Open Systems

[Low-Q]

Don't know if this is said before in this thread, but when we speak of open systems such as a heat pump, you put in x amount of work and get Y amount of heat out.
If the COP is 5, you get 5 times energy as heat out than the input work.
A heat engine have the opposite efficiency. A heat engine is in principle a heat pump that is somewhat reversed. You put heat in, and get work out. This efficiency will allways be lower than 1/COP of the heat pump if you are using the only two reservoirs available on the heat pump - the cold and the hot one.
In order to make this heat engine to work as wanted, you need a third reservoir. Say the heat pump is using outside air as reservoir 1. Reservoir 2 is the heat pump output. Reservoir 3 is the ground.
Now, connect one side of the heat engine to the cold or hot reservoir of the heat pump, and connect the other reservoir to the ground. Then this heat engine will potentially do more work than the work required to run the heat pump. I discussed this issue with competent people at physicsforums.com. It seems reasonable.


Vidar

[LibreEnergia]

Don't know if this is said before in this thread, but when we speak of open systems such as a heat pump, you put in x amount of work and get Y amount of heat out.
If the COP is 5, you get 5 times energy as heat out than the input work.
A heat engine have the opposite efficiency. A heat engine is in principle a heat pump that is somewhat reversed. You put heat in, and get work out. This efficiency will allways be lower than 1/COP of the heat pump if you are using the only two reservoirs available on the heat pump - the cold and the hot one.
In order to make this heat engine to work as wanted, you need a third reservoir. Say the heat pump is using outside air as reservoir 1. Reservoir 2 is the heat pump output. Reservoir 3 is the ground.
Now, connect one side of the heat engine to the cold or hot reservoir of the heat pump, and connect the other reservoir to the ground. Then this heat engine will potentially do more work than the work required to run the heat pump. I discussed this issue with competent people at physicsforums.com. It seems reasonable.


Vidar

The end result of such a scheme is  the equivalent of a heat engine operating with an efficiency dictated by the reservoirs at the highest and lowest temperatures. It does not and can not ever give rise to a self running system where the heat recovered by the heat pump is in excess of the heat required to drive the rest of the system.

The efficiency of the heat pump is at a maximum when the temperature of the hot and cold reservoir  is at a minimum. The opposite is true for the heat engine. Combine those two facts together and the result given above can be deduced.

[MarkE]

Don't know if this is said before in this thread, but when we speak of open systems such as a heat pump, you put in x amount of work and get Y amount of heat out.
If the COP is 5, you get 5 times energy as heat out than the input work.

No you don't.  For a COP 5 heat pump you get about 5/6ths useful heat energy moved as you put in.  1 part motive energy (usually electricity) moves 5 parts heat from a colder reservoir to a hotter reservoir.  The 5 parts heat moved is the useful result.  Absent the 5 parts heat in the colder reservoir the heat pump does not function.  The cold reservoir only has to get cold enough that it no longer rejects heat to the refrigerant, or the hot reservoir so hot that the refrigerant cannot reject heat to it and the heat pump stops working.

A heat engine have the opposite efficiency. A heat engine is in principle a heat pump that is somewhat reversed. You put heat in, and get work out. This efficiency will allways be lower than 1/COP of the heat pump if you are using the only two reservoirs available on the heat pump - the cold and the hot one.
In order to make this heat engine to work as wanted, you need a third reservoir. Say the heat pump is using outside air as reservoir 1. Reservoir 2 is the heat pump output. Reservoir 3 is the ground.
Now, connect one side of the heat engine to the cold or hot reservoir of the heat pump, and connect the other reservoir to the ground. Then this heat engine will potentially do more work than the work required to run the heat pump. I discussed this issue with competent people at physicsforums.com. It seems reasonable.

Vidar

Similar unworkable ideas have been proposed many times.  In order to make such a machine work, you have to extract more energy returning heat energy from the hot reservoir to a cold reservoir than you expend moving energy from a cold reservoir to the hot reservoir.  That means that you not only have to convert all heat from a single reservoir to useful work, you have to more than do that.  That puts you squarely at odds with the Second Law of Thermodynamics.  So, first you will have to find an exception to the Second Law.

[tinman]

   

There you go again with inffering things i never said

Surely you are not so naive as to believe that we could put another solar panel behind the first one and it would intercept the same light as the first.

But here is a little something for you to ponder. Take a large glass jug,and place one solar panel behind the other. In front of the first solar panel place a light bulb.Cover the glass jug so as no ambiant light can get into the jug/reflect onto the solar panels.Switch on the light bulb and messure the avaliable power from each solar panel. The second solar panel wont have any where near as much as the first solar panel(the one opposite the light bulb)Right?.
Now,fill jug with water,and redo test.

Your boundaries are junk. As can be seen in your diagram,you just place them where ever you think will help you debunk anyones idea's.
Please show me(and everyone here) that placeing a solar panel near a light bulb,and placing a load across that solar panel, reduces the output of the lightbulb,and/or increases the energy input to that light bulb.Show me this,and then we'll talk. You cannot. Insted you come up with all these mythical themes like boundaries.The !!boundary! is around each system-as i explained. When one system dose not impart or reflect on another,then that boundary is set.We have two sepperate system's,and one feeds off another without interference.

lets do a little more math here.
an incandecent bulb is about 5 to 6% efficient at converting electrical energy into light.
So a 5 watt bulb would put out around 300mW of light power.
A solar panel at best is about 17% efficient at converting light energy into electrical energy. So from our 300mW of light energy,we would see 17% of that converted into electrical energy.This means that we should only have around 51mW output from the solar panel at best if we cover the whole bulb in solar panel. So we need only achieve 1 volt over an 18 ohm load from our solar panel

[tinman]

  So, first you will have to find an exception to the Second Law.

http://www.wsj.com/articles/SB1028220616493488320