To determine the volume of air that the GE90 engine takes in per second, we can use the formula relating thrust, mass flow rate, exhaust velocity, and air density. The thrust \(F\) can be expressed as:
\[ F = \dot{m} \cdot v_e \]
Where:
- \(F\) is the thrust (in Newtons).
- \(\dot{m}\) is the mass flow rate (in kg/s).
- \(v_e\) is the exhaust velocity (in m/s).
Rearranging for the mass flow rate gives us:
\[ \dot{m} = \frac{F}{v_e} \]
Now, we convert the exhaust speed from knots to meters per second:
1 knot = 0.514444 m/s, therefore:
\[ 730.8 , \text{kts} \times 0.514444 , \frac{\text{m}}{\text{s}} = 375.5 , \text{m/s} \]
Next, we substitute the values into the mass flow rate equation. The thrust of the engine is 500 kN, which is:
\[ F = 500,000 , \text{N} \]
Now we can calculate the mass flow rate:
\[ \dot{m} = \frac{500,000 , \text{N}}{375.5 , \text{m/s}} \approx 1331.1 , \text{kg/s} \]
Now that we have the mass flow rate, to find the volume flow rate of air \(\dot{V}\), we can use the relationship:
\[ \dot{V} = \frac{\dot{m}}{\rho} \]
Where \(\rho\) is the air density. At sea level, the density of air is approximately \(1.225 , \text{kg/m}^3\).
Now we can calculate the volume flow rate:
\[ \dot{V} = \frac{1331.1 , \text{kg/s}}{1.225 , \text{kg/m}^3} \approx 1087.4 , \text{m}^3/\text{s} \]
Thus, the volume of air taken in by the engine per second is approximately:
\[ \boxed{1087.4} , \text{m}^3/\text{s} \]