The document presents a preliminary design for an energy recovery system in a ramjet engine. It proposes introducing a moving element to bleed some compressed air from the ramjet into a turbine. This would generate electricity or other energy conservation. Calculations are shown comparing thrust produced with and without air bleeding. The goal is to extract thrust from the turbine to compensate for thrust lost from air bleeding, reducing battery weight and improving missile efficiency through energy recovery.

1. PRELIMINARY DESIGN OF AN ENERGY
RECOVERY SYSTEM
MBDA PROJECTPresentation
BY
RajendraPatel
KishalayKumar sinha
JohnPaul
Under the guidance of:
Mr. S.A.HASHIM
Asst. Professor
Department of Aeronautical Engineering
1

2. Abstracts:
• A ramjet is a form of an air breathing jet engine.
• It uses engine’s forward motion to compress incoming air.
• Absence of rotary compressor.
• It cannot produce thrust at zero airspeed.
• It works most efficiently at supersonic speed around mach 3
and can operate till mach 6.
• Weapon designers are looking to use ramjet technology in
artillery shells to give added range (around 35 Km).
• The combustion chamber is closed except for an opening at
the exhaust nozzle. Burning fuel produces high pressure
gasses that escape through the nozzle at the rear, and gas
pressure at the front end of the engine pushes the rocket
forward.
2

3. What our project basically is……
• Introducing a moving element in ramjet.
• Bleeding some percentage of air which is highly
compressed into the turbine.
• Using the bleed air in energy conservation like
electricity generator etc.
• Comparing the thrust produced with and
without bleeding.
• Extraction of some thrust from the turbine to
compensate the loss.
3

4. Our project’s mission…..
• To reduce the use of heavy batteries carried by
them for their maintenance.
• To reduce the weight of missiles.
• Energy conservation.
• Innovative concept that can give ramjet a new
life.
4

5. Ramjet Engine:
5

6. PROJECT PREFACE:
• SUPERSONIC DIFFUSER DESIGN
• THRUST CALCULATION WITH & WITHOUT
BLEEDING
• COMBUSTION CHAMBER HEAT TARNSFER AND
ENERGY RECOVERY FROM COMBUSTOR TO
DIFFUSED BLEED AIR FOR TUTBINE INLET
• SUPERSONIC NOZZLE DESIGN
• TURBINE DESIGN
• SUBSONIC NOZZLE DESIGN
6

7. SINGLE SPIKE SUPERSONIC DIFFUSER
7

8. 1. SUPERSONIC DIFFUSER DESIGN
SINGLE WEDGE SPIKE SUPERSONIC DIFFUSER
DOUBLE WEDGE SPIKE SUPERSONIC DIFFUSER
DATA GIVEN
Sea level condition, at 20,000m from standard atmospheric conditions
Mach no, M1 = 3.0
Static temperature, T1 = 216.66 K
Static pressure, P1 = 0.055293 bar
Static density, ρ1 = 0.08909 kg/m³
Mass flow rate , = 20 kg/s
Flow deflection angle (wedge angle) = 50,5.50,60,6.50,70,7.50,80 ,8.50, 90 ,9.50,100 ,10.50
Combustion chamber inlet mach number(Mcci)=0.4
8

9. Calculations:
●To calculate Mach number:
M₁= 3 , Θ=50 , ẞ=23.130
We have,
Mn₁ =M ₁ SINẞ =1.178456
 Mn₂= 1+(ɤ-1)/2*(M₁sinẞ)² = 0.876218
ɤ(M₁sinẞ) ²-(ɤ-1)/2
M₂ =
Mn₂ = 2.815846
Sin(ẞ-Ө)
 M₃ = 1+(ɤ-1)/2*(M₂)² = 0.487049
ɤM₂ ²-(ɤ-1)/2
9

10. Table No. 1:
Calculation of M₂ & M₃:
M₁ Ө ẞ Mn₁ Mn₂ M₂ M₃
3 5 23.13 1.178456 0.876218 2.815846 0.487049
3 5.5 23.53 1.197688 0.862917 2.787967 0.489028
3 6 23.93 1.216861 0.85022 2.761757 0.490938
3 6.5 24.34 1.236452 0.837792 2.734666 0.492963
3 7 24.76 1.256455 0.825636 2.706734 0.495108
3 7.5 25.18 1.27639 0.814026 2.680358 0.497187
3 8 25.61 1.296729 0.802669 2.653132 0.499392
3 8.5 26.04 1.316996 0.791813 2.627365 0.501534
3 9 26.49 1.338125 0.780957 2.598517 0.504
3 9.5 26.93 1.358705 0.770809 2.573305 0.506214
3 10 27.38 1.37967 0.760879 2.547237 0.508565
3 10.5 27.84 1.401012 0.75117 2.520355 0.511056 10

11. Angle(Ө) Vs Mach number
0.475
0.48
0.485
0.49
0.495
0.5
0.505
0.51
0.515
1 2 3 4 5 6 7 8 9 10 11 12
MACH(M₃)
ANGLE(Ө)
M₃
11

12. ●To calculate pressures:
Total or stagnation pressure,
 P₀₁ = P₁ [1 + (ɤ-1)M₁²] ^ɤ/ɤ-1
2
 P₀₁ =2.031062 bar
 P₂=P₁ *[1+2ɤ*(M₁²sin²ẞ-1)] =0.080371 bar
ɤ+1
 P₀₂= P₂ [1 + (ɤ-1)M₂²] ^ɤ/ɤ-1 =0.730078 bar
2
 P₃= P₂* [1+ 2ɤ (M₂²-1) = 0.730078
ɤ+1
 P₀₃= P₁ *[1+(ɤ-1)*M₃²] ^ɤ/ɤ-1
2
= 0.85867 bar
12

13. Table No. 2:
Calculation for pressures and it’s losses:
Ө P₁(bar) p₂(bar) p₃(bar) p₀₁(bar) p₀₂(bar) p₀₃(bar) Loss of
pressure
(%)
5 0.055293 0.080371 0.730078 2.031062 2.234508 0.85867 57.72313
5.5 0.055293 0.083319 0.741669 2.031062 2.219963 0.873429 56.99645
6 0.055293 0.086305 0.753607 2.031062 2.209175 0.888597 56.24963
6.5 0.055293 0.089406 0.765147 2.031062 2.195467 0.903406 55.5205
7 0.055293 0.092623 0.776253 2.031062 2.178964 0.917816 54.81104
7.5 0.055293 0.09588 0.787658 2.031062 2.165895 0.932584 54.08395
8 0.055293 0.099256 0.798577 2.031062 2.150019 0.946898 53.37915
8.5 0.055293 0.102673 0.809771 2.031062 2.137298 0.961545 52.65804
9 0.055293 0.106292 0.819618 2.031062 2.116091 0.974847 52.00312
9.5 0.055293 0.109872 0.830511 2.031062 2.103598 0.989276 51.29269
10 0.055293 0.113576 0.840818 2.031062 2.088338 1.003144 50.60988
10.5 0.055293 0.117404 0.850501 2.031062 2.070458 1.016413 49.9565913

14. Ө Vs loss of pressure
46
48
50
52
54
56
58
60
1 2 3 4 5 6 7 8 9 10 11 12
LOSSOFPRESSURE(%)
ANGLE(Ө)
Loss of pressure(%)
14

15. ●To calculate area,velocity and densities :
 ρ₃= p₃ * 10⁵ =0.426752 kg/m^3
R * T₃
 V₃= M₃*(ɤRT₃)^1/2 =238.3599 m/s
 A₃ = ṁ =0.196617
ρ₃*v₃
 ρ₁/ρ₂ = (ɤ-1) *M₁² sin² ẞ+2 = 0.766723
(ɤ+1) M₁²sin²ẞ
 P₂/P₁ = (ɤ-1) M₁ ²* sin²ẞ+2 = 1.453552
(ɤ+1)M₁ ²*sin²ẞ
15

16. Table No. 3:
Ө ρ₃ ρcci ρ₁/ρ₂ P₂/P₁ V₃(m/s) A₃ Acci
5 0.426752 0.4429 0.766723 241.4614 238.3599 0.196617 0.22897
5.5 0.434288 0.451137 0.747607 244.0768 239.119 0.192592 0.224946
6 0.441803 0.459354 0.729445 246.683 239.91 0.188692 0.221014
6.5 0.449243 0.467534 0.711752 249.3466 240.7188 0.184943 0.217269
7 0.456585 0.475655 0.694534 252.0691 241.5486 0.181344 0.213709
7.5 0.463878 0.483728 0.678174 254.7871 242.4102 0.177859 0.210233
8 0.471042 0.491713 0.662254 257.5668 243.2955 0.174516 0.206937
8.5 0.478145 0.499637 0.647119 260.3448 244.2114 0.171279 0.20372
9 0.484984 0.507382 0.632066 263.2518 245.1525 0.168216 0.200775
9.5 0.491842 0.515105 0.618074 266.0952 246.1263 0.165214 0.197806
10 0.498508 0.522679 0.604459 269.0054 247.1298 0.162343 0.195005
10.5 0.504956 0.530079 0.591223 271.9836 248.1664 0.1596 0.192371
16

17. Ө Vs Acci
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.24
1 2 3 4 5 6 7 8 9 10 11 12
COMBUSTIONCHAMBERINLETAREA
ANGLE(Ө)
Acci
17

18. ●Calculation for temperatures:
 T₁ = 216.66 K(given)
 T₂=P₂* ρ ₁*T₁ = 241.4614 K
P₁* ρ₂
 T₃= T₂[1+2(ɤ-1) (ɤM₁²+1) (M₂²-1)] = 596.09 K
(ɤ+1)² * M₂²
 T ₀₁=T₁ *[1+(ɤ-1)M₁²] =606.648 K
2
 T ₀₂=T₁ *[1+(ɤ-1)M₂²]=624.3704 K
2
 T ₀₃=T₁ *[1+(ɤ-1)M₃²]=624.3704 K
2
 T₅=(P₅/P₀₄) ^ɤ/ɤ-1 *T₀₄ =1096.178 K
18

19. Table No:4
Ө T₂(K) T₃(K) T₀₁(K) T₀₂=T₀₃(K) T₅(K)
5 241.4614 596.09 606.648 624.3704 1096.178
5.5 244.0768 595.0459 606.648 623.5067 1090.853
6 246.683 594.3385 606.648 622.988 1085.5
6.5 249.3466 593.4462 606.648 622.2892 1080.386
7 252.0691 592.3792 606.648 621.4213 1075.512
7.5 254.7871 591.6318 606.648 620.8816 1070.619
8 257.5668 590.7112 606.648 620.175 1065.969
8.5 260.3448 590.0934 606.648 619.7794 1061.304
9 263.2518 588.8471 606.648 618.7624 1057.147
9.5 266.0952 588.3529 606.648 618.5063 1052.718
10 269.0054 587.6894 606.648 618.0892 1048.539
10.5 271.9836 586.8664 606.648 617.5218 1044.61
19

20. Angle(Ө) Vs Temperature(T₅)
1010
1020
1030
1040
1050
1060
1070
1080
1090
1100
1110
1 2 3 4 5 6 7 8 9 10 11 12
TEMPERATURE(T₅)
ANGLE(Ө)
T₅
20

21. ●Calculation for thrust:
Thrust without Bleed(T)=ṁ(Cj-Ci) ,ṁ=20kg
= 14.6741 KN
Thrust with Bleed(T)= ṁ(Cj-Ci) , ṁ=18kg
=13.20669 KN
Cj=√(2*R*(T₀₄-T₅) Ci=3*√Cp(T₀₄-T₅)
=1618.852 m/s = 885.1474 m/s
21

22. Table No: 5
Ө Ci (m/s) Cj (m/s) Thrust(wit
hout
bleeding)
KN
Thrust(wit
h bleeding)
KN
Loss in
thrust (KN)
Loss of
thrust(%)
5 885.1474 1618.852 14.6741 13.20669 1.46741 10
5.5 885.1474 1622.155 14.74014 13.26613 1.474014 10
6 885.1474 1625.468 14.8064 13.32576 1.48064 10
6.5 885.1474 1628.626 14.86958 13.38262 1.486958 10
7 885.1474 1631.631 14.92968 13.43671 1.492968 10
7.5 885.1474 1634.643 14.98991 13.49092 1.498991 10
8 885.1474 1637.499 15.04703 13.54232 1.504703 10
8.5 885.1474 1640.359 15.10423 13.59381 1.510423 10
9 885.1474 1642.905 15.15514 13.63963 1.515514 10
9.5 885.1474 1645.611 15.20928 13.68835 1.520928 10
10 885.1474 1648.162 15.26028 13.73425 1.526028 10
10.5 885.1474 1650.556 15.30817 13.77735 1.530817 10
22

23. Thrust Vs Angle(Ө)
12
12.5
13
13.5
14
14.5
15
15.5
1 2 3 4 5 6 7 8 9 10 11 12
THRUST(T)
ANGLE(Ө)
Thrust(without bleeding) KN
Thrust(with bleeding) KN
23

24. DOUBLE WEDGESPIKESUPERSONICDIFFUSER:
24

25.  Double wedge spike :
PERFORMANCE PARAMETERS:
25

26.  Mach No. Vs Loss Of Thrust:
graph:
0
0.5
1
1.5
2
2.5
3
3.5
1 2 3
LOSSOFTHRUST
MACH NO
MACH NO
LOSS OF THRUST
26

27. HEAT EXCHANGER:
• A heat exchanger may be defined as an equipment which
transfers energy from a hot fluid to cold fluid with maximum
rate and minimum investment and running cost.
• Eg.
 Intercoolers and preheaters.
 Condensers and evaporators in
refrigeration unit.
 Oil coolers of heat engines etc.
27

28. Types Of Heat Exchangers:
 On basis of Nature:
 Direct contact heat exchanger
Indirect contact heat exchanger
 On basis of Flow direction:
Parallel flow heat exchanger
Counter flow heat exchanger
Cross flow heat exchanger
28

29. FINS
 These are the elongated body parts that extend outside
of the body which helps in heat transfer.
 The heated surface transfer it’s part of heat to the
elongated fins.
 These are classified based on shapes as:
i) Rectangular
ii) Square
iii) Triangular
iv) Pentagonal etc.
29

30. Selection of fin materials:
 The temperature of combustion chamber is about 2400k.
 Fin should have high melting point based on which given material
can be selected:
1. Tantalum hafnium carbide(Ta₄HFC₅)
melting point: 3290K – 4488K
2. Titanium Carbide (Tic)
melting point: 3433K
OR
alloys of zirconium,Ti,Hf,Ta etc.
We have selected molybdenum for our fin material,
Melting point= 2896 k, boiling point= 4912k
Thermal conductivity at 2500k = 85.89 w/mk
30

31. •CALCULATION OF HEAT TRANSFER BY FINS:
Given data:
Tᵤ=624.4K Tᵥ= 2400K
Film temp(T)= Tᵤ+Tᵥ = 1512.2K = 1239.2°C
2
At 1200°C, from HMTD
Density(Þ)=0.239 kg/mᶟ
Kinematic viscosity(ʋ)=223.70*10^-⁶
Prandlt number(P᷊)=0.724
Thermal conductivity(k)=0.09153w/Mk
Diameter(D)=0.4m
Length of cylinder(L)=1m
Thickness of fin(t)=0.005m
Length of fin(l)=0.05m
Area of fin= 0.05*0.05=0.0025m²
31

32. CALCULATION:
P₀=P₃+1/2*Þ*V ² =7.32*10⁴ N/m ²
u= ʋ*1 bar = 3.05*10^-⁴ ,velocity(V) = 50m/s
P₀
Now reynolds number(Re)=U*D/u =6.54*10⁴
Nusselt number(Nux)=0.332*(Re)^0.5*(Pr)^0.333=76.27
We know, Nux=hx*D , hx=17.46w/m²k , h=2*hx=34.90w/m²k
K
Perimeter(P)=2*L=2m , m=(h*P /KA)^0.5 = 123.49
Also , m*l = 6.17
Heat transfer by 1 fin,
Q=m*K*A*(Tᵥ-Tᵤ)*tan(m*l)=1003.47W
Therefore heat transfer by 95 fins(Q₁)=95329.65W
32

33. Heat transfer by unfinned surface(Q₂)= h*A*(Tᵥ-Tᵤ)
=h*(ᴫ*D*L-95*t*l)*1775.6
= 76400.08W
CALCULATION FOR TEMERATURE OF FIN(Tf) = Tᵥ-Tᵤ + Tᵤ = 631.82K
cosh (ml)
Overall Heat transfer ,
Q= Q₁+Q₂ = 171729.73W
Again,
Calculation for temperature of air(T) = Q/(m*K*A*tanh(ml))+Tf
=944.79K
where k of fin material=85.89W/mK
33

34. •Table no.1
Tᵤ Tᵥ P₀(N
/m²)
Re
No.
Nux h
w/m
²k
No
of fin
Q
(with
fin)
W
Velo
city
(v)
m/s
Q
(with
out
fin)
Q
(over
all)
W
Tem
p of
fin(Tf
) K
Tem
p of
air
(T)
K
596.09 2400 73679.99 98810.89 93.72025 42.89107 95 107374.9 50 95387.62 202762.5 599.9298 932.6136
595.046 2400 74597.1 80032.65 84.34603 38.60096 96 102995 60 85878.89 188873.9 600.5054 939.1265
594.339 2400 75946.25 95060.13 91.92428 42.06914 97 108685.2 70 93612.55 202297.8 598.4438 945.8596
593.446 2400 77279.5 110547.3 99.13002 45.36685 98 114084.7 80 100980 215064.7 596.611 952.2748
592.379 2400 78593.25 126480 106.0332 48.5261 99 119264.6 90 108053.9 227318.5 594.8677 958.3528
591.632 2400 79960.8 142978.6 112.737 51.5941 100 124270.6 100 114909.7 239180.3 593.6164 964.5231
590.711 2400 81303.65 159917.8 119.2283 54.56484 101 129141.9 110 121563.2 250705.2 592.3157 970.363
590.093 2400 82697.9 177447.4 125.5931 57.4777 105 137840.5 120 127992.4 265832.9 591.403 981.9725
588.847 2400 83981.35 195218.2 131.732 60.28713 110 147993.2 130 134204.5 282197.7 589.9294 994.7665
588.353 2400 85739.85 229968.3 142.9766 65.43325 115 161232.4 150 145551.7 306784.2 589.124 1011.572
587.689 2400 87141 249308.2 148.8673 68.12912 120 171736.2 160 151449.7 323185.9 588.3386 1024.478
586.866 2400 88503.65 269032.1 154.644 70.77283 125 182412.6 170 157237.6 339650.2 587.4167 1037.13234

35. •2-D diagram of the engine:
35

36. •CALCULATION OF DIFFUSER PRESSURE PLACED BEFORE C.C.
Given,
Temperature(T)=596.09K
Cp value of dry air at 596.09K = 1045
Inlet velocity(V₁)= 195.77m/s
Inlet pressure(P₁)=0.732 bar
Inlet mach(M₁)=0.4
Exit Velocity(V₂)= 75m/s
D₂= 0.34
Mass flow rate(ṁ) = 2/0.65kg/s
 T₀₁= T₁ *[1+(ɤ-1)M₁²] = 615.27K
2
 P₀₁ = P₁ [1 + (ɤ-1)M₁²] ^ɤ/ɤ-1 = 0.8173 bar
2
 ṁ/A1= P₀₁ [ 2ɤ/ɤ+1)*(P₁/P₀₁)^2/ɤ-(P₁/P₀₁)^ɤ+1/ɤ]^1/2
(RT₀₁)^1/2
A₁ = 0.023m² , D₁=0.29m
Therefore diameter of the inlet diffuser(D)= D₂-D₁=0.34m-0.29m = 0.5m
36

37. FOR GIVEN DIA(d₄)=0.4
The bleed gap = D₂-D₁+0.6= 0.34-0.29+0.6
=0.11m
Total bleed gap = 2* 0.11 = 0.22 m
Therefore,
D₃= D₄-Total bleed gap
D₃ = 0.4-0.22 = 0.18m
Hence exit diameter of the diffuser at C.C.(D´) = D₄-D₁
= 0.4-0.29
= 0.11m
37

38. Next Work
• In the next report we will submit the complete
design of the Heat Exchanger and the Turbine.
38

39. work in progress
THANK YOU
39