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10/17/02 ME3324 Collaborative Problem 6 1Given: A ramjet engine operating at steady state with known states at the inlet (T1 =-300K, P1 = 101 kPa, Ad = 0.28 m2, Va = 250 m/s), combustor (AF = 15, LHV = 41,000kJ/kg, and comb), and exit (T4 = 2000K).DiffuserCombustorNozzle1234ADAnVaVjmfFind: The overall engine efficiency as a function of the combustor efficiency.Solution:Use equations developed in previous period to solve this problem. Start with theoverall efficiency defined by eqn. (14) in previous solution that is posted. LHVmVVVmLHV fuelpowerthrust fuelaajairo(1)Note that the jet velocity, air mass flow rate, and fuel mass flow rate are unknown.However, the ratio of the air mass flow rate to the fuel mass flow rate is simply the air-to-fuel ratio, or AF, and is given. Therefore, the only unknown in (14) is the jet velocity.This can be found in terms of known values through the analysis of the enginecomponents.It is easiest to work from the rear of the engine to the front, starting with the equationfor the jet exhaust velocity – eqn. (12) from before. 434hh2V (2)The velocity at state 4 is the jet velocity. Treating air as an ideal gas, the difference inenthalpy under the radical sign can be written as: 43p43TTChh (3)Since Cp and T4 are both given, the exit velocity becomes a matter of finding T3. Usingthe energy balance from the combustor (eqn. (8) from before) provides this quantity.AFLHVhmAFmLHVhmLHVmhhcomb2airaircomb2airfuelcomb23(4)10/17/02 ME3324 Collaborative Problem 6 2It is possible to find T3 by solving eqn. (4), since h is a function of temperature only foran ideal gas. Alternately, the Cp relation of eqn. (3) can be used to get eqn. (4) explicitlyin terms of temperatures. Making this substitution yields:AFCLHVTTpcomb23(5)Unfortunately, eqn. (5) still has T2 as an unknown value. This requires the analysis of thediffuser, given by eqn. (3) of the previous solution.)T(f2Vhh2Vhm2Vhm2211222222111(6)In this equation, h2, and hence T2, is the only unknown. Furthermore, state 2 does notdepend at all on the combustor efficiency. As such, it is a constant for all values ofcombustor efficiency, and hence all values of overall efficiency.Equation 6 can be re-written, assuming a constant specific heat over the range ofdiffuser temperatures (an initial exact analysis shows the temperature rise will only be onthe order of 30 K, so Cp = 1.0 kJ/kgK at T = 300K to T = 350K):p2112C2VTT (7)Combining equations (1) – (7) provides an explicit expression of the engine efficiency interms of known quantities and the combustor efficiency. This expression is:a4combcomb,pdif,p2a1noz,paoVTAFCLHVC2VTC2LHVAFVE x h a u s t v e l o c i t yF u e la d d i t i o nR a mc o m p r e s s i o nE n t e r i n gD u m p e di n e x h a u s t(8)Note the 1 unknown is the boxed quantity, the combustor efficiency, which will be variedbetween 0.9 and 1.0. Since the combustor efficiency is under the radical sign, it followsthat the overall efficiency will increase as the square root of the combustor efficiency.This means that an increase in the combustor efficiency will lead to a more marginalimprovement in the overall efficiency. For the constants defined in the problem, andminding the specific heat value for each engine component, the following plot isgenerated.10/17/02 ME3324 Collaborative Problem 6 300.010.020.030.040.050.060.070.080.090.10.7 0.75 0.8 0.85 0.9 0.95 1comboFigure 1. Overall Ramjet EfficiencyA few interesting points to notice from the plot. The overall engine efficiency isextremely low – less than 10% regardless of the combustor efficiency. This is an artifactof forcing the exit temperature to be 2000K and the relatively low velocities of themissile and the exhaust. Ramjets are generally more effective when operating at speedsover the speed of sound. However, flow compressibility effects need to be accounted forat such high velocities, making the analysis beyond the scope of this class.Another interesting aspect of the plot is that improving the combustor efficiency from75% to 100%, the overall engine efficiency only increases from 1% to 9%. This is afactor that can be seen in eqn. (8) where the overall efficiency varies only as the squareroot of the combustor efficiency. In practical terms, a huge improvement in a componentefficiency only results in a modest improvement in overall efficiency. It’s tough sleddingout


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U of M ME 3324 - Collaborative Problem

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