Easy Gravity wheel

[quantumtangles]

@ Vidar

That is most interesting and helpful.

In other words there is no benefit to using less dense working fluids.

I suspected this and said so in earlier posts about the invention.

I only used the example of less dense working fluid to answer critics who claimed working fluid had to be "lifted" to the top of tank A.

I always preferred the idea of using seawater as working fluid (due to its greater density).

However, what I have had difficulty calculating is the amount of energy required to force one cubic metre of seawater per second (density 1020kg/m3) into the base of a 30m high cylinder full of seawater.

one cubic metre per second is drawn from the top of the vessel at the same time as one cubic metre per second is crammed into the base of it.

You calculated required energy in joules, which makes sense.

But somewhere I recollect warnings about trying to convert joules to watts (I need to know the value in watts).

How would you go about calculating the precise energy in watts required to force one cubic metre of water per second into the base of a 30m high cylinder full of seawater? (mass out at the top equals mass in at the base).

Would the removal of 1m3/s from the top of the tank create a suction effect that would minimise the energy required to draw water into the base of the tank? Would adherence caused by covalent bonds in the water molecules 'glue' them together en bloc?

The pressure at the base of the 30m high cylinder is 300,186 Pascals gauge pressure (1020kg/m3 x 30m x 9.81 m/s/s) + Patmos of 101325 Pascals = 401,511 Pascals = 401.511kPa

I use compressed air in vessel B to force seawater back into the base of vessel A. But the amount of energy required is critical.

I took the direct approach of ensuring that the pressure in vessel B always exceeded the pressure at the base of vessel A (in other words I did the maths to ensure vessel B always maintained pressure of 500kPa).

So it is not a buoyancy question (as the working fluid is seawater just as the rest of the fluid in tank A is also seawater).

Rather it is a question about the energy required to cram one cubic metre per second of seawater into the base of a 30m high cylinder, a cylinder which is already full of seawater (note that one cubic metre per second of seawater is simultaneously being drawn out of the top of tank A by a pump assisted siphon so that mass in equals mass out).

Many thanks Vidar. I am getting closer to knowing the truth of the matter. if you have any thoughts whatever on how to calculate the energy in watts to cram this cubic metre of water per second into the base of cylinder A I would much appreciate them.

[Low-Q]

1 Joule is the same as 1W in a period of one second. In one hour you get 3600 Joule of energy at 1W/s. So the term Watt is not energy alone. Watt is energy only over a given period of time. So Joule should be the term to use in any context. Hope that clears it up:). To lift 9810N 30 meters require approx 30kJ of energy. Or 30kW/s. You need 30kW to lift 1m^3 of water 30 meters up pr second.

[quantumtangles]

@ LowQ

Many thanks.

If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.

The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.

Here is the maths for electrical output.

Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

So it seems we spend 30kW and get output of 225kW.

Interesting.

[Pirate88179]

@ LowQ

Many thanks.

If it really only takes 30kW to move one cubic metre of water up 30m, then my machine should work.

The same cubic metre of water falling 25m onto an impulse turbine generates about 200kW of electricity.

Here is the maths for electrical output.

Pwatts = h(25m) x g (9.81 m/s/s) x rho (1020kg/m3) x 0.9 (eff) x 1m3/s (flow)
= 225kW

So it seems we spend 30kW and get output of 225kW.

Interesting.

I believe the input would be 30 kw per second.  I think the time has to be factored in.

Bill

[Low-Q]

Just have in mind that Watt isn't energy unless you apply time. My guess is that the water falling 25 meters will for sure generate 225kW, but for how long? You talk about impulse turbine! That is exactly where the flaw is. You have a short burst of 225kW.

That is why I said that mesuring everything in Joules is better. Watt alone does not show how much energy there is because you miss time in the equation. Sorry. It will most probably not work. Do a recalculation in Joules. That water falling 25 meters creates less energy than it takes to push it up 30m.

This is btw a very good example why gravitywheels works inside the inventors mind, but does not work in real world. What do the inventor miss? Time!

Vidar