Question

The Boeing 787 Dreamliner is billed to be 20% more fuel efficient than the comparable Boeing 767 and will fly at Mach 0.85. The midsize Boeing 767 has a range of 12,000 km, a fuel capacity of 90,000 L, and flies at Mach 0.80. For Boeing 787, assume the speed of sound is 700 mph and calculate the projected volumetric flow rate of fuel for each of the two Dreamliner engines in m3/s.

Answer 1

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$_f$ = Range / fuel capacity

η$_f$ = ( 12,000 km / 90000 L ) ( 1000m / 1 km) ( 3.7854 L/gal × 264.2 gal/m² )

η$_f$  = 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

$V_{787$ = Mach number of 787 × speed of sound

given that; Mach number is 0.85 and speed of sound is 700 mph

we substitute

$V_{787$ = (0.85 × 700 mph) × ( 1 hr / 3600 s ) × ( 1609 m / 1 mile )

$V_{787$ = 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