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Aviation History
1941
1941 - 0904.PDF
• 286 EXHAUST (Continued) and the whole of the " tail " energy will be taken as avail- able to appear as kinetic energy in the efflux stream : subject to subsequent allowance for losses. Then, the mass of the charge, and therefore of the efflux gases, will be 70,000/250,000 = 0.28 lb. per sec. ; and on the 50 per cent, basis the energy in the " tail " ---- 35,oooft./Ib. per sec. In the equation v — V'zgs s = 35,000/0.28 = 125,000ft. Hence v = -v/64.4 x 125,000 = 2,840ft./sec. Thus by the rocket equation 2V/11, the effective cfliux efficiency is 800/2,840 = 0.281, say 28 per cent. And 0.28 x 35,000 = 9,800ft. lb. available. Thus "gain " in terms of engine i.h.p. = 9,800/70,000 = 14 per cent. • In the earlier articles reasons were given for discounting this figure owing to kisses which must be regarded as un- avoidable. As a provisional estimate, which is, in truth, little more than a " considered guess," the writer assumed that not more than half the energy theoretically available could be usefully directed ; in other words the efflux velocity as calculated must be reduced in the ratio -1/2 * 1. On this basis the effective efficiency as given by the rocket equatiorTTbecomes 800/2,000 = 0.4 or 40 per cent. But now the energy is only half, so that the available energy is, 0.4 x 17,500 = 7,000, that is to say 10 per cent, of the engine i.h.p. That is the writer's expectation of the practical limit of exhaust efflux auxiliary propulsion on the data above assumed. In order to add realism to the discussion, the diagram given in Fig. 1 may be considered as relating to an engine of approximately 5m. bore by 5m. stroke, containing 100 cu. in. stroke volume, giving a cylinder constant = 100/12 — S.^ units oi 12 cu. in. The compression ratio being 0.17, the total volume is 8.3 + 1.7 = 10, which is the volume given at points A and D in the figure, and used in the numerical calculations. Working under normal sea-level conditions, the initial induction pressure being taken at 13 lb./sq. in. the calculated mean pressure is 107. and the ft. lb. per working stroke 107 X 8.3 = 888 (more exact calculation gives the figure as 889). The detail results of calculation are given in Table 1, founded on gamma values, compression E to B, y — 1.36 and for expansion C to J, y = 1.25. TABLE I Data Relating to Diagram Plotted. Pressureper sq. in. (absolute) A B C D E F H J K L 13 M5 500 54-7 15 i 6.5 0 0-5 0 0 Volume12 cu. in units 10.0 i-7 !-7 10.0 16.6 28.2 8.3 IO.O 550 16.6 550 A H H K J J A Calculated Areas M.P. x Volume Giving Energy in ft. lbs. BC AE EJ AD AD BC D L K L J J (Without F A D F = 889 = + 760 = — 61 = - 250 = + 449 440 ~ 88u " •••>O;> Say 50"„ Supercharge = 173) Compression y — 1.36 Expansion y = 1.25 ABC D 889 = 19-5 Mean Cylinder pressure A 15 C D = 889 — 107 lb. sq. in. Having dealt with a special case by the same procedure as previously adopted, the way has been prepared for the problem to be treated in more general terms. Symbols :— V == velocity of flight : ft./sec. T = thrust in lb. W = work done in cylinder, ft.lb. 'sec. (indicated). APRIL IJTH, T941. 7) •=-- mechanical efficiency of engine. t = efficiency of propeller. R = ratio of " tail " area of diagram to area of cylinder diagram : i.e. energy relation. E = Energy ft. lb. exhibited in cylinder diagram per lb. of charge. m = mass per second in efflux gases. v == efflux velocity. s == " head " or height in feet corresponding to efflux velocity. VT Then, VT = TjeW or, W == — and RW = work done per sec. as represented by " tail " of diagram = RVT and m = W/E = VT/iyeE /. s = 1WT ,eE _ and efflux effi-Then v = V' zgs = V^H-4 RE = S ciency (by the rocket equation) = 2V/8 \/fUL .'. the work " 2Vavailable for efflux propulsion will be RW x —. 8-v/RE 4VE energy is - x W. Oi in terms of W the available efflux V A/R 4VE Applying this to the special case just dealt with, we take as before, V = 400ft./sec, E = 250,000ft./lb. and R = 0.5, then the work done by the exhaust efflux in terms of WT (the engine power) is :— 400 x 0.7 = 0.14 or, 14 per cent., which is the result for 4 x 500 /i > f t this particular case already given. In Table II and Fig. 2, corresponding values are given as calculated from the above equation, for data ranging from V = 200ft./sec. to 800ft./sec, and values of R from 0.2 to 0.7. The value of R depends upon the degree of supercharge ; R = 0.2 corresponds to the condition of no supercharge ; R = 0.5 corresponds to supercharge in the relation of 2 : 1 ; and R = 0.7 to a degree of supercharge approximately in the relation of 4 : 1. To make this clear a scale is given showing the pressure without supercharge as unity ; and the lower pressures at which the air enters the compressor corresponding to R values. Each graph in Fig. 2 relates to a definite value of V. Abscissae (upper scale) give " gain " in terms of indicated h.p., as calculated. The lower scale gives " gain " as a percentage, in terms of thrust h.p., the value of 7?e being taken = 62.5 per cent. The figures given are as calculated; there has been no deduction for losses of any kind. \T • 3 •4 .=! .6 •7 200 .044 •055 .063 .071 .077 .08^ 300 .066 .082 .094 .106 .116 •125 TABLE 400 .088 .109 .126 .141 •155 .167 500 .111 .136 •157 .176 •193 .209 n 600 •133 .164 .189 .211 .232 .250 700 •J55 .191 .221 .246 .271 .292 800 •177.218 .252 .281 .310 •334 13.0 11.8 9-3 6.5 4-3 2.9 If we denote the mechanical efficiency of efflux by the symbol n the energy available on the jet from the " tail " of the diagram will be «RE (instead of RE), then, since W = VT/™, HRW = »RVT,V and as before, KRVT ,eE = «RE, v = -\/ 's = X VT Hence " rocket efficiency " = 2V and net work available = wRW x 8 V «RE 2V == SV'MRE X\ uli w Thus the values given in Fig. 2, need to be multiplied by \ n to give the net available gain in terms of W, in order that allowance be made for the mechanical efficiency of the efflux jet. If, as tentatively assumed above, n = 0.5 the expected net gain in terms of thrust h.p. will be given by multiplying the figures in the lower scale by 0.71 f
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