Mass 500,000 kg

Wing Area 845 m2
Wingspan 79.75 m
Number of Engines 4
CLmax flaps retracted 1.9
CLmax flaps extended 2.6
Oswald Efficiency Factor 0.92
CD0 0.022
The jet engines of the aircraft have an inlet diameter of 2.96m. Furthermore, the aircraft is flying at a True Airspeed of 486 kts and the total thrust provided by the engines is 400 kN (which are not the correct answers to the previous questions). The aircraft is still flying at FL280.

Calculate the air flow per engine at this altitude (in kg/s).
Calculate the exhaust/jet velocity of the engines (in m/s).
Calculate the exhaust/jet velocity of the engines (in m/s).

1 answer

To calculate the air flow per engine and the exhaust/jet velocity for the given aircraft parameters, we will use the following formulas:

Step 1: Calculate the air density at FL280 (28,000 feet)

FL280 corresponds to an altitude of approximately 28,000 feet, which we can convert into meters:

\[ 28,000 , \text{feet} = 28,000 \times 0.3048 \approx 8,534.4 , \text{m} \]

At this altitude, the standard atmospheric conditions give us a temperature and pressure that we can use to approximate the air density. A rough estimate for air density at FL280 is around 0.905 kg/m³.

Therefore, we have:

\[ \rho \approx 0.905 , \text{kg/m}^3 \]

Step 2: Calculate the total cross-sectional area of the engine inlets

For a circular inlet, the cross-sectional area \( A \) can be calculated using the diameter \( D \):

\[ A = \pi \left( \frac{D}{2} \right)^2 \]

Given \( D = 2.96 , \text{m} \):

\[ A = \pi \left( \frac{2.96}{2} \right)^2 \approx \pi \left( 1.48 \right)^2 \approx \pi \times 2.1904 \approx 6.87 , \text{m}^2 \]

Step 3: Calculate the air flow per engine

The mass flow rate \( \dot{m} \) through one engine can be calculated by the formula:

\[ \dot{m} = \rho \cdot V_{\text{inlet}} \cdot A \]

However, we first need the velocity of airflow in the inlets, which can be estimated based on true airspeed. Assuming all inlets are equally sized and that the engine operates optimally at the true airspeed (\( V_{\text{TAS}} = 486 , \text{kts} \)), we convert this speed to m/s:

\[ 486 , \text{kts} \approx 486 \times 0.514444 \approx 249.2 , \text{m/s} \]

The total airflow through all engines at this speed would be:

\[ \dot{m_{\text{total}}} = \rho \cdot V_{\text{TAS}} \cdot 4A \] Since there are 4 engines, we can calculate airflow per engine and total:

Calculating for total airflow through all engines: \[ \dot{m_{\text{total}}} = 0.905 \times 249.2 \times (4 \cdot 6.87) \]

Airflow per engine would then be re-calculated

Step 4: Total Mass flow per engine:

To compute \( \dot{m}_{\text{engine}} \):

\[ \dot{m}{\text{engine}} = \frac{\dot{m{\text{total}}}}{4} \]

\[ \dot{m}_{\text{engine}} = \frac{0.905 \cdot 249.2 \cdot 6.87}{4} \]

Calculating values:

Total area = 4 * A = 4 * 6.87 = 27.48 m²

Thus

\[ \dot{m_{\text{total}}} \approx 0.905 \cdot 249.2 \cdot 27.48 \approx 6231.0049 , \text{kg/s} \]

Per engine:

\[ \dot{m}_{\text{engine}} = \frac{6231.0049}{4} \approx 1557.75 , \text{kg/s} \]

Step 5: Calculating exhaust/jet velocity

Using the thrust equation for an engine:

\[ Thrust = \dot{m} \cdot V_{e} + (P_{e} - P_{0})A_{e} \]

Assuming atmospheric pressure is 0 at high altitude near vacuum, the thrust mainly counts on velocity of ejection.

Given thrust of 400 kN for 4 engines, we can find \( V_e \) with the effective mass flow calculation.

Given thrust \( T = 400 , \text{kN} = 400,000 , \text{N} \),

Mass flow total = 6231.0049 kg/s.

Assuming simplified case with no altitude pressure connection:

\[ T = \dot{m} \cdot V_e \to V_e = \frac{T}{\dot{m}_{\text{total}}} \]

Calculating

\[ V_e = \frac{400,000}{6231.0049} \approx 64.2 , \text{m/s} \]

Thus, the results are:

  • Air flow per engine: \( \approx 1557.75 , \text{kg/s} \)
  • Exhaust/jet velocity: \( \approx 64.2 , \text{m/s} \)

Final Summary

Air Flow Per Engine: 1557.75 kg/s

Exhaust/Jet Velocity: 64.2 m/s