Question

The Boeing 787 Dreamliner is designed to be 20% more fuel efficient than the comparable Boeing 767 and flies at an average cruise speed of 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. 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

The volumetric flow rate of fuel for each dream liner engine is
`4.98*10^(-3)m^3/s`

Step-by-step explanation:

To realize the problem it is necessary to apply the concepts related to distance efficiency traveled by volume consumed, as well as calculation of the volume flow of the two airplanes.

The units are given in different references, so it is necessary to convert them to an international system.

In the case of the fuel economy of the Boeing 767, we know that

`\eta_f = (12000Km)/(90000L)`

In International System we have,

`\eta_f = (12000Km)/(90000L)(1000L)/(1m^3)(1000m)/(1km)`

`\eta_f = 133333.33m/m^3`

Since the statement we have Boing 767 has 20% less efficiency than Boing 787, that is, Boing 787 consumes 20% less fuel than Boeing 767 when they draw the same route, that is, consumes
`80\% * \eta_f`that would come being,

`0.8*\eta_f = 0.8* 133333.33=106666.6m/m^3`

If we separate this in the amount of engines they use (2) in the case of the Boeing 787 and assuming that both consume the same amount, it would have to

`\eta_(engine) = (\eta_f)/(2) = (106666.6)/(2)`

`\eta_(engine) = 53333.3m/m^3`

Calculating the speed of the Boeing 787 to the indicated Mach number we have to,

`v = 0.85*700mph`

`v = 0.85 *700((1hr)/(3600))((1609m)/(1mile))`

`v = 265.932m/s`

The fuel consumption of each engine of a Boeing 787 would be given by

`\dot{v} = (Velocity (m/s))/(Economy(m/m^3))\\\dot{v} = (265.932)/(53333.3)\\\dot{v} = 4.98*10^(-3)m^3/s`

Therefore the volumetric flow rate of fuel for each dream liner engine is
`4.98*10^(-3)m^3/s`