вторник, 16 февруари 2016 г.

Theory on the topic: "Two working substances with a common cold part "

I will try to theorize on topic:  "Two working substances with different boiling point with common cold  part"
I'll start with a simple understanding of the process of converting heat into mechanical energy. (And conversion of mechanical energy into heat)
Let us have a particle substance in the cylinder piston. Let the particle be at a temperature T1  higher than the boiling point of the substance diagram 1.



 Let the other fraction of the same substance at a temperature equal to the boiling point of the substance Tbp  stands from other parts of the piston (A). Assume an ideal option where the piston will mass and heat isolation - no heat loss. When a hot particle hits the  piston that in turn will hit the particle on the other side (B) and so the two particles will catch your energies, your temperature accordingly (C). If the temperature - both particles will have a temperature( T1 – Tbp  )/ 2. So I can make the following conclusion:
In Idel version maximum work that can be done given amount of working substance  in a converter  on heat into mechanical energy (whether in the cylinder piston or turbine) is equal to the equivalent of the arithmetical average its temperature and its boiling point
W= cm(T1 - Tbp)/2




If  I put a transformer  of heat into mechanical energy diagram 2, 2a (turbine on diagram) between the two tanks with a liquid substance, one has a temperature above the boiling point T1 , and the other has a temperature equal to the boiling point of a substance Tbp ,this turbine will perform work equivalent to (T1 – Tbp) / 2. If work done on the substance that comes out of the turbine with a compressor and reducing valve,equal work which is carried turbine then after valve parameters of substance must return to the starting position - the work is performed, the work done on it and temperature, volume and pressure at the beginning of the process and at the end of the processes are the same.
So when we use work to close the cycle of a working substance can not it remain useful energy - the mechanical force we have received, so we have to use on the working substance to close the cycle .
This cycle will call it - ZERO cycle.


Let us have two substances with different boiling point chart 3. The first working substance- A has a higher boiling point than second -  B.  For each substance have a two tanks, one being at a temperature higher than their boiling point – T1 , the other at a temperature equal to its boiling point - Tbp. The temperature of the substance B in the hot reservoir is equal to (T1a - Tbpa) / 2. Connect turbines and compressors to tanks. Both substances have zero cycle  - as the work carried out,  so the work we perform on them working substances. Will unite hot and cold tanks of both substances (hot to hot tank 6a;6b ,cold with cold tank5a;5b) to heat exchange with each other.
Let the amount of circulating substances are such that the cold part due to heat exchange the temperature of the two substances is equal to the boiling point of the substance at a high temperature - the first substance is liquefied. So after the heat exchange between the two substances first liquefies. This defeats the compressor 4 and the reducing valve 7a.  Compressor 4 and  valve   became pump diagram 4. 



 So the zero cycle of the first substance(ammonia on diagram 4) becomes familiar Rankine Cycle which we use in external combustion engine.
Due to heat exchange conducted between the two substances in  their the cold and warm tanks, compressor 3 will operate at  high temperatures  than turbine 2, but at the same temperature range, which preserves the balance of carried to the attached forces (Win / Wout) operation for the second working substance .

By combining hot and cold tanks of two Zero cycle, this cycle with high boiling point is converted in to Rankin cycle.

18.02.2016
For me remains unclear whether we can reach a temperature of the first working substance at outlet lower than (Tenv + Tbp)/2 (the average of ambient temperature and boiling point).Whether we can set at practice the following temperatures as chart 5:
200K of the heat exchanger 5 (close to the melting point of ammonia)
 210K heat exchanger 6
and keep them?


As  lower temperatures achieved (as long as they do not extend below the melting point :)) in the cold part - the higher the useful power will have the unit. Though 10 degrees difference do our work, I mean that after an initial "investment" - setting the cold part, will not be necessary "to spend a penny more."

20.02.2016

QUANTITIES OF HEAT AND WORK

 Rankine cycle -diagram 6; Zero cycle -diagram 6a; Unit with two zero cycle with a common cold part- diagram 6b
Accept that a working substance at a temperature equal to its boiling point no pressure in the evaporator and can not perform work, so that will denote the amount of heat that has this substance 0Q.
Will denote the amount of heat that gets working substance from the heater with 1Q.
In the simplified version of the energy conversion (charts 1 and 2)





 Rankine cycle -diagrama 6
After heater 1Q
After turbine 1 / 2Q. As a result of work done by the turbine 1 / 2Q turns into mechanical energy and then have a working turbine substance with 1 / 2Q heat
After the heat exchanger - 0Q. Cooled heat exchanger working substance to its boiling point and after it has heat 0Q
Heater heated working substance, giving it heat 1Q




Zero cycle -diagram 6a
After heater 1Q
After turbine 1/2 Q .  1/2 Q heat has become mechanical energy
 After the compressor - 1Q. Compressor performs work equal to 1 / 2Q to liquefy the working substance in which mechanical energy is converted into heat, so after reducing valve total amount of heat which has the working substance is 1 / 2Q +1 / 2Q = 1Q
 So working substance with 1Q heat energy enters the heater and after him have the same heat. Zero cycle - 0Q heat has turned into mechanical energy




Two zero cycle with a common cold part - chart 6b

The first working substance
After  heater 1Q
 After 1 turbine 1/2Q
 After heat exchange performed in a heat exchanger 5 0Q
1 / 2Q after heat exchange with the second working substance in a heat exchanger 6
1Q after the heater is increased heat energiya- 1 / 2Q +1 / 2 Q = 1Q

The second working substance
 1Q  after heat exchange in heat exchanger 5 with the first working substance 
 Compressor performs work such as mechanical energy equal to 1 / 2Q becomes heat -1Q + 1/2Q = 
 = 3 / 2Q after compressor
1Q after the heat exchange in heat exchanger 6
1 / 2Q after turbine 2
1Q after heatexchanger 5

1 / 2Q thermal energy is converted into mechanical energy of the unit with two working substances with a common cold part compared to Rankine cycle

понеделник, 1 февруари 2016 г.

Two working substances with a common cold part as to start the internal cooling

Reflecting on the method for converting heat into mechanical energy that the cold part I create by doing work I saw an untapped so far from me the opportunity for effective work - It would be better to replace expansion valve on "refrigerator"with a turbine  (or piston / cylinder; generally speaking converter heat into mechanical energy) diagram (1a)( 1).










 On both sides of the expansion valve system for redistributing heat ("refrigerator") there is a temperature difference (differential pressure) which is a prerequisite turbine to perform work. By doing this work converter of heat into mechanical energy (as well as all others who work in the method) achieved two important goals for us:
- Mechanical energy
- Cold
The mechanical energy is our goal, and the cold part we need in the process of internal cooling in the absence of such a natural.
Let the operating cycle of the n-th working substance creating cold part by using cooler. Replace the expansion valve with a turbine. We now have a further beneficial force that is a result of the conversion of heat into mechanical energy.
Replacing expansion valve with turbine converts refrigerant in working  substance. Now in the cold part (the last n-th cycle of the working substance with the lowest boiling point) have two working substances with a common cold part - the cold part of the system for redistributing heat.
In the cold part  the working substance liquefies because of the low temperature generated in the expansion of refrigerant due to the compressor. In the warm part of the working cycle of the refrigerant already liquid working substance is heated with the same amount of heat which is taken away from him at liquefaction. Thus, after the heat exchanger in the hot part will have a liquid working substance having a temperature  ½ of the difference between the temperature on entry into the turbine and its boiling point
 Working cycle on refrigerant becomes neutral - no change in temperatures, considering that the same amount of heat is removed and transferred from refrigerant gas at working substance. Therefore  the work of the compressor and turbine are the same -  as the mechanical power is converted into heat in the course of operation of the compressor on refrigerant,  so the same heat turbine is turned into mechanical power. Cycle on refrigerant agent is a neutral shade.
Energy balance  on n-th working substance + refrigerant (provided that the turbines convert 50% of the heat into mechanical energy) will be:

A zero cycle – the  "refrigerator"
A cycle in which 50% of the heat is converted into mechanical energy - the working substance
The outcome of the two cycles is a liquid  working substance with  a temperature 1/2 of the difference between its temperature before entering the turbine and then the heat exchanger in the hot part of the "fridge".
Thus, in the last cycle (n) working substance decelerated for some amount of heat as has become a mechanical energy,  whereby  cooled the previous(n-1) to allow the unit to operate.

Now I can not go back heat into the evaporator  - diagram 2



Variant of еngine with two working substances (ammonia and ethylene in the case) - chart 3

In optimal load of the turbine and compressor at best we can achieve a temperature difference between input and output = (Tenv - Tbp) / 2
where :  Tenv- temperature on environment; 
              Tbp - boiling point of the first working substance
The power will be expressed (ideally):

P = cm (Tenv - Tbp) / 2

where: c - specific heat capacity of the first working substance
            m - mass of first working substance circulating for a given time







Note again: As with all engine variants of the external combustion - internal cooling method cold/s part create it beforehand using external force.