Previous page Part III Applications
(Work in progress)
Some state
that the (turbulent) Fluid Space Drive
concept cannot work because the force excreted on the rear wall of the
spacecraft by gas molecules will be the same regardless of the distance
separating the rear hull of the spacecraft to the RMA’s propellers.
This is
illustrated in fig 1, there are 2 spacecrafts in microgravity environment, a and b, both have a free
floating Ram Mass Assembly (RAM) that consists of electric motor, counter
rotating propellers and power source (or connected to a power source).
In both
spacecrafts the RMA blows gas in the –X
direction, I postulate that as demonstrated by empirical observation and experimentation, the longer
the distance the gas molecules must travel in the –X direction to impact the –X
hull, the less the resulting force in the –X direction
If D1 < D2 then F1 > F2.
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = > +X
Fig 1
Doubters
state that regardless of the difference between D1 and D2, F1 and F2 MUST be equal so as not to conflict
with the law of conservation of linear momentum.
But what if
the distance separating the rear hull of the spacecraft to the RMA’s propellers
is INFINITE as illustrated in fig 2, will force F3 be equal to force F1?
Fig 2
It is not
very practical for a spacecraft to be of infinite length, but I propose the
next best thing, the Laminar Flow
Fluid Space Drive Proposal.
Instead of
proposing a pressured structure with a diameter sufficiently large that the
propellers (or blowers) accelerating the RMA so that a turbulent flow is encouraged,
the diameter of the tube/duct is decreased as illustrated in fig 3 from
diameter d1 (sufficient for easy generation of turbulence) to diameter d2 where
a laminar flow of the gas is encouraged.
Fig 3
At first
glance this would seem to be a counter productive design decision, for with a
perfect laminar flow the force excreted on the –X hull of the spacecraft will
remain the same regardless of the length of the spacecraft, in fact we can
assume that for the length of the tube/duct the volumetric flow rate of the gas
(Q) will be the same across the length of the tube/duct (Fig 3, Q1 = Q2)
Ram
Mass Assembly (RAM)
Fig 4
Basic
Conditions
It may be difficult to simulate the forces
generated in a closed system by a turbulent flow of gases (as in the Turbulent
Fluid Space Drive) but there is no great difficulty simulating the behavior of
a laminar flow of gases in a tube/duct, the Navier–Stokes equations can be used
as a statement of the conservation of momentum if we maintain certain
conditions such as:
The fluid in use is a gas. (Under certain condition not considered here the fluid may be a liquid)
The gas velocities are relatively slow (well
below the speed of sound)
The gas is considered a Newtonian
fluid.
The gas is considered moistly incompressible.
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 5
If we add
two 180º tubes/ducts at the end of the “main” tube/duct forces (F4 and F2) will be created, the behavior of a laminar
gas flow on 90º and 180º has been intensely researched and documented. (Note 1)
Fig 6
illustrates a system where the fluid flows uninterrupted around a “closed loop”, it is a simple assembly that poses no challenge to current
flow calculations methods.
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 6
If the
mechanism described in Fig 6 is put in a micro gravity environment and we abide
by the parameters described in box Certain Conditions, if there are no large changes
in the general velocity of the flow, the volumetric flow rate of the gases Q1 is equal to Q2, Q3 is equal to Q4 and flows Q5, Q6, Q7 and Q8 are equal therefore forces F4, F5, F6 and F7 are equal.
(This bird
is not going anywhere)
As the RMA
is detached from the assembly (by releasing the retaining cable) it gains
relative velocity and momentum without interchanging momentum with the assembly
(Fig 7), the moment the RMA collides with the assembly (or the retaining cable
is drawn back see figs 14-xx) it transfers its momentum to the assembly that
gains velocity in the +X direction (Fig 8)
Fig 7
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 8
What we have is a propeller mechanism (RMA)
unattached to the spacecraft assembly (has total freedom of movement in the X
axis), when the propellers are activated there are two results:
a) A gas flow is created with the volumetric
flows and forces as described in fig 5
b) The RMA gains movement in the +X direction
increasing velocity and momentum in the +X direction.
When the RMA collides with the +X hull of the
spacecraft assembly its momentum is transferred to the spacecraft assembly that
gains velocity in the +X direction
If you don't believe in this
description as of yet, It will be a waste of time for you to continue reading.
Improvements on the idea
It is not
necessary for the spacecraft assembly’s tubes/ducts to be so elongated, figs 1 to
8 are only to illustrate a simple mean of expressing and comparing the flows
and forces generated with previous published research and present a simple
model for flow and momentum calculations, a more efficient effect can be
obtained with the arraignment illustrated in figs 9 and 10.
Fig 9
Fig 10
Fig 11
Figs 10 and
11 illustrates the main parts of the proposed mechanism, a pressurized cylinder
structure 2 is attached to the spacecraft 1 we wish to propel FIG. 10, power is
supplied by a AIP 5 (air independent power unit in this case a radio isotopic
power cell)
Inside the
pressurized cylinder structure 2 (figs 10, 11 and 12) is a smaller cylinder 27.
The radius of cylinder 27 is dependent on the radius of pressurized cylinder
structure
This is so
that the front to rear gas flow volumetric flow rate Qfr inside smaller cylinder 27 (Fig 12) is equal
to the volumetric flow rate Qrf in the space between outside smaller cylinder 27 and inside pressurized
cylinder structure 2 (Fig 12).
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 12
Inside
cylinder 27 is a ram mass assembly (RMA) (fig 4) that is guided by positioning
struts 25 with low friction wheels that permit easy displacement in the
horizontal (X) axis (length of cylinder 27)
Inside the
RMA 34 are one or more motors that power one or more pair of counter rotating
rings 23, the rings have rotors 24 attached (fig 4)
An air
independent power system (AIP) 5 provides the power, transferred to the RMA’s
via retaining cable 30 that is rolled up in winch mechanism 32, the winch
mechanism 32 releases or pulls the retaining cable so there is no tension
(puling) or snarling between
the RMA and the spacecraft.
The Propulsion Cycle
Textured
arrow represents gas flow direction.
Blue arrows
represent a velocity or velocity change.
Yellow-red
arrows represent a force.
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 13
Cycle 0
Fig 13
illustrates the RMA’s position at the start of the propulsion cycle, the
spinning propellers create an uninterrupted gas flow around a “closed loop”
structure, the RMS is held in position by the retaining cable 30 (winch 32 is
“locked”), no thrust generated on the spacecraft.
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 14
Cycle 1
Retaining
cable 30 is released by the winch 32 permitting the RMA to accelerate in the +X
direction (Fig 14) it gains velocity and momentum relative to the spacecraft.
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 15
Cycle 2
As the RMS
nears the front (+X) end of the cylinder (Fig 15), the winch 32 forcefully pulls
back the retaining cable 30, the moment the cable is taut is equivalent to a
elastic collision between the spacecraft and the RMS. 2 equal opposing forces
are created; Ff (forward) and
Fb (backwards).
Force Ff pushes the spacecraft (via
retaining cable 30) incrementing its velocity in the +X direction.
Force Fb sends the RMS traveling in the –X
direction.
-X < = = = = = = = = = = = = = = = = = = =
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = > +X
Fig 16
Cycle 3
As the RMS
travels in the –X direction, the rotation of it’s rotors decrement its –X
velocity (relative to the spacecraft) till it is at a “standstill” for an
instant (Cycle 0) and it begins acceleration in the +X direction (Fig 14. Cycle
1), Cycles 1, 2 and 3 repeat until destination is reached.
Cycle 0 to 2
Fig 17
Fig 17
graphs the cycle
|
Cycle |
Spacecraft |
RMS |
|
C0 |
Moving at
a constant velocity in the +X direction |
Moving at
the same velocity the spacecraft is traveling, rotors are turned on but is held
in position by retaining cable (fig 13) |
|
C1b |
No change |
Winch
replaces retaining cable, RMS begins to accelerate
in the +X direction. |
|
C2 |
Winch
pulls back retaining cable (fig 15), Force Ff increments the spacecraft’s velocity in the
+X direction |
Winch
pulls back retaining cable (fig 15), force Fb propels the RMA in the –X
direction |
|
C1a |
Continues
moving in the +X direction with increased velocity |
Traveling
in the –X but as rotors are active, the RMS slows (relative to the velocity of
the spacecraft) |
|
C0 |
No change |
RMS slows
till it is again traveling at same speed as spacecraft |
|
C1b |
No change |
Winch
releases retaining cable, RMS begins to accelerate in the +X direction. |
|
|
Return to cycle C2 |
|
Note 1:
CFD SIMULATION
OF A LAMINAR FLOW ALONG A CURVED CONVERGING RECTANGULAR DUCT
Norasyikin
binti maunud
Me070854
Universiti
tenaga nasional
NUMERICAL PREDICTION OF AIR FLOW IN A SHARP 90º ELBOW.
Ruth
MOSSAD1
William
YANG2
M. Philip
SCHWARZ2
CSIRO
Process Science and Engineering,
ON THE CORE
FLOW AND TURBULENT BOUNDARY LAYER IN A CURVED DUCT
Phillip
Lawrence
INVESTIGATION
OF THREE-DIMENSIONAL UPWARD AND DOWNWARD DIRECTED GAS-LIQUID TWO-PHASE BUBBLY
FLOWS IN A 180O-BENT TUBE
Th. Frank,
R. Lechner, F. Menter
CFX
Development, ANSYS
D-83624