Propane driven Generator system

[RonL]

Not wanting to be guilty of thread hi-jacking I'll start a new one. This design is in the start of assembly stage, but i will not engage it until i have a few more answers, but i do have all equipment, except for plumbing and fittings, and pressure release valves.

The thread by HashsBack, using hydrogen sounds similar, and Q_Goest has provided information that is over my head at this time, but I'll be looking at that quite a bit.

The design I'll use is a 500 gallon propane tank, and a few other tanks, smaller but rated for higher pressure.

The first answer I'm looking for is time of cycle in the first step, which involves the 500Gal. tank (painted black) in full sun, and containing from 30 to 50 gallons of liquid propane. As the sun heats the tank pressure builds to design pressure of 125 PSI, the gas vapor releases and flows Thur pipe to an air motor that drives the generator. As pressure drops in the tank i think it will cool quickly, so a quick cycle pressure of around a 20 PSI delta?

If work is transferred to the generator, then what condition needs to exist at the air motor exhaust in order to liquefy and be injected back into the tank? ( I'm thinking too low of a temperature at the exhaust of the main air motor will be a detriment to the input)
A section of tubing or small tank, having a low pressure, created by a secondary air motor, or compressor(taking some energy from the generator) might be a, not too negative solution.
After a slowdown or stop condition, the sun continues to heat the tank and the cycle repeats.

In the event of a failure of the mechanical portion of the system, an electromagnetic safety breaker would release, isolating the main tank, and pressure release valves would protect the low pressure portion of the system. Since the tank is painted black it to has to have a max release, so some propane might be lost. All released gas can be flared at a safe distance above.
Night temperatures of 80 to 90 F. will drive this system to a small output, even without sunshine.

There is a fair amount more to my design, but any feed back from this first portion would be great.

Thanks
RonL

PS. Sorry this is not in metric, if i need to change it, I'll put more effort into that next time (just one of those old holdouts)

 

[mgb_phys]

You're basically building a solar heat engine using propane as the working fluid ?
Sounds like it could be interesting if you get a leak and a spark anywhere!

 

[RonL]

One of my biggest concerns is, the shaft seals on motors or compressors. I think compressor piston seals might not be too big a issue.
But yes leaks would not be good

And yes it would be a solar heat engine.

 

[Danger]

I know zilch about working fluids, so I'm curious as to why you chose propane. Do its properties make it more desireable than something non-inflammable?

 

[mgb_phys]

You want something with a lowish boiling point a very low freezing point and a particular heat of vaporourisation - you used to use ammonia which is even less fun!

 

[Danger]

I see. Thanks.

 

[RonL]

At first i considered liquid nitrogen, but pressure would be outrageous, cost of containment would be excessive, "Danger" would be all around.

Propane can be picked up almost anywhere, it is easy to use if additional heat is needed for the system, ( i can BBQ on the spot) but main thing is, at 130 F the pressure in the tank will be around 287 PSI, so in a shutdown case the pressure will stay within the tank pressure rating. (I'll have to check, but my tank looked like 450 PSI) I will do a hydro test before use.

At 0 pressure the liquid is -44 F. and at 80F the pressure is right at 100PSI, and at 100 degrees, the pressure goes to 175PSI.

These are very workable numbers, and materials for the most part, with careful selection, are off the shelf purchases.

One last thing, in a closed system, the electrical components can be cooled, using either the vapor, or liquid.

 

[Mech_Engineer]

How big is the tank dimension-wise? The reason I ask is because assuming the 500 gallon tank is about .6m radius by about 1.65m long, this will only give you about 2 square meters which are perpandicular to the sun.

Given that the sun's power density at sea level is around 125 w/m^2, this will give you a MAXIMUM total power output of about 250W during the summer at noon, not including losses to the environment through convection and other inefficiencies. Doesn't sound like a whole heck of a lot considering the size of the system you are proposing... It's possible there will be an obvious optimum that can be achieved between tank surface area, operating pressure, and power output.

Will you be using a system of batteries to store the juice?

 

[russ_watters]

Here's a phase diagram for propane: http://www.elyenergy.com/pdf/CO13.pdf

As the sun heats the tank pressure builds to design pressure of 125 PSI, the gas vapor releases and flows Thur pipe to an air motor that drives the generator. As pressure drops in the tank i think it will cool quickly, so a quick cycle pressure of around a 20 PSI delta?

This is a completely arbitrary choice. It is up to you. I'd think you'd want to keep a constant temperature and pressure and vary the flow, though. The efficiency of the system is set by the operating temperature/pressure.

With that in mind, 125 PSI doesn't help you any: the matching temperature is about 75F so you aren't above ambient (unless it's winter), so you can't run a heat engine that way.

If work is transferred to the generator, then what condition needs to exist at the air motor exhaust in order to liquefy and be injected back into the tank? ( I'm thinking too low of a temperature at the exhaust of the main air motor will be a detriment to the input)
A section of tubing or small tank, having a low pressure, created by a secondary air motor, or compressor(taking some energy from the generator) might be a, not too negative solution.

The key here is that there is nothing special going on here. It is an ordinary thermodynamic cycle, so the rules are the same as with any other. What you are describing is a just a plain ordinary Rankin cycle power plant: http://en.wikipedia.org/wiki/Rankine_cycle

So the two pieces you need to get the propane back into the tank are a compressor and a condenser.

And Mech E is right: This type of system (not using propane) is being considered for solar power plants. But to get good power output, you need a lot of surface area. And that means a black tank alone doesn't cut it: you need collectors.

Anyway, because of the danger of using propane as your working fluid, I'd recommend against it. You're much better off with steam if you can get the water hot enough with your collector.

 

[RonL]

The tank is 1.2m (48") X 2.4m (96") i don't remember if that length is including the bell ends, The propane dealer referred to it as 500 gallon, and i believe tanks are filled to 75 or 80% of their actual volume.
The bleed down time with a 5/8" air hose takes a while, from 120 psi.
The air motors i have are rated for 100 psi, 3000 rpm, and 4.8 hp, they have 1/2" npt ports.

At present i have 12 deep cycle marine batteries, 12v, 100ah, for a total of 1200 ah capacity.
Cycle time will depend on what happens when the pressure relief valve opens, and how fast the tank cools due to pressure drop. I don't expect it to be anywhere as long as a compressed air cycle, and getting the gas back to a liquid state, and injected back into the tank at its lower pressure, will be quite important.

 

[RonL]

Thanks Russ, i was posting when you submitted your post.

The point I'm thinking to make, is that in the hot sun, all my tools and things metal (especially black things) get so hot they can't be handled with a bare hand, my guess is beyond 140 F.
It seems that the reheat time might not be so great, if the cycle time is short.
I'll study what has been said so far, and get back tomorrow. I have a meeting to go to now.
Thanks again.

Ron

 

[Mech_Engineer]

Russ is correct, based on the properties of saturated propane, 120psi will only be about 67 *F, not high enough to exhaust heat to the environment through the condenser unless its a cold day.

Your pump is rated at 3580 W, but your total solar input will only be able to hit a maximum of 360 W. I'm afraid there's a terrible mismatch in the energy going through this system. You'll need at least 28.6 m^2 of solar collectors (at noon in the middle of summer) just to equal the power required by the pump, not including efficiencies or desired power output or energy lost to the environment...

Edit-
I think it should also be noted that it would take about 11.6 MJ to heat 500 gallons of saturated liquid propane to saturated gaseous propane. At 360W of input, this would take about 9 hrs to achieve.

 

[russ_watters]

The point I'm thinking to make, is that in the hot sun, all my tools and things metal (especially black things) get so hot they can't be handled with a bare hand, my guess is beyond 140 F.

Yes, a black, metal object can easily reach 140F in the sun. But then, a tool doesn't have anywhere near the mass of a 500 gal propane tank. As ME said, it'll take a long time to heat it up that much.

And you wouldn't want your propane at 140F anyway: at that temperature, it'll be at 350 psi. And if you run it through a regulator to lower the pressure, you lose energy in the throttling process.

 

[Mech_Engineer]

Ah so I just realized that you plan on only having about 50 gals of propane in the tank, so really a complete heat from saturated liquid to gas would take a little under an hour. But, that still doesn't change the fact that you're absorbing far less power than needs to be supplied to the pump...

 

[RonL]

I think when i gave the specs of the air motor, two options of how to use it were not mentioned.
1. The fast release of a high pressure vapor would spin it up to speed quickly, producing a fast cycle since the delta is only 20psi. The recovery time will be a function of how fast thermal energy transfers through the shell. (wish i could afford copper)
How cold would the inside of the tank become with a 20psi pressure drop?

2. The high pressure gas pushes the rotor at a slower speed, the results of a step up pulley system that drives an alternator at as much as a 10:1 ratio (250rpm on the motor, and 2500rpm on an alternator) a 200 amp alternator can be loaded to equal in resistance, the thermal movement into the tank.

Now I'm in the dark about how to return the gas vapor to liquid, for injection back into the tank. Or will it have to be pushed in as a gas under pressure? Well i guess its back to the books unless someone has a quick answer

 

[russ_watters]

How cold would the inside of the tank become with a 20psi pressure drop?

I posted the phase diagram above. If it starts at 200 psi and drops to 180, for example, the temperature will drop from about 100 F to about 90F

(wish i could afford copper)

The heat flux is so low, the material really doesn't matter at all.

Now I'm in the dark about how to return the gas vapor to liquid, for injection back into the tank. Or will it have to be pushed in as a gas under pressure? Well i guess its back to the books unless someone has a quick answer

I'm starting to get the feeling you didn't read my post...

 

[RonL]

I did read your post Russ, but quickly because i was running late for a meeting. The answer of how the air motor would be used, was given to Mech_Engineer before i went back and read it again

The chart helps very much, thanks. Seeing the delta from 200 down to 100 psi, indicates a change of about 35* F, if I'm seeing it right.
I do need to go back and study the post some more, and look at what Q_Goest posted in another thread.

As short as your post is, there is a lot in it to take in, for someone that does not deal with AC systems on a regular basis.
I found that the Texas Railroad Commission, has a free to the public, class for certifying people to handle propane, sure will be worth looking into.

Thanks for your help
Ron

 

[Mech_Engineer]

What this system needs is a good thermodynamic analysis. I happen to have a printout of a P-h chart for Propane from school, so I'll use it for the cycle analysis.

--ASSUMPTIONS--

We will call this a Rankine cycle, so we have 4 components-

- Heater (Tank)
- Turbine
- Condenser
- Pump

Based on the information provided, we can see what kind of power we would get out of this thing, but let's define some states and processes.

State (1): Superheated vapor at 116*F, 200 psi (After the heater, before the turbine)
State (2): Saturated vapor, 100psi (After the trubine)
State (3): Saturated liquid at 100psi (After condenser)
State (4): Liquid propane, 200psi (After pump)

Processes 1-2 and 3-4 will be assumed to be isentropic processes (100% efficient). This is of course not possible in the real world, but will do for our purposes for now.

Processes 2-3 and 4-1 will be assumed to be constant-pressure under the curve, which means there is no frictional losses in the flow. Also not possible, but a good start for us.

So, going around the cycle, we're able to solve for all of the states of the propane at each point in the system, and then by looking at this we can determine power output and maximum possible efficiency.

--FINDING STATES OF THE PROPANE--

We know that state (1) is saturated liquid at 200psi, and we know that the turbine's process will follow a constant-entropy line down to 100psi. So following that line we can fix the state at point (2).

Now that we know state (2), we can follow a constant pressure line to saturated liquid and have fixed state (3).

From state (3) we follow another constant-entropy line to 200psi, and have fixed state (4). And with that, we have fixed all of the states of the propane in all points in the system, and can look up it's properties.

--PROCESS NET POWER OUTPUT/INPUT--

So we know that process (4-1) (heating in the tank) will have an input power of 360W. Because we can look up the properties of propane at each of these states, we can determine the flow rate of the system. Entahlpy at state (4) is about 205 kJ/kg, and at state (1) it is 600 kJ/kg. The difference between these two is 495 kJ/kg, so when you divide 360W by 495 kJ/kg, you get a mass flow rate of 0.000727 kg/s (5.77 lb/hr).

Process (1-2) is the propane being expanded through the turbine. Enthalpy at (1) is 600 kJ/kg and at (2) is about 550 kJ/kg. The difference between these two values is 50 kJ/kg, so using the mass flow rate of the system we can see the net power output from the turbine would be about 36.3 W.

Now, process 2-3 is condensing the gas back into a liquid. Enthalpy at (2) is 550 kJ/kg, and (3) is about 200 kJ/kg, so the difference is 350 kJ/kg, which equates to a total energy dump of 254.5 W.

Process 3-4 is going through the pump to get back to 200psi. Enthalpy at state (3) is 200 kJ/kg, and state (4) is 205 kJ/kg, giving the pump a total power input needed of about 2.88 W.

--CONCLUSION--

So given that we now know everything about the system, what kind of MAXIMUM efficiency are we looking at?

Well, in terms of total usable power out divided by total power in, we get about 9.3% of the total input energy in as usable work out when we take into account the power used by the pump.

What does all of this mean? Given the ambient temperature you're stuck with (about 70*F) which limits the minimum pressure you can expand the working fluid to, you are going to need to go to much higher maximum pressure if you want to gain any sort of efficiency from this system, and a lot more surface area to maximize power input from the sun.

 

[RonL]

I'm starting to get the feeling you didn't read my post...

I hope you don't dismiss me as a crank too quickly, i do respect you and your answers, and read them all, in every thread.

My ideas, and questions are generated by the things i use on a daily basis, i generally do all my own maintenance and repairs which requires that i learn and understand why and how things work, trying to put some things together in unorthodox ways, so that the energy losses of one thing feeds something else, is what i do for fun, (when i can afford it). I have not made any great discovery, and probably won't, but i learn more each day which opens the door to needing to learn about other things, the list just keeps growing.

You mentioned steam, which i have been trying to avoid because of losses to the atmosphere, but a thought of a dual system, where cold propane liquid condenses the steam, might keep those losses within the system, comes to mind.

Having three sources of heat, 1.the sun 2. electric heat elements, and possible resistance losses of motor/generator(into the cold propane) 3. and the burning of some propane, might produce a continuous running system.

I would be happy with anything from 1-15 hp.

I designed and built a small tool trailer, that has a 12 volt 1200ah capacity and with inverter, and a step-up transformer, there is 240v modified ac @ an amperage of 21.
For some things needing true sine wave voltage, a 2hp motor (delta industrial 220v @ 18a) drives a 10KW generator head, it works well for anything up to just under 2hp.
Adding a heavy flywheel let's things draw from the generator at close to max. amps, but only for a short time.

This was all designed to have power at remote locations on my gravel pit, without using gas or diesel engines (of which i have quite a few). a sun fed generating system for these batteries, is my basic goal.

Hope this explains a little about me and what I'm up to

If you've read this far thanks.

RonL

 

[Mech_Engineer]

Here is the chart I used for looking up the properties of Propane-

 

[RonL]

Thanks Mech_Engineer
A lot to study and comprehend, i promise to learn metric units by summer's end, and I'll think of you when I'm working in the 103 heat, at my little gravel pit.
Thanks for all that work you have gone to, i have to be across town in a few minutes, so I'll study it later.

My last post might give you a better understanding of the power needed, and what I'm working on, hope it makes a little sense (it does work).
I try to always buy things that can be used for more than one project, so my fun time is not a waste of money, only once in a while

Thanks again
Ron

 

[Mech_Engineer]

Thanks Mech_Engineer
A lot to study and comprehend, i promise to learn metric units by summer's end, and I'll think of you when I'm working in the 103 heat, at my little gravel pit.
Thanks for all that work you have gone to, i have to be across town in a few minutes, so I'll study it later.

My last post might give you a better understanding of the power needed, and what I'm working on, hope it makes a little sense (it does work).
I try to always buy things that can be used for more than one project, so my fun time is not a waste of money, only once in a while

Thanks again
Ron

Well, the important thing to take note of is that you only have about 0.48 Hp coming in as solar power incident on the tank, and from that the turbine will only put out about 0.048 Hp.

 

[ken5285]

Mech_engineer, I was wondering where can I get a complete copy of the document for the thermodynamic properties of propane that you posted here?

 

[Mech_Engineer]

That is just a sheet i have left over from my intermediate thermodynamics class. If you want more detailed properties of propane, you will probably want to look at the NIST website, or search through some scientific journals.

 

[ken5285]

I am looking at finding properties for propane at elevated pressures, such as 150,000 psi and above 600K. I have been trying so hard but have been unsuccessful in locating any date. Would anyone be of help here? Its for my Master's research.

 

[Mech_Engineer]

I am looking at finding properties for propane at elevated pressures, such as 150,000 psi and above 600K. I have been trying so hard but have been unsuccessful in locating any date. Would anyone be of help here? Its for my Master's research.

150,000 PSI?

That's around 1050 MPa! You're talking pressures way out of the scope of any thermodynamic properties I have seen published for standard thermo classes... Considering that sheet I have only goes to 20 MPa and NIST only has values published up to 100MPa, I'm thinking you're SOL. I shudder to think of what it would take to design a thermophysical measurement rig that can handle 15,000psi, let alone 10 times that.

What the heck are you doing with Propane that you need thermophysical properties at that pressure and temperature?!

 

[ken5285]

15,000 psi is just the first phase of what i am about to do. I am with the University of Kansas and working on some research that will be a breakthrough. We are about to deal with 150,000 psi and I have built a test rig to withstand that pressure. If we can achieve 15,000 psi, that would be good and I would be able to push the limits to observe the potential work that can be extracted.

 

[stewartcs]

Perhaps this will help. I generated this from REFPROP (from NIST).

EDIT: Keep in mind the critical temperature is about 369 K.

CS

 

[stewartcs]

Here is the saturation table for it as well.

CS

 

[ken5285]

Thank you very much everyone here for the advice so far.

Was this a program that you had used to generate this graph? If, so is it possible for me to get my hands on it?

On the graph, what is the bottom axis and what properties is being shown on the right? I am guessing it will be temperature in Kelvins and the entropy shown on this graph.

 

[stewartcs]

Thank you very much everyone here for the advice so far.

Was this a program that you had used to generate this graph? If, so is it possible for me to get my hands on it?

On the graph, what is the bottom axis and what properties is being shown on the right? I am guessing it will be temperature in Kelvins and the entropy shown on this graph.

The program was REFPROP version 8.0 from NIST. They sell it on their website for $200 bucks (or at least it was when I bought it).

The horizontal axis is the Enthalpy as indicated. The values on the right are the specific volume and the entropy. Save the plot to your computer and zoom in. You'll be able to see the units then.

CS

 

[ken5285]

What basis equations does this program use? Does it model the real gas behavior throughout all regions? And If I happen to acquire the program would i be able to ploit for temperatures even higher than 600K, probably to at least 1000K?

Thank you again.

 

[stewartcs]

What basis equations does this program use? Does it model the real gas behavior throughout all regions? And If I happen to acquire the program would i be able to ploit for temperatures even higher than 600K, probably to at least 1000K?

Thank you again.

The program uses various equations of state. Here is the program information available for propane (with limits):

Propane - CH3CH2CH3

CAS#: 74-98-6
Molar mass: 44.096 kg/kmol
Triple point temperature: 85.53 K
Normal boiling point temperature: 231.04 K
Critical point temperature: 369.89 K
Critical point pressure: 4.2512 MPa
Critical point density: 220.48 kg/m³
Gas phase dipole at NBP: 0.084 debye
Acentric factor: 0.1521

Equation of State
Equation type: Helmholtz energy
Limits: 85.53 K to 625.0 K, 1000.0 MPa, 908.37 kg/m³
Lemmon, E.W., McLinden, M.O., Wagner, W. to be submitted to J. Phys. Chem. Ref. Data, 2007.

Below 350 K, the uncertainties in density are 0.01% in the liquid phase and 0.03% in the vapor phase (including saturated states for both phases). The liquid phase value also applies at temperatures greater than 350 K (to about 500 K) at pressures greater than 10 MPa. In the extended critical region, the uncertainties increase to 0.1% in density, except very near the critical point where the uncertainties in density increase rapidly as the critical point is approached. However, in this same region, the uncertainty in pressure calculated from density and temperature is 0.04%, even at the critical point.

The uncertainties in the speed of sound are 0.01% in the vapor phase at pressures up to 1 MPa, 0.03% in the liquid phase between 260 and 420 K and 0.1% in the liquid phase at temperatures below 260 K. The uncertainty in vapor pressure is 0.02% above 180 K, 0.1% between 120 and 180 K, and increases steadily below 120 K. Below 115 K, vapor pressures are less than 1 Pa and uncertainty values might be as low as 3% at the triple point. Uncertainties in heat capacities are 0.5% in the liquid phase, 0.2% in the vapor phase, and higher in the supercritical region.

Viscosity
Limits: 85.47 K to 600.0 K, 100.0 MPa, 881.91 kg/m³
Vogel, E., Kuechenmeister, C., Bich, E., and Laesecke, A., "Reference Correlation of the Viscosity of Propane," J. Phys. Chem. Ref. Data, 27(5):947-970, 1998.

The uncertainty in viscosity varies from 0.4% in the dilute gas between room temperature and 600 K, to about 2.5% from 100 to 475 K up to about 30 MPa, and to about 4% outside this range.

Thermal Conductivity
Limits: 85.47 K to 600.0 K, 100.0 MPa, 881.91 kg/m³
Marsh, K., Perkins, R., and Ramires, M.L.V., "Measurement and Correlation of the Thermal Conductivity of Propane from 86 to 600 K at Pressures to 70 MPa," J. Chem. Eng. Data, 47(4):932-940, 2002.

Uncertainty in thermal conductivity is 3%, except in the critical region and dilute gas which have an uncertainty of 5%.

Surface Tension
Limits: 85.47 K to 369.85 K
Baidakov, V.G. and Sulla, I.I. "Surface tension of propane and isobutane at near-critical temperatures," Russian Journal of Physical Chemistry, 59:551-554, 1985.

Melting Line
Limits: 85.48 K to 2000.0 K
Reeves, L.E., Scott, G.J., Babb, S.E., Jr. "Melting curves of pressure-transmitting fluids," J. Chem. Phys., 40(12):3662-6, 1964.

Coefficients have been modified (2004)

Hope this helps.

CS

 

[ken5285]

Thank you so much for the information that you provided. I really appreciate it. I will work with this information that you provided here. If I have any further questions I would try to contact you again.

 

[stewartcs]

Thank you so much for the information that you provided. I really appreciate it. I will work with this information that you provided here. If I have any further questions I would try to contact you again.

You're welcome.

Let us know how your project turns out.

Good luck.

CS