Ambient Temp ORC

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Russell Philips

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Mar 3, 2025, 3:08:10 PMMar 3
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Feed Pump (low pressure cool liquid -> high pressure cool liquid)  (low volume, trickle flow)
Hot Exchanger (cool liquid -> warm vapor) (tube in tube with braided twisted copper wire thermal mesh)
Vapor Expander (high pressure vapor -> low pressure vapor) (large expanded volume) 
Cold Exchanger (warm vapor -> cool liquid) (tube in tube with mesh)

5th component
Heat Exchanger (tube in tube with mesh) (recycles thermal energy up to 100%)

The dynamics change dramatically by using propane as the working fluid. The expansion factor is much lower than water, but ambient operation is enabled! This opens many, many options for source and sink.

If temperatures stay below the melting point of plastics, non-standard material selection is enabled (consider the advantages of self lubricating materials).

Let's use a steam engine instead of a turbine. (Perhaps even an AODD pump!)

This fully sealed, pressurized cycle offers the highest efficiency at ambient temperatures.

ctyankee

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Mar 4, 2025, 1:23:20 AMMar 4
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Which end do you want at ambient temperature? You can choose either one, the hot end or the cold end but not both.
The Rankine Cycle *REQUIRES* a difference of temperature between hot & cold to extract work from the heat flow.
If both sides are at the same temperature, then there is no heat flow, and consequently, no work is extracted.

Before we can talk about 'efficiency' we must first clarify whether you are referring to the theoretical Carnot Efficiency (CE), which can be calculated by the temperature difference between hot & cold, is *always* less than 1.0, and is never attained by any mechanical system, or the 'Economic Efficiency' (EE) which is how much total value one can get for a given capital & operating expense over a period of years.  For any system to make sense the EE needs to exceed unity, even though the CE never can.

Russell Philips

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Mar 14, 2025, 4:35:37 PMMar 14
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Untitled.jpg
I have been very interested in heat pump and ORC technologies since 2007 (also HHO & H2). The heat pump mostly is used to move heat from one place to another. It can also move heat from one ambient area (source) to another ambient area (sink) creating thermal potentials. These potentials can be exchanged with the ORC. The union of these cycles has been debated. So far my math models are energy deficit. This is where my interest in ambient ORC operation comes from. The stand alone ORC has serious applications in low temp operations. A build for real application must determine target use and likely source and sink temperatures. Temperatures and working pressure will rule materials in or out of scope. Solar PV is the competing technology to beat and is likely less expensive, especially when R&D is factored. ORC can run 24/7/365, can operate safely indoors and securely underground. It is bizarre to consider components operating at full capacity while being able to keep your hand on them temperature wise. This is the most green energy on the planet (save very future hydrogen). I am pushing the edge for low temp and low pressure operation and uses, trying to include self-lubrication and robustness where applicable. I have relatively recently become aware of Air Operated Double Diaphragm pumps (AODD). These are off the shelf and available for purchase today! Their use as a vapor expander does not include an output shaft (electrical output) but the twin diaphragms pump fluid very well. It is my opinion that this is probably the lowest hanging fruit for open source steam. I will be posting more on components and home builds. Oh, COP only goes above 1.0 when environmental energies are available and not considered in COP calculations. It is possible that both source and sink are environmental, probably neither being ambient. Many would appear content with successfully harvesting one free environmental energy.

Paul Passarelli

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Mar 14, 2025, 8:32:01 PMMar 14
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you wrote:

  "So far my math models are energy deficit."

That's good. It means your math is probably correct.  If it had come out any other way, I'd be confident saying there was an error.

--Paul

Solar & Thermal
Paul Passarelli
Chairman & President
Pa...@SolarAndThermal.com
(203)846-2500 (main)  (213)207-6691 (direct)


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Russell Philips

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Mar 15, 2025, 12:46:40 PMMar 15
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The average home air conditioner heat pump is 400% output or 4.0 COP. The first law of thermodynamics supports that processing this in a heat engine would result in 4.0 cop minus losses. Confidence that losses in combined cycles will result in less than 1.0 cop appears to be naysaying. Is the glass half empty or half full?

So back to ORC. The ORC's first great advantage is that working fluid flow is in a continuous one direction loop. There is no cyclic thermal mixing in the greater cycle. This is huge, and limits cyclic "mixing of thermal losses" to the vapor expander, where we have choice of type, cost and efficiency. Continuous flow also limits unswept volumes to the vapor expander itself. Continuous flow also enables pure tube in tube exchange for thermal recycling with zero change to unswept volume and loss. Cyclic hot air engines appear to have strong disadvantage to add regeneration without increase to unswept volume. Thermal recycling is very important to high efficiency. It is a very simple choice to add the 5th component heat exchanger to an ORC build.

Every AODD includes an insane amount of technology and precision (and they enjoy small unswept volumes). My first AODD for testing only cost $300. This gets extremely costly to match with the steam engine. Industrial turbines are able to squeeze out the most amount of work, but have industry costs and support. Small turbine efficiency drops significantly, giving rise to steam and AODD expanders for home scale.

Propane is considered slippery and extends the window of self-lubrication.

Paul Passarelli

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Mar 15, 2025, 1:51:30 PMMar 15
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The COP is an indication of the total heat moved from one reservoir to another.  The COP declines rapidly as the delta-T increases.
The Rankine Cycle always has an efficiency less than the theoretical Carnot Cycle.  
The maximum efficiency of a perfect Carnot Cycle engine, meaning the work it can extract from a given quantity of heat is given by:

  η = 1 - Tc/Th

The work that one can extract from heat moved through a heat pump, will always be less than the work supplied to that heat pump.
I can assure you that generations of Perpetual Motion pursuers have tried and failed in this approach.

If you're interested in the legitimate pursuit of low-cost sustainable energy, I would welcome your support.
If you insist on chasing the shadows of perpetual motion or something for nothing (which Includes "Brown's Gas" aka "HHO"), then I have no more time to waste.

Regards.

--Paul

Solar & Thermal
Paul Passarelli
Chairman & President
Pa...@SolarAndThermal.com
(203)846-2500 (main)  (213)207-6691 (direct)

Russell Philips

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Mar 15, 2025, 2:56:38 PMMar 15
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Paul I agree with every sentence in your last posting save "The work that one can extract from heat moved through a heat pump, will always be less than the work supplied to that heat pump" which is certainly true until a combined output scores as abundant. Theoretically this is sound, practically I remain unsure either way, although I yet have thinning hope because of pounding out combined modeling that i hoped would show better. 

The modeling for stand alone ORC is overwhelming - many low temperature options exist around my homestead. Again, solar PV may be less expensive, but i have come to far to stop now. My current electrical back up is a propane generator, which we know is subject to propane availability and wear.

I am demonstrating that I am interested in low-cost energy and will continue to post component build information. I am gathering copper for my next tube in tube heat exchanger which includes copper wire thermal mesh. The local metal recycler has several copper bins that have proven of merit. I will try soaking wire scraps in white vinegar for cleaning. Cheers!



Paul Passarelli

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Mar 15, 2025, 3:14:01 PMMar 15
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Russel,

  I wrote:

  If you insist on chasing the shadows of perpetual motion or something for nothing (which Includes "Brown's Gas" aka "HHO"), then I have no more time to waste.

Regards.

  Well, I stand by my statement.  This will be my last response.   

--Paul

Solar & Thermal
Paul Passarelli
Chairman & President
Pa...@SolarAndThermal.com
(203)846-2500 (main)  (213)207-6691 (direct)

Russell Philips

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Mar 18, 2025, 7:05:08 AMMar 18
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Model16.jpg
Here is one of the models that I was working on. I think it is impressive that 6 watts of input can generate 225 Joules of pumping output. Also, with only a high pressure side of 35.5psi and 25°C, PVC pipe can be used. This is far from finished.

Paul Passarelli

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Mar 18, 2025, 11:47:39 AMMar 18
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Russel,

  No. Sorry, that sketch is an entirely unrealistic work of 'fiction' (I'm being kind).  It's not a model of anything, You're not working on it, and you're clearly not learning anything from it either.

  Let's say it's all charged up to those initial conditions, then it spins for a few minutes.  At that point it's dead, inert, finito.  There is no COP, it's just a wound up rubber band pulling a balsa wood airplane.

  Please stop posting this nonsense.  Posting incorrect garbage decreases the average level of knowledge in the universe.  Some of it is so bad it's not even wrong, it's wronger than wrong.
 
--Paul

Solar & Thermal
Paul Passarelli
Chairman & President
Pa...@SolarAndThermal.com
(203)846-2500 (main)  (213)207-6691 (direct)

Russell Philips

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Mar 18, 2025, 2:24:08 PMMar 18
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As previously discussed, The Organic Rankine Cycle has been and will always be energy deficit. The COP, even including recycling, will be much less than 1.0.  I have Models 11 through 17.

Here are the updated and current Formulas and Calculations for Model 16 feed pump.

5.7 watts will pump 3.8cL or 22.4 grams of liquid butane. 

I will update Model 16 and repost as Version 2.8 as I can get through updating each group of component calcs. The numbers will ripple down change, but the flow dynamics are correct and important.

Sorry for any confusion, please verify the following:


Feed Pump
Given: A [70% efficient] feed pump use 5.733 watts per second to compress subcooled liquid Butane from 20 psi to 35.5 psi [+15.5 psi] [ASHRAE lists 5°C butane liquid density at 595.1 kg/m³].
Find: How many centiliters (and grams) are pumped per second?

1. Calculate the useful power output of the pump:
Since [watts x efficiency = useful watts], then (5.733 × 0.70  = 4.014 useful watts).
2. Convert watts to joules:
Since [1 watt per second = 1 joule], then (4.014 useful watts per second = 4.014 joules).
3. Convert +15.5 psi pressure difference to pascals:
Since [1 psi x 6894.76 = pascals], then (15.5 × 6894.76 = 106,868.8 pascals).
4. Calculate the volume flow rate per second:
Since [joules / pascal pressure difference = cubic meter flow rate],
then (4.014 / 106,868.8 = 0.0000375601 cubic meters per second).
5. Convert cubic meters to centiliters:
Since [1 cubic meter = 1,000 liters], then (0.0000375601 x 1,000 = .0375601 liters);
{or 3.76 centiliters per second}.
6. Convert cubic meters to grams:
Since [cubic meters x liquid density = kilograms], then (.0000375601 x 595.1 = .022352 kilograms);
{or 22.352 grams}.


ken...@aol.com

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Mar 18, 2025, 10:27:32 PMMar 18
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This is why I dropped out of the group.  The concept was a simple, practical, small-scale steam power plant.  As inevitably happens, people do some cursory study without the sort of in-depth research needed to develop a legitimate project and then decide that they will "revolutionize" the world with superior insights the less-clever professionals in the field have missed.

An energy balance that isn't showing enthalpy or entropy throughout the cycle is simply not credible.  

As an example, below are snippets of my Excel calculations for the "Williams Cycle" steam engine based upon their 1958 calculations ... which is shown to be unworkable, as expected.


Inline image






Russell Philips

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Mar 18, 2025, 11:32:58 PMMar 18
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Its good to hear from you Ken.

It appears that a 'practical, small-scale steam power plant' is not viable in small production runs, with solar PV as an opportunity cost.

How would you change the Formulas and Calculations for the Feed Pump to be more credible?

ctyankee

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Mar 19, 2025, 9:14:55 PMMar 19
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It's also why the group went silent for nearly a decade.  That and scammers like Cyclone Power.

Russell Philips

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Mar 19, 2025, 11:23:59 PMMar 19
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Boiling Tube
Given: A [98% efficient] counterflow tube in tube exchanger flash boils [from feed pump calculations: 22.352 grams per second] of saturated liquid at 35.5 psi [from ASHRAE butane tables at 5°C: vapor enthalpy 592.40 kJ/kg, liquid enthalpy 211.92 kJ/kg; and at 25°C: vapor volume .16165 m3/kg].
Find: What is the liquid to vapor latent energy in joules?
Find: What is the superheated vapor volume in centiliters?

1. Calculate the liquid to vapor latent energy in joules:
Since [kilograms per second x (vapor enthalpy - liquid enthalpy) = latent energy in kilojoules],
then (.022352 x (592.40 - 211.92) =  8.5045 kilojoules).
{or 8,504.5 joules}.

2. Compensate for exchanger heat losses by 2%:
Since [joules x 1.02 = required heat in], then (8,504.5 x 1.02 =  8,674.6 joules latent energy).

3. Calculate the superheated vapor volume in centiliters:
Since [22.352 grams = .022352 kilograms] and
Since [kilograms per second x  vapor volume m3/kg = volume in cubic meters],
then (.022352 x .16165 = .0036132 cubic meters per second).
{or 361.3 centiliters per second}.

ctyankee

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Mar 20, 2025, 3:19:07 PMMar 20
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Where is this going? I'm solving this to one significant figure on the back of a cocktail napkin. This is what I see:

In a perfect magical world, you put 8.5 KW of heat into the system and you get *maybe* 300W (not the 437W if gamma was = 1.0) of mechanical work out.  
That's 3.5% efficiency.
I seriously doubt that's enough to bring your vapor back to starting conditions.
In fact with friction losses, it's probably not enough to make any useful motion at all.

Russell Philips

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Mar 20, 2025, 5:59:32 PMMar 20
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AODD Pump - Expander
Given: An [63% efficient] AODD expands [from boiling tube calculations: 361.3 centiliters per second] of butane vapor from 35.5 psi to 20 psi [-15.5 psi work output].
Find: How many joules are generated.

1. Convert the work output of -15.5 psi to pascals (The negative sign indicates work done in this system):
Since [1 psi = 6894.76 pascals], then (-15.5 × 6894.76 = -106,868.8 pascals).
2. Convert the flow rate to cubic meters per second:
Since [1 centiliter / 100,000 = cubic meters], then (361.3 / 100,000 = .003613 cubic meters per second).
3. Calculate the work output in watts from the pressure drop:
Since [cubic meters x pascals = joules generated], then (0.003613 × -106,868.8 = -386.117 joules).
4. Compensate for 63% expander generator efficiency:
Since [total joules x efficiency = joules output], then (-386.117 x 0.63 = -243.3 joules output).

I will post a Model 18 jpg soon (four components only, no recycling), this image will include a finished and detailed COP caption.

Recognize that the Rankine cycle is a multi-system device, where each thermodynamic system tracks and details its efficiency.

Separate the 63% AODD component/ system efficiency from the greater cycle COP calculation.


Thermodynamic Systems (Components) in Working Fluid Flow Order:
Feed Pump (Rankine Cycle Start)
Hot Exchanger - Boiling Tube
AODD Pump - Expander
Cold Exchanger - Condensing Tube (Flows back to Feed Pump)
  

Russell Philips

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Mar 20, 2025, 8:07:19 PMMar 20
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Cold Exchanger - Condensing Tube
Given: A [98% efficient] counterflow tube in tube exchanger flash condenses [from feed pump calculations: 22.352 grams per second] of saturated butane vapor at 20 psi [from ASHRAE butane tables at 25°C: vapor enthalpy 620.98 kJ/kg, liquid enthalpy 259.65 kJ/kg; and at 5°C: liquid density 595.1 kg/m3].
Find: What is the vapor to liquid latent energy in joules?
Find: What is the saturated liquid volume in centiliters?
 
1. Calculate the vapor to liquid latent energy in joules:
Since 22.352 grams = .022352 kilograms] and

Since [kilograms per second x (vapor enthalpy - liquid enthalpy) = latent energy in kilojoules],
then (.022352 x (620.98 - 259.65) =  8.0764 kilojoules).
{or 8,076.4 joules}.

2. Compensate for exchanger heat losses by 2%:
Since [joules x 1.02 = required heat out], then (8,076.4 x 1.02 =  8,237.9 joules latent energy).
3. Calculate the saturated liquid volume in centiliters per second:

Since 22.352 grams = .022352 kilograms] and
Since [1 / liquid density 595.1 kg/m3 = liquid volume .0016804 m3/kg] and
Since [kilograms per second x  liquid volume m3/kg = volume in cubic meters],
then (.022352 × .0016804 = 0.000037560 cubic meters per second).

{or 3.76 centiliters per second}.

These calculation groups are correct to the best of my knowledge (version 2.8). I am willing to edit any errors and repost. Please verify. Better formulas, values or more detail may be developed - updating the model and version number.

Russell Philips

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Mar 20, 2025, 10:03:40 PMMar 20
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M18.jpg

The fluid flow or pressure flow, flows inside tubing and components. This is a fully sealed cycle, and pressurized to 20psi. The heat flow enters in through the boiling tube and flows with the pressure flow, until it exits through the condensing tube. Heat only flows along half of the cycle!!! Insulation reduces unwanted exchange.

My self taught reverse engineering of the heat pump had me realize, eventually, that pressure must flow inside tubing and components, while heat does not need to follow this rule - heat is not tightly coupled to pressure flow! I realized that any amount of heat can flow in or out of copper and is bound to energy ratio. The working principle of the heat pump is that we pay for inexpensive heating costs, yet reap expensive latent vaporization costs! In other words, in the multi-system heat pump - I learned that the systems have different amounts of energy. This sets up energy ratios. Correctly engineering the energy amounts is HOW the heat pump averages cop way above 1.0. Understanding that heat can flow with pressure flow is basic. Looking for when heat does not flow with pressure flow - opens our eyes. Please, please, try to be open minded, humble and teachable.

In the basic four component Model 18 above: heat flows with the pressure flow for half the cycle. We can engineer a way to alter the heat path for our benefit. Please trust me when I say that this is just the beginning. Model 16 introduces a central heat exchanger. Without going in-depth here: The summary is that heat does not cross the ambient boundary (if the heat exchanger is long enough or has enough thermal mesh). Cold flows in its own smaller circular heat path. Hot flows in its own smaller circular heat path. This recycles recoverable cold losses, and recycles recoverable hot losses. I have successfully tested this in a benchtop closed steam cycle, as can you. Now, do not shoot the messenger. With this new information, I hope your reaction is to ask which losses are recoverable and which are irrecoverable. 

The largest loss is our vapor expander in-efficiency and It appears that this is an irrecoverable loss. Please confirm!

Russell Philips

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Mar 22, 2025, 11:17:47 AMMar 22
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M16.jpg
Here is the updated Model 16 jpg.

The AODD is a working model available today and is probably the least expensive expander.

It lacks an output shaft for making electricity, but is tremendous at pumping water (consider irrigation).

This has been placed into the public domain for humanitarian purposes.

ctyankee

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Mar 22, 2025, 12:20:43 PMMar 22
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Russel,

  It's been more the ten years since Ken shared some sage advice with you.  In all that time, have you learned *NOTHING*?

  Please stop.

  Most of the people here are too kind to say to you point blank that this mess is wrong and annoying.
  Please don't interpret their silence as agreement.  I can assure you, it's not.

Russell Philips

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Mar 22, 2025, 9:23:14 PMMar 22
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Thank you for the kind cease and desist. 

The steam Rankine Cycle has four components: feed pump, hot exchange, steam expander, cold exchange.

If I may ask, What would be the necessary hot and cold temperature to allow an AODD expander to pump water?

Russell Philips

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Mar 23, 2025, 3:29:28 AMMar 23
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Wait, are ya all saying that an AODD can not be used as an expander?

ken...@aol.com

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Mar 23, 2025, 10:47:32 PMMar 23
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The AODD looks like a poor choice for an expander.  

Rankine Cycle efficiency improves as clearance volume diminishes, I have yet to see a diaphragm pump that wouldn't exhibit greatly excessive clearance if configured as an engine.
Expander means that we cut off working fluid admission and allow the working fluid to naturally expand throughout the stroke; the sooner steam admission is cut off, the greater the expansion.  One can reasonably argue that accurately controlling short cutoff in an engine with an extremely short stroke-to-bore ratio is problematic.  If your cycle doesn't permit relatively short cutoff, efficiency already looks questionable.


ken...@aol.com

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Mar 24, 2025, 11:06:04 AMMar 24
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Hi Russell,

Sorry for the delayed response, I didn't see your question.

For the feed pump, I would use the formula Power = Q * p * g * H * 0.65  with Q being flow rate, H the pressure head, p for fluid density, g for gravitational acceleration, and 0.65 to account for the mechanical efficiency of small piston pumps.  Fortunately, water is a smaller molecule than those used in organic cycles and the pumping effort is significantly less.  When burning biofuels, I really don't think you can beat water as a practical working fluid.

There's a few niche applications where biofuel steam would outperform solar, but I'm prohibited from mentioning those due to confidentiality promises.

Regards,

Ken



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