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Aviation History
1932
1932 - 0282.PDF
SUPPLEMENT TO FLIGHT 22 THE AIRCRAFT ENGINEER MARCH 25, 1932 impossible to design an airscrew to fulfil the required conditions of r.p.m. required both at climb and at top speed. Assuming that the existing gear ratios are 0.657/1 in the case of engine A, and 1/1 in the case of engine B, and that the airscrews are designed to give maximum efficiency at top speed, we will find at what speeds normal r.p.m. will be maintained. Engine A.—Gear ratio 0.657/1 : VM, ft./sec. D. ft, ... 176 9-88 0-672 0-00752 0-95 0-0096 0-4 95 ft./sec. 114 ft./sec. 180 8-93 1-115 0-01247 1-43 0 0159 0-925 199 ft./sec. 171 ft./sec. 1 At • normal r.p.m. KQ ... P D K„ ... V nD V 0-65 VM The value of VjriD at normal r.p.m. is obtained from the torque coefficient curve (not shown) which is got by interpolation in Fig. 2. Engine B.- V, ft./sec. D, ft. V WDM *' P D " KQ ... V wD V, ft./sec. 0-65 VM -Gear ratio 1/1 176 ... 10-58 ... 0-5105 ... 0-00571 ... 0-75 ... 0-00672 ... 0-37 116 114 264 9-55 0-851 0-00952 1-15 0-012 0-75 212 171 352 8-89 1-218 0-01363 1-53 0-01605 1-09 287 229 At normal r.p.m. These results show that in the first case with engine A normal r.p.m. are maintained at a speed lower than the best climbing speed, so will exceed normal r.p.m. on climb at full throttle and throttling will be necessary. In the second case normal r.p.m. are reached at a speed greater than climbing speed, consequently the airscrew will hold the r.p.m. down to a lower value on climb, and power will be lost. With engine B normal r.p.m. are maintained on climb in the first case only, as the maximum speed increases the speed at which normal r.p.m. are maintained becomes more and more in excess of the climbing speed, and the r.p.m. on climb continually decrease. We will take now the case of engine B, and by assuming various gear ratios examine the effect which gearing has on the speed at which normal r.p.m. are maintained, taking in each case 180 m.p.h. as the maximum speed of the machine, i.e., the speed at which the airscrew is designed for maximum efficiency and maximum r.p.m. It should be understood that at present we are con sidering only the question of r.p.m. and speed; the effect on the net or propulsive efficiency of the airscrew will be examined later. Engine B. Maximum forward speed = 180 m.p.h. V = 264 ft./sec. Gearratio. D, ft. .. V nD„ P D 0-5/1 13-51 1-202 1-52 1/1 9-55 0-851 115 5/1 79 2/1 6-75 2-5/1 6-04 0-694 0-602 0-538 0-975 0-86 0-78 K • nDs nDs VN, ft./sec. Vc|, ft./sec. 0-01345 0-01584 1075 200 215 171 0-00952 00112 0-74 282-5 209 171 0-00776 0-00913 0-57 345 197 171 0-00672 0-00791 0-465 400 186 171 0-00602 (t-00708 0-395 447 176 171 The subscript N denotes conditions at normal r.p.m. where these do not. occur at Vcj. On examination of this table we see that unless the airscrew is geared to rotate at over 2.5 times the engine speed, normal r.p.m. are not attained until the forward speed is greater than the required 171 ft./sec, and that the lower the airscrew r.p.m. the greater the forward speed becomes at which normal r.p.m. are maintained. From various considerations it would not pay in practice to gear the airscrew to give r.p.m. greater than those of the engine. We will examine first the effect of the airscrew diameter on the additional drag due to the increased resistance of the parts of the machine in the slipstream. Kr = Parasitic drag coefficient of the whole machine. K8 = Parasitic drag coefficient of the parts in the slipstream. Aw = Wing area (to which all coefficients are referred). V ft./sec. = Forward speed. V. ft./sec. = Total speed of slipstream. A sq. ft. = Cross-sectional area of slipstream. D ft. = Diameter of airscrew. Assuming that, owing to the contraction of the slip stream, its cross-sectional area is 0.7 of that of the air screw disc TTD2 A = 0-7 . 4 Total drag of machine = PAwV2Kr - pAwV2K3 + pAwVs2Ks. = pAw-V*Kr + pAwV*Ks ( ^ - 1). Additional drag due to slipstream /VB2 If b and whence outflow v„ T = PAWV2K8(~ factor and T lb. = V (b + 1) = ipAV2b(b + = JpAV2 (b + 1 -^•m *;—* -i) = total thrust, 2) I)2- It -! T pAV2 . . (xvi) we get Substituting in (xvi) we get Additional drag due to slipstream = pAwVjKaT i pAV 2 2AwK„T Additional thrust horsepower required 2AWKBT V A ' 550 If H.P.T = total thrust horsepower available, and H.P.s = thrust horsepower absorbed by additional drag due to slipstream TV 5o0 and HP. Z OVA| TV 550 260/
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