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 T_bp  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 – T_bp  )/ 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 T_bp ,this turbine will perform work equivalent to (T₁ – T_bp) / 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 - T_bp. 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