Work in progress
Many 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 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