# Organic Rankine cycle refrigerant expander?

1. Mar 11, 2007

### matgt

I am working on a project to design an organic Rankine cycle to convert low temperature thermal energy (waste heat or geothermal) into electrical energy. We are trying to select an expander to extract work from the pressure differential in the refrigerant. The refrigerant (ammonia or R134a) will be heated to about 120 F in the evaporator and cooled to about 80 F in the condenser. For both of these refrigerants there will be about a 1.6-1.7 pressure ratio between the expander inlet and exit. We don't really want to use a turbo expander For this small application and are trying to keep this from being an extremely sophisticated system (more to show the viability of a "cheap solution" than to use state of the art technology). We have been considering using a modified compressor (e.g. reciprocating, radial screw, rotary vane) running in reverse to extract the work. How feasible would this be to accomplish? Which would be the best for the application? Could you expect similar efficiencies as the compressor operation? Also which could we expect the highest efficiency from? I know this isn't information that people probably know off the top of their head but any help would be appreciated.

MT

2. Mar 11, 2007

### Q_Goest

Hi matgt,
Any compressor can be modified and made to run in reverse to extract energy. The most obvious and common expanders are centrifugal and reciprocating. But any 'motor' driven by expanding gas can also be considered as an expander, and can be used for refrigeration, even though their primary intent is to produce usable work. The question then regards isentropic efficiency.

The machine that comes to mind for your application is a vane type as you've already mentioned. You can get vane type air motors very inexpensively, and already made for extracting work from a fluid stream. If you're in the US, I'd suggest using ThomasNet to search for specific companies that make air motors:

The very first one on the list for example, has a very good web page that covers power output, air consumption, etc... From the information they provide, you can easily calculate isentropic efficiency.
http://www.kinequip.com/airmotors.asp?Cat_Id=9# [Broken]

Last edited by a moderator: May 2, 2017
3. Mar 11, 2007

### matgt

Thanks so much for the help Q_Goest. I for some reason hadn't been considering already available air motors. Those are basically what we are looking for. I do have some questions that I'm hoping someone can help me with. The expansion of a refrigerant gas is quite different than the expansion of air through a motor.

First we are trying to figure out how to decide on the pressure of the refrigerant at the expander inlet to extract the most work from the gas. With air normally higher pressure would mean more power. However with the refrigerant if the pressure is to high it will condense within the expander which isn't what you want. I know that inevitably there will almost always be some condensation but how much is too much and how can we calculate the amount of condensation? Also I know that the gas should be superheated when it enters the expander. Since the temperature at the expander inlet is basically set for us based on the heat source temperature the only way to further superheat the fluid is to decrease the pressure at expander inlet. This is a tradeoff because this then decreases the pressure difference across the expander and therefore the amount of work that can be extracted. How can we find the optimum level of superheating so that we extract as much work as possible without having too much condensation inside the motor?

4. Mar 11, 2007

### Q_Goest

Hi matgt.
It's not as hard as it sounds. Doing the thermodynamic calculations is very easy. The problem you run into is how to determine fluid properties at various states. By that I mean, how do you find the enthalpy or the entropy of the fluid at some pressure and temperature? Or if you have entropy and pressure, how do you determine vapor fraction or temperature or enthalpy? Or density. Or any fluid property?

With the right database, you can create a model of the process very easily using Excel or other program. Then you simply change a bunch of input variables and you can test out the system under various conditions in order to optimize the process. The trick is to understand how to use a commercially available fluid properties database.

Remember that the state of a fluid only requires 2 known's such as pressure and vapor fraction to define the state. If you know the fluid state you can use that to determine all the other unknowns such as temperature, enthalpy, internal energy, entropy, density, etc. You might also use entropy and vapor fraction, or entropy and temperature as your 2 known's. The 2 known's don't have to be pressure and temperature, they can be essentially any of the other properties as long as the fluid properties database is set up to call in the state given those variables.

A few years ago I helped a group of mechanical engineering students at a local university. My company had funded a project, and I was to teach them something very similar to what you're doing. The company I work for has a proprietary database that links to Excel, and that database allows one to bring virtually any fluid property into the spreadsheet so it can be used for calculations. Unfortunately the students weren't allowed to use ours, but the students did find one that seemed to be roughly equivalent on the market for a reasonable cost. The students evaluated one from NIST along with some others, but determined this particular database from NIST was the best for their purposes. It can be purchased here:
http://www.nist.gov/data/nist23.htm [Broken]

There are other databases out there. Looking around, I found another one here:
http://www.megawatsoft.com/R134aexcel.asp?cat=8 [Broken]

Do a Google search and you'll surely find a bunch more. You may even find your university already has something like this, just ask your professors. Whichever one you use, do some research on it before you buy it. You need to understand a bit about how to use it and verify it will do what you're expecting and what you need.

For the NIST database, there's some HELP information here:

Just to give you an example of how they work, in some cell you can insert something like this:
=Enthalpy($B$20,"TP",B48,A51,B51)
This call returns enthalpy given temperature and pressure.

=Enthalpy($B$20,"TD",$E$17,A42,C42)
This one returns enthalpy given temperature and density.
http://www.boulder.nist.gov/div838/theory/refprop/REFPROP.XLS [Broken]

So to answer your question, I'd suggest modeling the system in Excel using a fluid properties database add-in. Once you have the model, you should be able to change a few variables such as pressure at the expander inlet or temperature, and the spread sheet simply recalculates the process.

Last edited by a moderator: May 2, 2017
5. Mar 12, 2007

### matgt

So I'm having a little trouble figuring out how to calculate the isentropic efficiency of an air motor. The information that was obtained from the manufacturer gives:

@ 2000 rpm and 700 kPa (7 bar)
flow of free air: 480 m^3/hr (I'm guessing this means the air exiting the motor)
Power = 7 kW

To calculate the isentropic efficiency I would need to use

eff= (h1-h2)/(h1-h2s)
where 1 signifies the motor inlet and 2 signifies the exit

Assuming air is an ideal gas I can calculate h1 using the input pressure (700 kPa) and by assuming the temperature(1) is room temp (25 C). I can also calculate h2s by using entropy(2)=entropy(1). However I cannot find h2 without assuming a temperature at the air exit, correct? I feel like I need to use the volumetric flow rate but my brain isn't working and I can't figure out how.

6. Mar 12, 2007

### Q_Goest

You should verify this with the manufacturer, but I believe "flow of free air" means air under standard conditions (sometimes called 'normal' conditions). So 480 m^3/hr would equate to that much air at standard conditions. Note that different industries and countries use different conditions for "standard" pressure and temperature, so you should verify with the manufacturer what they are using. In the end, you want to know the mass flow rate which will help eliminate any confusion.

The enthalpy out can then be calculated from:
Power out = Mass flow * (Enthalpy in - Enthalpy out)
(ie: they give you power out, mass flow rate, and you can determine enthalpy in.)

If you post your assumptions and calculation, I'll check it against my own database.

Edit: I should also note that I've made a number of simplifying assumptions by equating power to enthalpy as I have. Although the above is not strictly correct, it is a good first pass at determining isentropic efficiency. It also allows you to compare various expanders to each other and those comparisons are relatively valid, if not entirely accurate. When the actual system is created, discharge temperature can be determined by test and a more accurate value of efficiency can then be determined. So I'd agree that one can't get an exact measure of efficiency from the data provided by the manufacturer, but at least you can get a good estimate which can be used for comparison to other machines and as a value for use in process modeling.

Last edited: Mar 12, 2007
7. Mar 12, 2007

### matgt

Okay thank you so much for all the help Goest. I called the manufacturer and got the standard conditions. In response to your prior post, I do have access to a computer program with most fluid property tables built in. I actually have already modeled the system to some extent but am trying to improve upon the model based on a more detailed understanding of the actual operation of various components. I'll post more about that later however. Here are my calculations you can glance over them if you want but I'm pretty sure they are correct. Going from volumetric flow to mass flow was my problem because I didn't know the definition of free air consumption. I put assumptions and comments in blue.

Standard Conditions
P_std=101.325[kPa]
T_std=273.15[K]
RH=0%

Entrance Conditions
P[1]=700[kPa] - Taken from one of the motor operation charts
T[1]=298.15[K] - Estimated that the temperature of pressurized gas when testing was performed would be at approximately room temp

Exit Conditions
P[2]=P_std - Motor exhausts the air to the atmosphere

Other Conditions
rpm=2000[rev/min] - not important
V_dot_std=480[m^3/hour]*(1/60)[hour/min]*(1/60)[min/s] - convert to m^3/s
W_dot_m=7[kW] - given motor power output at specified air pressure and rpm

Mass Flow Calculation
v_std=VOLUME(AIR,T=T_std,P=P_std) - specific volume lookup from property table
m_dot=V_dot_std/v_std - mass flow rate calculation

"Calculate h[1]"
h[1]=enthalpy(AIR,T=T[1]) - property table

Calculate h_s[2]
s[1]=entropy(AIR,P=P[1],H=h[1]) - property table
s_s[2]=s[1]
h_s[2]=enthalpy(AIR,S=s_s[2],P=P[2]) - property table

Calculate h[2]
h[2]=h[1]-W_dot_m/m_dot - assuming heat, potential, and kinetic energy change of control volume is negligible

Calculate Isentropic Efficiency
eff=(h[1]-h[2])/(h[1]-h_s[2])

I think all of my assumptions should be valid but let me know if I overlooked something. Thanks again for all of the help.

Edit: I meant to include the calculated values.

isentropic efficiency = 32%
mass flow rate = 0.17 kg/s

Last edited: Mar 12, 2007
8. Mar 12, 2007

### Q_Goest

Could you please verify for me that this is gage pressure?

Thanks...

Edit: Here's what I get:
30.6% isentropic efficiency (if inlet pressure is gage)
32.1% isentropic efficiency (if inlet pressure is absolute)
17.2 kg/s
So it seems like our numbers match.

Seems like a relatively inefficient expander. Note that this expander may have significantly different isentropic efficiency at other inlet pressures. If you now have a spreadsheet to do these calculations, you can plug in a variety of different expanders at different inlet pressures and find out how they compare. I'll save my spreadsheet for this for a while if you have any other questions about it.

Last edited: Mar 12, 2007
9. Mar 12, 2007

### matgt

Yeah I just went back and looked and although it doesn't say explicitly I believe that it is gage which as you said would bring the pressure down to 30.4% based on my calculations. I'll try and call the company tomorrow to be sure. Thank you so much for all the help.

10. Jul 17, 2007

### dukejaz

r-142a

anyone interested in low delta t power should check out >>>yourownpower.com<<<<< . yes, it is a large Carrier vapour compression air conditioner, running in reverse as a organic rankine cycle motor. note to russ watters.... the unit has its vane style compressor acting as a turbine when running as a motor. ANY large air conditioning unit will run backwards during stutdown, if the condenser is hot and evaporator cool. as this is damaging to scroll and reciprocating compressors, a backpressure release valve, or other means, is designed to prevent this. the Carrier/utc had only to close this valve; no major overhaul necessary. the working Th is 70C. the heat sink is a cold alaskan creek, at ~ 5C. two such units are now in opperation, producing 2 MW each. there are considerable drilling and then pumping costs but carrier expects a payback on all drilling and infrastructure in under 7 years. the large retrofitted a/c units will retail for ~ \$100,000, less than half of a new waste heat to power unit of comparable size. these older rooftop a/c units are rapidly being decomisioned, as li-br absorption, and more importently, silica gell adsorption units become more economic.

it seems clear to me that a solar thermal plus geo-sourced heat sink could provide these delta t's and overall btu's. with carnot at ~ 20% there will be left over heat of course. at chena, this is dedacted to both winter resort heating, (humans can use what carnot cannot!), and to an 'ice palace', year round igloo, that is kept frozen by a li-br absorption chiller, which also uses the geothermal heat as input energy.

11. Sep 13, 2007

### squil

12. Oct 19, 2007

### Nicolas Perez

Some of you already create a model in Excel or any other program that can show it to me?

Thanks