A small jet airplane is traveling at a speed of 180 m/s (about 400 mi/h). Each second, the plane's engine takes in 68 m3 of air through the front. This air has a mass of 70 kg at the altitude the plane is flying and is used to burn 2.9 kg of fuel each second. The energy released is used to compress the air and eject the air + combustion products out the rear of the engine at a speed of 490 m/s (about 1100 mi/h) relative to the plane. (a) What is the thrust of this jet engine (in N and lb)? [HINT: Think about this one in the reference frame of the plane. Consider momentum transfers.]

Question

contestada

Respuesta :

cjmejiab cjmejiab · Answer 1 · 2019-11-04T16:39:30+01:00

Answer:

T = 23121N

Explanation:

To solve this problem we need to define our variables,

We start with the dates from the jet, so

[tex]v_{jet} = 180 m/s[/tex]

[tex]\frac{dMa}{dt} =70 kg/s[/tex] (to the air intake)

[tex]\frac{dMf}{dt} = 2.9 kg/s[/tex] (rate fuel burn)

[tex]u_{gas}[/tex] = 490 m/s

First we calculate the rate of mass change,

[tex]\frac{dM_T}{dt} = 70 kg/s + 2.9 kg/s[/tex]

[tex]\frac{dM_T}{dT}= 72.9 kg/s[/tex]

So we can now calculate the thrust on the rocket,

[tex]T = \frac{dM}{dt}u - \frac{dMa}{dt}v[/tex]

[tex]T = (72.9 kg/s)(490 m/s)-(70 kg/s)(180 m/s)[/tex]

[tex]T = 23121N[/tex]

To calculate the power of the rocket we need the equation,

[tex]P = vT\\P = (180 m/s)(23121 N)\\P = 4.16*106 W\\P = 4.16 MW[/tex]