Andre wrote:What you need to know when a fluid or more particularly a gas circulates in a conduit, there is an effect of walls, that is to say that on the surface of the conduit there is friction, the fluid wants to stick to the wall and on this thin layer, the fluid circulates in a turbulent way, it is only in the middle of the duct that the circulation is laminar.
This turbulent circulation depends on several factors, nature of the smooth or rough wall, the speed of the flow, the viscosity of the fluid and the diameter of the duct.
In the case of the reactor we have the reactor walls and the rod walls, so two zones of thin turbulence, if the air gap is small, or the rough or contaminated walls there can be no laminar flow between these two layers.
I think we should have just two turbulent layers without leaving a thin laminar layer,
the exchange of temperature with the walls is much better when it is turbulent.
Andre
Hello
In a laminar flow, the speed of the fluid goes from zero speed against the wall at maximum speed as far as possible from the walls (for a reactor) (or at the center of a cylindrical duct), equal to 2 times the average speed (the average speed = that which is calculated in relation to the flow).
The flow paths are parallel to the axis
So, indeed, the molecules which pass "far" from the walls remain "far" throughout the path, and therefore take less heat than those which "brush" against one of the walls
In a turbulent flow, the speed is still zero
salary. the wall, but it's already getting very high very close to it, but in the middle of the walls we don't have much more than 1,2 times the average speed
example of calculation concerning the upstream and downstream pipes of the reactor:
Reactor: rod: 12.7; tube: 15; lg: 20 cm
Steam flow: 7 m3 / h (dynamic viscosity: approx 15 * 10-6kg / ms; density: approx 1.2 kg / m3 (steam temp: 100 ° C and pressure approx 0.95 atm))
Reactor pressure drop: 37.7 mbar (approx 37.7 cm of water)
section (rod / tube space): 50.03784 mm²
inner diameter of a tube of equivalent section: 7.98185 mm
I suppose the same flow downstream of the reactor (it is undoubtedly higher in reality since the volume increases with the temperature) that gives us a pressure drop of 7,39 mbar (for 20 cm of lg) or
5.1 times less than in the reactor for the same passage section
And in this case we have (except temperature) the same speed of the fluid: about 140 km / h
We can have fun doing a calculation to have the same linear pressure drop in the downstream pipe as in the reactor: this gives an internal pipe diameter of 5,788 mm, and tjs for 7 m3 / h, the speed (for the same linear pressure drop) would then be
266 km/h
The electrification by flow would be stronger in the downstream tube than in the reactor
With the 7,98 mm tube, we already have a much more turbulent regime than in the reactor, and it is even more so with the 5,7888 mm tube
On the other hand, if we take another example: it seems that it is better to have a gap of 0,5 mm:
stem: 15; tube 16 int. ; lg: 20 cm; pressure drop: 37.99 mbar
which allows a steam flow of 2 m3 / h (no more or we deviate too much from the possible vacuum with the suction of the engine)
diam for section eq: 5,57 mm, but linear pressure loss 8,6 times less: 4.41 mbar
for the same linear pressure drop we can go down to 3,6247mm inside (lives in this case: 194 km / h)
With our 12/14 tubes there is therefore no pressure drop, but no speed either, and perhaps no electrification by flow either in this tube
Let's return to the annular space of 0,5 mm anyway: in the case described above with the 16/15 reactor and 2 m3 / h steam, contrary to what one might think, there is no turbulent flow: Reynolds number: 1825 (less than 2000), while in the downstream tube diam 5,57 we have 10164 and in diam 3,6247 we have 15612
(kelp <2000
note that with Example 15 / 12,7; 7m3 / h, it's turbulent: 7150
7 m3 / h in diam 7,981: 24814
7 m3 / h in diam 5,788: 34219 for number of Reynolds
What to think of all this?
bolt