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# [Q] will 12,100 pounds of molten salt at a temperature of 1050 stored in a 3’ by 8’ stora

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The question is will 12,100 pounds of molten salt at a temperature of 1050 stored in a 3’ by 8’ storage tank run through a heat exchanger generate enough steam to run this generator for 4 hours?

If not then how much will be needed?

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P.O. Box 427

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Phone 847-541-5600

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7500KW (9375KVA) GENERAL ELECTRIC

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STEAM TURBINE GENERATOR UNIT

850PSIG @ 825°F W/150PSIG #A.E 50PSIG B.P.

3/60/4160V. 3600RPM 1959

TURBINE

Capacity 7500 KW

Manufacturer Gerneral Electric

Stages 11 (Multi Valve)

Initial Pressure 850 PSIG

Initial Temperature 825°F

Automatic Extraction 150 PSIG

Extraction Size 14"

Exhaust Pressure/Vac 50PSIG

Speed 3600 RPM

Steam Rate 18.7 LB/KW.HR.

Inlet Size 8"

Exhaust Size 20"

GENERATOR A.C.

Capacity 9375 KVA

Manufacturer General Electric

Voltage 3/60/4160V

Speed 3600 RPM

P.F. 80%

Amps 1301

EXCITER

Rating 40 KW

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Voltage 125 V.D.C.

Speed 3600 RPM

Also included are: Lube oil reservoir with coolers, trip & throttle valve, switchgear and controls, relating connections and piping.

IMMEDIATELY AVAILABLE

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In order to answer such a question, I think you'd need to know a lot about the heat exchange unit. What flow rate is needed to produce enough steam to run that generator?

I'm not an engineer either, but just trying to help. I think if you can find the flow rate, then it's just a matter of doing some simple math to determine how long the generator will run.

Of course, this assumes a constant temperature and pressure, which would have to be controlled. You'd need some serious insulation on the tank.

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From first blush, I would say no. More info is needed though.

If you run the generator at full capacity 7500 KW, then the heat required is 7500KW x 3412 Btuh/hr KW = 26 million Btuh/hr (disregarding the losses).

I don't know the specific heat of molten salt, but assuming 0.4 Btuh/lbm F (good guess),

and assuming you are raising the temperature of salt from 100 to 1050 F = 950 F, then

Heat = 12,100 lbm x 0.4 Btu/lbm F x 950 F = 4.6 million Btu

So you need

OUTPUT = 26 million Btuh

Generating at Source = 4.6 million Btu.

(This is disregrding losses, or constraints of heat exchanger, or constraints of piping, or heating of water or pumping of water.)

So, from first blush it looks like the heat storage falls shor of the mark.

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Thank Freeztar and Lawcat,

The molten salt is at 1050f already and yes the storage loses less than 1% in 30 days in a day it lose almost no heat, The salt will be reheated every 4 hours.

I thought that the heat exchanger took the heat from the salt and flashed steamed the water? From what I have read the salt is down to about a temp of 550f after generating the steam.

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If you are reheating molten salt from 550 to 1050 F, then your temperature differenceis dT = 500F.

Then, you are generating:

Heat = 12,100 lbs x 0.4 Btu/lbm F x 500 = 2.4 million Btu per 4 hours, or

Heat = 600,000 btuh/hr = 600,000 Btuh

For that heat input, you can theoretically get

OUTPUT = 600,000 Btuh / 3412 Btuh/KW = 177 KW

Yet, you need 7500 KW

That is 7500 KW / 177 KW = 42 times more.

That means that you need to increase the mass of molten salt from 12100 lbs to 513,000 lbs.

And I have no idea where you will get the source, other than coal, to heat so much salt in 4 hours.

Your molten salt storage is capable of producing 177 KW theoretically disregarding the losses.

Assuming system efficiency of 60%, then your salt storage is capable of producing about 100 KW. Enough to run a nice size hous, but not a 7500 KW Turbine.

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This has aready been done. Generating 100 Mega watts. I just divided by 10. I have done my research and have a partial understanding. The questions I ask is because I need the answer not to waste time. This is an attemp to put a theroy into practise to make money yes but also the Save the Environment Save the Earth. here is an excerpt,

Molten salt can be employed as a heat store to retain heat collected by a solar tower or solar trough so that it can be used to generate electricity in bad weather or at night. It was demonstrated in the Solar Two project from 1995-1999. The system is predicted to have an annual efficiency of 99%, although it is not clear if this is the second law efficiency.[3][4] The molten salt is a mixture of 60 percent sodium nitrate and 40 percent potassium nitrate, commonly called saltpetre. It is non-flammable and nontoxic, and has already been used in the chemical and metals industries as a heat-transport fluid, so experience with such systems exists in non-solar applications.

The salt melts at 221 °C (430 °F). It is kept liquid at 288 °C (550 °F) in an insulated "cold" storage tank. The liquid salt is pumped through panels in a solar collector where the focused sun heats it to 566 °C (1,051 °F). It is then sent to a hot storage tank. This is so well insulated that the thermal energy can be usefully stored for up to a week.

When electricity is needed, the hot salt is pumped to a conventional steam-generator to produce superheated steam for a turbine/generator as used in any conventional coal, oil or nuclear power plant. A 100-megawatt turbine would need tanks of about 30 feet (9.1 m) tall and 80 feet (24 m) in diameter to drive it for four hours by this design.

Several parabolic trough power plants under development in Spain[5] and solar power tower developer SolarReserve plan to use this thermal energy storage concept.

So unless you love coal so much including the CO2, MERCURY and other grabage that solar makes no sense well I'm not with you.

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The molten salt is at 1050f already and yes the storage loses less than 1% in 30 days in a day it lose almost no heat, The salt will be reheated every 4 hours. ...

if this is a solar plant, how do you get sun every 4 hours? :phones:

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if this is a solar plant, how do you get sun every 4 hours? :phones:

According to the numbers joel gave, two extra tanks should suffice to run while the other is re-heated. In other words, a three tank/solar cycle would perhaps suffice. (omitting costs)

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if this is a solar plant, how do you get sun every 4 hours? :phones:
Thats funny. But seriously in AZ the sun shines 300 days a year about and average of 12 hours a day. The problem as everyong knows is the extra 12 hours and 65 days. They came up with a storage using molten salt that has been around a long time that started with nuke. It is actually a better system then was and is used ( as read in a articule) but it won out based upon industry wanting it to versus better.

Anybody remember Beta, VHS? Beta was better but????? Or most recently HD versus Blu Ray. Blu ray is better.

Baseed upon my research Tower is better than trough. Sterling and PV are great for buildings. Utility grade needs 24/7. I had a design pop into my head and have been pursing it since Sept to see if it can work. I really need A REAL ENGINEER WITH EXPERENCE IN SOLAR OR CONVENTIAL PLANTS. THE BASICS ARE THE SAME JUST THE HEAT SOURCE. Actually I need some one familar with solar arrays or heliostats for those calculations and design. My partner Chris and I are willing to share the wealth when we get our 400 Mega watt going but this is our proving system.

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So unless you love coal so much including the CO2, MERCURY and other grabage that solar makes no sense well I'm not with you.

My post was not about hydrocarbon advocacy.

I simply tried to illustrate that the mass of 12,100 lbs is not enough for the job you seek; which was your question. Sorry for answering it.

look at this article: IEEE Spectrum: Largest Solar Thermal Storage Plant to Start Up

It explains that for 50,000 KW, it needs 28,500 tons of molten salt for 8 hours.

In your case you need 7500 KW, or 7.5 MW; about 1/6 of the article's capacity. that means that you would need 28,500/6 = 4700 tons for 8 hours; or 2300 tons for 4 hours. Yet, your 12,100 lbs is 6 tons.

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Thanks Freeztar,

That is absolulutly right. So that gives us 16 hours which makes us profitable. Add in carbon credits and that is another revenue stream. FaceBook is just barely starting to make money. I'm submitting a propsal to the DOE to try and raise funding while looking for investors. I orginally thought that 10 solar concentrators would do the trick but it appears I need 200 based upon the 1980 design I have. I think with a little rework we can reduce that but need the money to hire the engineers. Freeztar every body else I really do appreciate all of the help. If someone could help me understand exactly how the arrays work. I go back to my childhood and the magnifying glass. I saw the array that could melt steel. I figured that had to be hotter than 1050 degrees so if you concentrated that with 10 arrays on a vessel(solar receivre) design to capture that heat would definatly heat the salt to 1050' F in 4 hours???

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well Lawcat first let me apoligize. I'm writing between calls. I thjought you used this from Wikapieia

Thermal energy storage - Wikipedia, the free encyclopedia

"When electricity is needed, the hot salt is pumped to a conventional steam-generator to produce superheated steam for a turbine/generator as used in any conventional coal, oil or nuclear power plant. A 100-megawatt turbine would need tanks of about 30 feet (9.1 m) tall and 80 feet (24 m) in diameter to drive it for four hours by this design."

Thisis the one I used and was trying to confirm. That is where I got my numbers. All of your numbers I just reviewed confused me a little. I will definatly look at the link you provided. I remember that the Wikipedia numbers are close to real world and maybe you can explain you math step by step for me.

Thanks Lawcat

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My post was not about hydrocarbon advocacy.

I simply tried to illustrate that the mass of 12,100 lbs is not enough for the job you seek; which was your question. Sorry for answering it.

look at this article: IEEE Spectrum: Largest Solar Thermal Storage Plant to Start Up

It explains that for 50,000 KW, it needs 28,500 tons of molten salt for 8 hours.

In your case you need 7500 KW, or 7.5 MW; about 1/6 of the article's capacity. that means that you would need 28,500/6 = 4700 tons for 8 hours; or 2300 tons for 4 hours. Yet, your 12,100 lbs is 6 tons.

Which articule is correct the ieee or the Wickpedia? I have seen simialar numbers to the Wickapedia but will research futher. Anybody else have any ideas?

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Which articule is correct the ieee or the Wickpedia?

I don't know. I am not an expert in this emerging technology.

The wiki article states: "A 100-megawatt turbine would need tanks of about 30 feet (9.1 m) tall and 80 feet (24 m) in diameter to drive it for four hours by this design."

I take that "tanks" means multiple tanks. Each tank is 24 m in diameter, and 10 m high. That is each tank is: 2400 cm in diamater and 1000cm high.

EDIT: if the tank has 5m (500cm) buffer all around for containment and insulation, then the diameter of the tank is D = 24 - 5 -5 = 12 m (1200 cm).

The volume of such tank is: V = (3.14 x 1200^2 / 4) x 1000 = 1.2 billion cm3.

Let's presume that tanks are filled to 80% of volume. Then the volume of salt in the tank is: V = 1.2 billion cm3 x 0.80 = 1.0 billion cm3.

The density of sodium nitrate is: ro = 2.26 grams/ cm3

Then, the mass of salt is: m = V x ro = 1.0 billion cm3 x 2.26 cm3 = 2.26 billion grams.

That is, a tank will contan mass of salt: m = 2260 metric tons.

Presuming that we need only one tank for 100 MW = 100,000 KW.

Then, we need 2260 tons / 100,000 KW = 0.023 tons/KW of salt.

If you need to power a turbine of 7500KW, then

you need mass of salt: m = 0.023 tons/KW x 7500KW = 173 metric tons

173 metric tons is 173,000 kg

1 kg = 2.2 lbs, so 173,000 kg = 380,000 lbs

CONCLUSION:

If you follow Wiki, and you need only one tank there, then to power a 7500 KW turbine you need 380,000 lbs, and you have 12,100 lbs.

____

Remember also, you may need multiple "tanks" as the Wiki article suggests; so you may need more than 380,000 lbs of molten salt. I don't know the exact values because I do not know the parameters of the system, such as heat transfer, pumping efficiency, and the density and specific heat of the salt mixture.

EDIT: I revised the tank volume to account for a more conservative assumption of molten mass..

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That means that you need to increase the mass of molten salt from 12100 lbs to 513,000 lbs.

Obviously a Green solution (runs with government subsidies and customer wallet rape). NaNO3-KNO3 eutectic was used for high temp melting point fluid in undergrad labs. It is a monster oxidizer and can attack metal. It slowy decomposes at hight temps, liberating oxygen. Break a hot Thiele tube of tehs tuff over an epoxy lab bench and WHOOMP! Crisco was an economic replacement.

The system is predicted to have an annual efficiency of 99%

That's crap at face value, merely for the insulation losses. 566 C to 300 C, Carnot efficiency, 1-Tc/Th (kelvins) is 31.7%. Now subtract generation Carnot efficiency and real world losses

Power Chips Carnot Efficiency Calculator

Space Scuttle SRBs are 14% petroleum organic binder. Use soybeans and Save Al Gore!

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Obviously a Green solution (runs with government subsidies and customer wallet rape).

No doubt, the system is huge. I was just trying to tell the guy that a little 3x8' room ain't gonna cut it.

That's crap at face value, merely for the insulation losses. 566 C to 300 C, Carnot efficiency, 1-Tc/Th (kelvins) is 31.7%. Now subtract generation Carnot efficiency and real world losses

Great Point Uncle Al. I suspect that the system's efficiency is around 30%. (Easily could be lower).

Output = 7500 kW x 3412 Btuh/kW = 25.6 MBtuh

Heat required for 4 hours = Q = 25.6 MBtuh x 4 hours = 102.4 MBtu

Q at 30% eff. = 102.4 MBtu / 0.30 = 340 MBtu

Qsalt = m x c x dT = m x 0.65 Btu/lbm F x 500 F = 340,000,000 Btu

m salt = 1,000,000 lbs = 450,000 kg

V salt = 450,000,000 g / 2.26 g / cm3 = 200,000,000 cm3 x m3/1,000,000 cm3 = 200 m3

V salt = 200 m3 = 6 x 6 x 6 m.

If we desing insulation and containment buffer of 2 m around, then

V tank = 10 x 10 x 8 m. ( This is for 7500 kW, and is consistent with the article that says that 100 MW needs 24 m x 9 m tanks).

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• 4 weeks later...

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