ORIGINAL: Danielle Fong
by Danielle Fon.
a wick for ideas
One of these days, I’ll try to write up the rules of thermodynamics so that people stop getting mislead. This is not a particularly well edited essay, but it should go up somewhere. Apparently there was a considerable armchair debate about what we are trying to do at LightSail on SciAm.com. Rebuttal below:
Ok, I’ll bite.
First of all if you’re actually looking for rebuttals, it’s usually easier if you post it to my email or some website I own (e.g. daniellefong.com) and have notifications for. We get an awful lot of media coverage and I don’t monitor everything.
Second, the efficiency we’re targeting is 70%. Not 91%. I don’t know where people got that from. We include all of the practical losses that people have mentioned —
Third, it appears you are under some confusion about thermodynamics.
It is a slippery field, and I don’t blame you: both Bill Gates and his advisors made similar mistakes the first time through.
a) We’re not doing isentropic compression or expansion. The whole point of the water spray technique is to approximate an isothermal compression and expansion cycle — the water absorbs heat from the air rapidly. The correct first order approximation is that the heat capacity of the *mixture* is effectively added to the heat capacity of the air. Try deriving this from the 1st law, starting from T_water = T_air, and following the derivation of adiabatic compression without heat exchange to the outside that you see in any thermodynamics text.
Actual results have our output ∆T < 20 C and maximum hotspot ∆T = 60 C. Water spray actually cools. It is surprising how controversial this has been in the 21st century…
b) You’re using Carnot efficiency in an erroneous way. Compression and expansion are only part of the cycle. While it’s true using the generated heat alone in a heat cycle would grant you the efficiencies you describe, this is irrelevant, because we’re not doing that. There’s a whole other thermodynamic resource: *the compressed air* that this is wasting. So we’re not doing that.
Here’s an illustrative exercise.
A Carnot Cycle is perfectly reversible: run the cycle backwards, and 100% of the heat turns back into mechanical energy. How is this possible, one might ask, while at the same time being compatible with Carnot efficiency?
Several reasons: as a heat pump, the Carnot cycle turn W units of work into 1/(1 – Tc/Th) units of Th heat! There’s more heat, in joules, pumped than work put in.
This might seem to violate intuitions, but you can purchase heat pumps at any hardware store. You will notice that there are heat pumps and refrigerators with a coefficient of performance much greater than 1 widely available. This really works.
Now, draw a T-S diagram of a Carnot cycle for an ideal gas. It’s a rectangle in T-S space, the isothermal compression and expansion processes are horizontal lines, and the adiabatic processes are vertical.
Shrink the adiabatic processes to nothing, so that isothermal compression and isothermal expansion are at the same temperature. No heat is moved, and there is no net work. It is still a reversible Carnot cycle. But it doesn’t seem to do anything.
Why would you do a thermodynamic cycle if you get no net work energy out?
Answer: if you get energy out at a *better time*!
If you get 100% of the energy out that you put in, but at a different time, then this is an *IDEAL* energy storage cycle. You can’t get more efficient than that!
However, by your mathematics, you’ll have a 0% efficient heat engine.
The thermodynamic equations for a full heat cycle are *different* than for an energy storage cycle. You cannot just use them blindly. You have to go back to the first principles: the first and second law. (which, by the way, are never violated here — there is never entropy destruction in this or any other ideal reversible cycle).
—
All this said, this is an idealization. In fact there are losses in the process.
What we do with this heat is that we collect it so that we expand air at as high a temperature as we can.
We don’t get as much energy out as if the energy never went to heat, but it is a small boost if we can get it. About 10% relative energy storage efficiency (E_out/E_in) for a 30 C heat increase if we’ve got it, nothing to sneeze at.
But even if we lose 100% of the extra heat, and have to expand at ambient temperature, our efficiency only goes down by that same 10% relative efficiency. It is not bad.
Also, it’s not so hard to insulate a large tank.
—
In general, while I applaud the efforts of people to work out things for themselves, you have to be extra careful that you’re not deluding yourself. It is worse to take a well known equation, misapply it, and declare impossibility, than it is to say that you heard about something but haven’t worked to complete understanding from the fundamentals yet.
It’s not actually working something from first principles if you get them wrong…
Cheers,
Danielle Fong
LightSail Energy
by Danielle Fon.
a wick for ideas
 |
Ok, I’ll bite.
First of all if you’re actually looking for rebuttals, it’s usually easier if you post it to my email or some website I own (e.g. daniellefong.com) and have notifications for. We get an awful lot of media coverage and I don’t monitor everything.
Second, the efficiency we’re targeting is 70%. Not 91%. I don’t know where people got that from. We include all of the practical losses that people have mentioned —
- motor inefficiency (at our scale, typically 5% loss, not 10%, as some people seem to believe — it depends upon scale),
- friction,
- heat loss through our insulated tanks, etc.
Third, it appears you are under some confusion about thermodynamics.
It is a slippery field, and I don’t blame you: both Bill Gates and his advisors made similar mistakes the first time through.
a) We’re not doing isentropic compression or expansion. The whole point of the water spray technique is to approximate an isothermal compression and expansion cycle — the water absorbs heat from the air rapidly. The correct first order approximation is that the heat capacity of the *mixture* is effectively added to the heat capacity of the air. Try deriving this from the 1st law, starting from T_water = T_air, and following the derivation of adiabatic compression without heat exchange to the outside that you see in any thermodynamics text.
Actual results have our output ∆T < 20 C and maximum hotspot ∆T = 60 C. Water spray actually cools. It is surprising how controversial this has been in the 21st century…
b) You’re using Carnot efficiency in an erroneous way. Compression and expansion are only part of the cycle. While it’s true using the generated heat alone in a heat cycle would grant you the efficiencies you describe, this is irrelevant, because we’re not doing that. There’s a whole other thermodynamic resource: *the compressed air* that this is wasting. So we’re not doing that.
Here’s an illustrative exercise.
A Carnot Cycle is perfectly reversible: run the cycle backwards, and 100% of the heat turns back into mechanical energy. How is this possible, one might ask, while at the same time being compatible with Carnot efficiency?
Several reasons: as a heat pump, the Carnot cycle turn W units of work into 1/(1 – Tc/Th) units of Th heat! There’s more heat, in joules, pumped than work put in.
This might seem to violate intuitions, but you can purchase heat pumps at any hardware store. You will notice that there are heat pumps and refrigerators with a coefficient of performance much greater than 1 widely available. This really works.
Now, draw a T-S diagram of a Carnot cycle for an ideal gas. It’s a rectangle in T-S space, the isothermal compression and expansion processes are horizontal lines, and the adiabatic processes are vertical.
Shrink the adiabatic processes to nothing, so that isothermal compression and isothermal expansion are at the same temperature. No heat is moved, and there is no net work. It is still a reversible Carnot cycle. But it doesn’t seem to do anything.
Why would you do a thermodynamic cycle if you get no net work energy out?
Answer: if you get energy out at a *better time*!
If you get 100% of the energy out that you put in, but at a different time, then this is an *IDEAL* energy storage cycle. You can’t get more efficient than that!
However, by your mathematics, you’ll have a 0% efficient heat engine.
The thermodynamic equations for a full heat cycle are *different* than for an energy storage cycle. You cannot just use them blindly. You have to go back to the first principles: the first and second law. (which, by the way, are never violated here — there is never entropy destruction in this or any other ideal reversible cycle).
—
All this said, this is an idealization. In fact there are losses in the process.
- Friction, for example.
- Resistance in our motor coils.
- Air turbulence running through valves.
What we do with this heat is that we collect it so that we expand air at as high a temperature as we can.
We don’t get as much energy out as if the energy never went to heat, but it is a small boost if we can get it. About 10% relative energy storage efficiency (E_out/E_in) for a 30 C heat increase if we’ve got it, nothing to sneeze at.
But even if we lose 100% of the extra heat, and have to expand at ambient temperature, our efficiency only goes down by that same 10% relative efficiency. It is not bad.
Also, it’s not so hard to insulate a large tank.
—
In general, while I applaud the efforts of people to work out things for themselves, you have to be extra careful that you’re not deluding yourself. It is worse to take a well known equation, misapply it, and declare impossibility, than it is to say that you heard about something but haven’t worked to complete understanding from the fundamentals yet.
It’s not actually working something from first principles if you get them wrong…
Cheers,
Danielle Fong
LightSail Energy
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