Engineering Filter: Volume of water at high pressure and temperature. details inside
March 18, 2012 8:32 PM Subscribe
Can anyone tell me the density of water at 200GPa (pressure) and 210deg Celsius. The volume of water in questions is 1,500,000L or 1.5km3 at 0GPa and 0degC.
Im looking at electricity generation from geothermal. I've seen a few calculations online such as here. but my background in thermal sciences and maths is not great.
Im looking at electricity generation from geothermal. I've seen a few calculations online such as here. but my background in thermal sciences and maths is not great.
www.wolframalpha.com is great for these types of questions
posted by blargerz at 8:51 PM on March 18, 2012
posted by blargerz at 8:51 PM on March 18, 2012
IAPWS (The International Association for the Properties of Water and Steam) has the formulas that you need. The Main Thermodynamic Formulations page (it's in frames so I can't make a direct link) has links to programs that will give you the properties. Also the Releases and Guidelines page has links to the papers describing how to calculate the numbers yourself. I have windows .dll that calculates water density but at fairly ordinary temperatures and pressures compared to what you want.
posted by Confess, Fletch at 9:04 PM on March 18, 2012
posted by Confess, Fletch at 9:04 PM on March 18, 2012
I'm a little confused, you have three conflicting statements.
A) 200 GPa, 210 C
B) 0 GPa, 0 C
C) 0 GPA ( I assume the capital A is a typo), 'Ambient temp'
As I mention in my previous comment, I have dll that calculates the density for #3 that you list. But you'll have to use my links to get #1 and #2.
I can calculate #3 for you when I get to work tomorrow but I'll need to know exactly what you mean by 'ambient' and whether the pressure you list is absolute or gauge. If your pressure is gauge you'll also have to give me the air pressure that your using. I'll also need to know the impurity level of your water (typically about 0.001 g/cc) or should I assume that it's pure water?
Alternatively, if by some chance, your in the Boulder Colorado area you can give me a sample of your water and I can measure it in my Anton-Paar DMA 5000 M.
posted by Confess, Fletch at 9:24 PM on March 18, 2012 [1 favorite]
A) 200 GPa, 210 C
B) 0 GPa, 0 C
C) 0 GPA ( I assume the capital A is a typo), 'Ambient temp'
As I mention in my previous comment, I have dll that calculates the density for #3 that you list. But you'll have to use my links to get #1 and #2.
I can calculate #3 for you when I get to work tomorrow but I'll need to know exactly what you mean by 'ambient' and whether the pressure you list is absolute or gauge. If your pressure is gauge you'll also have to give me the air pressure that your using. I'll also need to know the impurity level of your water (typically about 0.001 g/cc) or should I assume that it's pure water?
Alternatively, if by some chance, your in the Boulder Colorado area you can give me a sample of your water and I can measure it in my Anton-Paar DMA 5000 M.
posted by Confess, Fletch at 9:24 PM on March 18, 2012 [1 favorite]
Best answer: Just to check, I don't think 1.5 km^3 is 1,500,000 L. It's probably just a mistake with the notation, but it's worth being careful. (I make it to be 1,500,000,000,000 L.)
posted by theyexpectresults at 9:46 PM on March 18, 2012
posted by theyexpectresults at 9:46 PM on March 18, 2012
Response by poster: Thanks Confess, Sorry for confusion, What im trying to find exactly is the following from my textbook:
Consider the Geothermal energy production well in the Otway basin of Victoria. One may assume that the development has a 30 year lifetime and during this period the steamfield operations will extract heat from the reservoir at depths of between 3000 and 4500m. It is anticipated that the extraction of energy will result in the cooling of the reservoir from 210°C to 110°C. Under these conditions how much electricity can be generated from an area of reservoir of 1km2 ? (HINT: assume the specific heat capacity of the reservoir rocks is 2.5MJ/m3/°C).
I have used the equation Q=deltaT.m.cp with cp given as 2.5MJ/m3/degC
Ive assumed the Litres of water available is (4500m-3000m)x1km2 area = 1,500,000L or 1.5km3 (this volume of water is assuming 0GPa).
Based on above Ive calculated that 375,000MJ or 104.25MWh is available.
In actual fact the pressure is around 200GPa at between 3000-4500m depth so the above energy availability is a minimum estimate. Im trying to work out the volume of water L at 200GPa based on above.
posted by Under the Sea at 9:47 PM on March 18, 2012
Consider the Geothermal energy production well in the Otway basin of Victoria. One may assume that the development has a 30 year lifetime and during this period the steamfield operations will extract heat from the reservoir at depths of between 3000 and 4500m. It is anticipated that the extraction of energy will result in the cooling of the reservoir from 210°C to 110°C. Under these conditions how much electricity can be generated from an area of reservoir of 1km2 ? (HINT: assume the specific heat capacity of the reservoir rocks is 2.5MJ/m3/°C).
I have used the equation Q=deltaT.m.cp with cp given as 2.5MJ/m3/degC
Ive assumed the Litres of water available is (4500m-3000m)x1km2 area = 1,500,000L or 1.5km3 (this volume of water is assuming 0GPa).
Based on above Ive calculated that 375,000MJ or 104.25MWh is available.
In actual fact the pressure is around 200GPa at between 3000-4500m depth so the above energy availability is a minimum estimate. Im trying to work out the volume of water L at 200GPa based on above.
posted by Under the Sea at 9:47 PM on March 18, 2012
Best answer: Yikes!, that's a very high pressure. I think is what you'll find is that the behavior of steam in that pressure/temperature range is not at all understood.
This page has a nice calculator for superheated steam but it stops at about 1/10000 of the pressure that your looking at.
Why are you assuming that the pressure is that high? It may be that high underground, but it'll be at a much more reasonable pressure once you get it where you can use it. Can't you just work using the pressure that you extract it at?
posted by Confess, Fletch at 10:10 PM on March 18, 2012
This page has a nice calculator for superheated steam but it stops at about 1/10000 of the pressure that your looking at.
Why are you assuming that the pressure is that high? It may be that high underground, but it'll be at a much more reasonable pressure once you get it where you can use it. Can't you just work using the pressure that you extract it at?
posted by Confess, Fletch at 10:10 PM on March 18, 2012
Yes, listen to Confess. Water at 200 GPa and 210°C (483°K) isn't steam, it's probably a very exotic high-pressure ice.
posted by RichardP at 10:16 PM on March 18, 2012 [1 favorite]
posted by RichardP at 10:16 PM on March 18, 2012 [1 favorite]
Response by poster: Thanks All, Agreed
I thought it would be interesting to consider the total geothermal energy availability in the heated water at that pressure/depth.
Looking back over your inputs and the data I can see how bad I am with this stuff. 200GPa is at 3000km not 3000m. The range im looking at it more likely 10GPa pressure difference to the surface.
posted by Under the Sea at 10:31 PM on March 18, 2012
I thought it would be interesting to consider the total geothermal energy availability in the heated water at that pressure/depth.
Looking back over your inputs and the data I can see how bad I am with this stuff. 200GPa is at 3000km not 3000m. The range im looking at it more likely 10GPa pressure difference to the surface.
posted by Under the Sea at 10:31 PM on March 18, 2012
104 MWh is umm.. zero in power generation terms. You're building and then running your 100MW generating plant for 1 hour, and then you're done... that's your first hint that you screwed up.
Heat volume avail - 1500m x 1000m x 1000m = 1.5*10^9m3 of heat retaining rock. at 2.5MJ/m3/°C x 100°C is 250MJ/m3. Total is 3.75x10^11MJ or 3.75x10^8 GJ, which is ~104,166,375GWh. At 30 year operation, that's ~262,800 hours, or ~396MW/hr. That sounds reasonable for a geothermal plant.
I don't know why you need water pressures etc. to finish answering the question, the water is simply a transport device for the heat in the rocks, and you can ignore it in this case.
Homework questions are discouraged on askmefi...but I guess that didn't really come up until the clarification.
posted by defcom1 at 8:49 AM on March 19, 2012
Heat volume avail - 1500m x 1000m x 1000m = 1.5*10^9m3 of heat retaining rock. at 2.5MJ/m3/°C x 100°C is 250MJ/m3. Total is 3.75x10^11MJ or 3.75x10^8 GJ, which is ~104,166,375GWh. At 30 year operation, that's ~262,800 hours, or ~396MW/hr. That sounds reasonable for a geothermal plant.
I don't know why you need water pressures etc. to finish answering the question, the water is simply a transport device for the heat in the rocks, and you can ignore it in this case.
Homework questions are discouraged on askmefi...but I guess that didn't really come up until the clarification.
posted by defcom1 at 8:49 AM on March 19, 2012
Sorry, 3.75*10^8 GJ is 104,166,345 MWh, managed to miscopy that. Also, please review your m3 / km3 conversions. the conversion factor is 1000^3 for volume, not 1000.
What I think will help you is checking your equations by unit factors (see wiki), you write down all your equation steps with units, and then cancel the units out, to make sure you're ending up with the units you think you are ending up with. It can be a bit tedious, but worked properly, makes it simple to find your error. Also, reality check your conclusions. When you see a number like 104MWh, that's not a realistic number for a 30-year output, it's much too low. One house in 30 years will use ~300kWh/month x 12months/year x 30 years x 1MWh/1000kWh = 108MWh, so you know somewhere you've taken a wrong turn, which in this case it was your conversion to km3.
Last point, don't use units you don't need. Energy was given in m3. Don't go to L, you're adding points of error (1000L in 1 m3, another conversion error you could make)
posted by defcom1 at 9:35 AM on March 19, 2012
What I think will help you is checking your equations by unit factors (see wiki), you write down all your equation steps with units, and then cancel the units out, to make sure you're ending up with the units you think you are ending up with. It can be a bit tedious, but worked properly, makes it simple to find your error. Also, reality check your conclusions. When you see a number like 104MWh, that's not a realistic number for a 30-year output, it's much too low. One house in 30 years will use ~300kWh/month x 12months/year x 30 years x 1MWh/1000kWh = 108MWh, so you know somewhere you've taken a wrong turn, which in this case it was your conversion to km3.
Last point, don't use units you don't need. Energy was given in m3. Don't go to L, you're adding points of error (1000L in 1 m3, another conversion error you could make)
posted by defcom1 at 9:35 AM on March 19, 2012
The unit factor method that defcom1 mentions is extremely, extremely useful. Especially when you break out units into their component parts. For instance, knowing that pascals are newtons per meter squared (and a newton is kilograms times meters per second squared, e.g., a mass times an acceleration) I can calculate the pressure at 4500m depth.
3000 kg m^-3 (density of rocks) x 9.8 m s^-2 (gravity) x 4500 m (depth) = 135 000 000 Pa, or 135 MPa.
That's just an example, since I don't think you actually need the pressure to solve the problem. Anyway, I use this method all the time in solving physics problems and highly recommend it!
posted by fermion at 9:58 PM on March 19, 2012
3000 kg m^-3 (density of rocks) x 9.8 m s^-2 (gravity) x 4500 m (depth) = 135 000 000 Pa, or 135 MPa.
That's just an example, since I don't think you actually need the pressure to solve the problem. Anyway, I use this method all the time in solving physics problems and highly recommend it!
posted by fermion at 9:58 PM on March 19, 2012
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posted by Under the Sea at 8:33 PM on March 18, 2012