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