Answer:
the projected volumetric flow rate of fuel for each of the two Dreamliner engines is 0.005 m³/s
Explanation:
Given the data in the question;
First we determine the fuel economy of Boeing 767
Range = 12,000 km
fuel capacity = 90,000 L
so, fuel economy of Boeing 767 will be
η = Range / fuel capacity
η = ( 12,000 km / 90000 L ) ( 1000m / 1 km) ( 3.7854 L/gal × 264.2 gal/m² )
η = 133,347.024 m/m³
Now, Boeing 787 is 20% more fuel efficient than Boeing 767
so fuel economy of Boeing 787 will be;
⇒ (1 - 20%) × fuel economy of Boeing 767
⇒ (1 - 0.2) × 133,347.024 m/m³
⇒ 0.8 × 133,347.024 m/m³
⇒ 106,677.6 m/m³
Hence, fuel economy of Boeing 787 dream line engine is
⇒ 106,677.6 m/m³ / 2 = 53,338.8 m/m³
Next, we find the velocity of Boeing 787
= Mach number of 787 × speed of sound
given that; Mach number is 0.85 and speed of sound is 700 mph
we substitute
= (0.85 × 700 mph) × ( 1 hr / 3600 s ) × ( 1609 m / 1 mile )
= 265.9319 m/s
Now, to get the Volume flow rate for each dream liner engine { Boeing 787 };
Volumetric flow rate = velocity of flight / fuel economy
we substitute
= 265.9319 m/s / 53,338.8 m/m³
= 0.0049857 ≈ 0.005 m³/s
Therefore, the projected volumetric flow rate of fuel for each of the two Dreamliner engines is 0.005 m³/s
