Using the Heat Conductors (Refrigerants):
In this vehicle drive system, lithium battery energy storage is by way of the electric current generator.
Here the statements are about lithium heat energy storage, whereby the heat is then used to generate electric power.
That means that drive energy will come from excess heat stored in lithium as well as burning heating oil, together with the energy stored in batteries that also happen to use lithium as one of their main component parts.
Graph 2
Due to the lack of space in a vehicle, only a limited amount of heat energy will be stored at any given time.
Q=mCpdT
Where Q (W) is the heat that the fluid loses/gains, m (kgs) is the mass flow rate of the fluid, Cp (Jkg.K) is the specific heat of the fluid, and dT (K) is the temperature difference between the outlet and the inlet.
From the Vapour Pressure Graph it is assumed that drive power of the propellant for the turbine will be between 5 bar and 15 bar, regardless of whether 1 or 4 heat amplification stages are used.
The Jacket Pump (JP) is also an heat amplifier, just like the other compression pumps. It is number 1.
That means that, whether or not ICG1 is built in, or V1 and V2 are turned off, the temperature of the Lithium in HS will fluctuate.
It can be as much as 50deg.K, if the Water/Glycerol circuit is used. The heat conductors have been selected, dependant on their retention period in the atmosphere. Except for water vapor or water molecules, they have the lowest documented atmospheric retention times.
When water vapor decays to hydrogen and oxygen, living organisms do not seem to mind.
The situation looks different for other substances, because over dose can occur.
The Specific Heat of the Heat Conducting Liquids in Graph 2:
propane = 1.8kj/kgK
n-butane = 2.5kj/kgK
n-pentane = 2.4kj/kgK
n-heptane = 2,1 kj/kgK
Those specific heat values are close enough together to assume that they will have very similar heat collecting capabilities or Enthalpies.
HFC R-134a seems to be in every house old refrigerator and is said to be a mixture of propane and n-butane. It's data sheets are also easier to find and that justifies it's use here for this description.
The diagram shows that:
In the process of collecting heat, the HFC R-134a will collect heat from the heat collector H-COL as follows:
value2 = 574kj/kg
value3 = 606kj/kg
value1 = value4 = 480kj/kg
input temperature = 12deg.C
output temperature = 60deg.C
Calculation:
value2 = 574e3kj/kg
value3 = 606e3kj/kg
value1 = 480e3kj/kg
value4 = 480e3kj/kg
Cooling capacity of HTB1 using R-134a:
Qhtb = value2 - value1 = 94kj/kg
The compressor is in ICG1 or Back End.
Spent compressor work is:
Qw = (value3 - value4) = (606kj/kg - 574kj/kg) = 32kj/kg
Heat energy delivered to ICG1 or Back End:
Qc = (value3 - value4) = 126kj/kg
Efficiency:
? = (value3 – value4)/(value2-value3) = 3,94
Note how the expansion factor of the heat conducting liquid results in an efficiency ratio that is greater than 1.
For the calculation, just use the same values for each stage. The size of components will depend on the time that is allocated in order to collect the heat.
From Graph 1, the required heat power is defined to be:
20KW. That allows 10KW for losses and 10KW for the wheels of the vehicle.
Due to the undefined design parameters of the Turbine, Generator Battery and Motor combination, it can not be specified how much of the heat losses in those components can be recollected.
Those paths are not shown in the drawing, but the same Water/Glycerol can be directed through them just like through the Jacket Pump, if the temperature is not expected to exceed the self ignition temperature of the compounds used.
Supply of R-134a ˜ n-butane ˜ n-pentane ˜ n-heptane Gas that is required will be:
Gas Flow Per Second = Rs = 20kw/126kj/kg = 0.15873 kg/s
Per Hour = Rs * 3600 = 571.428 Kg/h
The same gas is circulated continuously.
Compressor Power For Cmp-P4 in the Back End:
D) Per Second = Qp = Qw * Rs = 32kj/kg * 0.15873 kg/s = 5.0794kw
Per Hour = 32kj/kg * 571.428kg/h * 1h = 18.285696 MW
E) Per Second = Qh = Qhtb * Rs = 94kj/kg * 0.15873 kg/s = 14.92kw
Per Hour = 94kj/kg * 571.428kg/h * 1h = 53.714232 MW
F) Check Per Second = Qt = Qc * Rs = 126kj/kg * 0.15873 kg/s = 19.999980kw
Check Per Hour = 126kj/kg * 571.428kg/h * 1h = 71.999928 MW
Once heat has reached the Heat Storage (HS), it is reused several times.
One litre of heating oil is said to contain about 10KWh of heat.
From B) above it was assumed that only 2.75 KW of new energy was required , if all the heat is recollected at the Water/Glycerol point.
When the storage is full, the Burner is turned off.
Go down to 80KM/h and charge the batteries on the move.
Amount of Lithium Heat Storage Required:
Heat stored for a temperature change of 50deg.C = 50deg.K:
In deciding how much heat to store, due to space restrictions, one can define that 1/8 m3 Lithium is to be used as storage in HS.
Amount of Lithium = 1/8m3 * 535kg/ m3 = 66.88kg Lithium
Q=mCpdT = (66.88kg * 3.6 kj/kgK * 50K) = 12.038Mj = 12.038MW
Example:
Period Of Supply = 12.038MJ/5.0794KW = 2370 seconds = 39.5 minutes.
That means that the 1/8 m3 of Lithium is first brought up to the operating temperature, whereby it uses Cmp-P4 to store and reuse 39.5 minutes of energy per hour for the operation of the vehicle in question.
The other 8* 20.5 minutes of energy that belong to a full 8 hour day can be collected and stored in the batteries during the 16 hours when the vehicle is stationary.
Because only the temperature difference is relevant, it will not make any difference whether 120deg.C or 180deg.C maximum temperature is used for the propellent as long as a suitable type is selected for the job.
The said quantity of heat will serve as a energy reservoir that can be topped up together with the battery, even if the vehicle is stationary. It however does not define the range of a vehicle, because any extra heat that is required will be inserted through the use of the Burner and or the Front End heat collection mechanism.
If the above amount is to be collected in 1 hour = (60*60) seconds,then:
G) Amount to be collected per second:
Qs = Q/(60*60) = 12.038MW/3600 = 3.34 KW
H) Compare this:
G) = Qs = 3.34 KW with the values
B) = Qn = 2.75 KW and
D) = Qp = 5.0794kw above.
I) The most important thing here is that the Heat Storage (HS) can take on the heat that is presented to it in the Water/Glycerol heat conductor or directly through the Jacket Pump (JP). That means that, if no Water/Glycerol or Cmp-P4 is used, JP will have to compress the propellant itself. That would be Mode 4, which is not discussed directly in this text, because we are looking to collect heat from the chimney of the Burner and the Front End.
Note that only the top 50deg.C = 50deg.K that are relevant for the propulsion of the vehicle is included in the calculation, just like only the energy in the battery that can be used to produce propulsion is taken into account.
The heat collecting for HS is turned off by turning off Cmp-P4 when the heat storage has reached maximum temperature. At that time the Water/Glycerol heat conductor will have minimum temperature.
The path through ADV and Adj-4 would then be used.
The Water/Glycerol coolant has two different paths through Adj-V4 to the Jacket Pump (JP).
Note that the Front End and Intermediate part of the system could have been left out, but there are applications where they are an advantage.
When one calculates the amount of heat required to melt 1m3 of iron compared to that which is required to drive 100KM per day the situation does not look that good, but if the vehicles have a long life span, the calculation will look even better from the angle of wasted energy that can not be replaced.
Q=mCpdT
Number of Jules to melt Iron Q = 7870 * 0.45e3 * 1538
You will probably have to look again, if you want to find the 5KW to 20KW gas turbine, otherwise you will have to design one yourself or have one designed for you, but we are talking infrastructure development in Africa here, therefore that should not be a problem.
For stationary applications of the SHE, zero energy input at operation time can be achieved through the use of Hydro Electric or Solar Electric power.
In such cases, one method of implementation would be to replace the Burner with a microwave or a bank of microwaves.
The inputs and outputs of HS can also be reconfigured in order to accommodate a microwave.
The electrical energy from the Hydro Electric or Solar Electric source is then used to provide the external energy for the SHE.
The first design challenge of hydro electric current generation is how to guide fishes around or away from the turbine blades.
The second one is how to keep a defined supply of water constant for a whole year.
Building hydro or solar electric power generators around rivers, lakes, salt lakes or seas of any sort means that the hydro and or solar electric power can be used to multiply the amount of available heat energy, by using the initial amount of hydro or solar electric power to power the SHE and thereby capture an higher quantity of energy that will then be the power supply for the process that requires electric power.
Using water as the input source of the heat compared to air means that the heat collection process will be less energy intensive.
Note that the SHE for vehicles described in other parts of the description does not actually rely on fans to blow air over the Heat Collector (H-COL), instead it relies mainly on the forwards movement of the vehicle which causes an air flow over the H-COL.
The faster the vehicle travels the more air will pass over Heat Collector (H-COL) and more heat will become available.
When the vehicle is stationary in the day time, direct radiated heat from the sun will be the main source of energy.
It is stated here again that only the 50 Kelvin that is required in order to produce the 5 to 15 bar of propellant pressure is relevant, therefore the question of how many heat collection steps should be used and how large they should be are design time decisions, based on the substances used to implement the heat collecting and propelling functions.
There is no law that says the only water vapour or heated air can be used as a turbine propellant.
Even though iron has what seems to be a low heat capacity, it is also a good medium in which to store heat for stationary applications.
With:
1 litre water (m) = 1.000kg
Specific heat water (Cp) = 4.182kj/kgK
1 litre Iron (m) = 7.870kg
Specific Heat Iron (Cp) = 0.45kj/kgK
dt = 50K
Q=mCpdT
Water Q = 1.000kg* 4.182e3J/kg * 50K = 2.091E05J = 2.091E05W
Iron Q = 7.870kg * 0.45e3J/kg * 50K = 1.7708E05J = 1.7708E05W
Note that the real heat storage capacities of both substances per unit volume are closer than the specific heat values would suggest.
Furthermore, iron will not melt or boil in the temperature ranges being discussed here.
With a suitable liquid for the heat transferred from the Microwave to the Heat Storage (HS) in order to produce high storage temperatures, steam could be back in the race to be a turbine propellant for this type of application.
Note that the Heat Storage (HS) can be reconfigured to suit different situations.
Here is a partial list of heat conductors that have boiling points in the range required for medium power applications. Suppliers can tell you more.
Name Boiling Point °C
Butane 5 = R600
neoPentane 10
Diethyl ether 33
Trichlorofluoromethane 23.8
Dichloromethane 39.9 = R 30
Methylene chloride 40 = R30
Methyl Iodide 42
Methyl Acetate 56.9
Chlorinated solvent Chloroform 61
Methanol 65
1,1,1-trichloroethane 74
Chlorinated solvent Carbon tetrachloride 77
Ethanol 78.3
Isopropyl Alcohol 80.3
Chlorinated solvent Trichlorethylene 87
Water 100
Chlorinated solvent Perchlorethylene 121
Brominated solvent Ethylene Dibromide
(1,2-dibromoethane) 132
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Application Examples:
Burner (Extension 1)
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