Conserving Momentum for Rocket Engines

[David112234]

Homework Statement

Homework Equations

conservation of momentum
Thrust of rocket engine = -u (dm/dt)
u is the velocity of the gas being expelledAlso found this from Wikipedia, not sure if it is relevant

The Attempt at a Solution

For these kinds of problems I assume you need to apply conservation of momentum on infinitesimally small parts.

dp = (v+dv)(m-dm) + dm)(v-u) - mv +

at the end of this equation how do I factor in the intake?
I know dm will be the amount of fuel + air it shoots out back

and If this is the wrong approach and I supposed to use the equation form wiki?
If so what is mass rate of flow? is that the same as dv/dt?

 

[andrewkirk]

The wiki equation is based on conservation of momentum. If we rewrite it as
$$F_N=\dot m_{air}(v_e-v) + \dot m_{fuel}\,v_e$$
and note that all quoted velocities are in the reference frame of the plane, we see that the first term is the rate of increased backwards momentum the engine applies to the ejected air and the second term is the rate of increased backwards momentum the engine applies to the fuel.

By conservation momentum, the total of those two is the rate of increased forward momentum applied to the engine, and hence to the plane, by the ejected gases, which is the Thrust.

If the plane is flying at constant velocity, the Thrust will equal the Drag, which is the rate of loss of forwards momentum from collision of the plane with the air particles.

 

[David112234]

The wiki equation is based on conservation of momentum. If we rewrite it as
$$F_N=\dot m_{air}(v_e-v) + \dot m_{fuel}\,v_e$$
and note that all quoted velocities are in the reference frame of the plane, we see that the first term is the rate of increased backwards momentum the engine applies to the ejected air and the second term is the rate of increased backwards momentum the engine applies to the fuel.

By conservation momentum, the total of those two is the rate of increased forward momentum applied to the engine, and hence to the plane, by the ejected gases, which is the Thrust.

If the plane is flying at constant velocity, the Thrust will equal the Drag, which is the rate of loss of forwards momentum from collision of the plane with the air particles.

Could you help me derive that formula?

 

[andrewkirk]

Sure. Assuming the plane velocity and jet thrust is constant, multiply both sides by a small time increment $\delta t$. Then the LHS is the impulse the engine imparts onto the gas in that time. With a little bit of rearranging you can show that the RHS is the change in momentum of the fuel air ejected in that time. So the two sides must be equal because impulse is equal to change in momentum (Newton's 2nd law integrated over the time interval).

 

[David112234]

Sure. Assuming the plane velocity and jet thrust is constant, multiply both sides by a small time increment $\delta t$. Then the LHS is the impulse the engine imparts onto the gas in that time. With a little bit of rearranging you can show that the RHS is the change in momentum of the fuel air ejected in that time. So the two sides must be equal because impulse is equal to change in momentum (Newton's 2nd law integrated over the time interval).

Not sure what you mean, shouldn't I divide both sides by dt and the left side would be dp/dt which is force? Where do you get impulse?
Also before that, the equation I wrote shouldn't I add ( or subtract) something in order to include the intake? What do I do with the intake?

 

[andrewkirk]

Impulse is force integrated over time. We have assumed force is constant, so multiplying by $\delta t$ makes the LHS $F_N\,\delta t=\int_t^{t+\delta t}F_N(u)\,du$ which is the impulse imparted over the time interval $[t,t+\delta t]$.

 

[David112234]

Impulse is force integrated over time. We have assumed force is constant, so multiplying by $\delta t$ makes the LHS $F_N\,\delta t=\int_t^{t+\delta t}F_N(u)\,du$ which is the impulse imparted over the time interval $[t,t+\delta t]$.

I am not familiar with what $\delta t$ if possible I would like to avoid it, also why integrate at all?
Here is the method that am use to.
For a regular rocket
dp = (v+dv)(m-dm) + dm(v-u) - mv
it simplifies to
dp = m(dv) - u(dm)
dividing by dt

dp/dt = m(dv/dt) - u (dm/dt)
dp/dt = 0 since net Force is zero
m(dv/dt) = - u(dm/dt) = thrust

Now, for the jet engine in this problem, I don't know how to set up the original equation

 

[andrewkirk]

The difference between the jet and the rocket is that the former ejects two things - fuel and air - and they have different initial velocities, whereas the rocket ejects only one thing (fuel), and hence has only one initial velocity to consider.

The initial velocity of the jet's fuel is the same as that of the plane, which is zero in the plane's reference frame. The initial velocity of the air in that reference frame is the airspeed $v$ of the plane.

So you need to break up your m dv into two parts m1 dv1 and m2 dv2, one for the fuel and one for the air. These are
$(\dot m_{fuel}\delta t) (v_e-0) + (\dot m_{air}\delta t) (v_e-v)$ respectively.

 

[David112234]

The difference between the jet and the rocket is that the former ejects two things - fuel and air - and they have different initial velocities, whereas the rocket ejects only one thing (fuel), and hence has only one initial velocity to consider.

The initial velocity of the jet's fuel is the same as that of the plane, which is zero in the plane's reference frame. The initial velocity of the air in that reference frame is the airspeed $v$ of the plane.

So you need to break up your m dv into two parts m1 dv1 and m2 dv2, one for the fuel and one for the air. These are
$(\dot m_{fuel}\delta t) (v_e-0) + (\dot m_{air}\delta t) (v_e-v)$ respectively.

Still kind of confused,
so since it is also shooting out air I will add an dm_{air}(v-u)
dp = (v+dv)(m-dm) + dm(v-u)+dm + dm_{air}(v-u) - mv
but how does this take into consideration that it is also sucking in air at the front? If it is sucking in air it should contribute some forward momentum for the jet right? Like a person laying on the floor pulling on a rope.
You said the initial air speed is that of the airplane, can you elaborate on this please?

 

[andrewkirk]

You said the initial air speed is that of the airplane, can you elaborate on this please?

The plane is moving through the air at 290 m/s (the question is unclear, just saying the plane's speed is 290 m/s without specifying whether that is ground-speed or air-speed. We assume they meant air-speed), so the air is moving towards the jet at 290 m/s. That is $v$ in the wiki equation.

We don't need to think about the sucking in of air by the engine. It's not like pulling on a rope because, unlike a rope, the air is not attached to anything fixed. All we need to think about is the velocity of the air before and after it has interacted with the engine. Those velocities are 290 m/s and 620 m/s respectively.

 

[David112234]

The plane is moving through the air at 290 m/s (the question is unclear, just saying the plane's speed is 290 m/s without specifying whether that is ground-speed or air-speed. We assume they meant air-speed), so the air is moving towards the jet at 290 m/s. That is $v$ in the wiki equation.

We don't need to think about the sucking in of air by the engine. It's not like pulling on a rope because, unlike a rope, the air is not attached to anything fixed. All we need to think about is the velocity of the air before and after it has interacted with the engine. Those velocities are 290 m/s and 620 m/s respectively.

But even if it is not attached shouldn't it still cause the plane to go forward by Newtons 3rd law. 2 magnets one larger than the other would be a better example of what I mean rather that rope.
so do I subtract the velocities of the air befre and after the jet?
This only makes sense to me if the planes speed stays constant and that is the frame of reference
So my formula set up is correct?