A common problem Stirling engine designers face is how much volume should the power piston sweep in comparison with the displacer? This of course only applies to a Beta or Gamma engine; an alpha engine designer faces a related problem of what phase angle to use between the hot and cold pistons. The answer for an ideal Stirling engine is that a larger piston will always give you more power. This isn’t true for a real Stirling engine.

The problem centers around the ideal isothermal compression and expansion cycles. In a real engine you can’t realize perfect isothermal compression. It’s probably safe to say that the higher the cycle rate and the more powerful the engine is, the more difficult it is to get close to isothermal compression.

A ready example of the difficulty in achieving isothermal compression is a typical shop air compressor. Compression cylinders normally have cooling fins that get very hot when the compressor is operating coninuously.The air inside the compression cylinder is even hotter than the cylinder fins. Those cooling fins are used to get as close to isothermal compression as practical because isothermal compression uses the least amount of power.

An ideal Stirling engine uses isothermal compression and expansion cycles. Concentrating for the moment on the compression cycle, the gas has moved through the regenerator and cooling tubes (if used) and into the cold end of the displacer cylinder before the compression cycle begins. The compression cycle (besides increasing pressure) adds heat uniformly to all of the working gas. Remember this is after the gas has gone through cooling in the regenerator and the cooling tubes. During the compression cycle the only components to help cool the gas are the cylinder walls and the piston. Isothermal compression is just not the reality for real engines.

On one hand we have isothermal compression as an ideal best case. On the other hand we have the corresponding ideal worst case of adiabatic compression. In the adiabatic or isentropic compression case, no heat leaves the gas during compression. This would be the case if you had all material contacting the gas be a perfect insulator. The isentropic compression case is computed using the equation:

T2/T1 = (V1/V)^(k-1)
where k = cp/cv (about 1.4 for air at typical temperatures)
T1, T2 are beginning and ending absolute temperatures
V1/V2 is volume ratio (maximum volume to minimum volume)

If you have a volume of air at 0 degrees C (273K) and compress it down to half the volume, the air would double in pressure for isothermal compression and of course stay at 0 degrees C. For isentropic compression the temperature would increase to about 87 degrees C. Because the temperature is higher in the isentropic case, the pressure would be more than double the initial pressure.

Real world compression gives you a result somewhere between the isothermal and isentropic cases. Machinery’s Handbook (27th edition) suggests air compressors typically operate about midway between isothermal and adiabatic compression. A Stirling engine may perform better or worse than that depending on its design and cycle rate.

Also note that everything I’ve covered here that applies to the compression cycle applies equally to the expansion cycle. The only difference is that during expansion while the gas is doing work pushing the power piston, you’d like the gas to remain isothermal for the most power, but the gas will be cooling due to the expansion.

In my simulator, I include an output labeled isentropic compression heating that is listed as a percentage. This value is calculated using the engine volume ratio and assumes completely isentropic compression and expansion. What this means is that if you have a delta T (=Th – Tc) of 100 degrees and you see 25% for the isentropic compression heating, then 25% of the temperature swing (25 degrees) would be caused by compression heating and expansion cooling. The power output value listed by the simulator is based on strictly isothermal compression and expansion. To correct for the isentropic compression heating and expansion cooling losses, I subtract the 25% value (or whatever the simulator shows) from the power output value to get a more accurate value of the power output. This result approximates adiabatic rather than isothermal compression and expansion. Using this corrected value gives a conservative estimate although you might want to use something midway between isothermal and adiabatic if you have a reason to be optimistic.

To show how the results are affected, the plot below shows the power output based on both isothermal and adiabatic compression and expansion as the power piston varies in size and alters the % compression heating on my engine 3F.

You might be tempted to design your engine for isentropic compression heating values of up to 40% or even higher because the plot even for the worst case adiabatic compression has not reached a peak value. In reality that may not give you the most power. Typically as you increase the piston diameter or stroke to increase the compression, two things happen.

First, you’ll pick up more friction with the larger piston or longer stroke. As you approach the nearly level part of the power curve, you may end up with less net power due to friction.

Second, the higher the compression, the more likely you’ll experience power loss due to gas leaks past the piston. Gas leaks past the piston (or leaks anywhere) reduce the pressure ratio and will reduce engine power even at low compression. The losses become greater as the compression increases.

Once you have an operating engine you can determine the lowest temperature ratio the engine will operate at. This will give you more insight into how close you are to the adiabatic curve vs the isothermal. Of course the engine must overcome all sources of friction to run, so you aren’t seeing the real zero-power level, but you’ll get an idea. Another problem is that the effect of any gas leaks is magnified when the engine is operated at a minimum rpm condition. Yet another problem is actually measuring the hot and cold gas temperatures in the displacer. I end up measuring the external temperature of the displacer cylinder which is not an accurate measure of the gas temperature. For the engines of mine that I’ve tested, the minimum operating condition seems to be in the 30% to 50% compression heating region (based on external temperatures).

You can’t just blindly assume some ratio of piston diameter to displacer diameter, even for engines operating at the same temperature ratio. The dead volume can have a huge effect on the volume ratio and thus the appropriate piston diameter. A high temperature engine can easily work with a power piston displacement as large or even larger than the displacer swept volume while a low temperature engine may need a power piston displacement that is a small fraction of the displacer swept volume.

Let me summarize what has been covered and what remains in this efficiency analysis:

After a lot of thought I came up with an approach that I believe is sound to finish this efficiency analysis. In the process I discovered that my heat bookkeeping method would need to be revised. The changes are shown in the above table. The numbers I’ve computed so far weren’t changed, just how they are grouped.

The gross engine power is now grouped as part of the thermodynamic cycle input. The gross engine power is what remains from the thermodynamic cycle input heat after you’ve paid off all your thermodynamic cycle losses. The regenerator is where you save heat from one cycle to the next, so it also belongs as part of the thermodynamic cycle input.

Although I often think of the regenerator losses in comparison with a perfect regenerator, another way to look at it is comparing it with no regenerator. The more efficient your regenerator, the less heat you need to supply overall for the thermodynamic cycle input. A regenerator should never lose heat, an efficient one saves more heat than an inefficient one.

Here are the assumptions and equations:

(1) After elimination of other avenues for heat loss, the only remaining heat is that which goes into the thermodynamic power cycle. If we call the remaining heat Q_available then the thermodynamic cycle efficiency is:

Cycle efficiency = gross engine power / Q_available
Cycle efficiency = 1.6w/22.8w = 7.0%

(2) The heat required in an engine cycle can be computed using the internal energy change for an ideal gas and the PdV work. Air behaves as an ideal gas at the temperatures and pressures we are considering, so this should be a good assumption:

Q_required = U₂ – U₁ + PdV work = mCv(T₂-T₁) ∫₁² PdV

Where: U₂ – U₁ is the internal energy change in joules between two states
m = mass of gas
Cv= heat capacity at constant volume for the gas
T₂, T₁ = temperatures Kelvin of the two states
∫₁² PdV = PdV work between the two states

This computation is complicated by having three different volumes of gas (hot, cold, and regenerator) that are continually changing mass. I used a spreadsheet to compute these values from my simulation results using the following method:

U₂ – U₁ = Cv(Σm₂T – Σm₁T)

The summation includes each of the three masses multiplied by each of the three temperatures. In this case the three temperatures are fixed and the masses in each volume change. Then I sum together all the steps in the simulation where heating takes place (but not cooling) to arrive at a total internal energy increase required of 8.5 j/cycle or 29.8w at 3.5Hz. The internal energy change is the majority of the heating on a low temperature engine such as this one.

The total Q_required can be made more accurate by including the PdV work that occurs during the part of the cycle where heating occurs. Including the PdV work (from the simulator) increases the total heat required to: 9.3 j/cycle or 32.6w at 3.5Hz.

(3)The heat required minus the heat available is the heat provided by the regenerator. This certainly makes sense and is why we use a regenerator. The regenerator saves heat from one cycle to the next so that the engine requires less heat than it would without the regenerator.

Q_required – Q_available = Q_regenerator
32.6w – 22.8w = 9.8w

(4) If the regenerator was 0% efficient, then it provides no power. The thermodynamic cycle efficiency would just be the gross engine power divided by the heat required.

Cycle efficiency = gross engine power / Q_required (0% regeneration)
Cycle efficiency = 1.6w / 32.6w = 4.9% efficiency (without regeneration)

(5) If the regenerator was 100% efficient, then the thermodynamic cycle efficiency would be the Carnot efficiency. This would be a valid assumption if the engine used only reversible processes. Ideal Stirling engines are reversible, you can use them as efficient heat pumps as well as engines. Real Stirling engines are not 100% reversible, they have losses. I think this is a reasonably accurate assumption for the present purpose. Even if a 100% efficient regenerator resulted in 75% of Carnot efficiency, the error introduced in this analysis is 1.7w.

Cycle efficiency = Carnot cycle (100% regeneration)
Carnot Efficiency = 1- (Tc/Th) Where Th and Tc are absolute temperatures

On my cycle simulator the 1.6w gross engine power corresponds to Th and Tc temperatures of 205.3 C and 58.8C.

Carnot Efficiency = 1 – (58.8 + 273.15)/(205.3 + 273.15) = 30.6%

Thermodynamic cycle input = gross power out / cycle efficiency
= 1.6w/.306 = 5.2w

If the regenerator were 100% efficient, then Q_available would need to be 5.2w. Using assumption (3) then:

Q_required – Q_available = Q_regenerator =
32.6 – 5.2 = 27.4w (100% efficient)

Knowing the heat supplied by a 100% efficient regenerator lets us compute the actual regenerator efficiency:

Q_{regen(actual)} / Q_regen(100%) =
9.8/27.4 = 36% regenerator efficiency

Using the above computations the updated heat loss table is now:

Conclusion

Although I believe the methods and assumptions I used in this part of the analysis are reasonably sound, the regenerator efficiency seems low. I expected something around 80% and have difficulty believing it is below 50%.

Part of the problem may be caused by the possibly inaccurate 22.8w value arrived at by subtraction from the total. For example, a 5 watt increase in estimating the heat lost directly to the environment (a 12.5% error) would result in a 5 watt higher regenerator output and an efficiency of about 54%.

There is also the possibility that the 36% regenerator efficiency is close to correct and I need to design a better regenerator. Because I believe my analysis is not that far off, I think I have to accept this result.

Overall I was surprised by the amount of heat that goes directly from the heater (the light bulb) into the environment. When I say I’m surprised, I don’t mean by the computed result. When I ran the test I could feel considerable heat coming from the insulated cylinder. This large loss was in spite of my efforts to insulate the hot section.

I’ve also tried propane and kerosene (paraffin) burners. As heat sources they are fine for powering this enginge, but the exhaust gases carry away a significant amount of heat and complicate insulating the engine. Another problem is measuring the power (heat). At the low power level this engine uses, I only burn a few grams an hour. I’d need a laboratory grade scale to measure the input power accurately. My energy loss table would also need another entry for heater efficiency.

Rev 1 details:

Section (2) was replaced above. The original is shown below. Because the value for Q_required was changed, it required changing other sections, but only for the value change. The essential change was adding the effect of the PdV work to Q _required.

(2) The heat required in an engine cycle can be computed using the internal energy change for an ideal gas. Air behaves as an ideal gas at the temperatures and pressures we are considering, so this should be a good assumption:

Q_required = U₂ – U₁ = mCv(T₂-T₁)

Where: U₂ – U₁ is the internal energy change in joules between two states
m = mass of gas
Cv= heat capacity at constant volume for the gas
T₂, T₁ = temperatures Kelvin of the two states

This computation is complicated by having three different volumes of gas (hot, cold, and regenerator) that are continually changing mass. I used a spreadsheet to compute these values from my simulation results using the following method:

U₂ – U₁ = Cv(Σm₂T – Σm₁T)

The summation includes each of the three masses multiplied by each of the three temperatures. In this case the three temperatures are fixed and the masses in each volume change. Then I sum together all the steps in the simulation where heating takes place and arrive at a total heat required of 8.5 j/cycle or 29.8w at 3.5Hz.

Conduction losses

A Stirling engine of gamma configuration, such as the one being considered, uses a displacer to shuttle the operating gas back and forth between the hot end and the cold end. In this engine both the displacer cylinder and the displacer itself are constructed from stainless steel tubes. These tubes conduct heat lengthwise between the hot and cold ends. The illustration below shows the hot end gas as orange near the bottom and the cold end gas as blue near the top.

To compute these losses I would like to have reasonably accurate temperature measurements for the displacer cylinder and displacer. The only temperature I measure directly is the cold end of the displacer cylinder (90F, 32.2C). I don’t directly measure the hot end of the displacer cylinder, just the air temperature nearby (450F, 232C). I have no measure of the gas temperatures inside the displacer volume or of the displacer itself, so I have to make some estimates.

Based on my engine power computations, which I’ll go into further in the next part, I believe that Th for the operating gas is much lower than the measured external temperature and Tc is higher. These temperature differences are also necessary for heat transfer to take place. For this computation I use the values shown in the table below. I use 17 w/m-degK as the average thermal conductivity of stainless steel in the computations.

The rate of conduction heat transfer for the one-dimensional case is:

Qdot = -kA(dT/dx)

Where:
Qdot = heat transfer rate in watts
k = thermal conductivity of the material (w/m-degK)
A = cross sectional area of the material (m^2)
(dT/dx) = rate of change of temperature per unit length (degK/m)

This analysis assumes dT/dx is constant along the length of the tubes.

For the computations, I have reduced the conduction length of the tubes from the full length by one-half the stroke length on each end. I think this best represents the points where the temperatures were measured or computed.

Total conduction losses for the two components are 5.9 watts.

I’d like to make a few comments on the displacer design. The displacer was made relatively long for two reasons:

The longer length reduces dT/dx in the conduction equation to reduce power lost through thermal conduction. The long length was also used to provide more surface area to act as a more efficient regenerator for the operating gas moving back and forth through the annular space. This type of regenerator is not as efficient as a fine mesh screen, but is advantageous from the simplicity standpoint. I’ll look at the efficiency of the regenerator in another part of this analysis.

You may have also noticed the displacer cylinder has a .083” wall, much thicker than it needs to be. This was chosen to reduce the cross sectional area of the annular space the gas travels through to increase the velocity and improve heat transfer. It also reduces the dead volume. A more ideal choice would have been a tube with the same inside diameter and a thinner wall. For simplicity I chose a standard tube size and accepted the penalty (2.9w).

Stainless steel was used for the displacer cylinder and displacer to benefit from the 3.5x lower heat conductivity compared with ordinary steels. If ordinary steel had been used instead of stainless steel, the conduction losses would increase from 5.9 to 20.7 watts. If aluminum was used the losses would be above 35 watts.

So far we’ve accounted for 40.3 watts going directly to the environment and 5.9 watts that does no useful work before going back to the environment through the cold end of the engine. 46.2 watts total.

Part 3 will start to look at the power that goes into doing useful work.

Click on photo to enlarge

The above diagram of the ideal Stirling cycle shows how a displacer-type engine (gamma configuration) would implement the cycle. Note that the displacer and power piston operate independently. During the expansion and compression phases the power piston moves and the displacer is stationary. During the heating and cooling phases the displacer moves and the power piston is stationary.

One could implement the ideal cycle action by using cams to drive the displacer and power piston independently. More commonly the Stirling cycle is implemented imperfectly with the following arrangement:

Click on photo to enlarge

In the above implementation of the Stirling cycle, the power piston and the displacer are essentially driven by two cranks on a single crankshaft. The displacer leads the power piston by 90°. This simplification results in the loss theoretically of about one-quarter of the work.

The engine shown in both of the above diagrams is a schematic diagram of my engine 3d. It has some simplifications. During a cycle the working gas moves between the hot and cold ends by passing through the narrow opening between the displacer and the cylinder wall. My original engine used this design. Later I added a regenerator by boring four holes vertically through the displacer, filling them with regenerator material, and creating a better seal between the displacer and the cylinder wall to force the gas through the regenerator. This modification doubled the engines power output by improving the rate of heat transfer.

Why use the Stirling Cycle?

The Stirling Cycle is an implementation of the Carnot Cycle, the most efficient thermodynamic cycle possible for a heat engine. The theoretical limit is:

Efficiency = 1-Tc/Th

where:

            Th = Absolute temperature of the heat source (K or °R)

            Tc = Absolute temperature of the cooling sink

The following diagram shows the theoretical maximum efficiency for the Carnot Cycle for some low temperatures.

These theoretical efficiency numbers cannot be achieved with real engines. There are several barriers to achieving ideal efficiency including:

1. The isothermal expansion and compression would need to happen very slowly to maintain near constant temperature to allow for heat transfer.
2. The regenerator would need to transfer heat efficiently without friction. At any reasonable gas flow rate through the regenerator the gas will experience friction losses.
3. Ideal efficiency assumes all heat transfer is between the working gas and the appropriate heat source, cool sink, or regenerator. Any paths that take heat from the heat source to the cool sink and bypass the working gas are wasted energy and contribute to engine inefficiency.
4. All the usual sources of energy loss including friction on bearings, moving seals, and airflow.

Some Stirling engine developers have measured efficiencies approaching approximately half of the theoretical efficiency. The most efficient engines have heat sources operating at much higher temperatures (over 1000 deg F) and very high pressures (over 1000 psi).

More information on the Stirling Cycle:

Wikipedia Stirling Engine

One should keep all this ideal thermodynamic cycle discussion in perspective. Other thermodynamic cycles such as the Brayton cycle can also approach Carnot efficiency and have their advantages. For example, no one has figured out how to implement the Stirling cycle as continuous flow process, allowing use of axial flow compressors and turbines. The Brayton cycle is commonly used with turbine engines. Stirling engines are probably confined to lower-power systems (tens of kw)  using pistons. Megawatt scale systems require an array of Stirling engines.

Here is some basic information for those interested in converting solar radiation into power. At the average earth-sun distance of 92.9 million miles the solar radiation intensity is:

435 BTU/(ft²-hr) or 1370 watts/meter².  

After going through the atmosphere and reaching the ground the values drop to:

340 BTU/(ft²-hr) or 1070 W/m². 

The above value is what you have to work with at noon on a clear day at the equator. As you depart from those conditions the values decrease. A good ballpark value to use for 40 degrees North latitude appears to be around:

300 BTU/(ft²-hr) or 944 W/m². 

Just so you know there is more to learn about solar radiation, there is the total radiation which is the sum of direct normal radiation (sunlight), diffuse or sky radiation (when you stand in the shade it isn’t dark and your solar calculator still works), and reflected solar radiation (you probably noticed it’s warmer in the sun by the south side of a building).

The above numbers will get you started if you know what to do with them. I’ll talk about heating things up with solar power on a future post so you can figure how much heat you need to change the temperature of things like air, water, or a piece of metal.  

By the way, typical solar cells have about 10% conversion efficiency so you get roughly 94 W/m² out of them. State-of-the-art solar cells are approaching 15-20% efficiency.

To begin with I’ll skip phase changes and just look at heating things that stay in the same phase—solid, liquid, or gas.

The basic equation is:

Q=mc(deltaT)

 where:

Q = heat (units: cal, kcal, Joules, ft-lb, Btu, or kWhr)

m= mass (units: kg or lb)

c= specific heat for the particular material and temperature (units:kJoules/(kg-degC) or Btu/(lb-degF)

deltaT = change in temperature (units: degC, or degF)

To use the above equation correctly you need to make sure you use consistent units.

Below is a table that contains specific heats for various materials at 20 deg C. If you need other materials you can probably find them on the web or in handbooks for engineering, physics, or chemistry. Note that the values change with temperature and may change greatly when the substance goes through a phase change. These values are also for constant pressure. If you heat or cool at constant volume there are other numbers to use and I’ll discuss that at a later date.

I’ll give a few examples:

You want to raise the temperature of 1.5 pounds of aluminum by 50 degF then you’ll need:

Q=1.5lb x 0.22 Btu/(lb-degF) x 50 degF = 16.5 Btu of heat

You want to raise the temperature of 0.1kg of air by 100 degC

Q=0.1kg x 1.02kJ/(kg-degC) x 100degC = 10.2kJ

A few useful conversions:

1kw-hr = 3412 Btu

1 kw-hr = 3600 kJ,

1Btu=1.055kJ

Sources:

Marks’ Handbook for Mechanical Engineers, tenth edition

General Physics, 1984, Giancoli, Douglas