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PRELIMINARY DESIGN OF AN ENERGY RECOVERY SYSTEM MBDA PROJECT Presentation BY Rajendra PatelKishalay Kumar sinhaJohn Paul

Under the guidance of:Mr. S.A.HASHIMAsst. ProfessorDepartment of Aeronautical Engineering

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Abstracts:A ramjet is a form of an air breathing jet engine.It uses engines 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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What our project basically isIntroducing 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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Our projects 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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Ramjet Engine:

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PROJECT PREFACE:SUPERSONIC DIFFUSER DESIGNTHRUST CALCULATION WITH & WITHOUT BLEEDINGCOMBUSTION CHAMBER HEAT TARNSFER AND ENERGY RECOVERY FROM COMBUSTOR TO DIFFUSED BLEED AIR FOR TUTBINE INLETSUPERSONIC NOZZLE DESIGNTURBINE DESIGNSUBSONIC NOZZLE DESIGN

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SINGLE SPIKE SUPERSONIC DIFFUSER

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1. SUPERSONIC DIFFUSER DESIGN

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Calculations:To calculate Mach number:M= 3 , =50 , =23.130 We have,Mn =M SIN =1.178456

Mn= 1+(-1)/2*(Msin) = 0.876218 (Msin) -(-1)/2

M = Mn = 2.815846 Sin(-)

M = 1+(-1)/2*(M) = 0.487049 M -(-1)/2

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Table No. 1: Calculation of M & M:MMnMnMM3523.131.1784560.8762182.8158460.48704935.523.531.1976880.8629172.7879670.4890283623.931.2168610.850222.7617570.49093836.524.341.2364520.8377922.7346660.4929633724.761.2564550.8256362.7067340.49510837.525.181.276390.8140262.6803580.4971873825.611.2967290.8026692.6531320.49939238.526.041.3169960.7918132.6273650.5015343926.491.3381250.7809572.5985170.50439.526.931.3587050.7708092.5733050.50621431027.381.379670.7608792.5472370.508565310.527.841.4010120.751172.5203550.511056

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Angle() Vs Mach number11

To calculate pressures:Total or stagnation pressure, P = P [1 + (-1)M] ^/-1 2 P =2.031062 bar P=P *[1+2*(Msin-1)] =0.080371 bar +1 P= P [1 + (-1)M] ^/-1 =0.730078 bar 2P= P* [1+ 2 (M-1) = 0.730078 +1P= P *[1+(-1)*M] ^/-1 2 = 0.85867 bar12

Table No. 2:Calculation for pressures and its losses:P(bar)p(bar)p(bar)p(bar)p(bar)p(bar)Loss of pressure(%)50.0552930.0803710.7300782.0310622.2345080.8586757.723135.50.0552930.0833190.7416692.0310622.2199630.87342956.9964560.0552930.0863050.7536072.0310622.2091750.88859756.249636.50.0552930.0894060.7651472.0310622.1954670.90340655.520570.0552930.0926230.7762532.0310622.1789640.91781654.811047.50.0552930.095880.7876582.0310622.1658950.93258454.0839580.0552930.0992560.7985772.0310622.1500190.94689853.379158.50.0552930.1026730.8097712.0310622.1372980.96154552.6580490.0552930.1062920.8196182.0310622.1160910.97484752.003129.50.0552930.1098720.8305112.0310622.1035980.98927651.29269100.0552930.1135760.8408182.0310622.0883381.00314450.6098810.50.0552930.1174040.8505012.0310622.0704581.01641349.95659

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Vs loss of pressure14

To calculate area,velocity and densities := p * 10 =0.426752 kg/m^3 R * TV= M*(RT)^1/2 =238.3599 m/s A = =0.196617 *v

/ = (-1) *M sin +2 = 0.766723 (+1) Msin P/P = (-1) M * sin+2 = 1.453552 (+1)M *sin

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Table No. 3:cci/P/PV(m/s)AAcci50.4267520.44290.766723241.4614238.35990.1966170.228975.50.4342880.4511370.747607244.0768239.1190.1925920.22494660.4418030.4593540.729445246.683239.910.1886920.2210146.50.4492430.4675340.711752249.3466240.71880.1849430.21726970.4565850.4756550.694534252.0691241.54860.1813440.2137097.50.4638780.4837280.678174254.7871242.41020.1778590.21023380.4710420.4917130.662254257.5668243.29550.1745160.2069378.50.4781450.4996370.647119260.3448244.21140.1712790.2037290.4849840.5073820.632066263.2518245.15250.1682160.2007759.50.4918420.5151050.618074266.0952246.12630.1652140.197806100.4985080.5226790.604459269.0054247.12980.1623430.19500510.50.5049560.5300790.591223271.9836248.16640.15960.192371

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Vs Acci17

Calculation for temperatures:T = 216.66 K(given)T=P* *T = 241.4614 K P* T= T[1+2(-1) (M+1) (M-1)] = 596.09 K (+1) * MT =T *[1+(-1)M] =606.648 K 2T =T *[1+(-1)M]=624.3704 K 2

T =T *[1+(-1)M]=624.3704 K 2T=(P/P) ^/-1 *T =1096.178 K

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Table No:4T(K)T(K)T(K)T=T(K)T(K)5241.4614596.09606.648624.37041096.1785.5244.0768595.0459606.648623.50671090.8536246.683594.3385606.648622.9881085.56.5249.3466593.4462606.648622.28921080.3867252.0691592.3792606.648621.42131075.5127.5254.7871591.6318606.648620.88161070.6198257.5668590.7112606.648620.1751065.9698.5260.3448590.0934606.648619.77941061.3049263.2518588.8471606.648618.76241057.1479.5266.0952588.3529606.648618.50631052.71810269.0054587.6894606.648618.08921048.53910.5271.9836586.8664606.648617.52181044.61

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Angle() Vs Temperature(T)20

Calculation for thrust:Thrust without Bleed(T)=(Cj-Ci) ,=20kg = 14.6741 KNThrust with Bleed(T)= (Cj-Ci) , =18kg =13.20669 KN Cj=(2*R*(T-T) Ci=3*Cp(T-T) =1618.852 m/s = 885.1474 m/s21

Table No: 5Ci (m/s)Cj (m/s)Thrust(without bleeding) KNThrust(with bleeding) KNLoss in thrust (KN)Loss of thrust(%)5885.14741618.85214.674113.206691.46741105.5885.14741622.15514.7401413.266131.474014106885.14741625.46814.806413.325761.48064106.5885.14741628.62614.8695813.382621.486958107885.14741631.63114.9296813.436711.492968107.5885.14741634.64314.9899113.490921.498991108885.14741637.49915.0470313.542321.504703108.5885.14741640.35915.1042313.593811.510423109885.14741642.90515.1551413.639631.515514109.5885.14741645.61115.2092813.688351.5209281010885.14741648.16215.2602813.734251.5260281010.5885.14741650.55615.3081713.777351.53081710

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Thrust Vs Angle()23

DOUBLE WEDGE SPIKE SUPERSONIC DIFFUSER:

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Double wedge spike :PERFORMANCE PARAMETERS:

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Mach No. Vs Loss Of Thrust:graph:26

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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Types Of Heat Exchangers:On basis of Nature: Direct contact heat exchangerIndirect contact heat exchanger On basis of Flow direction:Parallel flow heat exchangerCounter flow heat exchangerCross flow heat exchanger28

FINSThese are the elongated body parts that extend outside of the body which helps in heat transfer. The heated surface transfer its 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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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(TaHFC) melting point: 3290K 4488K 2. Titanium Carbide (Tic) melting point: 3433KOR 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

CALCULATION OF HEAT TRANSFER BY FINS:Given data:T=624.4K T= 2400KFilm temp(T)= T+T = 1512.2K = 1239.2C 2At 1200C, from HMTDDensity()=0.239 kg/mKinematic viscosity()=223.70*10^-Prandlt number(P)=0.724Thermal conductivity(k)=0.09153w/MkDiameter(D)=0.4mLength of cylinder(L)=1mThickness of fin(t)=0.005mLength of fin(l)=0.05mArea of fin= 0.05*0.05=0.0025m 31

CALCULATION:P=P+1/2**V =7.32*10 N/m u= *1 bar = 3.05*10^- ,velocity(V) = 50m/s PNow reynolds number(Re)=U*D/u =6.54*10Nusselt number(Nux)=0.332*(Re)^0.5*(Pr)^0.333=76.27 We know, Nux=hx*D , hx=17.46w/mk , h=2*hx=34.90w/mk K Perimeter(P)=2*L=2m , m=(h*P /KA)^0.5 = 123.49Also , m*l = 6.17Heat transfer by 1 fin, Q=m*K*A*(T-T)*tan(m*l)=1003.47WTherefore heat transfer by 95 fins(Q)=95329.65W

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Heat transfer by unfinned surface(Q)= h*A*(T-T) =h*(*D*L-95*t*l)*1775.6 = 76400.08WCALCULATION FOR TEMERATURE OF FIN(Tf) = T-T + T = 631.82K cosh (ml) Overall Heat transfer ,Q= Q+Q = 171729.73WAgain,Calculation for temperature of air(T) = Q/(m*K*A*tanh(ml))+Tf =944.79K where k of fin material=85.89W/mK33

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Table no.1TTP(N/m)Re No.Nuxh w/mk No of finQ (with fin) WVelocity (v) m/sQ (without fin) Q (overall) WTemp of fin(Tf) KTemp of air(T)K596.09240073679.9998810.8993.7202542.8910795107374.95095387.62202762.5599.9298932.6136595.046240074597.180032.6584.3460338.60096961029956085878.89188873.9600.5054939.1265594.339240075946.2595060.1391.9242842.0691497108685.27093612.55202297.8598.4438945.8596593.446240077279.5110547.399.1300245.3668598114084.780100980215064.7596.611952.2748592.379240078593.25126480106.033248.526199119264.690108053.9227318.5594.8677958.3528591.632240079960.8142978.6112.73751.5941100124270.6100114909.7239180.3593.6164964.5231590.711240081303.65159917.8119.228354.56484101129141.9110121563.2250705.2592.3157970.363590.093240082697.9177447.4125.593157.4777105137840.5120127992.4265832.9591.403981.9725588.847240083981.35195218.2131.73260.28713110147993.2130134204.5282197.7589.9294994.7665588.353240085739.85229968.3142.976665.43325115161232.4150145551.7306784.2589.1241011.572587.689240087141249308.2148.867368.12912120171736.2160151449.7323185.9588.33861024.478586.866240088503.65269032.1154.64470.77283125182412.6170157237.6339650.2587.41671037.132

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2-D diagram of the engine:35

CALCULATION OF DIFFUSER PRESSURE PLACED BEFORE C.C. Given,Temperature(T)=596.09KCp value of dry air at 596.09K = 1045Inlet velocity(V)= 195.77m/sInlet pressure(P)=0.732 barInlet mach(M)=0.4Exit Velocity(V)= 75m/sD= 0.34Mass flow rate() = 2/0.65kg/s T= T *[1+(-1)M] = 615.27K 2P = P [1 + (-1)M] ^/-1 = 0.8173 bar 2

/A1= P [ 2/+1)*(P/P)^2/-(P/P)^+1/]^1/2 (RT)^1/2 A = 0.023m , D=0.29m

Therefore diameter of the inlet diffuser(D)= D-D=0.34m-0.29m = 0.5m

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FOR GIVEN DIA(d)=0.4The bleed gap = D-D+0.6= 0.34-0.29+0.6 =0.11mTotal bleed gap = 2* 0.11 = 0.22 m Therefore, D= D-Total bleed gap D = 0.4-0.22 = 0.18mHence exit diameter of the diffuser at C.C.(D) = D-D = 0.4-0.29 = 0.11m 37

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

work in progress THANK YOU39