Convert EAS to TAS: Equations and Solution for a Boeing 747 at 10,000m Altitude

[lufbrajames]

Homework Statement

Hi, I am really stuck on this problem, I've been finding equations to convert the EAS to TAS but none of them have temperature involved.




2) A Boeing 747 is cruising at an altitude of 10,000m the pilot looks at the air speed indicator (calibrated for ISA) which shows an EAS of 450 knts. If the sea level pressure is 101530 Pa and the temperature is 20 oC, calculate the True Air Speed (TAS) of the aircraft.

Homework Equations

the equation I've found is

EAS = TAS * sqrt(actual air density / standard air density)

The Attempt at a Solution

the values I've got don't fit into the equation ?

 

[Danger86514]

Try these formulas ... just a thought


TAS = 2 x EAS x (T + 460)/P

Reference: http://www.nar-associates.com/technical-flying/airspeed/airspeed_wide_screen.pdf

The barometric formula

Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/barfor.html#c3

Pressure Altitude

Reference: http://en.wikipedia.org/wiki/Pressure_altitude

 

[Mene]

Hello,

To convert EAS to TAS, you can use the following equation:

TAS = EAS * sqrt(ρ/ρ0)

Where:
- TAS is the true air speed (in meters per second)
- EAS is the equivalent air speed (in meters per second)
- ρ is the actual air density (in kilograms per cubic meter)
- ρ0 is the standard air density (in kilograms per cubic meter)

To calculate ρ, you can use the following equation:

ρ = P/(R * T)

Where:
- P is the pressure (in Pascals)
- R is the gas constant for air (287.058 J/kg*K)
- T is the temperature (in Kelvin)

For a Boeing 747 at 10,000m altitude, the standard air density can be calculated as follows:

ρ0 = P0/(R * T0) = (101530 Pa)/(287.058 J/kg*K * 293.15 K) = 0.413 kg/m^3

Now, to calculate ρ, we need to find the actual pressure and temperature at 10,000m altitude. For this, we can use the International Standard Atmosphere (ISA) model, which states that at 10,000m altitude:
- The pressure is 264.9 Pa
- The temperature is -56.5 oC (or 216.65 K)

Substituting these values into the equation for ρ, we get:

ρ = (264.9 Pa)/(287.058 J/kg*K * 216.65 K) = 0.000993 kg/m^3

Now, we can plug these values into the equation for TAS:

TAS = (450 knts) * sqrt(0.000993 kg/m^3 / 0.413 kg/m^3) = 235.7 m/s

Therefore, the true air speed of the Boeing 747 at 10,000m altitude is approximately 235.7 m/s.

I hope this helps! Let me know if you have any further questions.