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In article , Bryan
Martin wrote: in article , Chris W at wrote on 7/10/03 4:25 PM: I don't know if more information is needed or not but I thought I would ask. If a plane can maintain 174 knots TAS at 75% power at 8000' and the same plane can maintain 156 knots TAS at 55% power at the same 8000', what would the TAS be at 16,000' at 55% (it is my understanding that 16,000' is the maximum altitude a naturally aspirated engine can put out 55% power). If the same plane had turbo normalizing that allowed it to run at 75% power at 18,000 ft what would the TAS be? This is of course assuming a standard atmosphere. Convert all of those TAS numbers to CAS. The airplane will maintain about the same CAS at a particular percent power setting regardless of altitude. This is incorrect. If everything else is the same, the drag will be about the same if the CAS is constant. But, the power required is equal to the drag times the TAS. So, if we kept the CAS the same, and increased the altitude, the TAS would increase, which means the power required would also increase. If we kept the power the same, the CAS achieved would decrease as we increased the altitude. A quick look at the cruise perf charts in the POH of any certified aircraft will show that the CAS for a given power decreases with altitude. The following simplified approach assumes that the prop efficiency does not change as we change the operating condition, and that the drag coefficient also does not change. In the real world, the prop efficiency will change a bit, but we have no way to know how much unless we have a prop chart. And the angle of attack will be a bit higher as we increase the altitude, so the drag coefficient will also change a bit. If the mach number is high enough, the changes in mach will also change the drag coefficient. So the following approach becomes less and less accurate as we consider larger altitude changes. D = Drag rho = air density rho0 = air density at sea level standard day sigma = density ratio = rho/rho0 V = True Air Speed Cd = drag coefficient S = wing area T = thrust P = power available (equal to engine power times prop efficiency) I'll add subscripts 1 and 2 when we discuss specific cases. e.g. V1 = the speed for case 1. V2 = the speed for case 2, etc. D = 0.5 * rho * V^2 * Cd * S (eqn 1) rho = sigma * rho0 (eqn 2) If we substitute eqn 1 into eqn 2, we get: D = 0.5 * sigma * rho0 * V^2 * Cd * S (eqn 3) P = T * V (eqn 4) In a steady state condition, drag = thrust, so: P = D * V (eqn 5) If we substitute eqn 3 into eqn 5, we get: P = 0.5 * sigma * rho0 * V^3 * Cd * S (eqn 6) If we have two conditions with the same power (i.e. 55% at 8,000 ft and 55% at 16,000 ft), then: P1 = P2, and if we use eqn 6 we get: 0.5 * sigma1 * rho0 * V1^3 * Cd1 * S = 0.5 * sigma2 * rho0 * V2^3 * Cd2 * S (eqn 7) If we assume that Cd1 = Cd2, and we throw away the other stuff that is the same on each side, we get: sigma1 * V1^3 = sigma 2 * V2^3 (eqn 8) So: V2 = V1 * (sigma1/sigma2)^(1/3) (eqn 9) sigma = (1-0.00000687558 * altitude)^4.2559 (eqn 10) Note, eqn 10 is only valid for standard temperature, and only valid up to 11,000 m, or about 36,000 ft. The density ratio at 8,000 ft with standard temperature is 0.786 The density ratio at 16,000 ft with standard temperature is 0.689 So, if the speed at 8,000 ft with 55% power (V1) is 156 KTAS, then the speed at 16,000 ft with 55% power (V2) would be about 170 KTAS. Now, in the real world, the prop efficiency at 16,000 ft will differ a bit from the value at 8,000 ft, so this will change the speed at 16,000 ft, up or down. Also the Cd at 16,000 is probably a bit lower than it is at 8,000 ft, as we are probably a bit closer to the min drag speed. So the actual achieved speed at 16,000 ft might be a bit more than the predicted 170 KTAS. -- Kevin Horton - RV-8 Ottawa, Canada http://go.phpwebhosting.com/~khorton/rv8/ |
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