- #1
RGClark
- 86
- 0
In this post to sci.astro I suggested a method for air-breathing propulsion to orbit:
From: Robert Clark
Date: Sat, Jun 17 2006 2:43 pm
Email: "Robert Clark" <rgregorycl...@yahoo.com>
Groups: sci.astro, sci.space.policy, sci.physics, sci.mech.fluids, sci.engr.mech
Subject: Proposals for air breathing hypersonic craft. II
http://groups.google.com/group/sci.astro/msg/b9c19abd8f97a5d2
What I want to ask about is the portion of the post copied below where I derive a version of the rocket equation. I wanted some feedback on the validity of the derivation.
- Bob Clark
=====================================================
I shall argue that the method of not slowing the incoming air at all
but accelerating the fuel up to the relative air speed will result in a
marked improvement in fuel efficiency. Specifically, the exponential
increases in fuel according to velocity given by the rocket equation
will no longer be needed.
Let X be the mass of the rocket with fuel, v the rocket velocity, r
the ratio of air to fuel in mass, and e the nominal exhaust velocity of
combusting still air with still fuel.
For this method to work I will assume that the force produced by the
combustion of the air with the fuel can be fully communicated to the
craft. To derive the thrust equation take the craft including the fuel
to be a closed system and the air to be outside the system and take the
rest frame to be the Earth, or likewise the still air.
Now if we did not combust the ejected fuel with the air then by
momentum conservation we would have:
(X + dX)(v + dv) + 0(-dX) = Xv
In the first term on the left we add dX to X because dX is negative
since the mass is decreasing as fuel is consumed. So the first term
represents the mass of the rocket less the ejected fuel times the
increased velocity of the rocket. In the second term we are multiplying
the velocity with respect to ground of the ejected fuel times the mass
of the fuel ejected. Since we are ejecting the fuel at a speed to stay
at zero relative velocity to air, i.e., to the ground, this velocity
here is 0. The negative sign in front of dX again is because dX is
negative so -dX is the positive mass of the fuel.
This equation expanded out is Xv + Xdv+ vdX + dXdv = Xv. So the change
in momentum is Xdv + vdX + dXdv = 0, and the rate of change of momentum
is:
0 = Xdv/dt + vdX/dt + (dXdv)/dt = Xdv/dt + vdX/dt , because the term
with two differentials dXdv vanishes as dt ---> 0.
Now when we do combust the fuel with the air, then the rate of change
in momentum of the system is the force on the craft due to the
combustion of the air and fuel. This is the thrust produced by this
combustion which equals mass flow rate, air + fuel, times the nominal
exhaust velocity of the combustion of still air and still fuel:
Xdv/dt + vdX/dt = -ed(rX+X)/dt = -e(r+1)dX/dt , where the minus sign
comes from dX being negative.
Let c = e(r+1). Then the equation becomes Xdv/dt + (c + v)dX/dt = 0,
which is equivalent to:
d[(c +v)X]/dt = 0
This has solution (c + v)X = constant. Let X0 be the initial mass and
v0 the initial speed of the rocket. Then the solution is (c +v)X = (c +
v0)X0.
Therefore X0/X = (c + v)/(c + v0), i.e., the mass ratio of the fully
fueled rocket to the empty rocket is just a linear function of ending
velocity.
==============================================
From: Robert Clark
Date: Sat, Jun 17 2006 2:43 pm
Email: "Robert Clark" <rgregorycl...@yahoo.com>
Groups: sci.astro, sci.space.policy, sci.physics, sci.mech.fluids, sci.engr.mech
Subject: Proposals for air breathing hypersonic craft. II
http://groups.google.com/group/sci.astro/msg/b9c19abd8f97a5d2
What I want to ask about is the portion of the post copied below where I derive a version of the rocket equation. I wanted some feedback on the validity of the derivation.
- Bob Clark
=====================================================
I shall argue that the method of not slowing the incoming air at all
but accelerating the fuel up to the relative air speed will result in a
marked improvement in fuel efficiency. Specifically, the exponential
increases in fuel according to velocity given by the rocket equation
will no longer be needed.
Let X be the mass of the rocket with fuel, v the rocket velocity, r
the ratio of air to fuel in mass, and e the nominal exhaust velocity of
combusting still air with still fuel.
For this method to work I will assume that the force produced by the
combustion of the air with the fuel can be fully communicated to the
craft. To derive the thrust equation take the craft including the fuel
to be a closed system and the air to be outside the system and take the
rest frame to be the Earth, or likewise the still air.
Now if we did not combust the ejected fuel with the air then by
momentum conservation we would have:
(X + dX)(v + dv) + 0(-dX) = Xv
In the first term on the left we add dX to X because dX is negative
since the mass is decreasing as fuel is consumed. So the first term
represents the mass of the rocket less the ejected fuel times the
increased velocity of the rocket. In the second term we are multiplying
the velocity with respect to ground of the ejected fuel times the mass
of the fuel ejected. Since we are ejecting the fuel at a speed to stay
at zero relative velocity to air, i.e., to the ground, this velocity
here is 0. The negative sign in front of dX again is because dX is
negative so -dX is the positive mass of the fuel.
This equation expanded out is Xv + Xdv+ vdX + dXdv = Xv. So the change
in momentum is Xdv + vdX + dXdv = 0, and the rate of change of momentum
is:
0 = Xdv/dt + vdX/dt + (dXdv)/dt = Xdv/dt + vdX/dt , because the term
with two differentials dXdv vanishes as dt ---> 0.
Now when we do combust the fuel with the air, then the rate of change
in momentum of the system is the force on the craft due to the
combustion of the air and fuel. This is the thrust produced by this
combustion which equals mass flow rate, air + fuel, times the nominal
exhaust velocity of the combustion of still air and still fuel:
Xdv/dt + vdX/dt = -ed(rX+X)/dt = -e(r+1)dX/dt , where the minus sign
comes from dX being negative.
Let c = e(r+1). Then the equation becomes Xdv/dt + (c + v)dX/dt = 0,
which is equivalent to:
d[(c +v)X]/dt = 0
This has solution (c + v)X = constant. Let X0 be the initial mass and
v0 the initial speed of the rocket. Then the solution is (c +v)X = (c +
v0)X0.
Therefore X0/X = (c + v)/(c + v0), i.e., the mass ratio of the fully
fueled rocket to the empty rocket is just a linear function of ending
velocity.
==============================================