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 reasonable 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 efficiency (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.

The gross engine power includes not only the measured net shaft power, but also the internal engine losses. These losses are primarily the friction of the dynamic seals (piston seal and displacer shaft seal), bearing friction, and the power required to move the operating gas back and forth by the displacer.

Although I have measured the net output power at 1.16w, I have no measure of the gross engine power or of the internal engine losses. I can, however, compute an upper limit for the gross engine power from my Stirling engine simulator. As an absolute upper power limit we can use the measured temperatures of 450F for Th and 90F for Tc. Using these values yields the first computed row in the table below. This simulator makes some idealized assumptions that make it virtually impossible for the real engine to output more power than the simulator. I consider this value of 2.34w to be an overly generous estimate of the gross power generated by the engine.

For this engine operating at 3.5 Hz, my heat transfer computations indicate the closest the operating gas can get to the measured external temperatures is a delta T of about 20 degC. The second data row in the table uses these temperature limits to compute the maximum estimated gross power. The third data row lists what I estimate to be the minimum gross power that could support a measured output power of 1.16w. It leaves only .25w for all the engine losses.

Based both on these calculations and on my experiences measuring power and friction in this engine, I feel reasonably confident that the gross engine power falls in the range of 1.41w to 1.79w. Using the average I will call the gross engine power 1.6w +/- 0.2w. Now we can fill in another block in the table:

So far we’ve accounted for 47.8w and we have 21.2w remaining for the last two items.

Engine power computations

The following values were used in the engine simulator to compute power:
Power cylinder 31.75mm dia x 63.5mm stroke
Displacer 44.5mm dia x 63.5mm stroke
Dead volumes: hot, cold, regenerator = 10, 60, 31 cm^3
Engine cycles/sec = 3.5
Average pressure = .968 atm (to account for 900ft elevation)

The computed power output of the simulator is reduced by the isentropic compression heating value (also from the simulator) so that the first data row in the table is 2.73w (1-14.33%) = 2.34w.

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.

The data I’m using for this analysis is not from the video. The measured data for this analysis: 3.5 hz (210 rpm), 71 grams @3 inches torque, Th = 450F, Tc = 90F, halogen bulb power 69w, computed shaft power (from measurements) 1.16w, computed efficiency 1.7%

To start with let’s look at a diagram of the engine.

The halogen light bulb provides heat to the system. The primary goal of the supplied heat is to heat the working gas in the hot end of the displacer cylinder to power the engine. The heat, however, doesn’t just go where you’d like it to go. A large part of the heat from the light bulb dissipates directly to the environment without ever going into the displacer. The thermally conductive displacer and displacer cylinder also conduct heat from the hot end to the cold end of the displacer. Although the engine design uses these surfaces as a regenerator, the conductive losses are neither thermodynamically desirable nor necessary for power production.

Ideally, the regenerator would provide the all the heating and cooling of the working gas except that part which is thermodynamically necessary to generate the power output. This necessary heating is determined by the efficiency of the power cycle used.

The engine itself is not 100% efficient, even after accounting for the thermodynamic inefficiencies. Frictional and other losses in the engine mean that the engine is actually generating more power than you measure at the crankshaft.

In the course of this multi-part analysis I’ll address all of the above issues and do my best to determine where those 69 watts of input heat went.

Heat lost directly to the environment

The first part to start with is the insulated galvanized steel cylinder surrounding most of the displacer cylinder including the hot end. The part of the displacer cylinder that is not inside the insulating can is the cold end plus roughly the displacer stroke length. I measured temperatures over multiple points on the insulated cylinder and then averaged the side, bottom, and top temperatures.

With the average surface temperatures, the ambient temperature and the estimated emissivity of the bright galvanized metal of the insulated cylinder, I used the calculator (This website has some excellent thermal calculator applets) to compute the total power dissipation for natural convection and radiation.

When using the calculator, your choices are to use a rectangular surface that is either horizontal or vertical. You can also select whether to use a the top or bottom surface for a horizontal surface or a single side or both sides in the case of a vertical surface. I used a single side vertical surface for the side of the cylinder with the width equal to the circumference. The top and bottom surfaces were analyzed as squares with the areas equivalent to the circular area.

The values are shown below:

Ambient temperature 63 deg F (17.3 C)
Emissivity of bright galvanized metal 0.28
Altitude 900ft (273m)

The heat loss from the bottom and sides of the insulated cylinder are not necessary for power production. The top surface of the insulated cylinder is more difficult to assess. It connects firmly to the displacer and could act as a cooling fin for the cold end of the displacer or it could be receiving heat through the insulation. Measuring the temperature gradients in the vicinity aren’t conclusive, although some of the heat is definitely coming through the insulation and from the sides of the galvanized steel cylinder. I decide to split it about in half and attribute 1.7w as heat loss. The total heat lost directly to the environment is therefore 33 + 5.6 + 1.7 = 40.3w. The remaining 1.6w of the top surface is considered part of the heat used in producing power.

My next post will go into the conductive losses.

Stirling Engine Simulator

I recently added a Stirling engine simulator to this site. It’s shown on your right in the navigation bar under the Pages heading. The guide below should be helpful to those interested in using the simulator. In the future I’ll post more information on using the simulator and what you can learn from it. Although it’s a fairly simple simulator, you can learn a lot about Stirling engine design by trying out different designs and operating conditions.

Stirling Engine Simulator Guide

What is it?

The Stirling engine simulator is a cycle simulator for Beta or Gamma engine configurations. If you aren’t familiar with Beta and Gamma configurations see Stirlilng Engines (Wikipedia). These engines use a displacer to force the operating gas between hot and cold regions and a power piston to extract work. To use the simulator you enter the key dimensions, dead volumes, hot and cold gas temperatures, and average system pressure. The simulator also accepts inputs for the ideal gas constant for using gases other than air, and lets you specify a cycle rate to generate power output.

This simulator performs simple thermodynamic calculations for ideal gases but does not perform heat transfer computations. That means the simulator can compute the gas pressure versus crankshaft angle if you supply the configuration and temperature information. The simulator can also compute the work done per engine cycle (one revolution). Heat transfer computations would determine the rate at which the gas changes temperature. Those computations are both extremely complex and would require a huge amount of engine design detail.

The simulator can only determine power output at a specific RPM if you specify the hot and cold gas temperatures and the cycle rate. Low cycle rates will give the most reliable information. As cycle rate goes up, heat transfer limitation will decrease the hot gas temperature and increase the cold gas temperature. Those temperature changes are highly dependent on the heat transfer characteristics of your design and will probably be difficult for you to compute or even estimate with any accurately.

Simulator assumptions:

1. The simulator assumes ideal gas characteristics and computes the pressure in the engine based on the volume and temperature changes over one cycle.

2. At any instant the gas pressure is uniform in all regions. This assumption is not satisfied in real engines except when operating at very low speed. At higher speeds this simulator will indicate higher power than the real engine can produce, providing an upper limit on the possible power output of the engine using the specified inputs.

3. Related to (2) above, the simulator does not take into account friction of any kind including gas movement, motion of shafts, piston, displacer, and associated seals. You will have to make your own allowances for these friction sources and reduce the work and power outputs accordingly.

4. The simulator assumes the engine is not limited by heat transfer. You control the heat transfer assumptions by the temperatures and cycle rates you input. High cycle rates are very dependent on good heat transfer design.

5. Gas in the hot end of the displacer cylinder is always assumed to be at the hot gas temperature. Gas in the cold end of the displacer cylinder and in the power piston cylinder will be at the cold gas temperature. Dead volumes are as you have assigned them. More on dead volumes later.

6. The temperature in the regenerator is computed as Th-Tc/(ln(Th/Tc)) where Th and Tc are in Kelvin. The computed value is close to the average of the hot and cold temperature (Th + Tc)/2, but corrects for the decrease in density of the gas at elevated temperatures.

7. The phase angle between the displacer and the power piston is fixed at 90 degrees.

8. This simulator assumes an ideal gas. At high pressures (over 15 Atm) this simulator will be less accurate but should still be adequate for the estimation purposes for which it is intended.

Simulation Inputs

• Power Piston – The first pair of inputs is for the bore and stroke of the power piston in mm.

• Displacer - The second pair of inputs is for the diameter and stroke of the displacer. If there is a significant clearance between the outside diameter of the displacer and the inside diameter of the displacer cylinder as is sometimes the case on simple engines, use the diameter of the displacer and account for the annular volume between the displacer and displacer cylinder as dead volume.

• Dead Volumes – The next three inputs are dead volumes. In a Stirling engine the operating gas never leaves the engine but is compressed or expanded by the power piston motion and moved around by the displacer. Any volumes in the engine that are fixed and never have the piston or displacer entering them are dead volumes. Add up the cold and hot dead volumes and put them in their respective inputs. If the engine has a regenerator, include that volume in the regenerator dead volume. Be sure to account for all significant dead volumes because they have a large influence on the pressure variations and power output of the engine. Some examples of dead volumes include the following:

1. The volume between the top of the power piston and the top of the
cylinder when the piston is at TDC. In a normal engine this will be a cold volume.

2. Similar dead volumes remaining in the displacer cylinder when the displacer is at its extreme travel positions. These would be hot on the hot end of the displacer and cold on the cold end.

3. If the engine has a regenerator then the gas volume in the regenerator would be used for the regenerator dead volume. The volume taken up by the regenerator material can be subtracted out of the dead volume if it is significant. This is computed by dividing the mass of the regenerator material by the density of the material.

4. The volume contained in gas heating or cooling tubes.

5. If no regenerator is provided, the path the gas takes between hot and cold regions of the displacer could be considered the regenerator volume.

6. In the special case of a Beta engine design it is common to have a negative cold dead volume where the power piston and displacer overlap the same volume in the cylinder (but not at the same time of course). You can design a gamma engines with a negative cold dead volume too, but it is less common. Enter a negative value if it is appropriate.

• Engine cycles/sec – The cycle rate input is provided so you can conveniently compute the power output in watts at a given cycle rate. If you expect your engine to operate at 300 rpm you can enter 5 cycles/sec and see the power it would generate (before subtracting friction losses).

• Hot and cold gas temperatures – are the average values in the hot and cold regions respectively. You have to provide these values from your computations or estimates. A few things you should keep in mind. The hot gas temperature will always be colder than the highest external temperature measured. If you just heat the end of a displacer cylinder you should expect the gas temperature to be a lot less than you measure externally when the engine is running. Similarly the cold gas temperature will be hotter than you measure externally. As you increase the cycle rate of the engine, Th will decrease and Tc will increase for the same external temperatures.

• Average pressure – is the operating pressure for the engine. An unpressurized engine will operate at the default one atmosphere unless it will be operating at an altitude above sea level where the pressure is lower. Normally a pressurized Stirling engine will have both the operating gas volume and the crankcase pressurized. There are two reasons for this:

1. It is very difficult (impossible?) to make piston seals and displacer rod seals that don’t leak. It’s even more difficult to make good seals with low friction. By pressurizing the crankcase any leaks around piston and rod seals will average out over time, maintaining the operating gas pressure. That leaves only the output shaft seal leakage problem. That seal can be eliminated if the output shaft stays internal and runs an internal generator for example.

2. The second reason is that even if you could make the piston seal not leak, the piston would have atmospheric pressure on one side and very high pressure on the other side. The pressure difference would result in extremely high pressure on the piston and bearings.

• Gas constant – The default gas constant is set for air (.287kJ/kg-K). You can change the gas constant for other operating gases. Its only affect on the simulation will be to change the mass of operating gas required.

• Input errors – I have not put any error trapping on the simulator. All inputs must be numbers. If you see results of “Nan” (means “not a number”) then a non-numeric value has been entered. The simulator should recover if you correct the offending input. Allowable entries include numbers like the following 3, 300, 1.23, .004, 1.3e3, 1.3e-3. You can also enter negative numbers but they would only make sense for the temperatures (degrees C) and for the previously mentioned possible negative cold dead volume for some engine configurations.

Results

• The results table compiles some useful information on the engine cycle. The work is always shown for a single engine cycle. The power is (net work x cycle/sec). If you prefer horsepower the conversion is 1 hp = 746 W.

• I’ll go into more details about the results and how to use them in a future posts, but I want to point out one general rule for successful Stirling engine design. Keep the pressure ratio less than or equal to the temperature ratio. Even though the simulator will show increasing power with increasing pressure ratio, a pressure ratio higher than the temperature ratio will push your design into an area where the compression will heat the operating gas so high that little heat can be transferred to the gas from the heat source. The reverse problem happens after expansion. I’ll go into more detail on this in a future post.

• The simulator also generates a normalized plot of the engine cycle pressure-volume curve. This curve provides a lot of information visually. I’ll go into it more in a later post. The table below the plot lists the pressure and volume for all the plotted data points. The table also shows the work increment for each cycle step. The work is computed assuming the back side of the piston is at the average pressure. This results typically in two power pulses per engine cycle (as it does in a real engine). The first when the pressure is highest and the volume is expanding. The second power pulse is when the pressure is below the average and the volume is decreasing. In the second case the pressure on the back of the piston is pushing the piston up into the lower pressure in front of the piston.

This Excel 2003 spreadsheet stirling-cycle-ideal-v-02 lets you input the basic values for a Stirling cycle engine and calculates various quantities such as pressures, work in, work out, heating, cooling, and ideal efficiency.

Don’t use this spreadsheet with the idea that your engine will get close to the values it shows unless you have an engine that truly approximates the ideal Stirling cycle. Most Stirling engines that use a crankshaft to drive the power piston and displacer including the engines I’ve built so far cannot possibly put out more than about 77% of the work per cycle of an ideal Stirling engine. This is assuming piston, displacer, crankshaft and all related mechanisms are frictionless.

A large source of uncertainty when using this spreadsheet is in the minimum and maximum temperatures. The spreadsheet assumes these are the cold and hot temperatures of the working gas in the engine. These temperatures are difficult to measure in an operating engine. I usually measure the hot and cold temperatures as close to the gas as I can. Typically I measure the temperature of a metal plate that directly interfaces with the working gas or the water temperature adjacent to the metal plate. The temperature spread between the hot and cold gas may be only 50% of what I measure or even less depending on the heat transfer. In the future I’ll be working more on the heat transfer efficiency of my engines.

If I use the numbers for my large engine (engine 3d) in the spreadsheet and use the temperatures I actually measure on the metal plates, then I am getting about 11% of the net work out predicted by the spreadsheet (so far). Even if I were to use more realistic hot and cold temperatures, I might still be seeing only 25% of the spreadsheet-predicted work output. Where is the missing work?

Part of the work is lost to friction, something not covered by this spreadsheet at all. In another post I’ve talked about measuring the friction losses on your engine with a flywheel spin-down test. Other friction losses are due to pushing the operating gas around in the engine, especially through the regenerator. Any gas leaks in the engine will also reduce engine power.

This spreadsheet also does not take into account dead space. Any space that is not swept by the displacer or piston is dead space. This includes the space where the regenerator material is placed and also includes internal clearance space above and below the displacer. As a rough approximation the ratio of live volume to total volume (live + dead) will be the ratio of work put out by the engine compared with the ideal output. If your live and dead volumes are equal, then you’ll at best see about 50% of the work out predicted by the spreadsheet. This is after you’ve lost 77% to your engine not following the ideal cycle. My engine 3d has approximately 80 in^3 minimum volume and 41 in^3 dead space including the regenerator. That gives me 77% x (80/121)= 51%.

The idea Stirling cycle also assumes that the compression and expansion of the gas is an isothermal (constant temperature) process. In practice it is normally not possible to closely approximate the isothermal process. A slow turning engine typically can come closer than a high-rpm engine. If no heat is transferred during the compression or expansion cycles, it is an isentropic process. Real engines operate somewhere between isothermal and isentropic and can be simulated by using a polytropic process. I plan to add this feature to a future simulator.

With all these inaccuracies you may be wondering whether this spreadsheet is useful at all. The real value of this spreadsheet is to put an absolute upper bound on a Stirling engine’s work per cycle and efficiency. This spreadsheet also shows the importance of a regenerator, especially in low temperature engines (although it doesn’t assume any volume is required for it).

Using the spreadsheet

I’ve included brief instructions with the spreadsheet. If you aren’t familiar with the Stirling cycle, I suggest this primer as a good starting point.

Notice that this is a cycle spreadsheet so the output is work not power. If instead you’d like to know power then you need to compute the power based on the number of cycles per second. For example, at 10 hz (600 RPM) a joule output per cycle is equivalent to 10 watts.