficiency 0.98
Compressor Isentropic Efficiency GRAPHICS
Turbines Isentropic Efficiency 0.91
Mechanical Transmission Efficiency 0.98
Combustion Efficiency 0.96
Combustion Pressure Loss 1.50%
Air mass flow rate 8 Kg/s
Propeller Efficiency 0.95
Fuel Calorific Power 44000 KJ/Kg
Propelling Nozzle Efficiency 0.97
DT Power Turbine 200

Part II
Thermodynamic cycle
1 Difusser-Inlet Duct
T a = 288K

P a = 101335P a

p
Ca = M ∗ (γ ∗ R ∗ T ) (1)
s
J
Ca = 0.4 ∗ 1.4 ∗ 287 ∗ 288K = 136.0696m/s
Kg ∗ K

Ca2
T01 = T a + (2)
2 ∗ Cpa

2
 (136.0696 m
s )
2
T01 = 288K + J
= 297.211311K
2 ∗ 1005 Kg∗K

Ca2 1.4
P01 = Pa ∗ (1 + ηa ∗ ) 1.4−1 (3)
2 ∗ Cpa ∗ Ta
136.0696K 2 1.4−1
1.4
P01 = 101325P a ∗ (1 + 0.98 ∗ ) = 112883.2724P a
2 ∗ 1005 ∗ 288
That concludes the thermodynamic section for the intake

2 Compressor
For this section we need to look at the book
We are going to enter with the value given for the pressure ratio, 3.8.
After reading through the graphic we have encountered the next value, also
indicated on the Figure 1. It´s observable that the value corresponding to the
efficiency of the compressor is around 0.813
Now we can proceed with the calculations required for the thermodynamic cycle
onto the compressor section for the Turboprop engine.
γa −1
T01
T02 = T01 + ( × (πc γa ) − 1) (4)
ηC
297.211411K 1.4−1
T02 = 297.211411K + ( × (3.9 1.4 ) − 1) = 470.96269K
0.813
P02 = πc ∗ P01 (5)
P02 = 3.9 × 112883.2724P a = 440244.7624P a

3 Combustion Chamber
T03 = TM ax (6)
T03 = 1300K
P03 = P02 × (1 − CP L) (7)
P03 = 440244.7624P a × (1 − .015) = 433641.091P a

4 Gasogenous Turbine-High Pressure

Cpa
T04 = T03 − ( × (∆T2−1 )) (8)
Cpg × ηM
J
1005 kg∗K
T04 = 1300K−( J
×(470.96269K−297.211411K)) = 1144.787709K
1148 Kg∗K × 0.98

3
 Figure 1: The value corresponding to the Pressure ratio is 0.813 aprox.

′ 1
T04 = T03 − × (∆T3−4 ) (9)
ηT
′ 1
T04 = 1300K − × (1300K − 1144.787709K) = 1129.437043K
0.91
T′ γg
P04 = P03 × ( 04 ) γg −1 (10)
T03
1129.437043K 1.333−1
1.333
P04 = 433641.091P a × ( ) = 246957.2053P a
1300K

5 Power Turbine-Low Pressure

′
T05 = T04 − ∆TP T (11)
T05 = 1129.437043K − 200K = 929.437043K
′ 1 ′
T05 = T04 − ( × (T04 − T05 )) (12)
ηT

4
 ′ 1
T05 = 1144.787709K −( ×(1129.437043K −929.437043K)) = 939.4267345K
0.91
T ′ γgγ−1
g
P05 = P04 × ( 05
′ ) (13)
T04
939.4267345K 1.333
P05 = 246957.2053P a × ( ) 0.333 = 118136.749P a
1129.437043K

6 Propelling nozzle-Exhaust Duct

P05 118136.749P a
= = 1.16591906
Pa 101325P a
P05 1
= γg = 1.891103262
Pc 1 γ −1
(1 − ( ηP N × ( γgg +1 ))) γg −1
P05 P05
<
Pa Pc
Therefore, it is not a choking nozzle.
1 γg −1
T06 − T05 = ηP N ∗ T05 × (1 − ( P05 ) γg ) (14)
Pa

1 1.333−1
T06 − T05 = 0.97 ∗ 929.437043K × (1 − ( ) 1.333 ) = 33.918828K
1.16591906
p
C6 = 2 ∗ Cpg ∗ ∆T6−5 (15)
s
J m
C6 = 2 ∗ 1148 ∗ 33.918828K = 279.0656376
Kg ∗ K s
P6 = Pa (16)

7 Residual Thrust
Er = ṁa × (C6 − Ca ) (17)
Kg m m
Er = 8 × (279.0656376 − 136.0696 ) = 1143.968301N
s s s

8 Specific Thrust
Er
Es = (18)
ṁa
1143.968301N Ns
Es = = 142.9960376
8 Kg
s
Kga

5
9 Specific Fuel Consumption SFC
We will proceed to consult the Graphic, obtained from the book for theory for
gas turbines, by Saravanamuttoo. We are going to do the reading with the
difference between the Maximum temperature and the temperature 02. ∆T =
1300K − 470.96269K = 829.03731K
Kgf
fideal = 0.0228
Kga
fideal
factual = (19)
ηB
Kg
0.0228 Kgfa Kgf
factual = = 0.02375
0.96 Kga
factual
SF C = (20)
Es
Kg
0.02375 Kgfa Kgf 3600s Kgf
SF C = = 1.6608851 × = 0.59791866
142.9960376 KNga
s Ns h Nh

10 Propulsive efficiency
2
ηP = C6
(21)
1+ C a

2
ηP = 279.0656376 m
= 0.655543484
1+ 136.0696 m
s
s

11 Energy Conversion Efficiency

C62 − Ca2
ηE = (22)
2 ∗ factual ∗ HC
2 m2
279.0656376 ms − 136.0696 s
ηE = KJ
= 0.0171078 = 1.7107848%
2 ∗ 0.02375 ∗ 44000 kg f
∗ 1000J
KJ

12 Thrust Power THP

T HP = SHP × ηP R + ER × Ca (23)
m 1HP
T HP = 300HP ×0.95+1143.968301N ×136.0696 × = 493.6595913HP
s 746 Nsm

6
Figure 2: The lectures results in a value of 0.0228 as for the fuel-air ratio

7
13 Effective Power
T HP
EHP = (24)
ηP R
493.6595913HP
EHP = = 519.64062HP
0.95
Now we are going to proceed to calculate the Gasogenous Turbine, those calcu-
lations will be shown starting in the next page.

8
Part III
Turbine section-Gasogenous
turbine
Let´s make emphasis on the fact that the input data for the gasogenous
turbine is the output data for the combustion chamber thermodynamic cycle
section.
FROM THERMODYNAMIC CYCLE
Inlet Temperature 1300K
Inlet pressure 433641.091 Pa
Outlet pressure 246957.2053 Pa
Outlet Temperature 1144.787709K
Turbine Isentropic Efficiency 0.91
Combustion Gases mass flowrate 8 Kg
s
Proposed Data
Degree of Reaction 0.5
Stage Temperature Drop 130K
Flow coefficient 0.9
Axial Velocity 210 m
s
Height/cord ratio 3.5
Nozzle/Stator Efficiency 0.05
Peripheral Speed U = Cϕa = 233.3333 ms
Rotational speed 15700 RPM

14 Number of stages
∆Tturbine = T01 − T03 (25)
∆Tturbine = 1300K − 1144.78771K = 155.212291K
∆Tturbine
ηs = (26)
∆Ts
155.2122991K
ηs = = 1.1939 → 2
130K
∆Tturbine
∆Tsactual = (27)
ηs
155.212291K
∆Tsactual = = 77.6061455K
2

15 Temperature drop Coefficient

2 × Cpg × ∆Tsactual
ψ= (28)
U2

9
 J
2 ∗ 1148 KG∗K ∗ 77.6061455K
ψ= = 3.272762
233.3333 m
s
The value of psi must be in the range of 3 up to 5, so the current value is
within the range, so we can proceed with all of the following calculations.

16 Gas Angles
For a degree of reaction of 0.5, we know that β3 = α2 & α1 = α3 = β2 , but
yet the calculations for all of the angles will be shown, as a way of
corroboration of that statement.

1 ψ
β3 = arctan( × ( + Λ × 2)) (29)
2×ϕ 2
1 3.272762
β3 = arctan( ×( + 0.5 × 2)) = 55.6765315◦
2 × 0.9 2
1 ψ
β2 = arctan( × ( − Λ × 2)) (30)
2×ϕ 2
1 3.272762
β3 = arctan( ×( − 0.5 × 2)) = 19.4707936◦
2 × 0.9 2
1
α2 = arctan(tan β2 + ) (31)
ϕ
1
α2 = arctan(tan(55.6765315◦ ) + ) = 55.6765315◦
0.9
1
α1 = α3 = arctan(tan β3 − ) (32)
ϕ
1
α1 = α3 = arctan(tan(19.4707936◦ ) − ) = 19.4707936◦
0.9

17 mean radius
U
rm = (33)
2π × N
233.33333 m
s
rm = 1min
= 0.14192161m
2π × 15700RP M × 60seg

10
 18 Stages and planes
For the first part of this section, a table with the number of stages,
temperatures, and pressures, per section; will be presented as following. The
aforementioned table uses the following equations for the calculations of it’s
elementents:

T03 = T01 − ∆Tsactual (34)

∆Tsactual γgγ−1
g
P03 = P01 × (1 − ) (35)
ηT × T01
Stage T01 (K) T03 (K) P01 (P a) P03 (P a)
1 1300 12222.39385 433641.091 330499.755
2 1222.39385 1144.78771 330499.755 247426.23
Also, thorough all the planes, Can is constant.

18.1 Plane 2- Critical Plane Stage 1

T02 = T01 (36)
T02 = 1300K
Ca2
C2 = (37)
cos(α2 )
210 m
s m
C2 = ◦
= 372.429985
cos(55.6765315 ) s
C22
T2 = T01 − (38)
2 × Cpg
372.429985 m
s
T2 = 1300K − J
= 1239.58881K
2 × 1148 Kg∗K
C22
T2 − T2′ = λN × ( ) (39)
2 × C pg
372.4299852 m
T2 − T2′ = 0.05 × J
s
= 3.02055953K
2 × 1148 Kg∗K
T2′ = T2 − (T2 − T2′ ) (40)
T2′ = 1239.58881K − 3.02055953K = 1236.56825K
P01
P2 = γg (41)
( TT01′ ) γg −1
2

433641.091P a
P2 = 1.333 = 354947.518P a
1300K
( 1236.56825K ) 1.333−1

11
 P01 T01 γg
= ( ′ ) γg −1 (42)
P2 T2
P01 1300K 1.333
=( ) 1.333−1 = 1.22170481
P2 1236.56825K
P01 γg + 1 γgγ−1
g
=( ) (43)
Pc 2
P01 1.333 + 1 1.333−1
1.333
=( ) = 1.85242156
Pc 2
P01 P01
<
P2 Pc
It is a requirement for the last statement to be true, and as it is, we are
allowed to continue with the calculations.

P2
ρ2 = (44)
T2 × R
354947.518P a Kg
ρ2 = J
= 0.99771063 3
1239.58881K × 287 Kg∗K m

18.1.1 Anulus Area

ṁ
A2 = (45)
ρ2 × Ca2
8 Kg
A2 = s
= 0.03818265m2
0.99771063 Kg m
m3 × 210 s
A2 × N
h2 = (46)
U
1min
0.03818265 × 15700RP M × 60s
h2 = = 0.04281912m
233.33333 m
s
C2
M2 = p (47)
γg × T2 × R
372.429985 m
s
M2 = q = 0.54081633
J
1.333 × 1239.58881K × 287 Kg∗K

It is also a requirement for the Mach number in the plane 2 to be less than 0.8,
once again we have met the requirements and we are allowed to proceed.

12
 18.1.2Radii
h2
rr2 = rm − (48)
2
0.04281912m
rr2 = 0.14192161m − = 0.12051205m
2
h2
rr2 = rm − (49)
2
0.04281912m
rt2 = 0.14192161m + = 0.16333116m
2
rt2
= 1.35530985
rr2

18.2 Plane 1 Stage 1

Ca3
C1 = C3 = (50)
cos α3
210 m
s m
C1 = C3 = = 222.728049
cos 19.4707936◦ s
2
C1
T1 = T01 − (51)
2 × Cpg
(222.728049 m
s )
2
T1 = 1300 − J
= 1278.39188K
2 × 1148 Kg∗K
T1 γgγ−1
g
P1 = P01 × ( ) (52)
T01
1278.39188K 1.333−1
1.333
P1 = 433641.091P a × ( ) = 405500.296P a
1300
P1
ρ1 = (53)
T1 × R
405500.296P a Kg
ρ1 = J
= 1.10521119 3
1278.39188K × 287 Kg∗K m

18.2.1 Anulus Area

ṁ
A1 = (54)
ρ1 × Ca1
8 Kg
A1 = s
= 0.03446874m2
1.10521119 Kg m
m3 × 210 s
A1 × N
h1 = (55)
U
2 1min
0.03446874m × 15700RP M × 60s
h1 = = 0.03865423m
233.33333 m
s

13
 C1
M1 = p (56)
γg × T1 × R
222.728049 m
s
M1 = q = 0.41094
J
1.333 × 1278.39188K × 287 Kg∗K

18.2.2 Radii
h1
rr1 = rm − (57)
2
0.03865423m
rr1 = 0.14192161m − = 0.12259449m
2
h1
rt1 = rm + (58)
2
0.03865423m
rt1 = 0.14192161m + = 0.16124872m
2
rt1
= 1.31530153
rr1

18.3 Plane 3 Stage 1

Ca3
C1 = C3 = (59)
cos α3
210 m
s m
C1 = C3 = ◦
= 222.728049
cos 19.4707936 s
C32
T3 = T03 − (60)
2 × Cpg
(222.728049 m
s )
2
T3 = 1222.39385K − J
= 1200.78574K
2 × 1148 Kg∗K
T3 γgγ−1
g
P3 = P03 × ( ) (61)
T03
1200.78574K 1.333−1
1.333
P3 = 330499.755P a × ( ) = 307726.808P a
1222.39385K
P3
ρ3 = (62)
T3 × R
307726.808P a Kg
ρ3 = J
= 0.89293103 3
1200.78574K × 287 Kg∗K m

14
 18.3.1 Anulus Area
ṁ
A3 = (63)
ρ3 × Ca3
8 Kg
A3 = s
= 0.04266314m2
0.89293103 Kg m
m3 × 210 s
A3 × N
h3 = (64)
U
1min
0.04266314m2 × 15700RP M × 60s
h3 = = 0.04784366m
233.33333 m
s
C3
M3 = p (65)
γg × T1 × R
222.728049 m
s
M3 = q = 0.32862881
J
1.333 × 1200.78574K × 287 Kg∗K

18.3.2 Radii
h3
rr3 = rm − (66)
2
0.04784366m
rr3 = 0.14192161m − = 0.11799978m
2
h3
rt3 = rm + (67)
2
0.04784366m
rt3 = 0.14192161m + = 0.16584344m
2
rt3
= 1.40545552
rr3
rt
Plane A(m2 ) h(m) rr (m) rt (m) rr
1 0.03446874 0.03865423 0.12259449 0.16124872 1.31530153
2 0.03818265 0.04281912 0.12051205 0.16333116 1.35530985
3 0.04266314 0.04784366 0.11799978 0.16584344 1.40545552
It is a requirement for the first stage to get the values of rrrt within the range
(1.2,1.4). We can observe that all the values of the first plane are in the
aforementioned range, so we have met the requirement and so we are able to
continue the calculations.

15
 19 Rotor efficiency Stage 1
T2
T3′ = γg −2 (68)
(P
P3 )
2 γg

1239.58881K
T3′ = 354947.518P a 1.333−1 = 1196.16082K
( 307726.808P a)
1.333

Ca3
V3 = (69)
cos α3
210 m
s m
V3 = = 222.738049
cos(19.4707936◦ ) s
T3 − T3′
λR = V32
(70)
2×Cpg

1200.78574K − 1196.16082K
λR = (222.738049 m 2 = 0.21403625
s )
J
2×1148 Kg∗K

It is required that the rotor efficiency is bigger than the nozzle efficiency,
which is true for our values.

20 Variation of angles Stage 1

20.1 Root angles.
rm
α2r = arctan( × tan α2 ) (71)
rr2
0.14192161m
α2r = arctan( × tan 55.6765315◦ ) = 59.8966703◦
0.12051205m
rm
α3r = arctan( × tan α3 ) (72)
rr3
0.14192161m
α3r = arctan( × tan 19.4707936◦ ) = 23.0360898◦
0.11799978m
rm rr U
β2r = arctan( × tan α2 − 2 × ) (73)
rr2 rm Ca2
0.14192161m 0.12051205m 233.3333 m
β2r = arctan( ×tan(55.6765315◦ )− × s
) = 38.0028185◦
0.12051205m 0.14192161m 210 m
s
rm rr U
β3r = arctan( × tan α3 + 3 × ) (74)
rr3 rm Ca3
0.14192161m 0.11799978m 233.3333 m
β3r = arctan( ×tan(19.4707936◦ ))+ × s
) = 53.4517362◦
0.11799978m 0.14192161m 210 m
s

16
 20.2 Tip Angles
rm
α2t = arctan( × tan α2 ) (75)
rt2
0.14192161m
α2t = arctan( × tan 55.6765315◦ ) = 51.8414773◦
0.16333116m
rm
α3t = arctan( × tan α3 ) (76)
rt3
0.14192161m
α3t = arctan( × tan 19.4707936◦ ) = 16.8331094◦
0.16584344m
rm rt U
β2t = arctan( × tan α2 − 2 × ) (77)
rt2 rm Ca2
0.14192161m 0.16333116m 233.3333 m
β2t = arctan( ×tan(55.6765315◦ )− × s
) = −0.3471836◦
0.16333116m 0.14192161m 210 m
s
rm rt U
β3t = arctan( × tan α3 + 3 × ) (78)
rt3 rm Ca3
0.14192161m 0.16584344m 233.3333 m
β3t = arctan( ×tan(19.4707936◦ ))+ × s
) = 58.0098151◦
0.16584344m 0.14192161 210 m
s

21 Plane 2 Stage 1 Root Mach Number

Ca2
V2r = (79)
Cosβ2r
210 m
s m
V2r = = 266.504068
cos(38.0028185◦ ) s
Ca2
C2r = (80)
Cosα2r
210 m
s m
C2r = = 418.692829
Cos(59.8966703◦ ) s
C22r
T2r = T02 − (81)
2 × C pg
(418.692829 m
s )
2
T2r = 1300K − J
= 1223.64822K
2 × 1148 Kg∗K
V2r
MV2r = p (82)
γg × T2r × R
266.504068 m
s
MV2r = q = 0.38951082
J
1.333 × 1223.64822K × 287 Kg∗K

17
 Figure 3: Angle variations through stage 1

22 Height of vanes and blades, stage 1

For this section we need to make the reading for the value of the optimal
Pitch-to-chord ratio, from the image below (fig. 4) we can see that the value
for the reading is around 0.83, which is going to be the value used for the
calculations.
Stator Value Rotor
hs = h1 +h
2
2
Height hR =
h2 +h3
2
0.03865423m+0.04281912m 0.04281912m+0.04784366m
hs = 2 hR = 2
hs = 0.04073667m hR = 0.04533139m
Cs = (hhs) Chord CR = (hhR)
c c
Cs = 0.04073667m
3.5 CR = 0.045331394m
3.5
Cs = 0.01163905m CR = 0.01295183m
Ss = Cs × ( sc )OP T Pitch SR = CR × ( sc )OP T
Ss = 0.01163905m × (0.83) SR = 0.01295183m × (0.83)
Ss = 0.00966041m SR = 0.01075001m

18
 Figure 4: The reading is around 0.83

23 Number of blades and vanes, stage 1

Stator Value Rotor
ns = 2π×r
Ss
m
Uncorrected number nR = 2π×r
SR
m

2π×0.14192161m 2π×0.14192161m
ns = 0.00966041m nR = 0.01075001m
ns = 92.3066 nR = 82.9505
nsc = 94 Corrected number nRc = 83
Csc = (nhs) Chord Csc = (nhR)
c c
94
Csc = 3.5 CRc = 83m
3.5
Csc = 26.8571429m CRc = 23.7142857m
Ssc = Csc × ( sc )OP T Pitch SRc = CRc × ( sc )OP T
Ss = 26.8571429m × (0.83) SR = 23.7142857m × (0.83)
Ss = 22.2914286m SR = 19.6828571m

19
 24 Stage 2
The calculations for the stage number 2 are fairly the same as for the stage
number 1. Therefore is considered to be unnecessarily extensive for the current
project to show the calculations for the stage, the tables with the
corresponding values for all the stage are shown below.
Stage 2
Var Plane 1 Plane 2 Plane 3
Cn ( ms ) 222.738049 372.429985 222.738049
Tn (K) 1200.78574 1161.98266 1123.1759
T2 − T2′ (K) 3.02055953
Pn (P a) 307726.808 267013.954 229254.499
ρn ( Kg
m3 ) 0.89293102 0.80066789 0.71119184
An (m2 ) 0.04266314 0.04757933 0.05356535
hn (m) 0.04784366 0.05335681 0.06006971
Mn 0.32862881 0.55858439 0.3397925
rrn (m) 0.11799978 0.1152432 0.11188675
rtn (m) 0.16584344 0.16860001 0.17195646
r tn
r rn 1.40545552 1.46299318 1.53687958
T2′ (K) 1158.9621
P01
P2 1.23776211
P01
Pc 1.85242156
Rotor efficiency
Var Value
T3′ (K) 1118.5568
V3 ms 222.73805
λR 0.2139374
Variation of angles
Root angles
Angle name Angle Value
α2r 59.8966703◦
α3r 23.0360898◦
β2r 38.0028185◦
β3r 53.4517362◦
Tip angles
Angle name Angle Value
α2t 52.376756◦
α3t 17.157876◦
β2t 2.4691681◦
β3t 57.687478◦
Plane 2 root mach number

20
 var value
V2r ( m
s ) 282.733333
C2r ( m
s ) 433.099343
T2r (K) 1140.69741
MV2r 0.42799211
Height of Vanes and Blades
stator value rotor
0.05060024 height(m) 0.05671326
0.01445721 chord(m) 0.01620379
0.01199948 pitch(m) 0.01344915
Number of Vanes and Blades
Stator Value Rotor
74.3131 Number uncorrected 66.3030796
76 Number corrected 67
21.7142857 Cs corrected 19.1428571
18.0228571 Ss corrected 15.8885714
Angle variations graphics for the stage 2.

Figure 5: Angle variation on stage 2

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 25 Geometrical Arrangement
First off, we are going to display the drawings corresponding to the first stage
vane and blade. Upcoming, the general arrangement for both stages of the

Figure 6: Drawing of the vanes and blades.

gasogenous turbine.

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 23

Figure 7: General arrangement with a square to the scale of 0.005 m