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Author Topic: Energy losses, Thermodynamics and Efficiency  (Read 12905 times)

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #50 on: 24/10/2013 13:05:41 »
There has not been enough information given to actually do ANY calculations. But I will give my ideas a go. The system is 150kg cylindrical block (probably benefit from a different shape but for calc means a cylinder) at 8000kg/m3 this give a cylinder with a height of 1m (arbitrary) and a radius of 0.077m.

This is more like it, how positive and optimistic it is to actually be doing some calculations. What is not so nice though, is the armchair cynicism, can we really afford to be so offhand about things, I think if there is the slightest chance of something working we should get our noses to the grindstone until the idea has been completely, exhausted.

The volume of air would be 0.18m3 we would need to remove 54% of this.. 0.0972m3 per movement, 1 a second will give a vacuum requirement of 175m3/hr which requires 9kW (Power=flow.pressure difference). That's a very basic calc.. another works on pump down time (S=v/t . ln(Po/Pi) ) which gives 468M3/hr.. this would need a 25kW vacuum pump.
So depending on which you look at its still not possible. you will gain 7.35kW in total from each unit and need minimum of 9kW to operate it. (based on this system of course)


Have a look at the SMB-C06 roots vacuum pump manufactured by Vacuum Pumps America, it has a throughput of 600 m^^3/hr and operates at 3500 rpm, the motor works on 2.2 KW and the final vacuum produced is on the order of 10^^-4  Torr.  The fantastic thing is that the ‘Home Grid’ application needs only one hundredth thousandth of that degree of vacuum !!! What do you think of that, one hundredth thousandth of that degree of vacuum, several orders of magnitude lower than what is available ! So yes, it lays the field wide open as to what is possible and what is not. I am pretty sure that it can be done.
« Last Edit: 24/10/2013 13:09:50 by McQueen »
 

Offline SimpleEngineer

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #51 on: 24/10/2013 14:15:25 »
Of course, I used the simpleset of simple calcs for my run through, however it may have shown why it is a very difficult system to calculate..

That pump you found is to be used as a booster pump, meaning that there is another beefier pump that will take the main power for the draw.. but a direct mechanical linkage to a piston type evacuater built correctly MAY provide enough vacuum but I am no pneumatics expert.
 

Offline SimpleEngineer

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #52 on: 24/10/2013 16:42:02 »
Actually.. I reran the calcs.. dropped a few clangers in there.. apologies. Will try to fix tomorrow.
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #53 on: 25/10/2013 06:40:34 »
Actually.. I reran the calcs.. dropped a few clangers in there.. apologies. Will try to fix tomorrow.

While trolling the net I have come across several vacuum pumps that will fulfill the criteria of 600^^3m/hr, dry running and using just 0.4 KW, that will reach 1 Torr without any backing or booster pump involved. This still leaves a margin of a factor of ten, since the vacuum that is required is more like 100 Torr than 1 Torr.  With the pump drawing such low power, it is possible that a separate vacuum pump can be used for each paired tube system, increasing the efficiency and  making the continuous generation of 15 KW possible.
 

Offline SimpleEngineer

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #54 on: 25/10/2013 09:08:19 »
That is pretty much what I am finding with the errors in my calcs.. I am about a factor of 10 off in a few areas.. *sigh* the eternal struggle with calculations using different units.. We noticed this when i passed my calcs in front of my colleague who has designed vacuum systems.. (He is resolute that the system will lose too much energy through friction yet cannot negate the principle)
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #55 on: 25/10/2013 15:08:10 »
(He is resolute that the system will lose too much energy through friction yet cannot negate the principle)

Since you seem to be so obsessed with the idea that friction is going to clog up the works, I decided to investigate the problem. It seems that PTFE  ( polytetrafluoroethylene) has a sliding co-efficient of friction lower than ice, this applies to PTFE and steel and  almost any other material including itself.  The co-efficient of friction of PTFE (a type of teflon) is just 0.05, lower in fact than ice on ice, you can be sure that the pulley will also have a  PTFE component  in it, either as a coating for the bearings or as an integral material.  Could you kindly factor in the numbers and tell me how you think that friction will slow the whole system down and eventually bring it to a halt. Frankly the logic behind the statement continues to baffle me, given the large amount of K.E involved  both during descent and ascent of the counterweight. The descent being powered by gravity and the ascent by atmospheric pressure in the presence of a vacuum.
Here are a few more facts about PTFE
PTFE possesses the lowest friction coefficients of all solid materials; between 0.05 and 0.09:

    The static and dynamic friction coefficients are almost equal, so that there is no seizure or stick-slip action
    when increasing the load, the friction coefficient decreases until reaching a stable value
    The friction coefficient increases with the speed
    The friction coefficient remains constant at temperature variations.
Wear
The wear depends upon the condition of the other sliding surface and obviously depends upon the speed and loads. The wear is considerably reduced when adding suitable fillers to the PTFE (see filled PTFE).
 
« Last Edit: 25/10/2013 15:36:36 by McQueen »
 

Offline SimpleEngineer

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #56 on: 25/10/2013 16:31:07 »
See work up..
Dimensions
Weight = 150kg
Density (steel) = 8000kg/m3
Volume = 0.018m3
h=1m
r= 0.075m
A= 0.018m2
movement = 10m
Volume air = 0.18m3
Forces
F=ma
Downwards force = 1470 N
Maximum upwards force
If pressure above is 0
Pressure below weight = 100kPa
Force acting on weight = 100000*0.018  = 1800 N
Maximum Pressure required for return = 18kPa
Acceleration and times
Downstroke
 a= 9.8m/s2
 t= 1.43s
Upstroke
amax = 2.2m/s2
tmin=3s
Vacuum Calcs
cycle time of 5s  (using vacuum pot to ensure vacuum is instantaneous)
so 0.18m3 of air in 5 secs
minimum increase in pressure.. say 1kPa, If vacuum pot is at 1 torr. (133pa) then 0.18m3 of atmospheric pressure air would occupy 555.5M3 to give a rise of 1kPa... This require a vacuum pump that can do 111.1M3/sec on outline calc (555.5/5)
Using S=V/t x ln(Po/Pi) = 238 M3/s (V= 555.5, t= 5secs , Po= 1133Pa, Pi=133Pa)
Using Power = S x dP/mech eff (typically 0.85)   
=   130kW or 280kW

Friction force with PTFE will still be 60N with the differential force on this example being 330N, this is fairly significant.. that does not take into account the friction of the air movements etc.

Work up may be wrong only had 15mins to write it (10 mins was drawing a picture that was too large to be posted :( )
Point out changes please.. As the power is large.

Something forgotten is the increase volume of air at lower pressures.. I have tried to take this into account
 

Offline McQueen

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #57 on: 25/10/2013 16:49:34 »
Using Power = S x dP/mech eff (typically 0.85)   =   130kW or 280kW

All I can say is that using your calculations, the vacuum elevator would never have been built, would not have even been a pipe dream or a gleam in someone's eye. Don't forget that the vacuum pumps used in a vacuum elevator have to be pretty sophisticated to maintain a pressure differential that allows a constant velocity of 0.15m/sec, the final velocity for 'Home Grid' would be about 14m/sec! So I would surmise that the power needed for the vacuum pump would be far less than the 3 KW needed for the Vacuum elevator. I repeat a figure of about 0.5 KW, I could give links but that doesn't seem to bother you too much.
 

Offline SimpleEngineer

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #58 on: 28/10/2013 10:40:02 »
oh it does bother me.. I have rerun the calcs you see above, and now get a solid solution.. This is for the system described.. a better designed system could obviously offer a massive reduction to power or other means of recouping enrgy.. but as for now I only have the calculation steps above to go with.
 

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Re: Energy losses, Thermodynamics and Efficiency
« Reply #58 on: 28/10/2013 10:40:02 »

 

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