Stirling cycle

The Stirling cycle is a thermodynamic cycle involving isothermal compression, isochoric heating, isothermal expansion, and isochoric cooling.

When work is done on the working gas, the energy does not stay "in" the gas, because of the isothermal nature of the compression. Instead, the energy moves into the surroundings. When the gas does work, the energy does not come "out" of the gas, but comes from the surroundings. (Well, it very recently came from the surroundings into the gas and then immediately did work).

This means that all of the work done on the gas is immediately converted to heat leaving the container, and all of the work done by the gas was recently converted from heat entering the container. So we know how much heat is involved -- it is equal to the work done on or by the gas. For an engine with a flywheel, the heat goes in and out every cycle.

What is the difference between low and high pressure gas at the same temperature? The number of times the molecules hit the walls of the enclosure. With less volume, molecules are more likely to reach the walls.

How much gas is needed to handle a given amount of energy?
Assume there is a maximum allowable pressure. The problem is if we "handle" energy by moving it into the surroundings, then what amount of energy is ascribed to the gas?

Well, how much energy can we get out of the gas at the given temperature? That would be the amount of energy we should consider to be "in" the compressed gas. There is a pratical limit on how much the gas is allowed to expand. So we can't get more energy out than comes after isothermal expansion to that volume limit. Could we get more out if the expansion is not isothermal? Of course, because if the temperature is allowed to rise, more energy goes in.

Assuming the working gas cannot be released from confinement
So, a reasonable definition of the minimum amount of gas to handle a given amount of energy is the amount which gives that much work upon isothermal expansion from maximum pressure to maximum volume.

The amount of work available given minimum and maximum volumes is


 * $$w = \int_{V_\mathrm{min}}^{V_\mathrm{max}} P(v)\mathrm{d}v = nRT \int_{V_\mathrm{min}}^{V_\mathrm{max}} \frac{\mathrm{d}v}{V} = nRT \log(\frac{V_\mathrm{max}}{V_\mathrm{min}})$$

Or, given maximum (start) and minimum (end) pressures, the work is


 * $$w = \int_{P_\mathrm{min}}^{P_\mathrm{max}} V(p)\mathrm{d}p = nRT \int_{P_\mathrm{min}}^{P_\mathrm{max}} \frac{\mathrm{d}p}{P} = nRT \log(\frac{P_\mathrm{max}}{P_\mathrm{min}})$$

Note that the intergral dP represents the area to the left of the curve from smaller P to larger P. This traverses the curve in the opposite direction from the integral dV.

Numerical calculation to solve for result
If we suppose we already know $$n$$,


 * $$P_\mathrm{min} = \frac{nRT}{V_\mathrm{max}} $$


 * $$w = \int_{\frac{nRT}{V_\mathrm{max}}}^{P_\mathrm{max}} V(p)\mathrm{d}p = nRT \int_{\frac{nRT}{V_\mathrm{max}}}^{P_\mathrm{max}} \frac{\mathrm{d}p}{P} = nRT \log(\frac{P_\mathrm{max}V_\mathrm{max}}{nRT})$$

Thus, we must find $$n$$ such that for known $$w, R, T, P_\mathrm{max}$$, and $$V_\mathrm{max}$$,


 * $$nRT \log(\frac{P_\mathrm{max}V_\mathrm{max}}{nRT}) - w = 0$$

This can be done using numerical techniques.

The pressure times volume product is energy, but that number represents the amount of energy the gas would absorb if it started at absolute zero and absorbed heat reversibly (?) up to the current temperature.

Temperature is 1/2 the average energy of a degree of freedom of a gas molecule.

Well, lets see what happens as the amount of gas goes to zero. In order to move energy through the gas, you have to do work on the molecules of the gas. The volume has to change. But if the volume is nearly zero, there won't be any distance left for a force to act through.

So I guess the amount of gas required is limited by the minimum volume and the maximum pressure. Hmm.
 * At and below maximum pressure, the volume must be over the minimum.
 * $$V = nRT/P_\mathrm{max}$$


 * At and below minimum volume, the pressure must be over maximum pressure.

What if working gas is released and recaptured?
If the Stirling cycle releases air at atmospheric pressure during expansion, and recaptures it during compression, and if the machine has many compression tanks instead of a single cylinder, then maximum volume is not a constraint.

Anyway, the amount of work done is based on maximum and minimum pressure. In fact, we already calculated it:


 * $$w = \int_{P_\mathrm{min}}^{P_\mathrm{max}} V(p)\mathrm{d}p = nRT \int_{P_\mathrm{min}}^{P_\mathrm{max}} \frac{\mathrm{d}p}{P} = nRT \log(\frac{P_\mathrm{max}}{P_\mathrm{min}})$$


 * Earlier versions of the following calculations were wrong because the units program uses ln for log and log for log10.

The minimum pressure would be the ambient pressure, and the maximum would be dictated by the pressure tanks. A reasonable estimate given 300 atm maximum pressure and 300K temperature is
 * $$RT \log(300) = R 300\mathrm{K} \log(300) = 14.2 \mathrm{kJ}/\mathrm{mol}$$.

One gigawatt of power is produced by about 70,000 mols/s, 2 metric ton/s or 5.8 m^3/s of air expanding isothermally at 300K from 300 atm to 1 atm. A new 5.8 m^3 of air must begin expanding every second, but if there are N tanks and we transfer every second to a new tank, it can take N seconds for a given batch of 5.8 m^3 of air to expand to 1730 m^3.

The volume of the 1 atm tank has to be 1730 m^3, a cube 12 m on a side. But that will store energy at the rate of 1 gigawatt (1GJ/s). If the cold temperature is 270K, then about 100 MW of excess power is produced.

One megawatt of work comes 70 mol/sec, 2 kg/s, 5.8 liters / second of air, expanding to 1.7 m^3 every second. Releasing 1.7 m^3 per second of ambient air.

The molecular weight of dry air is about 0.029 kg / mol. That is close enough for rough calculations (used here to calculate 2 kg/s expanding air for 1 MW), even though this air is wet, which would tend to reduce the molecular weight since the molecular weight of water is 0.018 kg / mol.

/usr/bin/units version 1.80 units-1.80-12.src.rpm /usr/share/units.dat Version 1.34 16 June 2002

2084 units, 71 prefixes, 32 nonlinear units ln(300): Definition: 5.7037825 ln(300)/log(300) = ln(10) = 2.3025851

R 300K ln(300) -> 14227.182 kg m^2 / mol s^2 = 14.227182 kJ / mol 1 GW / (R 300K ln(300)) -> 70287.989 mol / s (29 g / mol) 1 GW / (R 300K ln(300)) -> 2038.3517 kg / s (R 300K / 300 atm) 1 GW / (R 300K ln(300)) -> 5.7676537 m^3 / s (R 300K / 1 atm) 1 GW / (R 300K ln(300)) -> 1730.2961 m^3 / s (1730.296)^(1/3) -> 12.005312

(R 300K / 300 atm) 1 MW / (R 300K ln(300)) -> 0.0057676537 m^3 / s = 5.7676537 liter/sec (R 300K / 1 atm) 1 MW / (R 300K ln(300)) -> 1.7302961 m^3 / s 1 MW / (R 300K ln(300)) -> 70.287989 mol / s

Links
Interesting discussion on halfbakery.