Change in TAS with constant Power and increasing altitude.

Chris

At you talking about throtttle setting?. To hold the same % POWER as
you climb, you have to advance the throttle until you get WOT from
which point the % power decreases as you continue the climb.

Big John
Point of the sword

On Thu, 10 Jul 2003 23:10:39 GMT, Chris W wrote:

On Thu, 10 Jul 2003 17:46:28 -0400, Bryan Martin
wrote:

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.

It's my understanding that, that only works for small changes in
altitude. And that the amount of power required to maintan the same
CAS goes up with altitude.

Chris W wrote in news:riirgv0rbvpnmooht9f0f4h26k26so0vb7@
4ax.com:

...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).

The 16K figure for 55% is not a given, as actual installation specifics can
make a big difference. But it's not an uncommon number.

Regardless, changing from 8000 to 16,000 at the same power setting would
raise the TAS by about 2 knots per 1000 feet, or from 156 knots TAS to 172
knots.

If the same plane had turbo normalizing that allowed it to run at
75% power at 18,000 ft what would the TAS be?

Same calculation. That would raise the 174 knots TAS at 8000 (75% power)
to 194 knots TAS at 18,000.

-----------------------------------------------
James M. Knox
TriSoft ph 512-385-0316
1109-A Shady Lane fax 512-366-4331
Austin, Tx 78721
-----------------------------------------------

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/